OF THE

UNIVERSITY
OF

HYDEAULICS

FOE
ENGINEERS AND ENGINEERING STUDENTS

HYDRAULICS

FOR

ENGINEERS AND ENGINEERING STUDENTS

F. 0. LEA,

D,Sc. (ENGINEERING, LONDON)

SENIOR WHITWOBTH SCHOLAR ; ASSOC. R. COL. SC. ; M. INST. C. E. J

TELFORD PRIZEMAN ; PROFESSOR OF CIVIL ENGINEERING

IN THE UNIVERSITY OF BIRMINGHAM.

THIRD EDITION

SECOND IMPRESSION

LONDON

EDWARD ARNOLD

41 & 43, MADDOX STREET, BOND STREET, W.

1919
{All Rights reserved]

Engineering
Library

PREFACE TO THE THIRD EDITION

OINCE the publication of the first edition of this book there
has been published a number of interesting and valuable
papers describing researches of an important character which add
materially to our knowledge of experimental hydraulics. While
the results of these supplement the matter in the original text
they have not made it necessary to modify the contents to any
considerable extent and instead, therefore, of attempting to
incorporate them in the original relevant chapters summaries of
the researches together with a few critical notes have been added
in the Appendix. This arrangement has the advantage that, from
the point of the student, he is able to obtain a grasp of a subject,
which is essentially an experimental one, without being over-
burdened. At a later reading he will find the Appendix useful
and of interest not only as an attempt to give some account of the
most recent researches but also as a reference to the original papers.
As proved to be the case when the original book was written,
so in the present volume the difficulty of selection, without going
far beyond the original purpose of the book and keeping the
volume within reasonable dimensions, has not been easy.

F. C. LEA.

BIRMINGHAM,
June 1916.

424806

a 3

CONTENTS.
CHAPTER I.

FLUIDS AT REST.

Introduction. Fluids and their properties. Compressible and incom-
pressible fluids. Density and specific gravity. Hydrostatics. Intensity
of pressure. The pressure at a point in a fluid is the same in all directions.
The pressure on any horizontal plane in a fluid must be constant. Fluids
at rest with free surface horizontal. Pressure measured in feet of water.
Pressure head. Piezometer tubes. The barometer. The differential gauge.
Transmission of fluid pressure. Total or whole pressure. Centre of
pressure. Diagram of pressure on a plane area. Examples . Page 1

CHAPTER II.

FLOATING BODIES.

Conditions of equilibrium. Principle of Archimedes. Centre of
buoyancy. Condition of stability of equilibrium. Small displacements.
Metacentre. Stability of rectangular pontoon. Stability of floating vessel
containing water. Stability of floating body wholly immersed in water.
Floating docks. Stability of floating dock. Examples . . Page 21

CHAPTER III.

FLUIDS IN MOTION.

Steady motion. Stream line motion. Definitions relating to flow of
water. Energy per pound of water passing any section in a stream line.
Bernoulli's theorem. Venturi meter. Steering of canal boats. Extension
of Bernouilli's theorem. Examples ... . . Page 37

viii CONTENTS

CHAPTER IV.

FLOW OF WATER THROUGH ORIFICES AND OVER WEIRS.

Velocity of discharge from an orifice. Coefficient of contraction for
sharp-edged orifice. Coefficient of velocity for sharp-edged orifice. Bazin's
experiments on a sharp-edged orifice. Distribution of velocity in the plane
of the orifice. Pressure in the plane of the orifice. Coefficient of discharge.
Effect of suppressed contraction on the coefficient of discharge. The form
of the jet from sharp-edged orifices. Large orifices. Drowned orifices.
Partially drowned orifice. Velocity of approach. Coefficient of resistance.
Sudden enlargement of a current of water. Sudden contraction of a
current of water. Loss of head due to sharp-edged entrance into a pipe or
mouthpiece. Mouthpieces. Borda's mouthpiece. Conical mouthpieces
and nozzles. Flow through orifices and mouthpieces under constant
pressure. Time of emptying a tank or reservoir. Notches and weirs.
Rectangular sharp-edged weir. Derivation of the weir formula from that
of a large orifice. Thomson's principle of similarity. Discharge through
a trianglar notch by the principle of similarity. Discharge through a
rectangular weir by the principle of similarity. Rectangular weir with
end contractions. Bazin's formula for the discharge of a weir. Bazin's
and the Cornell experiments on weirs. Velocity of approach. Influence of
the height of the weir sill above the bed of the stream on the contraction.
Discharge of a weir when the air is not freely admitted beneath the nappe.
Form of the nappe. Depressed nappe. Adhering nappes. Drowned or
wetted nappes. Instability of the form of the nappe. Drowned weirs with
sharp crests. Vertical weirs of small thickness. Depressed and wetted
nappes for flat-crested weirs. Drowned nappes for flat-crested weirs. Wide
flat- crested weirs. Flow over dams. Form of weir for accurate gauging.
Boussinesq's theory of the discharge over a weir. Determining by ap-
proximation the discharge of a weir, when the velocity of approach is
unknown. Time required to lower the water in a reservoir a given distance
by means of a weir. Examples . Page 50

CHAPTER V.

FLOW THROUGH PIPES.

Resistances to the motion of a fluid in a pipe. Loss of head by friction.
Head lost at the entrance to the pipe. Hydraulic gradient and virtual
slope. Determination of the loss of head due to friction. Reynold's
apparatus. Equation of flow in a pipe of uniform diameter and determi-
nation of the head lost due to friction. Hydraulic mean depth. Empirical

CONTENTS IX

formulae for loss of head due to friction. Formula of Darcy. Variation
of C in the formula v=G\^mi with service. Ganguillet and Kutter's
formula. Reynold's experiments and the logarithmic formula. Critical
velocity. Critical velocity by the method of colour bands. Law of
frictional resistance for velocities above the critical velocity. The de-
termination of the values of C given in Table XII. Variation of fc, in the
formula i=kv n , with the diameter. Criticism of experiments. Piezometer
fittings. Effect of temperature on the velocity of flow. Loss of head due
to bends and elbows. Variations of the velocity at the cross section of a
cylindrical pipe. Head necessary to give the mean velocity v m to the
water in the pipe. Practical problems. Velocity of flow in pipes. Trans-
mission of power along pipes by hydraulic pressure. The limiting diameter
of cast iron pipes. Pressures on pipe bends. Pressure on a plate in a pipe
filled with flowing water. Pressure on a cylinder. Examples . Page 112

CHAPTER VI.

FLOW IN OPEN CHANNELS.

Variety of the forms of channels. Steady motion in uniform channels.
Formula for the flow when the motion is uniform in a channel of uniform
section and slope. Formula of Chezy. Formulae of Prony and Eytelwein.
Formula of Darcy and Bazin. Ganguillet and Kutter's formula. Bazin's
formula. Variations of the coefficient 0. Logarithmic formula for flow in
channels. Approximate formula for the flow in earth channels. Distribu-
tion of velocity in the cross section of open channels. Form of the curve
of velocities on a vertical section. The slopes of channels and the velocities
allowed in them. Sections of aqueducts and sewers. Siphons forming
part of aqueducts. The best form of channel. Depth of flow in a circular
channel for maximum velocity and maximum discharge. Curves of velocity
and discharge for a channel. Applications of the formulae. Problems.
Examples . , V_ . Page 178

CHAPTER VII.

GAUGING THE FLOW OF WATER.

Measuring the flow of water by weighing. Meters. Measuring the flow
by means of an orifice. Measuring the flow in open channels. Surface
floats. Double floats. Rod floats. The current meter. Pitot tube. Cali-
bration of Pitot tubes. Gauging by a weir. The hook gauge. Gauging
the flow in pipes ; Venturi meter. Deacon's waste-water meter. Kennedy's
meter. Gauging the flow of streams by chemical means. Examples

Page 224

X CONTENTS

CHAPTER VIII.

. IMPACT OF WATER ON VANES.

Definition of vector. Sum of two vectors. Resultant of two velocities.
Difference of two vectors. Impulse of water on vanes. Relative velocity.
Definition of relative velocity as a vector. To find the pressure on a
moving vane, and the rate of doing work. Impact of water on a vane
when the directions of motion of the vane and jet are not parallel.
Conditions which the vanes of hydraulic machines should satisfy.
Definition of angular momentum. Change of angular momentum. Two
important principles. Work done on a series of vanes fixed to a wheel
expressed in terms of the velocities of whirl of the water entering and
leaving the wheel. Curved vanes. Pelton wheel. Force tending to move
a vessel from which water is issuing through an orifice. The propulsion
of ships by water jets. Examples ...... Page 261

CHAPTER IX.

WATER WHEELS AND TURBINES.

Overshot water wheels. Breast wheel. Sagebien wheels. Impulse
wheels. Poncelet wheel. Turbines. Reaction turbines. Outward flow
turbines. Losses of head due to frictional and other resistances in outward
flow turbines. Some actual outward flow turbines. Inward flow turbines.
Some actual inward flow turbines. The best peripheral velocity for
inward and outward flow turbines. Experimental determination of the
best peripheral velocity for inward and outward flow turbines. Value of e

to be used in the formula = eH. The ratio of the velocity of whirl V to

the velocity of the inlet periphery v. The velocity with which water
leaves a turbine. Bernoulli's equations for inward and outward flow
turbines neglecting friction. Bernoulli's equations for the inward and
outward flow turbines including friction. Turbine to develope a given
horse-power. Parallel or axial flow turbines. Regulation of the flow to
parallel flow turbines. Bernoulli's equations for axial flow turbines.
Mixed flow turbines. Cone turbine. Effect of changing the direction of
the guide blade, when altering the flow of inward flow and mixed flow
turbines. Effect of diminishing the flow through turbines on the velocity
of exit. Regulation of the flow by means of cylindrical gates. The Swain
gate. The form of the wheel vanes between the inlet and outlet cf
turbines. The limiting head for a single stage reaction turbine. Series
or multiple stage reaction turbines. Impulse turbines. The form of the
vanes for impulse turbines, neglecting friction. Triangles of velocity for
an axial flow impulse turbine considering friction. Impulse turbine for
high head, Pelton wheel. Oil pressure governor or regulator. Water
pressure regulators for impulse turbines. Hammer blow in a long turbine
supply pipe. Examples Page 283

CONTENTS XI

CHAPTER X.

PUMPS.

Centrifugal and turbine pumps. Starting centrifugal or turbine pumps.
Form of the vanes of centrifugal pumps. Work done on the water by the
wheel. Katio of velocity of whirl to peripheral velocity. The kinetic energy
of the water at exit from the wheel. Gross lift of a centrifugal pump.
Efficiencies of a centrifugal pump. Experimental determination of the
efficiency of a centrifugal pump. Design of pump to give a discharge Q.
The centrifugal head impressed on the water by the wheel. Head-velocity
curve of a centrifugal pump at zero discharge. Variation of the discharge
of a centrifugal pump with the head when the speed is kept constant.
Bernouilli's equations applied to centrifugal pumps. Losses in centrifugal
pumps. Variation of the head with discharge and with the speed of a
centrifugal pump. The effect of the variation of the centrifugal head and
the loss by friction on the discharge of a pump. The effect of the diminu-
tion of the centrifugal head and the increase of the friction head as the
flow increases, on the velocity. Discharge curve at constant head. Special

U 2
arrangements for converting the velocity head ^ , with which the water

leaves the wheel, into pressure head. Turbine pumps. Losses in the
spiral casings of centrifugal pumps. General equation for a centrifugal
pump. The limiting height to which a single wheel centrifugal pump can
be used to raise water. The suction of a centrifugal pump. Series or
multi-stage turbine pumps. Advantages of centrifugal pumps. Pump
delivering into a long pipe line. Parallel flow turbine pump. Inward flow
turbine pump. Reciprocating pumps. Coefficient of discharge of the
pump. Slip. Diagram of work done by the pump. The accelerations
of the pump plunger and the water in the suction pipe. The effect of
acceleration of the plunger on the pressure in the cylinder during the
suction stroke. Accelerating forces in the delivery pipe. Variation of
pressure in the cylinder due to friction. Air vessel on the suction pipe.
Air vessel on the delivery pipe. Separation during the suction stroke.
Negative slip. Separation in the delivery pipe. Diagram of work done
considering the variable quantity of water in the cylinder. Head lost at
the suction valve. Variation of the pressure in hydraulic motors due to
inertia forces. Worked examples. High pressure plunger pump. Tangye
Duplex pump. The hydraulic ram. Lifting water by compressed air.
Examples Page 392

CHAPTER XL

HYDRAULIC MACHINES.

Joints and packings used in hydraulic work. The accumulator. Dif-
ferential accumulator. Air accumulator. Intensifiers. Steam intensifiers.
Hydraulic forging press. Hydraulic cranes. Double power cranes.
Hydraulic crane valves. Hydraulic press. Hydraulic riveter. Brother-
hood and Rigg hydraulic engines. Examples . . . Page 485

Xll CONTENTS

CHAPTER XII.

RESISTANCE TO THE MOTION OF BODIES IN WATER.

Froude's experiments on the resistance of thin boards. Stream line
theory of the resistance offered to motion of bodies in water. Determination
of the resistance of a ship from that of the model. Examples . Page 507

CHAPTER XIII.

STREAM LINE MOTION.

Hele Shaw's experiments. Curved stream line motion. Scouring of
river banks at bends Page 517

APPENDIX

1. Coefficients of discharge ...,,. Page 521

2. The critical velocity in pipes. Effect of temperature Page 522

3. Losses of head in pipe bends Page 525

4. The Pitot tube * Page 526

5. The Herschel fall increaser . . . . . Page 529

6. The Humphrey internal combustion pump . . Page 531

7. The hydraulic ram . . , . , . Page 537

8. Circular Weirs Page 537

9. General formula for friction in smooth pipes . Page 539

10. The moving diaphragm method of measuring the

flow of water in open channels .... Page 540

11. The Centrifugal Pump . . . . \ . Page 542

ANSWERS TO EXAMPLES 6 Page 553

INDEX Page 557

HYDEAULICS.

CHAPTER I.

FLUIDS AT BEST.

1. Introduction.

The science of Hydraulics in its limited sense and as originally
understood, had for its object the consideration of the laws
regulating the flow of water in channels, but it has come to
have a wider significance, and it now embraces, in addition, the
study of the principles involved in the pumping of water and other
fluids and their application to the working of different kinds of
machines.

The practice of conveying water along artificially constructed
channels for irrigation and domestic purposes dates back into
great antiquity. The Egyptians constructed transit canals for
warlike purposes, as early as 3000 B.C., and works for the better
utilisation of the waters of the Nile were carried out at an even
earlier date. According to Josephus, the gardens of Solomon
were made beautiful by fountains and other water works. The
aqueducts of Borne*, some of which were constructed more than
2000 years ago, were among the "wonders of the world," and
to-day the city of Athens is partially supplied with water by
means of an aqueduct constructed probably some centuries before
the Christian era.

The science of Hydraulics, however, may be said to have only
come into existence at the end of the seventeenth century when
the attention of philosophers was drawn to the problems involved
in the design of the fountains, which came into considerable use
in Italian landscape gardens, and which, according to Bacon,
were of "great beauty and refreshment." The founders were
principally Torricelli and Mariotte from the experimental, and
Bernoulli from the theoretical, side. The experiments of Torri-
celli and of Mariotte to determine the discharge of water through
orifices in the sides of tanks and through short pipes, probably
* The Aqueducts of Rome. Frontinus, translated by Herschel.

L.H. 1

mark the v fitit*fejfcf<eiilpis':fcb.^etet*isime the laws regulating the
flow of water, and Torricelli's famous theorem may be said to
be the foundation of modern Hydraulics. But, as shown at the
end of the chapter on flow in channels, it was not until a century
later that any serious attempt was made to give expression to the
laws regulating the flow in long pipes and channels, and practi-
cally the whole of the knowledge we now possess has been
acquired during the last century. Simple machines for the
utilisation of the power of natural streams have been made for
many centuries, examples of which are to be found in an interest-
ing work Hydrostatiks and Hydrauliks written in English by
Stephen Swetzer in 1729, but it has been reserved to the workers
of the nineteenth century to develope all kinds of hydraulic
machinery, and to discover the principles involved in their correct
design. Poncelet's enunciation of the correct principles which
should regulate the design of the "floats" or buckets of water
wheels, and Fourneyron's application of the triangle of velocities
to the design of turbines, marked a distinct advance, but it must
be admitted that the enormous development of this class of
machinery, and the very high standard of efficiency obtained, is
the outcome, not of theoretical deductions, but of experience,
and the careful, scientific interpretation of the results of
experiments.

2. Fluids and their properties.

The name fluid is given, in general, to a body which offers
very small resistance to deformation, and which takes the shape
of the body with which it is in contact.

If a solid body rests upon a horizontal plane, a force is required
to move the body over the plane, or to overcome the friction
between the body and the plane. If the plane is very smooth
the force may be very small, and if we conceive the plane to be
perfectly smooth the smallest imaginable force would move the
body.

If in a fluid, a horizontal plane be imagined separating the
fluid into two parts, the force necessary to cause the upper
part to slide over the lower will be very small indeed, and
any force, however small, applied to the fluid above the plane
and parallel to it, will cause motion, or in other words will cause
a deformation of the fluid.

Similarly, if a very thin plate be immersed in the fluid in any
direction, the plate can be made to separate the fluid into two
parts by the application to the plate of an infinitesimal force,
and in the imaginary perfect fluid this force would be zero.

FLUIDS AT REST 3

Viscosity. Fluids found in nature are not perfect and are
said to have viscosity; but when they are at rest the conditions
of equilibrium can be obtained, with sufficient accuracy, on
the assumption that they are perfect fluids, and that therefore
no tangential stresses can exist along any plane in a fluid.
This branch of the study of fluids is called Hydrostatics; when
the laws of movement of fluids are considered, as in Hydraulics,
these tangential, or frictional forces have to be taken into
consideration.

3. Compressible and incompressible fluids.

There are two kinds of fluids, gases and liquids, or those which
are easily compressed, and those which are compressed with
difficulty. The amount by which the volumes of the latter are
altered for a very large variation in the pressure is so small that
in practical problems this variation is entirely neglected, and
they are therefore considered as incompressible fluids.

In this volume only incompressible fluids are considered, and
attention is confined, almost entirely, to the one fluid, water.

4. Density and specific gravity.

The density of any substance is the weight of unit volume at
the standard temperature and pressure.

The specific gravity of any substance at any temperature and
pressure is the ratio of the weight of unit volume to the weight
of unit volume of pure water at the standard temperature and
pressure.

The variation of the volume of liquid fluids, with the pressure,
as stated above, is negligible, and the variation due to changes of
temperature, such as are ordinarily met with, is so small, that in
practical problems it is unnecessary to take it into account.

In the case of water, the presence of salts in solution is of
greater importance in determining the density than variations
of temperature, as will be seen by comparing the densities of sea
water and pure water given in the following table.

TABLE I.

Useful data.

One cubic foot of water at 391 F. weighs 62-425 Ibs.

60 F. 62-36

One cubic foot of average sea water at 60 F. weighs 64 Iba.
One gallon of pure water at 60 F. weighs 10 Ibs.
One gallon of pure water has a volume of 277-25 cubic inches.
One ton of pure water at 60 F. has a volume of 35-9 cubic feet.

4 HYDRAULICS

Table of densities of pure water.

Temperature

Degrees Fahrenheit Density

32 -99987

391 1-000000

50 0-99973

60 0-99905

80 0-99664

104 0-99233

From the above it will be seen that in practical problems it
will be sufficiently near to take the weight of one cubic foot of
fresh water as 62*4 Ibs., one gallon as 10 pounds, 6*24 gallons in a
cubic foot, and one cubic foot of sea water as 64 pounds.

5. Hydrostatics.

A knowledge of the principles of hydrostatics is very helpful
in approaching the subject of hydraulics, and in the wider sense
in which the latter word is now used it may be said to include the
former. It is, therefore, advisable to consider the laws of fluids
at rest.

There are two cases to consider. First, fluids at rest under the
action of gravity, and second, those cases in which the fluids are
at rest, or are moving very slowly, and are contained in closed
vessels in which pressures of any magnitude act upon the fluid,
as, for instance, in hydraulic lifts and presses.

6. Intensity of pressure.

The intensity of pressure at any point in a fluid is the pressure
exerted upon unit area, if the pressure on the unit area is uniform
and is exerted at the same rate as at the point.

Consider any little element of area a, about a point in the fluid,
and upon which the pressure is uniform.

If P is the total pressure on a, the Intensity of Pressure p, is then

'--..

or when P and a are indefinitely diminished,

8P
pa3 5-

7. The pressure at any point in a fluid is the same in all
directions.

It has been stated above that when a fluid is at rest its resist-
ance to lateral deformation is practically zero and that on any
plane in the fluid tangential stresses cannot exist. From this
experimental fact it follows that the pressure at any point in the
fluid is the same in all directions.

FLUIDS AT REST 5

Consider a small wedge ABC, Fig. 1, floating immersed in a
fluid at rest.

Since there cannot be a tangential
stress on any of the planes AB, BC, or AC,
the pressures on them must be normal.

Let p, pi and p a be the intensities of
pressures on these planes respectively.

The weight of the wedge will be very Fig. 1.

small compared with the pressures on its
faces and may be neglected.

As the wedge is in equilibrium under the forces acting on
its three faces, the resolved components of the force acting on
AC in the directions of p and pi must balance the forces acting
on AB and BC respectively.

Therefore p 2 . AC cos = p . AB,

and p 2 AC sin = p^ BO.

But AB = AC cos 0, and BC = AC sin 0.

Therefore p = PI = p 2 .

8. The pressure on any horizontal plane in a fluid must
be constant.

Consider a small cylinder of a fluid joining any two points A
and B on the same horizontal plane in the fluid.

Since there can be no tangential forces acting on the cylinder
parallel to the axis, the cylinder must be in equilibrium under the
pressures on the ends A and B of the cylinder, and since these
are of equal area, the pressure must be the same at each end of
the cylinder.

9. Fluids at rest, with the free surface horizontal.

The pressure per unit area at any depth h below the free
surface of a fluid due to the weight of the fluid is equal to the
weight of a column of fluid of height h and of unit sectional area.

Let the pressure per unit area acting on the surface of the
fluid be p Ibs. If the fluid is in a closed vessel, the pressure p may
have any assigned value, but if the free surface is exposed to the
atmosphere, p will be the atmospheric pressure.

If a small open tube AB, of length h, and cross sectional area a,
be placed in the fluid, the weight per unit volume of which is
w Ibs., with its axis vertical, and its upper end A coincident with
the surface of the fluid, the weight of fluid in the cylinder must be
w.a.h Ibs. The pressure acting on the end A of the column
is pa Ibs.

HYDRAULICS

Since there cannot be any force acting on the column parallel
to the sides of the tube, the force of wah Ibs. + pa Ibs. must be
kept in equilibrium by the pressure of the external fluid acting on
the fluid in the cylinder at the end B.

The pressure per unit area at B, therefore,

wah

f , ,,

= (wh + p) Ibs.

The pressure per unit area, therefore, due to the weight of the
fluid only is wh Ibs.

In the case of water, w may be taken as 62'40 Ibs. per cubic
foot and the pressure per sq. foot at a depth of h feet is, therefore,
62'40/z, Ibs., and per sq. inch *433/& Ibs.

It should be noted that the pressure is independent of the form
of the vessel, and simply depends upon the vertical depth of the
point considered below the surface of the fluid. This can be
illustrated by the different vessels shown in Fig. 2. If these
were all connected together by means of a pipe, the fluid when
at rest would stand at the same level in all of them, and on any
horizontal plane AB the pressure would be the same.

Pr&sure an the Plane AB~w-& Ws per sq Foot.
Fig. 2.

If now the various vessels were sealed from each other
by closing suitable valves, and the pipe taken away without
disturbing the level CD in any case, the intensity of pressure on
AB would remain unaltered, and would be, in all cases, equal
to wh.

Example. In a condenser containing air and water, the pressure of the air is
2 Ibs. per sq. inch absolute. Find the pressure per sq. foot at a point 3 feet below
the free surface of the water.

j> = 2x 144 + 3x62-4
= 475 -2 Ibs. per sq. foot.

FLUIDS AT REST

10. Pressures measured in feet of water. Pressure head.

It is convenient in hydrostatics and hydraulics to express the
pressure at any point in a fluid in feet of the fluid instead of pounds
per sq. foot or sq. inch. It follows from the previous section that
if the pressure per sq. foot is p Ibs. the equivalent pressure in feet

of water, or the pressure head, is h = ft. and for any other fluid
having a specific gravity /o, the pressure per sq. foot for a head
h of the fluid is p = w.p.h, or h =

11. Piezometer tubes.

The pressure in a pipe or other vessel can conveniently be
measured by fixing a tube in the pipe and noting the height to
which the water rises in the tube.

Such a tube is called a pressure, or piezometer, tube.

The tube need not be made straight but may be bent into any
form and carried, within reasonable limits, any distance horizon-
tally.

The vertical rise h of the water will be always

where p is the pressure per sq. foot in the pipe.

If instead of water, a liquid of specific gravity p is used the
height h to which the liquid will rise in the tube is

w .p

Example. A tube having one end open to the atmosphere is fitted into a pipe
containing water at a pressure of 10 Ibs. per sq. inch above the atmosphere. Find
the height to which the water will rise in the tube.

The water will rise to such a height that the pressure at the end of the tube in
the pipe due to the column of water will be 10 Ibs. per sq. inch.

Therefore h

12. The barometer.

The method of determining the atmospheric
pressure by means of the barometer can now be
understood.

If a tube about 3 feet long closed at one end be
completely filled with mercury, Fig. 3, and then
turned into a vertical position with its open end
in a vessel containing mercury, the liquid in the
tube falls until the length h of the column is about
30 inches above the surface of the mercury in the
vessel.

Fig. 3.

HYDRAULICS

Since the pressure p on the top of the mercury is now zero, the
pressure per unit area acting on the section of the tube, level with
the surface of the mercury in the vessel, must be equal to the
weight of a column of mercury of height h.

The specific gravity of the mercury is 13'596 at the standard
temperature and pressure, and therefore the atmospheric pressure
per sq. inch, p ay is,

30" x 13-596 x 62-4 ,.
Pa = 10 IXM = 14 7 Ibs. per sq. inch.

12 x 144
Expressed in feet of water,
,147x144
62-4

= 33'92 feet.

This is so near to 34 feet that for the standard atmospheric
pressure this value will be taken throughout this book.

A similar tube can be conveniently used for measuring low
pressures, lighter liquids being used when a more sensitive gauge
is required.

13. The differential gauge.

A more convenient arrangement for measuring pressures, and
one of considerable utility in many hydraulic experiments, is
known as the differential gauge.

Let ABCD, Fig. 4, be a simple U tube
containing in the lower part some fluid of :
known density.

If the two limbs of the tube are open to
the atmosphere the two surfaces of the fluid
will be in the same horizontal plane.

If, however, into the limbs of the tube a
lighter fluid, which does not mix with the
lower fluid, be poured until it rises to C in
one tube and to D in the other, the two
surfaces of the lower fluid will now be at
different levels.

Let B and E be the common surfaces of
the two fluids, h being their difference of
level, and hi and h z the heights of the free
surfaces of the lighter fluid above E and B respectively.

Let p be the pressure of the atmosphere per unit area, and d
and di the densities of the lower and upper fluids respectively.
Then, since upon the horizontal plane AB the fluid pressure must
be constant,

p + dih^ = p + djii + dh 9
or di (Tia hi) = dh.

f
S

iJB

Fig. 4.

FLUIDS AT REST 9

If now, instead of the two limbs of the U tube being open to
the atmosphere, they are connected by tubes to closed vessels in
which the pressures are pi and p 2 pounds per sq. foot respectively,
and hi and h are the vertical lengths of the columns of fluid above
B and B respectively, then

= P! + d l . ^ + d . h,

An application of such a tube to determine the difference of
pressure at two points in a pipe containing flowing water is shown
in Fig. 88, page 116.

Fluids generally used in such U tubes. In hydraulic experiments
the upper part of the tube is filled with water, and therefore the
fluid in the lower part must have a greater density than water.
When the difference of pressure is fairly large, mercury is generally
used, the specific gravity of which is 13'596. When the difference
of pressure is small, the height h is difficult to measure with
precision, so that, if this form of gauge is to be used, it is desirable
to replace the mercury by a lighter liquid. Carbon bisulphide
has been used but its action is sluggish and the meniscus between
it and the water is not always well defined.
Nitro-benzine gives good results, its prin-
cipal fault being that the falling meniscus
does not very quickly assume a definite
shape.

The inverted air gauge. A more sen-
sitive gauge, than the mercury gauge,
can be made by inverting a U tube and
enclosing in the upper part a certain
quantity of air as in the tube BHC, Fig. 5.

Let the pressure at D in the limb DF
be PI pounds per square foot, equivalent
to a head hi of the fluid in the lower part
of the gauge, and at A in the limb AE let
the pressure be p 2 , equivalent to a head h 2 .
Let h be the difference of level of Gr and C.

Fig. 5.

Then if CHG contains air, and the weight of the air be
neglected, being very small, the pressure at C must equal the
pressure at Gr ; and since in a fluid the pressure on any horizontal
plane is constant the pressure at C is equal to the pressure at D,
and the pressure at A equal to the pressure at B. Again the
pressure at Gr is equal to the pressure at K.

Therefore h*-h = h 1

HYDRAULICS

If the fluid is water p may then be taken as unity ; for a given
difference of pressure the value of h will clearly be much greater
than for the mercury gauge, and it has the further advantage that
h gives directly the difference of pressure in feet of water. The
temperature of the air in the tube does not affect the readings, as
any rise in temperature will simply depress the two columns
without affecting the value of h.

The inverted oil gauge. A still more sensitive gauge can
however be obtained by using, in the
upper part of the tube, an oil lighter
than water instead of air, as shown
in Fig. 6.

Let pi and p 2 be the pressures in
the two limbs of the tube on a given
horizontal plane AB, hi and h 2 being
the equivalent heads of water. The
oil in the bent tube will then take up
some such position as shown, the
plane AD being supposed to coincide
with the lower surface C.

Then, since upon any horizontal
plane in a homogeneous fluid the
pressure must be constant, the pres-
sures at G- and H are equal and also
those at D and C.

Let PI be the specific gravity of
the water, and p of the oil.

Then pi hi-ph = pi (h^-h).

Therefore h (pi - p) = Pi Oa - hi)

-f

f*N

ft]

H *

B JrC

=5=

fcw<4

^<S

Fig. 6.

and -fe^r

Substituting for hi and h* the values

(1).

h =

and h<> =

-Pi

. (PI-P)

(2),
.(3).

From (2) it is evident that, if the density of the oil is not very
different from that of the water, h may be large for very small
differences of pressure. Williams, Hubbell and Fenkell* found
that either kerosene, gasoline, or sperm oil gave excellent results,
but sperm oil was too sluggish in its action for rapid work.
* Proceedings Am.S.C.E., Vol. xxvn. p. 384.

FLUIDS AT REST

Kerosene gave the best results. The author has used mineral oils
lighter than water of specific gravities varying from 0'78 to 0'96
and heavier than water of specific gravities from 1*1 to 1'2.

Temperature coefficient of the inverted oil gauge. Unlike the
inverted air gauge the oil gauge has a considerable temperature
coefficient, as will be seen from the table of specific gravities at
various temperatures of water and the kerosene and gasoline used
by Williams, Hubbell and Fenkell.

In this table the specific gravity of water is taken as unity
at 60 F.

Temperature F.
Specific gravity

Water

Kerosene

Gasoline

40
1-00092

60 100
1-0000 -9941

40 60 100
7955 -7879 '7725

40 60 80
72147 -71587 '70547

The calibration of the inverted oil gauge. An arrangement
similar to that shown in Fig. 6 can conveniently be used for
calibrating these gauges.

The difference of level of E and F clearly gives the difference
of head acting on the plane AD in feet of water, and this from

equation (1) equals .

Water is put into AB and FD so that the surfaces B and F
are on the same level, the common surfaces of the oil and the
water also being on the same level, this level being zero for the
oil. Water is then run out of FD until the surface F is
exactly 1 inch below E and a reading for h taken. The surface F
is again lowered 1 inch and a reading of h taken. This process
is continued until F is lowered as far as convenient, and then
the water in EA is drawn out in a similar manner. When E
and F are again level the oil in the gauge should read zero.

14. Transmission of fluid pressure.

If an external pressure be applied at any point in a fluid, it is
transmitted equally in all direc-
tions through the whole mass.
This is proved experimentally
by means of a simple apparatus
such as shown in Fig. 7.

If a pressure P is exerted upon
a small piston Q of a sq. inches Fig. 7.

12 HYDRAULICS

p
area, the pressure per unit area p = , arid the piston at B on the

same level as Q, which has an area A, can be made to lift a load W

p
equal to A ; or the pressure per sq. inch at R is equal to the

pressure at Q. The piston at R is assumed to be on the same level
as Q so as to eliminate the consideration of the small differences of
pressure due to the weight of the fluid.

If a pressure gauge is fitted on the connecting pipe at any
point, and p is so large that the pressure due to the weight of the
fluid may be neglected, it will be found that the intensity of
pressure is p. This result could have been anticipated from that
of section 8.

Upon this simple principle depends the fact that enormous
forces can be exerted by means of hydraulic pressure.

If the piston at Q is of small area, while that at E, is large,
then, since the pressure per sq. inch is constant throughout the
fluid,

W_A
P ~a f

or a very large force W can be overcome by the application of
a small force P. A very large mechanical advantage is thus
obtained.

It should be clearly understood that the rate of doing work
at W, neglecting any losses, is equal to that at P, the distance
moved through by W being to that moved through by P in
the ratio of P to W, or in the ratio of a to A.

Example. A pump ram has a stroke of 3 inches and a diameter of 1 inch. The
pump supplies water to a lift which has a ram of 5 inches diameter. The force
driving the pump ram is 1500 Ibs. Neglecting all losses due to friction etc.,
determine the weight lifted, the work done in raising it 5 feet, and the number
of strokes made by the pump while raising the weight.

Area of the pump ram = '7854 sq. inch.

Area of the lift ram= 19'6 sq. inches.

Therefore W = 1 -

Work done = 37,500 x 5 = 187,500 ft. Ibs.

Let N equal the number of strokes of the pump ram.
Then N x T 3 5 x 1500 Ibs. = 187,500 ft. Ibs.

and N = 500 strokes.

15. Total or whole pressure.

The whole pressure acting on a surface is the sum of all the
normal pressures acting on the surface. If the surface is plane all
the forces are parallel, and the whole pressure is the sum of these
parallel forces.

FLUIDS AT REST 13

Let any surface, which need not be a plane, be immersed
in a liuid. Let A be the area of the wetted surface, and h the
pressure head at the centre of gravity of the area. If the area
is immersed in a fluid the pressure on the surface of which is zero,
the free surface of the fluid will be at a height h above the centre
of gravity of the area. In the case of the area being immersed in
a fluid, the surface of which is exposed to a pressure p, and below
which the depth of the centre of gravity of the area is h , then

If the area exposed to the fluid pressure is one face of a body,
the opposite face of which is exposed to the atmospheric pressure,
as in the case of the side of a tank containing water, or the
masonry dam of Fig. 14, or a valve closing the end of a pipe as
in Fig. 8, the pressure due to the
atmosphere is the same on the two
faces and therefore may be neglected.

Let w be the weight of a cubic
foot of the fluid. Then, the whole
pressure on the area is

T"
i

If the surface is in a horizontal
plane the theorem is obviously true,
since the intensity of pressure is con-
stant and equals w . h.

In general, imagine the surface, Flg * 8t

Fig. 9, divided into a large number of small areas a, Ch, Oa ... .

Let 05 be the depth below the free surface FS, of any element
of area a ; the pressure on this element = w . x . a.

The whole pressure P = ^w .x.a.

But w is constant, and the sum of the moments of the elements
of the area about any axis equals the moment of the whole area*
about the same axis, therefore

2# . a = A . h,
and P = w . A . h.

16. Centre of pressure.

The centre of pressure of any plane
surface acted upon by a fluid is the
point of action of the resultant pressure
acting upon the surface.

Depth of the centre of pressure. Let
DEC, Fig. 9, be any plane surface
exposed to fluid pressure.

* See text-books on Mechanics.

14 HYDKAULICS

Let A be the area, and h the pressure head at the centre of
gravity of the surface, or if FS is the free surface of the fluid, h is
the depth below FS of the centre of gravity.
Then, the whole pressure

P = w.A.h.

Let X be the depth of the centre of pressure.
Imagine the surface, as before, divided into a number of small
areas a, Oi, 03, ... etc.

The pressure on any element a

= w . a . x t

and P = 2wax.

Taking moments about FS,

P . X = (way? + wciiX? + ...)

wAh

~ A/i '

When the area is in a vertical plane, which intersects the
surface of the water in FS, 2a# 2 is the " second moment " of the
area about the axis FS, or what is sometimes called the moment
of inertia of the area about this axis.

Therefore, the depth of the centre of pressure of a vertical
area below the free surface of the fluid

moment of inertia of the area about an axis in its own plane

and in the free surface
area x the depth of the centre of gravity
or, if I is the moment of inertia,

Moment of Inertia about any axis. Calling I the Moment
of Inertia about an axis through the centre of gravity, and I the
Moment of Inertia about any axis parallel to the axis through the
centre of gravity and at a distance h from it,

I-Io + ATz, 2 .

Examples. (1) Area is a rectangle breadth 6 and depth d.

P=w.b.d.h,

FLUIDS AT REST 15

If the free surface of the water is level with the upper edge of the rectangle,

(2) Area is a circle of radius B.

E 2 ,

~Th + h -
tk

If the top of the circle is just in the free surface or 7&=B,

X=B.

TABLE II.

Table of Moments of Inertia of areas.

Form of area

Moment of inertia about
an axis AB through the
C. of G. of the section

Rectangle

rf*l

jtr

Triangle

E^-

H-6--H

Circle

ifcz

Trd 4
64

Semicircle

J73&T
A B

About the axis AB

rr*
8

Parabola

H-.fr-^

l !*f^B

HYDRAULICS

17. Diagram of pressure on a plane area.

If a diagram be drawn showing the intensity of pressure on
a plane area at any depth, the whole pressure is equal to the volume
of the solid thus formed, and the centre of pressure of the area is
found by drawing a line through the centre
of gravity of this solid perpendicular to the
area.

For a rectangular area ABCD, having the
side AB in the surface of the water, the
diagram of pressure is AEFCB, Fig. 10. The
volume of AEFCB is the whole pressure and
equals %bd?w, b being the width and d the
depth of the area.

Since the rectangle is of constant width,
the diagram of pressure may bo represented
by the triangle BCF, Fig. 11, the resultant pressure acting
through its centre of gravity, and therefore at f d from the surface.

a, b -Intensity of pressure,
ojv a/ci.

Fig. 11. Fig. 12.

For a vertical circle the diagram of pressure is as shown in
Figs. 12 and 13. The intensity of pressure ab on any strip at a
depth \ is wh . The whole pressure is the volume of the truncated
cylinder efJch and the centre of pressure is found by drawing a
line perpendicular to the circle, through the centre of gravity
of this truncated cylinder.

Fig. 13.

FLUIDS AT REST

Another, and frequently a very convenient method of deter-
mining the depth of the centre of pressure, when the whole of the
area is at some distance below the surface of the water, is to
consider the pressure on the area as made up of a uniform pressure
over the whole surface, and a pressure of variable intensity.

Take again, as an example, the vertical circle the diagrams of
pressure for which are shown in Figs. 12 and 13.

At any depth h the intensity of pressure on the strip ad is

The pressure on any strip ad is, therefore, made up of a
constant pressure per unit area wh\ and a variable pressure whi ;
and the whole pressure is equal to the volume of the cylinder efgh,
Fig. 12, together with the circular wedge fkg.

The wedge fkg is equal to the whole pressure on a vertical
circle, the tangent to which is in the free surface of the water and

equals w . A . , and the centre of gravity of this wedge will be at

the same vertical distance from the centre of the circle as the
centre of pressure when the circle touches the surface. The whole
pressure P may be supposed therefore to be the resultant of two
parallel forces PI and P 2 acting through the centres of gravity of
the cylinder efgh, and of the circular wedge fkg respectively, the
magnitudes of PI and P 2 being the volumes of the cylinder and
the wedge respectively.

To find the centre of pressure on the circle AB it is only
necessary to find the resultant of two parallel forces

Pi = A.wh A and P 2 = i0.^

of which Pi acts at the centre c, and P 2 at a point Ci which is at
a distance from A of r.

Example. A masonry dam, Fig. 14,
has a height of 80 feet from the founda-
tions and the water face is inclined at
10 degrees to the vertical ; find the whole
pressure on the face due to the water per
foot width of the dam, and the centre of
pressure, when the water surface is level
with the top of the dam. The atmo-
spheric pressure may be neglected.

The whole pressure will be the force
tending to overturn the dam, since the
horizontal component of the pressure
on AB due to the atmosphere will be
counterbalanced by the horizontal com-
ponents of the atmospheric pressure on
the back of the dam. Since the pressure
on the face is normal, and the intensity
of pressure is proportional to the depth,

L. H.

R is the> reswLtcLTLt thrust
OIL the base, DB and, defy
E.
Fig. 14.

HYDRAULICS

the diagram of pressure on the face AB will be the triangle ABC, BC being equal
to wd and perpendicular to AB.

The centre of pressure is at the centre of gravity of the pressure diagram and is,
therefore, at $ the height of the dam from the base.

The whole pressure acts perpendicular to AB, and is equal to the area ABC

= %wd? x sec 10 per foot width

= \ . 62-4 x 6400 x 1-0154 = 202750 Ibs.

Combining P with W, the weight of the dam, the resultant thrust R on the base
and its point of intersection E with the base is determined.

Example. A vertical flap valve closes the end of a pipe 2 feet diameter ; the
pressure at the centre of the pipe is equal to a head of 8 feet of water. To determine
the whole pressure on the valve and the centre of pressure. The atmospheric
pressure may be neglected.

The whole pressure P =wirW . 8'

= 62-4. TT. 8 = 1570 Ibs.

Depth of the centre of pressure.

The moment of inertia about the free surface, which is 8 feet above the centre
of the valve, is

Therefore

x =1-f= 8 ' *"-

That is, f inch below the centre of the naive.

The diagram of pressure is a truncated cylinder efkh, Figs. 12 and 13, ef and hk
being the intensities of pressure at the top and bottom of the valve respectively.

Example. The end of a pontoon which floats in sea water is as shown in Fig. 15.
The level WL of the water is also shown. Find the whole pressure on the end of
the pontoon and the centre of pressure.

A 1

]

Fig. 15.

The whole pressure on BE

= 64 Ibs. x 1CK x 4-5' x 2-25'= 6480 Ibs.
The depth of the centre of pressure of BE is

$. 4-5 = 3'.
The whole pressure on each of the rectangles above the quadrants

= w. 5 = 320 Ibs.,
and the depth of the centre of pressure is feet.

The two quadrants, since they are symmetrically placed about the vertical
centre line, may be taken together to form a semicircle. Let d be the distance
below the centre of the semicircle of any element of area, the distance of the
element below the surface being h g .

FLUIDS AT REST 19

Then the intensity of pressure at depth 7?

= to . 2 + to . d.
And the whole pressure on the semicircle is P = w 2* + the whole pressure

on the semicircle when the diameter is in the surface of the water.

The distance of the centre of gravity of a semicircle from the centre of the
circle is

Therefore,

= 201R 2 + 42 -66 E 3 = 1256 + 666 Ibs.

The depth of the centre of pressure of the semicircle when the surface of the
water is at tho centre of the circle, is

2 ' '6ir

And, therefore, the whole pressure on the semicircle is the sum of two forces,
one of which, 1256 Ibs., acts at the centre of gravity, or at a distance of 3'06' from
AD, and the other of 666 Ibs. acts at a distance of 3 : 47' from AD.

Then taking moments about AD the product of the pressure on the whole area
into the depth of the centre of pressure is equal to the moments of all the forces
acting on the area, about AD. The depth of the centre of pressure is, therefore,

_ 6480 Ibs. x 3' + 320 Ibs. x 2 x f' + 1256 Ibs. x 3-06 + 666 Ibs. x 3-47'
= 2-93 feet.

EXAMPLES.

(1) A rectangular tank 12 feet long, 5 feet wide, and 5 feet deep is
filled with water.

Find the total pressure on an end and side of the tank.

(2) Find the total pressure and the centre of pressure, on a vertical/
sluice, circular in form, 2 feet in diameter, the centre of which is 4 feet
below the surface of the water. [M. S. T. Cambridge, 1901.]

(3) A masonry dam vertical on the water side supports water of
120 feet depth. Find the pressure per square foot at depths of 20 feet and
70 feet from the surface; also the total pressure on 1 foot length of the dam.

(4) A dock gate is hinged horizontally at the bottom and supported in
a vertical position by horizontal chains at the top.

Height of gate 45 feet, width 30 ft. Depth of water at one side of the
gate 32 feet and 20 feet on the other side. Find the tension in the chains.
Sea- water weighs 64 pounds per cubic foot.

(5) If mercury is 13| times as heavy as water, find the height of a
column corresponding to a pressure of 100 Ibs. per square inch.

(6) A straight pipe 6 inches diameter has a right-angled bend connected
to it by bolts, the end of the bend being closed by a flange.

The pipe contains water at a pressure of 700 Ibs. per sq. inch. Determine
the total pull in the bolts at both ends of the elbow.

HYDRAULICS

(7) The end of a dock caisson is as shown in Fig. 16 and the water
level is AB.

Determine the whole pressure and the centre of pressure.

L.'M

k 40.0- *!
Fig. 16.

(8) An U tube contains oil having a specific gravity of 1*1 in the lower
part of the tube. Above the oil in one limb is one foot of water, and above
the other 2 feet. Find the difference of level of the oil in the two limbs.

(9) A pressure gauge, for use in a stokehold, is made of a glass U tube
with enlarged ends, one of which is exposed to the pressure in the stokehold
and the other connected to the outside air. The gauge is filled with water
on one side, and oil having a specific gravity of 0*95 on the other the
surface of separation being in the tube below the enlarged ends. If the
area of the enlarged end is fifty times that of the tube, how many inches of
water pressure in the stokehold correspond to a displacement of one inch
in the surface of separation ? [Lond. Un. 1906.]

(10) An inverted oil gauge has its upper U filled with oil having a
specific gravity of 0*7955 and the lower part of the gauge is filled with
water. The two limbs are then connected to two different points on a pipe
in which there is flowing water.

Find the difference of the pressure at the two points in the pipe when
the difference of level of the oil surfaces in the limbs of the U is
15 inches.

(11) An opening in a reservoir dam is closed by a plate 3 feet square,
which is hinged at the upper horizontal edge ; the plate is inclined at an
angle of 60 to the horizontal, and its top edge is 12 feet below the surface
of the water. If this plate is opened by means of a chain attached to the
centre of the lower edge, find the necessary pull in the chain if its line of
action makes an angle of 45 with the plate. The weight of the plate is
400 pounds. [Lond. Un. 1905.]

(12) The width of a lock is 20 feet and it is closed by two gates at each
end, each gate being 12' long.

If the gates are closed and the water stands 16' above the bottom on one
side and 4' on the other side, find the magnitude and position of the resultant
pressure on each gate, and the pressure between the gates. Show also that
the reaction at the hinges is equal to the pressure between the gates. One
cubic foot of water =62-5 Ibs. [Lond. Un. 1905.]

CHAPTER II.

FLOATING BODIES.

18. Conditions of equilibrium.

When a body floats in a fluid the surface of the body in
contact with the fluid is subject to hydrostatic pressures, the
intensity of pressure on any element of the surface depend-
ing upon its depth below the surface. The resultant of the
vertical components of these hydrostatic forces is called the
buoyancy, and its magnitude must be exactly equal to the weight
of the body, for if not the body will either rise or sink. Again
the horizontal components of these hydrostatic forces must
be in equilibrium amongst themselves, otherwise the body will
have a lateral movement.

The position of equilibrium for a floating body is obtained
when (a) the buoyancy is exactly equal to the weight of the
body, and (6) the vertical forces the weight and the buoyancy
act in the same vertical line, or in other words, in such a way as
to produce no couple tending to make the body rotate.

Let G-, Fig. 17, be the centre of gravity of a floating ship and
BK, which does not pass through Gr, the line of action of the
resultant of the vertical buoyancy forces. Since the buoyancy

Fig. 17.

Fig. 18.

must equal the weight of the ship, there are two parallel forces
each equal to W acting through G- and along BK respectively,
and these form a couple of magnitude "Wo?, which tends to bring
the ship into the position shown in Fig. 18, that is, so that BK

HYDRAULICS

passes through Gr. The above condition (&) can therefore only be
realised, when the resultant of the buoyancy forces passes through
the centre of gravity of the body. If, however, the body is
displaced from this position of equilibrium, as for example a ship
at sea would be when made to roll by wave motions, there will
generally be a couple, as in Fig. 17, acting upon the body, which
should in all cases tend to restore the body to its position of
equilibrium. Consequently the floating body will oscillate about
its equilibrium position and it is then said to be in stable equi-
librium. On the other hand, if when the body is given a small
displacement from the position of equilibrium, the vertical forces
act in such a way as to cause a couple tending to increase the
displacement, the equilibrium is said to be unstable.

The problems connected with floating bodies acted upon by
forces due to gravity and the hydrostatic pressures only,
resolve themselves therefore into two,

(a) To find the position of equilibrium of the body.

(6) To find whether the equilibrium is stable.

19. Principle of Archimedes.

When a body floats freely in a fluid the weight of the body is
equal to the weight of the fluid displaced.

Since the weight of the body is equal to the resultant of the
vertical hydrostatic pressures, or to the buoyancy, this principle
will be proved, if the weight of the water displaced is shown to be
equal to the buoyancy.

Let ABC, Fig. 19, be a body floating in equilibrium, AC being
in the surface of the fluid.

Fig. 19.

Consider any small element ab of the surface, of area a and
depth h, the plane of the element being inclined at any angle to
the horizontal. Then, if w is the weight of unit volume of the
fluid, the whole pressure on the area a is wha, and the vertical
component of this pressure is seen to be wha cos 0.

FLOATING BODIES 23

Imagine now a vertical cylinder standing on this area, the top
of which is in the surface AC.

The horizontal sectional area of this cylinder is a cos 0, the
volume is ha cos and the weight of the water filling this volume
is wha cos 0, and is, therefore, equal to the buoyancy on the
area ab.

If similar cylinders be imagined on all the little elements
of area which make up the whole immersed surface, the total
volume of these cylinders is the volume of the water displaced,
and the total buoyancy is, therefore, the weight of this displaced
water.

If the body is wholly immersed as in
body is supposed to be made up of small
vertical cylinders intersecting the surface of
the body in the elements of area ab and ab',
which are inclined to the horizontal at angles
and 4> and having areas a and ai respectively,
the vertical component of the pressure on ab
will be wha cos and on ab' will be wh^a\ cos <.
But a cos must equal i cos <, each being Fi 8-

equal to the horizontal section of the small cylinder. The whole
buoyancy is therefore

2>wha cos ^whitti cos <,
and is again equal to the weight of the water displaced.

In this case if the fluid be assumed to be of constant density
and the weight of the body as equal to the weight of the fluid
of the same volume, the body will float at any depth. The
slightest increase in the weight of the body would cause it to
sink until it reached the bottom of the vessel containing the fluid,
while a very small diminution of its weight or increase in its
volume would cause it to rise immediately to the surface. It
would clearly be practically impossible to maintain such a body
in equilibrium, by endeavouring to adjust the weight of the body,
by pumping out, or letting in, water, as has been attempted in a
certain type of submarine boat. In recent submarines the lowering
and raising of the boat are controlled by vertical screw propellers.

20. Centre of buoyancy.

Since the buoyancy on any element of area is the weight of
the vertical cylinder of the fluid above this area, and that the
whole buoyancy is the sum of the weights of all these cylinders, it
at once follows, that the resultant of the buoyancy forces must
pass through the centre of gravity of the water displaced, and this
point is, therefore, called the Centre of Buoyancy.

HYDRAULICS

21. Condition of stability of equilibrium.

Let AND, Fig. 21, be the section made by a vertical plane
containing G the centre of gravity and B the centre of buoyancy
of a floating vessel, AD being the surface of the fluid when the
centre of gravity and centre of buoyancy are in the same vertical
line.

Fig. 21,

Fig. 22.

Let the vessel be heeled over about a horizonal axis, FE being
now the fluid surface, and let Bi be the new centre of buoyancy,
the above vertical sectional plane being taken to contain G, B,
and Bi. Draw BiM, the vertical through BI, intersecting the line
GB in M. Then, if M is above G- the couple W . a? will tend to
restore the ship to its original position of equilibrium, but if M is
below Gr, as in Fig. 22, the couple will tend to cause a further
displacement, and the ship will either topple over, or will heel over
into a new position of equilibrium.

In designing ships it is necessary that, for even large displace-
ments such as may be caused by the rolling of the vessel, the
point M shall be above G. To determine M, it is necessary to
determine G and the centres of buoyancy for the two positions
of the floating body. This in many cases is a long arid somewhat
tedious operation.

22. Small displacements. Metacentre.

When the angular displacement is small the point M is called
the Metacentre, and the distance of M from G can be calculated.

Assume the angular displacement in Fig. 21 to be small and
equal to 0.

Then, since the volume displacement is constant the volume of
the wedge ODE must equal CAF, or in Fig. 23 ; dC 3 DE must equal

FLOATING BODIES

Let G-i and G- 2 be the centres of gravity of the wedges
and CiC 2 DE respectively.

df B

Fig. 23.

The heeling of the ship has the effect of moving a mass of
water equal to either of these wedges from GK to Gr 2 , and this
movement causes the centre of gravity of the whole water
displaced to move from B to BI .

Let Z be the horizontal distance between GK and Gr 2 , when FE
is horizontal, and S the perpendicular distance from B to BiM.

Let V be the total volume displacement, v the volume of the
wedge and w the weight of unit volume of the fluid.

Then w.v.Z = w. V. S

= .V.BM.sin0.

Or, since is small, =w.V.BM.0 (1).

The restoring couple is

_ T7 . "V7" ~D/^ /Q / O\

But w . v . Z = twice the sum of the moments about the axis
C 2 Ci, of all the elements such as acdb which make up the wedge

Taking ab as x, bf is o?0, and if ac is 9Z, the volume of the
element is J# 2 # . 3Z.

The centre of gravity of the element is at \x from CiC a .

w s r/J

(3).

But, -5- is the Second Moment or Moment of Inertia of the
o

element of area aceb about C 2 Ci, and 2 / -=* is, therefore, the

Jo o

Moment of Inertia I of the water-plane area ACiDC 2 about dCa.
Therefore w .v .Z = w .1.0 ........................ (4).

26 HYDRAULICS

The restoring couple is then

If this is positive, the equilibrium is stable, but if negative it is
unstable.

Again since from (1)

wv.Z = w.Y
therefore w . Y . BM . 6 = wlO,

and

If BM is greater than BGr the equilibrium is stable, if less than
BGr it is unstable, and the body will heel over until a new position
of equilibrium is reached. If BGr is equal to BM the equilibrium
is said to be neutral.

The distance GrM is called the Metacentric Height, and varies
in various classes of ships from a small negative value to a positive
value of 4 or 5 feet.

When the metacentric height is negative the ship heels until
it finds a position of stable equilibrium. This heeling can be
corrected by ballasting.

Example. A ship has a displacement of 15,400 tons, and a draught of 27'5 feet.
The height of the centre of buoyancy from the bottom of the keel is 15 feet.

The moment of inertia of the horizontal section of the ship at the water line
is 9,400,000 feet 4 units.

Determine the position of the centre of gravity that the metacentric height shall
not be less than 4 feet in sea water.

9,400,000x64
~ 15,400x2240
= 17-1 feet.

Height of metacentre from the bottom of the keel is, therefore, 32*1 feet.
As long as the centre of gravity is not higher than 0*6 feet above the surface of
the water, the metacentric height is more than 4 feet.

23. Stability of a rectangular pontoon.

Let RFJS, Fig. 24, be the section of the pontoon and Gr its
centre of gravity.

Let YE be the surface of the water when the sides of the
pontoon are vertical, and AL the surface of the water when the
pontoon is given an angle of heel 0.

Then, since the weight of water displaced equals the weight of
the pontoon, the area AFJL is equal to the area YFJE.

Let B be the centre of buoyancy for the vertical position,
B being the centre of area of YFJE, and Bi the centre of buoyancy
for the new position, BI* being the centre of area of AFJL. Then
the line joining BGr must be perpendicular to the surface YE and

* In the Fig.,Bj is not the centre of area of AFJL, as, for the sake of clearness,
it is further removed from B than it actually should be,

FLOATING BODIES

is the direction in which the buoyancy force acts when the sides
of the pontoon are vertical, and BiM perpendicular to AL is the
direction in which the buoyancy force acts when the pontoon is
heeled over through the angle 0. M is the metacentre.

Fig. 24.

The forces acting on the pontoon in its new position are, W the
weight of the pontoon acting vertically through G and an equal and
parallel buoyancy force W through BI .

There is, therefore, a couple, W.HG, tending to restore the
pontoon to its vertical position.

If the line BiH were to the right of the vertical through Or, or
in other words the point M was below G, the pontoon would be in
unstable equilibrium.

The new centre of buoyancy BI can be found in several ways.
The following is probably the simplest.

The figure AFJL is formed by moving the triangle, or really
the wedge-shaped piece GEL to CYA, and therefore it may be
imagined that a volume of water equal to the volume of this wedge
is moved from G 2 to Gi . This will cause the centre of buoyancy
to move parallel to GiG 2 to a new position BI, such that

BBi x weight of pontoon = GiG 2 x weight of water in GEL.

Let 6 be half the breadth of the pontoon,
I the length,

D the depth of displacement for the upright position,
d the length LE, or AY,
and w the weight of a cubic foot of water.

Then, the weight of the pontoon

W = 2b.D.l.w

and the weight of the wedge CLE = -~- x I . w.

28 HYDRAULICS

Therefore HB, . 26 . p^^M,

and BR = ^GA.

Besolving BB> and GriGr 2 , which are parallel to each other, along
and perpendicular to BM respectively,

TiO d rK- d f 2 W\ ld &nan ^
Bl Q = 4D GlK ~ 4D V3 2 M = 3D = ^D~'

^ -R -D _ -o n &2K - M <L - d * - Vt&rfQ

J ^'a 1 K~3D26~6D~ 6D '

To find the distance of the point M from G- and the value of the
restoring couple. Since B X M is perpendicular to AL and BM to
VE, the angle BMBi equals 0.

Therefore QM = B X Q cot B = J^ cot = Jg .

Let z be the distance of the centre of gravity G from 0.
Then QG = QC -3 = BC-BQ -z

P 6 2 tan 2
2 6D
Therefore

And since HGr = GrM sin ^,

the righting couple,

D 6 2 tan 2

The distance of the metacentre from the point B, 13
QM + QB = B,Q cot + ^~~

" 3D 6D
Wlien is small, the term containing tan 2 is negligible, and

This result can be obtained from formula (4) given in
section 22.

I for the rectangle is T y (26) 3 = %W, and V = 2bDL

Therefore

If BG is known, the metacentric height can now be found.

FLOATING BODIES

Example. A pontoon has a displacement of 200 tons. Its length is 50 feet.
The centre of gravity is 1 loot above the centre of area of the cross section. Find
the breadth and depth of the pontoon so that for an angular displacement of 10 degrees
the metacentre shall not be less than 3 feet from the centre of gravity, and the free-
board shall not be less than 2 feet.

Referring to Fig. 24, G is the centre of gravity of the pontoon and is the
centre of the cross section KJ.

Then, GO = 1 foot,

F =2 feet,

GM = 3feet.

Let D be the depth of displacement. Then

D x 26 x 62-4 x 50 Ibs. =200 tons x 2240 Ibs.
Therefore D6 = 71'5 .......................................... (1).

The height of the centre of buoyancy B above the bottom of pontoon ia

BT = D.
Since the free-board is to be 2 feet,

Then

Therefore

But

B0 = l' and BG = 2*.
BM=5'.

Multiplying numerator and denominator by 6, and substituting from equation (1)

6 s & 3 tan 2 ,
= 5,

from which

therefore

and

214-5 ' 429
6(2-K-176) 2 ):

6 =10- 1ft.,

The breadth B = 20-2 ft.
depth =7-1 ft.

An*.

24. Stability of a floating vessel containing water.

If a vessel contains water with a free surface, as for instance
the compartments of a floating dock, such as is described on page
31, the surface of the water in these compartments will remain
horizontal as the vessel heels over, and the centre of gravity of
the water in any compartment will change its position in such
a way as to increase the angular displacement of the vessel.

In considering the stability
of such vessels, therefore, the
turning moments due to the
water in the vessel must be
taken into account.

As a simple case consider
the rectangular vessel, Fig. 25,
which, when its axis is vertical,
floats with the plane AB in the Fi - 25

HYDRAULICS

surface of the fluid, DE being the surface of the fluid in the
vessel.

When the vessel is heeled through an angle 0, the surface of
fluid in the vessel is KH.

The effect has been, therefore, to move the wedge of fluid OEH
to ODK, and the turning couple due to this movement is w . v . Z,
v being the volume of either wedge and Z the distance between
the centre of gravity of the wedges.

If 26 is the width of the vessel and I its length, v is -^ I tan 0,

Z is |5 tan 0, and the turning couple is w |6 3 1 tan 2 0.

If is small wvZ is equal to wI0, 1 being the moment of inertia
of the water surface KH about an axis through O, as shown in
section 22.

For the same width and length of water surface in the
compartment, the turning couple is the same wherever the
compartment is situated, for the centre of gravity of the wedge
OHE, Fig. 26, is moved by the same amount in all cases.

If, therefore, there are free fluid surfaces in the floating vessel,
for any small angle of heel 0, the tippling-moment due to these
surfaces is 2i0I0, I being in all cases the moment of inertia of the
fluid surface about its own axis of oscillation, or the axis through
the centre of gravity of the surface.

Fig. 26.

Fig. 27.

25. Stability of a floating body wholly immersed.

It has already been shown that a floating body wholly im-
mersed in a fluid, as far as vertical motions are concerned, can
only with great difficulty be maintained in equilibrium.

If further the body is made to roll through a small angle, the
equilibrium will be unstable unless the centre of gravity of the
body is below the centre of buoyancy. This will be seen at once
on reference to Fig. 27. Since the body is wholly immersed the
centre of buoyancy cannot change its position on the body itself,
as however it rolls the centre of buoyancy must be the centre of
gravity of the displaced water, and this is not altered in form by

FLOATING BODIES

any movement of the body. If, therefore, Gr is above B and the
body be given a small angular displacement to the right say, Gr
will move to the right relative to B and the couple will not restore
the body to its position of equilibrium.

On the other hand, if Gr is below B, the couple will act so as to
bring the body to its position of equilibrium.

26. Floating docks.

Figs. 28 and 29 show a diagrammatic outline of the pontoons
forming a floating dock, and in the section is shown the outline of
a ship on the dock.

-:.-*!

Fig. 29.

To dock a ship, the dock is sunk to a sufficient depth by
admitting water into compartments formed in the pontoons, and the
ship is brought into position over the centre of the dock.

Water is then pumped from the pontoon chambers, and the
dock in consequence rises until the ship just rests on the keel
blocks of the dock. As more water is pumped from the pontoons
the dock rises with the ship, which may thus be lifted clear of
the water.

Let Gri be the centre of gravity of the ship, G 2 of the dock and its
water ballast and G the centre of gravity of the dock and the
ship.

The position of the centre of gravity of the dock will vary

32 HYDRAULICS

relative to the bottom of the dock, as water is pumped from the
pontoons.

As the dock is raised care must be taken that the metacentre
is above Gr or the dock will " list."

Suppose the ship and dock are rising and that WL is the
water line.

Let B 2 be the centre of buoyancy of the dock and BI of the
portion of the ship still below the water line.

Then if Vi and Y 2 are the volume displacements below
the water line of the ship and dock respectively, the centre of
buoyancy B of the whole water displaced divides B 2 Bi, so that

The centre of gravity G- of the dock and the ship divides GiGr 2
in the inverse ratios of their weights.

As the dock rises the centre of gravity Gr of the dock and the
ship must be on the vertical through B, and water must be
pumped from the pontoons so as to fulfil this condition and as
nearly as possible to keep the deck of the dock horizontal.

The centre of gravity G^ of the ship is fixed, while the centre of
buoyancy of the ship BI changes its position as the ship is raised.

The centre of buoyancy B 2 of the dock will also be changing,
but as the submerged part of the dock is symmetrical about its
centre lines, B 2 will only move vertically. As stated above, B
must always lie on the line joining BI and B 2 , and as Gr is to be
vertically above B, the centre of gravity Gr 2 and the weight of
the pontoon must be altered by taking water from the various
compartments in such a way as to fulfil this condition.

Quantity of water to be pumped from the pontoons in raising the
dock. Let V be the volume displacement of the dock in its lowest
position, YO the volume displacement in its highest position. To
raise the dock without a ship in it the volume of the water to be
pumped from the pontoons is Y Y .

If, when the dock is in its highest position, a weight W is put
on to the dock, the dock will sink, and a further volume of water

cubic feet will be required to be taken from the pontoons to
w

raise the dock again to its highest position.

To raise the dock, therefore, and the ship, a total quantity of

water

+ Y-YO

w
cubic feet will have to be taken from the pontoons.

FLOATING BODIES S3

Example. A floating dock as shown dimensioned in Fig. 28 is made up of a
bottom pontoon 540 feet long x 96 feet wide x 14-75 feet deep, two side pontoons
380 feet long x 13 feet wide x 48 feet deep, the bottom of these pontoons being
2 feet above the bottom of the dock, and two side chambers on the top of the
bottom pontoon 447 feet long by 8 feet deep and 2 feet wide at the top and 8 feet at
the bottom. The keel blocks may be taken as 4 feet deep.

The dock is to lift a ship of 15,400 tons displacement and 27' 6" draught.

Determine the amount of water that must be pumped from the dock, to raise
the ship so that the deck of the lowest pontoon is in the water surface.

When the ship just takes to the keel blocks on the dock, the bottom of the
dock is 27-5' + 14-75' + 4' =46 -25 feet below the water line.

The volume displacement of the dock is then

14-75 x 540 x 96 + 2 x 44-25 x 13 x 380 + 447 x 8 x 5'= 1,219, 700 cubic feet.
The volume of dock displacement when the deck is just awash is

540 x 96 x 14-75 + 2 x 380 x 13' x (14-75 - 2) = 890,600 cubic feet.
The volume displacement of the ship is

15,400 x 2240 .

- =539,000 cubic feet,

and this equals the weight of the ship in cubic feet of water.

Of the 890,600 cubic feet displacement when the ship is clear of the water,
351,600 cubic feet is therefore required to support the dock alone.

Simply to raise the dock through 31'5 feet the amount of water to be pumped is
the difference of the displacements, and is, therefore, 329, 100 cubic feet.

To raise the ship with the dock an additional 539,000 cubic feet must be
extracted from the pontoons.

The total quantity, therefore, to be taken from the pontoons from the time the
ship takes to the keel blocks to when the pontoon deck is in the surface of the
water is

868,100 cubic feet =24,824 tons.

27. Stability of the floating dock.

As some of the compartments of the dock are partially filled
with water, it is necessary, in considering the stability, to take
account of the tippling-moments caused by the movement of the
free surface of the water in these compartments.

If Gr is the centre of gravity of the dock and ship on the
dock, B the centre of buoyancy, I the moment of inertia of the
section of the ship and dock by the water-plane about the axis of
oscillation, and Ii, I a etc. the moments of inertia of the water
surfaces in the compartments about their axes of oscillation, the
righting moment when the dock receives a small angle of
heel 0, is

The moment of inertia of the water-plane section varies
considerably as the dock is raised, and the stability varies
accordingly.

When the ship is immersed in the water, I is equal to the
moment of inertia of the horizontal section of the ship at the
water surface, together with the moment of inertia of the
horizontal section of the side pontoons, about the axis of
oscillation 0.

L. H. 3

34- HYDRAULICS

When the tops of the keel blocks are just above the surface
of the water, the water-plane is only that of the side pontoons,
and I has its minimum value. If the dock is L-shaped as in
Fig. 30, which is a very convenient form
for some purposes, the stability when
the tops of the keel blocks come to the
surface simply depends upon the moment
of inertia of the area AR about an axis
through the centre of AB. This critical
point can, however, be eliminated by

fitting an air box, shown dotted, on the Fig 30

outer end of the bottom pontoon, the

top of which is slightly higher than the top of the keel blocks.

Example. To find the height of the metacentre above the centre of buoyancy of
the dock of Fig. 28 when

(a) the ship just takes to the keel blocks,

(b) the keel is just clear of the water,

Take the moment of inertia of the horizontal section of the ship at the
water line as 9,400,000 ft. 4 units, and assume that the ship is symmetrically
placed on the dock, and that the dock deck is horizontal. The horizontal distance
between the centres of the side tanks is 111 ft.

(a) Total moment of inertia of the horizontal section is

9, 400,000 + 2 (380 x 13' x 55 -5 a + T^ x 380 x 13 3 ) = 9,400,000 + 30,430,000 + 139, 000.
The volume of displacement

= 539,000 + 1,219,700 cubic feet.
The height of the metacentre above the centre of buoyancy is therefore

(6) When the keel is just clear of the water the moment of inertia is
30,569,000.

The volume displacement is

540 x 96 x 14-75 + 380 x 2 x 13 x (14-75 -I- 4 - 2)

= 930,000 cubic feet.
Therefore BM = 32-8 feet.

= 70,269,000.

The volume displacement is 890,600 cubic feet.
Therefore BM = 79'8 feet.

The height, of the centre of buoyancy above the bottom of the dock can be
determined by finding the centre of buoyancy of each of the parts of the dock, and
of the ship if it is in the water, and then taking moments about any axis.

For example. To find the height h of the centre of buoyancy of the dock and
the ship, when the ship just comes on the keel blocks.

The centre of buoyancy for the ship is at 15 feet above the bottom of the keel.
The centre of buoyancy of the bottom pontoon is at 7 '375' from the bottom.
side pontoons 24-125'

,, ,, chambers 17'94'

FLOATING BODIES 35

Taking moments about the bottom of the dock

h (510,000 + 437,000 + 76,5,000 + 35,760)
= 540,000 x 33-75 + 765,000 x 7'375
+ 437,000 x 24-125 + 35,760 x 17 '95,
therefore ft =19 '7 feet.

For case (a) the metacentre is, therefore, 40*3' above the bottom of the dock. If
now the centre of gravity of the dock and ship is known the metacentrio height
can be found.

EXAMPLES.

(1) A ship when fully loaded has a total burden of 10,000 tons. Find
the volume displacement in sea water.

(2) The sides of a ship are vertical near the water line and the area of
the horizontal section at the water line is 22,000 sq. feet. The total weight
of the ship is 10,000 tons when it leaves the river dock.

Find the difference in draught in the dock and at sea after the weight
of the ship has been reduced by consumption of coal, etc., by 1500 tons.
Let 9 be the difference in draught.
Then 9 x 22,000= the difference in volume displacement
_ 10,000 x 2240 8500 x 2240

62-43 64

=6130 cubic feet.
Therefore 8 = -278 feet

=3*34 inches.

(3) The moment of inertia of the section at the water line of a boat
is 1200 foot 4 units; the weight of the boat is 11'5 tons.

Determine the height of the metacentre above the centre of buoyancy.

(4) A ship has a total displacement of 15,000 tons and a draught of
27 feet.

When the ship is lifted by a floating dock so that the depth of the bottom
of the keel is 16'5 feet, the centre of buoyancy is 10 feet from the bottom of
the keel and the displacement is 9000 tons.

The moment of inertia of the water-plane is 7,600,000 foot 4 units.

The horizontal section of the dock, at the plane 16*5 feet above the
bottom of the keel, consists of two rectangles 380 feet x 11 feet, the distance
apart of the centre lines of the rectangles being 114 feet.

The volume displacement of the dock at this level is 1,244,000 cubic feet.

The centre of buoyancy for the dock alone is 24-75 feet below the surface
of the water.

Determine (a) The centre of buoyancy for the whole ship and the dock.

(6) The height of the metacentre above the centre of buoyancy.

(5) A rectangular pontoon 60 feet long is to have a displacement of
220 tons, a free-board of not less than 3 feet, and the metacentre is not to
be less than 3 feet above the centre of gravity when the angle of heel
is 15 degrees. The centre of gravity coincides with the centre of figure.

Find the width and depth of the pontoon.

36 HYDRAULICS

(6) A rectangular pontoon 24 feet wide, 50 feet long and 14 feet deep,
has a displacement of 180 tons.

A vertical diaphragm divides the pontoon longitudinally into two
compartments each 12 feet wide and 50 feet long. In the lower part
of each of these compartments there is water ballast, the surface of the
water being free to move.

Determine the position of the centre of gravity of the pontoon that it
may be stable for small displacements.

(7) Define "metacentric height" and show how to obtain it graphically
or otherwise. A ship of 16,000 tons displacement is 600 feet long, 60 feet
beam, and 26 feet draught. A coefficient of ^ may be taken in the moment
of inertia term instead of fo to allow for the water-line section not being
a rectangle. The depth of the centre of buoyancy from the water line is
10 feet. Find the height of the metacentre above the water line and
determine the position of the centre of gravity to give a metacentric height
of 18 inches. [Lond. Un. 1906.]

(8) The total weight of a fully loaded ship is 5000 tons, the water line
encloses an area of 9000 square feet, and the sides of the ship are vertical
at the water line. The ship was loaded in fresh water. Find the change
in the depth of immersion after the ship has been sufficiently long at sea to
burn 500 tons of coal.

Weight of 1 cubic foot of fresh water 62 Ibs.
"Weight of 1 cubic foot of salt water 64 Ibs.

CHAPTER III.

FLUIDS IN MOTION.

28. Steady motion.

The motion of a fluid is said to be steady or permanent, when
the particles which succeed each other at any point whatever
have the same density and velocity, and are subjected to the same
pressure.

In practice it is probably very seldom that such a condition of
flow is absolutely realised, as even in the case of the water flowing
steadily along a pipe or channel, except at very low velocities, the
velocities of succeeding particles of water which arrive at any
point in the channel, are, as will be shown later, not the same
either in magnitude or direction.

For practical purposes, however, it is convenient to assume
that if the rate at which a fluid is passing through any finite area
is constant, then at all points in the area the motion is steady.

For example, if a section of a stream be taken at right angles
to the direction of flow of the stream, and the mean rate at which
water flows through this section is constant, it is convenient
to assume that at any point in the section, the velocity always
remains constant both in magnitude and direction, although the
velocity at different points may not be the same.

Mean velocity. The mean velocity through the section, or the
mean velocity of the stream, is equal to the quantity of flow per
unit time divided by the area of the section.

29. Stream line motion.

The particles of a fluid in motion are frequently regarded as
flowing along definite paths, or in thread-like filaments, and when
the motion is steady these filaments are supposed to be fixed in
position. In a pipe or channel of constant section, the filaments
are generally supposed to be parallel to the sides of the channel.
It will be seen later that such an ideal condition of flow is only
realised in very special cases, but an assumption of such flow if
not abused is helpful in connection with hydraulic problems.

HYDRAULICS

30. Definitions relating to flow of water.

Pressure head. The pressure head at a point in a fluid at rest
has been defined as the vertical distance of the point from the free

surface of the fluid, and is equal to , where p is the pressure per

sq. foot and w is weight per cubic foot of
the fluid. Similarly, the pressure head at
any point in a moving fluid at which the

pressure is p Ibs. per sq. foot, is - feet,

and if a vertical tube, called a piezometer
tube, Fig. 31, be inserted in the fluid, it
will rise in the tube to a height h t which
equals the pressure head above the atmo-
spheric pressure. If p is the pressure per
sq. foot, above the atmospheric pressure,

h = , but if p is the absolute pressure per
sq. foot, and p A the atmospheric pressure,

\L*A

Fig. 31.

W W

Velocity head. If through a small area around the point B,
the velocity of the fluid is v feet per second, the velocity head is

5- , g being the acceleration due to gravity in feet per second per

second.

Position head. If the point B is at a height z feet above any
convenient datum level, the position head of the fluid at B above
the given datum is said to be z feet.

31. Energy per pound of water passing any section in
a stream line.

The total amount of work that can be obtained from every
pound of water passing the point B, Fig. 31, assuming it can fall to
the datum level and that no energy is lost, is

w 2g

Proof. Work available due to pressure head. That the work
which can be done by the pressure head per pound is ^ foot

pounds can be shown as follows.

Imagine a piston fitting into the end of a small tube of cross
sectional area a, in which the pressure is h feet of water as in

FLUIDS IN MOTION 39

Fig. 32, and let a small quantity 3Q cubic feet of water enter the
tube and move the piston through a small dis-
tance dx.

Then dQ,=a.dx.

The work done on the piston as it enters
will be

w . h . a . dx = u

But the weight of dQ cubic feet is w . 9Q pounds, Fl 8- 32 -

and the work done per pound is, therefore, h, or foot pounds.

A pressure head h is therefore equivalent to h foot pounds of
energy per pound of water.

Work available due to velocity. When a body falls through
a height h feet, the work done on the body by gravity is h foot
pounds per pound. It is shown in books on mechanics that if the
body is allowed to fall freely, that is without resistance, the
velocity the body acquires in feet per second is

v = \i2ghj

* -L

2-g = h '

And since no resistance is offered to the motion, the whole of
the work done on the body has been utilised in giving kinetic

v 2
energy to it, and therefore the kinetic energy per pound is ^- .

In the case of the fluid moving with velocity v, an amount of

energy equal to -j foot pounds per pound is therefore available

before the velocity is destroyed.

Work available due to position. If a weight of one pound
falls through the height z the work done on it by gravity will be
z foot pounds, and, therefore, if the fluid is at a height z feet above
any datum, as for example, water at a given height above the
sea level, the available energy on allowing the fluid to fall to
the datum level is z foot pounds per pound.

32. Bernoulli's theorem.

In a steady moving stream of an incompressible fluid in which
the particles of fluid are moving in stream lines, and there is no
loss by friction or other causes

f) V*

+ cT + z
w 2g

is constant for all sections of the stream. This is a most important
theorem and should be carefully studied by the reader.

HYDRAULICS

It has been shown in the last paragraph that this expression
represents the total amount of energy per pound of water flowing
through any section of a stream, and since, between any two
points in the stream no energy is lost, by the principle of the
conservation of energy it can at once be inferred that this
expression must be constant for all sections of a steady flowing
stream. A more general proof is as follows.

Let DE, Fig. 33, be the path of a particle of the fluid.

Fig. 33.

Imagine a small tube to be surrounding DE, and let the flow
in this be steady, and let the sectional area of the tube be so small
that the velocity through any section normal to DE is uniform.

Then the amount of fluid that flows in at D through the area
AB equals the amount that flows out at E through the area OF.

Let p D and VDJ and p E and V E be the pressures and velocities at
D and E respectively, and A and a the corresponding areas of the
tube.

Let z be the height of D above some datum and z^ the height
of E.

Then, if a quantity of fluid ABAiBi equal to 3Q enters at D,
and a similar quantity CFCiFi leaves at E, in a time tit, the
velocity at D is

_3Q_
VD ~Ad*'

and the velocity at E is VE = ^

The kinetic energy of the quantity of fluid dQ entering at D

FLUIDS IN MOTION 41

and that of the liquid leaving at E

Since the flow in the tube is steady, the kinetic energy of the
portion ABCF does not alter, and therefore the increase of the
kinetic energy of the quantity dQ

The work done by gravity is the same as if ABBiAi fell to
i and therefore equals

w . 8Q (z - Zi).

The total pressure on the area AB is p D . A, and the work done
at D in time dt

and the work done by the pressure at B in time t

= pE #UE dt = PE dQ.

But the gain of kinetic energy must equal the work done, and
therefore

-nj (t>B 2 - ^D 2 ) = wdQ l (z- Zi) + p D 3Q - PB ^Q.
From which

^ + PE + + PD + tant>

2gr w 2g w

From this theorem it is seen that, if at points in a steady
moving stream, a vertical ordinate equal to the velocity head plus
the pressure head is erected, the upper extremities of these
ordinates will be in the same horizontal plane, at a height H

/VJ /*

equal to + ?r- + z above the datum level.
w 2g

Mr Froude* has given some very beautiful experimental illus-
trations of this theorem.

In Fig. 34 water is taken from a tank or reservoir in which
the water is maintained at a constant level by an inflowing
stream, through a pipe of variable diameter fitted with tubes
at various points. Since the pipe is short it may be supposed to
be frictionless. If the end of the pipe is closed the water will rise
in all the tubes to the same level as the water in the reservoir, but
if the end C is opened, water will flow through the pipe and the
water surfaces in the tubes will be found to be at different levels.

* British Assoc. Keport 1875.

HYDRAULICS

The quantity of water flowing per second through the pipe can be
measured, and the velocities at A, B, and C can be found by
dividing this quantity by the cross-sectional areas of the pipe at
these points.

Fig. 34.

If to the head of water in the tubes at A and B the ordinates
5^- and ^ be added respectively, the upper extremities of these

ordinates will be practically on the same level and nearly level
with the surface of the water in the reservoir, the small difference
being due to fractional and other losses of energy.

At C the pressure is equal to the atmospheric pressure, and
neglecting friction in the pipe, the whole of the work done by
gravity on any water leaving the pipe while it falls from the
surface of the water in the reservoir through the height H, which
is H ft. Ibs. per pound, is utilised in giving velocity of motion to
the water, and, as will be seen later, in setting up internal motions.

Neglecting these resistances,

Due to the neglected losses, the actual velocity measured will be
less than v c as calculated from this equation.

If at any point D in the pipe, the sectional area is less than the
area at C, the velocity will be greater than V G , and the pressure
will be less than the atmospheric pressure.

If v is the velocity at any section of the pipe, which is supposed
to be horizontal, the absolute pressure head at that section is

w w 2g w 2<7 2g'

p a being the atmospheric pressure at the surface of the water in
the reservoir.

At D the velocity -UD is greater than v and therefore p^ is less

FLUIDS IN MOTION

than p a . If coloured water be put into the vessel E, it will rise in
the tube DE to a height

2g'

If the area at the section is so small, that p becomes negative, the
!luid will be in tension, and discontinuity of flow will take place.

If the fluid is water which has been exposed to the atmosphere
and which consequently contains gases in solution, these gases
will escape from the water if the pressure becomes less than the
tension of the dissolved gases, and there will be discontinuity even
before the pressure becomes zero.

Figs. 35 and 36 show two of Froude's illustrations of the
theorem.

Fig. 35.

Fig. 36.

&t the section B, Fig. 36, the pressure head is hs and the
velocity head is

If a is the section of the pipe at A, and a t at B, since there
is continuity of flow,

and

If now a is made so that

the pressure head h A becomes equal to the atmospheric pressure,
and the pipe can be divided at A, as shown in the figure.

Professor Osborne Reynolds devised an interesting experiment,
to show that when the velocity is high, the pressure is small.

He allowed water to flow through a tube f inch diameter
under a high pressure, the tube being diminished at one section to
0'05 inch diameter.

44 HYDRAULICS

At this diminished section, the velocity was very high and the
pressure fell so low that the water boiled and made a hissing
noise.

33. Venturi meter.

An application of Bernoulli's theorem is found in the Venturi
meter, as invented by Mr Clemens Herschel*. The meter takes
its name from an Italian philosopher who in the last decade of the
18th century made experiments upon the flow of water through
conical pipes. In its usual form the Venturi meter consists of two
truncated conical pipes connected together by a short cylindrical
pipe called the throat, as shown in Figs. 37 and 38. The meter is
inserted horizontally in a line of piping, the diameter of the large
ends of the frustra being equal to that of the pipe.

Piezometer tubes or other pressure gauges are connected to
the throat and to one or both of the large ends of the cones.

Let a be the area of the throat.

Let 0,1 be the area of the pipe or the large end of the cone
at A.

Let a 2 be the area of the pipe or the large end of the cone
atC.

Let p be the pressure head at the throat.

Let pi be the pressure head at the up-stream gauge A.

Let p 2 be the pressure head at the dcrwn-stream. gauge C.

Let H and H a be the differences of pressure head at the throat
and large ends A and C of the cone respectively, or

H =P>-,

w w 9
and H, = -E.

W W

Let Q be the flow through the meter in cubic feet per sec.
Let v be the velocity through the throat.
Let v l be the velocity at the up-stream large end of cone A.
Let v 2 be the velocity at the down-stream large end of cone C.
Then, assuming Bernouilli's theorem, and neglecting friction,

+ + Sfc+SL

w 2g w 2g w 2g*

and H = ^.

If v 2 is equal to Vi, p 2 is theoretically equal to pi, but there is
always in practice a slight loss of head in the meter, the difference
pi ~ Pa being equal to this loss of head.

* Transactions Am.S.C.E., 1887.

FLUIDS IN MOTION

The velocity v is , and v l is - .
a a\

Therefore Q* (^ - ^] = %q . H,
\a efc /

and

-a

HYDRAULICS

Due to friction, and eddy motions that may be set up in the
meter, the discharge is slightly less than this theoretical value, or

v ch 2 a 2

(1)

*Jc being a coefficient which has to be determined by experiment.
For meters having a throat diameter not less than 2 inches and for
pipe line velocities not less than 1 foot per second a value of 0'985
for h will probably give discharges within an error of from 2 to 2*5
per cent. For smaller meters and lower velocities the error may
be considerable and special calibrations are desirable.

For a meter having a diameter of 25*5 inches at the throat and
54 inches at the large end of the cone, Herschel found the
following values for fc, given in Table III, so that the coefficient
varies but little for a large variation of H.

TABLE III.

Herschel

Coker

Hfeet

Discharge
in cu. ft.

995

0418

9494

992

0319

9587

985

0254

9572

9785

0185

9920

977

0096

1-2021

970

0084

1-3583

Professor Coker t, from careful experiments on an exceedingly
well designed small Yenturi meter, Fig. 38, the area of the throat
of which was "014411 sq. feet, found that for small flows the
coefficient was very variable as shown in Table III.

These results show, as pointed out by Professor Coker from an
analysis of his own and Herschel's experiments on meters of
various sizes, that in large Venturi meters, the discharge is very
approximately proportional to the square root of the head, but for
small meters it only follows this law for high heads.

Example. A Venturi meter having a diameter at the throat of 3G inches is
inserted in a 9 foot diameter pipe.

The pressure head at the throat gauge is 20 feet of water and at the pipe gauge
is 26 feet.

* See paper by Gibson, Proc. Inst. C.E. Vol. cxcix.
t Canadian Society of Civil Engineers, 1902.

FLUIDS IN MOTION

Find the discharge, and the velocity of flow through the throat.
The area of the pipe is 63'5 sq. feet.
throat 7-05

The difference in pressure head at the two gauges is 6 feet.

Therefore

x 32-2x6

= _*4o_ ^/sse
= 137 c. ft. per second.
The velocity of flow in the pipe is 2'15 ft. per sec.

through the throat is 19-4 ft. per sec.

34. Steering of canal boats.

An interesting application of Bernoulli's theorem is to show
the effect of speed and position on the steering of a canal boat.

When a boat is moved at a high velocity along a narrow
and shallow canal, the boat tends to leave behind it a hollow
which is filled by the water rushing past the boat as shown
in Figs. 39 and 40, while immediately in front of the boat the
impact of the bow on the still water causes an increase in the
pressure and the water is " piled up " or is at a higher level than
the still water, and what is called a bow wave is formed.

Fig. 39.

Fig. 41.

Fig. 40.

Let it be assumed that the water moves past the boat in
stream lines.

If vertical sections are taken at B and F, and the points E and
F are on the same horizontal line, by Bernoulli's theorem

PE + V = Pv + V
w 2g n 2g '

At B the water is practically at rest, and therefore v s is
zero, and

Pv _ PI + v
w w 2g'

The surface at E will therefore be higher than at F.

4S HYDRAULICS

When the boat is at the centre of the canal the stream lines on
both sides of the boat will have the same velocity, but if the boat
is nearer to one bank than the other, as shown in the figures, the
velocity v F ' of the stream lines between the boat and the nearer
bank, Fig. 41, will be higher than the velocity v v on the other
side. But for each side of the boat

PE = PF + v^ = pr + v^
w w 2# w 2g '

And since vy is greater than t? P , the pressure head p F is
greater than >r, or in other words the surface of the water at
the right side D of the boat will be higher than on the left side B.

The greater pressure on the right side D tends to push the
boat towards the left bank A, and at high speeds considerably
increases the difficulty of steering.

This difficulty is diminished if the canal is made sufficiently
deep, so that flow can readily take place underneath the boat.

35. Extension of Bernoulli's theorem.

In deducing this theorem it has been assumed that the fluid
is a perfect fluid moving with steady motion and that there are no
losses of energy, by friction of the surfaces with which the fluid
may be in contact, or by the relative motion of consecutive ele-
ments of the fluid, or due to internal motions of the fluid.

In actual cases the value of

***

w 2g

diminishes as the motion proceeds.

If hf is the loss of head, or loss of energy per pound of fluid,
between any two given points A and B in the stream, then more
generally

w 2g w 2g

EXAMPLES.

(1) The diameter of the throat of a Venturi meter is | inch, and of
the pipe to which it is connected If inches. The discharge through the
meter in 20 minutes was found to be 314 gallons.

The difference in pressure head at the two gauges was 49 feet.
Determine the coefficient of discharge.

(2) A Venturi meter has a diameter of 4 ft. in the large part and
1-25 ft. in the throat. With water flowing through it, the pressure head is
100 ft. in the large part and 97 ft. at the throat. Find the velocity in the
small part and the discharge through the meter. Coefficient of meter
taken as unity.

FLUIDS IN MOTION 49

(3) A pipe AB, 100 ft. long, has an inclination of 1 in 5. The head due
to the pressure at A is 45 ft., the velocity is 3 ft. per second, and the section
of the pipe is 3 sq. ft. Find the head due to the pressure at B, where the
section is 1 sq. ft. Take A as the lower end of the pipe.

(4) The suction pipe of a pump is laid at an inclination of 1 in 5, and
water is pumped through it at 6 ft. per second. Suppose the air in the
water is disengaged if the pressure falls to more than 10 Ibs. below
atmospheric pressure. Then deduce the greatest practicable length of
suction pipe. Friction neglected.

(5) Water is delivered to an inward-flow turbine under a head of 100 feet
(see Chapter IX). The pressure just outside the wheel is 25 Ibs. per
sq. inch by gauge.

Find the velocity with which the water approaches the wheel. Friction
neglected.

(6) A short conical pipe varying in diameter from 4' 6" at the large end
to 2 feet at the small end forms part of a horizontal water main. The
pressure head at the large end is found to be 100 feet, and at the small end
96-5 feet.

Find the discharge through the pipe. Coefficient of discharge unity.

(7) Three cubic feet of water per second flow along a pipe which as it
falls varies in diameter from 6 inches to 12 inches. In 50 feet the pipe
falls 12 feet. Due to various causes there is a loss of head of 4 feet.

Find (a) the loss of energy in foot pounds per minute, and in horse-
power, and the difference in pressure head at the two points 50 feet apart.
(Use equation 1, section 35.)

(8) A horizontal pipe in which the sections vary gradually has sections
of 10 square feet, 1 square foot, and 10 square feet at sections A, B, and C.
The pressure head at A is 100 feet, and the velocity 3 feet per second.
Find the pressure head and velocity at B.

Given that in another case the difference of the pressure heads at A
and B is 2 feet. Find the velocity at A.

(9) A Venturi meter in a water main consists of a pipe converging to
the throat and enlarging again gradually. The section of main is 9 sq. ft.
and the area of throat 1 sq. ft. The difference of pressure in the main and
at the throat is 12 feet of water. Find the discharge of the main per hour.

(10) If the inlet area of a Venturi meter is n times the throat area, and
v and p are the velocity and pressure at the throat, and the inlet pressure
is mp, show that

and show that if p and mp are observed, v can be found.

(11) Two sections of a pipe have an area of 2 sq. ft. and 1 sq. ft.
respectively. The centre of the first section is 10 feet higher than that of
the second. The pressure head at each of the sections is 20 feet. Find
the energy lost per pound of flow between the two sections, when 10 c. ft.
of water per sec. flow from the higher to the lower section.

L. 11. 4

CHAPTER TV.

FLOW OF WATER THROUGH ORIFICES AND
OVER WEIRS.

36. Flow of fluids through orifices.

The general theory of the discharge of fluids through orifices,
as for example the flow of steam and air, presents considerable
difficulties, and is somewhat outside the scope of this treatise.
Attention is, therefore, confined to the problem of determining the
quantity of water which flows through a given orifice in a given
time, and some of the phenomena connected therewith.

In what follows, it is assumed that the density of the fluid is
constant, the effect of small changes of temperature and pressure
in altering the density being thus neglected.

Consider a vessel, Fig. 42, filled with water, the free surface of
which is maintained at a constant level; in the lower part of the
vessel there is an orifice AB.

Fig. 42.

Let it be assumed that although water flows into the vessel so
as to maintain a constant head, the vessel is so large that at some
surface CD, the velocity of flow is zero.

Imagine the water in the vessel to be divided into a number of
stream lines, and consider any stream line EF.

Let the velocities at E and F be V E and V F , the pressure heads
h E and 7i F , and the position heads above some datum, Z E and z p ,
respectively.

FLOW THROUGH ORIFICES

Then, applying Bernoulli's theorem to the stream line EF,
If v f is zero, then

VK ^ Vv

+ ^T- + ZE = ft? + or +

7^ = Tip - & E + z
But from the figure it is seen that

is equal to h, and therefore

V E =

Since h% is the pressure head at E, the water would rise in
a tube having its end open at E, a height h E , and h may thus
be called following Thomson the fall of "free level for the
point E."

At some section GK near to the orifice the stream lines are all
practically normal to the section, and the pressure head will be
equal to the atmospheric pressure ; and if the orifice is small the fall
of free level for all the stream lines is H, the distance of the centre
of the section GK below the free surface of the water. If the
orifice is circular and sharp-edged, as in Figs. 44 and 45, the section
GK is at a distance, from the plane of the orifice, about equal to
its radius. For small vertical orifices, and horizontal orifices,
H may be taken as equal to the distance of the centre of the
orifice below the free surface.

The theoretical velocity of flow through the small section GK
is, therefore, the same for all the stream lines, and equal to the
velocity which a body will acquire, in falling, in a vacuum,
through a height, equal to the depth of the centre of the orifice
below the free surface of the water in the vessel.

The above is Thomson's proof of Torricelli's theorem, which
was discovered experimentally, by him, about
the middle of the 17th century.

The theorem is proved experimentally as
follows.

If the aperture is turned upwards, as in
Fig. 43, it is found that the water rises
nearly to the level of the water in the vessel,
and it is inferred, that if the resistance of the
air and of the orifice could be eliminated, the
jet would rise exactly to the level of the
surface of the water in the vessel.

HYDRAULICS

Other experiments described on pages 5456, also show that,
with carefully constructed orifices, the mean velocity through the
orifice differs from v/2#H by a very small quantity.

37. Coefficient of contraction for sharp-edged orifice.

If an orifice is cut in the flat side, or in the bottom of a vessel,
and has a sharp edge, as shown in Figs. 44 and 45, the stream lines
set up in the water approach the orifice in all directions, as shown
in the figure, and the directions of flow of the particles of water,
except very near the centre, are not normal to the plane of the
orifice, but they converge, producing a contraction of the jet.

Fig. 44.

Fig. 45.

At a small distance from the orifice the stream lines become
practically parallel, but the cross sectional area of the jet is
considerably less than the area of the orifice.

If <o is the area of the jet at this section and a the area of the

orifice the ratio - is called the coefficient of contraction and may

be denoted by c. Weisbach states, that for a circular orifice, the
jet has a minimum area at a distance from the orifice slightly less
than the radius of the orifice, and defines the coefficient of
contraction as this area divided by the area of the orifice. For a
circular orifice he gives to c the value 0'64. Recent careful
measurements of the sections of jets from horizontal and vertical
sharp-edged circular and rectangular orifices, by Bazin, the
results of some of which are shown in Table IV, show, however,
that the section of the jet diminishes continuously and in fact has
no minimum value. Whether a minimum occurs for square orifices
is doubtful.

The diminution in section for a greater distance than that
given by Weisbach is to be expected, for, as the jet moves away
from the orifice the centre of the jet falls, and the theoretical
velocity becomes \/2g (H + y),y being the vertical distance between
the centre of the orifice and the centre of the jet.

FLOW THROUGH ORIFICES 53

At a small distance away from the orifice, however, the stream
lines are practically parallel, and very little error is introduced in
the coefficient of contraction by measuring the stream near the
orifice.

Poncelet and Lesbros in 1828 found, for an orifice '20 m. square,
a minimum section of the jet at a distance of *3 m. from the orifice
and at this section c was '563. M. Bazin, in discussing these
results, remarks that at distances greater than 0'3 m. the section
becomes very difficult to measure, and although the vein appears
to expand, the sides become hollow, and it is uncertain whether
the area is really diminished.

Complete contraction. The maximum contraction of the jet
takes place when the orifice is sharp edged and is well removed
from the sides and bottom of the vessel. In this case the contrac-
tion is said to be complete. Experiments show, that for complete
contraction the distance from the orifice to the sides or bottom of
the vessel should not be less than one and a half to twice the least
diameter of the orifice.

Incomplete or sn/ppressed contraction. An example of incom-
plete contraction is shown in Fig. 46, the lower edge of the
rectangular orifice being made level with the bottom of the vessel.
The same effect is produced by placing a horizontal plate in
the vessel level with the bottom of the orifice. The stream
lines at the lower part of the orifice are normal to its plane
and the contraction at the lower edge is consequently suppressed.

Fig. 46. Fig. 47.

Similarly, if the width of a rectangular orifice is made equal
to that of the vessel, or the orifice abed is provided with side walls
as in Fig. 47, the side or lateral contraction is suppressed. In any
case of suppressed contraction the discharge is increased, but, as
will be seen later, the discharge coefficient may vary more than
when the contraction is complete. To suppress the contraction
completely, the orifice must be made of such a form that the
stream lines become parallel at the orifice and normal to its plane.

HYDRAULICS

Fig. 49.

Experimental determination of c. The section of the stream
from a circular orifice can be obtained with considerable accu-
racy by the apparatus shown in Fig. 49, which consists of a
ring having four radial set
screws of fine pitch. The
screws are adjusted until the
points thereof touch the jet.
M. Bazin has recently used an
octagonal frame with twenty-
four set screws, all radiating
to a common centre, to deter-
mine the form of the section
of jets from various kinds of
orifices. Fig. 48.

The screws were adjusted
until they just touched the jet. The frame was then placed upon
a sheet of paper and the positions of the ends of the screws
marked upon the paper. The forms of the sections could then
be obtained, and the areas measured with considerable accuracy.
Some of the results obtained are shown in Table IV and also in
the section on the form of the liquid vein.

38. Coefficient of velocity for sharp-edged orifice.

The theoretical velocity through the contracted section is, as
shown in section 36, equal to \/2#H, but the actual velocity
t?i is slightly less than this due to friction at the orifice. The

ratio - L = Jc is called the coefficient of velocity.

Experimental determination of k. There are two methods
adopted for determining k experimentally.

First method. The velocity is determined by measuring the
discharge in a given time under a given head, and the cross
sectional area o> of the jet, as explained in the last paragraph, is
also obtained. Then, if Vi is the actual velocity, and Q the
discharge per second,

and

Second method. An orifice, Fig. 50, is formed in the side of a
vessel and water allowed to flow from it. The water after leaving
the orifice flows in a parabolic curve. Above the orifice is fixed
a horizontal scale 011 which is a slider carrying a vertical scale,
to the bottom of which is clamped a bent piece of wire, with a sharp

FLOW THROUGH ORIFICES

point. The vertical scale can be adjusted so tliat the point touches
the upper or lower surface of the jet, and the horizontal and vertical
distances of any point in the axis of the jet from the centre of the
orifice can thus be obtained.

Fig. 50.

Assume the orifice is vertical, and let Vi be the horizontal
velocity of flow. At a time t seconds after a particle has passed
the orifice, the distance it has moved horizontally is

x = vj .................................... (1).

The vertical distance is

v = \gf ................................. (2).

Therefore y = \g 3

and Vl = x V jfy'

The theoretical velocity of flow is

Therefore

It is better to take two values of x and y so as to make
allowance for the plane of the orifice not being exactly perpen-
dicular.

If the orifice has its plane inclined at an angle to the
vertical, the horizontal component of the velocity is Vi cos and
the vertical component Vi sin 0.

At a time t seconds after a particle has passed the orifice, the
horizontal movement from the orifice is,

X = ViCOS0t ........................... (1),

and the vertical movement is,

y = v 1 *m0t + lgp .................... .(2).

After a time ti seconds Xi = v i co&0t l ........................... (3),

y l = vi sin 6^ + \gt? .................... (4).

56 HYDRAULICS

Substituting the value of t from (1) in (2) and t r from (3)
in (4),

and, m -

Prom (5),

# y-xtanO

Substituting for v* in (6),

tan^^'-^y. ..(8).

XX l (X - 0?i)

Having calculated tan 0, sec can be found from mathematical
tables, and from (7) Vi can be calculated. Then

39. Bazin's experiments on a sharp-edged orifice.

In Table IY are given values of k as obtained by Bazin from
experiments on vertical and horizontal sharp-edged orifices, for
various values of the head.

The section of the jet at various distances from the orifice was
carefully measured by the apparatus described above, and the
actual discharge per second was determined by noting the time
taken to fill a vessel of known capacity.

The mean velocity through any section was then

Q being the discharge per second and A the area of the section.

The fall of free level for the various sections was different, and
allowance is made for this in calculating the coefficient k in the
fourth column.

Let y be the vertical distance of the centre of any section
below the centre of the orifice ; then the fall of free level for that
section is H + y and the theoretical velocity is

The coefficients given in column 3 were determined by dividing
the actual mean velocity through different sections of the jet by
\/2#H, the theoretical velocity at the centre of the orifice.

Those in column 4 were found by dividing the actual mean
velocity through the section by */2g (H + y), the theoretical
velocity at any section of the jet.

The coefficient of column 3 increases as the section is taken
further from the jet, and in nearly all cases is greater than unity.

FLOW THROUGH ORIFICES

TABLE IV.

Sharp-edged Orifices Contraction Complete.

Table showing the ratio of the area of the jet to the area of
the orifice at definite distances from the orifice, and the ratio of
the mean velocity in the section to \/2^H and to \/2g . (H + 7/),
H being the head at the centre of the orifice and y the vertical
distance of the centre of the section of the jet from the centre of
the orifice.

Vertical circular orifice 0*20 m. ('656 feet) diameter, H = '990 m.
(3-248 feet).

Coefficient of discharge m, by actual measurement of the flow is

Distance of the section

from the plane of the

orifice in metres

0-08

0-13

0-17

0-235

0-335

0-515

Area of Jet

Area of Orifice

6079
5971
5951
5904
5830
5690

Mean Velocity Mean Velocity

0-983
1-001
1-004
1-012
1-025
1-050

998

999

1-003

1-007

1-010

Horizontal circular orifice 0*20 m. ('656 feet) diameter,
= '975m. (3198 feet).

m = 0-6035.

0-075
0-093
0-110
0-128
0-145
0-163

0-6003
0-5939
0-5824
0-5734
0-5658
0-5597

1-005
1-016
1-036
1-053
1-067
1-078

0-968
0-971
0-982
0-990
0-996
0-998

Vertical orifice '20 m. ('656 feet) square, H = '953 m. (3126 feet)
m = 0'6066.

0-151
0-175
0-210
0-248
0-302
0-350

0-6052
0-6029
0-5970
0-5930
0-5798
0-5783

1-002
1-006
1-016
1-023
1-046
1-049

997
1-000
1-007
1-010
1-027
1-024

The real value of the coefficient for the various sections is
however that given in column 4.

For the horizontal orifice, for every section, it is less than
unity, but for the vertical orifice it is greater than unity.

Bazin's results confirm those of Lesbros and Poncelet, who in

* See section 42 and Appendix 1.

58 HYDRAULICS

1828 found that the actual velocity through the contracted section
of the jet, even when account was taken of the centre of the
section of the jet being below the centre of the orifice, was
-sV greater than the theoretical value.

This result appears at first to contradict the principle of the
conservation of energy, and Bernoulli's theorem.

It should however be noted that the vertical dimensions of the
orifice are not small compared with the head, and the explanation
of the apparent anomaly is no doubt principally to be found in the
fact that the initial velocities in the different horizontal filaments
of the jet are different.

Theoretically the velocity in the lower part of the jet is greater
than \/2<jr (H + y), and in the upper part less than \/2g (H + y).

Suppose for instance a section of a jet, the centre of which is
1 metre below the free surface, and assume that all the filaments
have a velocity corresponding to the depth below the free surface,
and normal to the section. This is equivalent to assuming that
the pressure in the section of the jet is constant, which is probably
not true.

Let the jet be issuing from a square orifice of '2 m. ('656 feet)
side, and assume the coefficient of contraction is "6, and for
simplicity that the section of the jet is square.

Then the side of the jet is '1549 metres.

The theoretical velocity at the centre is \/2#, and the discharge
assuming this velocity for the whole section is

6 x -04 x *Jfy = '024 J2g cubic metres.

The actual discharge, on the above assumption, through any
horizontal filament of thickness dh, and depth h, is

3Q = 01549 x<ta
and the total discharge is

rl -0775

Q = 01549

The theoretical discharge, taking account of the varying heads
is, therefore, 1*004 times the discharge calculated on the assumption
that the head is constant.

As the head is increased this difference diminishes, and when
the head is greater than 5 times the depth of the orifice, is very
small indeed.

The assumed data agrees very approximately with that given
in Table IY for a square orifice, where the value of A; is given as
1-006.

FLOW THROUGH ORIFICES 59

This partly then, explains the anomalous values of fe, but it
cannot be looked upon as a complete explanation.

The conditions in the actual jet are not exactly those assumed,
and the variation of velocity normal to the plane of the section is
probably much more complicated than here assumed.

As Bazin further points out, it is probable that, in jets like
those from the square orifice, which, as will be seen later when the
form of the jet is considered, are subject to considerable deformation,
the divergence of some of the filaments gives rise to pressures less
than that of the atmosphere.

Bazin has attempted to demonstrate this experimentally, and
his instrument, Fig. 150, registered pressures less than that of the
atmosphere; but he doubts the reliability of the results, and
points out the extreme difficulty of satisfactorily determining the
pressure in the jet.

That the inequality of the velocity of the filaments is the
primary cause, receives support from the fact that for the
horizontal orifice, discharging downwards, the coefficient Jc is
always slightly less than unity. In this case, in any horizontal
section below the orifice, the head is the same for all the stream
lines, and the velocity of the filaments is practically constant.
The coefficient of velocity is never less than '96, so that the loss
due to the internal friction of the liquid is very small.

40. Distribution of velocity in the plane of the orifice.
Bazin has examined the distribution of the velocity in the

various sections of the jet by means of a fine Pitot tube (see
page 245). In the plane of the orifice a minimum velocity
occurs, which for vertical orifices is just above the centre, but at a
little distance from the orifice the minimum velocity is at the top
of the jet.

For orifices having complete contraction Bazin found the
minimum velocity to be '62 to '64 N/20H, and for the rectangular
orifice, with lateral contraction suppressed, 0'69 N/20H.

As the distance from the plane of the orifice increases, the
velocities in the transverse section of the jets from horizontal
orifices, rapidly become uniform throughout the transverse section.

For vertical orifices, the velocities below the centre of the jet
are greater than those in the upper part.

41. Pressure in the plane of the orifice.

M. Lager j elm stated in 1826 that if a vertical tube open at
both ends was placed with its lower end near the centre, and not
perceptibly below the plane of the inner edge of a horizontal

60 HYDRAULICS

orifice made in the bottom of a large reservoir, the water rose in
the tube to a height equal to that of the water in the reservoir,
that is the pressure at the centre of the orifice is equal to the head
over the orifice even when flow is taking place.

M. Bazin has recently repeated this experiment and found,
that the water in the tube did not rise to the level of the water in
the reservoir.

If Lager j elm's statement were correct it would follow that the
velocity at the centre of the orifice must be zero, which again does
not agree with the results of Bazin's experiments quoted above.

42. Coefficient of discharge.

The discharge per second from an orifice, is clearly the area
of the jet at the contracted section GK multiplied by the mean
velocity through this section, and is therefore,

Q = c.fc. as/2011.
Or, calling m the coefficient of discharge,

This coefficient m is equal to the product c.Jc. It is the only
coefficient required in practical problems and fortunately it can
be more easily determined than the other two coefficients c and k.

Experimental determination of the coefficient of discharge.
The most satisfactory method of determining the coefficient of
discharge of orifices is to measure the volume, or the weight of
water, discharged under a given head in a known time.

The coefficients emoted in the Tables from M. Bazin*, were
determined by finding\accurately the time required to fill a vessel
of known capacity.

The coefficient of discharge m } has been determined with
a great degree of accuracy for sharp-edged orifices, by Poncelet
and Lesbrost, WeisbachJ, Bazin and others . In Table IY
Bazin's values for m are given.

The values as given in Tables Y and VI may be taken as
representative of the best experiments.

For vertical, circular and square orifices, and for a head of
about 3 feet above the centre of the orifice, Mr Hamilton Smith,
junr.H, deduces the values of m given in Table YL

* Annales des Fonts et Chaussees, October, 1888.

f Flow through Vertical Orifices.

j Mechanics of Engineering.

% Experiments upon the Contraction of the Liquid Vein. Bazin translated by
Trautwine. Also see Appendix and the Bulletins of the University of Wisconsin.

|| The Flow of Water through Orifices and over Weirs and through open Conduits
and Pipes, Hamilton Smith, junr., 1886.

FLOW THROUGH ORIFICES

TABLE V.

Experimenter

Particulars of orifice

Coefficient of
discharge 771

Bazin

Vertical square orifice side of square 0'6562 ft.

0-606

Poncelet and

f\.ar\K

Lesbros

(J DUO

Bazin

Vertical Kectangular orifice '656 ft. high x 2'624
ft. wide with side contraction suppressed

0-627

Vertical circular orifice 0*6562 ft. diameter

0-598

Horizontal

0-6035

0-3281

0-6063

TABLE VI.

Circular orifices.

Diameter of
orifice in ft.

0-0197
0-627

0-0295
0-617

0-039
0-611

0-0492
0-606

0-0984
0-603

0-164
0-600

0-328
0-599

0-6562
0-598

0-9843
0-597

Square orifices.

Side of square
in feet

0-0197
0-631

0-0492
0-612

0-0984
0-607

0-197
0-605

0-5906
0-604

0-9843
0-603

TABLE VII.

Table showing coefficients of discharge for square and rect-
angular orifices as determined by Poncelet and Lesbros.

Width of orifice -6562 feet

Width of orifice
j-968 feet

Head of water

above the top

of the orifice

Depth of orifice in feet

in feet

0328

0656

0984

1640

3287

6562

0656

6562

0328

701

660

630

607

0656

694

659

634

615

596

572

643

1312

683

658

640

623

603

582

642

595

2624

670

656

638

629

610

589

640

601

3937

663

653

636

630

612

593

638

603

6562

655

648

633

630

615

598

635

605

1-640

642

638

630

627

617

604

630

607

3-281

632

633

628

626

615

605

626

605

4-921

615

619

620

611

602

623

602

6-562

611

612

613

607

601

620

602

9-843

609

610

608

606

603

601

615

601

62 HYDRAULICS

The heads for which Bazin determined the coefficients in
Tables IY and V varied only from 2'6 to 3'3 feet, but, as will be
seen from Table VII, deduced from results given by Poncelet and
Lesbros* in their classical work, when the variation of head is not
small, the coefficients for rectangular and square orifices vary
considerably with the head.

43. Effect of suppressed contraction on the coefficient
of discharge.

Sharp-edged orifice. When some part of the contraction of a
transverse section of a jet issuing from an orifice is suppressed,
the cross sectional area of the jet can only be obtained with
difficulty.

The coefficient of discharge can, however, be easily obtained,
as before, by determining the discharge in a given time. The
most complete and accurate experiments on the effect of contrac-
tion are those of Lesbros, some of the results of which are quoted
in Table VIII. The coefficient is most constant for square or
rectangular orifices when the lateral contraction is suppressed. The
reason being, that whatever the head, the variation in the section
of the jet is confined to the top and bottom of the orifice, the
width of the stream remaining constant, and therefore in a greater
part of the transverse section the stream lines are normal to the
plane of the orifice.

According to Bidone, if x is the fraction of the periphery of a
sharp-edged orifice upon which the contraction is suppressed, and
m the coefficient of discharge when the contraction is complete,
then the coefficient for incomplete contraction is,

mi = m (1 + *15#),
for rectangular orifices, and

m l = m(l + *13aj)
for circular orifices.

Bidone's formulae give results agreeing fairly well with
Lesbros' experiments.

His formulae are, however, unsatisfactory when x approaches
unity, as in that case mi should be nearly unity.

If the form of the formula is preserved, and m taken as '606,
for mi to be unity it would require to have the value,
mi = m (1 + *65oj).

For accurate measurements, either orifices with perfect con-
traction or, if possible, rectangular or square orifices with the
lateral contraction completely suppressed, should be used. It will

* Experiences hydrauliques sur Us lois de Cecoulement de Veau a travers lea
orifices, etc., 1832. ' Poncelet and Lesbros.

FLOW THROUGH ORIFICES

generally be necessary to calibrate the orifice for various heads,
but as shown above the coefficient for the latter kind is more
likely to be constant.

TABLE VIII.

Table showing the effect of suppressing the contraction on the
coefficient of discharge. Lesbros*.

Square vertical orifice 0*656 feet square.

Head of water
above the upper
edge of the orifice

Sharp-edged

Side con-
traction
suppressed

Contraction
suppressed at
the lower edge

Contraction
suppressed at
the lower and
side edges

0-06562

0-572

0-599

0-1640

0-585

0-631

0-608

0-3281

0-592

0-631

0-615

0-6562

0-598

0-632

0-621

0-708

1-640

0-603

0-631

0-623

0-680

3-281

0-605

0-628

0-624

0-676

4-921

0-602

0-627

0-624

0-672

6-562

0-601

0-626

0-619

0-668

9-843

0-601

0-624

0-614

0-665

Fig. 51. Section of jet from
circular orifice.

44. The form of the jet from sharp-edged orifices.

From a circular orifice the jet emerges like a cylindrical rod
and retains a form nearly cylindrical for some distance from the
orifice.

Fig. 51 shows three sections of a jet from a vertical circular
orifice at varying distances from the
orifice, as given by M. Bazin.

The flow from square orifices is
accompanied by an interesting and
curious phenomenon called the in-
version of the jet.

At a very small distance from
the orifice the section becomes as
shown in Fig. 52. The sides of the
jet are concave and the corners are
cut off by concave sections. The
section then becomes octagonal as in
Fig. 53 and afterwards takes the form of a square with concave
sides and rounded corners, the diagonals of the square being
perpendicular to the sides of the orifice, Fig. 54.

Figs. 52 54. Section of jet from
square orifice.

Experiments liydrauliques sur les lois de Vecoulement de Veau.

HYDRAULICS

45. Large orifices.

Table VII shows very clearly that if the depth of a vertical orifice
is not small compared with the head, the coefficient of discharge
varies very considerably with the head, and in the discussion of
the coefficient of velocity &, it has already been shown that the
distribution of velocity in jets issuing from such orifices is not
uniform. As the jet moves through a large orifice the stream
lines are not normal to its plane, but at some section of the stream
very near to the orifice they are practically normal.

If now it is assumed that the pressure is constant and equal to
the atmospheric pressure and that the shape of this section is
known, the discharge through it can be calculated.

Rectangular orifice. Let efgh, Fig. 55, be the section by a
vertical plane EF of the stream issuing from a vertical rectangular
orifice. Let the crest E of the stream be at a depth h below
the free surface of the water in the vessel and the under edge
F at a depth h^.

Fig. 55.

At any depth h, since the pressure is assumed constant in the
section, the fall of free level is h, and the velocity of flow through
the. strip of width dh is therefore, k\/2gh, and the discharge is

If k be assumed constant for all the filaments the total discharge
in cubic feet per second is

Q =

hr,

Here at once a difficulty is met with. The dimensions h , hi
and b cannot easily be determined, and experiment shows that
they vary with the head of water over the orifice, and that they
cannot therefore be written as fractions of H , Hj, and B.

FLOW THROUGH ORIFICES

By replacing h , hi and b by H , Hi and B an empirical
formula of the same form is obtained which, by introducing a
coefficient c, can be made to agree with experiments. Then

or replacing |c by n t

(1).

The coefficient n varies with the head H , and for any orifice
the simpler formula _

Q=m.a.v/2^H .............................. (2),

a being the area of the orifice and H the head at the centre,
can be used with equal confidence, for if n is known for the
particular orifice for various values of H , m will also be known.

From Table VII probable values of ra for any large sharp-
edged rectangular orifices can be interpolated.

Rectangular sluices. If the lower edge of a sluice opening is
some distance above the bottom of the channel the discharge
through it will be practically the same as through a sharp-edged
orifice, but if it is flush with the bottom of the channel, the
contraction at this edge is suppressed and the coefficient of
discharge will be slightly greater as shown in Table VIII.

46. Drowned orifices.

When an orifice is submerged as in Fig. 56 and the water in
the up-stream tank or reservoir is moving so slowly that its velocity
may be neglected, the head causing velocity of flow through any
filament is equal to the difference of the up- and down-stream
levels. Let H be the difference of level of the water on the two
sides of the orifice.

Pfe, 63.

L. H.

HYDRAULICS

Consider any stream line FE which passes through the orifice
at B. The pressure head at E is equal to h z , the depth of E below
the down-stream level. If then at F the velocity is zero,

or taking a coefficient of velocity k

which, since H is constant, is the same for all filaments of the
orifice.

If the coefficients of discharge and contraction are c and m
respectively the whole discharge through the orifice is then

Q = cka v 2<?H = wi . a . v 2yH.
*The coefficient m may be taken as 0'6.
47. Partially drowned orifice.

H the orifice is partially drowned, as in
Fig. 57, the discharge may be considered in
two parts. Through the upper part AC the
discharge, using (2) section 45, is

.-.-

and through the lower part BC

Kg. 57.

48. Velocity of approach.

It is of interest to consider the effect of the
water approaching an orifice having what is
called a velocity of approach, which will be equal to the velocity
of the water in the stream above the orifice.

In Fig. 56 let the water at F approaching the drowned orifice
have a velocity VF.

Bernoulli's equation for the stream line drawn is then

and

which is again constant for all filaments of the orifice.
Then Q = m.c

* Bulletins of University of Wisconsin, Nos. 216 and 270.

SUDDEN ENLARGEMENT OF A STREAM 67

49. Effect of velocity of approach on the discharge
through a large rectangular orifice.

If the water approaching the large orifice, Fig. 55, has
a velocity of approach v\ 9 Bernoulli's equation for the stream line
passing through the strip at depth h y will be

w 2g w

p a being the atmospheric pressure, or putting in a coefficient of
velocity,

The discharge through the orifice is now,

t>,2

50. Coefficient of resistance.

In connection with the flow through orifices, and hydraulic
plant generally, the term " coefficient of resistance " is frequently
used. Two meanings have been attached to the term. Some-
times it is defined as the ratio of the head lost in a hydraulic
system to the effective head, and sometimes as the ratio of the
head lost to the total head available. According to the latter
method, if H is the total head available and h/ the head lost,
the coefficient of resistance is

51. Sudden enlargement of a current of water.

It seems reasonable to proceed from the consideration of flow
through orifices to that of the flow through mouthpieces, but
before doing so it is desirable that the effect of a sudden
enlargement of a stream should be considered.

Suppose for simplicity that a pipe as
in Fig. 58 is suddenly enlarged, and that
there is a continuous sinuous flow along
the pipe. (See section 284.)

At the enlargement of the pipe, the
stream suddenly enlarges, and, as shown
in the figure, in the corners of the large
pipe it may be assumed that eddy motions
are set up which cause a loss of energy.

68 HYDRAULICS

Consider two sections aa and dd at such a distance from 66
that the flow is steady.

Then, the total head at dd equals the total head at aa minus
the loss of head between aa and dd, or if h is the loss of head due
to shock, then

fe + .a + f + fc

w zg w 2g

Let A and A^ be the area at aa and dd respectively.
Since the flow past aa equals that past dd,

t Then, assuming that each filament of fluid at aa has the
velocity v a , and v d at dd, the momentum of the quantity of water

which passes aa in unit time is equal to A^ a 2 , and the momentum
of the water that passes dd is

the momentum of a mass of M pounds moving with a velocity
v feet per second being ~M.v pounds feet.
The change of momentum is therefore,

The forces acting on the water between aa and dd to produce
this change of momentum, are

paAa acting on aa, pd^d acting on dd,

and, if p is the mean pressure per unit area on the annular ring
66, an additional force p(Ad A).

There is considerable doubt as to what is the magnitude of the
pressure p, but it is generally assumed that it is equal to p, for
the following reason.

The water in the enlarged portion of the pipe may be looked
upon as divided into two parts, the one part having a motion of
translation, while the other part, which is in contact with the
annular ring, is practically at rest. (See section 284.)

If this assumption is correct, then it is to be expected that the
pressure throughout this still water will be practically equal at all
points and in all directions, and must be equal to the pressure in
the stream at the section 66, or the pressure p is equal to p a .

Therefore

- A a ) - PA a - W

y
A v
from which (p* - p a ) A<i = w (v a - v<i

SUDDEN ENLARGEMENT OF A STREAM 69

and since A. a v a

,1 f 'Pa Pd Isa^a ^a,

therefore - + .

w w g g

Adding -?j- to both sides of the equation and separating
into two parts,

4. = 4.,

w 2g w 2g

or h the loss of head due to shock is equal to

According to St Yenant this quantity should be increased by

*1 2

an amount equal to ~ ~ , but this correction is so small that as
a rule it can be neglected.

52. Sudden contraction of a current of water.

Suppose a pipe partially closed by means of a diaphragm as in
Fig. 59.

As the stream approaches the diaphragm -
which is supposed to be sharp-edged
it contracts in a similar way to the stream
passing through an orifice on the side of ^ _ _
a vessel, so that the minimum cross sec-
tional area of the flow will be less than the Fig. 59.

area of the orifice*.

The loss of head due to this contraction, or due to passing
through the orifice is small, as seen in section 39, but due to
the sudden enlargement of the stream to fill the pipe again, there
is a considerable loss of head.

Let A be the area of the pipe and a of the orifice, and let c be
the coefficient of contraction at the orifice.

Then the area of the stream at the contracted section is ca, and,
therefore, the loss of head due to shock

* The pressure at the section cc will be less than in the pipe to the left of the
diaphragm. From Bernoulli's equation an expression similar to eq. 1 p. 46 can be
obtained for the discharge through the pipe, and such a diaphragm can be used as
a meter. Proc. Inst. C.E. Vol. cxcvii.

70 HYDRAULICS

If the pipe simply diminishes in diameter as in Fig. 58, the
section of the stream enlarges from the contracted area ca to fill
the pipe of area a, therefore the loss of head in this case is

Or making St Yenant correction

* Value of the coefficient c. The mean value of c for a sharp-edged
circular orifice is, as seen in Table IV, about 0'6, and this may be
taken as the coefficient of contraction in this formula.

Substituting this value in equation (1) the loss of head is

found to be -~ , and in equation (2), -^ , v being the velocity in

the small pipe. It may be taken therefore as -~ . Further
experiments are required before a correct value can be assigned.

53. Loss of head due to sharp-edged entrance into a pipe
or mouthpiece.

When water enters a pipe or mouthpiece from a vessel through
a sharp-edged entrance, as in Fig. 61, there is first a contraction, and
then an enlargement, as in the second case considered in section 52.

The loss of head may be, therefore, taken as approximately -~

and this agrees with the experimental value of ~ - given by

Weisbach.

This value is probably too high for small pipes and too low for
large pipes t.

54. Mouthpieces. Drowned Mouthpieces,

If an orifice is provided with a short pipe or mouthpiece, through
which the liquid can flow, the discharge may be very different
from that of a sharp-edged orifice, the difference depending upon
the length and form of the mouthpiece. If the orifice is cylindrical
as shown in Fig. 60, being sharp at the inner edge, and so short
that the stream after converging at the inner edge clears the
outer edge, it behaves as a sharp-edged orifice.

J Short external cylindrical mouthpieces. If the mouthpiece is
cylindrical as ABFE, Fig. 61, having a sharp edge at AB and
a length of from one and a half to twice its diameter, the jet

* Proc. Inst. C.E. Vol. cxcvn.

f See M. Bazin, Experiences nouvelles sur la distribution des vitesses dans
les tuyaux. { See Bulletins Nos. 216 and 270 University of Wisconsin.

Shorter mouthpieces are unreliable.

FLOW THROUGH MOUTHPIECES

contracts to CD, and then expands to fill the pipe, so that at EF
it discharges full bore, and the coefficient of contraction is then
unity. Experiment shows, that the coefficient of discharge is

Fig. 60.

Fig. 61.

from 0'80 to 0'85, the coefficient diminishing with the diameter
of the tube. The coefficient of contraction being unity, the
coefficients of velocity and discharge are equal. Good mean
values, according to Weisbach, are 0*815 for cylindrical tubes,
and 0'819 for tubes of prismatic form.

These coefficients agree with those determined on the assump-
tion that the only head lost in the mouthpiece is that due to
sudden enlargement, and is

2g '
v being the velocity of discharge at EF.

Applying Bernoulli's theorem to the sections CD and EF, and

taking into account the loss of head of -- , and p a as the atmo-
spheric pressure,

PCD ^CD 2 = Pa.tf_. '5^ 2 p. Pa

"" -<7 w 2g + 2g ^ + - '

= H.

Therefore

and v-

The area of the jet at EF is a, and therefore, the discharge
per second is

Or m, the coefficient of discharge, is 0'812.
The pressure head at the section CD. Taking the area at CD
as 0'606 the area at EF,

tto~ l*65v.

72 HYDRAULICS

Therefore P=& + ^-2^ = _ _

w w 2g 2g w 2g

or the pressure at C is less than the atmospheric pressure.

If a pipe be attached to the mouthpiece, as in Fig. 61, and the
lower end dipped in water, the water should rise to a height of about

o feet above the water in the vessel.

55. Borda's mouthpiece.

A short cylindrical mouthpiece projecting into the vessel, as in
Fig. 62, is called a Borda's mouthpiece, and is of interest, as the
coefficient of discharge upon certain assumptions can be readily
calculated. Let the mouthpiece be so short
that the jet issuing at EF falls clear of GH.
The orifice projecting into the liquid has
the effect of keeping the liquid in contact
with the face AD practically at rest, and
at all points on it except the area BF the
hydrostatic pressure will, therefore, simply
depend upon the depth below the free

surface AB. Imagine the mouthpiece produced to meet the
face BC in the area IK. Then the hydrostatic pressure on AD,
neglecting EF, will be equal to the hydrostatic pressure on BC,
neglecting IK.

Again, BC is far enough away from EF to assume that the
pressure upon it follows the hydrostatic law.

The hydrostatic pressure on IK, therefore, is the force which
gives momentum to the water escaping through the orifice, over-
comes the pressure on EF, and the resistance of the mouthpiece.

Let H be the depth of the centre of the orifice below the free
surface and p the atmospheric pressure. Neglecting frictional
resistances, the velocity of flow v t through the orifice, is vfylL

Let a be the area of the orifice and w the area of the transverse
section of the jet. The discharge per second will be w . w V20H Ibs.

The hydrostatic pressure on IK is

pa + waS. Ibs.

The hydrostatic pressure on EF is pa Ibs.

The momentum given to the issuing water per second, is

M = -.

Therefore pa + <o 2#H = pa + walL,

and w = a.

FLOW THROUGH MOUTHPIECES

The coefficient of contraction is then, in this case, equal to
one half.

Experiments by Borda and others, show that this result is
justified, the experimental coefficient being slightly greater
than J.

56. Conical mouthpieces and nozzles.

These are either convergent as in Fig. 63, or divergent as in
Fig. 64.

Fig. 63.

Fig. 64.

Calling the diameter of the mouthpiece the diameter at the
outlet, a divergent tube gives a less, and a convergent
tube a greater discharge than a cylindrical tube of the
same diameter.

Experiments show that the maximum discharge for a
convergent tube is obtained when the angle of the cone
is from 12 to 13J degrees, and it is then 0'94 . a . J2gh.
If, instead of making the convergent mouthpiece conical,
its sides are curved as in Fig. 65, so that it follows as
near as possible the natural form of the stream lines, the
coefficient of discharge may, with high heads, approxi-
mate very nearly to unity.

Weisbach*, using the method described on page 55
to determine the velocity of flow, obtained, for this
mouthpiece, the following values of k. Since the mouth-
piece discharges full the coefficients of velocity k and
discharge m are practically equal.

Fig. 65.

Head in feet
k and m

0-66
959

1-64
967

11-48
975

55-8
994

338
994

According to Freeman t, the fire-hose nozzle shown in Fig. 66
has a coefficient of velocity of '977.

* Mechanics of Engineering.

f Transactions Am. Soc. C.E., Vol. xxi.

74 HYDRAULICS

If the mouthpiece is first made convergent, and then divergent,

Fig. 66.

as in Fig. 67, the divergence being sufficiently gradual for the
stream lines to remain in contact with the tube, the coefficient of
contraction is unity and there is but a
small loss of head. The velocity of efflux
from EF is then nearly equal to \/2#H
and the discharge is ra . a . >/2#H, a being
the area of EF, and the coefficient m
approximates to unity.

It would appear, that the discharge
could be increased indefinitely by length-
ening the divergent part of the tube and
thus increasing a, but as the length

Fig. 67.

increases, the velocity
decreases due to the friction of the sides of the tube, and further,
as the discharge increases, the velocity through the contracted
section CD increases, and the pressure head at CD consequently
falls.

Calling p a the atmospheric pressure, pi the pressure at CD,
and Vi the velocity at CD, then

w 2g

and

w w g

If ~- is greater than H + , pi becomes negative.

As pointed out, however, in connection with Froude's apparatus,
page 43, if continuity is to be maintained, the pressure cannot be
negative, and in reality, if water is the fluid, it cannot be less
than -- the atmospheric pressure, due to the separation of the air
from the water. The velocity v\ cannot, therefore, be increased
indefinitely.

FLOW THROUGH MOUTHPIECES 75

Assuming the pressure can just become zero, and taking the
atmospheric pressure as equivalent to a head of 34 ft. of water, the
maximum possible velocity, is

and the maximum ratio of the area of EF to CD is

34ft.

TT~*

Practically, the maximum value of Vi may be taken as

and the maximum ratio of EF to CD as

The maximum discharge is

The ratio given of EF to CD may be taken as the maximum
ratio between the area of a pipe and the throat of a Venturi meter
to be used in the pipe.

5 7. Plow through orifices and mouthpieces under constant
pressure.

The head of water causing flow through an orifice may be
produced by a pump or other mechanical means, and the discharge
may take place into a vessel, such as the condenser of a steam
engine, in which the pressure is less than that of the atmosphere.

For example, suppose water to be discharged from a cylinder
A, into a vessel B, Fig. 68, through
an orifice or mouthpiece by means
of a piston loaded with P Ibs., and
let the pressure per sq. foot in B
be po Ibs.

Let the area of the piston be
A square feet. Let h be the height
of the water in the cylinder above
the centre of the orifice and 7i of
the water in the vessel B. The
theoretical effective head forcing water through the orifice may
be written

76 HYDRAULICS

If P is large h and h will generally be negligible.

At the orifice the pressure head is 7& + , and therefore for

any stream line through the orifice, if there is no friction,

+ ft, + a P .*

2g w A.W

The actual velocity will be less than v, due to friction, and if Jc
is a coefficient of velocity, the velocity is then

v = Jc.*/2gH.,
and the discharge is Q = m . a\/2gIL.

In practical examples the cylinder and the vessel will generally
be connected by a short pipe, for which the coefficient of velocity
will depend upon the length.

If it is only a few feet long the principal loss of head will be
at the entrance to the pipe, and the coefficient of discharge will
probably vary between 0'65 and 0*85.

The effect of lengthening the pipe will be understood after the
chapter on flow through pipes has been read.

Example. Water is discharged from a pump into a condenser in which the
pressure is 3 Ibs. per sq. inch through a short pipe 3 inches diameter.
The pressure in the pump is 20 Ibs. per sq. inch.

Find the discharge into the condenser, taking the coefficient of discharge 0'75.
The effective head is

H _ 20x144 3x144

62-4 02-4

= 39 2 feet.

Therefore, Q= -75 x -7854 x ^ x ^64-4 x 39-2 cubic feet per seo.
= 1*84 cubic ft. per see.

58. Time of emptying a tank or reservoir.

Suppose a reservoir to have a sharp-edged horizontal orifice
as in Fig. 44. It is required to find the time taken to empty
the reservoir.

Let the area of the horizontal section of the reservoir at any
height h above the orifice be A sq. feet, and the area of the

orifice a sq. feet, and let the ratio be sufficiently large that the

velocity of the water in the reservoir may be neglected.

When the surface of the water is at any height h above the
orifice, the volume which flows through the orifice in any time ot
will be ma \/2gh . dt.

FLOW THROUGH MOUTHPIECES 77

The amount dh by which the surface of water in the reservoir
falls in the time dt is

g, _ ma \J2ghdt

ma \/2gh '
The time for the water to fall from a height H to H! is

H A ^_ = 1 _ ( H Adh
H, ma \l2gh a \/2g J H,

\/2g

If A is constant, and m is assumed constant, the time required
for the surface to fall from a height H to Hi above the orifice is

_ 1 f H Adh
ma \/2g J H, h%

ma
and the time to empty the vessel is

ma \/2gr '

or is equal to twice the time required for the same volume of
water to leave the vessel under a constant head H.

Time of emptying a lock with vertical drowned sluice. Let the
water in the lock when the sluice is closed be at a height H,
Fig. 56, above the down-stream level.

Then the time required is that necessary to reduce the level in
the lock by an amount H.

When the flow is taking place, let x be the height of the water
in the lock at any instant above the down-stream water.

Let A be the sectional area of the lock, at the level of the
water in the lock, a the area of the sluice, and m its coefficient of
discharge.

The discharge through the sluice in time dt ia

9Q = m . a \l2gx . dt.

If da? is the distance the surface falls in the lock in time fit, then
Ada? = ma \/2gxdt t

or ot =

To reduce the level by an amount H,

o ma

78 HYDRAULICS

If m and A are constant,

2A N/H

ma \/2gr "

Example. A reservoir, 200 yards long and 150 yards wide at the bottom, and
having side slopes of 1 to 1, has a depth of water in it of 25 feet. A short pipe
3 feet diameter is used to draw off water from the reservoir.

Find the time taken for the water in the reservoir to fall 10 feet. The
coefficient of discharge for the pipe is 0-7.

When the water has a depth h the area of the water surface is

A = (600 + 2/i) (450 + 2/i).
The area of the pipe is a=7'068 sq. feet.

Therefore . - ' - /* (+) (+*)
0-70 V20- 7-068J 15 fci

= -^ P 5 2 x 270000*4 + 1 x 2100** +
39'b Ll5

- * (610200 + 93800 + 3606)

= 17,850 sees.
= 4-95 hours.

Example. A horizontal boiler 6 feet diameter and 30 feet long is half full of
water.

Find the time of emptying the boiler through a short vertical pipe 3 inches
diameter attached to the bottom of the boiler.

The pipe may be taken as a mouthpiece discharging full, the coefficient of
velocity for which is 0'8.

Let r be the radius of the boiler.

When the water has any depth h above the bottom of the boiler the area A is

=30x2 s /r 2 -(r-*) 2
= 30x2 N /2r*-* 2 .

The area of the pipe is 0-049 sq. feet.
2x30

8x0-049^
\2r-h)*dh

_, t

Therefore t=

= 127-4x9-5
= 1210 sees.

EXAMPLES.

(1) Find the velocity due to a head of 100 ft.

(2) Find the head due to a velocity of 500 ft. per see.

(3) Water issues vertically from an orifice under a head of 40 ft. To
what height will the jet rise, if the coefficient of velocity is 0'97 ?

(4) What must be the size of a conoidal orifice to discharge 10 c. ft.
per second under a head of 100 ft.? w='925.

FLOW THROUGH ORIFICES AND MOUTHPIECES 79

(5) A jet 3 in. diameter at the orifice rises vertically 50 ft. Find its
diameter at 25 ft. above the orifice.

(6) An orifice 1 sq. ft. in area discharges 18 c. ft. per second under a
head of 9 ft. Assuming coefficient of velocity =0*98, find coefficient of
contraction.

(7) The pressure in the pump cylinder of a fire-engine is 14,400 Ibs.
per sq. ft. ; assuming the resistance of the valves, hose, and nozzle is such
that the coefficient of resistance is 0*5, find the velocity of discharge, and
the height to which the jet will rise.

(8) The pressure in the hose of a fire-engine is 100 Ibs. per sq. inch;
the jet rises to a height of 150 ft. Find the coefficient of velocity.

(9) A horizontal jet issues under a head of 9 ft. At 6 ft. from the
orifice it has fallen vertically 15 ins. Find the coefficient of velocity.

(10) Required the coefficient of resistance corresponding to a coefficient
of velocity =0-97.

(11) A fluid of one quarter the density of water is discharged from a
vessel in which the pressure is 50 Ibs. per sq. in. (absolute) into the
atmosphere where the pressure is 15 Ibs. per sq. in. Find the velocity of
discharge.

(12) Find the diameter of a circular orifice to discharge 2000 c. ft. per
hour, under a head of 6 ft. Coefficient of discharge 0'60.

(13) A cylindrical cistern contains water 16 ft. deep, and is 1 sq. ft. in
cross section. On opening an orifice of 1 sq. in. in the bottom, the water
level fell 7 ft. in one minute. Find the coefficient of discharge.

(14) A miner's inch is defined to be the discharge through an orifice in
a vertical plane of 1 sq. in. area, under an average head of 6| ins. Find
the supply of water per hour in gallons. Coefficient of discharge 0'62.

(15) A vessel fitted with a piston of 12 sq. ft. area discharges water
under a head of 10 ft. What weight placed on the piston would double the
rate of discharge?

(16) An orifice 2 inches square discharges under a head of 100 feet
T338 cubic feet per second. Taking the coefficient of velocity at 0'97, find
the coefficient of contraction.

(17) Find the discharge per minute from a circular orifice 1 inch
diameter, under a constant pressure of 34 Ibs. per sq. inch, taking 0*60 as
the coefficient of discharge.

(18) The plunger of a fire-engine pump of one quarter of a sq. ft. in
area is driven by a force of 9542 Ibs. and the jet is observed to rise to a
height of 150 feet. Find the coefficient of resistance of the apparatus.

(19) An orifice 8 feet wide and 2 feet deep has 12 feet head of water
above its centre on the up-stream side, and the backwater on the other
side is at the level of the centre of the orifice. Find the discharge if

80 HYDRAULICS

(20) Ten c. ft. of water per second flow through a pipe of 1 sq. ft. area,
which suddenly enlarges to 4 sq. ft. area. Taking the pressure at 100 Ibs.
per sq. ft. in the smaller part of the pipe, find (1) the head lost in shock,
(2) the pressure in the larger part, (3) the work expended in forcing the
water through the enlargement.

(21) A pipe of 3" diameter is suddenly enlarged to 5" diameter. A U
tube containing mercury is connected to two points, one on each side of the
enlargement, at points where the flow is steady. Find the difference in
level in the two limbs of the U when water flows at the rate of 2 c. ft. per
second from the small to the large section and vice versd. The specific
gravity of mercury is 13'6. Lond. Un.

(22) A pipe is suddenly enlarged from 2 inches in diameter to 3
inches in diameter. Water flows through these two pipes from the smaller
to the larger, and the discharge from the end of the bigger pipe is two
gallons per second. Find :

(a) The loss of head, and gain of pressure head, at the enlarge-
ment.

(&) The ratio of head lost to velocity head in small pipe.

(23) The head and tail water of a vertical-sided lock differ in level
12 ft. The area of the lock basin is 700 sq. ft. Find the time of emptying
the lock, through a sluice of 5 sq. ft. area, with a coefficient 0*5. The
sluice discharges below tail water level.

(24) A tank 1200 sq. ft. in area discharges through an orifice 1 sq. ft.
in area. Calculate the time required to lower the level in the tank from
50 ft. to 25 ft. above the orifice. Coefficient of discharge 0'6.

(25) A vertical-sided lock is 65 ft. long and 18 ft. wide. Lift 15 ft.
Find the area of a sluice below tail water to empty the lock in 5 minutes.
Coefficient 0'6.

(26) A reservoir has a bottom width of 100 feet and a length of 125
feet.

The sides of the reservoir are vertical.

The reservoir is connected to a second reservoir of the same dimensions
by means of a pipe 2 feet diameter. The surface of the water in the first
reservoir is 17 feet above that in the other. The pipe is below the surface
of the water in both reservoirs. Find the time taken for the water in the
two reservoirs to become level. Coefficient of discharge 0'8.

59. Notches and Weirs.

When the sides of an orifice are
produced, so that they extend be-
yond the free surface of the water,
as in Figs. 69 and 70, it is called a
notch.

Notches are generally made tri-
angular or rectangular as shown
in the figures and are largely used
for gauging the flow of water.

FLOW OVER WETRS

For example, if the flow of a small stream is required, a dam is
constructed across the stream and the water allowed to pass
through a notch cut in a board or metal plate.

Fig. 70. Rectangular Notch.

They can conveniently be used for measuring the compensation
water to be supplied from collecting reservoirs, and also to gauge
the supply of water to water wheels and turbines.

The term weir is a name given to a structure used to dam up
a stream and over which the water flows.

The conditions of flow are practically the same as through
a rectangular notch, and hence such notches are generally called
weirs, and in what follows the latter term only is used. The top
of the weir corresponds to the horizontal edge of the notch and is
called the sill of the weir.

The sheet of water flowing over a weir or through a notch is
generally called the vein, sheet, or nappe.

The shape of the nappe depends upon the form of the sill and
sides of the weir, the height of the sill above the bottom of the
up-stream channel, the width of the up-stream channel, and the
construction of the channel into which the nappe falls.

The effect of the form of the sill and of the down-stream
channel will be considered later, but, for the present, attention
will be confined to weirs with sharp edges, and to those in which
the air has free access under the nappe so that it detaches itself
entirely from the weir as shown in Fig. 70.

60. Rectangular sharp-edged weir.

If the crest and sides of the weir are made sharp-edged, as
shown in Fig. 70, and the weir is narrower than the approaching
channel, and the sill some distance above the bed of the stream,
there is at the sill and at the sides, contraction similar to that at
a sharp-edged orifice.

The surface of the water as it approaches the weir falls, taking
a curved form, so that the thickness h S) Fig. 70, of the vein over
the weir, is less than H, the height, above the sill, of the water at

L. H.

82 HYDRAULICS

some distance from the weir. The height H, which is called the
head over the weir, should be carefully measured at such a distance
from it, that the water surface has not commenced to curve.
Fteley and Stearns state, that this distance should be equal to
2| times the height of the weir above the bed of the stream.

For the present, let it be assumed that at the point where H is
measured the water is at rest. In actual cases the water will
always have some velocity, and the effect of this velocity will have
to be considered later. H may be called the still water head over
the weir, and in all the formulae following it has this meaning.

Side contraction. According to Fteley and Stearns the amount
by which the stream is contracted when the weir is sharp-edged
is from 0'06 to 0'12H at each side, and Francis obtained a mean of
O'lH. A wide weir may be divided into several bays by parti-
tions, and there may then be more than two contractions, at each
of which the effective width of the weir will be diminished, if
Francis' value be taken, by O'lH.

If L is the total width of a rectangular weir and N the number
of contractions, the effective width Z, Fig. 70, is then,

(L-O'INH).

When L is very long the lateral contraction may be neglected.

Suppression of the contraction. The side contraction can be
completely suppressed by making the approaching channel with
vertical sides and of the same width as the weir, as was done for
the orifice shown in Fig. 47. The width of the stream is then
equal to the width of the sill.

61. Derivation of the weir formula from that of a large
orifice.

If in the formula for large orifices, p. 64, h is made equal to
zero and for the effective width of the stream the length I is
substituted for 6, and k is unity, the formula becomes

If instead of hi the head H, Fig. 70, is substituted, and
a coefficient C introduced,

The actual width I is retained instead of L, to make allowance
for the end contraction which as explained above is equal to O'lH
for each contraction. If the width of the approaching channel is
made equal to the width of the weir I is equal to L.

With N contractions I =^L - 01NH),
and Q = f C v/2^ . (L - O'INH) Hi

If C is given a mean value of 0'625, and L and H are in feet,
the discharge in cubic feet per second is

Q = 3'o3(L- O'INH) H 1 (2).

FLOW OVER WEIRS 83

This is the well-known formula deduced by Francis* from
a careful series of experiments on sharp-edged weirs.

The formula, as an empirical one, is approximately correct and
gives reliable values for the discharge.

The method of obtaining it from that for large orifices is,
however, open to very serious objection, as the velocity at F on
the section EF, Fig. 70, is clearly not equal to zero, neither is the
direction of flow at the surface perpendicular to the section EF,
and the pressure on EF, as will be understood later (section 83)
is not likely to be constant.

That the directions and the velocities of the stream lines are
different from those through a section taken near a sharp-edged
orifice is seen by comparing the thickness of the jet in the two
cases with the coefficient of discharge.

For the sharp-edged orifice with side contractions suppressed,
the ratio of the thickness of the jet to the depth of the orifice is not
very different from the coefficient of discharge, being about 0*625,
but the thickness EF of the nappe of the weir is very nearly 0'78H,
whereas the coefficient of discharge is practically 0'625, and the
thickness is therefore 1*24 times the coefficient of discharge.

It appears therefore, that although the assumptions made in
calculating the flow through an orifice may be justifiable, providing
the head above the top of the orifice is not very small, yet when
it approaches zero, the assumptions are not approximately true.

The angles which the stream lines make with the plane of EF
cannot be very different from 90 degrees, so that it would appear,
that the error principally arises from the assumption that the
pressure throughout the section is uniform.

Bazin for special cases has carefully measured the fall of the
point F and the thickness EF, and if the assumptions of constant
pressure and stream lines perpendicular to EF are made, the
discharge through EF can be calculated.

For example, the height of the point E above the sill of the
weir for one of Bazin's experiments was 0'112H and the thickness
EF was 0'78H. The fall of the point F is, therefore, O'lOSH.
Assuming constant pressure in the section, the discharge per foot
width of the weir is, then,

0-108H

= 53272^. H*.

Lowell, Hydraulic Experiments, New York, 1858.

HYDRAULICS

The actual discharge per foot width, by experiment, was

q = 0-433 x/2<7.H*,

so that the calculation gives the discharge 1*228 greater than the
actual, which is approximately the ratio of the thickness EF to
the thickness of the stream from a sharp-edged orifice having
a depth H. The assumption of constant pressure is, therefore,
quite erroneous.

62. Thomson's principle of similarity.

" When a frictionless liquid flows out of similar and similarly
placed orifices in similar vessels in which the same kind of liquid
is at similar heights, the stream lines in the different flows are
similar in form, the velocities at similar points are proportional to
the square roots of the linear dimensions, and since the areas of
the stream lines are proportional to the squares of the linear
dimensions, the discharges are proportional to the linear dimensions
raised to the power of *."

Let A and B, Figs. 71 and 72, be exactly similar vessels with
similar orifices, and let all the dimensions of A be n times those
of B. Let c and Ci be similarly situated areas on similar stream
lines.

Fig. 71. Fig. 72.

Then, since the dimensions of A are n times those of B, the
fall of free level at c is n times that at Ci. Let v be the velocity
at c and Vi at GI.

Then, since it has been shown (page 51) that the velocity in
any stream line is proportional to the square root of the fall of
free level,

.*. v : Vi :: *Jn : 1.

Again the area at c is n a times the area at Ci and, therefore,
the discharge through c
the discharge through d = n
which proves the principle.

* British Association Keports 1858, 1876 and 1885.

FLOW OVER WEIRS 85

63. Discharge through a triangular notch by the
principle of similarity.

Let ADC, Figs. 73 and 74, be a triangular notch.

Let the depth of the flow through the notch at one time be H
and at another n . H.

Suppose the area of the stream in the two cases to be divided
into the same number of horizontal elements, such as ab and aj)i.

Then clearly the thickness of ab will be n times the thickness
of aj)i .

Let a$i be at a distance x from the apex B, and ab at a
distance nx ; then the width of ab is clearly n times the width of
aA, and the area of ab will therefore be n* times the area of aj>i.

Again, the head above ab is n times the Jiead above afo and
therefore the velocity through ab will be >Jn times the velocity
through aA and the discharge through ab will be n* times
that through Oi&i.

More generally Thomson expresses this as follows :

" If two triangular notches, similar in form, have water flowing
through them at different depths, but with similar passages of
approach, the cross sections of the jets at the notches may be
similarly divided into the same number of elements of area, and
the areas of corresponding elements will be proportional to the
squares of the lineal dimensions of the cross sections, or pro-
portional to the squares of the heads."

As the depth h of each element can be expressed as a fraction
of the head H, the velocities through these elements are propor-
tional to the square root of the head, and, therefore, the discharge
is proportional to H^.

Therefore Q oo H*,

or Q = C.H',

C being a coefficient which has to be determined by experiment.

From experiments with a sharp-edged notch having an angle
at the vertex of 90 degrees, he found C to be practically constant
for all heads and equal to 2 '535. Then, H being measured in feet,
the discharge in cubic feet per second is

Q = 2-535.H* (3).

86 HYDRAULICS

64. Flow through a triangular notch.

The flow through a triangular notch is frequently given as

in which B is the top width of the notch and n an experimental coefficient.

It is deduced as follows :

Let ADC, Fig. 74, be the triangular notch, H being the still water head over
the apex, and B the width at a height H above the apex. At any depth h the

width b of the strip a^ is " ' .

If the velocity through this strip is assumed to be v = k^/2gh, the width of the

stream through o 1 & 1 , - - - , and the thickness dh t the discharge through it is
H

The section of the jet just outside the orifice is really less than the area EFD.
The width of the stream through any strip Oj&j is less than a^, the surface is lower
than EF, and the apex of the jet is some distance above B.

The diminution of the width of Oj&j has been allowed for by the coefficient c, and
the diminution of depth might approximately be allowed for by integrating between
fc=0 and /j = H, and introducing a third coefficient Cj.

Then -

Replacing cc^k by n

Qrr^.nVV.BH* ....................................... (4).

Calling the angle ADC, 0,

and Q = T 8 7

When B is 90 degrees, B is equal to 2H, and

Taking a mean value for n of 0-5926

Q = 2-535 . IT* for a right-angled notch,
and Q = 1-464^ for a 60 degrees notch,

which agrees with Thomson's formula for a right-angled notch.

The result is the same as obtained by the method of similarity, but the method
of reasoning is open to very serious objection, as at no section of the jet are all the
stream lines normal to the section, and k cannot therefore be constant. The
assumption that the velocity through any strip is proportional to Jh is also open
to objection, as the pressure throughout the section can hardly be uniform.

65. Discharge through a rectangular weir by the
principle of similarity.

The discharge through a rectangular weir can also be obtained
by the principle of similarity.

FLOW OVER WEIRS

Consider two rectangular weirs each of length L, Figs. 75
and 76, and let the head over the sill be H in the one case and
Hi, or nH, in the other. Assume the approaching channel to be
of such a form that it does not materially alter the flow in either
case.

, ^ K- L

Fig. 7

Fig. 75.

To simplify the problem let the weirs be fitted with sides
projecting up stream so that there is no side contraction.

Then, if each of the weirs be divided into any number of equal
parts the flow through each of these parts in any one of the weirs
will be the same.

Suppose the first weir to be divided into N equal parts. If

then, the second weir is divided into

N.H

equal parts, the parts

in the second weir will be exactly similar to those of the first.

By the principle of similarity, the discharge through each of
the parts in the first weir will be to the discharge in the second

as 7 , and the total discharge through the first weir is to the

discharge through the second as
N.H*

Kj n*'

Instead of two separate weirs the two cases may refer to the
same weir, and the discharge for any head H is, therefore, pro-
portional to * H* ; and since the flow is proportional to L

Q = C.L.H*,

in which C is a coefficient which should be constant.

66. Rectangular weir with end contractions.

If the width of the channel as it approaches the weir is greater
than the width of the weir, contraction takes place at each side,
and the effectual width of the stream or nappe is diminished ; the
amount by which the stream is contracted is practically inde-
pendent of the width and is a constant fraction of H, as explained
above, or is equal to JtH, Jc being about 0*1.
* gee Example 3, page 260.

88 HYDRAULICS

Let the total width of each, weir be now divided into three
parts, the width of each end part being equal to n . k . H. The
width of the end parts of the transverse section of the stream will
each be (n - 1) k . H, and the width of central part L - 2?iA;H.

The flow through the central part of the weir will be equal to

Now, whatever the head on the weir, the end pieces of the
stream, since the width is (n 1) &H and A; is a constant, will be
similar figures, and, therefore, the flow through them can be
expressed as

The total flow is, therefore,

Q = C (L - 2wfcH) H* + 20j (n -
If now Ci is assumed equal to C

Q = 0(L-2fcH)H*.

If instead of two there are N contractions, due to the weir
being divided into several bays by posts or partitions, the formula
becomes

Q = 0(L-N01.H)H*.

This is Francis' formula, and by Thomson's theory it is thus
shown to be rational.

67. Bazin's* formula for the discharge of a weir.

The discharge through a weir with no side contraction may be
written

the coefficient ra being equal to

Taking Francis' value for C as 3'33, ra is then 0*415.
From experiments on sharp-crested weirs with no side con-
traction Bazin deduced for rat the value

n ., n - -00984
ra = 405 + ^ .
1

In Table IX, and Fig. 77, are shown Bazin's values for ra for
different heads, and also those obtained by Rafter at Cornell upon
a weir similar to that used by Bazin, the maximum head in the
Cornell experiments being much greater than that in Bazin's
experiments. In Fig. 77 are also shown several values of ra, as
calculated by the author, from Francis' experimental data.

* Annales des Fonts et Chaussees, 18881898.

t " Experiments on flow over Weirs," Am.S,C.E, Vol. xxvu,

Bazin.

0-164

0-328

0-656

0-984

1-312

1-64

1-968

0-448

0-432

0-421

0-417

0-414

0-412

0-409

0-00984

FLOW OVER WEIRS 89

TABLE IX.

Values of the coefficient m in the formula Q = wL \/2gr H^
Weir, sharp-crested, 6'56 feet wide with free overfall and lateral
contraction suppressed, H being the still water head over the weir,
or the measured head h* corrected for velocity of approach.

Head in feet

"* .v w iw i - T j-

Rafter.
Head in feet m G

0-1 0-4286 3-437

0-5 0-4230 3-392

1-0 0-4174 3-348

1-5 0-4136 3-317

2-0 0-4106 3-293

2-5 0-4094 3-283

3-0 0-4094 3-283

3-5 0-4099 3-288

4-0 0-4112 3-298

4-5 0-4125 3-308

5-0 0-4133 3-315

5-5 0-4135 3-316

6-0 0-4136 3-317

68. Bazin's and the Cornell experiments on weirs.

Bazin's experiments were made on a weirt 6'56 feet long
having the approaching channel the same width as the weir, so
that the lateral contractions were suppressed, and the discharge
was measured by noting the time taken to fill a concrete trench of
known capacity.

The head over the weir was measured by means of the hook
gauge, page 249. Side chambers were constructed and connected
to the channel by means of circular pipes O'l m. diameter.

The water in the chambers was very steady, and its level
could therefore be accurately gauged. The gauges were placed
5 metres from the weir. The maximum head over the weir in
Bazin's experiments was however only 2 feet.

The experiments for higher heads at Cornell University were
made on a weir of practically the same width as Bazin's, 6'53 feet,
the other conditions being made as nearly the same as possible ;
the maximum head on the weir was 6 feet.

* See page 90.

f Annales des Pouts et Chaussees, p. 445, Vol. 11. 1801.

HYDRAULICS

The results of these experiments, Fig. 77, show that the
coefficient m diminishes and then increases, having a minimum
value when H is between 2*5 feet and 3 feet.

cxp

s: no.

> -w*

3 '/IQ

5 *^

v \

\ i

\ \

\ %

\ \

\ i

^ ^/7,O

*3 T^

J 7

rFSi

,^-^j

^^fei

v ^/

ss^

J-- ^ <

J 4?

^~*>

1 J

Head in, Feet.

Mean, coefHcuenb curves for Sharp -edged, Weirs
+ jBo^t/uy Kccpervmjents
o Corneli

A Fronds' (Deduced by the author)

Fig. 77.

It is doubtful, however, although the experiments were made
with great care and skill, whether at high heads the deduced
coefficients are absolutely reliable.

To measure the head over the weir a 1 inch galvanised pipe
with holes Jinch diameter and opening downwards, 6 inches
apart, was laid across the channel. To this pipe were connected
f inch pipes passing through the weir to a convenient point below
the weir where they could be connected to the gauges by rubber
tubing. The gauges were glass tubes f inch diameter mounted
on a frame, the height of the water being read on a scale
graduated to 2mm. spaces.

69. Velocity of approach.

It should be clearly understood that in the formula given, it
has been assumed in giving values to the coefficient m, that H is
the height above the sill of the weir of the still water surface,

FLOW OVER WETT5S 91

In actual cases the water where the head is measured will have
some velocity, and due to this, the discharge over the weir will be
increased.

If Q is the actual discharge over a weir, and A is the area of
the up-stream channel approaching the weir, the mean velocity in

the channel is v = -f .
.A.

There have been a number of methods suggested to take into
account this velocity of approach, the best perhaps being that
adopted by Hamilton Smith, and Bazin.

This consists in considering the equivalent still water head H,
over the weir, as equal to

a being a coefficient determined by experiment, and h the
measured head.

The discharge is then

(5),

Expanding (5), and remembering that =r-, is generally a small

quantity,

The velocity v depends upon the discharge Q to be determined
and is equal to -^ .

Therefore Q = mL hJSgh 1 + -, .................. (6).

From five sets of experiments, the height of the weir above the
bottom of the channel being different for each set, Bazin found
the mean value of a to be 1*66.

This form of the formula, however, is not convenient for use,
since the unknown Q appears upon both sides of the equation.

If, however, the discharge Q is expressed as

the coefficient n for any weir can be found by measuring Q and h.

It will clearly be different from the coefficient m, since for m
to be used h has to be corrected.

From his experimental results Bazin calculated n for various
heads, some of which are shown in Table X.

92 HYDRAULICS

Substituting this value of Q in the above formula,

(7).

Let few 3 be called Jc.
Then Q = m

/ Z-T 2 7i 2 \

( 1 + ^f )
\ A. /

Or, when the width of channel of approach is equal to the
width of the weir, and the height of the sill, Fig. 78, is p feet above
the bed of the channel, and h the measured head,

and

2 (8).

Fig. 78.

The mean value given to the coefficient k by Bazin is 0'55,
so that

This may be written

in which

,
= m f 1 +

Substituting for m the value given on page 88,

mi may be called the absolute coefficient of discharge.

The coefficient given in the Tables.

It should be clearly understood that in determining the values
of m as given in the Tables and in Fig. 77 the measured head h
was corrected for velocity of approach, and in using this

FLOW OVER WEIRS 93

coefficient to determine Q, h must first be corrected, or Q
calculated from formula 9.

Rafter in determining the values of m from the Cornell ex-
periments, increased the observed head h by x- only, instead of

by 1-66 g.

Fteley and Stearns*, from their researches on the flow over
weirs, found the correction necessary for velocity of approach to
be from

1-45 to T5 |^.

Hamilton Smith t adopts for weirs with end contractions
suppressed the values

T33 to T40 1^,

and for a weir with two end contractions,
1*1 to 1*25 1~.

TABLE X.
Coefficients n and m as calculated by Bazin from the formulae

and Q =

h being the head actually measured and H the head corrected for
velocity of approach.

Head
h in feet

Height of sill
p in feet

Coefficient
n

Coefficient
m

0-164

0-656
6-560

0-458
0-448

0-448

0-984

0-656
6-560

0-500
0-421

0-417

1-640

0-656
6-560

0-500
0-421

0-4118

An example is now taken illustrating the method of deducing
the coefficients n and m from the result of an experiment, and the
difference between them for a special case.

Example. In one of Bazin's experiments the width of the weir and the
approaching channel were both 6*56 feet. The depth of the channel approaching
the weir measured at a point 2 metres up stream from the weir was 7'544 feet and
the head measured over the weir, which may be denoted by ft, was 0-984 feet. The
measured discharge was 21-8 cubic ft. per second.

* Transactions Am.S.G.E., Vol. xn.
t Hydraulics.

HYDRAULICS

The velocity at the section where h was measured, and which may be called the
velocity of approach was, therefore,

Q 21-8

7-544x6-56' "7-544x6-56
=0-44 feet per second.
If now the formula for discharge be written

and n is calculated from this formula by substituting the known values of
Q, L and h

n= 0-421.
Correcting h for velocity of approach,

= 9888.
Then

from which m = - ' 8 =fKiig.

6 -56 V20.-9888

It will seem from Table X that when the height p of the sill of the weir above
the stream bed is small compared with the head, the difference may be much
larger than for this example.

When the head is 1-64 feet and larger than p, the coefficient n is eighteen
per cent, greater than m. In such cases failure to correct the coefficient will lead
to considerable inaccuracy.

70. Influence of the height of the weir sill above the bed
of the stream on the contraction.

The nearer the sill is to the bottom of the stream, the less the
contraction at the sill, and if the depth is small compared with H,
the diminution on the contraction may considerably affect the
flow.

When the sill was 1'15 feet above the bottom of a channel,

of the same width as the weir, Bazin found the ratio ^ (Fig. 85)
to be 0'097, and when it was 3'70 feet, to be 0*112. For greater

heights than these the mean value of ^ was 0'13.

71. Discharge of a weir when the air is not freely
admitted beneath the nappe. Form of the nappe.

Francis in the Lowell experiments, found that, by making the
width of the channel below the weir equal to the width of the
weir, and thus preventing free access of air to the underside of the
nappe, the discharge was increased. Bazin*, in the experiments
already referred to, has investigated very fully the effect upon
the discharge and upon the form of the nappe, of restricting the
free passage of the air below the nappe. He finds, that when the
flow is sufficient to prevent the air getting under the nappe, it may
assume one of three distinct forms, and that the discharge for
* Annales des Fonts et Chaussees, 1891 and 1898.

FLOW OVER WEIRS

one of them may be 28 per cent, greater than when the air is
freely admitted, or the nappe is "free." Which of these three
forms the nappe assumes and the amount "by which the discharge
is greater than for the "free nappe," depends largely upon the
head over the weir, and also upon the height of the weir above
the water in the down-stream channel.

The phenomenon is, however, very complex, the form of the
nappe for any head depending to a very large extent upon
whether the head has been decreasing, or increasing, and for a
given head may possibly have any one of the three forms, so that
the discharge is very uncertain. M. Bazin distinguishes the forms
of nappe as follows :

(1) Free nappe. Air under nappe at atmospheric pressure,
Figs. 70 and 78.

(2) Depressed nappe enclosing a limited volume of air at a
pressure less than that of the atmosphere, Fig. 79.

(3) Adhering nappe. No air enclosed and the nappe adher-
ing to the down-stream face of the weir, Fig. 80. The nappe in this
case may take any one of several forms.

Top of ChanrvdK

Fig. 80.

(4) Drowned or wetted nappe, Fig. 81. No air enclosed but
the nappe encloses a mass of turbulent water which does not move
with the nappe, and which is said to wet the nappe.

Fig. 79.
Drowned or wetted nappe, Fig. 81.

Fig. 81.

HYDRAULICS

72. Depressed nappe.

The air below the nappe being at less than the atmospheric
pressure the excess pressure on the top of the nappe causes it to
be depressed. There is also a rise of water in the down-stream
channel under the nappe.

The discharge is slightly greater than for a free nappe. On a
weir 2*46 feet above the bottom of the up-stream channel, the
nappe was depressed for heads below 0*77 feet, and at this head
the coefficient of discharge was 1'08 mi, mi being the absolute
coefficient for the free nappe.

73. Adhering nappes.

As the head for this weir approached 0*77 feet the air was
rapidly expelled, and the nappe became vertical as in Fig. 80, its
surface having a corrugated appearance. The coefficient of dis-
charge changed from 1*08 mi to l'28mi. This large change in
the coefficient of discharge caused the head over the weir to fall
to 0'69 feet, but the nappe still adhered to the weir.

74. Drowned or wetted nappes.

As the head was further increased, and approached 0*97 feet,
the nappe came away from the weir face, assuming the drowned
form, and the coefficient suddenly fell to 119 mi. As the head
was further increased the coefficient diminished, becoming 112
when the head was above 1*3 feet.

The drowned nappes are more stable than the other two, but
whereas for the depressed and adhering nappes the discharge is
not affected by the depth of water in the down-stream channel,
the height of the water may influence the flow of the drowned
nappe. If when the drowned nappe falls into the down stream
the rise of the water takes place at a distance from the foot of the
nappe, Fig. 81, the height of the down-stream water does not affect
the flow. On the other hand if the rise encloses the foot of the
nappe, Fig. 82, the discharge is affected. Let h- 2 be the difference

Fig. 82.

FLOW OVER WEIRS 97

of level of the sill of the weir and the water below the weir. The,
coefficient of discharge in the first case is independent of h^ but is
dependent upon p the height of the sill above the bed of the up-
stream channel, and is

(11).

Bazin found that the drowned nappe could not be formed if h
is less than 0*4 p and, therefore, ? cannot be greater than 2*5.
Substituting for m x its value

from (10) page 92

m = 0-470 + 0-0075^ .................. (12).

In the second case the coefficient depends upon h*, and is,

l -06 -t-0'16 --0'05 ............ (13),

for which, with a sufficient degree of approximation, may be
substituted the simpler formula,

(14).

The limiting value of m is 1*2 mi, for if h^ becomes greater
than h the nappe is no longer drowned.

Further, the rise can only enclose the foot of the nappe when
h$ is less than (f p - h). As h 2 passes this value the rise is pushed
down stream away from the foot of the nappe and the coefficient
changes to that of the preceding case.

75. Instability of the form of the nappe.

The head at which the form of nappe changes depends upon
whether the head is increasing or diminishing, and the depressed
and adhering nappes are very unstable, an accidental admission
of air or other interference causing rapid change in their form.
Further, the adhering nappe is only formed under special circum-
stances, and as the air is expelled the depressed nappe generally
passes directly to the drowned form.

If, therefore, the air is not freely admitted below the nappe
the form for any given head is very uncertain and the discharge
cannot be obtained with any great degree of assurance.

With the weir 2'46 feet above the bed of the channel and 6'56
feet long Bazin obtained for the same head of 0*656 feet, the four
kinds of nappe, the coefficients of discharge being as follows :
L. H. 7

98 HYDRAULICS

Free nappe, 0'433

Depressed nappe, 0'460
Drowned nappe, level of water down stream

0*41 feet below the crest of the weir, 0*497

Nappe adhering to down-stream face, 0'554
The discharge for this weir while the head was kept constant,
thus varied 26 per cent.

76. Drowned weirs with sharp crests*.

When the surface of the water down stream is higher than the
sill of the weir, as in Fig. 83, the weir is said to be drowned.

Fig. 83.

Bazin gives a formula for deducing the coefficients for such a
weir from those for the sharp-edged weirs with a free nappe, which
in its simplest form is,

A 2 being the height of the down-stream water above the sill of
the weir, h the head actually measured above the weir, p the
height of the sill above the up-stream channel, and Wi the
coefficient ((10), p. 92) for a sharp-edged weir. This expression
gives the same value within 1 or 2 per cent, as the formulae (13)
and (14).

Example. The head over a weir is 1 foot, and the height of the sill above the
up-stream channel is 5 feet. The length is 8 feet and the surface of the water
in the down-stream channel is 6 inches above the sill. Find the discharge.

From formula (10), page 92, the coefficient 7% for a sharp-edged weir with free
nappe is

* Attempts have been made to express the discharge over a drowned weir as
equivalent to that through a drowned orifice of an area equal to Lft 2 , under a head
h-h%, together with a discharge over a weir of length L when the head is h - }i%.

The discharge is then

n V^IA (h-h^+m JZgl* (h-h$ 9
n and m being coefficients. Du Buat gave the formula

and Monsieur Mary Q = 8/ig \/2g (h - /' 2 + head due to velocity of stream).

FLOW OVER WEIRS 99

Therefore m = -4215 [1 -05 (1 + -021) 0-761]

= 3440.
Then Q = -344 x 8 </2g . it

= 22-08 cubic ft. per second.

77. Vertical weirs of small thickness.

Instead of making the sill of a weir sharp-edged, it may
have a flat sill of thickness c. This will frequently be the case in
practice, the weir being constructed of timbers of uniform width
placed one upon the other. The conditions of flow for these weirs
may be very different from those of a sharp-edged weir.

The nappes of such weirs present two distinct forms, according
as the water is in contact with the crest of the weir, or becomes
detached at the up-stream edge and leaps over the crest without
touching the down-stream edge. In the second case the discharge
is the same as if the weir were sharp-edged. When the head h
over the weir is more than 2c this condition is realised, and may
obtain when h passes f c. Between these two values the nappe is
in a condition of unstable equilibrium ; when h is less than f c the
nappe adheres to the sill, and the coefficient of discharge is

0185

^),

any external perturbation such as the entrance of air or the
passage of a floating body causing the detachment.

If the nappe adheres between f c and 2c the coefficient m varies
from *98wi to l'07mi, but if it is free the coefficient m^^m^.
When H = Jc, m is '79rai. If therefore the coefficients for a
sharp-edged weir are used it is clear the error may be con-
siderable.

The formula for m gives approximately correct results when
the width of the sill is great, from 3 to 7 feet for example.

If the up-stream edge of the weir is rounded the discharge is
increased. The discharge* for a weir having a crest 6*56 feet
wide, when the up-stream edge was rounded to a radius of 4 inches,
was increased by 14 per cent., and that of a weir 2*624 feet wide
by 12 per cent.

The rounding of the corners, due to wear, of timber weirs of
ordinary dimensions, to a radius of 1 inch or less, will, therefore,
affect the flow considerably.

78. Depressed and wetted nappes for flat-crested weirs.

The nappes of weirs having flat sills may be depressed, and
may become drowned as for sharp-edged weirs.

* Amiales de* Fonts et Chausstes, Vol. u. 1896.

100 HYDRAULICS

The coefficient of discharge for the depressed nappes, whether
the nappe leaps over the crest or adheres to it, is practically the
same as for the free nappes, being slightly less for low heads and
becomes greater as the head increases. In this respect they differ
from the sharp-crested weirs, the coefficients for which are always
greater for the depressed nappes than for the free nappes.

79. Drowned nappes for flat-crested weirs.

As long as the nappe adheres to the sill the coefficient ra may
be taken the same as when the nappe is free, or

/' ^ 0'185/A
w = Wj (0 70 + - J .

When the nappe is free from the sill and becomes drowned,
the same formula

as for sharp-crested weirs with drowned nappes, may be used.
For a given limiting value of the head h these two formulae give
the same value of m . When the head is less than this limiting
value, the former formula should be used. It gives values of m
slightly too small, but the error is never more than 3 to 4 per cent.
When the head is greater than the limiting value, the second
formula should be used. The error in this case may be as
great as 8 per cent.

80. Wide flat-crested weirs.

When the sill is very wide the surface of the water falls
towards the weir, but the stream lines, as they pass over the weir,
are practically parallel to the top of the weir.

Let H be the height of the still water surface, and h the depth
of the water over the weir, Fig. 84.

' _^-=^^ ^~**77^%77S7777?7\ c^j!'

Fig. 84.

Then, assuming that the pressure throughout the section of the
nappe is atmospheric, the velocity of any stream line is

v = \/20 (H - h),
and if L is the length of the weir, the discharge is

Q = J&JLh x/CHT^TO (16).

FLOW OVER WEIRS" J01

For the flow to be permanent (see 'page 106) XJ> mist be a

maximum for a given value of h, or -~ must equal zero.

QLrit

Therefore

From which 2 (H - ft) - h = 0,

and h = f H.

Substituting for h in (16)

= 0-385L 2^H . H = 3-08L x/H . H.

The actual discharge will be a little less than this due to
friction on the sill, etc.

Bazin found for a flat-crested weir 6*56 feet wide the coefficient
mwasO'373, or C = 2'991.

Lesbros' experiments on weirs sufficiently wide to approximate
to the conditions assumed, gave '35 for the value of the co-
efficient m.

In Table XI the coefficient C for such weirs varies from 2'66
to 310.

81. Plow over dams.

Weirs of various forms. M. Bazin has experimentally investi-
gated the flow over weirs having (a) sharp crests and (&) flat
crests, the up- and down-stream faces, instead of both being vertical,
being

(1) vertical on the down-stream face and inclined on the
up-stream face,

(2) vertical on the up-stream face and inclined on the down-
stream face,

(3) inclined on both the up- and down-stream faces,
and (c) weirs of special sections.

The coefficients vary very considerably from those for sharp-
crested vertical weirs, and also for the various kinds of weirs.
Coefficients are given in Table XI for a few cases, to show the
necessity of the care to be exercised in choosing the coefficient for
any weir, and the errors that may ensue by careless evaluation of
the coefficient of discharge.

For a full account of these experiments and the coefficients
obtained, the reader is referred to Bazin's* original papers, or to
Rafter's t paper, in which also will be found the results of experi-

* Annalex des Fonts et Chaussges, 1898.

t Transactions oj tlie Am.S.C.E., Vol. xuv., 1900.

102

HYDRAULICS

*i -ci^ -t it<-\

i ..'

TABLE XL

Values of the coefficient C in the formula Q = CL . h*, for weirs
of the sections shown, for various values of the observed head h.

Bazin.

Section of
weir

Head in feet

0-3

0-5

1-0

1-3

2-0

3-0

4-0

5-0

6-0

I>31$

2-66

2-90

3-10

I
i

3-61

3-80

4-01

3-91

-r^\

4-02

4-15

4-18

4-15

3-46

3-57

3-86

3-80

*T^

1 ^V

3-46

3-49

3-59

3-63

jg^v;

3-08

3-19

3-22

FLOW OVER WEIRS

103

TABLE XI (continued).
Bazin.

Section of
weir

66'

wr a

Head in feet

0-3

3-10

2-75

0-5

3-27

3-05

1-0 1-3

3-73

3-52

3-90

3-73

2-0 3-0 4-0 5-0 6-0

Rafter.

Section of
weir

Head in feet

0-3 0-5 10 13 2-0 3-0 4-0 5-0 6-0

3-35

314

3-68

3-42

3-83

3-52

3 ! 77

3-61

3-68

3-66

3-70

3-66

3-71

3-64

3-71

3-63

2-95

3-16

3-27

3-45

3-56

3-61

3-65

3-67

HYDRAULICS

ments made at Cornell University on the discharge of weirs, similar
to those used by Bazin and for heads higher than he used, and
also weirs of sections approximating more closely to those of
existing masonry dams, used as weirs. From Bazin's and Rafter's
experiments, curves of discharge for varying heads for some of
these actual weirs have been drawn up.

82. Form of weir for accurate gauging.

The uncertainty attaching itself to the correction to be applied
to the measured head for velocity of approach, and the difficulty
of making proper allowance for the imperfect contraction at the
sides and at the sill, when the sill is near the bed of the channel
and is not sharp-edged, and the instability of the nappe and
uncertainty of the form for any given head when the admission of
air below the nappe is imperfect, make it desirable that as far as
possible, when accurate gaugings are required, the weir should
comply with the following four conditions, as laid down by
Bazin.

(1) The sill of the weir must be made as high as possible
above the bed of the stream.

(2) Unless the weir is long compared with the head, the
lateral contraction should be suppressed by making the channel
approaching the weir with vertical sides and of the same width as
the weir.

(3) The sill of the weir must be made sharp-crested.

(4) Free access of air to the sides and under the nappe of
the weir must be ensured.

83. Boussinesq's* theory of the discharge over a weir.

As stated above, if air is freely admitted below the nappe of
a weir there is a contraction of the stream at the sharp edge of the
sill, and also due to the falling curved surface.

If the top of the sill is well removed from the bottom of the
channel, the amount by which the arched under side of the nappe
is raised above the sill of the weir is assumed by Boussinesq and
this assumption has been verified by Bazin's experiments to be
some fraction of the head H on the weir.

Let CD, Fig. 85, be the section of the vein at which the
maximum rise of the bottom of the vein occurs above the sill, and
let e be the height of D above S.

Let it be assumed that through the section CD the stream
lines are moving in curved paths normal to the section, and that
they have a common centre of curvature 0.

* Comptes Bendus, 1867 and 1889.

FLOW OVER WEIRS

105

Let H be the height of the surface of the water up stream
above the sill. Let R be the radius of the stream line at any
point B in CD at a height x above S, and RI and R 2 the radii of
curvature at D and C respectively. Let Y, YI and Y 2 be the
velocities at E, D, and C respectively.

Fig. 85.

Consider the equilibrium of any element of fluid at the point
E, the thickness of which is SR and the horizontal area is a. If w
is the weight of unit volume, the weight of the element is w . aSR.

Since the element is moving in a circle of radius R the centri-

fugal force acting on the element is wa

Y 2 3B

Ibs.

The force acting on the element due to gravity is wa 8R Ibs.
Let p be the pressure per unit area on the lower face of the
element and p + &p on the upper face.

Then, equating the upward and downward forces,

From which

f * N JVT3

(p + op) a + waoii = pa+

-1 +

(1).

wdR gR, "

Assuming now that Bernoulli's theorem is applicable to the
stream line at EF,

Differentiating, and remembering H is constant,

J" VdY =Q

1 dp = 1 VdV
w dx g . dx *

106 HYDRAULICS

And since

therefore
or

- ,
dx dli '

V 2 ' YdY

Integrating, YR = constant.

Therefore YR = ViRi = V 2 R..

At the upper and lower surfaces of the vein the pressure is
atmospheric, and therefore,

Since YR = YiRi, and R from the figure is (Ri + x - e), therefore,

.................. (2).

The total flow over the weir is

- e
.................. (3).

Now if the flow over the weir is permanent, the thickness h Q of
the nappe must adjust itself, so that for the given head H the
discharge is the maximum possible.

The maximum flow however can only take place if each
filament at the section GrF has the maximum velocity possible to
the conditions, otherwise the filaments will be accelerated; and
for a given discharge the thickness h is therefore a minimum, or
for a given value of h the discharge is a maximum. That is, when

Q is a maximum, -jS^ = 0.

CLn,Q

If therefore RI can be written as a function of h Q} the value of
ho, which makes Q a maximum, can be determined by differ-

entiating (3) and equating ^ to zero.

Then, since

-,
and

FLOW OVER WEIRS 107

Therefore, h = (H - e) (1 - n*),

and Bi = w(l + n) (H-e).

Substituting this value of RI in the expression for Q,

which, since Q is a maximum when ~rjr = 0, & n( l ^ is a function

of n, is a maximum when -jr = Q.
Differentiating and equating to zero,

the solution of which gives

71 = 0-4685,

and therefore, Q - 0'5216 N/2^ (H -

= 0-5216 ^(l-g

= 0'5216 l - * x/2 . H/

= ra
the coefficient m being equal to

0-5216 (l - g) 1 .

M. Bazin has found by actual measurement, that the mean
value for 4-, when the height of the weir is at considerable
distance from the bottom of the channel, is 0'13.

Then, l- f - 0-812,

and m = 0'423.

It will be seen on reference to Fig. 77, that this value is very
near to the mean value of m as given by Francis and Bazin, and
the Cornell experiments. Giving to g the value 32'2,

Q = 3-39 H^ per foot length of the weir.

If the length of the weir is L feet and there are no end con-
tractions the total discharge is

and if there are N contractions

Q = 3'39(L-N01H)Hi

108 HYDRAULICS

The coefficient 3'39 agrees remarkably well with the mean
value of C obtained from experiment.

The value of a theory must be measured by the closeness of
the results of experience with those given by the theory, and in
this respect Boussinesq's theory is the most satisfactory, as it not
only, in common with the other theories, shows that the flow is
proportional to H*, but also determines the value of the
constant C.

84. Solving for Q, by approximation, when the velocity
of approach is unknown.

A simple method of determining the discharge over a weir
when the velocity of approach is unknown, is, by approximation,
as follows.

Let A be the cross-sectional area of the channel.

First find an approximation to Q, without correcting for
velocity of approach, from the formula

Q = mLJi *j2gh.
The approximate velocity of approach is, then,

= a ,

and H is approximately

A nearer approximation to Q can then be obtained by sub-
stituting H for h } and if necessary a second value for v can be
found and a still nearer approximation to H.

In practical problems this is, however, hardly necessary.

Example. A weir without end contractions has a length of 16 feet. The head
as measured on the weir is 2 feet and the depth of the channel of approach below
the sill of the weir is 10 feet. Find the discharge.

m= 0-405 + = . 4099.

Therefore C = 328.

Approximately, Q=3-28 2^.16

= 148 cubic feet per second.

The velocity v = -^ fg= '77 ft. per sec. ,

and i^?= -0147 feet.

A second approximation to Q is, therefore,

Q = 3-28 (2-0147)1 16

= 150 cubic feet per second.

A third value for Q can be obtained, but the approximation is sufficiently near
for all practical purposes.

In this case the error in neglecting the velocity of approach altogether, is
probably less than the error involved in taking m as 0-4099.

FLOW OVER WEIRS 109

85. Time required to lower the water in a reservoir a
given distance by means of a weir.

A reservoir has a weir of length L feet made in one of its sides,
and having its sill H feet below the original level of the water in
the reservoir.

It is required to find the time necessary for the water to fall to
a level H feet above the sill of the weir. It is assumed that the
area of the reservoir is so large that the velocity of the water as
it approaches the weir may be neglected.

When the surface of the water is at any height h above the sill
the flow in a time dt is

Let A be the area of the water surface at this level and dh the
distance the surface falls in time dt.

Then,
and

The time required for the surface to fall (H-H ) feet is,
therefore,

t&j*'**

L./H

The coefficient may be supposed constant and equal to 3*34.
If then A is constant

A f H dh

_
"CLWSo

To lower the level to the sill of the weir, H must be made
equal to and t is then infinite.

That is, on the assumptions made, the surface of the water
never could be reduced to the level of the sill of the weir. The
time taken is not actually infinite as the water in the reservoir is
not really at rest, but has a small velocity in the direction of the
weir, which causes the time of emptying to be less than that
given by the above formula. But although the actual time is
not infinite, it is nevertheless very great.

O A

When Ho is JH, *

When Ho is T VH, t

So that it takes three times as long for the water to fall from
to T VH as from H to iH.

110 HYDKAULTCS

Example 1. A reservoir has an area of 60,000 sq. yards. A weir 10 feet long
has its sill 2 feet below the surface. Find the time required to reduce the level of
the water 1' 11".

Therefore ' (3 ' 46 " ' 7 8) '

= 89,000 sees.
= 24-7 hours.

So that, neglecting velocity of approach, there will be only one inch of water on
the weir after 24 hours.

Example 2. To find hi the last example the discharge from the reservoir in
15 hours.

Therefore 54,000=^ (-^ - -^) .

From which N / H o=' 421 ,

H = 0-176 feet.
The discharge is, therefore,

(2-0-176) 540,000 cubic feet
= 984,960 cubic feet.

EXAMPLES.

(1) A weir is 100 feet long and the head is 9 inches. Find the discharge
in c. ft. per minute. C = 3'34.

(2) The discharge through a sharp-edged rectangular weir is 500
gallons per minute, and the still water head is 2| inches. Find the effective
length of the weir, m = '43.

(3) A weir is 15 feet long and the head over the crest is 15 inches.
Find the discharge. If the velocity of approach to this weir were 5 feet
per second, what would be the discharge ?

(4) Deduce an expression for the discharge through a right-angled
triangular notch. If the head over apex of notch is 12 ins., find the
discharge in c. ft. per sec.

(5) A rectangular weir is to discharge 10,000,000 gallons per day
(1 gallon =10 Ibs.), with a normal head of 15 ins. Find the length of the
weir. Choose a coefficient, stating for what kind of weir it is applicable,
or take the coefficient C as 3'33.

(6) What is the advantage in gauging, of using a weir without end
contractions ?

(7) Deduce Francis' formula by means of the Thomson principle of
similarity.

Apply the formula to calculate the discharge over a weir 10 feet wide
under a head of 1-2 feet, assuming one end contraction, and neglecting the
effect of the velocity of approach.

FLOW OVER WEIRS 111

(8) A rainfall of ^ * nc h P er hour is discharged from a catchment area
of 5 square miles. Find the still water head when this volume flows over
a weir with free overfall 30 feet in length, constructed in six bays, each
5 feet wide, taking O415 as Bazin's coefficient.

(9) A district of 6500 acres (1 acre =43,560 sq. ft.) drains into a large
storage reservoir. The maximum rate at which rain falls in the district is
2 ins. in 24 hours. When rain falls after the reservoir is full, the water
requires to be discharged over a weir or bye-wash which has its crest at
the ordinary top-water level of the reservoir. Find the length of such a
weir for the above reservoir, under the condition that the water in the
reservoir shall never rise more than 18 ins. above its top-water level.

The top of the weir may be supposed flat and about 18 inches wide
(see Table XI).

(10) Compare rectangular and V notches in regard to accuracy and
convenience when there is considerable variation in the flow.

In a rectangular notch 50" wide the still water surface level is 15" above
the sill.

If the same quantity of water flowed over a right-angled V notch, what
would be the height of the still water surface above the apex ?

If the channels are narrow how would you correct for velocity of
approach in each case? Lon. Un. 1906.

(11) The heaviest daily record of rainfall for a catchment area was
found to be 42*0 million gallons. Assuming two-thirds of the rain to reach
the storage reservoir and to pass over the waste weir, find the length of
the sill of the waste weir, so that the water shall never rise more than two
feet above the sill.

(12) A weir is 300 yards long. What is the discharge when the head
is 4 feet ? Take Bazin's coefiicient

-00984

(13) Suppose the water approaches the weir in the last question in a
channel 8' 6" deep and 500 yards wide. Find by approximation the dis-
charge, taking into account the velocity of approach.

(14) The area of the water surface of a reservoir is 20,000 square
yards. Find the time required for the surface to fall one foot, when the
water discharges over a sharp-edged weir 5 feet long and the original head
over the weir is 2 feet.

(15) Find, from the following data, the horse-power available in a given
waterfall :

Available height of faU 120 feet.

A rectangular notch above the fall, 10 feet long, is used to measure
the quantity of water, and the mean head over the notch is found to be
15 inches, when the velocity of approach at the point where the head
is measured is 100 feet per minute. Lon. Un. 1905.

CHAPTER V.

FLOW THROUGH PIPES.

86. Resistances to the motion of a fluid in a pipe.

When a fluid is made to flow through a pipe, certain resistances
are set up which oppose the motion, and energy is consequently
dissipated. Energy is lost, by friction, due to the relative motion
of the water and the pipe, by sudden enlargements or contractions
of the pipe, by sudden changes of direction, as at bends, and by
obstacles, such as valves which interfere with the free flow of the
fluid.

It will be necessary to consider these causes of the loss of
energy in detail.

Loss of head. Before proceeding to do so, however, the student
should be reminded that instead of loss of energy it is convenient
to speak of the loss of head.

It has been shown on page 39 that the work that can be
obtained from a pound of water, at a height z above datum,
moving with a velocity v feet per second, and at a pressure head

, is + =r- + z foot pounds.
w w 2g

If now water flows along a pipe and, due to any cause, Ti foot
pounds of work are lost per pound, the available head is clearly
diminished by an amount h.

In Fig. 86 water is supposed to be flowing from a tank through
a pipe of uniform diameter and of considerable length, the end B
being open to the atmosphere.

Fig. 86. Loss of head by friction in a pipe.

FLOW THROUGH PIPES 113

Let - be the head due to the atmospheric pressure.

Then if there were no resistances and assuming stream line
flow, Bernoulli's equation for the point B is

*>B ' f) uy i

w 2g w

from which ^- = Z P - Z B = H,

or v B = \/2<;H.

The whole head H above the point B has therefore been
utilised to give the kinetic energy to the water leaving the pipe at
B. Experiment would show, however, that the mean velocity of
the water would have some value v less than V B , and the kinetic

energy would be ~- .

A head h = -pr rr- = H ^~

2g 2g 2g

has therefore been lost in the pipe.

By carefully measuring H, the diameter of the pipe d, and the
discharge Q in a given time, the loss of head h can be determined.

For v = *

= -- ~ *

and therefore h = H

The head h clearly includes all causes of loss of head, which,
in this case, are loss at the entrance of the pipe and loss by
friction.

87. Loss of head by friction.

Suppose tubes 1, 2, 3 are fitted into the pipe AB, Fig. 86, at
equal distance apart, and with their lower ends flush with the inside
of the pipe, and the direction of the tube perpendicular to the
direction of flow. If flow is prevented by closing the end B of the
pipe, the water would rise in all the tubes to the level of the water
in the reservoir.

Further, if the flow is regulated at B by a valve so that the
mean velocity through the pipe is v feet per second, a permanent
regime being established, and the pipe is entirely full, the mean
velocity at all points along the pipe will be the same ; and there-
fore, if between the tank and the point B there were no resistances
offered to the motion, and it be assumed that all the particles
L.H. 8

114 HYDRAULICS

have a velocity equal to the mean velocity, the water would again
rise in all the tubes to the same height, but now lower than the

i; 2
surface of the water in the tank by an amount equal to ~-.

It is found by experiment, however, that the water does not
rise to the same height in the three tubes, but is lower in 2 than
in 1 and in 3 than in 2 as shown in the figure. As the fluid moves
along the pipe there is, therefore, a loss of head.

The difference of level h 2 of the water in the tubes 1 and 2 is
called the head lost by friction in the length of pipe 1 2. In any
length I of the pipe the loss of head is h.

This head is not wholly lost simply by the relative movement
of the water and the surface of the pipe, as if the water were
a solid body sliding along the pipe, but is really the sum of the
losses of energy, by friction along the surface, and due to relative
motions in the mass of water.

It will be shown later that, as the water flows along the pipe,
there is relative motion between consecutive filaments in the pipe,
and that, when the velocity is above a certain amount, the water
has a sinuous motion along the pipe. Some portion of this head h z
is therefore lost by the relative motion of the filaments of water,
and by the eddy motions which take place in the mass of the
water.

When the pipe is uniform the loss of head is proportional
to the length of the pipe, and the line CB, drawn through the tops
of the columns of water in the tubes and called the hydraulic
gradient, is a straight line.

It should be noted that along CB the pressure is equal to that
of the atmosphere.

88. Head lost at the entrance to the pipe.

For a point E just inside the pipe, Bernoulli's equation is

+ - + head lost at entrance to the pipe = h + ,
w Zg w

being the absolute pressure head at B.

The head lost at entrance has been shown on page 70 to be
about -Q^- 3 and therefore,

E p a ,

A ^ .
w w 2g

That is, the point C on the hydraulic gradient vertically above

1 * 5?j 2
E, is -~ below the surface FD.

FLOW THROUGH PIPES

115

If the pipe is bell-mouthed, there will be no head lost at entrance,
and the point C is a distance equal to ^~ below the surface.

89. Hydraulic gradient and virtual slope.

The line CB joining the tops of the columns of water in the
tube, is called the hydraulic gradient, and the angle i which it
makes with the horizontal is called the slope of the hydraulic
gradient, or the virtual slope. The angle i is generally small, and

sin i may be taken therefore equal to i, so that y = t.

In what follows the virtual slope y is denoted by .

More generally the hydraulic gradient may be defined as the
line, the vertical distance between which and the centre of the
pipe gives the pressure head at that point in the pipe. This line
will only be a straight line between any two points of the pipe,
when the head is lost uniformly along the pipe.

If the pressure head is measured above the atmospheric
pressure, the hydraulic gradient in Fig. 87 is AD, but if above
zero, AiDi is the hydraulic gradient, the vertical distance between

v 144
AD and AiDi being equal to , p a being the atmospheric

pressure per sq. inch.

Fig. 87. Pipe rising above the Hydraulic Gradient.

If the pipe rises above the hydraulic gradient AD, as in Fig. 87,
the pressure in the pipe at C will be less than that of the atmosphere
by a head equal to CB. If the pipe is perfectly air-tight it will
act as a siphon and the discharge for a given length of pipe will
not be altered. But if a tube open to the atmosphere be fitted at

116

HYDRAULICS

the highest point, the pressure at C is equal to the atmospheric
pressure, and the hydraulic gradient will be now AC, and the flow
will be diminished, as the available head to overcome the resist-
ances between B and C, and to give velocity to the water, will only
be CF, and the part of the pipe CD will not be kept full.

In practice, although the pipe is closed to the atmosphere, yet
air will tend to accumulate and spoil the siphon action.

As long as the point C is below the level of the water in the
reservoir, water will flow along the pipe, but any accumulation of
air at C tends to diminish the flow. In an ordinary pipe line it is
desirable, therefore, that no point in the pipe should be allowed to
rise above the hydraulic gradient.

90. Determination of the loss of head due to friction.
Reynolds' apparatus.

Fig. 88 shows the apparatus as used by Professor Reynolds* for
determining the loss of head by friction in a pipe.

Fig. 88. Beynolds' apparatus for determining loss of bead by friction in a pipe.

A horizontal pipe AB, 16 feet long, was connected to the water
main, a suitable regulating device being inserted between the
main and the pipe.

At two points 5 feet apart near the end B, and thus at a distance
sufficiently removed from the point at which the water entered
the pipe, that any initial eddy motions might be destroyed and a
steady regime established, two holes of about 1 mm. diameter were
pierced into the pipe for the purpose of gauging the pressure, at
these points of the pipe.

Short tubes were soldered to the pipe, so that the holes
communicated with these tubes, and these were connected by

* Phil. Trans. 1883, or Vol. n. Scientific Papers, Keynolds.

FLOW THROUGH PIPES 117

indiarubber pipes to the limbs of a siphon gauge Gr, made of glass,
and which contained mercury or bisulphide of carbon. Scales
were fixed behind the tubes so that the height of the columns
in each limb of the gauge could be read.

For very small differences of level a cathetometer was used*.
When water was made to flow through the pipe, the difference in
the heights of the columns in the two limbs of the siphon measured
the difference of pressure at the two points A and B of the pipe,
and thus measured the loss of head due to friction.

If s is the specific gravity of the liquid, and H the difference
in height of the columns, the loss of head due to friction in feet of
water is h = H (s - 1).

The quantity of water flowing in a time t was obtained by
actual measurement in a graduated flask.

Calling v the mean velocity in the pipe in feet per second, Q
the discharge in cubic feet per second, and d the diameter of the
pipe in feet,

The loss of head at different velocities was carefully measured,
and the law connecting head lost in a given length of pipe, with
the velocity, determined.

The results obtained by Keynolds, and others, using this
method of experimenting, will be referred to later.

91. Equation of flow in a pipe of uniform diameter
and determination of the head lost due to friction.

Let dl be the length of a small element of pipe of uniform
diameter, Fig. 89.

Fig. 89.

Let the area of the transverse section be o>, P the length of
the line of contact of the water and the surface on this section, or
the wetted perimeter, a the inclination of the pipe, p the pressure
per unit area on AB, and p dp the pressure on CD.
* p. 258, Vol. i. Scientific Paper*, Eeynolda.

118 HYDRAULICS

Let v be the mean velocity of the fluid, Q the flow in cubic
feet per second, and w the weight of one cubic foot of the fluid.
The work done by gravity as the fluid flows from AB to CD
= Qw . dz = . v . w . 9a.

The work done on ABCD by the pressure acting upon the area
AB

= p . w . v f t. Ibs. per sec.

The work done by the pressure acting upon CD against the
flow

= (p dp) . o> . v ft. Ibs. per sec.

The frictional force opposing the motion is proportional to the
area of the wetted surface and is equal to F . P . dl, where F is some
coefficient which must be determined by experiment and is the
frictional force per unit area. The work done by friction per sec.
is, therefore, F , P . 9Z . v.

The velocity being constant, the velocity head is the same at
both sections, and therefore, applying the principle of the con-
servation of energy,

p.w.t; + a>.i;.i0.92! = (p dp) w . V + F . P . dl . V.

Therefore w . w . dz = - dp . w + F . P . 9Z,

, dp Y.P.dl

or dz = - + .

W W . <*>

Integrating this equation between the limits of z and z ly p and
PI being the corresponding pressures, and I the length of the pipe,

= Pl p. F.P I

1 W W W o> *

FP I

Therefore, + z

w w w w

FPZ
The quantity is equal to h/ of equation (1), page 48, and is

the loss of head due to friction. The head lost by friction is
therefore proportional to the area of the wetted surface of the pipe
Pl } and inversely proportional to the cross sectional area of the
pipe and to the density of the fluid.

92. Hydraulic mean depth.

The quantity p is called the hydraulic radius, or the hydraulic

mean depth.

If then this quantity is denoted by m, the head h lost by
friction, is

~ w .m*

FLOW THROUGH PIPES 119

The quantity F, which has been called above the friction per
unit area, is found by experiments to vary with the density,
viscosity, and velocity of the fluid, and with the diameter and
roughness of the internal surface of the pipe.

In Hydraulics, the fluid considered is water, and any variations
in density or viscosity, due to changes of temperature, are generally
negligible. F, therefore, may be taken as proportional to the
density, or to the weight w per cubic foot, to the roughness of the
pipe, and as some function, f(v) of the mean velocity, and f(d) of
the diameter of the pipe.

Then, h

in which expression ^ may be called the coefficient of friction.

It will be seen later, that the mean velocity v is different from
the relative velocity u of the water and the surface of the pipe,
and it probably would be better to express F as a function of u,
but as u itself probably varies with the roughness of the pipe and
with other circumstances, and cannot directly be determined, it
simplifies matters to express F, and thus h, as a function of v.

93. Empirical formulae for loss of head due to friction.

The difficulty of correctly determining the exact value of
f(v) /(d), has led to the use of empirical formulae, which have
proved of great practical service, to express the head h in terms of
the velocity and the dimensions of the pipe.

The simplest* formula assumes that the friction simply varies as
the square of the velocity, and is independent of the diameter of
the pipe, or /(?;) f(d) = av*.

^ .............................. (1),

or writing p- 2 for a,

from which is deduced the well-known t Chezy formula,

or v = C "Jmi.

Another form in which formula (1) is often found is

\v*l

* See Appendix 9, t See also pages 231-233.

120

HYDRAULICS

or since m = 7 for a circular pipe full of water,

2g.d

(3),

in which for a of (1) is substituted f- .

The quantity 2g was introduced by Weisbach so that h is
expressed in terms of the velocity head.

Adopting either of these forms, the values of the coefficients C
and / are determined from experiments on various classes of pipes.

It should be noticed that C = */ -% .

Values of these constants are shown in Tables XII to XIV for
different kinds and diameters of pipes and different velocities.

TABLE XII.

Values of C in the formula v = C *Jmi for new and old cast-iron

pipes.

New cast-iron pipes

Old cast-iron pipes

Velocities in ft. per second

Diameter of pipe

100

102

101

104

106

105

109

112

12"

100

108

112

117

15"

102

110

117

122

18"

105

112

119

125

24"

111

120

126

131

101

104

30"

118

126

131

136

103

106

109

36"

124

131

136

140

103

108

111

114

42"

130

136

140

144

105

111

114

117

48"

135

141

145

148

106

112

115

118

60"

142

147

150

152

For method of determining the values of C given in the tables,
see page 132.

On reference to these tables, it will be seen, that C and / are
by no means constant, but vary very considerably for different
kinds of pipes, and for different values of the velocity in any
given pipe.

FLOW THROUGH PIPES

121

The fact that varies with the velocity, and the diameter of
the pipe, suggests that the coefficient is itself some function of
the velocity of flow, and of the diameter of the pipe, and that
does not, therefore, equal av*.

TABLE XIII.

Values of / in the formula

, 4/yj

New cast-iron pipes

Old cast-iron pipes

Velocities in
ft. per second

Diam. of pipe

0071

0067

0064

0062

0152

0139

0128

0122

007

0063

006

0057

0135

0117

0108

0103

0067

0058

0055

0051

0122

0105

010

0092

12"

0064

0056

0051

0048

0108

0096

0089

0084

15"

0062

0053

0048

0043

0099

0087

0081

0078

18"

0058

0051

0045

0041

0087

0078

0073

0069

24"

0053

0045

0040

0037

0076

0067

0063

0060

80"

0046

0040

0037

0035

0067

0061

0057

0055

36"

0042

0037

0035

0033

0061

0056

0052

0050

42"

0038

0035

0033

0031

0058

0052

005

0048

48"

0036

0032

0031

0029

0057

0051

0049

0046

60"

0032

0030

0029

0028

TABLE XIV.

Values of C in the formula v = C Jmi for steel riveted pipes.

Velocities in ft. per second

Diameter of pipe

11"

102

107

115

llf"
15"

93
109

99
112

102
114

105
117

38"

113

42"

102

106

108

111

48"

105

72"*

110

111

72"

101

105

110

103"

114

109

106

104

* See pages 124 and 137.

122 HYDRAULICS

94. Formula of Darcy.

In 1857 Darcy* published an account of a series of experiments
on flow of water in pipes, previous to the publication of which, it
had been assumed by most writers that the friction and consequently
the constant C was independent of the nature of the wetted surface
of the pipe (see page 232). He, however, showed by experiments
upon pipes of various diameters and of different materials,
including wrought iron, sheet iron covered with bitumen, lead,
glass, and new and old cast-iron, that the condition of the internal
surface was of considerable importance and that the resistance was
by no means independent of it.

He also investigated the influence of the diameter of the pipe
upon the resistance. The results of his experiments he expressed
by assuming the coefficient a in the formula

7 O'l 2

h = . ^
m

was of the form a - a + - ,

r being the radius of the pipe.

For new cast-iron, and wrought-iron pipes of the same
roughness, Darcy's values of and ft when transferred to English
units are,

a = 0-000077,
= 0'000003235.

For old cast-iron pipes Darcy proposed to double these values.
Substituting the diameter d for the radius r, and doubling /?, for
new pipes,

>- 0-000077

= 0-00000647

Substituting for m its value ^ and multiplying and dividing

For old cast-iron pipes,

& = 0-00001294

0-01 4 - -

l ( l+ I2d) 2g 'd
* Eeclierclies Experiment ales.

FLOW THROUGH PIPES 123

Or, *-^ 8 V l^ffl

As the student cannot possibly retain, without unnecessary
labour, values of / and C for different diameters it is convenient
to remember the simple forms,

for new pipes, and

for old pipes.

According to Darcy, therefore, the coefficient in the
formula varies only with the diameter and roughness of the pipe.

The values of C as calculated from his experimental results, for
some of the pipes, were practically constant for all velocities, and
notably for those pipes which had a comparatively rough internal
surface, but for smooth pipes, the value of varied from 10 to
20 per cent, for the same pipe as the velocity changed. The
experiments of other workers show the same results.

The assumption that p>f(v)f(d)=av* in which a is made to
vary only with the diameter and roughness, or in other words, the
assumption that h is proportional to v 2 is therefore not in general
justified by experiments.

95. As stated above, the formulae given must be taken as
purely empirical, and though by the introduction of suitable
constants they can be made to agree with any particular experi-
ment, or even set of experiments, yet none of them probably
expresses truly the laws of fluid friction.

The formula of Chezy by its simplicity has found favour, and
it is likely, that for some time to come, it will continue to be used,
either in the form v = C Vrai, or in its modified form

In making calculations, values of C or f y which most nearly suit
any given case, can be taken from the tables.

96. Variation of C in the formula v = C >/mi with service.

It should be clearly borne in mind, however, that the dis-
charging capacity of a pipe may be considerably diminished after
a few years' service.

Darcy's results show that the loss of head in an old pipe may
be double that in a new one, or since the velocity v is taken as

124 HYDRAULICS

proportional to the square root of h, the discharge of the old pipe
for the same head will be j=. times that of the new pipe, or about

30 per cent. less.

An experiment by Sherman *on a 36-inch cast-iron main showed
that after one year's service the discharge was diminished by
23 per cent., but a second year's service did not make any further
alteration.

Experiments by Kuichlingt on a 36-inch cast-iron main showed
that the discharge during four years diminished 36 per cent., while
experiments by Fitzgerald % on a cast-iron main, coated with tar,
which had been in use for 16 years, showed that cleaning increased
the discharge by nearly 40 per cent. Fitzgerald also found that
the discharge of the Sudbury aqueduct diminished 10 per cent, in
one year due to accumulation of slime.

The experiments of Marx, Wing, and Hoskins on a 72-inch steel
main, when new, and after two years' service, showed that there
had been a change in the condition of the internal surface of the
pipe, and that the discharge had diminished by 10 per cent, at low
velocities and about 5 per cent, at the higher velocities.

If, therefore, in calculations for pipes, values of C or / are used
for new pipes, it will in most cases be advisable to make the pipe
of such a size that it will discharge under the given head at least
from 10 to 30 per cent, more than the calculated value.

97. Ganguillet and Kutter's formula. Bazin formula.

Ganguillet and Kutter endeavoured to determine a form for
the coefficient C in the Chezy formula v = C Jmi, applicable
to all forms of channels, and in which C is made a function of the
virtual slope i, and also of the diameter of the pipe.

They gave C the value,

(10).

Vra

This formula is very cumbersome to use, and the value of the
coefficient of roughness n for different cases is uncertain; tables
and diagrams have however been prepared which considerably
facilitate its use. A simpler form has been suggested for channels
by Bazin (see page 185) which, by changing the constants, can be
used for pipes ||.

* Trans. Am.S.C.E. Vol. XLIV. p. 85. f Trans. Am.S.C.E. Vol. XLIV. p. 56.
J Trans. Am.S.C.E. Vol. xuv. p. 87. See Table No. XIV. || Proc. Inst. C.E. 1919.

FLOW THROUGH PIPES 125

Values of n in Ganguillet and Kutter's formula.
Wood pipes = "01, may be as high as '015.

Cast-iron and steel pipes = '011, '02.
Grlazed earthenware = '013.

98. Reynolds' experiments and the logarithmic formula.

The formulae for loss of head due to friction previously given
have all been founded upon a probable law of variation of h
with v, but no rational basis for the assumptions has been adduced.

It has been stated in section 93, that on the assumption that h
varies with -u 2 , the coefficient C in the formula

is itself a function of the velocity.

The experiments and deductions of Reynolds, and of later
workers, throw considerable light upon this subject, and show that
h is proportional to v n , where n is an index which for very small
velocities* as previously shown by Poiseuille by experiments on
capillary tubes is equal to unity, and for higher velocities may
have a variable value, which in many cases approximates to 2.

As Darcy's experiments marked a decided advance, in showing
experimentally that the roughness of the wetted surface has an
effect upon the loss due to friction, so Reynolds' work marked
a further step in showing that the index n depends upon the state
of the internal surface, being generally greater the rougher the
surface.

The student will be better able to follow Reynolds, by a brief
consideration of one of his experiments.

In Table XY are shown the results of an experiment made
by Reynolds with apparatus as illustrated in Fig. 88.

In columns 1 and 5 are shown the experimental values of

i = j , and v respectively.

The curves, Fig. 90, were obtained by plotting v as abscissae
and i as ordinates.

For velocities up to 1*347 feet per second, the points lie very close
to a straight line and i is simply proportional to the velocity, or

i = hv (11),

&i being a coefficient for this particular pipe.

Above 2 feet per second, the points lie very near to a continuous
curve, the equation to which is

i = Jcv n (12).

Phil. Trans. 1883.

126 HYDRAULICS

Taking logarithms,

log i = log k + n log v.

Curve N?2 is the part Aft of

Curve N?l drawn to laraer
ScaleL *

Velocity.

Fig. 90.

The curve, Fig. 90 a, was determined by plotting log i as
ordinate and logv as abscissae. Eeynolds calls the lines of this
figure the logarithmic homologues.

Calling log i, #, and log v, x, the equation has the form

which is an equation to a straight line, the inclination of which to
the axis of x is

= tan" 1 ^,

or n = tan 0.

Further, when x = 0, y = Jc, so that the value of Jc can readily be
found as the ordinate of the line when x or log v = 0, that is,
when v = 1.

Up to a velocity of 1*37 feet per second, the points lie near to
a line inclined at 45 degrees to the axis of v, and therefore, n is
unity, or as stated above, i - kv.

The ordinate when v is equal to unity is 0*038, so that for the
first part of the curve ~k = '038, and i = 'OoSv.

FLOW THROUGH PIPES

127

Above the velocity of 2 feet per second the points lie about
a second straight line, the inclination of which to the axis of v is

= tan' 1 T70.

Therefore log i = 1 '70 log v + Je.

The ordinate when v equals 1 is 0*042, so that

fc = 0-042,
and t

-3-0
20

-1-0

-8

-7

-5

--Z

vetoctty

3 4 S 6 755/0

Fig. 90 a. Logarithmic plottings of i and v to determine the index n in
the formula for pipes, i = kv n .

In the table are given values of i as determined experimentally
and as calculated from the equation i = Jc . v n .

The quantities in the two columns agree within 3 per confc.

123

HYDRAULICS

TABLE XV.

Experiment on Resistance in Pipes.
Lead Pipe. Diameter 0'242". Water from Manchester Main.

Slope

. h
~T

Velocity ft. per
second

Experimental value

Calculated from
i=kv n

0086

0092

038

209

0172

038

451

0258

0261

038

690

0345

0347

038

914

0430

0421

038

1-109

0516

0512

038

1-349

0602

. .

...

1-482

0682

, t

1-573

0861

. .

1-671

1033

1-775

1206

1-857

-1378

1352

042

1-70

1-987

1714

1610

042

1-70

2-203

3014

2944

042

1-70

3-141

4306

4207

042

1-70

3-93

8185

8017

042

1-70

5-66

1-021

1-033

042

1-70

6-57

1-433

1-476

042

1-70

8-11

2-455

2-404

042

1-70

10-79

3-274

3-206

042

1-70

12-79

3-873

3-899

042

1-70

14-29

NOTE. To make the columns shorter, only part of Keynolds' results are given.

99. Critical velocity.

It appears, from Reynolds' experiment, that up to a certain
velocity, which is called the Critical Velocity, the loss of head is
proportional to v, but above this velocity there is a definite change
in the law connecting i and v.

By experiments upon pipes of different diameters and the
water at variable temperatures, Reynolds found that the critical
velocity, which was taken as the point of intersection of the two
straight lines, was

0388P

the value of P being

(13),

1+ 0*0336 T + -000221T 2
T being the temperature in degrees centigrade and D the diameter
of the pipe.

FLOW THROUGH PIPES 129

100. Critical velocity by the method of colour bands.

The existence of the critical velocity has been beautifully
shown by Reynolds, by the method of colour bands, and his
experiments also explain why there is a sudden change in the law
connecting i and v.

"Water was drawn through tubes (Figs. 91 and 92), out of
a large glass tank in which the tubes were immersed, and in
which the water had been allowed to come to rest, arrangements
being made as shown in the figure so that a streak or streaks of
highly coloured water entered the tubes with the clear water."

Fig. 91.

Fig. 92.

The results were as follows :

" (1) When the velocities were sufficiently low, the streak
of colour extended in a beautiful straight line through the tube "
(Fig. 91).

"(2) As the velocity was increased by small stages, at
some point in the tube, always at a considerable distance from the
trumpet-shaped intake, the colour band would all at once mix up
with the surrounding water, and fill the rest of the tube with
a mass of coloured water" (Fig. 92).

This sudden change takes place at the critical velocity.

That such a change takes place is also shown by the apparatus
illustrated in Fig. 88; when the critical velocity is reached there is
a violent disturbance of the mercury in the U tube.

There is, therefore, a definite and sudden change in the con-
dition of flow. For velocities below the critical velocity, the flow
is parallel to the tubes, or is " Stream Line " flow, but after the
critical velocity has been passed, the motion parallel to the tube is
accompanied by eddy motions, which cause a definite change to
take place in the law of resistance.

Barnes and Coker* have determined the critical velocity by
noting the sudden change of temperature of the water when its
motion changes. They have also found that the critical velocity,
as determined by noting the velocity at which stream-line flow

* Proceedings of the Royal Society , Vol. LXXIV. 1904; Phil. Transactions,
Eoyal Society, Vol. xx. pp. 4561.

L. H. 9

130 HYDRAULICS

breaks up into eddies, is a nrncli more variable quantity than
that determined from the points of intersection of the two lines
as in Fig. 90. In the former case the critical velocity depends
upon the condition of the water in the tank, and when it is
perfectly at rest the stream lines may be maintained at much
higher velocities than those given by the formula of Reynolds.
If the water is not perfectly at rest, the results obtained by both
methods agree with the formula.

Barnes and Coker have called the critical velocity obtained by
the method of colour bands the upper limit, and that obtained by
the intersection of the logarithmic homologues the lower critical
velocity. The first gives the velocity at which water flowing from
rest in stream-line motion breaks up into eddy motion, while the
second gives the velocity at which water that is initially disturbed
persists in flowing with eddy motions throughout a long pipe, or
in other words the velocity is too high to allow stream lines to be
formed.

That the motion of the water in large conduits is in a similar
condition of motion is shown by the experiment of Mr Gr. H.
Benzenberg* on the discharge through a sewer 12 feet in diameter,
2534 ft. long.

In order to measure the velocity of water in the sewer, red
eosine dissolved in water was suddenly injected into the sewer,
and the time for the coloured water to reach the outlet half a
mile away was noted. The colour was readily perceived and it
was found that it was never distributed over a length of more than
9 feet. As will be seen by reference to section 130, the velocities
of translation of the particles on any cross section at any instant
are very different, and if the motion were stream line the colour
must have been spread out over a much greater length.

101. Law of frictional resistance for velocities above the
critical velocity.

As seen from Reynolds' formula, the critical velocity except
for very small pipes is so very low that it is only necessary in
practical hydraulics to consider the law of frictional resistance for
velocities above the critical velocity.

For any particular pipe,

i = Jcv n ,

and it remains to determine k and n.

From the plottings of the results of his own and Darcy's

* Transactions Am.S.C.E. 1893; and also Proceedings Am.S.C.E., Vol. xxvu.
p. 1173.

FLOW THROUGH PIPES

131

experiments, Reynolds found that the law of resistance " for all
pipes and all velocities " could be expressed as

AD 3 . /BD V

~P rl = (~P~ v )

"DWT\W nM T)2

Transposing, i ' '

AP".D 3

(15),

and

K ~7

A D 3 - n *

D is diameter of pipe, A and B are constants, and P is obtained
from formula (13).

Taking the temperature in degrees centigrade and the metre
as unit length,

A = 67,700,000,
B - 396,

_
"

L + -0036T + -000221T 3 '

B re . V n . P 2 ~ n y . v*

67,700,000 D 3 - = D 33 "

.(16),

in which

y 67,700,000'
Values of y when the temperature is 10 C.

1-75

0-000265

1-85

0-000388

1-95

0-000587

2-00

0-000704

The values for A and B, as given by Reynolds, are, however,
only applicable to clean pipes, and later experiments show that
although

- DP >

it is doubtful whether

p = 3 n y

as given by Eeynolds, is correct.

Value of n. For smooth pipes n appears to be nearly 1*75.
Reynolds found the mean value of n for lead pipes was T723.

Saph and Schoder*, in an elaborate series of experiments
carried out at Cornell University, have determined for smooth

* Transactions of the American Society of Civil Engineer*, May, 1903. See
exercise 31, page 172.

132 HYDRAULICS

brass pipes a mean value for n of 1'75. Coker and Clements
found that n for a brass pipe "3779 inches diameter was 1'731. In
column 5 of Table XVI are given values of n, some taken from
Saph and Schoder's paper, and others as determined by the
author by logarithmic plotting of a large number of experiments.

It will be seen that n varies very considerably for pipes of
different materials, and depends upon the condition of the surface
of a given material, as is seen very clearly from Nos. 3 and 4.
The value for n in No. 3 is 1*72, while for No. 4, which is the
same pipe after two years' service, the value of n is 1'93. The
internal surface had no doubt become coated with a deposit of
some kind.

Even very small differences in the condition of the surface,
such as cannot be seen by the unaided eye, make a considerable
difference in the value of n, as is seen by reference to the values
for galvanised pipes, as given by Saph and Schoder. For large
pipes of riveted steel, riveted wrought iron, and cast iron, the
value of n approximates to 2.

The method, of plotting the logarithms of i and v determined
by experiment, allows of experimental errors being corrected
without difficulty and with considerable assurance.

102. The determination of the values of C given in
Table XII.

The method of logarithmic plotting has been employed for
determining the values of C given in Table XII.

If values of C are calculated by the substitution of the
experimental values of v and i in the formula

many of the results are apparently inconsistent with each other
due to experimental errors.

The values of C in the table were, therefore, determined as
follows.

Since i = kv n

and in the Chezy formula

v = C *Jmi,

mC*'

v*
therefore p- 2 = kv n

and 21ogC = 21ogv- (log ra + log fc + w log v) (17).

The index n and the coefficient k were determined for a
number of cast-iron pipes.

FLOW THROUGH PIPES 133

Yalues of C for velocities from 1 to 10 were calculated. Curves
were then plotted, for different velocities, having C as ordinates
and diameters as abscissae, and the values given in the table were
deduced from the curves.

The values of C so interpolated differ very considerably, in
some cases, from the experimental values. The difficulties
attending the accurate determination of i and v are very great,
and the values of C, for any given pipe, as calculated by substi-
tuting in the Chezy formula the losses of head in friction and the
velocities as determined in the experiments, were frequently
inconsistent with each other.

As, for example, in the pipe of 3'22 ins. diameter given in
Table XVI which was one of Darcy^s pipes, the variation of C as
calculated from Ji and v given by Darcy is from 78'8 to 100.

On plotting log/I and log-u and correcting the readings so
that they all lie on one line and recalculating C the variation was
found to be only from 95'9 to 101.

Similar corrections have been made in other cases.

The author thinks this procedure is justified by the fact that
many of the best experiments do not show any such inconsistencies.

An attempt to draw up an interpolated table for riveted pipes
was not satisfactory. The author has therefore in Table XIV
given the values of C as calculated by formula (17), for various
velocities, and the diameters of the pipes actually experimented
upon. If curves are plotted from the values of C given in
Table XIV, it will be seen that, except for low velocities, the
curves are not continuous, and, until further experimental evidence
is forthcoming for riveted pipes, the engineer must be content
with choosing values of C which most nearly coincide, as far as
he can judge, with the case he is considering.

103. Variation of k, in the formula i = kv n , with the
diameter.

It has been shown in section 98 how the value of &, for a
given pipe, can be obtained by the logarithmic plotting of i and v.

In Table XVI, are given values of &, as determined by the
author, by plotting the results of different experiments. Saph
and Schoder found that for smooth hard-drawn brass pipes
of various sizes n varied between 1*73 and 1'77, the mean value
being 1*75.

By plotting logd as abscissae and log A; as ordinates, as in
Fig. 93, for these brass pipes the points lie nearly in a straight line
which has an inclination with the axis of d, such that

tan = - 1*25

184

HYDRAULICS

and the equation to the line is, therefore,

log k = log y-p log d,
where p = 1*25,

and log y = log k

when d \.

From the figure

y = G'000296 per foot length of pipe.

-01

Equation to line
log.lo-Log y -WSLog d

twvO 725

02 -03 -04- OG 08 ho -2O -3 \5 -6 -8 WO* t ' < t,

Logd, X

0031

Fig. 93. Logarithmic plottings of fc and d to determine the index p in the formula

On the same figure are plotted logd and logfc, as deduced
from experiments on lead and glass pipes by various workers. It
will be seen that all the points lie very close to the same line.

For smooth pipes, therefore, and for velocities above the
critical velocity, the loss of head due to friction is given by

the mean value for y being 0'000296, for n, T75, and for p T25.

From which, v = 104i' 672 ^ 715 ,

or log v = 2-017 + 0-572 log t + 0715 log d.

FLOW THROUGH PIPES 135

The value of p in this formula agrees with that given by
Reynolds in his formula

. yv n

Professor Unwin* in 1886, by an examination of experiments
on cast-iron pipes, deduced the formula, for smooth cast-iron
pipes,

-,, , . . '0007?; 2

and for rough pipes, i - , ri .

M. Flamantt in 1892 examined carefully the experiments
available on flow in pipes and proposed the formula,

yy 175

for all classes of pipes, and suggested for y the following values :
Lead pipes \

Glass \ '000236 to '00028,

Wrought-iron (smooth) J
Cast-iron new "000336,

in service '000417.

If the student plots from Table XVI, log d as ordinates, and
log k as abscissae, it will be found, that the points all lie between
two straight lines the equations to which are

log k = log '00069 - 1'25 log d,
and log k = log '00028 - 1'25 log d.

Further, the points for any class of pipes not only lie between
these two lines, but also lie about some line nearly parallel to
these lines. So that p is not very different from 1'25.
From the table, n is seen to vary from 1*70 to 2'08.
A general formula is thus obtained,

, -00028 to '00069^' torflg I

d 1 *

The variations in y, n, and p are, however, too great to admit
of the formula being useful for practical purposes.
For new cast-iron pipes,

, -000296 to 000418i? r84tor97 Z

h= -~a~

If the pipes are lined with bitumen the smaller values of y and
n may be taken.

* Industries, 1886.

f Ann-ales des Fonts et Chausstes, 1892, Vol. n.

136

HYDRAULICS

For new, steel, riveted pipes,

, -0004to'00054t; r93to2 - 08 Z

h r^ ...... . ...

Fig. 94 shows the result of plotting log k and log d for all
the pipes in Table XVI having a value of n between 1'92 and 1*94.
They are seen to lie very close to a line having a slope of 1*25,
and the ordinate of which, when d is 1 foot, is '000364.

Therefore h = -~^ or t> = 59i 518 cZ' 647

very approximately expresses the law of resistance for particular
pipes of wood, new cast iron, cleaned cast iron, and galvanised
iron.

Locjk.

Logarithmic plottings of log h

and log d from, Table 76,

to detefyiine the indea> p ttttfie

-, vihvn. IL Is Obouutl 93

Fig. 94.

Taking a pipe 1 foot diameter and the velocity as 3 feet per
second, the value of i obtained by this formula agrees with that
from Darcy's formula for clear cast-iron pipes within 1 per cent.

Use of the logarithmic formula for practical calculations. A
very serious difficulty arises in the use of the logarithmic
formula, as to what value to give to n for any given case, and
consequently it has for practical purposes very little advantage
over the older and simpler formula of Chezy.

TABLE XVI.

Experimenter

Kind of pipe

Diameter
(in ins.)

Velocity in
ft. per sec.
from to

Value of n
in formula
i = kv n

Value of k
in formula
i = kv n

Noble

Wood

3-46 4-415

1-73

0001254

2-28 4-68

1-75

000083

Marx, Wing )

72-5

1 4

1-72

000061

and Hoskins }

72-5

1 5-5

1-93

000048

Galtner Kitcham

Riveted

1-88

00245

H. Smith

Wrought

1-81

000515

iron or steel

11|

1-90

000470

1-94

000270

Kinchling

505 1-254

2-0

000099

Herschel

2-10 4-99

1-93

00011

2 5 (?)

2-0

000090

Marx, Wing )

1 4

1-99

000055

and Hoskins j

1 5-5

1-85

000077

Herschel

103

1 4-5

2-08

000036

Darcy

Cast iron

3-22

28910-71

1-97

00156

new

5-39

48 15-3

1-97

OOQ79

7-44

67316-17

1-956

00062

1-779

000323

Williams

16-25

1-858

000214

Lampe

16-5

2-48 3-09

1-80

000267

19-68

1-38 3-7

1-84

00022

Sherman

4 7

000062

Stearns

1-243 3-23

1-92

0000567

Hubbell&Fenkell

00003

Darcy

Cast iron

1-4136

167 2-077

1-99

0098

old and

3-1296

403 3-747

1-94

0035

tuberculated

9-575

1-00712-58

1-98

0009

Sherman

2-71 5-11

1-1 4-5

000105

Fitzgerald

1-176 3-533

2-04

000083

1-135 3-412

2-00

OOC085

Darcy

Cast-iron

1-4328

371_ 3-69

1-85

0041

old pipes

3-1536

633 5-0

1-97

00185

cleaned

11-68

8 10-368

2-0

000375

Fitzgerald

3-67 5-6

2-02

000082

395 7-245

1-94

000059

Darcy

Sheet-iron

1-055

098 8-225

1-73

0074

3-24

32812-78

1-81

00154

7-72

59119-72

1-78

00059

11-2

1-29610-52

1-81

00039

Gas

113 3-92

1-83

0278

1-55

205 8-521

1-86

00418

1-91

0072

Saph and Schoder

Galvanised

364

1-96

0352

494

1-91

0181

623

1-86

0132

824

1-80

0095

1-048

1-93

0082

Hard-drawn

15 pipes

* m rrK

00025 to

brass

up to 1-84

1 75

00035

Reynolds

Lead

1-732

Darcy

1-761

0126

1-61

1-783

00425

138

HYDRAULICS

TABLE XVII.

Showing reasonable values of y, and n, for pipes of various
kinds, in the formula,

,_n

Reasonable

values for

Clean cast-iron pipes

00029 to -000418

1-80 to 1-97

00036

1-93

Old cast-iron pipes

00047 to -00069

1-94 to 2-04

00060

Riveted pipes

00040 to -00054

1-93 to 2-08

00050

Galvanised pipes

00035 to -00045

1-80 to 1-96

00040

1-88

Sheet-iron pipes cover-
ed with bitumen

00030 to -00038

1-76 to 1-81

00034

1-78

Clean wood pipes
Brass and lead pipes

00056 to -00063

1-72 to 1-75

00060
00030

1-75
1-75

When further experiments have been performed on pipes, of
which the state of the internal surfaces is accurately known, and
special care taken to ensure that all the loss of head in a given
length of pipe is due to friction only, more definiteness may be
given to the values of y, n, and p.

Until such evidence is forthcoming the simple Chezy formula
may be used with almost as much confidence as the more
complicated logarithmic formula, the values of C or / being taken
from Tables XII XIV. Or the formula h = kv n may be used,
values of k and n being taken from Table XVI, which most nearly
fits the case for which the calculations are to be made.

104. Criticism of experiments.

The difficulty of differentiating the loss of head due to friction
from other sources of loss, such as loss due to changes in direction,
change in the diameter of the pipe and other causes, as well as the
possibilities of error in experiments on long pipes of large diameter,
makes many experiments that have been performed of very little
value, and considerably increases the difficulty of arriving at
correct formulae.

The author has found in many cases, when log i and log d were
plotted, from the records of experiments, that, although the results
seemed consistent amongst themselves, yet compared with other
experiments, they seemed of little value.

FLOW THROUGH PIPES

139

The value of n for one of Couplet's* experiments on a lead and
earthenware pipe being as low as 1*56, while the results of an
experiment by Simpson t on a cast-iron pipe gave n as 2'5. In the
latter case there were a number of bends in the pipe.

In making experiments for loss of head due to friction, it is
desirable that the pipe should be of uniform diameter and as
straight as possible between the points at which the pressure head
is measured. Further, special care should be taken to ensure the
removal of all air, and it is most essential that a perfectly steady
flow is established at the point where the pressure is taken.

105. Piezometer fittings.

It is of supreme importance that the
piezometer connections shall be made
so that the difference in the pressures
registered at any two points shall be
that lost by friction, and friction only,
between the points.

This necessitates that there shall
be no obstructions to interfere with the
free flow of the water, and it is, there-
fore, very essential that all burrs shall
be removed from the inside of the pipe.

In experiments on small pipes in
the laboratory the best results are no
doubt obtained by cutting the pipe
completely through at the connection
as shown in Fig. 95, which illustrates
the form of connection used by Dr
Coker in the experiments cited on
page 129. The two ends of the pipe are not more than
of an inch apart.

Fig. 96 shows the method adopted by Marx, Wing and Hoskins
in their experiments on a 72-inch wooden pipe to ensure a correct
reading of the pressure.

The gauge X was connected to the top of the pipe only while
Y was connected at four points as shown.

Small differences were observed in the readings of the two
gauges, which they thought were due to some accidental circum-
stance affecting the gauge X only, as no change was observed
in the reading of Y when the points of communication to Y were
changed by means of the cocks.

* Hydraulics, Hamilton Smith, Junr.

t Proceedings of the Institute of Civil Engineers, 1855.

Fig. 95.

140

HYDRAULICS

106. Effect of temperature on the velocity of flow.

Poiseuille found that by raising the temperature of the water
from 50 C. to 100 C. the discharge of capillary tubes was
doubled.

Fig. 96. Piezometer connections to a wooden pipe.

Reynolds* showed that for pipes of larger diameter, the effect
of changes of the temperature was very marked for velocities
below the critical velocity, but for velocities above the critical
velocity the effect is comparatively small.

The reason for this is seen, at once, from an examination of
Reynolds'* formula. Above the critical velocity n does not differ
very much from 2, so that P 2 ~" is a small quantity compared with
its value when n is 1.

Saph and Schodert, for velocities above the critical velocity,
found that, as the temperature rises, the loss of head due to
friction decreases, but only in a small degree. For brass pipes of
small diameter, the correction at 60 F. was about 4 per cent, per

* Scientific Papers, Vol. n.

t See also Barnes and Coker, Proceedings of the Royal Society, Vol. LXX. 1904 ;
Coker and Clements, Transactions of the Royal Society, Vol. cci. Proceedings
Am.S.C.E. Vol. xxix.

FLOW THROUGH PIPES 141

10 degrees F. With galvanised pipes the correction appears to
be from 1 per cent, to 5 per cent, per 10 degrees F.

Since the head lost increases, as the temperature falls, the
discharge for any given head diminishes with the temperature,
but for practical purposes the correction is generally negligible.

107. *Loss of head due to bends and elbows.

The loss of head due to bends and elbows in a long pipe is
generally so small compared with the loss of head due to friction
in the straight part of the pipe, that it can be neglected, and
consequently the experimental determination of this quantity has
not received much attention.

Weisbacht, from experiments on a pipe 1J inches diameter,
with bends of various radii, expressed the loss of head as

. *923r\ v*

+ --

r being the radius of the pipe, R the radius of the bend on the
centre line of the pipe and v the velocity of the water in feet per
second. If the formula be written in the form

7^ ^_

the table shows the values of a for different values of ^ ,

1 -157

2 -250

5 -526

St Tenant J has given as the loss of head & B at a bend,
TIB = '00152 j~ y/ 1 v 2 =0'l^ g y^ nearly,

Z being the length of the bend measured on the centre line of the
bend and d the diameter of the pipe.
When the bend is a right angle

L /I = * /I

RV R 2 V R*
When | = 1, '5, '2,

V R~

111, '702

See page 525. t Mechanics of Engineering.

$ Comptes Rendus, 1862.

142

HYD-RAULICS

Recent experiments by Williams, Hubbell and Fenkell* on cast-
iron pipes asphalted, by Saph and Schoder on brass pipes, and
others by Alexander t on wooden pipes, show that the loss of head
in bends, as in a straight pipe, can be expressed as

n being a variable for different kinds of pipes, while

..-;:'' . . *sr

y being a constant coefficient for any pipe.

For the cast-iron pipes of Hubbell and Fenkell, y, n, m, and p
have approximately the following values.

Diameter of pipe

12"

0040

0-83

1-78

1-09

16"

1-86

30"

2-0

When v is 3 feet per second and jr is i, the bend being a right

angle, the loss of head as calculated by this formula for the

-lo-i '2068u a , , ,, OA . , . *238v 2
12-inch pipe is ~ - , and for the 30-inch pipe -^ .

For the brass pipes of Saph and Schoder, 2 inches diameter,
Alexander found,

and for varnished wood pipes when =5- is less than 0'2,

h* = "008268 (5) "to 1 ",
and when ^ is between 0'2 and 0*5,

He further found for varnished wood pipes that, a bend of
radius equal to 5 times the radius of the pipe gives the minimum
loss of head, and that its resistance is equal to a straight pipe 3'38
times the length of the bend.

Messrs Williams, Hubbell and Fenkell also state at the end of
their elaborate paper, that a bend having a radius equal to 2J

* Proc. Amer. Soc. Civil Engineers, Vol. xxvn. f Proc. Inst. Civil Engineers,
Vol. CLIX. See also Bulletin No. 576 University of Wisconsin.

FLOW THROUGH PIPES 143

diameters, offers less resistance to the flow of water than those of
longer radius. It should not be overlooked, however, that although
the loss of head in a bend of radius equal to * 2 diameters of the
pipe is less than for any other, it does not follow that the loss of
head per unit length of the pipe measured along its centre line
has its minimum value for bends of this radius.

108. Variations of the velocity at the cross section of a
cylindrical pipe.

Experiments show that when water flows through conduits of
any form, the velocities are not the same at all points of any
transverse section, but decrease from the centre towards the
circumference.

The first experiments to determine the law of the variation of
the velocity in cylindrical pipes were those of Darcy, the pipes
varying in diameter from 7'8 inches to 19 inches. A complete
account of the experiments is to be found in his Recherches
Experimentales dans les tuyaux.

The velocity was measured by means of a Pitot tube at five
points on a vertical diameter, and xx ^^^^^^^

the results plotted as shown in
Fig. 97.

Calling V the velocity at the
centre of a pipe of radius R, u the
velocity at the circumference, v m
the mean velocity, v the velocity
at any distance r from the centre,
and i the loss of head per unit
length of the pipe, Darcy deduced the formulae

1-33

and v m =

When the unit is the metre the value of Jc is 11 '3, and 20*4 when
the unit is the English foot.

Later experiments commenced by Darcy and continued by
Bazin, on the distribution of velocity in a semicircular channel,
the surface of the water being maintained at the horizontal
diameter, and in which it was assumed the conditions were similar
to those in a cylindrical pipe, showed that the velocity near the
surface of the pipe diminished much more rapidly than indicated
by the formula of Darcy.

* See Appendix 3.

144 HYDRAULICS

Bazin substituted therefore a new formula,

........................ (1),

or snce

It was open to question, however, whether the conditions of flow
in a semicircular pipe are similar to those in a pipe discharging
full bore, and Bazin consequently carried out at Dijon* experi-
ments on the distribution of velocity in a cement pipe, 2'73 feet
diameter, the discharge through which was measured by means
of a weir, and the velocities at different points in the transverse
section by means of a Pitot tubet.

From these experiments Bazin concluded that both formulae (1)
and (2) were incorrect and deduced the three formulae

(3),
...... (4),

Y - v = VRi SS^l-^/l- '95 () 2 } ............... (5),

the constants in these formulae being obtained from Bazin's by
changing the unit from 1 metre to the English foot.

Equation (5) is the equation to an ellipse to which the sides of
the pipes are not tangents but are nearly so, and this formula
gives values of v near to the surface of the pipe, which agree much
more nearly with the experimental values, than those given by
any of the other formulae.

Experiments of Williams, Hubbell and Fenkell*. An elaborate
series of experiments by these three workers have been carried out
to determine the distribution of velocity in pipes of various
diameters, Pitot tubes being used to determine the velocities.

The pipes at Detroit were of cast iron and had diameters of 12,
16, 30 and 42 inches respectively.

The Pitot tubes were calibrated by preliminary experiments
on the flow through brass tubes 2 inches diameter, the total

* " Memoire de 1' Academic des Sciences de Paris, Kecueil des Savants Etrangeres,"
Vol. xxxn. 1897. Proc. Am.S.C.E. Vol. xxvn. p. 1042.

+ See page 241.

J "Experiments at Detroit, Mich., on the effect of curvature on the flow of
water in pipes," Proc. Am.S.C.E. Vol. xxvn. p. 313.

See page 246.

FLOW THROUGH PIPES 145

discharge being determined by weighing, and the mean velocity
thus determined. From the results of their experiments they
came to the conclusion that the curve of velocities should be an
ellipse to which the sides of the pipe are tangents, and that the
velocity at the centre of the pipe Y is l'I9v m , v m being the mean
velocity.

These results are consistent with those of Bazin. His experi-

y
mental value for for the cement pipe was T1675, and if the

constant *95, in formula (5), be made equal to 1, the velocity curve
becomes an ellipse to which the walls of the pipe are tangents.

The ratio can be determined from any of Bazin's formulae.

Substituting ^p for >/E5 in (1), (3), (4) or (5), the value of
v at radius r can be expressed by any one of them as

Then, since the flow past any section in unit time is v,?rR a , and
that the flow is also equal to

Zvrdr.v,

f E f v2o /rM
therefore v m 7rR 2 = 27r I JV P^VP )| r ^ r *

/ v \ ^ftr 3

Substituting for f (^ ) > ifa value -5-3- from equation (1), and

\.Q// K>

integrating,

^i" 1 + "C~ W

and by substitution of ft ^J from equation (4),

= 1 + (8)

V m C

so that the ratio is not very different when deduced from the
v m

simple formula (2) or the more complicated formula (4).
When C has the values

= 80, 100, 120,

from (8) = 1-287, 1'23, T19.

V m

The value of C, in the 30-inch pipe referred to above, varied
between 109'6 and 123'4 for different lengths of the pipe, and
L. H. 10

146 HYDRAULICS

the mean value was 116, so that there is a remarkable agreement
between the results of Bazin, and Williams, Hubbell and Fenkell.

The velocity at the surface of a pipe. Assuming that the
velocity curve is an ellipse to which
the sides of the pipe are tangents, as MB
in Fig. 98, and that Y=l'19v m , the
velocity at the surface of the pipe
can readily be determined.

Let u = the velocity at the surface
of the pipe and v the velocity at any
radius r.

Let the equation to the ellipse be Fi S- 98

in which x = v - u,

and b = Y - u.

Then, if the semi-ellipse be revolved about its horizontal axis,
the volume swept out by it will be f*rB, a 6, and the volume of
discharge per second will be
/R

irR 2 t? m = I Zirrdr . V = 7rR a . U +

and u = "621 -y m .

Using Bazin's elliptical formula, the values of for

= 80, 100, 120,
are - = '552, '642, '702.

The velocities, as above determined, give the velocity of
translation in a direction parallel to the pipe, but as shown by
Reynolds' experiments the particles of water may have a much
more complicated motion than here assumed.

109. Head necessary to give the mean velocity v m to
the water in the pipe.

It is generally assumed that the head necessary to give a mean

velocity v m to the water flowing in a pipe is |p-, which would be

correct if all the particles of water had a common velocity v m .

If, however, the form of the velocity curve is known, and on the
assumption that the water is moving in stream lines with definite
velocities parallel to the axis of the pipe, the actual head can
be determined by calculating the mean kinetic energy per Ib. of

v *
water flowing in the pipe, and this is slightly greater than -- .

FLOW THROUGH PIPES 147

As before, let v be the velocity at radius r.
The kinetic energy of the quantity of water which flows past
any section per second

R -y2

w.%Trrdr.v . ~-,
o 20'

w being the weight of 1 c. ft. of water.
The kinetic energy per lb., therefore,

w . 2-n-rdrv
o

The simplest value for / ( ^ ) is that of Bazin's formula (1)
above, from which

21'5

and

Substituting these values and integrating, the kinetic energy
per Ib. is , and when

C is 80, 100,
a is 112, T076.

On the assumption that the velocity curve is an ellipse to which
the walls of the pipe are tangents the integration is easy, and the
value of a is 1'047.

Using the other formulae of Bazin the calculations are tedious
and the values obtained differ but slightly from those given.

The head necessary to give a mean velocity v m to the water in

the pipe may therefore "be taken to be ^~ , the value of a being

about 1*12. This value* agrees with the value of 1*12 for a,
obtained by M. Boussinesq, and with that of M. J. Delemer who
finds for a the value 1*1346.

110. Practical problems.

Before proceeding to show how the formulae relating to the
loss of head in pipes may be used for the solution of various
probjems, it will be convenient to tabulate them.
* Flamant's Hydraulique.

102

148 HYDRAULICS

NOTATION.
h = loss of head due to friction in a length I of a straight pipe.

i = the virtual slope = j .
I/

v = the mean velocity of flow in the pipe.
d = the diameter.
m = the hydraulic mean depth

A Tp*p A fj

= Wetted Perimeter = P = 4 when the pipe is c y lindrioal and ful1

Formula 1. h ^ = 4M

Cm C d

This may be written y = ,

or v

The values of C for cast-iron and steel pipes are shown in
Tables XII and XIV.

Formula 2. ^ = 2^5'

^- in this formula being equal to ^ of formula (1).

Values of /are shown in Table XIII.

Either of these formulae can conveniently be used for
calculating h, v, or d when /, and Z, and any two of three
quantities h, v y and d t are known.

Formula 3. As values of C and / cannot be remembered for
variable velocities and diameters, the formulae of Darcy are
convenient as giving results, in many cases, with sufficient
accuracy. For smooth clean cast-iron pipes

12<2/20. d
r=19 Vl2JTI^

= 394 N/l2iVl^-
For rough and dirty pipes

1 \ k?l

IZdJZg.d*
or vssm ^-*Ja

= 278 v /j2| ri x.

FLOW THROUGH PIPES 149

If d is the unknown, Darcy's formulae can only be used to solve
for d by approximation. The coefficient 1 1 + T^JJ is first neglected

and an approximate value of d determined. The coefficient can
then be obtained from this approximate value of d with a greater
degree of accuracy, and a new value of d can then be found, and
so on. (See examples.)

Formula 4. Known as the logarithmic formula.

, yi

d" '

h . y.v"

=t=

Values of y, n, and p are given on page 138.
By taking logarithms

log h = log y + n log v + log I p log d,
from which h can be found if Z, v, and d are known.
If h t l f and d are known, by writing the formula as

n log v - log h log I - log y + p log eZ,
v can be found.

If h, I, and v are known, d can be obtained from
p log d - log y + n log v + log I - log h.

This formula is a little more cumbersome to use than either (1) or
(2) but it has the advantage that y is constant for all velocities.
Formula 5. The head necessary to give a mean velocity v to

the water flowing along the pipe is about ~ , but it is generally

v 9
convenient and sufficiently accurate to take this head as 5-, as

was done in Fig. 87. Unless the pipe is short this quantity is
negligible compared with the friction head.

Formula 6. The loss of head at the sharp-edged entrance to a

\/jj^

pipe is about -g and is generally negligible.

Formula 7. The loss of head due to a sudden enlargement in
a pipe where the velocity changes from v l to t? a is ^ Vl ~ V2 ' .

Formula 8. The loss of head at bends and elbows is a very
variable quantity. It can be expressed as equal to in which
a varies from a very small quantity to unity.

Problem 1. The difference in level of the water in two reservoirs is h feet,
Fig. 99, and they are connected by means of a straight pipe of length I and
diameter d; to find the discharge through the pipe.

150

HYDRAULICS

Let Q be the number of cubic feet discharged per second. The head h is utilised
in giving velocity to the water and in overcoming resistance at the entrance to the
pipe aud the fractional resistances.

Fig. 99. Pipe connecting two reservoirs.

Let v be the mean velocity of the water. The head necessary to give the water

l*12y 2
this mean velocity may be taken as ~ , and to overcome the resistance at the

entrances

Then

Or using in the expression for friction, the coefficient 0,
A=-0174v 3 + -0078t; 2 +^

= -025t; 2 +

C 2 d'

If - is greater than 300 the head lost due to friction is generally great compared
a

w th the othjr quantities, and these may be neglected.
4 fto 2 4lv' 2

Then h== >

and

C /Jh
2~VT*

As the velocity is not known, the coefficient C cannot be obtained from the
table, but an approximate value can be assumed, or Darcy's value

0=394

= 278

for clean pipes,
if the pipe is dirty,

and

can be taken.

An approximation to v which in many cases will be sufficiently near or will be
as near probably as the coefficient can be known is thus obtained. From the
table a value of C for this velocity can be taken and a nearer approximation to
v determined.

Then Q=^d 2 .v.

The velocity can be deduced directly from the logarithmic formula h=^^,
provided y and n are known for the pipe.

FLOW THROUGH PIPES 151

The hydraulic gradient is EF.

At any point C distant x from A the pressure head is equal to the distance
between the centre of the pipe and the hydraulic gradient. The pressure head
just inside the end A of the pipe is h -- - , and at the end B the pressure head
must be equal to /IB- The head lost due to friction is h, which, neglecting the
small quantity - , is equal to the difference of level of the water in the two
tanks.

Example 1. A pipe 3 inches diameter 200 ft. long connects two tanks, the
difference of level of the water in which is 10 feet, and the pressure is atmospheric.
Find the discharge assuming the pipe dirty.

Using Darcy's coefficient
V=278

3l
= 3-88 ft. per sec.

For a pipe 3 inches diameter, and this velocity, C from the table is about 69, so
that the approximation is sufficiently near.

OOOGivi-w. I
Taking h= - jf^ ,

v=3-88 ft. per sec.,

- # '
gives v = 3'85 ft. per sec.

Example 2. A pipe 18 inches diameter brings water from a reservoir 100 feet
above datum. The total length of the pipe is 15,000 feet and the last 5000 feet
are at the datum level. For tbis 5000 feet the water is drawn off by service pipes at
the uniform rate of 20 cubic feet per minute, per 500 feet length. Find the pressure
at the end of the pipe.

The total quantity of flow per minute is
00x
oOU

Area of the pipe is 1'767 sq. feet.
The velocity in the first 10,000 feet is, therefore,
200

The head lost due to friction in this length, is

4./. 10,000.1-888*
2p.l-5 ---
In the last 5000 feet of the pipe the velocity varies uniformly. At a distance

x feet from the end of the pipe the velocity is .

In a length dx the head lost due to friction is

4./. l-888 2 .a; 2 ds
20.T5.5000 2 '
and the total loss by friction is

4/. 1-888 2 /"MM 2 _4/. (l-888) a 5000
~2<7.1-5.5000 2 Jo ' 20.1-5 ' 3 '

The total head lost due to friction in the whole pipe is, therefore,

152 HYDRAULICS

Taking / as -0082, H = 14-3 feet.

Neglecting the velocity head and the loss of head at entrance, the pressure head
at the end of the pipe is (100 - H) feet = 85-7 feet.

Problem 2. Diameter of pipe to give a given discharge.

Bequired the diameter of a pipe of length I feet which will discharge Q cubic feet
per second between the two reservoirs of tbe last problem.
Let v be the mean velocity and d the diameter of the pipe.

and

Therefore,

Squaring and transposing,
I
If I is long compared with d,

and

(1),

0-040G.Q 2 d
~~

JL=o /**

7T ,. V 4Z '

Since v and d are unknown C is unknown, and a value for C must be pro-
visionally assumed.

Assume C is 100 for a new pipe and 80 for an old pipe, and solve equation (3)
ford.

From (1) find v, and from the tables find the value of G corresponding to the
values of d and v thus determined.

If C differs much from the assumed value, recalculate d and v using this second
value of C, and from the tables find a third value for C. This will generally be
found to be sufficiently near to the second value to make it unnecessary to calculate
d and v a third time.

The approximation, assuming the values of G in the tables are correct, can be
taken to any degree of accuracy, but as the values of G are uncertain it will not as
a rule be necessary to calculate more than two values of d.

yv n l
Logarithmic formula. If the formula 'h, -^ be used, d can be found direct,

from

p log d=n log u + log7 + log I - log h.

Example 3. Find the diameter of a steel riveted pipe, which will discharge
14 cubic feet per second, the loss of head by friction being 2 feet per mile. It is
assumed that the pipe has become dirty and that provisionally C = 110.

From equation (3)

5 _ 2-55. 14 /528Q

or $ log d- log 16 -63,

therefore d = 3-08 feet.

For a thirty-eight inch pipe Kuichling found C to be 113.

The assumption that C is 110 is nearly correct and the diameter may be taken
as 37 inches.

Using the logarithmic formula

FLOW THROUGH PIPES

153

and substituting for v the value 2-

000450^

h /_\ 1-95

( - ) d 5 ' 15

from which

5 -15 log d = log -000 45-1 -95 log 0-7854 + 1-95 log 14 + log 2640,
and d = 3-07 feet.

Short pipe. If the pipe is short so that the velocity head and the head lost at
entrance are not negligible compared with the loss due to friction, the equation

. -0406Q 2 d 6-5ZQ 3

when a value is given to C, can be solved graphically by plotting two curves

and

040GQ 2 6-5ZQ 2

~~~ ~

The point of intersection of the two curves will give the
diameter d.

It is however easier to solve by approximation in the
following manner.

Neglect the term in d and solve as for a long pipe.

Choose a new value for C corresponding to this ap-
proximate diameter, and the velocity corresponding to it,
and then plot three points on the curve y = d 5 , choosing
values of d which are nearly equal to the calculated value
of d, and two points of the straight line

0406Q 2 d
2/i=-

Fig. 100.

The curve y = d 5 between the three points can easily *
be drawn, as hi Fig. 100, and where the straight line cuts
the curve, gives the required diameter.

Example 4. One hundred and twenty cubic feet of water are to be taken
per minute from a tank through a cast-iron pipe 100 feet long, having a square-
edged entrance. The total head is 10 feet. Find the diameter of the pipe.

Neglecting the term in d and assuming C to be 100,

and

Therefore

100.100.10
d= -4819 feet.
2

v=-

10-9 ft. per seo.

From Table XII, the value of C is seen to be about 106 for these values of
d and v.

A second value for d 6 is

from which d= '476'.

The schedule shows the values of d 5 and y for values of d not very different
from the calculated value, and taking C as 106.

d -4 -5 -6

d 6 -01024 -03125 -0776

y l -0297 -0329

The line and curve plotted in Fig. 100, from this schedule, intersect atp for which

d= -4*98 feet.

154 HYDRAULICS

It is seen therefore that taking 106 as the value of C, neglecting the term in d,
makes an error of -022' or -264".

This problem shows that when the ratio -r is about 200, and the virtual slope is
even as great as j^, for all practical purposes, the friction head only need be con-
sidered. For smaller values of the ratio the quantity '0250 2 may become im-
portant, but to what extent will depend upon the slope of the hydraulic gradient.

The logarithmic formula may be used for short pipes but it is a little more
cumbersome.

Using the logarithmic formula to express the loss of head for short pipes with
square-edged entrance,

-*+

When suitable values are given to 7 and n, this can be solved by plotting the
two curves

and

the intersection of the two curves giving the required value of d.

Problem 3. To find what the discharge between the reservoirs of problem (1)
would be, if for a given distance Z a the pipe i

of diameter d is divided into two branches j |

laid side by side having diameters d-, and rf. t< Z^ --- >K- --- L --- H
Fig- 101. 4> * j

Assume all the head is lost in friction. A _ *< ^s/^ *%* '

Let Qj be the discharge in cubic feet. j y (

Then, since both the branches BC and BD * -- >v> - -* - , -~

are connected at B and to the same reservoir, j x> -- ^ - 1 1)

the head lost in friction must be the same in I ,

BC as in BD, and if there were any number ; ~
of branches connected at B the head lost in -pj g 101

them all would be the same.

The case is analogous to that of a conductor joining two points between which
a definite difference of potential is maintained, the conductor being divided between
the points into several circuits in parallel.

The total head lost between the reservoirs is, therefore, the head lost in AB
together with the head lost in any one of the branches.

Let v be the velocity in AB, v 1 in BC and v z in BD.

Then vd^^v^ + v^ .................................... (1),

and the difference of level between the reservoirs

h=?2L + ^ l (2).

And since the head lost in BC is the same as in BD, therefore,

f (3).

If provisionally Gj be taken as equal to C 2 ,

FLOW THROUGH PIPES

155

Therefore,

and

v.d?

.(4).

From (2), v can be found by substituting for Vj from (4), and thus Q can
be determined.

If AB, BC, and CD are of the same diameter and ^ is equal to ,, then

and h

Problem 4. Pipes connecting three reservoirs. As in Fig. 102, let three pipes
AB, BC, and BD, connect three reservoirs A, C, D, the level of the water in each
of which remains constant.

Let t>j, v 2 , and t> 8 be the velocities in AB, BC, and BD respectively, Q
and Q 3 the quantities flowing along these pipes in cubic feet per sec., l lt J a , and'
the lengths of the pipes, and d^ , d a aud d s their diameters.

Fig. 105!.

Let t , 2 > and z 3 be the heights of the surfaces of the water in the reservoirs,
and z the height of the junction B above some datum.
Let h Q be the pressure head at B.

Assume all losses, other than those due to friction in the pipes, to be negligible.
The head lost due to friction for the pipe AB is

(1),

(2),

and for the pipe BC,

the upper or lower signs being taken, according as to whether the flow is from, or
towards, the reservoir C.

For the pipe BD the head lost is

(3).

Since the flow from A and C must equal the flow into D, or else the flow
from A must equal the quantity entering C and D, therefore,

Q 1 iQ 2 =Q 8 ,

or t^AtvV-tyV .................................... (4).

There are four equations, from which four unknowns may be found, if it is
further known which sign to take in equations (2) and (4). There are two cases to
consider.

156 HYDRAULICS

Case (a). Given the levels of the surfaces of the water in the reservoirs and
of the junction B, and the lengths and diameters of the pipes, to find the quantity
flowing along each of the pipes.

To solve this problem, it is first necessary to obtain by trial, whether water flows
to, or from, the reservoir C.

First assume there is no flow along the pipe BC, that is, the pressure head /? at
B is equal to z z - z .

Then from (1), substituting for v l its value

from which an approximate value for Q x can be found. By solving (3) in the same
way, an approximate value for Q 3 , is,

(6).

If Q 3 is found to be equal to Q a , the problem is solved ; but if Q 3 is greater than
Qu the assumed value for h 9 is too large, and if less, h is too small, for a diminu-
tion in the pressure head at B will clearly diminish Q 3 and increase Qj, and will
also cause flow to take place from the reservoir C along CB. Increasing the
pressure head at B will decrease Q 1} increase Q 3 , and cause flow from B to C.

This preliminary trial will settle the question of sign in equations (2) and (4)
and the four equations may be solved for the four unknowns, v lt v%, v 9 and h . It
is better, however, to proceed by "trial and error."

The first trial shows whether it is necessary to increase or diminish h 9 and new
values are, therefore, given to h until the calculated values of v lt v 2 and v, satisfy
equation (4).

Case (b). Given Qj, Q 2 , Q 3 , and the levels of the surfaces of the water in
the reservoirs and of the junction B, to find the diameters of the pipes.

In this case, equation (4) must be satisfied by the given data, and, therefore,
only three equations are given from which to calculate the four unknowns d lt
dg, d 3 and h . For a definite solution a fourth equation must consequently be
found, from some other condition. The further condition that may be taken is
that the cost of the pipe lines shall be a minimum.

The cost of pipes is very nearly proportional to the product of the length and
diameter, and if, therefore, Iid l + l 2 d 2 + l s d s is made a minimum, the cost of the
pipes will be as small as possible.

Differentiating, with respect to k Q , the condition for a minimum is, that

Substituting in (1), (2) and (3) the values for v lt v a and v a ,

differentiating and substituting in (7) t .

FLOW THROUGH PIPES 157

Putting the values of Q a , Q 2 , and Q 3 in (1), (2), (3), and (8), there are four
equations as before for four unknown quantities.

It will be better however to solve by approximation.

Give some arbitrary value to say d. 2 , and calculate /? from equation (2).

Then calculate d\ and d a by putting h n in (1) and (3), and substitute in
equation (8).

If this equation is satisfied the problem is solved, but if not, assume a second
value for d a and try again, and so on until such values of d l1 d! 2 , d s are obtained
that (8) is satisfied.

In this, as in simpler systems, the pressure at any point in the pipes ought not
to fall below the atmospheric pressure.

Flow through a pipe of constant diameter when the flow is diminishing at a
uniform rate. Let I be the length of the pipe and d its diameter.

Let h be the total loss of head in the pipe, the whole loss being assumed to be
by friction.

Let Q be the number of cubic feet per second that enters the pipe at a section A,
and Q! the number of cubic feet that passes the section B, I feet from A, the
quantity Q - Q x being taken from the pipe, by branches, at a uniform rate of

Q~Q* cubic feet per foot.

Then, if the pipe is assumed to be continued on, it is seen from Fig. 103, that
if the rate of discharge per foot length of the
pipe is kept constant, the ^ whole of Q will be
discharged in a length of pipe,

'(Q-Qi)'

The discharge past any section, x feet from
C, will be

*~ L ~ l *' Fig. 103.

The velocity at the section is

Assuming that in an element of length dx the loss of head due to friction is
and substituting for v x its value

the loss of head due to friction in the length I is

x n dx

t _[ L 7 /iQ\
"/M 7 VSBPj

/i
_ _y_

n +
If Qi is zero, I is equal to L, and

The result is simplified by taking for 9& the value

and assuming C constant.
Then

158 HYDRAULICS

Problem 5. *Pumping water tJirough long pipes. Kequired the diameter of a
long pipe to deliver a given quantity of water, against a given effective head, in
order that the charges on capital outlay and working expenses shall be a minimum.

Let I be the length of the pipe, d its diameter, and h feet the head against which
Q cubic feet of water per second is to be pumped.

Let the cost per horse-power of the pumping plant and its accommodation
be N, and the cost of a pipe of unit diameter n per foot length.

Let the cost of generating power be m per cent, of the capital outlay in the
pumping station, and the interest, depreciation, and cost of upkeep of the pumping
plant, taken together, be r per cent, of the capital outlay, and that of the pipe line
?*! per cent. ; r^ will be less than r. The horse-power required to lift the water
against a head h and to overcome the frictional resistance of the pipe is

60. Q. 62-4 ( 4t;*J
Hr - 33,000 < h +^

Let e be the ratio of the average effective horse-power to the total horse-povrer,
including the stand-by plant. The total horse-power of the plant is then
T 0-1186Q

The cost of the pumping plant is N times this quantity.
The total cost per year, P, of the station, is
m+r N.Q/

Assuming that the cost of the pipe line is proportional to the diameter and to
the length, the capital outlay for the pipe is, nld, and the cost of upkeep and

.. . .
interest is

is to be a minimum.

Differentiating with respect to d and equating to zero,

That is, d is independent of the length I and the head against which the water
is pumped.

Taking C as 80, e as 0-6 and v "*" ; as 50, then

WTj

If ( m+r ) N is 100,

3-68 x 50
80 x 80 x -6
= 0-603 VOT

d= -675^/01

Mr,

Problem 6. Pipe with a nozzle at the end. Suppose a pipe of length I and
diameter D has at one end a nozzle of diameter d, through which water is dis-
charged from a reservoir, the level of the water in which is h feet above the centre
of the nozzle.

Required the diameter of the nozzle so that the kinetic energy of the jet is
a maximum,

* See also example 61, page 177.

FLOW THROUGH PIPES 159

Let V be the velocity of the water in the pipe.

Then, since there is continuity of flow, v the velocity with which the water

V.D 2

leaves the nozzle is ^ .
a*

The head lost by friction in the pipe is

4/ V 2 l 4/r 2 Z . d*
2g.D~ 2#D 5 *

The kinetic energy of the jet per Ib. of flow as it leaves the nozzle is - .

Therefore *~* .............................. -f-*

from which by transposing and taking the square root,

/ frD.fc \i

The weight of water which flows per second =j d 2 . v . w where M> = the weight of

a cubic foot of water.

Therefore, the kinetic energy of the jet, is

This is a maximum when -rr=6.
M

Therefore

~ 4 * * 2 5*

from which D 5 + 4/Zd 4 = 12/Zd 4 ,

and D

If the nozzle is not circular but has an area a, then since in the circular nozzle
of the same area

jd2=a,

v u i 16a2

from which d*= p.

Therefor, D'=i^,

and

J-

V /t

By substituting the value of D 8 from (5) in (1) it is at once seen that, for
mazimum kinetic energy, the head lost in friction is

Problem 7. Taking the same data as in problem 6, to find the area of the
nozzle that the momentum of the issuing jet is a maximum.

The momentum of the quantity of water Q which flows per second, as it leaves

the nozzle, is W ' ^ V Ibs. feet. The momentum M is, therefore,
9

Substituting for * from equation (1), problem 6,

160 HYDRAULICS

Differentiating, and equating to zero,

/
4 /D5

If the nozzle has an area a, D 5 = -

and

a = -392

Substituting for D 5 in equation (1) it is seen that when the momentum is a
maximum half the head h is lost in friction.

Problem 6 has an important application, in determining the ratio of the size
of the supply pipe to the orifice supplying water to a Pelton Wheel, while problem 7
gives the ratio, in order that the pressure exerted by the jet on a fixed plane
perpendicular to the jet should be a maximum.

Problem 8. Loss of head due to friction in a pipe, the diameter of which varies
uniformly. Let the pipe be of length I and its diameter vary uniformly from d
to d l .

Suppose the sides of the pipe produced until they meet in P, Fig. 104.

The diameter of the pipe at any distance x from the small end is

The loss of head in a small element of length dx is - 8 , v being the velocity
when the diameter is d.

Fig. 104.

If Q is the flow in cubic ft. per second

Q 4 Q

The total loss of head h in a length I is
64Q 2 . dx

64 .

= /"*

16Q 2 . S 5

/ J_ __ 1_ \

\S 4 (8 + /) 4 /

Substituting the value of S from equation (1) the loss of head due to friction
can be determined.

Problem 9. Pipe line consisting of a number of pipes of different diameters. In
practice only short conical pipes are used, as for instance in the limbs of a Venturi
meter.

If it is desirable to diminish the diameter of a long pipe line, instead of using
a pipe the diameter of which varies uniformly with the length, the line is made up
of a number of parallel pipes of different diameters and lengths.

FLOW THROUGH PIPES

161

Let 7 lf Z 2 , 1 3 ... be the lengths and d lt d z ,d 3 ... the diameters respectively, of
the sections of the pipe.

The total loss of head due to friction, if C be assumed constant, is

*i**a+*-).

The diameter d of the pipe, which, for the same total length, would give the
same discharge for the same loss of head due to friction, can be found from ^he
equation

The length L of a pipe, of constant diameter D, which will give the same
discharge for the same loss of head by friction, is

Problem 10. Pipe acting as a siphon. It is sometimes necessary to take a
pipe line over some obstruction, such as a hill, which necessitates the pipe rising,
not only above the hydraulic gradient as in Fig. 87, but even above the original
level of the water in the reservoir from which the supply is derived.

Let it be supposed, as in Fig. 105, that water is to be delivered from the reservoir
B to the reservoir C through the pipe BAG, which at the point A rises fy feet above
the level of the surface of the water in the upper reservoir.

Fig. 105.

Let the difference in level of the surfaces of the water in the reservoirs
be fc 2 feet.

Let h a be the pressure head equivalent to the atmospheric pressure.

To start the flow in the pipe, it will be necessary to fill it by a pump or other
artificial means.

Let it be assumed that the flow is allowed to take place and is regulated so that
it is continuous, and the velocity v is as large as possible.

Then neglecting the velocity head and resistances other than that due to friction,

4/v 8 L /Zgdh*

*- V or "=v in?

L and d being the length and diameter of the pipe respectively.

The hydraulic gradient is practically the straight line DE.

Theoretically if AF is made greater than h at which is about 34 feet, the pressure
at A becomes negative and the flow will cease.

Practically AF cannot be made much greater than 25 feet.

To find the maximum velocity possible in the rising limb AB, so that the pressure
head at A shall just be zero.

Let v m be this velocity. Let the datum level be the surface of the water in C.

L. II,

162 HYDRAULICS

Then
But

Therefore

If the pressure head is not to be less than 10 feet of water,

If v m is less than v, the discharge of the siphon will he determined by this
limiting velocity, and it will be necessary to throttle the pipe at C by means of a
valve, so as to keep the limb AC full and to keep the " siphon " from being broken.

In designing such a siphon it is, therefore, necessary to determine whether the
flow through the pipe as a whole under a head h 2 is greater, or less than, the flow
in the rising limb under a head h a h^.

If AB is short, or h^ so small that v m is greater than v, the head absorbed by
friction in AB will be

2nd '

If the end C of the pipe is open to the atmosphere instead of being connected to
a reservoir, the total head available will be h s instead of 7^.

111. Velocity of flow in pipes.

The mean velocity of flow in pipes is generally about 3 feet
per second, but in pipes supplying water to hydraulic machines it
may be as high as 10 feet per second, and in short pipes much
higher velocities are allowed. If the velocity is high, the loss of
head due to friction in long pipes becomes excessive, and the risk
of broken pipes and valves through attempts to rapidly check
the flow, by the sudden closing of valves, or other causes, is
considerably increased. On the other hand, if the velocity is too
small, unless the water is -very free from suspended matter,
sediment* tends to collect at the lower parts of the pipe, and
further, at low velocities it is probable that fresh water sponges
and polyzoa will make their abode on the surface of the pipe, and
thus diminish its carrying capacity.

112. Transmission of power along pipes by hydraulic
pressure.

Power can be transmitted hydraulically through a considerable
distance, with very great efficiency, as at high pressures the per
centage loss due to friction is small.

Let water be delivered into a pipe of diameter d feet under a
head of H feet, or pressure of p Ibs. per sq. foot, for which the

equivalent head is H = - feet.

* An interesting example of this is quoted on p. 82 Trans. Am.S.C.E.

Vol. XLIV.

FLOW THROUGH PIPES 163

Let the velocity of flow be v feet per second, and the length of
the pipe L feet.

The head lost due to friction is

2g.d ................... >

and the energy per pound available at the end of the pipe is,
therefore,

The efficiency is

The fraction of the given energy lost is

m = H-

For a given pipe the efficiency increases as the velocity
diminishes.

If / and L are supposed to remain constant, the efficiency is

v*
constant if -TFT is constant, and since v is generally fixed from

other conditions it may be supposed constant, and the efficiency
then increases as the product dH. increases.

If W is the weight of water per second passing through the
pipe, the work put into the pipe is W . H foot Ibs. per second, the
available work per second at the end of the pipe is W (H - Ji) t and
the horse-power transmitted is

W.(H-fr) WH n ,
~" = 550" (1 " m) '

Since

Tj
*the horse-power = -- (H -

From (1) mH

therefore, v = 4*01 i\J ~j^ >

and the horse-power = 0'357 J j^ dK* (1 - m).
- * See example 60, page 177.

164 HYDRAULICS

If p is the pressure per sq. inch

and the horse-power = 1'24 ^ -^ <2*p* (1 - ra).

From this equation if m is given and L is known the diameter d
to transmit a given horse-power can be found, and if d is known the
longest length L that the loss shall not be greater than the given
fraction m can be found.

The cost of the pipe line before laying is proportional to its
weight, and the cost of laying approximately proportional to its
diameter.

If t is the thickness of the pipe in inches the weight per foot
length is 37*5^^ Ibs., approximately.

Assuming the thickness of the pipe to be proportional to the
pressure, i.e. to the head H,

= fcp=&H,

and the weight per foot may therefore be written

w - fad . H.

The initial cost of the pipe per foot will then be

C=feJWH = K.d.H,

and since the cost of laying is approximately proportional to d,
the total cost per foot is

p=K.d.n+K l d.

And since the horse-power transmitted is

HP = '357 ^/^ <#H* (1 - m),

for a given*horse-power and efficiency, the initial cost per horse-
power including laying will be a minimum when

0-357 d*H* (1 - m)

is a maximum.

In large works, docks, and goods yards, the hydraulic trans-
mission of power to cranes, capstans, riveters and other machines
is largely used.

A common pressure at which water is supplied from the pumps
is 700 to 750 Ibs. per sq. inch, but for special purposes, it is
sometimes as high as 3000 Ibs. per sq. inch. These high pressures
are, however, frequently obtained by using an intensifier (Ch. XI)
to raise the ordinary pressure of 700 Ibs. to the pressure required.
* See example 61, page 177.

FLOW THROUGH PIPES 165

The demand for hydraulic power for the working of lifts, etc.
has led to the laying down of a network of mains in several of the
large cities of Great Britain. In London a mean velocity of 4 feet
per second is allowed in the mains and the pressure is 750 Ibs.
per sq. inch. In later installations, pressures of 1100 Ibs. per
sq. inch are used.

113. The limiting diameter of cast-iron pipes.

The diameter d for a cast-iron pipe cannot be made very large
if the pressure is high.

If p is the safe internal pressure per sq. inch, and s the safe
stress per sq. inch of the metal, and TI and r a the internal and
external radii of the pipe,

For a pressure p = 1000 Ibs. per sq. inch, and a stress s of
3000 Ibs. per sq. inch, r a is 5'65 inches when n is 4 inches, or the
pipe requires to be 1'65 inches thick.

If, therefore, the internal diameter is greater than 8 inches, the
pipe becomes very thick indeed.

The largest cast-iron pipe used for this pressure is between
7" and 8" internal diameter.

Using a maximum velocity of 5 feet per second, and a pipe
7 J inches diameter, the maximum horse-power, neglecting friction,
that can be transmitted at 1000 Ibs. per sq. inch by one pipe is
4418x1000x5

-55Q-

= 400.

The following example shows that, if the pipe is 13,300 feet
long, 15 per cent, of the power is lost and the maximum power
that can be transmitted with this length of pipe is, therefore,
320 horse-power.

Steel mains are much more suitable for high pressures, as the
working stress may be as high as 7 tons per sq. inch. The greater
plasticity of the metal enables them to resist shock more readily
than cast-iron pipes and slightly higher velocities can be used.

A pipe 15 inches diameter and \ inch thick in which the
pressure is 1000 Ibs. per sq. inch, and the velocity 5 ft. per second,
is able to transmit 1600 horse-power.

Example. Power is transmitted along a cast-iron main 7$ inches diameter at
a pressure of 1000 Ibs. per sq. inch. The velocity of the water is 5 feet per second.

Find the maximum distance the power can be transmitted so that the efficiency
is not less thanS5/ .

* Swing's Strength of Materials.

166 HYDRAULICS

d = 0-625feet,

therefore h= 0-15x2300

= 345 feet.
Then 34y= 4x 0-0104 x 25 .j.

2g x 0-625
345 x 64-4 x 0-625

from which L =

0-0104x100
13,300 feet.

114. Pressures on pipe bends.

If a bent pipe contain a fluid at rest, the intensity of pressure
being the same in all directions,
the resultant force tending to move
the pipe in any direction will be
the pressure per unit area multiplied
by the projected area of the pipe
on a plane perpendicular to that
direction.

If one end of a right-angled
elbow, as in Pig 106 be bolted to
a pipe full of water at a pressure p

pounds per sq. inch by gauge, and on the other end of the elbow
is bolted a flat cover, the tension in the bolts at A will be the
same as in the bolts at B. The pressure on the cover B is clearly
'7854pd 2 , d being the diameter of the pipe in inches. If the elbow
be projected on to a vertical plane the projection of ACB is daefc,
the projection of DEF is dbcfe. The resultant pressure on the
elbow in the direction of the arrow is, therefore, p . abed = *7854pd 2 .

If the cover B is removed, and water flows through the pipe
with a velocity v feet per second, the horizontal momentum of the
water is destroyed and there is an additional force in the direction
of the arrow equal to '7854wcV/144<7.

When flow is taking place the vertical force tending to lift the
elbow or to shear the bolts at A is A

\ v\
If the elbow is less than a right \\

angle, as in Fig. 108, the total \VLl''

tension in the bolts at A is ^ "*"

T = p (daehgc - aefgc) + '^* " (1 - cos 0),
and since the area aehgcb is common to the two projected areas,

FLOW THROUGH PIPES

167

Consider now a pipe bent as shown in Fig. 109, the limbs AA
and FF being parallel, and the water being supposed at rest.

The total force acting in the direction AA is

P = p {dcghea - aefgcb + d'cg'tie'd - a'ef'g'c'b'},
which clearly is equal to 0.

If now instead of the fluid being at rest it has a uniform
velocity, the pressure must remain constant, and since there is no
change of velocity there is no change of momentum, and the re-
sultant pressure in the direction parallel to AA is still zero.

There is however a couple acting upon the bend tending to
rotate it in a clockwise direction.

Let p and q be the centres of gravity of the two areas daehgc
and aefgcb respectively, and m and n the centres of gravity of
d'de'h'g'c and aefgcb'.

Through these points there are parallel forces acting as shown
by the arrows, and the couple is

M = P' . mn P . pq.

The couple P . pq is also equal to the pressure on the semicircle
adc multiplied by the distance between the centres of gravity of
adc and efg, and the couple P' . mn is equal to the pressure on a'd'c
multiplied by the distance between the centres of gravity of a'd'c
and efg.

Then the resultant couple is the pressure on the semicircle efg
multiplied by the distance between the centres of gravity of efg
and efg.

If the axes of FF and AA are on the same straight line the
couple, as well as the force, becomes zero.

It can also be shown, by similar reasoning, that, as long as the
diameters at F and A are equal, the velocities at these sections
being therefore equal, and the two ends A and F are in the same
straight line, the force and the couple are both zero, whatever the
form of the pipe. If, therefore, as stated by Mr Froude, "the

168 HYDRAULICS

two ends of a tortuous pipe are in the same straight line, there is
no tendency for the pipe to move."

115. Pressure on a plate in a pipe filled with flowing water.

The pressure on a plate in a pipe filled with flowing water, with
its plane perpendicular to the direction of flow, on certain assump-
tions, can be determined.

Let PQ, Fig. 110, be a thin plate of area a and let the sectional
area of the pipe be A.

The stream as it passes the edge of p. a &

the plate will be contracted, and the
section of the stream on a plane gd will
be c(A-a), c being some coefficient of
contraction.

It has been shown on page 52 that
for a sharp-edged orifice the coefficient
of contraction is about 0'625, and when
part of the orifice is fitted with sides so that the contraction is
incomplete and the stream lines are in part directed perpendi-
cular to the orifice, the coefficient of contraction is larger.

If a coefficient in this case of 0'66 is assumed, it will probably
be not far from the truth.

Let Vi be the velocity through the section gd and V the mean
velocity in the pipe.

The loss of head due to sudden enlargement from gd to ef is

Let the pressures at the sections a&, gd, ef be p, pi and p 2 pounds
per square foot respectively.

Bernoulli's equations for the three sections are then,

to 20 w 2g

and ^ + -S = ^ + +( \/ ) (2) -

Adding (1) and (2)

(P ffA (Yi-V) a t
Vw w 2gr

The whole pressure on the plate in the direction of motion is then

2#
V 2 / A ^

FLOW THROUGH PIPES

169

If a = J A,

Y 2

P = 4iiva 7p nearly.

0-46. a. Y 2

116. Pressure on a cylinder.

If instead of a thin plate a cylinder be placed in the pipe,
with its axis coincident with the axis of the pipe, Fig. Ill, there
are two enlargements of the section of the water.

As the stream passes the up-stream edge of the cylinder, it
contracts to the section at cd, and then enlarges to the section
ef. It again enlarges at the down-stream end of the cylinder
from the section ef to the section gh.

<JU

zz^z^^

E~cuz--z. ~-r^r--.

>-=-

===r *

Fig. 111.

Let 0i, 2 , 03, 04 be the velocities at a&, cd, ef and
spectively, v 4 and 0! being equal.

Between cd and e/ there is a loss of head

(02 - 3 ) 2
2<7 '
and between e/ and gh there is a loss of

fo-0i) a
*]
The Bernouilli's equations for the sections are

re-

w w

Adding (2) and (3),

P? + ^L = ^ + !!L
w 2g w 2g

.ax

.(2),
(3).

LZB =

fa -

170 HYDRAULICS

If the coefficient of contraction at cd is c, the area at cd

A-a

m-i Vi.A. Vi A

Then V 2 = 7-1 r and v s =

/A \ BMJWA "3 A

c . (A a) A-a

Therefore

v wvS |Y a \ 2 / A A \ 2 )

(pi - p 4 ) = -g I ( v ^i^ J + ( (^ _ a ) ~ (A - a)/ J '

and the pressure on the cylinder is

=i-t .a.

EXAMPLES.

(1) A new cast-iron pipe is 2000 ft. long and 6 ins. diameter. It is to
discharge 50 c. ft. of water per minute. Find the loss of head in friction
and the virtual slope.

(2) What is the head lost per mile in a pipe 2 ft. diameter, discharging
6,000,000 gallons in 24 hours ? /= -007.

(3) A pipe is to supply 40,000 gallons in 24 hours. Head of water
above point of discharge =36 ft. Length of pipe = 2^ miles. Find its
diameter. Take C from Table XII.

(4) A pipe is 12 ins. in diameter and 3 miles in length. It connects
two reservoirs with a difference of level of 20 ft. Find the discharge per
minute in c. ft. Use Darcy's coefficient for corroded pipes.

(5) A water main has a virtual slope of 1 in 900 and discharges 35 c. ft.
per second. Find the diameter of the main. Coefficient / is 0'007.

(6) A pipe 12 ins. diameter is suddenly enlarged to 18 ins., and then to
24 ins. diameter. Each section of pipe is 100 feet long. Find the loss of
head in friction in each length, and the loss due to shock at each enlarge-
ment. The discharge is 10 c. ft. per second, and the coefficient of friction
/= -0106. Draw, to scale, the hydraulic gradient of the pipe.

(7) Find an expression for the relative discharge of a square, and a
circular pipe of the same section and slope.

(8) A pipe is 6 ins. diameter, and is laid for a quarter mile at a slope
of 1 in 50; for another quarter mile at a slope of 1 in 100; and for a third
quarter mile is level. The level of the water is 20 ft. above the inlet end,
and 9 ft. above the outlet end. Find the discharge (neglecting all losses
except skin friction) and draw the hydraulic gradient. Mark the pressure
in the pipe at each quarter mile.

(9) A pipe 2000 ft. long discharges Q c. ft. per second. Find by how
much the discharge would be increased if to the last 1000 ft. a second pipe
of the same size were laid alongside the first and the water allowed to flow
equally well along either pipe.

FLOW THROUGH PIPES 171

(10) A reservoir, the level of which is 50 ft. above datum, discharges
into a second reservoir 30 ft. above datum, through a 12 in. pipe, 5000 ft.
in length ; find the discharge. Also, taking the levels of the pipe at the
upper reservoir, and at each successive 1000 ft., to be 40, 25, 12, 12, 10, 15,
fb. above datum, write down the pressure at each of these points, and
sketch the position of the line of hydraulic gradient.

(11) It is required to draw off the water of a reservoir through a
pipe placed horizontally. Diameter of pipe 6 ins. Length 40 ft. Ef-
fective head 20 ft. Find the discharge per second.

(12) Given the data of Ex. 11 find the discharge, taking into account
the loss of head if the pipe is not bell-mouthed at either end.

(18) A pipe 4 ins. diameter and 100 ft. long discharges \ c. ft. per
second. Find the head expended in giving velocity of entry, in overcoming
mouthpiece resistance, and in friction.

(14) Kequired the diameter of a pipe having a fall of 10 ft. per mile,
and capable of delivering water at a velocity of 3 ft. per second when dirty.

(15) Taking the coefficient / as O'Ol (l + ^Y find how much water

would be discharged through a 12-inch pipe a mile long, connecting two
reservoirs with a difference of level of 20 feet.

(16) Water flows through a 12-inch pipe having a virtual slope of 3 feet
per 1000 feet at a velocity of 3 feet per second.

Find the friction per sq. ft. of surface of pipe in Ibs.

Also the value of / in the ordinary formula for flow in pipes.

(17) Find the relative discharge of a 6-inch main with a slope of
1 in 400, and a 4-inch main with a slope of 1 in 50.

(18) A 6-inch main 7 miles in length with a virtual slope of 1 in 100
is replaced by 4 miles of 6-inch main, and 3 miles of 4-inch main. Would
the discharge be altered, and, if so, by how much ?

(19) Find the velocity of flow in a water main 10 miles long, con-
necting two reservoirs with a difference of level of 200 feet. Diameter of
main 15 inches. Coefficient /=0'009.

(20) Find the discharge, if the pipe of the last question is replaced for
the first 5 miles by a pipe 20 inches diameter and the remainder by a pipe
12 inches diameter.

(21) Calculate the loss of head per mile in a 10-inch pipe (area of cross
section 0'54 sq. ft.) when the discharge is 2^ c. ft. per second.

(22) A pipe consists of J a mile of 10 inch, and a mile of 5 -inch pipe,
and conveys 2| c. ft. per second. State from the answer to the previous
question the loss of head in each section and sketch a hydraulic gradient.
The head at the outlet is 5 ft.

(23) What is the head lost in friction in a pipe 3 feet diameter
discharging 6,000,000 gallons in 12 hours?

(24) A pipe 2000 feet long and 8 inches diameter is to discharge 85 c. ft.
per minute. What must be the head of water ?

172 HYDRAULICS

(25) A pipe 6 inches diameter, 50 feet long, is connected to the bottom
of a tank 50 feet long by 40 feet wide. The original head over the open
end of the pipe is 15 feet. Find the time of emptying the tank, assuming
the entrance to the pipe is sharp-edged.

If ft = the head over the exit of the pipe at any moment,

v^ -5v z 4fv*5W
"20 + 20 + 20x0-5'

from which,

In time dt, the discharge is

28-27

v T44

In time dt the surface falls an amount dh.

Integrating,

_2000 (1-5 + 400/) / _ 79000 (1-5 +400/)

I -- - tu \/ J.O - 7- - - SGCS-

0-196 \/20 A/20

(26) The internal diameter of the tubes of a condenser is 0*654 inches.
The tubes are 7 feet long and the number of tubes is 400. The number of
gallons per minute flowing through the condenser is 400. Find the loss of
head due to friction as the water flows through the tubes. /=0'006.

(27) Assuming fluid friction to vary as the square of the velocity, find
an expression for the work done in rotating a disc of diameter D at an
angular velocity a in water.

(28) What horse-power can be conveyed through a 6-in. main if the
working pressure of the water supplied from the hydraulic power station is
700 Ibs. per sq. in.? Assume that the velocity of the water is limited to
3 ft. per second.

(29) Eighty-two horse-power is to be transmitted by hydraulic pressure
a distance of a mile. Find the diameter of pipe and pressure required for
an efficiency of '9 when the velocity is 5 ft. per sec.

The frictional loss is given by equation

Mi.*.

20 d

(30) Find the inclination necessary to produce a velocity of 4| feet per
second in a steel water main 31 inches diameter, when running full and
discharging with free outlet, using the formula

. -0005 v 1 " 94
dn*

(31) The following values of the slope i and the velocity v were
determined from an experiment on flow in a pipe '1296 ft. diam.

i -00022 -00182 -00650 -02389 -04348 -12315 -22408
v -205 -606 1-252 2-585 3'593 6-310 8-521

FLOW THROUGH PIPES 173

Determine k and n in the formula

i=kv n .

Also determine values of C for this pipe for velocities of *5, 1, 8, 5 and
7 feet per sec.

(32) The total length of the Coolgardie steel aqueduct is 307 miles
and the diameter 30 inches. The discharge per day may be 5,600,000
gallons. The water is lifted a total height of 1499 feet.

Find (a) the head lost due to friction,

(6) the total work done per minute in raising the water.

(33) A pipe 2 feet diameter and 500 feet long without bends furnishes
water to a turbine. The turbine works under a head of 25 feet and uses
10 c. ft. of water per second. What percentage of work of the fall is lost
in friction in the pipe ?

Coefficient /= "007 ( 1 +

(34) Eight thousand gallons an hour have to be discharged through
each of six nozzles, and the jet has to reach a height of 80 ft.

If the water supply is 1 miles away, at what elevation above the
nozzles would you place the required reservoir, and what would you
make the diameter of the supply main ?

Give the dimensions of the reservoir you would provide to keep a
constant supply for six hours. Lond. Un. 1903.

(35) The pipes laid to connect the Vyrnwy dam with Liverpool are
42 inches diameter. How much water will such a pipe supply in gallons
per diem if the slope of the pipe is 4^ feet per mile ?

At one point on the line of pipes the gradient is 6| feet per mile, and the
pipe diameter is reduced to 39 inches ; is this a reasonable reduction in the
dimension of the cross section ? Lond. Un. 1905.

(36) Water under a head of 60 feet is discharged through a pipe
6 inches diameter and 150 feet long, and then through a nozzle the area of
which is one-tenth the area of the pipe. Neglecting all losses except friction,
find the velocity with which the water leaves the nozzle.

(37) Two rectangular tanks each 50 feet long and 50 feet broad are
connected by a horizontal pipe 4 inches diameter, 1000 feet long. The
head over the centre of the pipe at one tank is 12 feet, and over the other
4 feet when flow commences.

Determine the time taken for the water in the two tanks to come to the
same level. Assume the coefficient C to be constant and equal to 90.

(38) Two reservoirs are connected by a pipe 1 mile long and 10"
diameter; the difference in the water surface levels being 25 ft.

Determine the flow through the pipe in gallons per hour and find by
how much the discharge would be increased if for the last 2000 ft. a second
pipe of 10" diameter is laid alongside the first. Lond. Un. 1905.

(39) A pipe 18" diameter leads from a reservoir, 300 ft. above the
datum, and is continued for a length of 5000 ft. at the datum, the length
being 15,000 ft. For the last 5000 ft. of its length water is drawn off by

174 HYDRAULICS

service pipes at the rate of 10 c. ft. per min. per 500 ft. uniformly. Find
the pressure at the end of the pipe. Lond. Un. 1906.

(40) 350 horse-power is to be transmitted by hydraulic pressure a
distance of 1^ miles.

Find the number of 6 ins. diameter pipes and the pressure required for
an efficiency of 92 per cent. /='01. Take v as 3 ft. per sec.

(41) Find the loss of head due to friction in a water main L feet long,
which receives Q cubic feet per second at the inlet end and delivers

=- cubic feet to branch mains for each foot of its length.

What is the form of the hydraulic gradient ?

(42) A reservoir A supplies water to two other reservoirs B and C.
The difference of level between the surfaces of A and B is 75 feet, and
between A and C 97*5 feet. A common 8-inch cast-iron main supplies for
the first 850 feet to a point D. A 6-inch main of length 1400 feet is then
carried on in the same straight line to B, and a 5 -inch main of length
630 feet goes to 0. The entrance to the 8-inch main is bell-mouthed, and
losses at pipe exits to the reservoirs and at the junction may be neglected.
Find the quantity discharged per minute into the reservoirs B and C.
Take the coefficient of friction (/) as '01. Lond. Un. 1907.

(43) Describe a method of finding the "loss of head" in a pipe due to
the hydraulic resistances, and state how you would proceed to find the
loss as a function of the velocity.

(44) A pipe, I feet long and D feet in diameter, leads water from a
tank to a nozzle whose diameter is d, and whose centre is h feet below
the level of water in the tank. The jet impinges on a fixed plane
surface. Assuming that the loss of head due to hydraulic resistance is
given by

show that the pressure on the surface is a maximum when

(45) Find the flow through a sewer consisting of a cast-iron pipe
12 inches diameter, and having a fall of 3 feet per mile, when discharging
full bore. c=100.

(46) A pipe 9 inches diameter and one mile long slopes for the first
half mile at 1 in 200 and for the second half mile at 1 in 100. The pressure
head at the higher end is found to be 40 feet of water and at the lower
20 feet.

Find the velocity and flow through the pipe.

Draw the hydraulic gradient and find the pressure in feet at 500 feet
and 1000 feet from the higher end.

(47) A town of 250,000 inhabitants is to be supplied with water. Half
the daily supply of 32 gallons per head is to be delivered in 8 hours.

The service reservoir is two miles from the town, and a fall of 10 feet
per mile can be allowed in the pipe.

What must be the size of the pipe ? = 90.

FLOW THROUGH PIPES 175

(48) A water pipe is to be laid in a street 800 yards long with houses
on both sides of the street of 24 feet frontage. The average number of
inhabitants of each house is 6, and the average consumption of water for
each person is 30 gallons in 8 hrs. On the assumption that the pipe is laid
in four equal lengths of 200 yards and has a uniform slope of jfoj, and that
the whole of the water flows through the first length, three-fourths through
the second, one half through the third and one quarter through the fourth,
and that the value of C is 90 for the whole pipe, find the diameters of the
four parts of the pipe.

(49) A pipe 3 miles long has a uniform slope of 20 feet per mile, and is
18 inches diameter for the first mile, 30 inches for the second and 21
inches for the third. The pressure heads at the higher and lower ends of
the pipe are 100 feet and 40 feet respectively. Find the discharge through
the pipe and determine the pressure heads at the commencement of the
30 inches diameter pipe, and also of the 21 inches diameter pipe.

(50) The difference of level of two reservoirs ten miles apart is 80 feet.
A pipe is required to connect them and to convey 45,000 gallons of water
per hour from the higher to the lower reservoir.

Find the necessary diameter of the pipe, and sketch the hydraulic
gradient, assuming /=0'01.

The middle part of the pipe is 120 feet below the surface of the upper
reservoir. Calculate the pressure head in the pipe at a point midway
between the two reservoirs.

(51) Some hydraulic machines are served with water under pressure
by a pipe 1000 feet long, the pressure at the machines being 600 Ibs. per
square inch. The horse-power developed by the machine is 300 and the
friction horse -power in the pipes 120. Find the necessary diameter of the

I v 2

pipe, taking the loss of head in feet as 0*03 -5 x ^- and "43 Ib. per square

a zg

inch as equivalent to 1 foot head. Also determine the pressure at which
the water is delivered by the pump.

What is the maximum horse -power at which it would be possible to
work the machines, the pump pressure remaining the same ? Lond. Un.
1906.

(52) Discuss Reynolds' work on the critical velocity and on a general
law of resistance, describing the experimental apparatus, and showing the
connection with the experiments of Poiseuille and D'Arcy. Lond. Un.
1906.

(53) In a condenser, the water enters through a pipe (section A) at the
bottom of the lower water head, passes through the lower nest of tubes,
then through the upper nest of tubes into the upper water head (section B).
The sectional areas at sections A and B are 0'196 and 0'95 sq. ft. respec-
tively ; the total sectional area of flow of the tubes forming the lower nest
is 0-814 sq. ft., and of the upper nest 0'75 sq. ft., the number of tubes being
respectively 353 and 326. The length of all the tubes is 6 feet 2 inches.
When the volume of the circulating water was 1-21 c. ft. per sec., the
observed difference of pressure head (by gauges) at A and B was 6'5 feet.
Find the total actual head necessary to overcome frictional resistance, and

176 HYDRAULICS

the coefficient of hydraulic resistance referred to A. If the coefficient of
friction (4/) for the tubes is taken to be '015, find the coefficient of hydraulic
resistance for the tubes alone, and compare with the actual experiment.
Lond. Un. 1906. (C r = head lost divided by vel head at A.)

(54) An open stream, which is discharging 20 c. ft. of water per
second is passed under a road by a siphon of smooth stoneware pipe, the
section of the siphon being cylindrical, and 2 feet in diameter. When the
stream enters this siphon, the siphon descends vertically 12 feet, it
then has a horizontal length of 100 feet, and again rises 12 feet. If all the
bends are sharp right-angled bends, what is the total loss of head in the
tunnel due to the bends and to the friction ? C = 117. Lond. Un. 1907.

(55) It has been shown on page 159 that when the kinetic energy of a
jet issuing from a nozzle on a long pipe line is a maximum,

Hence find the minimum diameter of a pipe that will supply a Pelton
Wheel of 70 per cent, efficiency and 500 brake horse-power, the available
head being 600 feet and the length of pipe 3 miles.

(56) A fire engine supplies water at a pressure of 40 Ibs. per square
inch by gauge, and at a velocity of 6 feet per second into a pipe 8 inches
diameter. The pipe is led a distance of 100 feet to a nozzle 25 feet above
the pump. If the coefficient/ (of friction) in the pipe be '01, and the actual
lift of the jet is f of that due to the velocity of efflux, find the actual height
to which the jet will rise, and the diameter of the nozzle to satisfy the
conditions of the problem.

(57) Obtain expressions (a) for the head lost by friction (expressed in
feet of gas) in a main of given diameter, when the main is horizontal, and
when the variations of pressure are not great enough to cause any important
change of volume, and (b) for the discharge in cubic feet per second.

Apply your results to the following example:

The main is 16 inches diameter, the length of the main is 300 yards,
the density of the gas is 0'56 (that of air=l), and the difference of pressure
at the two ends of the pipe is inch of water ; find :
(a) The head lost in feet of gas.
(fc) The discharge of gas per hour in cubic feet.

Weight of 1 cubic foot of air=0'08 lb.; weight of 1 cubic foot of water
62-4 Ibs. ; coefficient / (of friction) for the gas against the walls of the pipe
0-005. Lond. Un. 1905.

(See page 118 ; substitute for w the weight of cubic foot of gas.)

(58) Three reservoirs A, B and are connected by a pipe leading
from each to a junction box P situated 450' above datum.

The lengths of the pipes are respectively 10,000', 5000' and 6000' and the
levels of the still water surface in A, B and are 800', 600' and 200' above
datum.

Calculate the magnitude and indicate the direction of mean velocity in
each pipe, taking v = WQ\ / mi t the pipes being all the same diameter,
namely 15". Lond. Un. 1905.

FLOW THROUGH PIPES 177

(59) A pipe 3' 6" diameter bends through 45 degrees on a radius of
25 feet. Determine the displacing force in the direction of the radial line
bisecting the angle between the two limbs of the pipe, when the head of
water in the pipe is 250 feet.

Show also that, if a uniformly distributed pressure be applied in the
plane of the centre lines of the pipe, normally to the pipe on its outer
surface, and of intensity

49ftd 2
R+l-7*
per unit length, the bend is in equilibrium.

E= radius of bend in feet. d = diameter of pipe.
h = head of water in the pipe.

(60) Show that the energy transmitted by a long pipe is a maximum
when one-third of the energy put into the pipe is lost in friction.

The energy transmitted along the pipe per second is

7T 7T 4 "fl) ^Z

p being the pressure per sq. foot at the inlet end of pipe.
Differentiating and equating to zero

dv
or, head lost by friction = J .

(61) For a given supply of water delivered to a pipe at a given
pressure, the cost of upkeep of the pipe line may be considered as made up
of the capital charges on initial cost, plus repairs, plus the cost of energy
lost in the pipe line. The repairs will be practically proportional to the
original cost, i.e. to the capital charges. The original cost of the pipe line
may be assumed proportional to the diameter and to the length. The
annual capital charges P are, therefore, proportional to Id, or

P=mld.

If W is the weight of water pumped per annum, the energy lost per
year is proportional to

20. d"

or, since v is proportional to W divided by the area of the pipe, the total
annual cost PI may be written,

For P! to be a minimum, - should be zero.

Therefore -=ml-5m 1 - = 0,

That is, the annual cost due to charges and repairs should be equal to
5 times the cost due to loss of energy.

If the cost of pipes is assumed proportional to d 2 , P x is a minimum
when the annual cost is \ of the cost of the energy lost.

L. H. 12

CHAPTER VI.

FLOW IN OPEN CHANNELS.

117. Variety of the forms of channels.

The study of the flow of water in open channels is much irore
complicated than in the case of closed pipes, because of tne
infinite variety of the forms of the channels and of the different
degrees of roughness of the wetted surfaces, varying, as they do,
from channels lined with smooth boards or cement, to the irregular
beds of rivers and the rough, pebble or rock strewn, mountain
stream.

Attempts have been made to find formulae which are applicable
to any one of these very variable conditions, but as in the case of
pipes, the logarithmic formulae vary with the roughness of the
pipe, so in this case the formulae for smooth regular shaped channels
cannot with any degree of assurance be applied to the calculation
of the flow in the irregular natural streams.

118. Steady motion in uniform channels.

The experimental study of the distribution of velocities of
water flowing in open channels reveals the fact that, as in the
case of pipes, the particles of water at different points in a cross
section of the stream may have very different velocities, and the
direction of flow is not always actually in the direction of the flow
of the stream.

The particles of water have a sinuous motion, and at any point
it is probable that the condition of flow is continually changing.
In a channel of uniform section and slope, and in which the total
flow remains constant for an appreciable time, since the same
quantity of water passes each section, the mean velocity v in the
direction of the stream is constant, and is the same for all the
sections, and is simply equal to the discharge divided by the area
of the cross section. This mean velocity is purely an artificial
quantity, and does not represent, either in direction or magnitude,
the velocity of the particles of water as they pass the section.

FLOW IN OPEN CHANNELS

179

Experiments with current meters, to determine the distribution
of velocity in channels, show, however, that at any point in the
cross section, the component of velocity in a direction parallel to
the direction of flow remains practically constant. The considera-
tion of the motion is consequently simplified by assuming that
the water moves in parallel fillets or stream lines, the velocities in
which are different, but the velocity in each stream line remains
constant. This is the assumption that is made in investigating
so-called rational formulae for the velocity of flow in channels,
but it should not be overlooked that the actual motion may be
much more complicated.

119. Formula for the flow when the motion is uniform
in a channel of uniform section and slope.

On this assumption, the conditions of flow at similarly situated
points C and D in any two cross sections AA and BB, Figs. 112
and 113, of a channel of uniform slope and section are exactly the
same ; the velocities are equal, and since C and D are at the same
distance below the free surface the pressures are also equal. For
the filament CD, therefore,

PC + W 5 = PD + V
w 2g w 2g*

and therefore, since the same is true for any other filament,

t w 2g
is constant for the two sections.

Fig. 112.

Let v be the mean velocity of the stream, i the fall per foot
length of the surface of the water, or the slope, dl the length
between AA and BB, to the cross sectional area EFGrH of the
stream, P the wetted perimeter, i.e. the length EF + FGr + GrH,
and w the weight of a cubic foot of water.

Let p = m be called the hydraulic mean depth.

Let dz be the fall of the surface between A A and BB. Since
the slope is small dz = i.dl.

12 2

180 HYDRAULICS

If Q cubic feet per second fall from AA to BB, the work done
upon it by gravity will be :

Then, since 3 + -

\w 2g

is constant for the two sections, the work done by gravity must
be equal to the work done by the frictional and other resistances
opposing the motion of the water.

As remarked above, all the filaments have not the same velocity,
so that there is relative motion between consecutive filaments,
and since water is not a perfect fluid some portion of the work
done by gravity is utilised in overcoming the friction due to this
relative motion. Energy is also lost, due to the cross currents or
eddy motions, which are neglected in assuming stream line flow,
and some resistance is also offered to the flow by the air on the
surface of the water.

The principal cause of loss is, however, the frictional resistance
of the sides of the channel, and it is assumed that the whole of
work done by gravity is utilised in overcoming this resistance.

Let F . v be the work done per unit area of the sides of the
channel, v being the mean velocity of flow. F is often called the
frictional resistance per unit area, but this assumes that the relative
velocity of the water and the sides of the channel is equal to the
mean velocity, which is not correct.

The area of the surface of the channel between AA and BB
isP.8Z.

Then, wwidl = J?vPdl,

CO . F

therefore P l== w*

or vni = .

F is found by experiment to be a function of the velocity and
also of the hydraulic mean depth, and may be written

b being a numerical coefficient.

Since for water w is constant may be replaced by Tc and

therefore, mi = Jc.f (v) f (m) .

The form of f(v) f(m) must be determined by experiment.

120. Formula of Chezy.

The first attempts to determine the flow of water in channels

FLOW IN OPEN CHANNELS 181

with precision were probably those of Chezy made on an earthen
canal, at Coupalet in 1775, from which he concluded that

and therefore mi = kv~ (1).

Writing C for 4=

v = C ,Jmij

which is known as the Chezy formula, and has already been given
in the chapter on pipes.

121. Formulae of Prony and Eytelwein.

Prony adopted the same formula for channels and for pipes, and
assumed that F was a function of v and also of v a , and therefore,

mi = av + bv*.

By an examination of the experiments of Chezy and those of
Du Buat* made in 1782 on wooden channels, 20 inches wide and
less than 1 foot deep, and others on the Jard canal and the river
Hayne, Prony gave to a and b the values

a = '000044,
b = '000094.
This formula may be written

mi=(-~ ) + b)v\
1

or v =

V v

The coefficient C of the Chezy formula is then, according to Prony,
a function of the velocity v.

If the first term containing v be neglected, the formula is the
same as that of Chezy, or

v = 103 *Jmi.

Eytelwein by a re-examination of the same experiments
together with others on the flow in the rivers Rhine t and Weser +,
gave values to a and b of

a = '000024,

6 = '00011 14.

Neglecting the term containing a,
v = 95 \lrni.

* Principes d'hydraulique. See also pages 231 233.

f Experiments by Funk, 1803-6.

$ Experiments by Brauings, 1790-92.

182 HYDRAULICS

As in the case of pipes, Prony and Eytelwein incorrectly
assumed that the constants a and 6 were independent of the
nature of the bed of the channel.

122. Formula of Darcy and Bazin.

After completing his classical experiments on flow in pipes
M. Darcy commenced a series of experiments upon open channels
afterwards completed by M. Bazin to determine, how the
frictional resistances varied with the material with which the
channels were lined and also with the form of the channel.

Experimental channels of semicircular and rectangular section
were constructed at Dijon, and lined with different materials.
Experiments were also made upon the flow in small earthen
channels (branches of the Burgoyne canal), earthen channels lined
with stones, and similar channels the beds of which were covered
with mud and aquatic herbs. The results of these experiments,
published in 1858 in the monumental work, Recherches Hydrau-
liqueSj very clearly demonstrated the inaccuracy of the assump-
tions of the old writers, that the frictional resistances were
independent of the nature of the wetted surface.

From the results of these experiments M. Bazin proposed for
the coefficient &, section 120, the form used by Darcy for pipes,

*=(+),

\ m/'

a and being coefficients both of which depend upon the nature
of the lining of the channel.

Thus, mi = ( a. + j-u 3

\ mj

. 1

The coefficient in the Chezy formula is thus made to vary
with the hydraulic mean depth m, as well as with the roughness
of the surface.

It is convenient to write the coefficient k as

Taking the unit as 1 foot, Bazin's values for a and /?, and
values of k are shown in Table XVIII.

It will be seen that the influence of the second term increases
very considerably with the roughness of the surface.

123. Ganguillet and Kutter, from an examination of Bazin's

FLOW IN OPEN CHANNELS

183

experiments, together with some of their own, found that the
coefficient C in the Chezy formula could be written in the form

6 + vra/

in which a is a constant for all channels, and 6 is a coefficient of
roughness.

TABLE XVIII.

Showing the values of a, /?, and Jc in Bazin's formula for
channels.

Planed boards and smooth
cement

0000457

0000045

0000157 (l I' 98N 1

\ m J

Rough boards, bricks and
concrete

0000580

0000133

000058 (l+\

\ Tfl J

Ashlar masonry

0000730

00006

(QO\
1 + )
mj

Earth

0000854

00035

0000854 (l+^Y

Gravel (Ganguillet and
Kutter)

0001219

00070

0001219(l+5-I5)

\ /

The results of experiments by Humphreys and Abbott upon
the flow in the Mississippi* were, however, found to give results
inconsistent with this formula and also that of Bazin.

They then proposed to make the coefficient depend upon the
slope of the channel as well as upon the hydraulic mean depth.

From experiments which they conducted in Switzerland, upon
the flow in rough channels of considerable slope, and from an
examination of the experiments of Humphreys and Abbott on the
flow in the Mississippi, in which the slope is very small, and
a large number of experiments on channels of intermediate slopes,
they gave to the coefficient C, the unit being 1 foot, the value

0'00281

1+41-6

00281 \ n '

in which n is a coefficient of roughness of the channel and has the
values given in Tables XIX and XIX A.

* Report on the Hydraulics of the Mississippi River, 1861 j Flow of water in
fivers and canals, Trautwine and Bering, 1893,

184 HYDRAULICS

TABLE XIX.

Showing values of n in the formula of Ganguillet and Kutter.
Channel H

Very smooth, cement and planed boards 009 to '01

Smooth, boards, bricks, concrete ... ... ... ... '012 to '013

Smooth, covered with slime or tuberculated -015

Hough ashlar or rubble masonry '017 to -019

Very firm gravel or pitched with stones -02

Earth, in ordinary condition free from stones and weeds ... -025

Earth, not free from stones and weeds -030

Gravel in bad condition '035 to '040

Torrential streams with rough stony beds -05

TABLE XIX A.

Showing values of n in the formula of Ganguillet and Kutter,
determined from recent experiments.

Rectangular wooden flume, very smooth -0098

Wood pipe 6 ft. diameter -0132

Brick, washed with cement, basket shaped sewer, 6'x6'8". nearly

. new -0130

Brick, washed with cement, basket shaped sewer, 6'x6'8", one

year old -0148

Brick, washed with cement, basket shaped sewer, 6'x6'8", four

years old -0152

Brick, washed with cement, circular sewer, 9 ft. diameter, nearly

new -0116

Brick, washed with cement, circular sewer, 9 ft. diameter, four

years old -0133

Old Croton aqueduct, lined with brick -015

New Croton aqueduct* '012

Sudbury aqueduct ... ... ... ... ... ... ... -01

Glasgow aqueduct, lined with cement -0124

Steel pipe, wetted, clean, 1897 (mean) -0144

Steel pipe, 1899 (mean) -0155

This formula has found favour with English, American and
German engineers, but French writers favour the simpler formula
of Bazin.

It is a peculiarity of the formula, that when m equals unity

then C = - and is independent of the slope ; and also when m is

large, C increases as the slope decreases.

It is also of importance to notice that later experiments upon
the Mississippi by a special commission, and others on the flow of
the Irrawaddi and various European rivers, are inconsistent with

New York Aqueduct Commission,

FLOW IN OPEN CHANNELS 185

the early experiments of Humphreys and Abbott, to which
Ganguillet and Kutter attached very considerable importance in
framing their formula, and the later experiments show, as described
later, that the experimental determination of the flow in, and the
slope of, large natural streams is beset with such great difficulties,
that any formula deduced for channels of uniform section and
slope cannot with confidence be applied to natural streams, and
vice versa.

The application of this formula to the calculation of uniform
channels gives, however, excellent results, and providing the value
of n is known, it can be used with confidence.

It is, however, very cumbersome, and does not appear to give
results more accurate than a new and simpler formula suggested
recently by Bazin and which is given in the next section.

124. M. Bazin's later formula for the flow in channels.

M. Bazin has recently (Annales des Pouts et Chaussees, 1897,
Vol. IV. p. 20), made a careful examination of practically all the
available experiments upon channels, and has proposed for the
coefficient C in the Chezy formula a form originally proposed by
Ganguillet and Kutter, which he writes

in which a is constant for all channels and {! is a coefficient of
roughness of the channel.

Taking 1 metre as the unit a = '0115, and writing y for ,

c=-2Z_ a)j

or when the unit is one foot,

(2),

the value of y in (2) being 1'Slly, in formula (1).

The values of y as found by Bazin for various kinds of channels
are shown in Table XX, and in Table XXI are shown values of

186 HYDRAULICS

C, to the nearest whole number, as deduced from Bazin's
coefficients for values of m from '2 to 50.

For the channels in the first four columns only a very few
experimental values for C have been obtained for values of m
greater than 3, and none for m greater than 7'3. For the earth
channels, experimental values for C are wanting for small values
of m, so that the values as given in the table when m is greater
than 7*3 for the first four columns, and those for the first three
columns for m less than 1, are obtained on the assumption, that
Bazin's formula is true for all values of m within the limits of the
table.

TABLE XX.

Values of y in the formula,

c 157 - 5

unit metre unit foot

Very smooth surfaces of cement and planed boards ... -06 -1085

Smooth surfaces of boards, bricks, concrete '16 *29

Ashlar or rubble masonry '46 '83

Earthen channels, very regular or pitched with stones,

tunnels and canals in rock *85 1/54

Earthen channels in ordinary condition 1/30 2'35

Earthern channels presenting an exceptional resistance,
the wetted surface being covered with detritus,

stones or weed, or very irregular rocky surface 1'7 3'17

125. Glazed earthenware pipes.

Vellut* from experiments on the flow in earthenware pipes has
given to C the value

in which
or

This gives values of C, not very different from those given by
Bazin's formula when y is 0'29.

In Table XXI, column 2, glazed earthenware pipes have been
included with the linings given by Bazin.

FLOW IN OPEN CHANNELS

187

TABLE XXI.

Values of in the formula v = C *J- mi calculated from Bazin's
formula, the unit of length being 1 foot,

157-5

1 +

Channels

Smooth

Earth canals

Hydraulic
mean
depth

Very smooth
cement and
planed
boards

boards, brick,
concrete,
glazed
earthenware

Smooth
but dirty
brick,
concrete

Ashlar
masonry

in very good
condition,
and canals
pitched with

Earth canals
in ordinary
condition

Earth canals
exceptionally
rough

pipes

stones

7 = '1085

y = -29

7 = -50

7 = -83

-y = l-54

7 = 2-35

7=3-17

127

131

103

135

108

137

112

139

116

141

119

101

1-0

142

122

105

1-3

144

126

109

1-5

145

128

112

1-75

146

130

114

2-0

147

132

116

2-5

148

134

119

103

3'0

149

136

122

107

4-0

150

138

126

111

5-0

151

140

129

115

6-0

151

142

131

118

8-0

152

144

134

122

102

10-0

153

145

136

125

106

12-0

109

15-0

113

20-0

117

103

30-0

123

110

100

50-0

129

119

108

126. Bazin's method of determining a and J&.
The method used by Bazin to determine the values of a and /?
is of sufficient interest and importance to be considered in detail.

He first calculated values of -j= and

vra

from experimental

data, and plotted these values as shown in Fig. 114, -= as

abscissae, and - - as ordinates.
v

188

HYDRAULICS

As will be seen on reference to the figure, points have been
plotted for four classes of channels, and the points lie close to four
straight lines passing through a common point P on the axis
of y.

The equation to each of these lines is
y = a + fix,

FLOW IN OPEN CHANNELS 189

or - = a + T-- ,

v vm

ct being the intercept on the axis of y, or the ordinate when r= is

Jm
zero, and /?, which is variable, is the inclination of any one of

these lines to the axis of x : for when /= is zero, - - = a, and

vm v

transposing the equation,

\frni

which is clearly the tangent of the angle of inclination of the line
to the axis of x.

It should be noted, that since - = p , the ordinates give

actual experimental values of ~ , or by inverting the scale, values

of C. Two scales for ordinates are thus shown.

In addition to the points shown on the diagram, Fig. 114,
Bazin plotted the results of some hundreds of experiments for all
kinds of channels, and found that the points lay about a series of
lines, all passing through the point P, Fig. 114, for which a is '00635,

and the values of - , i.e. y, are as shown in Table XX.
Bazin therefore concluded, that for all channels

v vm

the value of ft depending upon the roughness of the channel.

For very smooth channels in cement and planed boards, Bazin
plotted a large number of points, not shown in Fig. 114, and the
line for which y = '109 passes very nearly through the centre of
the zone occupied by these points.

The line for which y is 0*29 coincides well with the mean of
the plotted points for smooth channels, but for some of the points
y may be as high as 0*4.

It is further of interest to notice, that where the surfaces and
sections of the channels are as nearly as possible of the same
character, as for instance in the Boston and New York aqueducts,
the values of the coefficient C differ by about 6 per cent., the
difference being probably due to the pointing of the sides and
arch of the New York aqueduct not being so carefully executed
as for the Boston aqueduct. By simply washing the walls of the
latter with cement, Fteley found that its discharge was increased
20 per cent.

190 HYDRAULICS

y is also greater for rectangular-shaped channels, or those
which approximate to the rectangular form, than for those of
circular form, as is seen by comparing the two channels in wood
W and P, and also the circular and basket-shaped sewers.

M. Bazin also found that y was slightly greater for a very
smooth rectangular channel lined with cement than for one of
semicircular section.

In the figure the author has also plotted the results of some
recent experiments, which show clearly the effect of slime and
tuberculations, in increasing the resistance of very smooth channels.
The value of y for the basket-shaped sewer lined with brick,
washed with cement, rising from '4 to '642 during 4 years' service.

127. Variations in the coefficient C.

For channels lined with rubble, or similar materials, some of
the experimental points give values of C differing very considerably
from those given by points on the line for which y is 0'83, Fig. 114,
but the values of C deduced from experiments on particular
channels show similar discrepancies among themselves.

On reference to Bazin's original paper it will be seen that, for
channels in earth, there is a still greater variation between the
experimental values of C, and those given by the formula, but the
experimental results in these cases, for any given channel, are
even more inconsistent amongst themselves.

An apparently more serious difficulty arises with respect to
Bazin's formula in that C cannot be greater than 157*5. The
maximum value of the hydraulic mean depth m recorded in
any series of experiments is 74*3, obtained by Humphreys and
Abbott from measurements of the Mississippi at Carroll ton in 1851.
Taking y as 2'35 the maximum value for C would then be 124.
Humphreys and Abbott deduced from their experiments values
of C as large as 254. If, therefore, the experiments are reliable
the formula of Bazin evidently gives inaccurate results for excep-
tional values of m.

The values of C obtained at Carrollton are, however, incon-
sistent with those obtained by the same workers at Yicksburg,
and they are not confirmed by later experiments carried out at
Carrollton by the Mississippi commission. Further the velocities
at Carrollton were obtained by double floats, and, according to
Gordon*, the apparent velocities determined by such floats should
be at least increased, when the depth of the water is large, by ten
per cent.

Bazin has applied this correction to the velocities obtained by

* Gordon, Proceedings Inst. Civil Eng., 1893.

FLOW IN OPEN CHANNELS 191

Humphreys and Abbott at Vicksburg and also to those obtained
by the Mississippi Commission at Carrollton, and shows, that the
maximum value for C is then, probably, only 122.

That the values of C as deduced from the early experiments on
the Mississippi are unreliable, is more than probable, since the
smallest slope, as measured, was only '0000034, which is less than
j inch per mile. It is almost impossible to believe that such small
differences of level could be measured with certainty, as the
smallest ripple would mean a very large percentage error, and
it is further probable that the local variations in level would be
greater than this measured difference for a mile length. Further,
assuming the slope is correct, it seems probable that the velocity
under such a fall would be less than some critical velocity similar
to that obtained in pipes, and that the velocity instead of being
proportional to the square root of the slope i, is proportional
to i. That either the measured slope was unreliable, or that the
velocity was less than the critical velocity, seems certain from the
fact, that experiments at other parts of the Mississippi, upon the
Irrawaddi by Gordon, and upon the large rivers of Europe, in no
case give values of C greater than 124.

The experimental evidence for these natural streams tends,
however, clearly to show, that the formulae, which can with
confidence be applied to the calculation of flow in channels of
definite form, cannot with assurance be used to determine the
discharge of rivers. The reason for this is not far to seek, as
the conditions obtaining in a river bed are generally very far
removed from those assumed, in obtaining the formula. The
assumption that the motion is uniform over a length sufficiently
great to be able to measure with precision the fall of the surface,
must be far from the truth in the case of rivers, as the irregu-
larities in the cross section must cause a corresponding variation
in the mean velocities in those sections.

In the derivation of the formula, frictional resistances only
are taken into account, whereas a considerable amount of the
work done on the falling water by gravity is probably dissipated
by eddy motions, set up as the stream encounters obstructions in
the bed of the river. These eddy motions must depend very
much on local circumstances and will be much more serious in
irregular channels and those strewn with weeds, stones or other
obstructions, than in the regular channels. Another and probably
more serious difficulty is the assumption that the slope is uniform
throughout the whole length over which it is measured, whereas
the slope between two cross sections may vary considerably
between bank and bank. It is also doubtful whether locally

192 HYDRAULICS

there is always equilibrium between the resisting and accelerating
forces. In those cases, therefore, in which the beds are rocky or
covered with weeds, or in which the stream has a very irregular
shape, the hypotheses of uniform motion, slope, and section, will
not even be approximately realised.

128. Logarithmic formula for the flow in channels.

In the formulae discussed, it has been assumed that the f rictional
resistance of the channel varies as the square of the velocity, and
in order to make the formulae fit the experiments, the coefficient C
has been made to vary with the velocity.

As early as 1816, Du Buat* pointed out, that the slope i
increased at a less rate than the square of the velocity, and
half a century later St Tenant proposed the formula

mi = '000401 lA

To determine the discharge of brick-lined sewers, Mr Santo
Crimp has suggested the formula

and experiments show that for sewers that have been in use some
time it gives good results. The formula may be written

. 0-00006*; 2 '

- .--_. _____ T

- 1 '^J.

m !34

An examination of the results of experiments, by logarithmic
plotting, shows that in any uniform channel the slope

. bo*
*=^>

k being a numerical coefficient which depends upon the roughness
of the surface of the channel, and n and p also vary with the
nature of the surface.

Therefore, in the formula,

From what follows it will be seen that n varies between 1*75
and 2'1, while p varies between 1 and 1'5.

Jcv n
Since m is constant, the formula i = ^ may be written i = fo n ,

&
b being equal to ^ .

Therefore log i = log b + n log v.

* Principes d'Hydr antique, Vol. r. p. 29, 1810.

FLOW IN OPEN OTTAXNELS

193

In Fig. 115 are shown plotted the logarithms of i and v
obtained from an experiment by Bazin on the flow in a semi-
circular cement-lined pipe. The points lie about a straight line,
the tangent of the inclination of which to the axis of v is 1'96
and the intercept on the axis of i through v = 1, or log v = 0, is
0000808.

Fig. 115. Logarithmic plottings of i and v to determine the index n in

the formula for channels, i = -.
in"

For this experimental channel, therefore,

i = '00008085 v.

In the same figure are shown the plottings of logi and logv for
the siphon-aqueduct* of St Elvo lined with brick and for which
m is 278 feet. In this case n is 2 and b is '000283. Therefore

i = -000283v 9 .

If, therefore, values of v and i are determined for a channel,
while m is kept constant, n can be found.

Annales des Fonts et Chaussees, Vol. iv. 1897.

L, H.

194 HYDRAULICS

To determine the ratio - . The formula,

m j
may be written in the form,

k\*

or log m = log (- J + - log v.

By determining experimentally m and v, while the slope i is
kept constant, and plotting log m as ordinates and log v as
abscissae, the plottings lie about a straight line, the tangent of the

inclination of which to the axis of v is equal to - . and the

P
intercept on the axis of m is equal to

In Fig. 116 are shown the logarithmic plottings of m and v for
a number of channels, of varying degrees of roughness.

The ratio - varies considerably, and for very regular channels

increases with the roughness of the channel, being about 1*40 for
very smooth channels, lined with pure cement, planed wood or
cement mixed with very fine sand, 1*54 for channels in unplaned
wood, and 1*635 for channels lined with hard brick, smooth
concrete, or brick washed with cement. For channels of greater

roughness, - is very variable and appears to become nearly equal
to or even less than its value for smooth channels. Only in one
case does the ratio - become equal to 2, and the values of m and

v for that case are of very doubtful accuracy.

As shown above, from experiments in which m is kept constant,

*??

n can be determined, and since by keeping i constant - can be

found, n and p can be deduced from two sets of experiments.

Unfortunately, there are wanting experiments in which m is
kept constant, so that, except for a very few cases, n cannot
directly be determined.

There is, however, a considerable amount of experimental data
for channels similarly lined, and of different slopes, but here

FLOW IN OPEN CHANNELS

195

Log.

Fig. 116. Logarithmic plottings of m and v to determine the

ratio - in the formula i= - .
p mP

TABLE XXII.
Particulars of channels, plottings for which are shown in Fig. 116.

Semicircular channel, very smooth, lined with wood
,, ,, cement mixed with

n
P
1-45

1-36

4.
5.
6,

Rectangular channel, very smooth, lined with cement
, wood, 1' 1" wide
smooth , ,, ,, slope -00208
, -0043

1-44
1-38
1-54
1-54

7.
8.
9.

> > > "01)49
'00824
New Croton aqueduct, smooth, lined with bricks (Report New York
Water Supply)

1-54
1-54

1-74

10.

11.

Glasgow aqueduct, smooth, lined with concrete (Proc. I. C. E. 1896)
Sudbury ,, ,, ,, brick well pointed (Tr. Am.
S.C.E. 1883)

1-635
1-635

12.

Boston sewer, circular, smooth, lined with brick washed with cement
(Tr. Am.S. C. E. 1901)

1-635

13.

15!
15a.
156.

Rectangular channel, smooth, lined with brick
> wood ... ... ...
,, ,, ,, ,, small pebbles
Rectangular sluice channel lined with hammered ashlar

1-635
1-655
1-49
1-36
1-36

16.

1-29

17.

Torlonia tunnel, rock, partly lined

1-49

18.

Ordinary channel lined with stones covered with mud and weeds ...

1-18
94

20.

River Weser

1-615

21.

1-65

22.

2*1

23.

Earth channel. Gros bois ... ...

1-49

24.

Cavour canal

25.

1-37

132

196

HYDRAULICS

again, as will appear in the context, a difficulty is encountered, as
even with similarly lined channels, the roughness is in no two
cases exactly the same, and as shown by the plottings in Fig. 116,
no two channels of any class give exactly the same values

for - , but for certain classes the ratio is fairly constant.

Taking, for example, the wooden channels of the group (Nos. 4

to 8), the values of - are all nearly equal to 1'54.

The plottings for these channels are again shown in Fig. 117.
The intercepts on the axis of m vary from 0'043 to 0'14.

1-0
09
08
07

Lea v

Fig. 117. Logarithmic plottings to determine the ratio - for smooth channels.
Let the intercepts on the axis of m be denoted by y t then,

FLOW IN OPEN CHANNELS

197

1 1

& p p

If k and p are constant for these channels, and log* and
log y are plotted as abscissae and ordinates, the plottings should lie
about a straight line, the tangent of the inclination of which to the

axis of i is - , and when log y = 0, or y is unity, the abscissa i = &,

i.e. the intercept on the axis of i is k.

In Fig. 118 are shown the plottings of log i and log y for these
channels, from which p=l'14 approximately, and k = '00023.

Therefore, n is approximately 1*76, and taking - as 1'54

00023u 176

-OU5

' '

\ s

)

tan/ (L

_za

y fa-

OOO23.

V2 -0005, -OO1 '002 -005 '01

Log. i/

Fig. 118. Logarithmic plottings to determine the value of p for smooth
channels, in the formula i = .

Since the ratio - is not exactly 1*54 for all these channels, the

values of n and p cannot be exactly correct for the four channels,
but, as will be seen on reference to Table XXIII, in which are
shown values of v as observed and as calculated by the formula,
the calculated and observed values of v agree very nearly.

198

HYDRAULICS

TABLE XXIII.

Values of v, for rectangular channels lined with wood, as
determined experimentally, and as calculated from the formula

; = '00023

ri4'

Slope '00208

Slope -0049

Slope -00824

v ob-

v calcu-

v ob-

v calcu-

v ob-

v calcu-

m in

served

lated

m in

served

lated

m in

served

lated

metres

per sec.

0-1381

0-962

0-972

0-1042

1-325

1-314

0882

1-594

1-589

1609

1-076

1-07

1224

1-479

1-459

1041

1-776

1-764

1832

1-152

1-165

1382

1-612

1-58

1197

1-902

1-932

1976

1-259

1-223

1535

1-711

1-690

1313

2-053

2-051

2146

1-324

1-290

1668

1-818

1-782

1420

2-186

2-158

2313

1-374

1-354

1789

1-898

1-858

1543

2-268

2-275

2441

1-440

1-402

1913

1-967

1-947

1649

2-357

2-377

2578

1-487

1-452

2018

2-045

2-014

1744

2-447

2-460

2681

1-552

1-49

2129

2-102

2-089

1842

2-518

2-553

2809

1-587

1-552

2215

2-179

2-143

1919

2-612

2-618

As a further example, which also shows how n and p increase
with the roughness of the channel, consider two channels built in
hammered ashlar, for which the logarithmic plottings of m and v

are shown in Fig. 116, Nos. 15 a and 15 &, and - is 1'36.

The slopes of these channels are '101 and '037. By plotting
log* and log y, p is found to be 1'43 and k '000149. So that for

these two channels

. '000149^ r98

m 1 ' 43

The calculated and observed velocities are shown in Table XXXI
and agree remarkably well.

Very smooth channels. The ratio - for the four very smooth

channels, shown in Fig. 116, varies between 1'36 and 1'45, the
average value being about 1*4. On plotting logy and log* the
points did not appear to lie about any particular line, so that p
could not be determined, and indicates that k is different for the
four channels. Trial values of n = 1*75 and p = 1'25 were taken, or

k.v

and values of k calculated for each channel.

FLOW IN OPEN CHANNELS 199

Velocities as determined experimentally and as calculated for
three of the channels are shown in Table XXIII from which it will
be seen that k varies from '00006516 for the channel lined with
pure cement, to '0001072 for the rectangular shaped section lined
with carefully planed boards.

It will be seen, that although the range of velocities is con-
siderable, there is a remarkable agreement between the calculated
and observed values of v, so that for very smooth channels the
values of n and p taken, can be used with considerable confidence.

Channels moderately smooth. The plottings of log m and logv
for channels lined with brick, concrete, and brick washed with
cement are shown in Fig. 116, Nos. 9 to 13.

It will be seen that the value of - is not so constant as for the

P
two classes previously considered, but the mean value is about

1'635, which is exactly the value of - for the Sudbury aqueduct.

For the New Croton aqueduct - is as high as 1'74, and, as shown
in Fig. 114, this aqueduct is a little rougher than the Sudbury.

The variable values of -- show that for any two of these

channels either n> or p, or both, are different. On plotting logi
and logi; as was done in Fig. 115, the points, as in the last case,
could not be said to lie about any particular straight line, and the
value of p is therefore uncertain. It was assumed to be 1'15, and

therefore, taking - as 1'635, n is 1*88.

Since no two channels have the same value for - , it is to be

P
expected that the coefficient Jc will not be constant.

In the Tables XXIV to XXXIII the values of v as observed
and as calculated from the formula

._/b r88
~ m ri >
and also the value of Jc are given.

It will be seen that Jc varies very considerably, but, for the
three large aqueducts which were built with care, it is fairly
constant.

The effect of the sides of the channel becoming dirty with
time, is very well seen in the case of the circular and basket-
shaped sewers. In the one case the value of k, during four years'
service, varied from '00006124 to '00007998 and in the other from
'00008405 to '0001096. It is further of interest to note, that when

200 HYDRAULICS

m and v are both unity and k is equal to '000067, the value of i is
the same as given by Bazin's formula, when 7 is '29, and when k is
'0001096, as in the case of the dirty basket-shaped sewer, the value
of y is '642, which agrees with that shown for this sewer on
Fig. 114

Channels in masonry. Hammered ashlar and rubble. Attention
has already been called, page 198, to the results given in
Table XXXI for the two channels lined with hammered ashlar.

The values of n and p for these two channels were determined
directly from the logarithmic plottings, but the data is insufficient
to give definite values, in general, to n, p, and k.

In addition to these two channels, the results for one of
Bazin's channels lined with small pebbles, and for other channels
lined with rubble masonry and large pebbles are given. The

ratio - is quoted at the head of the tables where possible. In the

other cases n and p were determined by trial.

The value of n, for these rough channels, approximates to 2,
and appears to have a mean value of about 1'96, while p varies
from 1'36 to 1'5.

Earthen channels. A. very large number of experiments have
been made on the flow in canals and rivers, but as it is generally

impracticable to keep either i or m constant, the ratio - can only

be determined in a very few cases, and in these, as will be seen
from the plottings in Fig. 116, the results are not satisfactory, and

appear to be unreliable, as - varies between '94 and 2*18. It seems

probable that p is between 1 and 1*5 and n from 1*96 to 2' 15.
Logarithmic formulae for various classes of channels.
Very smooth channels, lined with cement, or planed boards,

fl,V75

i = ('000065 to '00011) ^ .

Smooth channels, lined with brick well pointed, or concrete,
t = '000065 to '00011 ~^.

Channels lined with ashlar masonry, or small pebbles,

-w 1 ' 96
t = '00015^-4.

Channels lined with rubble masonry, large pebbles, rock, and
exceptionally smooth earth channels free from deposits,

t-m
t = '00023

FLOW IN OPEN CHANNELS 201

Earth channels,

k varies from '00033 to '00050 for channels in ordinary condition
and from '00050 to '00085 for channels of exceptional resistance.

129. Approximate formula for the flow in earth
channels.

The author has by trial found n and p for a number of
channels, and except for very rough channels, n is not very
different from 2, and p is nearly 1'5. The approximate formula

v = C v m% i y

may, therefore, be taken for earth channels, in which C is about
50 for channels in ordinary condition.

In Table XXXIII are shown values of v as observed and
calculated from this formula.

The hydraulic mean depth varies from '958 to 14*1 and for all
values between these external limits, the calculated velocities
agree with the observed, within 10 per cent., whereas the variation
of C in the ordinary Chezy formula is from 40 to 103, and
according to Bazin's formula, C would vary from about 60 to 115.
With this formula velocities can be readily calculated with the
ordinary slide rule.

TABLE XXIV.

Very smooth channels.
Planed wood, rectangular, 1'575 wide.

i = -0001072 -^
w 1 - 5 '

log & = 4'0300.

v ft. per sec. v ft. per sec.
m feet observed calculated

2372 3'55 3'57

2811 4-00 4-03

3044 4-20 4-26

3468 4-67 4'68

3717 4-94 4-94

3930 5-11 5-12

4124 5-26 5-30

4311 5-49 5-47

202 HYDRAULICS

TABLE XXIV (continued).

Pure cement, semicircular.

. _ fa; 1 " 75
~m r23 '

-w 173
00006516 ~ 5i

log & = 5-8141.

m v observed v calculated

503 3-72 3-66

682 4-59 4-55

750 4-87 4-87

915 5-57 5-62

1-034 6-14 6-14

Cement and very fine sand, semicircular.

log & = 5-8802.

v ft. per sec. v ft. per sec.

ire feet observed calculated

379 2-87 2-74

529 3-44 8-49

636 3-87 3-98

706 4-30 4-30

787 4-51 4-59

839 4-80 4-84

900 4-94 5-10

941 5-20 5-26

983 5-38 5-43

1-006 5-48 5-53

1-02 5-55 5-58

1-04 5-66 5-66

TABLE XXY.

Boston circular sewer, 9 ft. diameter.

Brick, washed with cement, i = CTHHT (Horton).

i = '00006124^5,

log v = '6118 log m + '5319 log i + 2'2401,

v ft. per sec. v ft. per sec.
TO feet observed calculated

928 2-21 2-34

1-208 2-70 2-76

1-408 3-03 3-03

1-830 3-48 3-56

1-999 3-73 3-75

2-309 4-18 4-10

FLOW IN OPEN CHANNELS 203

TABLE XXY (continued).
The same sewer after 4 years' service.

i = '00007998^,
log v = '6118 logm + '5319 logi + 2*1795.

m v observed v calculated
1-120 2-38 2-29

1-606 2-82 2-78

1-952 3-16 3-22

2-130 3-30 3-39

TABLE XXVI.
New Croton aqueduct. Lined with concrete.

v 1 ' 88
i = -000073^,

logv = '6118 log m + '5319 log i+ 2'200.

v ft. per sec.

m feet

observed

calculated

1-000

1-37

1-250

1-59

1-57

1-499

1-79

1-76

1-748

1-95

1-93

2-001

2-11

2-10

2-250

2-27

2-26

2-500

2-41

2-40

2-749

2-52

2-55

2-998

2-65

2-68

3-251

2-78

2-82

3-508

2-89

2-96

3-750

3-00

3-08

3-838

3-02

3-12

TABLE XXVII.
Sudbury aqueduct. Lined with well pointed brick.

i = -00006427^,
log v = '6118 log m + '5319 log i + 2'23.

v ft. per sec. v ft. per sec.
TO feet observed calculated

4987 1-135 1-142

6004 1-269 1-279

8005 1-515 1-525

1-000 1-755 1-752

1-200 1-948 1-954

1-400 2-149 2-147

1-601 2-332 2-331

1-801 2-513 2-511

2-001 2-651 2-672

2-201 2-844 2-832

2-336 2-929 2-937

204 HYDRAULICS

TABLE XXVIII.

Rectangular channel lined with brick (Bazin).

* = '000107^.
m 11 -

v ft. per sec. v ft. per sec.
m feet observed calculated

1922 2-75 2-90

2838 3-67 3'68

3654 4-18 4-30

4235 4-72 4'71

4812 5-10 5-09

540 5-34 5-46

5823 5-68 577

6197 6-01 5-94

6682 6-15 6-22

6968 6-47 6'39

7388 6-60 6-62

7788 6-72 6'83

Glasgow aqueduct. Lined with concrete.

i = '0000696 ^pa,
log v = *6118 log m + '5319 log i + 2'2113.

v ft. per sec.

m feet

observed

calculated

1-227

1-87

1-89

1-473

2-07

2-11

1-473

2-106

2-11

1-489

2-214

2-13

1-499

2-13

2-14

1-499

2-15

2-14

1-548

2-18

2-22

1-597

2-21

2-23

1-607

2-23

1-610

2-22

2-24

1-620

2-24

1-627

2-25

2-27

1-738

2-26

2-33

1-811

2-47

2-40

TABLE XXIX.

Charlestown basket-shaped sewer 6' x 6' 8".
Brick, washed with cement, i -s^inf (Horton).

i= '00008405^,

logv = '6118 log m + '5319 log i + 21678.

v ft. per sec. v ft. per sec.
m feet observed calculated

688 1-99 2-05

958 2-46 2-52

1-187 2-82 2-87

1-539 3-44 3-36

FLOW IN OPEN CHANNELS

205

TABLE XXIX (continued).
The same sewer after 4 years' service,

v 1 ' 8 *
; = -0001096

log v = '6118 log m + *5319 log i + 21065.

m feet
1-342
1-508
1-645

v ft. per sec.
observed

2-66
2-86
3-04

v ft. per sec.
calculated

2-68
2-88
3-04

TABLE XXX.

Left aqueduct of the Solani canal, rectangular in section, lined
with rubble masonry (Cunningham),

000225
000206
000222
000207
000189?

,.1-96

t = '00026 ^j.
m 14

v ft. per sec. v ft. per seo.
m feet observed calculated

Right aqueduct,

6-43

6-81

7-21

7-643

7-94

* = '0002213

3'46
3-49
3-70
3'87
4-06

^
m 1

000195
000225
000205
000193
000193
000190

3-42
5'86
6-76
7-43
7'77
7-96

v observed
2-43
3'61
3'73
3-87
3-93
4-06

3'50
3'47
3'84
3'83
3'83

v calculated
2'26
3'58
3'76
3'89
4'04
4'06

Torlonia tunnel, partly in hammered ashlar, partly in solid

rock,

i= '00104,

-.ITO

i = -00022

m 1

1-932
2-172
2-552
2-696
3-251
3-438
3-531
3-718

v observed

v calculated

8-382

3-45

3-625

3-73

4-232

4-16

4-324

4-32

5-046

4-90

4-965

5-08

4-908

5-18

5-358

6-37

206

HYDRAULICS

TABLE XXXI.

Channel lined with hammered ashlar,

2,1-36,

i = -000149 ^L

log fc = 41740.

t'=-101

v ft. per sec.

m feet

observed

calculated

324

12-30

467

16-18

580

18-68

18-97

562

21-09

20-8

=037

v ft. per sec.

m feet

observed

calculated

424

9-04

9-02

620

11-46

11-86

745

13-55

13-52

852

15-08

14-93

Channel lined with small pebbles, i = '0049 (n = 1'96, p
will give equally good results).

1-32

000152

log k = 41913.

m feet

250
357
450
520
588
644
700
746
785
832
871
910

v ft. per sec.
observed

2-16
2-95
3-40
3-84
4-14
4-43
4-64
4-88
5-12
5-26
5-43
5-57

v ft. per sec.
calculated

2-34
2-97
3-47
3-82
4-15
4-43
4-66
4-88
5-05
5-25
5-43
5-58

FLOW IN OPEN CHANNELS

207

TABLE XXXII.

Channel lined with large pebbles (Bazin),

i = -000229^
m 16 '

m feet

291
417
510
587
656
712
772
823
867
909
946
987

log & = 4-3605.

v ft. per sec.
observed

1-79
2-43
2-90
3-27
3-56
3-85
4-03
4-23
4-43
4-60
4-78
4-90

v ft. per sec.
calculated

1-84
2-44
2-90
3-18
3-45
3-67
3-91
4-33
4-53
4-69
4-84
5-00

TABLE XXXIII.

Velocities as observed, and as calculated by the formula
v=C^mN. = 50.

Ganges Canal.

000155
000229
000174
000227
000291

m feet

5-40
8-69
7-82
9-34
4-50

v ft. per sec.
observed

v ft. per sec.
calculated

2-4

2-34

3-71

3'80

2-96

3-08

4-02

4-00

2-82

2-63

0005503
0005503
0002494
0002494

0001183
0001782
0001714
0002180

River Weser.

m v observed

8-93
13-35
14-1
10-5

6-29
7-90
5-69
4-75

Missouri,
n v observed

10-7
12-3
15-4
17-7

3-6
4-38
5-03
6-19

v calculated

6-0

8-18
5-70
4-78

v calculated

3-23
4-37
4-80

208

II YDRAULICS

00029
00029
00033
00033

Cavour Canal,
m v observed

7-32
5-15
5-63
4.74

3-70
3-10
3-40
3-04

v calculated

3-80
2-92
3-14
2-91

Earth channel (branch of Burgoyne canal).
Some stones and a few herbs upon the surface.

0*48.

000957
000929
000993
000986
000792
000808
000858
000842

v ft. per sec. v ft. per sec.
m feet observed calculated

958
1-181
1-405
1-538

958
1-210
1-436
1-558

1-243
1-702
1-797
1-958
1-233
1-666
1-814
1-998

1-30
1-66
1-94
2-06
1-25
1-56
1-79
2-08

130. Distribution of the velocity in the cross section
of open channels.

The mean velocity of flow in channels and pipes of small cross
sectional area can be determined by actually measuring the weight
or the volume of the water discharged, as shown in Chapter VII,
and dividing the volume discharged per second by the cross
section of the pipe. For large channels this is impossible, and
the mean velocity has to be determined by other means, usually
by observing the velocity at a large number of points in the same
transverse section by means of floats, current meters*, or Pi tot
tubes t. If the bed of the stream is carefully sounded, the cross
section can be plotted and divided into small areas, at the centres
of which the velocities have been observed. If then, the observed
velocity be assumed equal to the mean velocity over the small
area, the discharge is found by adding the products of the areas
and velocities.

Or Q = 2a.i>.

M. Bazint, with a thoroughness that has characterised his
experiments in other branches of hydraulics, has investigated the
distribution of velocities in experimental channels and also in
natural streams.

In Figs. 119 and 120 respectively are shown the cross sections
of an open and closed rectangular channel with curves of equal

See page 238.

Bazin, Eecherches Hydraulique.

t See page 241.

FLOW IN OPEN CHANNELS

209

velocity drawn on the section. Curves showing the distribution
of velocities at different depths on vertical and horizontal sections
are also shown.

Curves of equal Velocity
fbr Rectangular Channel/.

Fig. 119.

on
Vertical Sections.

i f /'' N \i

' / -i i

i / VeLodties orb \. |

[/ Horizontal Sections. NJ

! !

a, 5 e < e>

? f f

// x" x'" x^

^ // / X /

I ' ' /' / /

S .'/ / x ' <7

50 ill / /

11 \\ < :

/7? //?

9
K
L

\\\. \ \

V\ \ N \

c <i. &

\ V X N N N '

N^.

V \l\_ NV -W ^ ^

^--

Fig. 120.

Tt will be seen that the maximum velocity does not occur in
the free surface of the water, but on the central vertical section
at some distance from the surface, and that the surface velocity
may be very different from the mean velocity. As the maximum
velocity does not occur at the surface, it would appear that in

L. H. H

HYDRAULICS

assuming the wetted perimeter to be only the wetted surface of
the channel, some error is introduced. That the air has not the
same influence as if the water were in contact with a surface
similar to that of the sides of the channel, is very clearly
shown by comparing the curves of equal velocity for the closed
rectangular channel as shown in Fig. 119 with those of Fig. 120.
The air resistance, no doubt, accounts in some measure for the
surface velocity not being the maximum velocity, but that it does
not wholly account for it is shown by the fact that, whether the
wind is blowing up or down stream, the maximum velocity is still
below the surface. M. Flamant* suggests as the principal reason
why the maximum velocity does not occur at the surface, that
the water is less constrained at the surface, and that irregular
movements of all kinds are set up, and energy is therefore
utilised in giving motions to the water not in the direction of
translation.

Depth on any vertical at which the velocity is equal to the mean
velocity. Later is discussed, in detail, the distribution of velocity
on the verticals of any cross section, and it will be seen, that if u
is the mean velocity on any vertical section of the channel, the
depth at which the velocity is equal to the mean velocity is about
0'6 of the total depth. This depth varies with the roughness of
the stream, and is deeper the greater the ratio of the depth to
the width of the stream. It varies between *5 and '55 of the depth
for rivers of small depth, having beds of fine sand, and from *55
to '66 in large rivers from 1 to 3j- feet deep and having strong
bedst.

As the banks of the stream are approached, the point at which
the mean velocity occurs falls nearer still to the bed of the stream,
but if it falls very low there is generally a second point near the
surface at which the velocity is also equal to the mean velocity.

When the river is covered with ice the maximum velocity of
the current is at a depth of '35 to '45 of the total depth, and the
mean velocity at two points at depths of '08 to '13 and '68 to '74
of the total depth J.

If, therefore, on various verticals of the cross section of a stream
the velocity is determined, by means of a current meter, or Pitot
tube, at a depth of about *6 of the total depth from the surface,
the velocity obtained may be taken as the mean velocity upon the
vertical.

'* Hydrauliqne.

t Le Genie Civil, April, 1906, " Analysis of a communication by Murphy to
the Hydrological section of the Institute of Geology of the United States."
J Cunningham, Experiments on the Ganges Canal.

FLOW IN OPEN CHANNELS

211

The total discharge can then be found, approximately, by
dividing the cross section into a number of rectangles, such as
abcdy Fig. 120 a, and multiplying the area of the rectangle by the
velocity measured on the median line at 0'6 of its depth.

cu d

Fig. 120 a.

The flow of the Upper Nile has recently been determined in
this way.

Captain Cunningham has given several formulae, for the mean
velocity u upon a vertical section, of which two are here quoted.

(1),
(2),

V being the velocity at the surface, v 3. the velocity at f of the depth,
v at one quarter of the depth, and so on.

131. Form of the curve of velocities on a vertical
section.

M. Bazin* and Cunningham have both taken the curve of
velocities upon a vertical section as a parabola, the maximum
velocity being at some distance h m below the free surface of the
water.

Let V be the velocity measured at the centre of a current and
as near the surface as possible. This point will really be at 1 inch
or more below the surface, but it is supposed to be at the surface.

Let v be the velocity on the same vertical section at any depth
h t and H the depth of the stream.

Bazin found that, if the stream is wide compared to its depth,
the relationship between v, Y, h, and i the slope, is expressed by
the formula,

=v-ft()vm ax

k being a numerical coefficient, which has a nearly constant value
of 36'2 when the unit of length is one foot.

* Recherches Hydraulique, p. 228 ; Annales des Fonts et Chaussges, 2nd Vol..

1875.

142

212 HYDRAULICS

To determine the depth on any vertical at which the velocity is
equal to the mean velocity. Let u be the mean velocity on any
vertical section, and h u the depth at which the velocity is equal to
the mean velocity.

The discharge through a vertical strip of width dl is

v .dh.
o

/H / i
Therefore uTL

and A = V-<sH* (2).

Substituting u and h u in (1) and equating to (2),

and h u

This depth, at which the velocity is equal to the mean velocity,
is determined on the assumption that Jc is constant, which is only
true for sections very near to the centre of streams which are
wide compared with their depth.

It will be seen from the curves of Fig. 120 that the depth at
which the maximum velocity occurs becomes greater as the sides
of the channel are approached, and the law of variation of velocity
also becomes more complicated. M. Bazin also found that the
depth at the centre of the stream, at which the maximum velocity
occurs, depends upon the ratio of the width to the depth, the
reason apparently being that, in a stream which is wide compared
to its depth, the flow at the centre is but slightly affected by the
resistance of the sides, but if the depth is large compared with the
width, the effect of the sides is felt even at the centre of the
stream. The farther the vertical section considered is removed
from the centre, the effect of the resistance of the sides is
increased, and the distribution of velocity is influenced to a
greater degree. This effect of the sides, Bazin expressed by
making the coefficient k to vary with the depth h m at which
the maximum velocity occurs.

The coefficient is then,

36'2

Further, the equation to the parabola can be written in terms
of v m , the maximum velocity, instead of V.

FLOW IN OPEN CHANNELS 213

Thu3 , ^_36-27-p (3).

The mean velocity u, upon the vertical section, is then,
= i [*vdh
36'2

= m ~

Therefore

36'2

TT2 / 1 \

1 \f W

When v = u, Ji = h u>

-i it c J. fljn fT/u ^'ibu'l'm

and therefore, o ~ TJ = xfa TTT~

o 3 M H

The depth h m at which the velocity is a maximum is generally
less than *2H, except very near the sides, and h u is, therefore, not
very different from *6H, as stated above.

Ratio of maximum velocity to the mean velocity. From
equation (4),

v m =u +

/i_M 2 V3 H

V H/

In a wide stream in which the depth of a cross section is fairly
constant the hydraulic mean depth m does not differ very much
from H, and since the mean velocity of flow through the section is
C \/m? and is approximately equal to u, therefore,

36-2 /I h m h m *\
h m \ 2 \3 H HV*

Assuming h m to vary from to "2 and C to be 100, varies

from 1'12 to 1'09. The ratio of maximum velocity to mean
velocity is, therefore, probably not very different from 1*1.

132. The slopes of channels and the velocities allowed
in them.

The discharge of a channel being the product of the area and
the velocity, a given discharge can be obtained by making the
area small and the velocity great, or vice versa. And since the
velocity is equal to Cvwt, a given velocity can be obtained by

214 HYDRAULICS

varying either m or i. Since m will in general increase with the
area, the area will be a minimum when i is as large as possible.
But, as the cost of a channel, including land, excavation and
construction, will, in many cases, be almost proportional to its
cross sectional area, for the first cost to be small it is desirable
that i should be large. It should be noted, however, that the
discharge is generally increased in a greater proportion, by an
increase in A, than for the same proportional increase in i.

Assume, for instance, the channel to be semicircular.

The area is proportional to d?, and the velocity v to \/d . i.

Therefore Q oc d? *Jdi.

IfjZ_is kept constant and i doubled, the discharge is increased
to \/2Q, but if d is doubled, i being kept constant, the discharge
will be increased to 5'6Q. The maximum slope that can be given
will in many cases be determined by the diif erence in level of the
two points connected by the channel.

When water is to be conveyed long distances, it is often
necessary to have several pumping stations en route, as sufficient
fall cannot be obtained to admit of the aqueduct or pipe line being
laid in one continuous length.

The mean velocity in large aqueducts is about 3 feet per
second, while the slopes vary from 1 in 2000 to 1 in 10,000. The
slope may be as high as 1 in 1000, but should not, only in excep-
tional circumstances, be less than 1 in 10,000.

In Table XXXIY are given the slopes and the maximum
velocities in them, of a number of brick and masonry lined
aqueducts and earthen channels, from which it will be seen that
the maximum velocities are between 2 and 5J feet per second,
and the slopes vary from 1 in 2000 to 1 in 7700 for the brick and
masonry lined aqueducts, and from 1 in 300 to 1 in 20,000 for the
earth channels. The slopes of large natural streams are in some
cases even less than 1 in 100,000. If the velocity is too small
suspended matter is deposited and slimy growths adhere to the sides.

It is desirable that the smallest velocity in the channel shall be
such, that the channel is "self-cleansing," and as far as possible
the growth of low forms of plant life prevented.

In sewers, or channels conveying unfiltered waters, it is
especially desirable that the velocity shall not be too small, and
should, if possible, not be less than 2 ft. per second.

TABLE XXXIY.

Showing the slopes of, and maximum velocities, as determined
experimentally, in some existing channels.

FLOW IN OPEN CHANNELS

215

Smooth aqueducts

Slope

Maximum velocity

New Croton aqueduct -0001326

3 ft. per second

Sudbury aqueduct '000189

2-94

Glasgow aqueduct '000182

2-25

Paris Dhuis '000130

Avre, 1st part -0004

2nd part '00033

Manchester Thirlmcre '000315

Naples -00050

4-08

Boston Sewer '0005

3'44

000333

4-18 ...

Earth channels.

Slope Maximum velocity Lining

Ganges canal -000306 4-16 ft.

per second earth

Escher -003 4'08

Linth -00037 5'53

( gravel and

Cavour '00033 3'42

\ some stones

Simmen '0070 3'74

earth

Chazilly cut '00085 1'70

( earth, stony,

MarseiUes canal '00043 1'70

( few weeds

Chicago drainage canal

(of the bottom of the canal) '00005 3

> j>

TABLE XXXV.

Showing for varying values of the hydraulic mean depth m, the
minimum slopes, which brick channels and glazed earthenware
pipes should have, that the velocity may not be less than 2 ft.
per second.

m feet slope

1 i

Q 93

275

510

775

1058

1380

1 ,

, 2040

1-0

, 2760

mfeet

slope

1-25

1 i

n 3700

1-5

4700

1-75

5710

2-0

6675

2-5

9000

3'0

11200

4-0

15850

The slopes are calculated from the formula

157-5

The value of y is taken as 0'5 to allow for the channel becoming
dirty. For the minimum slope for any other velocity v, multiply

(2\ 2
-j . For example, the minimum slope

for a velocity of 3 feet per second when m is 1, is 1 in 1227.

210 HYDRAULICS

Velocity of flow in, and slope of earth channels. If the velocity
is high, in earth channels, the sides and bed of the channel are
eroded, while on the other hand if it is too small, the capacity of
the channel will be rapidly diminished by the deposition of sand
and other suspended matter, and the growth of aquatic plants.
Du Buat gives '5 foot per second as the minimum velocity that
mud shall not be deposited, while Belgrand allows a minimum
of '8 foot per second.

TABLE XXXVI.

Showing the velocities above which, according to Du Buat,
and as quoted by Rankine, erosion of channels of various materials
takes place.

Soft clay 0-25 ft. per second

Fine sand 0'50

Coarse sand and gravel as large as peas 0*70

Gravel 1 inch diameter 2'25

Pebbles 1| inches diameter 3'33

Heavy shingle 4-00

Soft rock, brick, earthenware 4'50

Rock, various kinds 6*00 and upwards

133. Sections of aqueducts and sewers.

The forms of sections given to some aqueducts and sewers are
shown in Figs. 121 to 131. In designing such aqueducts and
sewers, consideration has to be given to problems other than the
comparatively simple one of determining the size and slope to
be given to the channel to convey a certain quantity of water.
The nature of the strata through which the aqueduct is to be
cut, and whether the excavation can best be accomplished by
tunnelling, or by cut and cover, and also, whether the aqueduct
is to be lined, or cut in solid rock, must be considered. In many
cases it is desirable that the aqueduct or sewer should have such
a form that a man can conveniently walk along it, although its
sectional area is not required to be exceptionally large. In
such cases the section of the channel is made deep and narrow.
For sewers, the oval section, Figs. 126 and 127, is largely
adopted because of the facilities it gives in this respect, and it has
the further advantage that, as the flow diminishes, the cross
section also diminishes, and the velocity remains nearly constant
for all, except very small, discharges. This is important, as at
small velocities sediment tends to collect at the bottom of the
sewer.

134. Siphons forming part of aqueducts.

It is frequently necessary for some part of an aqueduct to be
constructed as a siphon, as when a valley has to be crossed or the

FLOW IN OPEN CHANNELS

217

aqueduct taken under a stream or other obstruction, and the
aqueduct must, therefore, be made capable of resisting con-
siderable pressure. As an example the New Croton aqueduct
from Croton Lake to Jerome Park reservoir, which is 33' 1 miles

Fig. 121.

Fig. 122.

Fig. 123.

Fig. 127.

Fig. 128.

Fig. 129.

Fig. 130.

Fig. 131.

218 HYDRAULICS

long, is made up of two parts. The first is a masonry conduit of
the section shown in Fig. 121, 23'9 miles long and having a slope
of '0001326, the second consists almost entirely of a brick lined
siphon 6'83 miles long, 12' 3" diameter, the maximum head in
which is 126 feet, and the difference in level of the two ends is
6*19 feet. In such cases, however, the siphon is frequently made
of steel, or cast-iron pipes, as in the case of the new Edinburgh
aqueduct (see Fig. 131) which, where it crosses the valleys, is
made of cast-iron pipes 33 inches diameter.

135. The best form of channel.

The best form of channel, or channel of least resistance, is
that which, for a given slope and area, will give the maximum
discharge.

Since the mean velocity in a channel of given slope is propor-

i
tional to p , and the discharge is A . v, the best form of channel for

a given area, is that for which P is a minimum.

The form of the channel which has the minimum wetted peri-
meter for a given area is a semicircle, for which, if r is the radius,

the hydraulic mean depth is ~.

More convenient forms, for channels to be excavated in rock
or earth, are those of the rectangular or trapezoidal section,
Fig. 133. For a given discharge, the best forms for these
channels, will be those for which both A and P are a minimum ;
that is, when the differentials dA and dP are respectively equal to
zero.

Rectangular channel. Let L be the width and Ji the depth,
Fig. 132, of a rectangular channel ; it is required to find the ratio

y that the area A and the wetted perimeter P may both be a

minimum, for a given discharge.

A-Lfc,

therefore 8A = /t . 8L + L3ft = (1),

P-L+2&,

therefore dP = dL + 2dh = (2).

Substituting the value of 3L from (2) in (1),

~L = 2h.

2tf h
Therefore m = ~4fa = 2'

Since L = 2h t the sides and bottom of the channel touch a circle
having h as radius and the centre of which is in the free surface
of the water.

FLOW IN OPEN CHANNELS

219

Earth channels of trapezoidal form. In Fig. 133 let

Z be the bottom width,

h the depth,

A the cross sectional area FBCD,

P the length of FBCD or the wetted perimeter,

i the slope,

and let the slopes of the sides be t horizontal to one vertical; CG
is then equal to th and tan CDGr = t.

-H

Fig. 132.

Fig. 133.

Let Q be the discharge in cubic feet per second.
Then A.

(3),
(4),

and

For the channel to be of the best form dP and dA. both equal
zero.

From (3) A = hl+th 2 ,

and therefore dA. = hdl + ldh + 2thdh = Q (6).

From (4) P = I + 2hJt 2 + l
and dP = dl + 2<J& + ldh = (7).

Substituting the value of dl from (7) in (6)

l = 2h>J^l-2th (8).

Therefore,

l-2ht

Let be the centre of the water surface FD, then since from (8)

I + th = Wf + 1,
therefore, in Fig. 133 CD = EG - OD.

220 HYDRAULICS

Draw OF and OE perpendicular to CD and BC respectively.

Then, because the angle OFD is a right angle, the angles CDG
and FOD are equal ; and since OF = OD cos FOD, and DG = OE,
and DG = CDcosCDG, therefore, OE = OF; and since OEC and
OFC are right angles, a circle with as centre will touch the sides
of the channel, as in the case of the rectangular channel.

136. Depth of flow in a channel of given form that,
(a) the velocity may be a maximum, (b) the discharge may
be a maximum.

Taking the general formula

. k.v*

l = ~^~

i P

and transposing, v = j-

For a given slope and roughness of the channel v is, therefore,
proportional to the hydraulic mean depth and will be a maximum
when m is a maximum.

That is, when the differential of ^ is zero, or

(1).

For maximum discharge, A.V is a maximum, and therefore,
P

A /A\".

A . ( p ] is a maximum.

Differentiating and equating to zero,

Q... ...(2).

n n

Affixing values to n and p this differential equation can be
solved for special cases. It will generally be sufficiently accurate
to assume n is 2 and p = 1, as in the Chezy formula, then

n + p_S
n ~2>
and the equation becomes

3PdA-AdP = ........................... (3).

137. Depth of flow in a circular channel of given
radius and slope, when the velocity is a maximum.

Let r be the radius of the channel, and 2< the angle subtended
by the surface of the water at the centre of the channel, Fig. 134.

FLOW IN OPEN CHANNELS 221

Then the wetted perimeter

and dP = 2rd<f>.

The area A = r 2 0-r 2 sin0 cos =
and dA. = r*d<l>- r 2 cos 20 d0.

Substituting these values of dP and dA. in equation (3),
section 136,

tan 20 = 20.

The solution in this case is obtained
directly as follows,

A_r /- sin 20\
P~2V 1 ~ 2+ }*

This will be a maximum when sin 20
is negative, and

sin 20

is a maximum, or when Fig. 134.

d /sin20\
d+\ 2<f> /~ U '
.'. 20 cos 20 -sin 20 = 0,
and tan 20 = 2^.

The solution to this equation, for which 20 is less than 360, is

20 = 257 27'.

Then A = 2'73V,

P = 4'494r,
m = '608r,
and the depth of flow d = T626r.

138. Depth of flow in a circular channel for maximum
discharge.

Substituting for dP and dA in equation (3), section 136,

6^0(^0 - 6^0 cos 20d0 - 2r 3 0d0 + r 3 sin 20d0 = 0,
from which 40 - 60 cos 20 + sin 20 = 0,

and therefore = 154.

Then A = 3'08r 2 ,

P = 5'30r,

and the depth of flow d = l'899r.

Similar solutions can be obtained for other forms of channels,
and may be taken by the student as useful mathematical exercises
but they are not of much practical utility.

222

HYDRAULICS

139. Curves of velocity and discharge for a given
channel.

The depth of flow for maximum velocity, or discharge, can be
determined very readily by drawing curves of velocity and dis-
charge for different depths of flow in the channel. This method
is useful and instructive, especially to those students who are not
familiar with the differential calculus.

As an example, velocities and discharges, for different depths
of flow, have been calculated for a large aqueduct, the profile of
which is shown in Fig. 135, and the slope i of which is (V0001326.
The velocities and discharges are shown by the curves drawn in
the figure.

Fig. 135.

Values of A and P for different depths of flow were first deter-
mined and m calculated from them.

The velocities were calculated by the formula

v = C *Jmi,

using values of C from column 3, Table XXI.

It will be seen that the velocity does not vary very much for
all depths of flow greater than 3 feet, and that neither the velocity
nor the discharge is a maximum when the aqueduct is full ; the
reason being that, as in the circular channel, as the surface of the
water approaches the top of the aqueduct the wetted perimeter
increases much more rapidly than the area.

The maximum velocity is obtained when m is a maximum
and equal to 3'87, but the maximum discharge is given, when the
depth of flow is greater than that which gives the greatest

FLOW IN OPEN CHANNELS 223

velocity. A circle is shown on the figure which gives the same
maximum discharge.

The student should draw similar curves for the egg-shaped
sewer or other form of channel.

140. Applications of tne formula.

Problem 1. To find the flow in a channel of given section and slope.

This is the simplest problem and can be solved by the application of either the
logarithmic formula or by Bazin's formula.

The only difficulty that presents itself, is to affix values to k, n, and p in the
logarithmic formula or to y in Bazin's formula.

(1) By the logarithmic formula.

First assign some value to fc, n, and p by comparing the lining of the channel
with those given in Tables XXIV to XXXIII. Let w be the cross sectional area of
the water.

k v n

Then since i = ,

mP '

log v = - log i + log m - - log fe,

and Q = b).v,

or logQ = logw + -logi+^logm log k.

(2) By the Chczy formula, using Bazin's coefficient.

The coefficient for a given value of m must be first calculated from the formula

or taken from Table XXI.
Then

and

Example. Determine the flow in a circular culvert 9 ft. diameter, lined with
smooth brick, the slope being 1 in 2000, and the channel half full.

Area _ d_

-Wetted perimeter -4-
(1) By the logarithmic formula

Therefore, log = j log -0005 + log 2-25 - - log '00007,
v = 4'55 ft. per sec.,
w = 7 Ll^ = 31-8 sq.ft.,

Q = 145 cubic feet per sec.
(2) By the Chezy formula, using Bazin's coefficient,

-43 ft. per sec.
Q = 31-8 x 4-43 = 141 cubic ft. per sec.

224 HYDRAULICS

Problem 2. To find the diameter of a circular channel of given slope, for which
the maximum discharge is Q cubic feet per second.

The hydraulic mean depth m for maximum discharge is '573r (section 138) and
A = 3-08r 2 .

Then the velocity is v=-757C*JrT,

and Q = 2-37 Cr*^,

1 /O 2 "

therefore r - \/^-.,

and the diameter D = 1-42 . .

The coefficient C is unknown, hut by assuming a value for it, an approximation
to D can be obtained ; a new value for C can then be taken and a nearer approxi-
mation to D determined ; a third value for C will give a still nearer approximation
to D.

Example. A circular aqueduct lined with concrete has a diameter of 5' 9" and
a slope of 1 foot per mile.

To find the diameter of two cast-iron siphon pipes 5 miles long, to be parallel
with each other and in series with the aqueduct, and which shall have the same
discharge; the difference of level between the two ends of the siphon being 12-5 feet.

The value of m for the brick lined aqueduct of circular section when the
discharge is a maximum is 573r = l-64 feet.

The area A = 3-08^= 25 sq. feet.

Taking C as 130 from Table XXI for the brick culvert and 110 for the cast-iron
pipe from Table XII, then

TO ,
Therefore

d=4-00 feet.

Problem 3. Having given the bottom width I, the slope t, and the side slopes t
of a trapezoidal earth channel, to calculate the discharge for a given depth.

First calculate m from equation (5), section 135.

From Table XXI determine the corresponding value of C, or calculate C from
Bazin's formula,

then v = C

and Q=A.v.

A convenient formula to remember is the approximate formula for ordinary
earth channels

For values of m greater than 2, v as calculated from this formula is very nearly
equal to v obtained by using Bazin's formula.

, -00037V 2 ' 1

The formula * = - rl

m 15

may also be used.

FLOW IN OPEN CHANNELS 225

Example. An ordinary earth channel has a width 1= 10 feet, a depth d = 4i'eet,
and a slope i = ^oVi7' Side slopes 1 to 1. To find Q.
A =56 sq. ft.,
P = 21-312 ft.,
= 2-628 ft.,

v = l'91 ft. per sec.,
Q = 107 cubic ft. per sec.

From the formula

v=l-8S ft. per sec.,
Q = 105-3 cubic ft. per sec.
From the logarithmic formula

r = l'9ft. per sec.,

Q = 106 4 cubic feet per sec.

Problem 4. Having given the flow in a canal, the slope, and the side slopes, to
find the dimensions of the profile and the mean velocity of flow,
(a) When the canal is of the best form.
(6) When the depth is given.

In the first case m = - , and from equations (8) and (4) respectively, section 136

Therefore
Substituting - for m

and A 2 = fc 4 (2

But t? = j =

Therefore C 3 i =

and fc 5 = - = - .... ........ (1).

A value for C should be chosen, say 0=70, and h calculated, from which a mean
value for m = - can be obtained.

A nearer approximation to h can then be determined by choosing a new value of C,
from Table XXI corresponding to this approximate value of m, and recalculating
h from equation (1).

Example. An earthen channel to be kept in very good condition, having a slope
of 1 in 10,000, and side slopes 2 to 1, is required to discharge 100 cubic feet
per second ; to find the dimensions of the chaunel ; take C = 70.

L. H. 15

226 HYDRAULICS

20,000
Then ft 5 -

20,000
~ -49 x 6-1
= 6700,

and ft =5-4 feet.

Therefore m = 2-l.

From Table XXI, = 82 for this value of m, therefore a nearer approximation
to ft is now found from

., 20,000 20,000

S'J-

10,000
from which h = 5'22 ft. and m = 2'61.

The approximation is now sufficiently near for all practical purposes and may
be taken as 5 feet.

Problem 5. Having given the depth d of a trapezoidal channel, the slope i, and
the side slopes t, to find the bottom width I for a given discharge.
First using the Chezy formula,

v = C*Jmi

and

The mean velocity

Therefore ^ + ^

In this equation the coefficient C is unknown, since it depends upon the value
of m which is unknown, and even if a value for C be assumed the equation cannot
very readily be solved. It is desirable, therefore, to solve by approximation.

Assume any value for m, and find from column 4, Table XXI, the corresponding
value for C, and use these values of m and C.

Then, calculate v from the formula

Since T =V >

and

Therefore dl + td' 2 = - (1).

From this equation a value of I can be obtained, which will probably not be the
correct value.

With this value of I calculate a new value for m, from the formula

For this value of m obtain a new value of G from the table, recalculate u, and
by substitution in formula (1) obtain a second value for I.

On now again calculating m by substituting for d in formula (2), it will generally
be found that m differs but little from in previously calculated ; if so, the approxi-
mation has proceeded sufficiently far, and d as determined by using this value of m
will agree with the correct value sufficiently nearly for all practical purposes.

The problem can be solved in a similar way by the logarithmic formula

The indices u &nd p may be taken as 2-1, and 1'5 respectively, and k as '00037.

FLOW IN OPEN CHANNELS 227

Example. The depth of an ordinary earth channel is 4 feet, the side slopes
1 to 1, the slope 1 in GOOO and the discharge is to be 7000 cubic feet per minute.
Find the bottom width of the channel.
Assume a value for m, say 2 feet.
From the logarithmic formula

2 -1 log v = log i+ 1-5 log m- 4-5682 ........................... (3),

v = 1-122 feet per sec.

Thcn A

But

Substituting this value for I in equation (2)

Becalculating v from formula (3)

v = 1-556.
Then A = 75 feet,

1= 14-75 feet,
and m = 2-88 feet.

The first value of Zis, therefore, too large, and this second value is too small.
Third values were found to be v = l'455,

A = 80- 2,
1 = 16-05,
m=2-935.
This value of I is again too large.

A fourth calculation gave v = 1-475,

A=79-2,

J=15'8,

m=2-92.

The approximation has been carried sufficiently far, and even further than is
necessary, as for such channels the coefficient of roughness k cannot be trusted to
an accuracy corresponding to the email difference between the third and fourth
values of I.

Problem 6. Having given the bottom width I, the slope t and the side slopes of
a trapezoidal channel, to find the depth d for a given discharge.

This problem is solved exactly as 5 above, by first assuming _a_value for m, and
calculating an approximate value for v from the formula v = C^Jmi.

Then, by substitution in equation (1) of the last problem and solving the
quadratic,

oy substituting this value for d in equation (2), a new value for m can be found,
and hence, a second approximation to d, and so on.

Using the logarithmic formula the procedure is exactly the same as for
problem 5.

Problem 1 *. Having a natural stream BC, Fig. 135 a, of given slope, it is required
to determine the point C, at which a canal, of trapezoidal section, which is to
deliver a definite quantity of water to a Riven point A at a given level, shall be
made to join the stream so that the cost of the canal is a minimum.

* The solution here given is practically the same as that given by M. Flamant
in his excellent treatise Hydraulique.

152

223 HYDRAULICS

Let I be the slope of the stream, i of the canal, h the height above some datum
of the surface of the water at A, and ft, of the
water in the stream at B, at some distance L
from C.

Let L be also the length and A the !T> \s~*

sectional area of the canal, and let it be j j ~*

assumed that the section of the canal is of the A *- ' j

most economical form, or m = - .

& -fig- loo a.

The side slopes of the canal will be fixed

according to the nature of the strata through which the canal is cut, and may be
supposed to be known.

Then the level of the water at C is

h 7?i
Therefore L = . .

Let I be the bottom width of the canal, and t the slope of the sides. The cross
section is then dl + td?, and

_A_ dl + tcP
~~

Substituting 2m for d,

I = 4/ A/ 2 + 1 - 4tm,

and therefore

4m A/ t 2 H
from which tn 2 = .-

The coefficient C in the formula v = G*Jrni may be assumed constant.
Then t? 2 =C 2 w,

and v 4 =C 4 i 2 i 2 .

For t? substituting ~ , and for w 2 the above value,

_ C 4 Ai 2
A 4 "^

and

Therefore

The cost of the canal will be approximately proportional to the product of the
length L and the cross sectional area, or to the cubical content of the excavation.
Let k be the price per cubic yard including buying of land, excavation etc. Let x
be the total cost.

Then * = &. L. A

This will be a minimum when -^-=0.
di

Differentiating therefore, and equating to zero,

I^IK-I,

and i = ?I.

The most economical slope is therefore $ of the slope of the natural stream.

If instead of taking the channel of the best form the depth is fixed, the,
slope 1 = ^.1.

FLOW IN OPEN CHANNELS 229

There have been two assumptions made in the calculation, neither of which is
rigidly true, the first being that the coefficient C is constant, and the second that
the price of the canal is proportional to its cross sectional area.

It will not always be possible to adopt the slope thus found, as the mean
velocity must be maintained within the limits given on page 216, and it is not
advisable that the slope should be less than 1 in 10,000.

EXAMPLES.

(1) The area of flow in a sewer was found to be 0*28 sq. feet; tb.3
wetted perimeter 1 '60 feet; the inclination 1 in 38*7. The mean velocity
of flow was 6'12 feet per second. Find the value of G in the formula

(2) The drainage area of a certain district was 19*32 acres, the whole
area being impermeable to rain water. The maximum intensity of the
rainfall was 0*360 ins. per hour and the maximum rate of discharge regis-
tered in the sewer was 96% of the total rainfall.

Find the size of a circular glazed earthenware culvert having a slope of

1 in 50 suitable for carrying the storm water.

(3) Draw a curve of mean velocities and a curve of discharge for an
egg-shaped brick sewer, using Bazin's coefficient. Sewer, 6 feet high by
4 feet greatest width ; slope 1 in 1200.

(4) The sewer of the previous question is required to join into a main
outfall sewer. To cheapen the junction with the main outfall it is thought
advisable to make the last 100 feet of the sewer of a circular steel pipe
3 feet diameter, the junction between the oval sewer and the pipe being
carefully shaped so that there is no impediment to the flow.

Find what fall the circular pipe should have so that its maximum
discharge shall be equal to the maximum discharge of the sewer. Having
found the slope, draw out a curve of velocity and discharge.

(5) A canal in earth has a slope of 1 foot in 20,000, side slopes of

2 horizontal to 1 vertical, a depth of 22 feet, and a bottom width of
200 feet; find the volume of discharge.

Bazin's coefficient -y=2*35.

(6) Give the diameter of a circular brick sewer to run half -full for a
population of 80,000, the diurnal volume of sewage being 75 gallons per
head, the period of maximum flow 6 hours, and the available fall 1 in 1000.

Inst. C. E. 1906.

(7) A channel is to be cut with side slopes of 1 to 1 ; depth of water,

3 feet; slope, 9 inches per mile: discharge, 6,000 cubic feet per minute.
Find by approximation dimensions of channel.

(8) An area of irrigated land requires 2 cubic yards of water per hour
per acre. Find dimensions of a channel 3 feet deep and with a side slope
of 1 to 1. Fall, 1 feet per mile. Area to be irrigated, 6000 acres. (Solve
by approximation.) y=2*35.

(9) A trapezoidal channel in earth of the most economical form has a
depth of 10 feet and side slopes of 1 to 1. Find the discharge when the
slope is 18 inches per mile. y=2*35.

230 HYDRAULICS

(10) A river has the following section : top -width, 800 feet ; depth of
water, 20 feet ; side slopes 1 to 1 ; fall, 1 foot per mile. Find the discharge,
using Bazih's coefficient for earth channels.

(11) A channel is to be constructed for a discharge of 2000 cubic feet
per second ; the fall is 1^ feet per mile ; side slopes, 1 to 1 ; bottom width,
10 times the depth. Find dimensions of channel. Use the approximate

formula, v=

(12) Find the dimensions of a trapezoidal earth channel, of the most
economical form, to convey 800 cubic feet per second, with a fall of 2 feet
per mile, and side slopes, 1 to 1. (Approximate formula.)

(13) An irrigation channel, with side slopes of l to 1, receives 600
cubic feet per second. Design a suitable channel of 3 feet depth and
determine its dimensions and slope. The mean velocity is not to exceed
2| feet per second. y=2'35.

(14) A canal, excavated in rock, has vertical sides, a bottom width of
160 feet, a depth of 22 feet, and the slope is 1 foot in 20,000 feet. Find the
discharge, y = 1*54.

(15) A length of the canal referred to in question (14) is in earth. It
has side slopes of 2 horizontal to 1 vertical; its width at the water line
is 290 feet and its depth 22 feet.

Find the slope this portion of the canal should have, taking y as 2'35.

(16) An aqueduct 95| miles long is made up of a culvert 50 miles
long and two steel pipes 3 feet diameter and 45 miles long laid side by side.
The gradient of the culvert is 20 inches to the mile, and of the pipes 2 feet
to the mile. Find the dimensions of a rectangular culvert lined with well
pointed brick, so that the depth of flow shall be equal to the width of the
culvert, when the pipes are giving their maximum discharge.

Take for the culvert the formula

._ -000061 yw
m '
and for the pipes the formula

, -00050. v 2

(17) The Ganges canal at Taoli was found to have a slope of 0*000146
and its hydraulic mean depth m was 7'0 feet ; the velocity as determined
by vertical floats was 2'80 feet per second; find the value of C and the
value of y in Bazin's equation.

(18) The following data were obtained from an aqueduct lined with
brick carefully pointed :

m i v

in metres in metres per sec.

229 0-0001326 '336

381 '484

533 '596

686 "691

838 '769

991 '848

1-143 -913

1-170 -922

FLOW IN OPEN CHANNELS 231

Plot -j-= as ordinates, as abscissae ; find values of a and /3 in Bazin's

Vm v

formula, and thus deduce a value of y for this aqueduct.

(19) An aqueduct 107 miles long consists of 13 miles of siphon, and
the remainder of a masonry culvert 6 feet 10 ^ inches diameter with a gradient
of 1 in 8000. The siphons consist of two lines of cast-iron pipes 43 inches
diameter having a slope of 1 in 500. Determine the discharge.

(20) An aqueduct consists partly of the section shown in Fig. 131,
page 217, and partly (i.e. when crossing valleys) of 33 inches diameter cast-
iron pipe siphons.

Determine the minimum slope of the siphons, so that the aqueduct
may discharge 15,000,000 gallons per day, and the slope of the masonry
aqueduct so that the water shall not be more than 4 feet 6 inches deep in
the aqueduct.

(21) Calculate the quantity delivered by the water main in question (30) ,
page 172, per day of 24 hours.

This amount, representing the water supply of a city, is discharged into
the sewers at the rate of one-half the total daily volume in 6 hours, and is
then trebled by rainfall. Find the diameter of the circular brick outfall
sewer which will carry off the combined flow when running half full, the
available fall being 1 in 1500. Use Bazin's coefficient for brick channels.

(22) Determine for a smooth cylindrical cast-iron pipe the angle
subtended at the centre by the wetted perimeter, when the velocity of flow
is a maximum. Determine the hydraulic mean depth of the pipe under
these conditions. Lond. Un. 1905.

(23) A 9-inch drain pipe is laid at a slope of 1 in 150, and the value of
c is 107 (v=cvW). Find a general expression for the angle subtended at
the centre by the water line, and the velocity of flow; and indicate how the
general equations may be solved when the discharge is given. Lond. Un.
1906.

141. Short account of the historical development of the pipe and channel formulae.
It seems remarkable that, although the practice of conducting water along pipes
and channels for domestic and other purposes has been carried on for many
centuries, no serious attempt to discover the laws regulating the flow seems
to have been attempted until the eighteenth century. It seems difficult to realise
how the gigantic schemes of water distribution of the ancient cities could have been
executed without such knowledge, but certain it is, that whatever information they
possessed, it was lost during the middle ages.

It is of peculiar interest to note the trouble taken by the Roman engineers in
the construction of their aqueducts. In order to keep the slope constant they
tunnelled through hills and carried their aqueducts on magnificent arches. The
Claudian aqueduct was 38 miles long and had a constant slope of five feet per mile.
Apparently they were unaware of the simple fact that it is not necessary for a pipe
or aqueduct connecting two reservoirs to be laid perfectly straight, or else they
wished the water at all parts of the aqueducts to be at atmospheric pressure.

Stephen Schwetzer in his interesting treatise on hydrostatics and hydraulics
published in 1729 quotes experiments by Marriott showing that, a pipe 1400 yards
long, 1| inches diameter, only gave of the discharge which a hole If inches diameter
in the side of a tank would give under the same head, and also explains that the
motion of the liquid in the pipes is diminished by friction, but he is entirely
ignorant of the laws regulating the flow of fluids through pipes. Even as late as

232 HYDRAULICS

1786 Du Buat* wrote, "We are yet in absolute iterance of the laws to which the
movement of water is subjected."

The earliest recorded experiments of any valne on long pipes are those of
Couplet, in which he measured the flow through the pipes which supplied the
famous fountains of Versailles in 1732. In 1771 Abb Bossut made experiments on
flow in pipes and channels, these being followed by the experiments of Du Buat, who
erroneously argued that the loss of head due to friction in a pipe was independent
of the internal surface of the pipe, and gave a complicated formula for the velocity
of flow when the head and the length of the pipe were known.

In 1775 M. Chezy from experiments upon the flow in an open canal, came to
the conclusion that the fluid friction was proportional to the velocity squared, and
that the slope of the channel multiplied by the cross sectional area of the stream,
was equal to the product of the length of the wetted surface measured on the cross
section, the velocity squared, and some constant, or

iA=Pat> 2 (1),

t being the slope of the bed of the channel, A the cross sectional area of the stream,
P the wetted perimeter, and a a coefficient.

From this is deduced the well-known Chezy formula

Prony f, applying to the flow of water in pipes the results of the classical experi-
ments of Coulomb on fluid friction, from which Coulomb had deduced the law that
fluid friction was proportional to av + bv 2 , arrived at the formula

\ v
This is similar to the Chezy formula, ( - + /3 j being equal to ^.

By an examination of the experiments of Couplet, Bossut, and Du Buat, Prony
gave values to a and which when transformed into British units are,

a = -00001733,
/S= -00010614.

For velocities, above 2 feet per second, Prony neglected the term containing the
first power of the velocity and deduced the formula

He continued the mistake of Du Buat and assumed that the friction was in-
dependent of the condition of the internal surface of the pipe and gave the following
explanation : " When the fluid flows in a pipe or upon a wetted surface a film of
fluid adheres to the surface, and this film may be regarded as enclosing the mass
of fluid in motion J." That such a film encloses the moving water receives support
from the experiments of Professor Hele Shaw. The experiments were made upon
such a smafl scale that it is difficult to say how far the results obtained are indica-
tive of the conditions of flow in large pipes, and if the film exists it does not seem
to act in the way argued by Prony.

The value of t in Prony's formula was equal to , H including, not only the

loss of head due to friction but, as measured by Couplet, Bossut and Du Buat,
it also included the head necessary to give velocity to the water and to overcome
resistances at the entrance to the pipe.

Eytelwein and also Aubisson, both made allowances for these losses, by sub-

tracting from H a quantity ^- , and then determined new values for a and b in the
formula

Le Discours prgliminaire de ses Principes d'hydraiiJique.

t See also Girard's Movement des fluids dans Ics tubes capillaires, 1817.

J Traite d'hydraulique. Engineer, Aug. 1897 and May 18U8.

FLOW IN OPEN CHANNELS 233

They gave to a and 6 the following values.

Eytelwein a = -000023534,

6= 000085434.

Anbisson* o =-000018837,

6= -000104392.

By neglecting the term containing v to the first power, and transforming the
terms, Aubisson's formula reduces to

Young, in the Encyclopaedia Britannica, gave a complicated formula for v when
FT and d were known, but gave the simplified formula, for velocities such as
are generally met with in practice,

St Tenant made a decided departure by making - proportional to V instead of

to r 2 as in the Chezy formula.

When expressed "in English feet as units, his formula becomes

v= 206 (mi) A,
Weisbach by an examination of the early experiments together with ten others by

himself and one by M. Gueynard gave to the coefficient a in the formula h=

the value

that is, he made it to vary with the velocity.

Then, mi

the values of a and being a =0*0144,

0=0-01716.

From this formula tables were drawn up by Weisbach, and in England by
Hawkesley, which were considerably used for calculations relating to flow of
water in pipes.

Darcy, as explained in Chapter V, made the coefficient a to vary with the
diameter, and Hagen proposed to make it vary with both the velocity and the
diameter.

Lis formula then became m* = i

The formulae of Ganguillet and Kutter and of Basin have been given in
Chapters V and VI.

Dr Lampe from experiments on the DanUig mains and other pipes proposed

the formula

thus modifying St Tenant's formula and anticipating the formulae of Beynoldi,
Flamant and Unwin, in which,

7**

*~*> 9
u and f being variable coefficients.

* Iruite

CHAPTER VII.

*GAUGING THE FLOW OF WATER

142. Measuring the flow of water by weighing.

In the laboratory or workshop a flow of water can generally
be measured by collecting the water in tanks, and either by
direct weighing, or by measuring the volume from the known
capacity of the tank, the discharge in a given time can be
determined. This is the most accurate method of measuring
water and should be adopted where possible in experimental
work.

In pump trials or in measuring the supply of water to boilers,
determining the quantity by direct weighing has the distinct
advantage that the results are not materially affected by
changes of temperature. It is generally necessary to have two
tanks, one of which is filling while the other is being weighed
and emptied. For facility in weighing the tanks should stand
on the tables of weighing machines.

143. Meters.

Linert meter. An ingenious direct weighing meter suitable for
gauging practically any kind of liquid, is constructed as shown in
Figs. 136 and 137.

It consists of two tanks A 1 and A 2 , each of which can swing
on knife edges BB. The liquid is allowed to fall into a shoot F,
which swivels about the centre J, and from which it falls into
either A 1 or A 2 according to the position of the shoot. The tanks
have weights D at one end, which are so adjusted that when a
certain weight of water has run into a tank, it swings over into
the dotted position, Fig. 136, and flow commences through a
siphon pipe 0. When the level of the liquid in the tank has
fallen sufficiently, the weights D cause the tank to come back to
its original position, but the siphon continues in action until the
tank is empty. As the tank turns into the dotted position

* See Appendix 10.

GAUGING THE FLOW OF WATER

235

it suddenly tilts over the shoot F, and the liquid is discharged
into the other tank. An indicator H registers the number of
times the tanks are filled, and as at each tippling a definite weight
of fluid is emptied from the tank, the indicator can be marked
off in pounds or in any other unit.

Fig. 13C. Fig. 137.

Linert direct weighing meter.

144. Measuring the flow by means of an orifice.

The coefficient of discharge of sharp-edged orifices can be
obtained, with considerable precision, from the tables of Chapter IV,
or the coefficient for any given orifice can be determined for
various heads by direct measurement of the flow in a given time,
as described above. Then, knowing the coefficient of discharge at
various heads a curve of rate of discharge for the orifice, as in
Fig. 138, may be drawn, and the orifice can then be used to
measure a continuous flow of water.

The orifice should be made in the side or bottom of a tank. If
in the side of the tank the lower edge should be at least one and
a half to twice its depth above the bottom of the tank, and the
sides of the orifice whether horizontal or vertical should be at
least one and a half to twice the width from the sides of the tank.
The tank should be provided with baffle plates, or some other
arrangement, for destroying the velocity of the incoming water
and ensuring quiet water in the neighbourhood of the orifice. The
coefficient of discharge is otherwise indefinite. The head over the
orifice should be observed at stated intervals. A head-time curve
having head as ordinates and time as abscissae can then be plotted
as in Fig. 139.

From the head-discharge curve of Fig. 138 the rate of discharge
can be found for any head h, and the curve of Fig. 139 plotted.
The area of this curve between any two ordinates AB and CD,

236

HYDRAULICS

which is the mean ordinate between AB and CD multiplied by the
time t y gives the discharge from the orifice in time t.

The head h can be measured by fixing a scale, having its zero
coinciding with the centre of the orifice, behind a tube on the side
of the tank.

if^L

Fig. 138.

B Tbne

Fig. 139.

' X v

Fig. 140.

145. Measuring the flow in open channels.

Large open channels : floats. The oldest and simplest method
of determining approximately the discharge in an open channel is
by means of floats.

A part of the channel as straight as possible is selected, and in
which the flow may be considered as uniform.

The readings should be taken on a calm day as a down-stream
wind will accelerate the floats and an up-stream wind retard them.

Two cords are stretched across the channel, as near to the
surface as possible, and perpendicular to the direction of flow. The
distance apart of the cords should be as great as possible consistent
with uniform flow, and should not be less than 150 feet. From a
boat, anchored at a point not less than 50 to 70 feet above stream,
so that the float shall acquire before reaching the first line a
uniform velocity, the float is allowed to fall into the stream and

GAUGING THE FLOW OF WATER 237

the time carefully noted by means of a chronometer at which it
passes both the first and second line. If the velocity is slow, the
observer may walk along the bank while the float is moving from
one cord to the other, but if it is greater than 200 feet per minute
two observers will generally be required, one at each line.

A better method, and one which enables any deviation of the
float from a path perpendicular to the lines to be determined, is,
for two observers provided with box sextants, or theodolites, to be
stationed at the points A and B, which are in the planes of the
two lines. As the float passes the line AA at D, the observer
at A signals, and the observer at B measures the angle ABD
and, if both are provided with watches, each notes the time.
When the float passes the line BB at E, the observer at B signals,
and the observer at A measures the angle BAE, and both
observers again note the time. The distance DE can then be
accurately determined by calculation or by a scale drawing, and
the mean velocity of the float obtained, by dividing by the time.

To ensure the mean velocities of the floats being nearly equal
to the mean velocity of the particles of water in contact with
them, their horizontal dimensions should be as small as possible,
so as to reduce friction, and the portion of the float above the
surface of the water should be very small to diminish the effect of
the wind.

As pointed out in section 130, the distribution of velocity in
any transverse section is not by any means uniform and it is
necessary, therefore, to obtain the mean velocity on a number of
vertical planes, by finding not only the surface velocity, but also
the velocity at various depths on each vertical.

146. Surface floats.

Surface floats may consist of washers of cork, or wood, or
other small floating bodies, weighted so as to just project above
the water surface. The surface velocity is, however, so likely to
be affected by wind, that it is better to obtain the velocity a
short distance below the surface.

147. Double floats.

To measure the velocity at points below the surface double
floats are employed. They consist of two bodies connected by
means of a fine wire or cord, the upper one being made as small
as possible so as to reduce its resistance.

Gordon*, on the Irrawaddi, used two wooden floats connected
by a fine fisning line, the lower float being a cylinder 1 foot long,

* Proc. inst. C. E. t 1&93.

238 HYDRAULICS

and 6 inches diameter, hollow underneath and loaded with clay to
sink it to any required depth ; the upper float, which swam on the
surface, was of light wood 1 inch thick, and carried a small flag.

The surface velocity was obtained by sinking the lower float
to a depth of 3J feet, the velocity at this depth being not very
different from the surface velocity and the motion of the float more
independent of the effect of the wind.

Fig. 141. Gurley's current meter.

Subsurface velocities were measured by increasing the depths
of the lower float by lengths of 3$ feet until the bottom was
reached.

GAUGING THE FLOW OF WATER 239

Gordon has compared the results obtained by floats with those
obtained by means of a current meter (see section 149). For
small depths and low velocities the results obtained by double
floats are fairly accurate, but at high velocities and great depths,
the velocities obtained are too high. The error is from to 10
per cent.

Double floats are sometimes made with two similar floats, of
the same dimensions, one of which is ballasted so as to float at any
required depth and the other floats just below the surface. The
velocity of the float is then the mean of the surface velocity
and the velocity at the depth of the lower float.

148. Rod floats.

The mean velocity, on any vertical, may be obtained ap-
proximately by means of a rod float, which consists of a long rod
having at the lower end a small hollow cylinder, which may be
filled with lead or other ballast so as to keep the rod nearly
vertical.

The rod is made sufficiently long, and the ballast adjusted, so
that its lower end is near to the bed of the stream, and its upper
end proj ects slightly above the water. Its velocity is approximately
the mean velocity in the vertical plane in which it floats.

149. The current meter.

The discharge of large channels or rivers can be obtained most
conveniently and accurately by determining the velocity of flow
at a number of points in a transverse section by means of a current
meter.

The arrangement shown in Fig. 141 is a meter of the anemo-
meter type. . A wheel is mounted on a vertical spindle and has
five conical buckets. The spindle revolves in bearings, from
which all water is excluded, and which are carefully made so
that the friction shall remain constant. The upper end of the
spindle extends above its bearing, into an air-tight chamber, and
is shaped to form an eccentric. A light spring presses against
the eccentric, and successively makes and breaks an electric
circuit as the wheel revolves. The number of revolutions of the
wheel is recorded by an electric register, which can be arranged
at any convenient distance from the wheel. When the circuit is
made, an electro-magnet in the register moves a lever, at the end
of which is a pawl carrying forward a ratchet wheel one tooth
for each revolution of the spindle. The frame of the meter, which
is made of bronze, is pivoted to a hollow cylinder which can be
clamped in any desired position to a vertical rod. At the right-

240 HYDRAULICS

hand side is a rudder having four light metal wings, which
balances the wheel and its frame. When the meter is being used
in deep waters it is suspended by means of a fine cable, and to
the lower end of the rod is fixed a lead weight. The electric
circuit wires are passed through the trunnion and so have no
tendency to pull the meter out of the line of current. When
placed in a current the meter is free to move about the horizontal
axis, and also about a vertical axis, so that it adjusts itself to
the direction of the current.

The meters are rated by experiment and the makers recommend
the following method. The meter should be attached to the bow
of a boat, as shown in Fig. 142, and immersed in still water not
less than two feet deep. A thin rope should be attached to the
boat, and passed round a pulley in line with the course in which
the boat is to move. Two parallel lines about 200 feet apart
should be staked on shore and at right angles to the course of the
boat. The boat should be without a rudder, but in the boat with
the observer should be a boatman to keep the boat from running

Fig. 142.

into the shore. The boat should then be hauled between the two
ranging lines at varying speeds, which during each passage should
be as uniform as possible. With each meter a reduction table is
supplied from which the velocity of the stream in feet per second
can be at once determined from the number of revolutions recorded
per second of the wheel.

The Haskell meter has a wheel of the screw propeller type
revolving upon a horizontal axis. Its mode of action is very
similar to the one described.

Comparative tests of the discharges along a rectangular canal
as measured by these two meters and by a sharp-edged weir which
had been carefully calibrated, in no case differed by more than
5 per cent, and the agreement was generally much closer*.

* Murphy on current Meter and Weir discharges, Proceedings Am.S.C.E.,
VoL xxvii., p. 779.

GAUGING THE FLOW OF WATER

241

150. *Pitot tube.

Another apparatus which can be used for determining the
velocity at a point in a flowing stream, even when the stream is of
small dimensions, as for example a small pipe, is called a Pitot
tube.

In its simplest form, as originally proposed by Pitot in 1732,
it consists of a glass tube, with a
small orifice at one end which may
be turned to receive the impact of
the stream as shown in Fig. 143.
The water in the tube rises to a
height h above the free surface of
the water, the value of h depending
upon the velocity v at the orifice of
the tube. If a second tube is placed

Fig. 143. Pitot tube.

beside the first with an orifice parallel to the direction of flow,
the water will rise in this tube nearly to the level of the free
surface, the fall hi being due to a slight diminution in pressure
at the mouth of the tube, caused probably by the stream lines
having their directions changed at the mouth of the tube. A
further depression of the free surface in the tube takes place,
if the tube, as EF, is turned so that the orifice faces down stream.

Theory of the Pitot tube. Let v be the velocity of the stream
at the orifice of the tube in ft. per sec. and a the area of the
orifice in sq. ft.

The quantity of water striking the orifice per second is wav
pounds.

The momentum is therefore - . a . v* pounds feet.

If the momentum of this water is entirely destroyed, the
pressure on the orifice which, according to Newton's second law of
motion is equal to the rate of change of momentum, is

P =

wav

and the pressure per unit area is

wv 2

9
The equivalent head

&=-- =- .
wg 9

According to this theory, the head of water in the tube, due to
the impact, is therefore twice ~- , the head due to the velocity v, and

' * See page 526.

L. H.

242

HYDRAULICS

the water should rise in the tube to a height above the surface
equal to h. Experiment shows however that the actual height
the water rises in the tube is practically equal to the velocity
head and, therefore, the velocity v of a mass of water w . a . v Ibs.
is not destroyed by the pressure on the area of the tube. The
head h is thus generally taken (see Appendix 4) as

, cv*

c being a coefficient for the tube, which experiment shows for
a properly formed tube is constant and practically equal to
unity.

f* '?*^ C 75

Similarly for given tubes hi =-~- and 7& 2 = -ST

The coefficients are determined by placing the tubes in streams
the velocities of which are known, or by attaching them to some
body which moves through still water with a known velocity, and
carefully measuring h for different velocities.

Fig. 144.

Fig. 145.

Darcy* was the first to use the Pitot tube as an instrument of
precision. His improved apparatus as used in open channels con-
sisted of two tubes placed side by side as in Fig. 144, the orifices
in the tubes facing up-stream and down-stream respectively. The

* Recherche* Hydrauliques, etc., 1857.

GAUGING THE FLOW OF WATER 243

two tubes were connected at the top, a cock C 1 being placed in the
common tube to allow the tubes to be opened or closed to the
atmosphere. At the lower end both tubes could be closed at the
same time by means of cock C. When the apparatus is put into
flowing water, the cocks C and C 1 being open, the free surface
rises in the tube B a height hi and is depressed in D an amount
hi. The cock C 1 is then closed, and the apparatus can be taken
from the water and the difference in the level of the two columns,

h = hi + hq,
measured with considerable accuracy.

If desired, air can be aspirated from the tubes and the columns
made to rise to convenient levels for observation, without moving
the apparatus. The difference of level will be the same, whatever
the pressure in the upper part of the tubes.

Fig. 145 shows one of the forms of Pitot tubes, as experimented
upon by Professor Gardner Williams*, and used to determine
the distribution of velocities of the water flowing in circular pipes.

The arrangement shown in Fig. 146, is a modified form of the
apparatus used by Freeman t to determine the distribution of
velocities in a jet of water issuing from a fire hose under con-
siderable pressure. As shown in the sketch, the small orifice
receives the impact of the stream and two small holes Q are drilled
in the tube T in a direction perpendicular to the flow. The lower
part of the apparatus OY, as shown in the sectional plan, is made
boat-shaped so as to prevent the formation of eddies in the
neighbourhood of the orifices. The pressure at the orifice is
transmitted through the tube OS, and the pressure at Q through
the tube QR. To measure the difference of pressure, or head,
in the two tubes, OS and QE were connected to a differential
gauge, similar to that described in section 13 and very small
differences of head could thus be obtained with great accuracy.

The tube shown in Fig. 145 has a cigar-shaped bulb, the
impact orifice being at one end and communicating with the
tube OS. There are four small openings in the side of the bulb,
so that any variations of pressure outside are equalised in the
bulb. The pressures are transmitted through the tubes OS and
TR to a differential gauge as in the case above.

In Fig. 147 is shown a special stuffing-box used by Professor
Williams, to allow the tube to be moved to the various positions in

* For other forms of Pitot tubes as used by Professor Williams, E. S. Cole and
others, see Proceedings of the Am.S.C.E., Vol. xxvn.
f Transactions of the Am.S.C.E., Vol. xxi.

162

244

HYDRAULICS

the cross section of a pipe, at which it was desired to determine
the velocity of translation of the water*.

Mr E. S. Colet has used the Pitot tube as a continuous meter,
the arrangement being shown in Fig. 148. The tubes were con-
nected to a U tube containing a mixture of carbon tetrachloride
and gasoline of specific gravity 1*25. The difference of level of
the two columns was registered continuously by photography.

Gauge

Fig. 147.

Fig. 146.

Fig. 148.

The tubes shown in Figs. 149 150, were used by Bazin to
determine the distribution of velocity in the interior of jets issuing

* See page 144.

t Proc. A.M.S.C.E., Vol. xxvii. See also experiments by Murphy and Torranee
in same volume.

GAUGING THE FLOW OF WATER

245

from orifices, and in the interior of the nappes of weirs. Each
tube consisted of a copper plate 1*89 inches wide, by 1181 inch
thick, sharpened on the upper edge and having two brass tubes
'0787 inch diameter, soldered along the other edge, and having
orifices "059 inch diameter, 0'394 inch apart. The opening in tube
A was arranged perpendicular to the stream, and in B on the face
of the plate parallel to the stream.

Fig. 149.

Fig. 150.

151. Calibration of Pitot tubes.

Whatever the form of the Pitot tube, the head h can be
expressed as

k being called the coefficient of the tube. This coefficient k in
special cases may have to be determined by experiment, but, as
remarked above, for tubes carefully made and having an impinging
surface which is a surface of revolution it is unity.

To calibrate the tubes used in the determination of the distri-
bution of velocities in open channels, Darcy* and Bazin used three
distinct methods.

(a) The tube was placed in front of a boat which was drawn
through still water at different velocities. The coefficient was
T034. This was considered too large as the bow of the boat
probably tilted a little, as it moved through the water, thus tilting
the tube so that the orifice was not exactly vertical.

(6) The tube was placed in a stream, the velocity of which
was determined by floats. The coefficient was 1*006.

(c) Readings were taken at different points in the cross
section of a channel, the total flow Q through which was carefully
measured by means of a weir. The water section was divided

246 HYDRAULICS

into areas, and about the centre of each a reading of the tube
was taken. Calling a the area of one of these sections, and h the
reading of the tube, the coefficient

and was found to be "993.

Darcy* and Bazin also found that by changing the position of
the orifice in the pressure tube the coefficients changed con-
siderably.

Williams, Hubbell and Fenkell used two methods of calibration
which gave very different results.

The first method was to move the tubes through still water at
known velocities. For this purpose a circumferential trough,
rectangular in section, 9 inches wide and 8 inches deep was built of
galvanised iron. The diameter of its centre line, which was made
the path of the tube, was 11 feet 10 inches. The tube to be rated
was supported upon an arm attached to a central shaft which was
free to revolve in bearings on the floor and ceiling, and which also
supported the gauge and a seat for the observer. The gauge was
connected with the tube by rubber hose. The arm carrying the
tube was revolved by a man walking behind it, at as uniform a
rate as possible, the time of the revolution being taken by means
of a watch reading to ^ of a second. The velocity was main-
tained as nearly constant as possible for at least a period of
5 minutes. The value of k as determined by this method was '926
for the tube shown in Fig. 145.

In the second method adopted by these workers, the tube was
inserted into a brass pipe 2 inches in diameter, the discharge
through which was obtained by weighing. Readings were taken
at various positions on a diameter of the pipe, while the flow in the
pipe was kept constant. The values of \/2gh t which may be called
the tube velocities, could then be calculated, and the mean value
V m of them obtained. It was found that, in the cases in which the
form of the tube was such that the volume occupied by it in the pipe
was not sufficient to modify the flow, the velocity was a maximum
at, or near, the centre of the pipe. Calling this maximum velocity

V c , the ratio W 2 for a given set of readings was found to be '81.

Vc
Previous experiments on a cast-iron pipe line at Detroit having

shown that the ratio Jt 2 was practically constant for all velocities,

a similar condition was assumed to obtain in the case of the brass
Recherches Hydrauli^ues,

GAUGING THE FLOW OF WATER 247

pipe. The tube was then fixed at the centre of the pipe, and
readings taken for various rates of discharge, the mean velocity
U, as determined by weight, varying from J to 6 feet per second.

For the values of h thus determined, it was found that

was practically constant. This ratio was '729 for the tube shown
in Fig. 145.

Then since for any reading h of the tube, the velocity v is

the actual mean velocity
r

7 _ ratio of U to Y c _'729_ Q0
""'-"

But
Therefore

For the tube shown in Fig. 146, some of the values of Tc ae
determined by the two methods differed very considerably.

It will be seen that the value of k determined by moving the
tube through still water, according to the above results, differs
from that obtained from the running water in a pipe. Other
experiments, however, on tubes the coefficients for which were
obtained by moving through still water and by being placed in
jets of water issuing from sharp-edged orifices, show that the
coefficient is unity in both cases. Professor Gregory* using a
tube (Fig. 373, Appendix 4), consisting of an impact tube i inch
diameter surrounded by a tapering tube of larger diameter in which
were drilled the static openings at a mean distance of 12'5 inches
from the impact opening, found that the coefficient was unity
when moved through still water, or when it was placed in flowing
water in a pipe. With tubes having impact openings of the form
shown in Fig. 144, or in Figs. 371 and 372, and the pressure
openings well removed from the influence of eddy motions it may
be taken that the coefficient is unity, and a properly designed Pitot
tube can with care, therefore, be used with confidence to measure
velocities of flow.

152. Gauging by a weir.

When a stream is so small that a barrier or dam can be easily
constructed across it, or when a large quantity of water is required
to be gauged in the laboratory, the flow can be determined by
means of a notch or weir.

* See Appendix 4, p. 528 ; Trans. 4m. S,M.E. 1904,

248

HYDRAULICS

The channel as it approaches the weir should be as far as
possible uniform in section, and it is desirable for accurate
gauging*, that the sides of the channel be made vertical, and the
width equal to the width of the weir. The sill should be sharp-
edged, and perfectly horizontal, and as high as possible above the
bed of the stream, and the down-stream channel
should be wider than the weir to ensure atmospheric
pressure under the nappe. The difference in level
of the sill and the surface of the water, before it
begins to slope towards the weir, should be ac-
curately measured. This is best done by a Boyden
hook gauge.

153. The hook gauge.

A simple form of hook gauge as made by G-urley
is shown in Fig. 151. In a rectangular groove formed
in a frame of wood, three or four feet long, slides
another piece of wood S to which is attached a scale
graduated in feet and hundredths, similar to a level
staff. To the lower end of the scale is connected a
hook H, which has a sharp point. At the upper end
of the scale is a screw T which passes through a lug,
connected to a second sliding piece L. This sliding
piece can be clamped to the frame in any position
by means of a nut, not shown. The scale can then
be moved, either up or down, by means of the milled
nut. A vernier V is fixed to the frame by two small
screws passing through slot holes, which allow for a
slight adjustment of the zero. At some point a few
feet up-stream from the weir*, the frame can be
fixed to a post, or better still to the side of a box
from which a pipe runs into the stream. The level
of the water in the box will thus be the same as the
level in the stream. The exact level of the crest of
the weir must be obtained by means of a level and a
line marked on the box at the same height as the
crest. The slider L can be moved, so that the hook
point is nearly coincident with the mark, and the
final adjustment made by means of the screw T.
The vernier can be adjusted so that its zero is
coincident with the zero of the scale, and the slider
again raised until the hook approaches the surface of the water.
By means of the screw, the hook is raised slowly, until, by piercing

v.m-

* See section 82.

GAUGING THE FLOW OF WATER

249

Fig. 152. Bazin's Hook Gauge.

250

HYDRAULICS

the surface of the water, it causes a distortion of the light reflected
from the surface. On moving the hook downwards again very
slightly, the exact surface will be indicated when the distortion
disappears.

A more elaborate hook gauge, as used by Bazin for his experi-
mental work, is shown in Fig. 152.

For rough gaugings a post can be driven into the bed of the
channel, a few feet above the weir, until the top of the post is
level with the sill of the weir. The height of the water surface

Fig. 154. .Recording Apparatus Kent Venturi Meter,

GAUGING THE FLOW OF WATER

251

above the top of the post can then be measured by any convenient
scale.

154. Gauging the flow in pipes; Venturi meter.

Such methods as already described are inapplicable to the
measurement of the flow in pipes, in which it is necessary that
there shall be no discontinuity in the flow, and special meters have
accordingly been devised.

For large pipes, the Venturi meter, Fig. 153, is largely used in
America, and is coming into favour in this country.

The theory of the meter has already been discussed (p. 44),
and it was shown that the discharge is proportional to the square
root of the difference H of the head at the throat and the head in
the pipe, or

Jc* being a coefficient.

For measuring the pressure heads at the two ends of the cone,
Mr W. G. Kent uses the arrangement shown in Fig. 154.

JFig. 155. Recording drum of the Kent Venturi Meter.
* See page 46.

252

HYDRAULICS

The two pressure tubes from the meter are connected to a U tube
consisting of two iron cylinders containing mercury. Upon the
surface of the mercury in each cylinder is a float made of iron and
vulcanite; these floats rise or fall with the surfaces of the mercury.

Fig. 156. Integrating drum of the Kent Venturi Meter.

When no water is passing through the meter, the mercury in the
two cylinders stands at the same level. When flow takes place
the mercury in the left cylinder rises, and that in the right
cylinder is depressed until the difference of level of the surfaces

GAUGING THE FLOW OF WATER

253

of the mercury is equal to j^j, s being the specific gravity of the

mercury and H the difference of pressure head in the two
cylinders. The two tubes are equal in diameter, so that the rise
in the one is exactly equal to the fall in the other, and the move-
ment of either rack is proportional to H. The discharge is
proportional to \/H, and arrangements are made in the recording
apparatus to make the revolutions of the counter proportional to
\/H. To the floats, inside the cylinders, are connected racks, as
shown in Fig. 154, gearing with small pinions. Outside the
mercury cylinders are two other racks, to each of which vertical
motion is given by a pinion fixed to the same spindle as the pinion
gearing with the rack in the cylinder. The rack outside the left
cylinder has connected to it a light pen carriage, the pen of which

Fig. 157. Kent Venturi Meter. Development of Integrating drum.

makes a continuous record on the diagram drum shown in
Fig. 155. This drum is rotated at a uniform rate by clockwork,
and on suitably prepared paper a curve showing the rate of
discharge at any instant is thus recorded. The rack outside the
right cylinder is connected to a carriage, the function of which is
to regulate the rotations of the counter which records the total
flow. Concentric with the diagram drum shown in Fig. 155, and
within it, is a second drum, shown in Fig. 156, which also rotates
at a uniform rate. Fig. 157 shows this internal drum developed.
The surface of the drum below the parabolic curve FEGr is recessed.
If the right-hand carriage is touching the drum on the recessed

254

HYDRAULICS

portion, the counter gearing is in action, but is put out of action
when the carriage touches the cylinder on the raised portion
above FGr. Suppose the mercury in the right cylinder to fall a
height proportional to H, then the carriage will be in contact
with the drum, as the drum rotates, along the line CD, but the
recorder will only be in operation while the carriage is in
contact along the length CE. Since FGr is a parabolic curve the
fraction of the circumference CE = ra . \/H, ra being a constant,
and therefore for any displacement H of the floats the counter for
each revolution of the drum will be in action for a period propor-
tional to \/H. When the float is at the top of the right cylinder,
the carriage is at the top of the drum, and in contact with the
raised portion for the whole of a revolution and no flow is
registered. When the right float is in its lowest position the
carriage is at the bottom of the drum, and flow is registered
during the whole of a revolution. The recording apparatus can
be placed at any convenient distance less than 1000 feet from
the meter, the connecting tubes being made larger as the distance
is increased.

155. Deacon's waste- water meter.

An ingenious and very simple meter designed by Mr Gr. F.
Deacon principally for detecting the leakage of water from pipes
is as shown in Fig. 158.

Fig. 158. Deacoii waste-water meter.

The body of the meter which is made of cast-iron, has fitted
into it a hollow cone C made of brass. A disc D of the same diameter
as the upper end of the cone is suspended in this cone by means of
a fine wire, which passes over a pulley not shown; the other end
of the wire carries a balance weight.

GAUGING THE FI,OW OF WATER 255

When no water passes through the meter the disc is drawn to
the top of the cone, but when water is drawn through, the disc is
pressed downwards to a position depending upon the quantity of
water passing. A pencil is attached to the wire, and the motion
of the disc can then be recorded upon a drum made to revolve by
clockwork. The position of the pencil indicates the rate of flow
passing through the meter at any instant.

When used as a waste-water meter, it is placed in a by-pass
leading from the main, as shown diagrammatically in Fig, 159.

"tBr

S.V.

s.v

Fig. 159.

The valve A is closed and the valve C opened. The rate of
consumption in the pipe AD at those hours of the night when the
actual consumption is very small, can thus be determined, and an
estimate made as to the probable amount wasted.

If waste is taking place, a careful inspection of the district
supplied by the main AD may then be made to detect where the
waste is occurring.

156. Kennedy's meter.

This is a positive meter in which the volume of water passing
through the meter is measured by the displacement of a piston
working in the measuring cylinder.

The long hollow piston P, Fig. 160, fits loosely in the cylinder
Co, but is made water-tight by means of a cylindrical ring of
rubber which rolls between the piston and the inside of the
cylinder, the friction being thus reduced to a minimum. At each
end of the cylinder is a rubber ring, which makes a water-tight
joint when the piston is forced to either end of the cylinder, so
that the rubber roller has only to make a joint while the piston is
free to move.

The water enters the meter at A, Fig. 161 5, and for the
position shown of the regulating cock, it flows down the passage
D and under the piston. The piston rises, and as it does so the
rack E/ turns the pinion S, and thus the pinion p which is keyed
to the same spindle as S. This spindle also carries loosely
a weighted lever W, which is moved as the spindle revolves by
either of two projecting fingers. As the piston continues to
ascend, the weighted lever is moved by one of the fingers until its

256

HYDRA LTL1CS

centre of gravity passes the vertical position, when it suddenly
falls on to a buffer, and in its motion moves the lever L, which
turns the cock, Fig. 161 6, into a position at right angles to that

Rubber Seating

Rubber Rclting
Pcuckmg

Robber Seating

rig. ico.

GAUGING THE FLOW OF WATER

257

shown. The water now passes from A through the passage C,
and thus to the top of the cylinder, and as the piston descends'

Fig. 161o.

L. II.

258

HYDRAULICS

the water that is below it passes to the outlet B. The motion of
the pinion S is now reversed, and the weight W lifted until it
again reaches the vertical position, when it falls, bringing the
cock C into the position shown in the figure, and another up

Fig. 161 c.

stroke is commenced. The oscillations of the pinion p are trans-
ferred to the counter mechanism through the pinions p t and p 2 ,
Fig. 161 a, in each of which is a ratchet and pawl. The counter
is thus rotated in the same direction whichever way the piston
moves.

157. Gauging the flow of streams by chemical means.

Mr Stromeyer* has very successfully gauged the quantity of
water supplied to boilers, and also
the flow of streams by mixing
with the stream during a definite
time and at a uniform rate, a
known quantity of a concentrated
solution of some chemical, the
presence of which in water, even
in very small quantities, can be
easily detected by some sensitive
reagent. Suppose for instance
water is flowing along a small
stream. Two stations at a known
distance apart are taken, and the
time determined which it takes
the water to traverse the dis-
tance between them. At a stated
time, by means of a special ap-
paratus Mr Stromeyer uses the
arrangement shown in Fig. 162
sulphuric acid, or a strong salt
solution, say, of known strength, is run into the stream at a known

* Transactions of Naval Architects, 1896; Proceedings Inst. C.E., Vol. CLX. and
" Jaugeages par Titrations " by Collet, Mellet and Liitschg. Swiss Bureau of
Hydrography.

Fig. 162.

GAUGING THE FLOW OF WATEB, 259

rate, at the upper station. While the acid is being put into the
stream, a small distance up-stream from where the acid is introduced
samples of water are taken at definite intervals. At the lower
station sampling is commenced, at a time, after the insertion of the
acid at the upper station is started, equal to that required by the
water to traverse the distance between the stations, and samples
are then taken, at the same intervals, as at the upper station.
The quantity of acid in a known volume of the samples taken
at the upper and lower station is then determined by analysis.
In a volume Y of the samples, let the difference in the amount of
sulphuric acid be equivalent to a volume v & of pure sulphuric
acid. If in a time t, a volume Y of water, has flowed down the
stream, and there has been mixed with this a volume v of pure
sulphuric acid, then, if the acid has mixed uniformly with the
water, the ratio of the quantity of water flowing down the stream
to the quantity of acid put into the stream, is the same as the
ratio of the volume of the sample tested to the difference of the
volume of the acid in the samples at the two stations, or

Mr Stromeyer considers that the flow in the largest rivers can
be determined by this method within one per cent, of its true value.

In large streams special precautions have to be taken in
putting the chemical solution into the water, to ensure a uniform
mixture, and also special precautions must be adopted in taking
samples.

For other important information upon this interesting method
of measuring the flow of water the reader is referred to the papers
cited above.

An apparatus for accurately gauging the flow of the solution
is shown in Fig. 162. The chemical solution is delivered into
a cylindrical tank by means of a pipe I. On the surface of the
solution floats a cork which carries a siphon pipe SS, and a balance
weight to keep the cork horizontal. After the flow has been
commenced, the head h above the orifice is clearly maintained
constant, whatever the level of the surface of the solution in the
tank.

172

260

HYDKAULICS

EXAMPLES.

(1) Some observations are made by towing a current meter, with the
following results :

Speed in ft. per sec.
1
5
Find an equation for the meter.

Revs, of meter per min.

560

(2) Describe two methods of gauging a large river, from observations
in vertical and horizontal planes; and state the nature of the results
obtained.

If the cross section of a river is known, explain how the approximate
discharge may be estimated by observation of the mid-surface velocity
alone.

(8) The following observations of head and the corresponding discharge
were made in connection with a weir 6'53 feet wide.

Head in feet ... ... O'l

Discharge in cubic feet per

sec. per foot width

0-17

0-5
1-2

1-0

3-35

1-5

6-1

2-0
9-32

2-5

13-03

3-0

17-03

3-5

21-54

4-0

26-4

Assuming the law connecting the head h with the discharge Q as

Q=mL.7i n ,
find ra and n. (Plot logarithms of Q and h.)

(4) The following values of Q and h were obtained for a sharp-edge
weir 6'53 feet long, without lateral contraction. Find the coefficient of
discharge at various heads.

Head h ...
Q per foot-
length ...

1-56

2-37

1-0
3-35

2-0

9-32

2-5 I 3-0

13-03 17-03

3-5
21-54

4-0

26-4

4-5
31-62

5-0

37-09

5-5

42-81

(5) The following values of the head over a weir 10 feet long were
obtained at 5 minutes intervals.

Head in feet '35 -36 '37 '37 '38 -39 '40 -41 -42 -40 '39 -41
Taking the coefficient of discharge C as 3'36, find the discharge in
one hour.

(6) A Pitot tube was calibrated by moving it through still water in a
tank, the tube being fixed to an arm which was made to revolve at
constant speed about a fixed centre. The following were the velocities of
the tube and the heads measured in inches of water.

Velocities ft. per sec. 1*432
Head in inches

of water '448

1-738
663

2-275
1-02

2-713
1-69

3-235
2-07

3-873

2-88 5-40

4-983

5-584

6-142

6-97 18-51

Determine the coefficient of the tube.

For examples on Venturi meters see Chapter II,

CHAPTER VIII.

IMPACT OF WATER ON VANES.

158. Definition of a vector. A right line AB, considered as
having not only length, but also direction, and sense, is said to be
a vector*. The initial point A is said to be the origin.

It is important that the difference between sense and direction
should be clearly recognised.

Suppose for example, from any point A, a line AB of
definite length is drawn in a northerly direction, then the
direction of the line is either from south to north or north to
south, but the sense of the vector is definite, and is from A to B,
that is from south to north.

The vector AB is equal in magnitude to the vector BA, but
they are of opposite sign or,

AB = -BA.

The sense of the vector is indicated by an arrow, as on AB,
Fig. 163.

Any quantity which has magnitude, direction, and sense, may
be represented by a vector.

D
c ,

Fig. 163.

For example, a body is moving with a given velocity in a
given direction, sense being now implied. Then a line AB drawn
parallel to the direction of motion, and on some scale equal in

* Sir W. Hamilton, Quaternions.

262

HYDRAULICS

length to the velocity of the body is the velocity vector ; the sense
is from A to B.

159. * Sum of two vectors.

If a and /?, Fig. 163, are two vectors the sum of these vectors
is found, by drawing the vectors, so that the beginning of ft is at
the end of a, and joining the beginning of a to the end of ft.
Thus y is the vector sum of a and ft.

160. Resultant of two velocities.

When a body has impressed upon it at any instant two
velocities, the resultant velocity of the body in magnitude and
direction is the vector sum of the two impressed velocities. This
may be stated in a way that is more definitely applicable to the
problems to be hereafter dealt with, as follows. If a body is
moving with a given velocity in a given direction, and a second
velocity is impressed upon the body, the resultant velocity is the
vector sum of the initial and impressed velocities.

Example. Suppose a particle of water to be moving along a vane DA, Fig. 164,
with a velocity V r , relative to the vane.

If the vane is at rest, the particle will leave it at A with this velocity.

If the vane is made to move in the direction EF with a velocity v, and the
particle has still a velocity V r relative to the vane, and remains in contact with the
vane until the point A is reached, the velocity of the water as it leaves the vane at
A, will be the vector sum 7 of a and p, i.e. of V r and v, or is equal to u.

161. Difference of two vectors.

The difference of two vectors a and ft is found by drawing both
vectors from a common origin A, and joining the end of ft to the
end of a. Thus, CB, Fig. 165, is the difference of the two vectors
a and ft or y = a /3, and BC is equal to ft - a, or ft - a = - y.

162. Absolute velocity.

By the terms " absolute velocity " or " velocity " without the
adjective, as used in this chapter, it should be clearly understood,
is meant the velocity of the moving water relative to the earth, or
to the fixed part of any machine in which the water is moving.

Ilenrici and Turner, Vectors and Rotors.

IMPACT OP WATER ON VANES 263

To avoid repetition of the word absolute, the adjective is
frequently dropped and " velocity " only is used.

163. When a body is moving with a velocity U, Fig. 166, in
any direction, and has its velocity changed to U' in any other
direction, by an impressed force, the change in velocity, or the
velocity that is impressed on the body, is the vector difference of
the final and the initial velocities. If AB is U, and AC, U', the
impressed velocity is BC.

By Newton's second law of motion, the resultant impressed
force is in the direction of the change of velocity, and if W is the
weight of the body in pounds and t is the time taken to change
the velocity, the magnitude of the impressed force is

P = (change of velocity) Ibs.
Qt

This may be stated more generally as follows.
The rate of change of momentum, in any direction, is equal to
the impressed force in that direction, or

r> W <fc>u,
P = .37 Ibs.

g dt

In hydraulic machine problems, it is generally only necessary
to consider the change of momentum of the mass of water that
acts upon the machine per second. W in the above equation then
becomes the weight of water per second, and t being one second,

P = (change of velocity).
y

164. Impulse of water on vanes.

It follows that when water strikes a vane which is either
moving or at rest, and has its velocity changed, either in magni-
tude or direction, pressure is exerted on the vane. .

As an example, suppose in one second a mass of water, weighing
W Ibs. and moving with a velocity U feet per second, strikes a
fixed vane AD, and let it glide upon the vane at A, Fig. 167, and
leave at D in a direction at right angles to its original direction
of motion. The velocity of the water is altered in direction but
not in magnitude, the original velocity being changed to a velocity
at right angles to it by the impressed force the vane exerts upon
the water.

The change of velocity in the direction AC is, therefore,

equal to U, and the change of momentum per second is .U
foot Ibs.

264.

HYDRAULICS

Since W Ibs. of water strike the vane per second, the pressure
P, acting in the direction CA, required to hold the vane in position
is, therefore,

Fig. 167.

Again, the vane has impressed upon the water a velocity U in
the direction DF which it originally did not possess.
The pressure PI in the direction DF is, therefore,

The resultant reaction of the vane in magnitude and direction
is, therefore, E, the resultant of P and PI.

This resultant force could have been
found at once by finding the resultant
change in velocity. Set out ac, Fig. 168,
equal to the initial velocity in magnitude
and direction, and ad equal to the , final
velocity. The change in velocity is the
vector difference cd, or cd is the velocity
that must be impressed on a particle of
water to change its velocity from ac to
ad.

Fig. 168.

The impressed velocity cd is V = \/U 2 + U 2 , and the total
impressed force is

-^To N/2W ,

IMPACT OF WATER ON VANES 265

It at once follows, that if a jet of water strikes a fixed plane
perpendicularly, with a velocity U, and glides along the plane, the

normal pressure on the plane is U.

Example. A stream of water 1 sq. foot in section and having a velocity of
10 feet per second glides on to a fixed vane in a direction making an angle of
30 degrees with a given direction AB.

The vane turns the jet through an angle of 90 degrees.

Find the pressure on the vane in the direction parallel to AB and the resultant
pressure on the vane.

In Fig. 167, AC is the original direction of the jet and DF the final direction.
The vane simply changes the direction of the water, the final velocity being equal
to the initial velocity.

The vector triangle is acd, Fig. 168, ac and ad being equal.

The change of velocity in magnitude and direction is cd, the vector difference of
ad and ac ; resolving cd parallel to, and perpendicular to AB, ce is the change of
velocity parallel to AB.

Scaling off ce and calling it v lt the force to be applied along BA to keep the
vane at rest is,

But cd=j2.10

and ce = cdcosl5

-J2. 10. 0-9659;

therefore, PBA= "o o x 13 ' 65

Oif*9)

= 264 Ibs.
The pressure normal to AB is

- ^ . 10 sin 15 =72 Ibs.

, .. x . 10.62-4 , 100 J2. 62-4 (
The resultant is B= Q0 cd= ^r = 274 Ibs.

O4'4 O&'a

165. Relative velocity.

Before going on to the consideration of moving vanes it is
important that the student should have clear ideas as to what is
meant by relative velocity.

A train is said to have a velocity of sixty miles an hour when,
if it continued in a straight line at a constant velocity for one
hour, it would travel sixty miles. What is meant is that the train
is moving at sixty miles an hour relative to the earth.

If two trains run on parallel lines in the same direction, one
at sixty and the other at forty miles an hour, they have a
relative velocity to each other of 20 miles an hour. If they move
in opposite directions, they have a relative velocity of 100 miles
an hour. If one of the trains T is travelling in the direction AB,
Fig. 169, and the other Ti in the direction AC, and it be supposed
that the lines on which they are travelling cross each other at A,

266

HYDRAULICS

and the trains are at any instant over each other at A, at the end

of one minute the two trains will be at B and C respectively, at

distances of one mile and two-thirds of a

mile from A, Relatively to the train T

moving along AB, the train TI moving

along AC has, therefore, a velocity equal

to BC, in magnitude and direction, and

relatively to the train TI the train T has

a velocity equal to CB. But AB and AC

may be taken as the vectors of the two

velocities, and BC is the vector difference

of AC and AB, that is, the velocity of

vector difference of AC and AB.

Fig. 169.
relative to T is the

166. Definition of relative velocity as a vector.

If two bodies A and B are moving with given velocities v and
i in given directions, the relative velocity of A to B is the vector
difference of the velocities v and Vi .

Thus when a stream of water strikes a moving vane the
magnitude and direction of the relative velocity of the water and
the vane is the vector difference of the velocity of the water and
the edge of the vane where the water meets it.

167. To find the pressure on a moving vane, and the
rate of doing work.

A jet of water having a velocity U strikes a flat vane, the
plane of which is perpendicular to the direction of the jet, and
which is moving in the same direction as the jet with a velocity v,

Fig. 170.

Fig. 171.

The relative velocity of the water and the vane is U - v, tho
vector difference of U and v, Fig. 170. If the water as it strikes
the vane is supposed to glide along it as in Fig. 171, it will do

IMPACT OF WATER ON VANES 267

so with a velocity equal to (U v), and as it moves with the vane
it will still have a velocity v in the direction of motion of the
vane. Instead of the water gliding along the vane, the velocity
U-v may be destroyed by eddy motions, but the water will still
have a velocity v in the direction of the vane. The change in
velocity in the direction of motion is, therefore, the relative
velocity U-v, Fig. 170.

For every pound of water striking the vane, the horizontal

change in momentum is - - , and this equals the normal pressure

P on the vane, per pound of water striking the vane.
The work done per second per pound is

9
The original kinetic energy of the jet per pound of water

U 2
striking the vane is -~- , and the efficiency of the vane is, therefore,

U 2 '

which is a maximum when v is |TJ, and e = J. An application of
such vanes is illustrated in Fig. 185, page 292.

Nozzle and single vane. Let the water striking a vane issue
from a nozzle of area a, and suppose that there is only one vane.

Let the vane at a given instant be supposed at A, Fig. 172. At
the end of one second the front of the jet, if perfectly free to
move, would have arrived at B and the vane at C. Of the water
that has issued from the jet, therefore, only the quantity BC will
have hit the vane.

! U ---- V-~ - -H! i

T C[

j<_ . JJ H

Fig. 172.
The discharge from the nozzle is

W = 62'4.a.U,

and the weight that hits the vane per second is

W.QJ-'u)

U
The change of momentum per second is

268 HYDRAULICS

and the work done is, therefore,

U.g

Or the work done per Ib. of water issuing from the nozzle is

U.g

hypothetical

case and has no practical

This is purely
importance.

Nozzle and a number of vanes. If there are a number of
vanes closely following each other, the whole of the water issuing
from the nozzle hits the vanes, and the work done is

W(U-v)v

The efficiency is

2v (U - v)
IP

and the maximum efficiency is

It follows that an impulse water wheel, with radial blades, as
in Fig. 185, cannot have an efficiency of more than 50 per cent.

168. Impact of water on a vane when the directions of
motion of the vane and jet are not parallel.

Let U be the velocity of a jet of water and AB its direction,
Fig. 173.

Fig. 173.

Let the edge A of the vane AC be moving with a velocity v ;
the relative velocity V r of the water and the vane at A is DB.
From the triangle DAB it is seen that, the vector sum of the
velocity of the vane and the relative velocity of the jet and the
vane is equal to the velocity of the jet; for clearly U is the vector
sum of v and V P .

If the direction of the tip of the vane at A is made parallel to
DB the water will glide on to the vane in exactly the same way

IMPACT OF WATER ON VANES 269

as if it were at rest, and the water were moving in the direction
DB. This is the condition that no energy shall be lost by shock.

When the water leaves the vane, the relative velocity of the
water and the vane must be parallel to the direction of the
tangent to the vane at the point where it leaves, and it is equal to
the vector difference of the absolute velocity of the water, and
the vane. Or the absolute velocity with which the water leaves
the vane is the vector sum of the velocity of the tip of the vane
and the relative velocity of the water to the vane.

Let CGr be the direction of the tangent to the vane at C. Let
CE be Vij the velocity of C in magnitude and direction, and let CF
be the absolute velocity Ui with which the water leaves the vane.

Draw EF parallel to CGr to meet the direction OF in F, then
the relative velocity of the water and the vane is EF, and the
velocity with which the water leaves the vane is equal to OF.

If Vi and the direction CGr are given, and the direction in which
the water leaves the vane is given, the triangle CEF can be
drawn, and OF determined.

If on the other hand Vi is given, and the relative velocity v r is
given in magnitude and direction, CF can be found by measuring
off along EF the known relative velocity v r and joining CF.

If Vi and Ui are given, the direction of the tangent to the vane
is then, as at inlet, the vector difference of Ui and VL

It will be seen that when the water either strikes or leaves the
vane, the relative velocity of the water and the vane is the vector
difference of the velocity of the water and the vane, and the actual
velocity of the water as it leaves the vane is the vector sum of the
velocity of the vane and the relative velocity of the water and
the vane.

Example. The direction of the tip of the vane at the outer circumference of a
wheel fitted with vanes, makes an angle of 165 degrees with the direction of motion
of the tip of the vane.

The velocity of the tip at the outer circumference is 82 feet per second.

The water leaves the wheel in such a direction and with such a velocity that the
radial component is 13 feet per second.

Find the absolute velocity of the water in direction and magnitude and the
relative velocity of the water and the wheel.

To draw the triangle of velocities, set out AB equal to 82 feet, and make the
angle ABC equal to 15 degrees. BC is then parallel to the tip of the vane.

Draw EC parallel to AB, and at a distance from it equal to 13 feet and
intersecting BG in C.

Then AC is the vector sum of AB and BC, and is the absolute velocity of the
water in direction and magnitude.

Expressed trigonometrically

AC 2 = (82 - 13 cot 15) 2 + 13 2

= 38-6* + 13 8 and AC = 36-7 ft. per sec.

sin BAG =^ = -354.
AC/

Therefore BAG = 20 45'.

270

HYDRAULICS

169. Conditions which the vanes of hydraulic machines
should satisfy.

In all properly designed hydraulic machines, such as turbines,
water wheels, and centrifugal pumps, in which water flowing in
a definite direction impinges on moving vanes, the relative velocity
of the water and the vanes should be parallel to the direction of
the vanes at the point of contact. If not, the water breaks into
eddies as it moves on to the vanes and energy is lost.

Again, if in such machines the water is required to leave the
vanes with a given velocity in magnitude and direction, it is only
necessary to make the tip of the vane parallel to the vector
difference of the given velocity with which the water is to leave
the vane and the velocity of the tip of the vane.

Example (1). A jet of water, Fig. 174, moves in a direction AB making an angle
of 30 degrees with the direction of motion AC of a vane moving in the atmosphere.
The jet has a velocity of 30 ft. per second and the vane of 15 ft. per second. To find
(a) the direction of the vane at A so that the water may enter without shock; (6) the
direction of the tangent to the vane where the water leaves it, so that the absolute
velocity of the water when it leaves the vane is in a direction perpendicular to AC ;
(c) the pressure on the vane and the work done per second per pound of water
striking the vane. Friction is neglected.

'U,

Change orV&oribf irv the
direction, ofmotiori.

Fig. 174.

The relative velocity V r of the water and the vane at A is CB, and for no shock
the vane at A must be parallel to CB.

Since there is no friction, the relative velocity V r of the water and the vane
cannot alter, and therefore, the triangle of velocities at exit is ACD or FAjCj .

The point D is found, by taking C as centre and CB as radius and striking the
arc BD to cut the known direction AD in D.

The total change of velocity of the jet is the vector difference DB of the initial
and final velocities, and the change of velocity in the direction of motion is BE.
Calling this velocity V, the pressure exerted upon the vane in the direction of
motion is

Ibs. per Ib. of water striking the vane.
9

The work done per Ib. is, therefore, ft. Ibs. and the efficiency, since there is
no loss by friction, or shock, is

Hgr-

IMPACT OF WATER ON VANES 271

The change in the kinetic energy of the jet is equal to the ivork done by the jet.
The kinetic energy per Ib. of the original jet is and the final kinetic energy is

2<7 '

The work done is, therefore, -= ~- ft. Ibs. and the efficiency is

It can at once be seen from the geometry of the figure that
Vv _ U 2 Uj 2
g ~2g"2g'

For AB 2 =AC 2 +CB 2 + 2AC.CG,

and since CD = CB and

therefore, AB 2 - AD 2 = 2 AC (AC + CG)

But

A , ,
therefore,

If the water instead of leaving the vane in a direction perpendicular to v, leaves
it with a velocity Uj having a component V x parallel to v t the work done on the
vane per pound of water is

If Uj be drawn on the figure it will be seen that the change of velocity in the

V- V

direction of motion is now (V- VJ, the impressed force per pound is - - 1 , and

/ V V \
the work done is, therefore, ( * ) ^ ft. Ibs. per pound.

As before, the work done on the vane is the loss of kinetic energy of the jet, and
therefore,

9 20

The work done on the vane per pound of water for any given value of Uj , is,
therefore, independent of the direction of U 1 .

Example (2). A series of vanes such as AB, Fig. 175, are fixed to a (turbine)
wheel which revolves about a fixed centre C, with an angular velocity u.

The radius of B is R and of A, r. Within the wheel are a number of guide
passages, through which water is directed with a velocity U, at a definite inclination
6 with the tangent to the wheel. The air is supposed to have free access to the
wheel.

To draw the triangles of velocity, at inlet and outlet, and to find the directions
of the tips of the vanes, so that the water moves on to the vanes without shock and
leaves the wheel with a given velocity U,. Friction neglected.

In this case the velocity relative to the vanes is altered by the whirling of the
water as it moves over the vanes. It will be shown later that the head impressed

- *-S-+3?-S-

The tangent AH to the vane at A makes an angle <f> with the tangent AD to the
wheel, so that CD makes an angle with AD. The triangle of velocities ACD at
inlet is, therefore, as shown in the figure and does not need explanation.

To draw the triangle of velocities at exit, set out BGr equal to vj and perpen-

272

HYDRAULICS

dicular to the radius BO, and with B and G as centres, describe circles with U x and
v r as radii respectively, intersecting in B. Then GE is parallel to the tangent to
the vane at B.

(See Impulse turhines.)

Work done on the wheel. Neglecting friction etc. the work done per pound of
water passing through the wheel, since the pressure is constant, being equal to the
atmospheric pressure, is the loss of kinetic energy of the water, and is

The work done on the wheel can also be found from the consideration of the
change of the angular momentum of the water passing through the wheel. Before
going on however to determine the work per pound by this method, the notation
that has been used is summarised and several important principles considered.

Notation used in connection with vanes, turbines and centrifugal
pumps. Let U be the velocity with which the water approaches
the vane, Fig. 175, and v the velocity, perpendicular to the radius
AC, of the edge A of the vane at which water enters the wheel.

Let Y be the component of U in the direction of v,

u the component of U perpendicular to v,

Y r the relative velocity of the water and vane at A,

Vi the velocity, perpendicular to BC, of the edge B of the vane
at which water leaves the wheel,

Ui the velocity with which the water leaves the wheel,

Yi the component of Ui in the direction of v it

IMPACT OF WATER ON VANES 273

U] the component of Ui perpendicular to Vi, or along BC,
v r the relative velocity of the water and the vane at B.
Velocities of whirl. The component velocities V and Vi are

called the velocities of whirl at inlet and outlet respectively.

This term will frequently be used in the following chapters.

170. Definition of angular momentum.

If a weight of W pounds is moving with a velocity U, Figs. 175
and 176, in a given direction, the perpendicular distance of which
is S feet from a fixed centre C, the angular momentum of W is

. U . S pounds feet.
9

171. Change of angular momentum.

If after a small time t the mass is moving with a velocity Ui in
a direction, which is at a perpendicular distance Si from C, the

angular momentum is now UiSij the change of angular

momentum in time t is

and the rate of change of angular momentum is

Fig. 176. Fig. 177.

172. Two important principles.

(1) Work done by a couple, or turning moment. When a
body is turned through an angle a measured in radians, under the
action of a constant turning moment, or couple, of T pounds feet,
the work done is Ta foot pounds.

If the body is rotating with an angular velocity w radians
per second, the rate of doing work is To> foot pounds per second,

and the horse-power is -=^ .

L. H. 18

HYDRAULICS

Suppose a body rotates about a fixed centre C, Fig. 177, and
a force P Ibs. acts on the body, the perpendicular distance from
C to the direction of P being S.

The moment of P about C is

T = P.S.

If the body turns through an angle <o in one second, the
distance moved through by the force P is o> . S, and the work
done by P in foot pounds is

P<oS=To>.

And since one horse-power is equivalent to 33,000 foot pounds
per minute or 550 foot pounds per second the horse-power is

HP T

(2) The rate of change of angular momentum of a "body
rotating about a fixed centre is equal to the couple acting upon
the body. Suppose a weight of W pounds is moving at any instant
with a velocity U, Fig. 176, the perpendicular distance of which
from a fixed centre C is S, and that forces are exerted upon W
so as to change its velocity from U to Ui in magnitude and
direction.

The reader may be helped by assuming the velocity U is
changed to Ui by a wheel such as that shown in Fig. 175.

Suppose now at the point A the velocity 'U is destroyed in a
time oti then a force will be exerted at the point A equal to

P_W U
~ g 'tt'

and the moment of this force about C is P . S.

At the end of the time dt, let the weight W leave the wheel
with a velocity Ui. During this time dt the velocity Ui might
have been given to the moving body by a force

P _WU 1
'~ g dt

acting at the radius Si.

The moment of Pi is PI Si ; and therefore if the body has been
acting on a wheel, Fig. 175, the reaction of the wheel causing the
velocity of W to change, the couple acting on the wheel is

(1).

When US is greater than UiSi, the body has done work on the
wheel, as in water wheels and turbines. When UiSi is greater
than US, the wheel does work on the body as in centrifugal pumps.

Let the wheel of Fig. 175 have an angular velocity w.

IMPACT OF WATER ON VANES 275

In a time 3t the angle moved through by the couple is wdt,
and therefore the work done in time dt is

W
T.oO* = eodJS-TLSO .................. (2).

Suppose now W is the weight of water in pounds per second
which strikes the vanes of a moving wheel of any form, and this
water has its velocity changed from U to Ui, then by making dt
in either equation (1) or (2) equal to unity, the work done per
second is

and the work done per second per pound of water entering the
wheel is

This result, as will be seen later (page 337), is entirely inde-
pendent of the change of pressure as the water passes through the
wheel, or of the direction in which the water passes.

173. Work done on a series of vanes fixed to a wheel
expressed in terms of the velocities of whirl of the water
entering and leaving the wheel.

Outward flow turbine. If water enters a wheel at the inner
circumference, as in Fig. 175, the flow is said to be outward.
On reference to the figure it is seen that since r is perpendicular
to V, and S to U, therefore

r_TJ

s~v

and for a similar reason

R Ux

STV/

Again the angular velocity of the wheel

therefore the work done per second is

and the work done per pound of flow is

Yt?

y 9

Inward flow turbine. If the water enters at the outer cir-
cumference of a wheel with a velocity of whirl V, and leaves at
the inner circumference with a velocity of whirl Vi, the velocities

182

276

HYDRAULICS

of the inlet and outlet tips of the vanes being v and
the work done on the wheel is still

Yt>

respectively

9 9
The flow in this case is said to be inward.

Parallel flow or axial flow turbine. If vanes, such as those
shown in Fig. 174, are fixed to a wheel, the flow is parallel to the
axis of the wheel, and is said to be axial.

For any given radius of the wheel, Vi is equal to v, and the
work done per pound is

which agrees with the result already found on page 271.

174. Curved vanes. Pelton wheel.

Let a series of cups, similar to Figs. 178 and 179, be moving
with a velocity v, and a stream with a greater velocity U in the
same direction.

The relative velocity is

V r =(U-).

Neglecting friction, the relative velocity Y r will remain con-
stant, and the water will, therefore, leave the cup at the point B
with a velocity, Y r , relative to the cup.

Fig. 178.

Fig. 179.

If the tip of the cup at B, Fig. 178, makes an angle with the
direction of v, the absolute velocity with which the water leaves
the cup will be the vector sum of v and Y r , and is therefore Ui.
The work done on the cups is then

IMPACT OF WATER ON VANES 277

per Ib. of water, and the efficiency is
U 2 Ui 2

For Ui, the value

H! = >J{v - (U - f>) cos BY + (U - v) 2 sin &

can be substituted, and the efficiency thus determined in terms of
v, U and 0.

Pelton wheel cups. If is zero, as in Fig. 178, and U v is
equal to v, or U is twice v, Ui clearly becomes zero, and the water
drops away from the cup, under the action of gravity, without
possessing velocity in the direction of motion.

The whole of the kinetic energy of the jet is thus absorbed
and the theoretical efficiency of the cups is unity.

The work done determined from consideration of the change of
momentum. The component of Ui, Fig. 178, in the direction of
motion, is

v(U v) cos 0,

and the change of momentum per pound of water striking the
vanes is, therefore,

9
The work done per Ib. is

and the efficiency is

U 2
When is 0, cos is unity, and

which is a maximum, and equal to unity, when v is -^ .

175. Force tending to move a vessel from which water
is issuing through an orifice.

When water issues from a vertical orifice of area a sq. feet,
in the side of a vessel at rest, in which the surface of the water is
maintained at a height h feet above the centre of the orifice, the

278 HYDRAULICS

pressure on the orifice, or the force tending to move the vessel
in the opposite direction to the movement of the water, is

F=2w.a.fclbs.,
w being the weight of a cubic foot of water in pounds.

The vessel being at rest, the velocity with which the water
leaves the orifice, neglecting friction, is

and the quantity discharged per second in cubic feet is

The momentum given to the water per second is
-., _ w . a . v*
9

But the momentum given to the water per second is equal to
the impressed force, and therefore the force tending to move the
vessel is

or is equal to twice the pressure that would be exerted upon a
plate covering the orifice. When a fireman holds the nozzle of a
hose-pipe through which water is issuing with a velocity v t there
is, therefore, a pressure on his hand equal to

2wav' 2 _ wav*

20 g

If the vessel has a velocity V backwards, the velocity U of the
water relative to the earth is

and the pressure exerted upon the vessel is

9
The work done per second is

. -x-r wav V (v V) P . ,,
F . V = ^ '- foot Ibs.,

or = Y(t? " V) foot Ibs.

9
per Ib. of flow from the nozzle.

V (v - V)
The efficiency is e = ~^~~

2YQ-V)

tf
which is a maximum, when

v = 2Y

and =i-

IMPACT OF WATEE ON VANES 279

176. The propulsion of ships by water jets.

A method of propelling ships by means of jets of water issuing
from orifices at the back of the ship, has been used with some
success, and is still employed to a very limited extent, for the
propulsion of lifeboats.

Water is taken by pumps carried by the ship from that
surrounding the vessel, and is forced through the orifices. Let
v be the velocity of the water issuing from the orifice relative

to the ship, and Y the velocity of the ship. Then ~ is the

head h forcing water from the ship, and the available energy
per pound of water leaving the ship is h foot pounds.

The whole of this energy need not, however, be given to the
water by the pumps.

Imagine the ship to be moving through the water and having
a pipe with an open end at the front of the ship. The water in
front of the ship being at rest, water will enter the pipe with a

Y 2
velocity V relative to the ship, and having a kinetic energy ~-

per pound. If friction and other losses are neglected, the work
that the pumps will have to do upon each pound of water to eject
it at the back with a velocity v is, clearly,

v Y 2

As in the previous example, the velocity of the water issuing
from the nozzles relative to the water behind the ship is v Y,

and the change of momentum per pound is, therefore, . If a
is the area of the nozzles the propelling force on the ship is

y
and the work done is

9
The efficiency is the work done on the ship divided by the

work done by the engines, which equals wav(~-~^\ and,

,, - \47 ty'

therefore,

_2YQ-Y)

280 HYDRAULICS

which can be made as near unity as is desired by making v and
V approximate to equality.

But for a given area a of the orifices, and velocity v, the nearer
v approximates to V the less the propelling force F becomes, and
the size of ship that can be driven at a given velocity V for the
given area a of the orifices diminishes.

If vis 2V, e = |.

EXAMPLES.

(1) Ten cubic feet of water per second are discharged from a stationary
jet, the sectional area of which is 1 square foot. The water impinges nor-
mally on a flat surface, moving in the direction of the jet with a velocity
of 2 feet per second. Find the pressure on the plane in Ibs., and the work
done on the plane in horse-power.

(2) A jet of water delivering 100 gallons per second with a velocity of
20 feet per second impinges perpendicularly on a wall. Find the pressure
on the wall.

(3) A jet delivers 160 cubic feet of water per minute at a velocity of
20 feet per second and strikes a plane perpendicularly. Find the pressure
on the plane (1) when it is at rest ; (2) when it is moving at 5 feet per
second in the direction of the jet. In the latter case find the work done
per second in driving the plane.

(4) A fire-engine hose, 3 inches bore, discharges water at a velocity of
100 feet per second. Supposing the jet directed normally to the side of a
building, find the pressure.

(5) Water issues horizontally from a fixed thin-edged orifice, 6 inches
square, under a head of 25 feet. The jet impinges normally on a plane
moving in the same direction at 10 feet per second. Find the pressure on
the plane in Ibs., and the work done in horse-power. Take the coefficient
of discharge as "64 and the coefficient of velocity as '97.

(6) A jet and a plane surface move in directions inclined at 30, with
velocities of 30 feet and 10 feet per second respectively. What is the
relative velocity of the jet and surface ?

(7) Let AB and BC be two lines inclined at 30. A jet of water moves
in the direction AB, with a velocity of 20 feet per second, and a series of
vanes move in the direction CB with a velocity of 10 feet per second. Find
the form of the vane so that the water may come on to it tangentially, and
leave it in the direction BD, perpendicular to CB.

Supposing that the jet is 1 foot wide and 1 inch thick before impinging,
find the effort of the jet on the vanes.

IMPACT OF WATER ON VANES 281

(8) A curved plate is mounted on a slide so that the plate is free to
move along the slide. It receives a jet of water at an angle of 30 with a
normal to the direction of sliding, and the jet leaves the plate at an angle

of 120 with the same normal. Find the force which must be applied to
the plate in the direction of sliding to hold it at rest, and also the normal
pressure on the slide. Quantity of water flowing is 500 Ibs. per minute
with a velocity of 35 feet per second.

(9) A fixed vane receives a jet of water at an angle of 120 with a
direction AB. Find what angle the jet must be turned through in order
that the pressure on the vane in the direction AB may be 40 Ibs., when the
flow of water is 45 Ibs. per second at a velocity of 30 feet per second.

(10) Water under a head of 60 feet is discharged through a pipe 6 inches
diameter and 150 feet long, and then through a nozzle, the area of which
is one-tenth the area of the pipe.

Neglecting all losses but the friction of the pipe, determine the pressure
on a fixed plate placed in front of the nozzle.

(11) A jet of water 4 inches diameter impinges on a fixed cone, the
axis coinciding with that of the jet, and the apex angle being 30 degrees,
at a velocity of 10 feet per second. Find the pressure tending to move the
cone in the direction of its axis.

(12) A vessel containing water and having in one of its vertical sides
a circular orifice 1 inch diameter, which at first is plugged up, is
suspended in such a way that any displacing force can be accurately
measured. On the removal of the plug, the horizontal force required to
keep the vessel in place, applied opposite to the orifice, is 3'6 Ibs. By the
use of a measuring tank the discharge is found to be 31 gallons per minute,
the level of the water in the vessel being maintained at a constant height
of 9 feet above the orifice. Determine the coefficients of velocity, con-
traction and discharge.

(13) A train carrying a Ramsbottom's scoop for taking water into the
tender is running at 24 miles an hour. What is the greatest height at
which the scoop will deliver the water ?

(14) A locomotive going at 40 miles an hour scoops up water from a
trough. The tank is 8 feet above the mouth of the scoop, and the delivery
pipe has an area of 50 square inches. If half the available head is wasted
at entrance, find the velocity at which the water is delivered into the tank,
and the number of tons lifted in a trench 500 yards long. What, under
these conditions, is the increased resistance ; and what is the minimum
speed of train at which the tank can be filled ? Lond. Un. 1906.

If air is freely admitted into the tube, as in Fig. 179 A, the water will

282 HYDRAULICS

move into the tube with a velocity v relative to the
tube equal to that of the train. (Compare with
Fig. 167.) The water will rise in the tube with a
diminishing velocity. The velocity of the train being
58'66 ft. per sec., and half the available head being
lost, the velocity at inlet is

The velocity at a height h feet is

179i ' W4TP^78

= 34-8 ft. per sec.
If the tube is full of water the velocity at inlet is 34'8 ft. per sec.

(15) A stream delivering 3000 gallons of water per minute with a
velocity of 40 feet per second, by impinging on vanes is caused freely to
deviate through an angle of 10, the velocity being diminished to 35 feet
per second. Determine the velocity impressed on the water and the
pressure on the vanes due to impact.

(16) Water flows from a 2-inch pipe, without contraction, at 45 feet per
second.

Determine the maximum work done on a machine carrying moving
plates in the following cases and the respective efficiencies :

(a) When the water impinges on a single flat plate at right angles and
leaves tangentially.

(5) Similar to (a) but a large number of equidistant flat plates are
interposed in the path of the jet.

(d) When a large number of cups are used as in a Pelton wheel.

(17) In hydraulic mining, a jet 6 inches in diameter, discharged under
a head of 400 feet, is delivered horizontally against a vertical cliff face.
Find the pressure on the face. What is the horse-power delivered by the
jet?

(18) If the action on a Pelton wheel is equivalent to that of a jet on a
series of hemispherical cups, find the efficiency when the speed of the
wheel is five-eighths of the speed of the jet.

(19) If in the last question the jet velocity is 50 feet per second,
and the jet area 0*15 square foot, find the horse-power of the wheel.

(20) A ship has jet orifices 3 square feet in aggregate area, and dis-
charges through the jets 100 cubic feet of water per second. The speed of
the ship is 15 feet per second. Find the propelling force of the jets, the
efficiency of the propeller, and, neglecting friction, the horse-power of the
engines.

CHAPTER IX.

WATER WHEELS AND TURBINES.

Water wheels can be divided into two classes as follows.

(a) Wheels upon which the water does work partly by
impulse but almost entirely by weight, the velocity of the water
when it strikes the wheel being small. There are two types of
this class of wheel, Overshot Wheels, Figs. 180 and 181, and
Breast Wheels, Figs. 182 and 184.

(6) Wheels on which the water acts by impulse as when
the wheel utilises the kinetic energy of a stream, or if a head h is
available the whole of the head is converted into velocity before
the water comes in contact with the wheel. In most impulse
wheels the water is made to flow under the wheel and hence
they are called Undershot Wheels.

It will be seen that in principle, there is no line of demarcation
between impulse water wheels and impulse turbines, the latter
only differing from the former in constructional detail.

177. Overshot water wheels.

This type of wheel is not suitable for very low or very high
heads as the diameter of the wheel cannot be made greater than
the head, neither can it conveniently be made much less.

Figs. 180 and 181 show two arrangements of the wheel, the
only difference in the two cases being that in Fig. 181, the top of
the wheel is some distance below the surface of the water in the
up-stream channel or penstock, so that the velocity v with which
the water reaches the wheel is larger than in Fig. 180. This has
the advantage of allowing the periphery of the wheel to have a
higher velocity, and the size and weight of the wheel is conse-
quently diminished.

The buckets, which are generally of the form shown in the
figures, or are curved similar to those of Fig. 182, are con-
nected to a rim M coupled to the central hub of the wheel by

284

HYDRAULICS

suitable spokes or framework. This class of wheel has been
considerably used for heads varying from 6 to 70 feet, but is now
becoming obsolete, being replaced by the modern turbine, which
for the same head and power can be made much more compact,
and can be run at a much greater number of revolutions per unit
time.

E D K

Fig. 180. Overshot Water Wheel.

Fig. 181. Overshot Water Wheel.

The direction of the tangent to the blade at inlet for no shock
can be found by drawing the triangle of velocities as in Figs. 180
and 181. The velocity of the periphery of the wheel is v and the
velocity of the water U. The tip of the blade should be parallel
to V r . The mean velocity U, of the water, as it enters the wheel

WATER WHEELS 285

in Fig. 181, will be v + k \/2(/H, v being the velocity of approach
of the water in the channel, H the fall of the free surface and k
a coefficient of velocity. The water is generally brought to the
wheel along a wooden flume, and thus the velocity U and the
supply to the wheel can be maintained fairly constant by a simple
sluice placed in the flume.

The best velocity v for the periphery is, as shown below,
theoretically equal to |U cos 0, but in practice the velocity v is
frequently much greater and * experiment shows that the best
velocity v of the periphery is about 0'9 of the velocity U of the
water.

If U is to be about 1'lv the water must enter the wheel at
a depth not less than

U 2 = r2^

2^ 2g
below the water in the penstock.

If the total fall to the level of the water in the tail race is h,
the diameter of the wheel may, therefore, be between h and

i l'2v*

fls Ct ~ *

Since U is equal to v 2^H, for given values of U and of h, the
larger the wheel is made the greater must be the angular distance
from the top of the wheel at which the water enters.

With the type of wheel and penstock shown in Fig. 181, the
head H is likely to vary and the velocity U will not, therefore, be
constant. If, however, the wheel is designed for the required power
at minimum flow, when the head increases, and there is a greater
quantity of water available, a loss in efficiency will not be
important.

The horse-power of the wheel. Let D be the diameter of the
wheel in feet which in actual wheels is from 10 to 70 feet.

Let N be the number of buckets, which in actual wheels is
generally from 2J to 3D.

Let Q be the volume of water in cubic feet of water supplied
per second.

Let <o be the angular velocity of the wheel in radians, and n
the number of revolutions per sec.

Let b be the width of the wheel.

Let d, which equals r a - TI , be the depth of the shroud, which
on actual wheels is from 10" to 20".

* Theory and test of an Overshot Water Wheel, by C. E. Weidner, Wisconsin, 1913.

286 HYDRAULICS

Whatever the form of the buckets the capacity of each bucket is

bd . -^- , nearly.
The number of buckets which pass the stream per second is

If a fraction k of each bucket is filled with water

llie fraction Jc in actual wheels is from ^ to .
If h is the fall of the water to the level of the tail race and &
the efficiency of the wheel, the horse-power is

550 '
and the width b for a given horse-power, HP, is

6 =

1100HP

= 17'6

of centrifugal forces. As the wheel revolves, the surface
of the water in the buckets, due to centrifugal forces, takes up a
curved form.

Consider any particle of water of mass w Ibs. at a radius r
equal to CB from the centre of the wheel and in the surface of

Fig. 181 a.

the water. The forces acting upon it are w due to gravity and

the centrifugal force - w 2 r acting in the direction CB,

being the
angular velocity of the wheel. The resultant BGr (Fig. 181 a) of

WATER WHEELS 287

these forces must be normal to the surface. Let BG- be produced
to meet the vertical through the centre in A. Then

AC AC w
CB r w 2

(D T

g
AC = 5.

That is the normal AB always cuts the vertical through C in
a fixed point A, and the surface of the water in any bucket lies
on a circle with A as centre.

Losses of energy in overshot wheels.

(a) The whole of the velocity head - is lost in eddies in the

buckets.

In addition, as the water falls in the bucket through the
vertical distance EM, its velocity will be increased by gravity,
and the velocity thus given will be practically all lost by eddies.

Again, if the direction of the tip of the bucket is not parallel to
V r the water will enter with shock, and a further head will be
lost. The total loss by eddies and shock may, therefore, be
written

U 2

or hi + h - ,

k and hi being coefficients and hi the vertical distance EM.

(6) The water begins to leave the buckets before the level of
the tail race is reached. This is increased by the centrifugal
forces, as clearly, due to these forces, the water will leave the
buckets earlier than it otherwise would do. If h m is the mean
height above the tail level at which the water leaves the buckets,
a head equal to h m is lost. By fitting an apron GrH in front of the
wheel the water can be prevented from leaving the wheel until it
is very near the tail race.

(c) The water leaves the buckets with a velocity of whirl
equal to the velocity of the periphery of the wheel and a further

head ~- is

(d) If the level of the tail water rises above the bottom of
the wheel there will be a further loss due to, (1) the head h equal to
the height of the water above the bottom of the wheel, (2) the
impact of the tail water stream on the buckets, and (3) the
tendency for the buckets to lift the water on the ascending side of
the wheel.

288 HYDRAULICS

In times of flood there may be a considerable rise of the
down-stream, and h may then be a large fraction of h. If on
the other hand the wheel is raised to such a height above the tail
water that the bottom of the wheel may be always clear, the
head h m will be considerable during dry weather now, and the
greatest possible amount of energy will not be obtained from the
water, just when it is desirable that no energy shall be wasted.

If h is the difference in level between the up and down-stream
surfaces, the maximum hydraulic efficiency possible is

..'-(";?*) ..................... ,,

and the actual hydraulic efficiency will be

h - i m

. e= ; - 5 g Sf

k, fa and Jc being coefficients.

The efficiency as calculated from equation (1), for any given
value of h m , is a maximum when

Y r 2 v* .

-~ H - is a minimum.

From the triangles EKF and KDF, Fig. 180,
(U cos - v) a + (U sin 0)* = Y r 2 .
Therefore, adding v 2 to both sides of the equation,
Y r 2 + v* = IP cos 2 6 - 2Uv cos 6 + 2v z + U 2 sin 2 0,

which is a minimum for a given value of U, when 2Uv cos 6 2i> 2
is a maximum. Differentiating and equating to zero this, and
therefore the efficiency, is seen to be a maximum, when

v - -ff cos 0.
ft

The actual efficiencies obtained from overshot wheels vary
from 60 to 89* per cent.

178. Breast wheel.

This type of wheel, like the overshot wheel, is becoming
obsolete. Fig. 182 shows the form of the wheel, as designed by
Fairbairn.

The water is admitted to the wheel through a number of
passages, which may be opened or closed by a sluice as shown in
the figure. The directions of these passages may be made so that
the water enters the wheel without shock. The water is retained

* Theory and test of Overshot Water Wheel. Bulletin No. 529 University of
Wisconsin.

WATER WHEELS

289

in the bucket, by the breast, until the bucket reaches the tail race,
and a greater fraction of the head is therefore utilised than in
the overshot wheel. In order that the air may enter and leave
the buckets freely, they are partly open at the inner rim. Since
the water in the tail race runs in the direction of the motion of
the bottom of the wheel there is no serious objection to the tail
race level being 6 inches above the bottom of the wheel.

The losses of head will be the same as for the overshot wheel
except that h m will be practically zero, and in addition, there will
be loss by friction in the guide passages, by friction of the water
as it moves over the breast, and further loss due to leakage
between the breast and the wheel.

Fig. 182. Breast Wheel.

According to Rankine the velocity of the rim for overshot and
breast wheels, should be from 4J to 8 feet per second, and the
velocity U should be about 2v.

The depth of the shroud which is equal to r 2 - n is from 1 to
If feet. Let it be denoted by d. Let H be the total fall and let
it be assumed that the efficiency of the wheel is 65 per cent. Then,
L. H. 19

290

HYDRAULICS

the quantity of water required per second in cubic feet for a
given horse-power N is

N.550

" 62-4xHxO'G5

From | to | of the volume of each bucket, or from | to of the
total volume of the buckets on the
loaded part of the wheel is filled with
water.

Let 6 be the breadth of the buckets.
If now v is the velocity of the rim, and
an arc AB, Fig. 183, is set off on the
outer rim .equal to v, and each bucket
is half full, the quantity of water
carried down per second is

JABCD.fe.
Therefore

/~ i ~ \

vdb.

2r 2

Equating this value of Q to the above value, the width 6 is

27ND

D being the outer diameter of the wheel.

Breast wheels are used for falls of from 5 to 15 feet and the
diameter should be from 12 to 25 feet. The width may be as
great as 10 feet.

Example. A breast wheel 20 feet diameter and 6 feet wide, working on a fall
of 14 feet and having a depth of shroud of 1' 3", has its buckets full The mean
velocity of the buckets is 5 feet per second. Find the horse-power of the wheel,
assuming the efficiency 70 per cent.

= 26-1.

The dimensions of this wheel should be compared with those calculated for an
inward flow turbine working under the same head and developing the same horse-
power. See page 339.

179. Sagebien wheels.

These wheels, Fig. 184, have straight buckets inclined to the
radius at an angle of from 30 to 45 degrees.

The velocity of the periphery of the wheel is very small, never
exceeding 2^ to 3 feet per second, so that the loss due to the water
leaving the wheel with this velocity and due to leakage between
the wheel arid breast is small.

WATER WHEELS

291

An efficiency of over 80 per cent, has been obtained with
these wheels.

The water enters the wheel in a horizontal direction with
a velocity U equal to that in the penstock, and the triangle of
velocities is therefore ABC.

If the bucket is made parallel to Y r the water enters without
shock, while at the same time there is no loss of head due to
friction of guide passages, or to contraction as the water enters or
leaves them ; moreover the direction of the stream has not to be
changed.

Fig. 184. Sagebien Wheel.

The inclined straight bucket has one disadvantage ; when the
lower part of the wheel is drowned, the buckets as they ascend are
more nearly perpendicular to -the surface of the tail water than
when the blades are radial, but as the peripheral speed is very
low the resistance due to this cause is not considerable.

180. Impulse wheels.

In Overshot and Breast wheels the work is done principally
by the weight of the water. In the wheels now to be considered
the whole of the head available is converted into velocity before
the water strikes the wheel, and the work is done on the wheel
by changing the momentum of the mass of moving water, or in
other words, by changing the kinetic energy of the water.

192

292

HYDRAULICS

Undershot wheel with flat blades. The simplest case is when
wheel with radial blades, similar to that shown in Fig. 185, is
into a running stream.
If b is the width of the wheel, d the depth of the stream under
the wheel, and U the velocity in feet per second, the weight of
water that will strike the wheel per second is b . d . w U Ibs., and
the energy available per second is

U 3

b . d . w 2~ foot Ibs.

Let v be the mean velocity of the blades.

The radius of the wheel being large the blades are similar to
a series of flat blades moving parallel to the stream and the water
leaves them with a velocity v in the direction of motion.

As shown on page 268, the best theoretical value for the
velocity v of such blades is U and the maximum possible
efficiency of the wheel is 0'5.

-f

Fig. 185. Impulse Wheel.

By placing a gate across the channel and making the bed near
the wheel circular as in Fig. 185, and the width of the wheel
equal to that of the channel, the supply is more under control, and
loss by leakage is reduced to a minimum.

The conditions are now somewhat different to those assumed
for the large number of flat vanes, and the maximum possible
efficiency is determined as follows.

Let Q be the number of cubic feet of water passing through
the wheel per second. The mean velocity with which the water
leaves the penstock at ab is U = k v 2a/&. Let the depth of the

WATER WHEELS 293

stream at rib be t. The velocity with which the water leaves the
wheel at the section cd is v, the velocity of the blades. If the
width of the stream at cd is the same as at ab and the depth
is h 0y then,

Ji Q x v = t x Uj

i W
h = .

Since IT is greater than v, h is greater than t, as shown in
the figure.

The hydrostatic pressure on the section cd is ^hfbw and on
the section ab it is %t* bw.

The change in momentum per second is

and this must be equal to the impressed forces acting on the mass
of water flowing per second through ab or cd.

These impressed forces are P the driving pressure on the wheel
blades, and the difference between the hydrostatic pressures acting
on cd and ab.

If, therefore, the driving force acting on the wheel is P Ibs.,
then,

P + Ihfbw - & 2 bw = 2^ (U - i>).
Substituting for Ji , , the work done per second is

Or, since Q = b . t . U,

The efficiency is then,

-tQ _ _

2\v

t?(U-tQ _ t_ /U _ v\

which is a maximum when

2u 2 U a - 4y 3 U + ^^U 2 + gtv* = 0.

The best velocity, v, for the mean velocity of the blades, has
been found in practice to be about 0'4U, the actual efficiency is
from 30 to 35 per cent., and the diameters of the wheel are
generally from 10 to 23 feet.

Floating wheels. To adapt the wheel to the rising and
lowering of the waters of a stream, the wheel may be mounted on

294

HYDRAULICS

a frame which may be raised or lowered as the stream rises, or the
axle carried upon pontoons so that the wheel rises automatically
with the stream.

181. Poncelet wheel.

The efficiency of the straight blade impulse wheels is very
small, due to the large amount of energy lost by shock, and to the
velocity with which the water leaves the wheel in the direction of
motion.

The efficiency of the wheel is doubled, if the blades are of such
a form, that the direction of the blade at entrance is parallel to
the relative velocity of the water and the blade, as first suggested
by Poncelet, and the water is made to leave the wheel with no
component in the direction of motion of the periphery of the
wheel.

Fig. 186 shows a Poncelet wheel.

tangle of
Velocities
atEcut,

Fig. 186. Undershot Wheel.

Suppose the water to approach the edge A of a blade with a
velocity U making an angle with the tangent to the wheel at A.

Then if the direction of motion of the water is in the direction
AC, the triangle of velocities for entrance is ABC.

The relative velocity of the water and the wheel is V r , and ii
the blade is made sufficiently deep that the water does not overflow
the upper edge and there is no loss by shock and by friction, a
particle of water will rise up the blade a vertical height

h _yj

1 20 '

WATER WHEELS 295

It then begins to fall and arrives at the tip of the blade with the
velocity V r relative to the blade in the inverse direction BE.

The triangle of velocities for exit is, therefore, ABE, BE being
equal to BC.

The velocity with which the water leaves the wheel is then

It has been assumed that no energy is lost by friction or by
shock, and therefore the work done on the wheel is

and the theoretical hydraulic efficiency* is

IP W

-1 Ul "
'

This will be a maximum when Ui is a minimum.

Now since BE = BC, the perpendiculars EF and CD, on to
AB and AB produced, from the points E and C respectively, are
equal. And since AC and the angle 6 are constant, CD is constant
for all values of v, and therefore FE is constant. But AE, that is
Ui, is always greater than FE except when AE is perpendicular
to AD. The velocity Ui will have its minimum value, therefore,
when AE is equal to FE or Ui is perpendicular to v.

The triangles of velocities are then as in Fig. 187, the point B
bisects AD, and

For maximum efficiency, therefore,

* In what follows, the terms theoretical hydraulic efficiency and hydraulic
efficiency will be frequently used. The maximum work per Ib. that can be utilised
by any hydraulic machine supplied with water under a head H, and from which

it?
the water exhausts with a velocity u is H - . The ratio

is the theoretical hydraulic efficiency. If there are other hydraulic losses in the
machine equivalent to a head h/ per Ib. of flow, the hydraulic efficiency is

The actual efficiency of the machine is the ratio of the external work done per Ib.
of water by the machine to H.

296 HYDRAULICS

The efficiency can also be found by considering the change of
momentum.

The total change of velocity impressed on the water is CE, and
the change in the direction of motion is
therefore FD, Fig. 186.

And since BE is equal to BC, FB is
equal to BD, and therefore,

FD = 2(Ucos0-t>).

The work done per Ib. is, then,

2(Ucosfl-i?)

9 ' V '

and the efficiency is

TJ, 2(Ut; cos - v*

& TT

U 2 ........................ *r

Differentiating with respect to v and equating to zero,

Ucos0-2i;=0,
or v = |U cos 0.

The velocity Uj with which the water leaves the wheel, is then
perpendicular to v and is

Ui = Usin0.

Substituting for v its value JU cos in (2), the maximum efficiency
is cos 2 0.

The same result is obtained from equation (1), by substituting
forU^Usinfl.

The maximum efficiency is then

A common value for is 15 degrees, and the theoretical
hydraulic efficiency is then 0*933.

This increases as diminishes, and would become unity if
could be made zero.

If, however, is zero, U and v are parallel and the tip of the
blade will be perpendicular to the radius of the wheel.

This is clearly the limiting case, which practically is not
realisable, without modifying the construction of the wheel. The
necessary modification is shown in the Pelton wheel described on
page 377.

The actual efficiency of Poncelet wheels is from 55 to 65 per
cent.

WATER WHEELS 297

Form of the bed. Water enters the wheel at all points between
Q and R, and for no shock the bed of the channel PQ should be
made of such a form that the direction of the stream, where it
enters the wheel at any point A between R and Q, should make
a constant angle 6 with the radius of the wheel at A.

With as centre, draw a circle touching the line AS which
makes the given angle with the radius AO. Take several
other points on the circumference of the wheel between R and
Q, and draw tangents to the circle STY. If then a curve
PQ is drawn normal to these several tangents, and the stream
lines are parallel to PQ, the water entering any part of the
wheel between R and Q, will make a constant angle with the
radius, and if it enters without shock at A, it will do so at all
points. The actual velocity of the water U, as it moves along the
race PQ, will be less than \/2grH, due to friction, etc. The
coefficient of velocity Jc v in most cases will probably be between
0'90 and 0'95, so that taking a mean value for Jc v of 0'925,

U = 0'925 V2<7H.

The best value for the velocity v taking friction into account.
In determining the best velocity for the periphery of the wheel no
allowance has been made for the loss of energy due to friction in
the wheel.

If Y r is the relative velocity of the water and wheel at entrance,
it is to be expected that the velocity relative to the wheel at exit
will be less than Y r , due to friction and interference of the rising
and falling particles of water.

The case is somewhat analogous to that of a stone thrown
vertically up in the atmosphere with a velocity v. If there were
no resistance to its motion, it would rise to a certain height,

and then descend, and when it again reached the earth it would
have a velocity equal to its initial velocity v. Due to resistances,
the height to which it rises will be less than hi, and the velocity
with which it reaches the ground will be even less than that due
to falling freely through this diminished height.

Let the velocity relative to the wheel at exit be riV r , n being
a fraction less than unity.

The triangle of velocities at exit will then be ABB, Fig. 188.
The change of velocity in the direction of motion is GrH, which
equals

(Ucos0-t>).

298 HYDRAULICS

If the velocity at exit relative to tlie wheel is only riV r , there
must have been lost by friction etc., a head equal to

The work done on the wheel per Ib. of water is, therefore,

-p)} V P
-2jV- n >'

tr c

Fig. 188.

Let (1 - w 2 ) be denoted by /, then since

V r 2 = BH 2 + CH 2 = (U cos B - vY + U 2 sin 2 0,
the efficiency

Differentiating with respect to v and equating to zero,
2 (1 +ri) Ucos^ -4 (1 + ri) v + 2U/cos 0-2vf=0 t
from which

If /is now supposed to be 0'5, i.e. the head lost by friction, etc.
is ^^ , n is 0'71 and

v = -56U cos 0.
If /is taken as 0*75,

v - 0'6U cos 0.

Dimensions of Poncelet wheels. The diameter of the wheel
should not be less than 10 feet when the bed is curved, and not
less than 15 feet for a straight bed, otherwise there will be con-
siderable loss by shock at entrance, due to the variation of the
angle which the stream lines make with the blades between R
and Q, Fig. 186. The water will rise on the buckets to a height

WATER WHEELS 299

V r 2

nearly equal to -^- , and since the water first enters at a point R,

the blade depth d must, therefore, be greater than this, or the
water will overflow at the upper edge. The clearance between
the bed and the bottom of the wheel should not be less than f ".
The peripheral distance between the consecutive blades is taken
from 8 inches to 18 inches.

Horse-power of Poncelet wheels. If H is the height of the
surface of water in the penstock above the bottom of the wheel,
the velocity U will be about

and v may be taken as

0'55 x 0-92 V2^H = 0'5 JZgK.

Let D be the diameter of the wheel, and b the breadth, and let
t be the depth of the orifice RP. Then the number of revolutions
per minute is

0-5-X/205
n - f^ .

7T.D

The coefficient of contraction c for the orifice may be from 0'6,
if it is sharp-edged, to 1 if it is carefully rounded, and may be
taken as 0'8 if the orifice is formed by a flat-edged sluice.

The quantity of water striking the wheel per second is, then,

If the efficiency is taken as 60 per cent., the work done per
second is 0'6 x 62'4QH ft. Ibs.
The horse-power N is then

550

182. Turbines.

Although the water wheel has been developed to a considerable
degree of perfection, efficiencies of nearly 90 per cent, having been
obtained, it is being almost entirely superseded by the turbine.

The old water wheels were required to drive slow moving
machinery, and the great disadvantage attaching to them of
having a small angular velocity was not felt. Such slow moving
wheels are however entirely unsuited to the driving of modern
machinery, and especially for the driving of dynamos, and they
are further quite unsuited for the high heads which are now
utilised for the generation of power.

Turbine wheels on the other hand can be made to run at either
low or very high speeds, and to work under any head varying

300 HYDRAULICS

from 1 foot to 2000 feet, and the speed can be regulated with
much greater precision.

Due to the slow speeds, the old water wheels could not develope
large power, the maximum being about 100 horse-power, whereas
at Niagara Falls, turbines of 10,000 horse-power have recently
been installed.

Types of Turbines.

Turbines are generally divided into two classes; impulse, or
free deviation turbines, and reaction or pressure turbines.

In both kinds of turbines an attempt is made to shape the
vanes so that the water enters the wheel without shock ; that is
the direction of the relative velocity of the water and the vane is
parallel to the tip of the vane, and the direction of the leaving
edge of the vane is made so that the water leaves in a specified
direction.

In the first class, the whole of the available head is converted
into velocity before the water strikes the turbine wheel, and the
pressure in the driving fluid as it moves over the vanes remains
constant, and equal to the atmospheric pressure. The wheel and
vanes, therefore, must be so formed that the air has free access
between the vanes, and the space between two consecutive vanes
must not be full of water. Work is done upon the vanes, or in
other words, upon the turbine wheel to which they are fixed, in
virtue of the change of momentum or kinetic energy of the
moving water, as in examples on pages 270 2.

Suppose water supplied to a turbine, as in Fig. 258, under an
effective head H, which may be supposed equal to the total head
minus losses of head in the supply pipe and at the nozzle. The
water issues from the nozzle with a velocity U = j2gH i and the
available energy per pound is

Work is done on the wheel by the absorption of the whole, or
part, of this kinetic energy.

If Ui is the velocity with which the water leaves the wheel,
the energy lost by the water per pound is

and this is equal to the work done on the wheel together with
energy lost by friction etc. in the wheel.

In the second class, only part of the available head is con-
verted into velocity before the water enters the wheel, and the

TURBINES 801 .

velocity and pressure both vary as the water passes through the
wheel. It is therefore essential, that the wheel shall always be
kept full of water. Work is done upon the wheel, as will be seen
in the sequence, partly by changing the kinetic energy the water
possesses when it enters the wheel, and partly by changing its
pressure or potential energy.

Suppose water is supplied to the turbine of Fig. 191, under
the effective head H ; the velocity U with which the water enters
the wheel, is only some fraction of v/2^H, and the pressure head
at the inlet to the wheel will depend upon the magnitude of U
and upon the position of the wheel relative to the head and tail
water surfaces. The turbine wheel always being full of water,
there is continuity of flow through the wheel, and if the head
impressed upon the water by centrifugal action is determined, as
on page 335, the equations of Bernoulli * can be used to determine
in any given case the difference of pressure head at the inlet and
outlet of the wheel.

If the pressure head at inlet is and at outlet , and the

w w '

velocity with which the water leaves the wheel is Ui, the work
done on the wheel (see page 338) is

i ~ + 2^ ~ w per pound of water >

or work is done on the wheel, partly by changing the velocity
head and partly by changing the pressure head. Such a turbine
is called a reaction turbine, and the amount of reaction is measured
by the ratio

P-Pl
w_ w

~H~-

Clearly, if p is made equal to pi, the limiting case is reached,
and the turbine becomes an impulse, or free-deviation turbine.

It should be clearly understood that in a reaction turbine no
work is done on the wheel merely by hydrostatic pressure, in the
sense in which work is done by the pressure on the piston of a
steam engine or the ram of a hydraulic lift.

183. Reaction turbines.

The oldest form of turbine is the simple reaction, or Scotch
turbine, which in its simplest form is illustrated in Fig. 189.

A vertical tube T has two horizontal tubes connected to it, the
outer ends of which are bent round at right angles to the direction

* fciee page 334

302

HYDRAULICS

of length of the tube, or two holes and Oi are drilled as in the
figure.

Water is supplied to the central tube at such a rate as to keep
the level of the water in the tube
constant, and at a height h above
the horizontal tubes. Water escapes
through the orifices and Oi and
the wheel rotates in a direction
opposite to the direction of flow of
the water from the orifices. Tur-
bines of this class are frequently
used to act as sprinklers for distri-
buting liquids, as for example for
distributing sewage on to bacteria
beds.

A better practical form, known as the Whitelaw turbine, is
shown in Fig. 190.

Fig. 189. Scotch Turbine.

Fig. 190. Whitelaw Turbine.

To understand the action of the turbine it is first necessary to
consider the effect of the whirling of the water in the arm upon

TURBINES 303

the discharge from the wheel. Let v be the velocity of rotation
of the orifices, and h the head of water above the orifices.

Imagine the wheel to be held at rest and the orifices opened ;
then the head causing velocity of flow relative to the arm is
simply h, and neglecting friction the water will leave the nozzle
with a velocity

t? = \/2gh.

Now suppose the wheel is filled with water and made to rotate
at an angular velocity w, the orifices being closed. There will
now be a pressure head at the orifice equal to h plus the head
impressed on the water due to the whirling of each particle of
water in the arm.

Assume the arm to be a straight tube, Fig. 189, having a cross
sectional area a. At any radius r take an element of thickness dr.

The centrifugal force due to this element is

s - w . a . o>V3r
dr = - .

The pressure per unit area at the outer periphery is, therefore,

1 f R
p = -

a Jo g

and the head impressed on the water is

p = o> 2 R 2
w 2g'

Let v be the velocity of the orifice, then v = o>R, and therefore

p _ v*
w~2g'

If now the wheel be assumed frictionless and the orifices are
opened, and the wheel rotates with the angular velocity <o, the
head causing velocity of flow relative to the wheel is

Let Y r be the velocity relative to the wheel with which the
water leaves the orifice.

The velocity relative to the ground, with which the water
leaves the wheel, is V r v, the vector sum of V r and v.

304 HYDRAULICS

The water leaves the wheel, therefore, with a velocity relative
to the ground of /* = V r v, and the kinetic energy lost is

^ per pound of water.

The theoretical hydraulic efficiency is then,
7 ~'

V r 2 -t; 2
2v

Vr + V*

Since from (2), V r becomes more nearly equal to v as v
increases, the energy lost per pound diminishes as v increases,
and the efficiency E, therefore, increases with v.

The efficiency of the reaction wheel when friction is considered.
As before,

Assuming the head lost by friction to be -- 1 - , the total head

*9
must be equal to

The work done on the wheel, per pound, is now

;

and the hydraulic efficiency is

, r /x

20 20*

Substituting for h from (4) and for /*, V r v,

(l + ^Vr 2 -^ 2
Let V r = nv,

then 6= (l+Aj)w f -l'

Differentiating and equating to zero,

TURBINES

305

From which

Or the efficiency is a maximum when

and

Fig. 191. Outward Flow Turbine.

L. H.

306 HYDRAULICS

184. Outward flow turbines.

The outward flow turbine was invented in 1828 by Four-
neyron. A cylindrical wheel W, Figs. 191, 192, and 201, having
a number of suitably shaped vanes, is fixed to a vertical axis.
The water enters a cylindrical chamber at the centre of the
turbine, and is directed to the wheel by suitable fixed guide
blades Gr, and flows through the wheel in a radial direction
outwards. Between the guide blades and the wheel is a cylindri-
cal sluice R which is used to control the flow of water through
the wheel.

Fig. 191 a.

This method of regulating the flow is very imperfect, as when
the gate partially closes the passages, there must be a sudden
enlargement as the water enters the wheel, and a loss of head
ensues. The efficiency at "part gate" is consequently very
much less than when the flow is unchecked. This difficulty is
partly overcome by dividing the wheel into several distinct
compartments by horizontal diaphragms, as shown in Fig. 192,
so that when working at part load, only the efficiency of one
compartment is affected.

The wheels of outward flow turbines may have their axes,
either horizontal or vertical, and may be put either above, or
below, the tail water level.

The "suction tube" If placed above the tail water, the
exhaust must take place down a " suction pipe," as in Fig. 201,
page 317, the end of which must be kept drowned, and the pipe
air-tight, so that at the outlet of the wheel a pressure less than
the atmospheric pressure may be maintained. If hi is the height
of the centre of the discharge periphery of the wheel above the
tail water level, and p a is the atmospheric pressure in pounds per
square foot, the pressure head at the discharge circumference is

fc-fc-84-fc.

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307

The wheel cannot be more than 84 feet above the level of the tail
water, or the pressure at the outlet of the wheel will be negative,
and practically, it cannot be greater than 25 feet.

It is shown later that the effective head, under which the
turbine works, whether it is drowned, or placed in a suction tube,
is H, the total fall of the water to the level of the tail race.

Fig. 192. Fourneyron Outward Flow Turbine.

The use of the suction tube has the advantage of allowing the
turbine wheel to be placed at some distance above the tail water
level, so that the bearings can be readily got at, and repairs can
be more easily executed.

By making the suction tube to enlarge as it descends, the
velocity of exit can be diminished very gradually, and its final

202

308 HYDRAULICS

value kept small. If the exhaust takes place direct from the
wheel, as in Fig. 192, into the air, the mean head available is the
head of water above the centre of the wheel.

Triangles of velocities at inlet and outlet. For the water to
enter the wheel without shock, the relative velocity of the water
and the wheel at inlet must be parallel to the inner tips of the
vanes. The triangles of velocities at inlet and outlet are shown
in Figs. 193 and 194

V ---------- *

Fig. 194.

Let AC, Fig. 193, be the velocity U in direction and magnitude
of the water as it flows out of the guide passages, and let AD be
the velocity v of the receiving edge of the wheel. Then DC is V r
the relative velocity of the water and vane, and the receiving
edge of the vane must be parallel to DC. The radial component
GC, of AC, determines the quantity of water entering the wheel
per unit area of the inlet circumference. Let this radial velocity
be denoted by u. Then if A is the peripheral area of the inlet
face of the wheel, the number of cubic feet Q per second entering
the wheel is

Q = A. W ,

or, if d is the diameter and b the depth of the wheel at inlet, and
t is the thickness of the vanes, and n the number of vanes,

Q. = (nd n.i).b.u.

Let D be the diameter, and AI the area of the discharge peri-
phery of the wheel.

The peripheral velocity v t at the outlet circumference is

v.T>

TURBINES 309

Let 1*1 be the radial component of velocity of exit, then what-
ever the direction with which the water leaves the wheel the
radial component of velocity for a given discharge is constant.

The triangle of velocity can now be drawn as follows :

Set off BE equal to Vi, Fig. 194, and BK radial and equal
to UL

Let it now be supposed that the direction EF of the tip of the
vane at discharge is known. Draw EF parallel to the tip of the
vane at D, and through K draw KF parallel to BE to meet EF
in F.

Then BF is the velocity in direction and magnitude with which
the water leaves the wheel, relative to the ground, or to the fixed
casing of the turbine. Let this velocity be denoted by Ui. If,
instead of the direction EF being given, the velocity TJi is given
in direction and magnitude, the triangle of velocity at exit can be
drawn by setting out BE and BF equal to Vi and Ui respectively,
and joining EF. Then the tip of the blade must be made parallel
toEF.

For any given value of Ui the quantity of water flowing
through the wheel is

Q = AiUi cos ft = AiWi.

Work done on the wheel neglecting friction, etc. The kinetic
energy of the water as it leaves the turbine wheel is

2^- per pound,

and if the discharge is into the air or into the tail water this
energy is of necessity lost. Neglecting friction and other losses,
the available energy per pound of water is then

H-^Lfootlbs.,
and the theoretical hydraulic efficiency is

and is constant for any given value of Ui, and independent of the
direction of Ui. This efficiency must not be confused with the
actual efficiency, which is much less than E.

The smaller Ui, the greater the theoretical hydraulic efficiency,
and since for a given flow through the wheel, Ui will be least
when it is radial and equal to t*i, the greatest amount of work
will be obtained for the given flow, or the efficiency will be a
maximum, when the water leaves the wheel radially. If the

310 HYDRAULICS

water leaves with a velocity Ui in any other direction, the
efficiency will be the same, but the power of the wheel will be
diminished. If the discharge takes place down a suction tube,
and there is no loss between the wheel and the outlet from the
tube, the velocity head, lost then depends upon the velocity Ui
with which the water leaves the tube, and is independent of the
velocity or direction with which the water leaves the wheel.

The velocity of whirl at inlet and outlet. The component of
U, Fig. 193, in the direction of v is the velocity of whirl at inlet,
and the component of Ui, Fig. 194, in the direction of v i9 is the
velocity of whirl at exit.

Let V and Vi be the velocities of whirl at inlet and outlet
respectively, then

and Vi = Ui sin /? = u t tan ft.

Work done on the wheel. It has already been shown,
section 173, page 275, that when water enters a wheel, rotating
about a fixed centre, with a velocity U, and leaves it with velocity
Ui, the component Vi of which is in the same direction as Vi, the
work done on the wheel is

per pound,

9 9
and therefore, neglecting friction,

_TT W

This is a general formula for all classes of turbines and should
be carefully considered by the student.
Expressed trigonometrically,

vU cos ^Mjtanff _ TT _ UL re>\

^^ _Li _. .. ( u) .

g 9 %g

If F is to the left of BK, V! is negative.

Again, since the radial flow at inlet must equal the radial flow
at outlet, therefore

AUsin0 = AiTJiCos0 ..................... (3).

When Ui is radial, Vi is zero, and Ui equals v l tan a.

-H-' ........................... (4),

, . , TJ i

from which - =H -- ^ - ..................... ( 5 )>

andfrom(3) AU sin 6 = Aj^ tan <* . .................... (6).

TURBINES 311

If the tip of the vane is radial at inlet, i.e. V r is radial,

and

V a v*

(8).

In actual turbines is from '02H to '07H.

Example. An outward flow turbine wheel, Fig. 195, has an internal diameter of
6-249 feet, and an external diameter of 6-25 feet, and it makes 250 revolutions per
minute. The wheel has 32 vanes, which may be taken as f inch thick at inlet and
1 inches thick at outlet. The head is 141*5 feet above the centre of the wheel and
the exhaust takes place into the atmosphere. The effective width of the wheel face
at inlet and outlet is 10 inches. The quantity of water supplied per second is
215 cubic feet.

Neglecting all frictional losses, determine the angles of the tips of the vanes at
inlet and outlet so that the water shall leave radially.

The peripheral velocity at inlet is

v = TT x 5-249 x Ytf 1 = 69 ft- per see.,
and at outlet v, = TT x 6-25 x 3f = 82 ft.

Fig. 195.

The radial velocity of flow at inlet is

215

TT x 5-249 x H - if
= 18-35 ft. per sec.
The radial velocity of flow at exit is

215

Therefore,

= 16-5 ft. per seo.
^=4-23 ft.

312

HYDRAULICS

Then

and

= 141-5 -4-23
9

= 137-27 ft.
T7 137-27 x 32-2
V= 69~

: 64 ft. per sec.

To draw the triangle of velocities at inlet set out v and u at right angles.

Then since V is 64, and is the tangential component of U, and n is the radial
component of U, the direction and magnitude of U is determined.

By joining B and C the relative velocity V r is obtained, and BC is parallel to the
tip of the vane.

The triangle of velocities at exit is DEF, and the tip of the vane must be parallel
toEF.

Fig. 196.

Pig. 197.

The angles 0, <f>, and a can be calculated; for

tan 0=-

- 3-670

and

and, therefore,

= 105 14',
a = 11 23'.

It will be seen later how these angles are modified when friction is considered.

Fig. 198 shows the form the guide blades and vaues of the wheel would
probably take.

The path of the water through the wheel. The average radial velocity through
the wheel may be taken as 17-35 feet.

The time taken for a particle of water to get through the wheel is, therefore,

The angle turned through by the wheel in this time is 0-39 radians.

Set off the arc AB, Fig. 198, equal to -39 radian, and divide it into four equal
parts, and draw the radii ea,fb, gc and Ed.

Divide AD also into four equal parts, and draw circles through A 1 , A 2 , and A v

Suppose a particle of water to enter the wheel at A in contact with a vane and
suppose it to remain in contact with the vane during its passage through the wheel.
Then, assuming the radial velocity is constant, while the wheel turns through tbe
arc A.e the water will move radially a distance AA A and a particle that came on to

TURBINES

313

the vane at A will, therefore, be in contact with the vane on the arc through A t .
The vane initially passing through A will be now in the position el, al being
equal to hJ and the particle will therefore be at 1. When the particle arrives on
the arc through Ag the vane will pass through /, and the particle will consequently
be at 2, 62 being equal to win. The curve A4 drawn through Al 2 etc. gives the
path of the water relative to the fixed casing.

Fig. 198.

185. Losses of head due to frictional and other resistances
in outward flow turbines.

The losses of head may be enumerated as follows :

(a) Loss by friction at the sluice and in the .penstock or
supply pipe.

If v is the velocity, and h a the head lost by friction in
the pipe,

h a = ~
2gm

(6) As the water enters and moves through the guide
passages there will be a loss due to friction and by sudden changes
in the velocity of flow.

This head may be expressed as

being a coefficient.

* Bee page 119.

314

HYDRAULICS

(c) There is a loss of head at entrance due to shock as
the direction of the vane at entrance cannot be determined
with precision.

This may be written

he = Jcil 2^>

that is, it is made to depend upon V r the relative velocity of the
water, and the tip of the vane.

(d) In the wheel there is a loss of head h d) due to friction,
which depends upon the relative velocity of the water and the
wheol. This relative velocity may be changing, and on any small
element of surface of the wheel the head lost will diminish, as the
relative velocity diminishes.

It will be seen on reference to Figs. 193 and 194, that as the
velocity of whirl YI is diminished the relative velocity of flow v r at
exit increases, but the relative velocity V r at inlet passes through
a minimum when V is equal to v, or the tip of the vane is radial.
If V is the relative velocity of the water and the vane at any
radius, and b is the width of the vane, and 'dl an element of
length, then,

& 2 being a third coefficient.

If there is any sudden change of velocity as the water passes
through the wheel there will be a further loss, and if the turbine
has a suction tube there may be also a small loss as the water
enters the tube from the wheel.

The whole loss of head in the penstock and guide passages may
be called H/ and the loss in the wheel h/. Then if U is the

Rotor

Boyden MSfuser

fixed

Fig. 199,

TURBINES 315

velocity with which the water leaves the turbine the effective
head is

U 2
H- 2T-&/-H/.

In well designed inward and outward flow turbines

varies from O'lOH to '22H and the hydraulic efficiency is, therefore,
from 90 to 78 per cent.

The efficiency of inward and outward flow turbines including
mechanical losses is from 75 to 88 per cent.

Calling the hydraulic efficiency e, the general formula (1),
section 184, may now be written

9 9

= '78to'9H.

Outward flow turbines were made by Boy den* about 1848 for
which he claimed an efficiency of 88 per cent. The workmanship
was of the highest quality and great care was taken to reduce
all losses by friction and shock. The section of the crowns of the
wheel of the Boyden turbine is shown in Fig. 199. Outside of
the turbine wheel was fitted a "diffuser" through which, after
leaving the wheel, the water moved radially with a continuously
diminishing velocity, and finally entered the tail race with a
velocity much less, than if it had done so direct from the wheel.
The loss by velocity head was thus diminished, and Boyden
claimed that the diffuser increased the efficiency by 3 per cent.

186. Some actual outward flow turbines.

Double outward flow turbines. The general arrangement of an
outward flow turbine as installed at Chevres is shown in Fig. 200.
There are four wheels fixed to a vertical shaft, two of which
receive the water from below, and two from above. The fall
varies from 27 feet in dry weather to 14 feet in time of flood.

The upper wheels only work in time of flood, while at other
times the full power is developed by the lower wheels alone, the
cylindrical sluices which surround the upper wheels being set in
such a position as to cover completely the exit to the wheel.

The water after leaving the wheels, diminishes gradually in
velocity, in the concrete passages leading to the tail race, and the
loss of head due to the velocity with which the water enters the

* JLoivell Hydraulic Experiments, J. B. Francis, 1855.

316

HYDRAULICS

tail race is consequently small. These passages serve the same
purpose as Boyden's diffuser, and as the enlarging suction tube,
in that they allow the velocity of exit to diminish gradually.

Fig. 200. Double Outward Flow Turbine. (Escher Wyss and Co.)

Outward flow turbine with horizontal axis. Fig. 201 shows a
section through the wheel, and the supply and exhaust pipes, of an
outward flow turbine, having a horizontal axis and exhausting
down a " suction pipe." The water after leaving the wheel enters
a large chamber, and then passes down the exhaust pipe, the
lower end of which is below the tail race.

The supply of water to the wheel is regulated by a horizontal
cylindrical gate S, between the guide blades Gr and the wheel. The
gate is connected to the ring R, which slides on guides, outside
the supply pipe P, and is under the control of the governor.

The pressure of the water in the supply pipe is prevented from
causing end thrust on the shaft by the partition T, and between
T and the wheel the exhaust water has free access.

Outward flow turbines at Niagara Falls. The first turbines
installed at Niagara Falls for the generation of electric power,

TURBINES

317

were outward flow turbines of the type shown in Figs. 202 and
203.

There are two wheels on the same vertical shaft, the water
being brought to the chamber between the wheels by a vertical
penstock 7' 6" diameter. The water passes upwards to one wheel
and downwards to the other.

Fig. 201. Outward Flow Turbine with Suction Tube.

As shown in Fig. 202 the water pressure in the chamber is
prevented from acting on the lower wheel by the partition MN,
but is allowed to act on the lower side of the upper wheel, the
upper partition HK having holes in it to allow the water free access
underneath the wheel. The weight of the vertical shaft, and of
the wheels, is thus balanced, by the water pressure itself.

The lower wheel is fixed to a solid shaft, which passes through
the centre of the upper wheel, and is connected to the hollow
shaft of the upper wheel as shown diagrammatically in Fig. 202.
Above this connection, the vertical shaft is formed of a hollow

318

HYDRAULICS

tube 38 inches diameter, except where it passes through the
bearings, where it is solid, and 11 inches diameter.

A thrust block is also provided to carry the unbalanced
weight.

The regulating sluice is external to the wheel. To maintain a
high efficiency at part gate, the wheel is divided into three separate
compartments as in Fourneyron's wheel.

Gonrvee&on, for
HoUUvw and
Solid Shaft

Water adsrvilted,
uruler th& upper
wheel to support

eight
of the- shaft

Fig. 202. Diagrammatic section of Outward Flow Turbine at Niagara Falls.

A vertical section through the lower wheel is shown in Fig.
203, and a part sectional plan of the wheel and guide blades in
Fig. 195.

(Further particulars of these turbines and a description of the
governor will be found in Cassier's Magazinej Yol. III., and in
Turbines Actuelle) Buchetti, Paris 1901.

187. Inward flow turbines.

In an inward flow turbine the water is directed to the wheel
through guide passages external to the wheel, and after flowing
radially finally leaves the wheel in a direction parallel to the axis.

Like the outward flow turbine it may work drowned or with a
suction tube.

The water only acts upon, the blades during the radial
movement.

TURBINES

319

As improved by Francis*, in 1849, the wheel was of the form
shown in Fig. 204 and was called by its inventor a "central vent
wheel."

The wheel is carried on a vertical shaft, resting on a footstep,
and supported by a collar bearing placed above the staging S.

* Lowell Hydraulic Experiments, F. B. Francis, 1855.

320

HYDRAULICS

Above tlie wheel is a heavy casting C, supported by bolts
from the staging S, which acts as a guide for the cylindrical
sluice F, and carries the bearing B for the shaft. There are
40 vanes in the wheel shown, and 40 fixed guide blades, the former
being made of iron one quarter of an inch thick and the latter
three-sixteenths of an inch.

Fig. 204. Francis' Inward flow or Central vent Turbine.

The triangles of velocities at inlet and outlet, Fig. 205, are
drawn, exactly as for the outward flow turbine, the only difference
being that the velocities v, U, V, Y r and u refer to the outer

TURBINES

321

periphery, and v,, Ui, V : , -y r and %! to the inner periphery of the
wheel.

The work done on the wheel is

iVi -, ,, ,,

--- - ft. Ibs. per lb.,
y y

and neglecting friction,

g 9 2g'

For maximum efficiency, for a given flow through the wheel,
Ui should be radial exactly as for the outward flow turbine.

Fig. 205.

The student should work the following example.

The outer diameter of an inward flow turbine wheel is 7*70 feet, and the inner
diameter 6-3 feet, the wheel makes 55 revolutions per minute. The head is
14-8 feet, the velocity at inlet is 25 feet per sec., and the radial velocity may be
assumed constant and equal to 7*5 feet. Neglecting friction, draw the triangles of
velocities at inlet and outlet, and find the directions of the tips of the vanes at
inlet and outlet so that there may be no shock and the water may leave radially.

Loss of head by friction. The losses of head by friction are
similar to those for an outward flow turbine (see page 313) and
the general formula becomes

When the flow is radial at exit,

The value of e varying as before between 078 and 0*90.

Example (1). An inward flow turbine working under a head of 80 feet has
radial blades at inlet, and discharges radially. The angle the tip of the vane
makes with the tangent to the wheel at exit is 30 degrees and the radial velocity
is constant. The ratio of the radii at inlet and outlet is 1-75. Find the velocity
of the inlet circumference of the wheel Neglect friction.

L. H.

322 HYDRAULICS

Since the discharge is radial, the velocity at exit la

Then

= pjg tan 30.

v z tan a 30

and since the blades are radial at inlet V is equal to r,

therefore v*=g . 80 v ^22!

1-75 2 2 '

from which

32x80

1-0543 '
;49'3 ft. per see.

Trijcungle cfPelociti^y

Fig. 206.

Example (2) The outer diameter of the wheel of an inward flow turbine of
200 horse-power is 2-46 feet, the inner diameter is 1-968 feet. The wheel makes
300 revolutions per minute. The effective width of the wheel at inlet = 1-15 feet. The
head is 39 '5 feet and 59 cubic feet of water per second are supplied. The radial
velocity with which the water leaves the wheel may be taken as 10 feet per second.

Determine the theoretical hydraulic efficiency E and the actual efficiency e l of
the turbine, and design suitable vanes.

200x550

39-5x59x62-5

Theoretical hydraulic efficiency

The radial velocity of flow at inlet,

= 6-7 feet per sec.

TURBINES 323

The peripheral velocity

v = 2-46 . TT x 8 ^}- = 38-6 feet.

The velocity of whirl V. Assuming a hydraulic efficiency of 85 %, from
the formula

v _ 39-5 x 32-2 x -85

38-6

= 28-0 feet per sec.
The angle 9. Since w = 6'7 ft. per sec. and V = 28'0 ft. per sec.

tan = ^=0-239,

= 13 27'.
The angle <f>* Since V is less than v, <f> is greater than 90.

and = 152.

For the water to discharge radially with a velocity of 10 feet per seo.

and a = 18 nearly.

The theoretical vanes are shown in Fig. 206.
Example (3). Find the values of <j> and a on the assumption that e is 0-80.

Thomson's inward flow turbine. In 1851 Professor James
Thomson invented an inward flow turbine, the wheel of which
was surrounded by a large chamber set eccentrically to the wheel,
as shown in Figs. 207 to 210.

Between the wheel and the chamber is a parallel passage, in
which are four guide blades Gr, pivoted on fixed centres C and
which can be moved about the centres C by bell crank levers,
external to the casing, and connected together by levers as shown
in Fig. 207. The water is distributed to the wheel by these guide
blades, and by turning the worm quadrant Q by means of the
worm, the supply of water to the wheel, and thus the power of
the turbine, can be varied. The advantage of this method of
regulating the flow, is that there is no sudden enlargement from
the guide passages to the wheel, and the efficiency at part load
is not much less than at full load.

Figs. 209 and 210 show an enlarged section and part sectional
elevation of the turbine wheel, and one of the guide blades Gr.
The details of the wheel and casing are made slightly different
from those shown in Figs. 207 and 208 to illustrate alternative
methods.

The sides or crowns of the wheel are tapered, so that the
peripheral area of the wheel at the discharge is equal to the
peripheral area at inlet. The radial velocities of flow at inlet
and outlet are, therefore, equal.

212

324 HYDRAULICS

The inner radius r in Thomson's turbine, and generally in
turbines of this class made by English makers, is equal to one-half
the external radius E.

Fig. 207. Guide blades and casing of Thomson Inward Flow Turbine.

The exhaust for the turbine shown takes place down two
suction tubes, but the turbine can easily be adapted to work below
the tail water level.

As will be seen from the drawing the vanes of the wheel are
made alternately long and short, every other one only continuing
from the outer to the inner periphery.

TURBINES

325

The triangles of velocities for the inlet and outlet are shown in
Pig. 211, the water leaving the wheel radially.

The path of the water through the wheel, relative to the fixed
casing, is also shown and was obtained by the method described
on page 312.

Inward flow turbines with adjustable guide blades, as made by
the continental makers, have a much greater number of guide
blades (see Fig. 233, page 352).

Fig. 208. Section through wheel and casing of Thomson Inward Flow Turbine.

188. Some actual inward flow turbines.

A later form of the Francis inward flow turbine as designed by
Pictet and Co., and having a horizontal shaft, is shown in Fig. 212.

The wheel is double and is surrounded by a large chamber
from which water flows through the guides G to the wheel W.
After leaving the wheel, exhaust takes place down the two suction
tubes S, thus allowing the turbine to be placed well above the
tail water while utilising the full head.

The regulating sluice F consists of a steel cylinder, which
slides in a direction parallel to the axis between the wheel and
guides.

326

HYDKAULICS

irvGwi/cLe.

Fig. 209. Fig- 210.

Detail of wheel and guide blade of Thomson Inward Flow Turbine.

- - v

. 211.

TURBINES

327

The wheel is divided into five separate compartments, so that
at any time only one can be partially closed, and loss of head by
contraction and sudden enlargement of the stream, only takes
place in this one compartment.

328

HYDRAULICS

The sluice F is moved by two screws T, which slide through
stuffing boxes B, and which can be controlled by hand or by the
governor B.

Inward flow turbine for low falls and variable head. The
turbine shown in Fig. 213 is an example of an inward flow turbine
suitable to low falls and variable head. It has a vertical axis and
works drowned. The wheel and the distributor surrounding the
wheel are divided into five stages, the two upper stages being
shallower than the three lower ones, and all of which stages can

Fig. 213. Inward Flow Turbine for a low and variable fall. (Pictet and Co.)

TURBINES 329

be opened or closed as required by the steel cylindrical sluice CO
surrounding the distributor.

When one of the stages is only partially closed by the sluice,
a loss of efficiency must take place, but the efficiency of this one
stage only is diminished, the stages that are still open working
with their full efficiency. With this construction a high efficiency
of the turbine is maintained for partial flow. With normal flows,
and a head of about 6*25 feet, the three lower stages only are
necessary to give full power, and the efficiency is then a
maximum. In times of flood there is a large volume of water
available, but the tail water rises so that the head is only about
4'9 feet, the two upper stages can then be brought into operation
to accommodate a larger flow, and thus the same power may be
obtained under a less head. The efficiency is less than when the
three stages only are working, but as there is plenty of water
available, the loss of efficiency is not serious.

The cylinder C is carried by four vertical spindles S, having
racks R fixed to their upper ends. Gearing with these racks, are
pinions p, Fig. 213, all of which are worked simultaneously by the
regulator, or by hand. A bevel wheel fixed to the vertical shaft
gears with a second bevel wheel on a horizontal shaft, the velocity
ratio being 3 to 1.

189. The best peripheral velocity for inward and outward
flow turbines.

When the discharge is radial, the general formula, as shown on
page 315, is

= eH>0-78toO'90H ....... .............. (1).

If the blades are radial at inlet, for no shock, v should be equal
to Y, and

or v = V = G'624 to

This is sometimes called the best velocity for v, but it should be
clearly understood that it is only so when the blades are radial at
inlet.

190. Experimental determination of the best peripheral
velocity for inward and outward flow turbines.

For an outward flow turbine, working under a head of 14 feet,
with blades radial at inlet, Francis* found that when v was

0'626

Lowell, Hydraulic Experiments,

3 30 HYDRAULICS

the efficiency was a maximum and equal to 79'87 per cent. The
efficiency however was over 78 per cent, for all values of v
between 0'545 *j2gK and '671 \/2#H. If 3 per cent, be allowed
for the mechanical losses the hydraulic efficiency may be taken
as 82'4 per cent.

~Vv
From the formula = *S24H, and taking V equal to v,

v = '64 V2^H,

so that the result of the experiment agrees well with the formula.
For an inward flow turbine having vanes as shown in Fig. 205,
the total efficiency was over 79 per cent, for values of v between
0-624 \/2#H and 0'708 \/2#H, the greatest efficiency being 79'7
per cent, when v was 0'708 v2#H and again when v was
637

It will be seen from Fig. 205 that although the tip of the vane
at the convex side is nearly radial, the general direction of the
vane at inlet is inclined at an angle greater than 90 degrees to
the direction of motion, and therefore for no shock Y should be
less than v.

When v was '708 N/20H, V, Fig. 205, was less than v. The
value of Y was deduced from the following data, which is also
useful as being taken from a turbine of very high efficiency.

Diameter of wheel 9'338 feet.

Width between the crowns at inlet 0'999 foot.

There were 40 vanes in the wheel and an equal number oi
fixed guides external to the wheel.

The minimum width of each guide passage was 0'1467 foot and
the depth T0066 feet,

The quantity of water supplied to the wheel per second was
112*525 cubic feet, and the total fall of the water was 13'4 feet.
The radial velocity of flow u was, therefore, 3*86 feet per second.

The velocity through the minimum section of the guide passage
was 19 feet per second.

When the efficiency was a maximum, v was 20'8 feet per sec.
Then the radial velocity of flow at inlet to the wheel being
3'86 feet, and U being taken as 19 feet per second, the triangle
of velocities at inlet is ABC, Fig. 205, and Y is 18'4 feet per sec.

If it is assumed that the water leaves the wheel radially, then

eH= = H'85 feet.
9

1 1*85
The efficiency e should be =88'5 per cent., which is 9 per

cent, higher than the actual efficiency.

TURBINES 331

The actual efficiency however includes not only the fluid losses
but also the mechanical losses, and these would probably be from
2 to 8 per cent., and the actual work done by the turbine on the
shaft is probably between 80 and 86*5 per cent, of the work done
by the water.

191. Value of e to be used in the formula = eH.

In general, it may be said that, in using the formula = eH,

the value of e to be used in any given case is doubtful, as even
though the efficiency of the class of turbines may be known, it is
difficult to say exactly how much of the energy is lost mechanically
and how much hydraulically.

A trial of a turbine without load, would be useless to deter-
mine the mechanical efficiency, as the hydraulic losses in such a
trial would be very much larger than when the turbine is working
at full load. By revolving the turbine without load by means of
an electric motor, or through the medium of a dynamometer, the
work to overcome friction of bearings and other mechanical losses
could be found. At all loads, from no load to full load, the
frictional resistances of machines are fairly constant, and the
mechanical losses for a given class of turbines, at the normal load
for which the vane angles are calculated, could thus approximately
be obtained. If, however, in making calculations the difference
between the actual and the hydraulic efficiency be taken as, say,
5 per cent., the error cannot be very great, as a variation of 5 per
cent, in the value assumed for the hydraulic efficiency e t will only
make a difference of a few degrees in the calculated value of
the angle <.

The best value for e, for inward flow turbines, is probably 0'80,
and experience shows that this value may be used with confidence.

Example. Taking the data as given in the example of section 184, and assuming
an efficiency for the turbine of 75 per cent., the horse-power is
215 x 62-4 x 141-5 x -75 x 60

33,000

=2600 horse-power.

If the hydraulic efficiency is supposed to be 80 per cent., the velocity of
whirl V should be

Sg.H^O-8.32.1415

v 69

=52 feet per sec.

Then tan - 18 ' 35 - - 1835

and 0=132 47'.

Now suppose the turbine to be still generating 2600 horse-power, and to have
an efficiency of 80 per cent., and a hydraulic efficiency of 85 per cent.

332

HYDRAULICS

Then the quantity of water required per second, is
215 x 0-75

0-8

: 200 cubic feet per sec.

and the radial velocity of flow at inlet will be

1835x200 .
u= =17'1 ft. per sec.

. -85.32.141-5

Then

tan

69
17-1

"55-4- 69 :
= 128. 24'.

:55 - 4 ft. per sec.

-17-1
13-6

192. The ratio of the velocity of whirl V to the velocity
of the inlet periphery v.

Experience shows that, consistent with Vu satisfying the general

formula, the ratio ^ may vary between very wide limits without
considerably altering the efficiency of the turbine.

Table XXXVII shows actual values of the ratio , taken

from a number of existing turbines, and also corresponding values

Fig. 214.

TURBINES

833

of /s = ff > V being calculated from = 0'8H. The corresponding

variation in the angle <, Fig. 214, is from 20 to 150 degrees.

For a given head, v may therefore vary within wide limits,
which allows a very large variation in the angular velocity of the
wheel to suit particular circumstances.

TABLE XXXVII.

Showing the heads, and the velocity of the receiving circum-
ference v for some existing inward and outward, and mixed flow
turbines.

Katio

Hfeet

v feet
per sec.

N/20H

H.P.

V being calculated
Vv

/O TT

^ offtL

from = -8H

Inward flow :

Niagara Falls*
Rheinfelden

146
14-8

70
22

96-8
30-7

0-72
0-71

5000
840

0-555
0-565

By Theodor )
BeU and Co. j

28-4

42-6

0-91

0-44

60-4

32-2

62-3

0-52

0-77

Pictet and Co.

183-7

51-1

76-8

0-47

300

0-85

134-5

46-6

65-6

0-505

300

0-79

6-25

16-6

0-83

0-48

25-75

0-58

700

0-69

38-5

50-3

0-77

200

0-52

Ganz and Co.

112

64-3

84-6

0-54

0-74

225

64-7

120

0-54

682

0-58

Rioter and Co.

10-66

15-2

0-585

0-69

Outward flow :

Niagara Falls
Pictet and Co.

141-5
130-5

69
69

95-2
91-6

0-725)
0-750)

5000

0-55
0-53

Ganz and Co.

95-1

38-7

78-0

0-495

290

0-81

223

55-6

120-0

0-46

1200

0-87

* Escher Wyss and Co.

For example, if a turbine is required to drive alternators
direct, the number of revolutions will probably be fixed by the
alternators, while, as shown later, the diameter of the wheel is
practically fixed by the quantity of water, which it is required to
pass through the wheel, consistent with the peripheral velocity of
the wheel, not being greater than 100 feet per second, unless, as
in the turbine described on page 373, special precautions are
taken. This latter condition may necessitate the placing of two
or more wheels on one shaft.

334 HYDRAULICS

Suppose then, the number of revolutions of the wheel to be
given and d is fixed, then v has a definite value, and V must be
made to satisfy the equation

Vv
=eH.
9

Fig. 214 is drawn to illustrate three cases for which Yv is
constant. The angles of the vanes at outlet are the same for all
three, but the guide angle and the vane angle </> at inlet vary
considerably.

193. The velocity with which water leaves a turbine.
In a well-designed turbine the velocity with which the water

leaves the turbine should be as small as possible, consistent with
keeping the turbine wheel and the down-take within reasonable
dimensions.

In actual turbines the head lost due to this velocity head
varies from 2 to 8 per cent. If a turbine is fitted with a
suction pipe the water may be allowed to leave the wheel itself
with a fairly high velocity and the discharge pipe can be made
conical so as to allow the actual discharge velocity to be as small
as desired. It should however be noted that if the water leaves
the wheel with a high velocity it is more than probable that there
will be some loss of head due to shock, as it is difficult to ensure
that water so discharged shall have its velocity changed gradually.

194. Bernouilli's equations applied to inward and out-
ward flow turbines neglecting friction.

Centrifugal head impressed on the water by the wheel. The
theory of the reaction turbines is best considered from the point
of view of Bernoulli's equations ; but before proceeding to discuss
them in detail, it is necessary to consider the " centrifugal head "
impressed on the water by the wheel.

This head has already been considered in connection with the
Scotch turbine, page 303.

Let r, Fig. 216, be the internal radius of a wheel, and R the
external radius.

At the internal circumference let the wheel be covered with a
cylinder c so that there can be no flow through the wheel, and let
it be supposed that the wheel is made to revolve at the angular
velocity w which it has as a turbine, the wheel being full of water
and surrounded by water at rest, the pressure outside the wheel
being sufficient to prevent the water being whirled out of the
wheel. Let d be the depth of the wheel between the crowns.
Consider any element of a ring of radius r and thickness dr, and
subtending a small angle 6 at the centre 0, Fig. 216.

TURBINES

335

The weight of the element is

wr . dr .d,
and the centrifugal force acting on the element is

wr Q . dr . d . wV ,,
Ibs.

Let p be the pressure per unit area on the inner face of the
element and p + dp on the outer.

wr .dr.d. wV

Then

g.r.e.d

Fig. 215.

Fig. 216.

The increase in the pressure, due to centrifugal forces, between
r and B, is, therefore,

t PC _ U) /T>2 2\ _ V V l

For equilibrium, therefore, the pressure in the water surround-
ing the wheel must be p c .

If now the cylinder c be removed and water is allowed to flow
through the wheel, either inwards or outwards, this centrifugal
head will always be impressed upon the water, whether the wheel
is driven by the water as a turbine, or by some external agency,
and acts as a pump.

Bernoulli's equations. The student on first reading these
equations will do well to confine his attention to the inward flow
turbine, Fig. 217, and then read them through again, confining his
attention to the outward flow turbine, Fig. 191.

336

HYDRAULICS

Let p be the pressure at A, the inlet to the wheel, or in the
clearance between the wheel and the guides, pi the pressure at
the outlet B, Fig. 217, and p a the atmospheric pressure, in pounds
per square foot. Let H be the total head, and H the statical
head at the centre of the wheel. The triangles of velocities are
as shown in Figs. 218 and 219.

Then at A

(i).

Between B and A the wheel impresses upon the water the
centrifugal head

v being greater than
outward flow.

_
2g 2g>

for an inward flow turbine and less for the

Fig. 217.

Consider now the total head relative to the wheel at A and B.
The velocity head at A is - and the pressure head is , and

at B the velocity and pressure heads are - and respectively.

If no head were impressed on the water as it flows through
the wheel, the pressure head plus the velocity head at A and B
would be equal to each other. But between A and B there is
impressed on the water the centrifugal head, and therefore,

_-
w 2g 2g 2g w 2g

TURBINES 337

This equation can be used to deduce the fundamental equation,

Y!?_!^ = ^. ...(3).

9
From the triangles ODE and ADE, Fig. 218,

Y r 2 =(Y-<y) 2 + ^ 2 andY 2 + ^ 2 = IP,
and from the triangle BFGr, Fig 219,

Vr = (vi - Yi) 2 + U? and Yi 2 + u* = Ui 2 .
Therefore by substitution in (2),

Pi , fa-Yi)' + v* _t?i a , u? = P i (V-vY { u*

2(7

From which

w g 2g w 2g g '

and

g g w w 2g 2g

JT2

Substituting for - + ~- from (1)

.(5).

Fig. 218.

Wheel in suction tube. If the centre of the wheel is 7i
above the surface of the tail water, and Uo is the velocity with
which the water leaves the down-pipe, then

Substituting f or ^ + i- in (6),

SI.ftSl.Hi+fis-***.-^

g g w w zg

= H-P.

L. U.

338 HYDRAULICS

IfVisO, ^H-^U.

9 20

The wheel can therefore take full advantage of the head H
even though it is placed at some distance above the level of the
tail water.

Drowned wheel. If the level of the tail water is CD, Fig. 217,
or the wheel is drowned, and Jh is the depth of the centre of the
wheel below the tail race level,

fc-n,+*,

w w '

and the work done on the wheel per pound of water is again

vV ViV! _ W ,
--- = JbL s~~ = fi.
99 20

vV
IfVjisO, g =h '

From equation (5),

vV t?iYi = p Pi , H* Hi 3
9 9 w w 2g 2g '

so that the work done on the wheel per pound is the difference
between the pressure head plus the velocity head at entrance and
the pressure head plus velocity head at exit.

In an impulse turbine p and pi are equal, and the work done
is then the change in the kinetic energy of the jet when it strikes
and when it leaves the wheel.

A special case arises when p is equal to p. In this case a
considerable clearance may be allowed between the wheel and the
fixed guide without danger of leakage.

Equation (2), for this case, becomes

lL = ^L 2 + ^_^L

20 2g 20 2g'

and if at exit v r is made equal to Vi, or the triangle BFG,
Fig. 219, is isosceles,

V,' = ^
20 20'

and the triangle of velocities at entrance is also isosceles.
The pressure head at entrance is

and at exit is either + fei , or - h .

TURBINES 339

Therefore, since the pressures at entrance and exit are equal,

U 2

2^ = H -^ = H,

or else H + 7io = H.

The water then enters the wheel with a velocity equal to that
due to the total head H, and the turbine becomes a free-deviation
or impulse turbine.

195. Bernoulli's equations for the inward and outward
flow turbines including friction.

If H/ is the loss of head in the penstock and guide passages,
hf the loss of head in the wheel, h t the loss at exit from the wheel
and in the suction pipe, and Ui the velocity of exhaust,

+-*+*-* ........................ a),

and = + fc e -fci ........................... (3),

w w

from which = H-(^ + h f + H/+k) .................. (4).

If the losses can be expressed as a fraction of H, or equal to KH,
then

= (l-K)H=eH

= 0'78H to 0-90II*.

196. Turbine to develop a given horse-power.

Let H be the total head in feet under which the turbine works.

Let n be the number of revolutions of the wheel per minute.

Let Q be the number of cubic feet of water per second required
by the turbine.

Let E be the theoretical hydraulic efficiency.

Let e be the hydraulic efficiency.

Let e m be the mechanical efficiency.

Let 61 be the actual efficiency including mechanical losses.

Let Ui be the radial velocity with which the water leaves the
wheel.

Let D be the diameter of the wheel in feet at the inlet circum-
ference and d the diameter at the outlet circumference.

Let B be the width of the wheel in feet between the crowns
at the inlet circumference, and b be the width between the crowns
at the outlet circumference.

Let N be the horse-power of the turbine.
* See page 315.

22-2

340 HYDRAULICS

The number of cubic feet per second required is

N.33 ? 000
* eiH. 62'4. 60 '

A reasonable value for e\ is 75 per cent.

The velocity U with which the water leaves the turbine, since

U 2

is U =\/20(l-E)Hft. per sec ................ (2).

If it be assumed that this is equal to u\ 9 which would of
necessity be the case when the turbine works drowned, or
exhausts into the air, then, if t is the peripheral thickness of the
vanes at outlet and m the number of vanes,

If Uo is not equal to Ui, then

(ird-mt) 1^1 = 0, ........................ (3).

The number of vanes m and the thickness t are somewhat
arbitrary, but in well-designed turbines t is made as small as
possible.

As a first approximation mt may be taken as zero and (3)
becomes

wd6wi = Q .............................. (4).

For an inward flow turbine the diameter d is fixed from
consideration of the velocity with which the water leaves the
wheel in an axial direction.

If the water leaves at both sides of the wheel as in Fig. 208,
and the diameter of the shaft is d , the axial velocity is

UQ - - j - ft. per sec.

The diameter d can generally be given an arbitrary value, or
for a first approximation to d it may be neglected, and u may be
taken as equal to %. Then

j /^Q> fi. ffc\

a = A/^ it \oj.

From (4) and (5) b and d can now be determined.

A ratio for -T having been decided upon, D can be calculated,

and if the radial velocity at inlet is to be the same as at outlet,
and i is the thickness of the vanes at inlet,

v^ / j /\ 7 /^5\

(TT -m ; - Ui -V* m

TURBINES 341

For rolled brass or wrought steel blades, t may be very small,
and for blades cast with the wheel, by shaping them as in Fig. 227,
to is practically zero. Then

7TUL)

If now the number of revolutions is fixed by any special
condition, such as having to drive an alternator direct, at some
definite speed, the peripheral velocity is

P er sec

Then = eR.

and if e is given a value, say 80 per cent.,

V= -^ ft. per sec (8).

Since u, V, and v are known, the triangle of velocities at inlet
can be drawn and the direction of flow and of the tip of vanes
at inlet determined. Or and <, Fig. 214, can be calculated from

(9)

and tan< = y ........................ (10).

Then IT, the velocity of flow at inlet, is

irn f .
At exit t?i = -gQ- it. per sec.,

and taking u^ as radial and equal to u, the triangle of velocities
can be drawn, or a calculated from

tan a = - .
v\

If Ho is the head of water at the centre of the wheel and H/ the
head lost by friction in the supply pipe and guide passages, the
pressure head at the inlet is

Example. An inward flow turbine is required to develop 300 horse-power under
a bead 6U feet, and to run at 250 revolutions per minute.
To determine the leading dimensions of the turbine.
Assuming c^ to be 75 per cent.,

300 x 33,000
^~ -75x60x62-4x60
= 58-7 cubic feet per sec.

342 HYDRAULICS

Assuming E is 95 per cent., or five per cent, of the head is lost by velocity
of exit and Uj = u,

| = -05.60

and w= 13-8 feet per sec.

Then from (5), page 340,

= 1-65 feet,
say 20 inches to make allowance for shaft and to keep even dimension.

Then from (4) , b = ^ = -82 foot

JL'OD

= 9 inches say.
Taking - as 1-8, D=3-0 feet, and

v = TT . 3 . *f = 39-3 feet per sec.,
and B = 5 inches say.

Assuming e to be 80 per cent.,

T7 -80 x 60 x 32 ,

- 3^3 - per sec<

13-8

and 0=19 30',

13-8

and 0=91 15'.

13-8x1-8

and a =32 18'.

The velocity U at inlet is

= 41-3 ft. per sec.
The absolute pressure head at the inlet to the wheel is

2- = H +^ -- - -- hf, the head lost by friction in the down pipe

= H + 34 -26-5 -ft/.

The pressure head at the outlet of the wheel will depend upon the height of the
wheel above or below the tail water.

197. Parallel or axial flow turbines.

Fig. 220 shows a double compartment axial flow turbine, the
guide blades being placed above the wheel and the flow through
the wheel being parallel to the axis. The circumferential section
of the vanes at any radius when turned into the plane of the
paper is as shown in Fig. 221. A plan of the wheel is also shown.

The triangles of velocities at inlet and outlet for any radius
are similar to those for inward and outward flow turbines, the
velocities v and v i9 Figs. 222 and 223, being equal.

TURBINES

The general formula now becomes

343

U, 1

For maximum efficiency for a given flow, the water should
leave the wheel in a direction parallel to the axis, so that it has
no momentum in the direction of v.

Fig. 220. Double Compartment Parallel Flow Turbine.

Figs. 221, 222, 223.
Then, taking friction and other losses into account,

344

HYDRAULICS

The velocity v will be proportional to the radius, so that if the
water is to enter and leave the wheel without shock, the angles 0,
<, and a must vary with the radius.

The variation in the form of the vane with the radius is shown
by an example.

A Jonval wheel has an internal diameter of 5 feet and an
external diameter of 8' 6". The depth of the wheel is 7 inches.
The head is 15 feet and the wheel makes 55 revolutions per
minute. The flow is 300 cubic feet per second.

To find the horse-power of the wheel, and to design the wheel
vanes.

Let TI be any radius, and r and r 2 the radii of the wheel at the
inner and outer circumference respectively. Then

r = 2'5 feet and v = 2-n-r |f = 14*4 feet per sec.,
TI = 3'75 feet and Vi = 2irri f = 21'5 feet per sec.,
r- 2 = 4*25 feet and v 2 = 27rr 2 f = 24*5 feet per sec.
The mean axial velocity is

300
u = = 8 ' 15 f

Fig. 224. Triangles of velocities at inlet and outlet at three different
radii of a Parallel Flow Turbine.

Taking e as 0'80 at each radius,

T7 _0'8. 32'2.15 385
V = 14'4

= 26'7 ft. per sec.,

JL"J? ~a? j.~x ~x

Vi = ni^c = 17'9 ft. per sec.,

V 2 = oTTc = 15*7 ft. per sec.

Inclination of the vanes at inlet. The triangles of velocities
for the three radii r, ri, r 2 are shown in Fig. 224. For example,
at radius r, ADC is the triangle of velocities at inlet and ABC the

TURBINES 345

triangle of velocities at outlet. The inclinations of the vanes at
inlet are found from

O.-l K

tan <ft = . _ . > fr m which < = 3330',

8'15

= 113 50,

8'15
tan < 2 = 15 . 7 _ 24-5 > from wnicn ^ = 137 6'.

The inclination of the guide blade at each of the three radii.
8 ' 15

from which = 17,

tan^ji^ and ^ = 24 30',

tan<9 2 = f^ and 2 = 27 30'.
lo /

The inclination of the vanes at exit.

tan a, = = 20 48',

Zi O

tan -,= |^> = 18 22'.

<u4 O

If now the lower tips of the guide blades and the upper tips
of the wheel vanes are made radial as in the plan, Fig. 221, the
inclination of the guide blade will have to vary from 17 to
27 degrees or else there will be loss by shock. To get over this
difficulty the upper edge only of each guide blade may be made
radial, the lower edge of the guide blade and the upper edge of
each vane, instead of being radial, being made parallel to the
upper edge of the guide. In Fig. 225 let r and R be the radii
of the inner and outer crowns of the wheel and also of the guide
blades. Let MN be the plan of the upper edge of a guide blade
and let DGr be the plan of the lower edge, DGr being parallel to
MN. Then as the water runs along the guide at D, it will leave
the guide in a direction perpendicular to OD. At Gr it will leave
in a direction HGr perpendicular to OGr. Now suppose the guide
at the edge DGr to have an inclination ft to the plane of the paper.
If then a section of the guide is taken by a vertical plane XX
perpendicular to DG , the elevation of the tip of the vane on this
plane will be AL, inclined at ft to the horizontal line AB, and AC

346

HYDRAULICS

will be the intersection of the plane XX with the plane tangent
to the tip of the vane.

Now suppose DE and GH to be the projections on the plane
of the paper of two lines lying on the tangent plane AC and
perpendicular to OD and OG respectively. Draw EF and HK
perpendicular to DE and GH respectively, and make each of
them equal to BC. Then the angle EDF is the inclination of the
stream line at D to the plane of the paper, and the angle HGK is
the inclination of the stream line at G to the plane of the paper.
These should be equal to and a .

flarb of lower edge of guide,
blade'& of upper edge* of*

Fig. 225. Plan of guide blades and vanes of Parallel Flow Turbines.

Let y be the perpendicular distance between MN and DGr.
Let the angles GOD and GOH be denoted by < an a respectively.

Since EF, BC and HK are equal,

ED tan B = y tan /? (1),

and GH tan a = # tan /3 (2).

But

and

Therefore
and

Again,

cos (a +

= cos(a + )tan/3 ..................... (3),

tan 2 = cos a tan p ........................... (4).

sin a =

(5).

There are thus three equations from which a, <f> and P can be
determined.

Let x and y be the coordinates of the point D, being the
intersection of the axes.

TURBINES

347

Then

and from (5)

cos (a

cos a

- /ITZ

-V ] R 2 '

Substituting for cos (a + <) and cos a and the known values of
tan and tan 2 in the three equations (35), three equations are
obtained with x, y, and ft as the unknowns.

Solving simultaneously

x = 1*14 feet,

y = 2'23 feet,

and tan p = 0'67,

from which p = 34.

Fig. 226.

Fig. 228.

The length of the guide blade is thus found, and the constant
slope at the edge DG so that the stream lines at D and Gr shall
have the correct inclination.

If now the upper edge of the vane is just below DG, and the
tips of the vane at D and G- are made as in Figs. 226 228, < and

HYDRAULICS

< 2 being 33 30' and 137 6' respectively, the water will move on to
the vane without shock.

The plane of the lower edge of the vane may now be taken as
D'G-', Fig. 225, and the circular sections DD', PQ, and GGT at the
three radii, r, r 1} and r 2 are then as in Figs. 226 228.

198. Regulation of the flow to parallel flow turbines.

To regulate the flow through a parallel flow turbine, Fontaine
placed sluices in the guide passages, as in Fig. 229, connected to
a ring which could be raised or lowered by three vertical rods
having nuts at the upper ends fixed to toothed pinions. When

Fig. 229. Fontaine's Sluices.

Fig. 230. Adjustable guide blades for Parallel Flow Turbine.

the sluices required adjustment, the nuts were revolved together
by a central toothed wheel gearing with the toothed pinions
carrying the nuts. Fontaine fixed the turbine wheel to a hollow
shaft which was carried on a footstep above the turbine. In some
modern parallel flow turbines the guide blades are pivoted, as in
Fig. 230, so that the flow can be regulated. The wheel may be
made with the crowns opening outwards, in section, similar to
the Grirard turbine shown in Fig. 254, so that the axial velocity
with which the water leaves the wheel may be small.

The axial flow turbine is well adapted to low falls with variable
head, and may be made in several compartments as in Fig. 220.
In this example, only the inner ring is provided with gates. In
dry weather flow the head is about .3 feet and the gates of the
inner ring can be almost closed as the outer ring will give the full

TURBINES 349

power. During times of flood, and when there is plenty of water,
the head falls to 2 feet, and the sluices of the inner ring are
opened. A larger supply of water at less head can thus be
allowed to pass through the wheel, and although, due to the shock
in the guide passages of the inner ring, the wheel is not so efficient,
the abundance of water renders this unimportant.

Example. A double compartment Jonval turbine has an outer diameter of
12' 6" and an inner diameter of 6 feet.

The radial width of the inner compartment is 1' 9" and of the outer compart-
ment r 6". Allowing a velocity of flow of 3-25 ft. per second and supposing the
minimum fall is V 8", and the number of revolutions per minute 14, find the horse-
power of the wheel when all the guide passages are open, and find what portion of
the inner compartment must be shut off so that the horse-power shall be the same
under a head of 3 feet. Efficiency 70 per cent.

Neglecting the thickness of the blades,

the area of the outer compartment = - (12-5 a -9-5 2 ) = 52 < 6 sq. feet.

inner = (9'5 2 -6 2 )=42-8 sq. feet.

Total area = 95 -4 sq. feet.

The weight of water passing through the wheel is

W=95-4 x 62-4 x 3-25 Ibs. per sec.

= 19,3001bs. per seo.
and the horse-power is

Hp = l<l800xl*6x0.7

550

Assuming the velocity of flow constant the area required when the head
is 3 feet is

40-8x33,000

k ~GOx62-5x3x-7
= 55-6sq. feet,
or the outer wheel will nearly develop the horse-power required.

199. Bernouilli's equations for axial flow turbines.

The Bernouilli's equations for an axial flow turbine can be
written down in exactly the same way as for the inward and
outward flow turbines, page 335, except that for the axial flow
turbine there is no centrifugal head impressed on the water
between inlet and outlet.

Then, + F = -' + ^ + A /.

w 2g w 2g

from which, since v is equal to v if

p Y 2 -2Yi; + i; 2 <u? p, . ^-2V 1 t?+Vi a , V , R

+ ~~ + ~~

p V 2 v u pi _,_ i
therefore -+5 --- + ^- = + -Q-
w 2g g 2g w 2g

, p u 2 u, 2 P!

and. --- = H n --- ^

g g w 2g 2g w

350

HYDRAULICS

But in Fig. 220,

and

~\7"<j) V it TL 2

Therefore, -^- -^ = H-^ -Hr-fc,

i/ I/ ^!/

If Ui is axial and equal to w, as in Fig. 223,

-p. -p- ,

1 ^ i/ fit

200. Mixed flow turbines.

By a modification of the shape of the vanes of an inward flow
turbine, the mixed flow turbine is obtained. In the inward and
outward flow turbine the water only acts upon the wheel while it
is moving in a radial direction, but in the mixed flow turbine the
vanes are so formed that the water acts upon them also, while
flowing axially.

Fig. 231. Mixed Flow Turbine.

Fig. 231 shows a diagrammatic section through the wheel of
a mixed flow turbine, the axis of which is vertical. The water

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351

enters the wheel in a horizontal direction and leaves it vertically,
but it leaves the discharging edge of the vanes in different
directions. At the upper part B it leaves the vanes nearly
radially, and at the lower part A, axially. The vanes are spoon-
shaped, as shown in Fig. 232, and should be so formed, or in other
words, the inclination of the discharging edge should so vary,
that wherever the water leaves the vanes it should do so with no
component in a direction perpendicular to the axis of the turbine,
i.e. with no velocity of whirl. The regulation of the supply to
the wheel in the turbine of Fig. 231 is effected by a cylindrical
sluice or speed gate between the fixed guide blades and the wheel.

Fig. 232. Wheel of Mixed Flow Turbine.

Fig. 233 shows a section through the wheel and casing of a
double mixed flow turbine having adjustable guide blades to
regulate the flow. Fig. 234 shows a half longitudinal section of
the turbine, and Fig. 235 an outside elevation of the guide blade
regulating gear. The guide blades are surrounded by a large

352

HYDRAULICS

vortex chamber, and the outer tips of the guide blades are of
variable shapes, Fig. 233, so as to diminish shock at the entrance
to the guide passages. Each guide blade is really made in two
parts, one of which is made to revolve about the centre C, while
the outer tip is fixed. The moveable parts are made so that the
flow can be varied from zero to its maximum value. It will be

Fig. 233. Section through wheel and guide blades of Mixed Flow Turbine.

noticed that the mechanism for moving the guide blades is
entirely external to the turbine, and is consequently out of the
water. A further special feature is that between the ring R
and each of the guide blade cranks is interposed a spiral spring.
In the event of a solid body becoming wedged between two of
the guide blades, and thus locking one of them, the adjustment of
the other guide blades is not interfered with, as the spring con-
nected to the locked blade by its elongation will allow the ring
to rotate.

As with the inward and outward flow turbine, the mixed
flow turbine wheel may either work drowned, or exhaust into a
"suction tube."

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353

For a given flow, and width of wheel, the axial velocity
with which the water finally flows away from the wheel being the
same for the two cases, the diameter of a mixed flow turbine can
be made less than an inward flow turbine. As shown on page 340,
the diameter of the inward flow turbine is in large measure fixed

Fig. 234. Half-longitudinal section of Mixed Flow Turbine.

by the diameter of the exhaust openings of the wheel. For the
same axial velocity, and the same total flow, whether the turbine
is an inward or mixed flow turbine, the diameter d of the exhaust
openings must be about equal. The external diameter, therefore,
of the latter will be much smaller than for the former, and the
L.IL 23

354

HYDRAULICS

general dimensions of the turbine will be also diminished. For
a given head H, the velocity v of the inlet edge being the same in
the two cases, the mixed flow turbine can be run at a higher
angular velocity, which is sometimes an advantage in driving
dynamos.

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355

Form oftlie vanes. At the receiving edge, the direction of the
blade is found in the same way as for an inward flow turbine.

ABC, Fig. 236, is the triangle of velocities, and BO is parallel
to the tip of the blade. This triangle has been drawn for the data
of the turbine shown in Figs. 233 235 ; v is 46'5 feet per second,
and from

Y = 33'5 feet per second.
The anglo < is 139 degrees.

'-\ w

/"" v.'A" -""-

<fc Triangle of Velocities
^ at receiving edge.

Fig. 236.

The best form for the vane at the discharge is somewhat
difficult to determine, as the exact direction of flow at any point
on the discharging edge of the vane is .not easily found. The
condition to be satisfied is that the water must leave the wheel
without any component in the direction of motion.

The following construction gives approximately the form of
the vane.

Make a section through the wheel as in Fig. 237. The outline
of the discharge edge FGrH is shown. This edge of the vane is
supposed to be on a radial plane, and the plan of it is, therefore,
a radius of the wheel, and upon this radius the section is taken.

It is now necessary to draw the form of the stream lines, as
they would be approximately, if the water entered the wheel
radially and flowed out axially, the vanes being removed.

Divide 04, Fig. 237, at the inlet, into any number of equal
parts, say four, and subdivide by the points a, 6, d, e.

Take any point A, not far from c, as centre, and describe
a circle MM a touching the crowns of the wheel at M and MI.
Join AM and AMi.

Draw a flat curve Mi Mi touching the lines AM and AM a in M
and MI respectively, and as near as can be estimated, perpendicular

233

356

HYDRAULICS

to the probable stream lines through a, 6, d, e, which can be
sketched in approximately for a short distance from 04.

Taking this curve MMi as approximately perpendicular to the
stream lines, two points / and g near the centres of AM and
are taken.

Fig. 237.

Let the radius of the points g and / be r and r L respectively.
If any point Ci on MMi is now taken not far from A, the
peripheral area of Mci is nearly 2wrMci, and the peripheral area
of MiCi is nearly SSwriM^.

On the assumption that the mean velocity through MiM is
constant, the flow through Md will be equal to that through
when,

TUKBTNES 857

If, therefore, MMi is divided at the point Ci so that

the point d will approximately be on the stream line through c.

If now when the stream line cci is carefully drawn in, it is
perpendicular to MMi, the point Ci cannot be much in error.

A nearer approximation to Ci can be found by taking new values
for r and n, obtained by moving the points / and g so that they
more nearly coincide with the centres of CiM and CiMi. If the
two curves are not perpendicular, the curve MMi and the point Ci
are not quite correct, and new values of r and n will have to be
obtained by moving the points / and g. By approximation Ci can
be thus found with considerable accuracy.

By drawing other circles to touch the crown of the wheels, the
curves M 2 M 3 , M 4 M 5 etc. normal to the stream lines, and the points
Ca, c 3 , etc. on the centre stream line, can be obtained.

The curve 22, therefore, divides the stream lines into equal
parts.

Proceeding in a similar manner, the curves 11 and 33 can be
obtained, dividing the stream lines into four equal parts, and
these again subdivided by the curves aa, 66, dd, and ee, which
intersect the outlet edge of the vane at the points F, Gr, H and e
respectively.

To determine the direction of the tip of the vane at points on the
discharging edge. At the points F, Gr, H, the directions of the
stream lines are known, and the velocities U F) U Q , UH can be found,
since the flows through 01, 12, etc. are equal, and therefore

at = u G ~R 2 mn = i .

O7T

Draw a tangent FK to the stream line at F. This is the inter-
section, with the plane of the paper, of a plane perpendicular to
the paper and tangent to the stream line at F.

The point F in the plane of FK is moving perpendicular to the
plane of the paper with a velocity equal to w.R , w being the
angular velocity of the wheel, and R the radius of the point F.

If a circle be struck on this plane with K as centre, this circle
may be taken as an imaginary discharge circumference of an
inward flow turbine, the velocity v of which is u>R , and the tip of
the blade is to have such an inclination, that the water shall
discharge radially, i.e. along FK, with a velocity up. Turning this
circle into the plane of the paper and drawing the triangle of
velocities FST, the inclination a r of the tip of the blade at F in
the piano FK is obtained.

358

HYDRAULICS

At Gr the stream line is nearly vertical, but wRg can be set out
in the plane of the paper, as before, perpendicular to U Q and the
inclination a , on this plane, is found.

At H, dtn is found in the same way, and the direction of the
vane, in definite planes, at other points on its outlet edge, can be
similarly found.

Fig. 233.

Fig. 239.

Sections of the vane "by planes 0Gb, and OiHd. These are
shown in Figs. 238 and 239, and are determined as follows.

Imagine a vertical plane tangent to the tip of the vane at
inlet. The angle this plane makes with the tangent to the wheel
at b is the angle <, Fig. 236. Let BC of the same figure be the

TURBINES 359

plan of a horizontal line lying in this plane, and BD the plan of
the radius of the wheel at 6. The angle between these lines is y.

Let ft be the inclination of the plane OG-fe to the horizontal.

From D, Fig. 236, set out DE, inclined to BD at an angle /?,
and intersecting AB produced in E; with D as centre* and DE
as radius draw the arc EG intersecting DB produced in G.
Join CG.

The angle CG-D is the angle 71, which the line of intersection,
of the plane 0Gb, Fig. 237, with the plane tangent to the inlet tip
of the vane, makes with the radius 0&; and the angle CGF is
the angle on the plane 0GB which the tangent to the vane
makes with the direction of motion of the inlet edge of the
vane.

In Fig. 238 the inclination of the inlet tip of the blade is yi as
shown.

To determine the angle a at the outlet edge, resolve U Q , Fig.
237, along and perpendicular to OG, U O Q being the component
along OG.

Draw the triangle of velocities DEF, Fig. 238.

The tangent to the vane at D is parallel to FE.

In the same way, the section on the plane Hd, Fig. 237, may be
determined; the inclination at the inlet is y 2 , Fig. 239.

Mixed flow turbine working in open stream. A. double turbine
working in open stream and discharging through a suction tube
is shown in Fig. 240. This is a convenient arrangement for
moderately low falls. Turbines, of this class, of 1500 horse-
power, having four wheels on the same shaft and working under
a head of 25 feet, and making 150 revolutions per minute, have
recently been installed by Messrs Escher Wyss at Wangen an der
Aare in Switzerland.

201. Cone turbine.

Another type of inward flow turbine, which is partly axial and
partly radial, is shown in Fig. 241, and is known as the cone
turbine. It has been designed by Messrs Escher Wyss to meet
the demand for a turbine that can be adapted to variable flows.

The example shown has been erected at Gusset near Lyons and
makes 120 revolutions per minute.

The wheel is divided into three distinct compartments, the
supply of water being regulated by three cylindrical sluices S, Si
and S 2 . The sluices S and Si are each moved by three vertical
spindles such as A and AI which carry racks at their upper ends.
These two sluices move in opposite directions and thus balance
each other. The sluice S 2 is normally out of action, the upper

360

HYDRAULICS

compartment being closed. At low heads this upper compartment
is allowed to come into operation. The sluice S 2 carries a rack
which engages with a pinion P, connected to the vertical shaft T.

Feet 6

Fig. 240.

The shaft T is turned by hand by means of a worm and
wheel W. When it is desired to raise the sluice S a , it is revolved
by means of the pinion P until the arms F come between collars
D and E on the spindles carrying the sluice Si, and the sluice S 2
then rises and falls with Si. The pinion, gearing with racks on A
and AI, is fixed to the shaft M, which is rotated by the rack R
gearing with the bevel pinion Q. The rack R, is rotated by two
connecting rods, one of which C is shown, and which are under
the control of the hydraulic governor as described on page 378.
The wheel shaft can be adjusted by nuts working on the
square-threaded screw shown, and is carried on a special collar
bearing supported by the bracket B. The weight of the shaft is
partly balanced by the water-pressure piston which has acting
underneath it a pressure per unit area equal to that in the supply
chamber. The dimensions shown are in millimetres.

TURBINES

361

Fig. 241. Cone Turbine.

362

HYDRAULICS

202. Effect of changing the direction of the guide blade,
when altering the flow of inward flow and mixed flow
turbines.

As long as the velocity of a wheel remains constant, the
backward head impressed on the water by the wheel is the same,
and the pressure head, at the inlet to the wheel, will remain
practically constant as the guides are moved. The velocity of
flow U, through the guides, will, therefore, remain constant;
but as the angle 0, which the guide makes with the tangent to the
wheel, diminishes the radial component u y of U, diminishes.

Fig. 242.

Let ABC, Fig. 242, be the triangle of velocities for full opening,
and suppose the inclination of the tip of the blade is made parallel
to BC. On turning the guides into the dotted position, the incli-
nation being <'i, the triangle of velocities is ABCi, and the relative
velocity of the water and the periphery of the wheel is now Bd
which is inclined to the vane, and there is, consequently, loss due
to shock.

It will be seen that in the dotted position the tips of the guide
blades are some distance from the periphery of the wheel and it is
probable that the stream lines on leaving the guide blades follow
the dotted curves SS, and if so, the inclination of these stream
lines to the tangent to the wheel will be actually greater than <'i,
and BCi will then be more nearly parallel to BC. The loss may
be approximated to as follows :

As the water enters the wheel its radial component will remain
unaltered, but its direction will be suddenly changed from Bd to
BC, and its magnitude to BC 2 ; dC 2 is drawn parallel to AB.
A velocity equal to dCa has therefore to be suddenly impressed on
the water.

On page 68 it has been shown that on certain assumptions the

TURBINES 363

head lost when the velocity of a stream is suddenly changed
from Vi to v* is

that is, it is equal to the head due to the relative velocity of
Vi and v 2 .

But CiC 2 is the relative velocity of BCi and BC 2 , and therefore
the head lost at inlet may be taken as

k being a coefficient which may be taken as approximately unity.

203. Effect of diminishing the flow through turbines on
the velocity of exit.

If water leaves a wheel radially when the flow is a maximum,
it will not do so for any other flow.

The angle of the tip of the blade at exit is unalterable, and if
u and u Q are the radial velocities of flow, at full and part load
respectively, the triangles of velocity are DEF and DEFi, Fig. 243.

For part flow, the velocity with which the water leaves the
wheel is Ui. If this is greater than u, and the wlieel is drowned,
or the exhaust takes place into the air, the theoretical hydraulic
efficiency is less than for full load, but if the discharge is down a
suction tube the velocity with which the water leaves the tube is
less than for full flow and the theoretical hydraulic efficiency is
greater for the part flow. The loss of head, by friction in the
wheel due to the relative velocity of the water and the vane,
which is less than at full load, should also be diminished, as also,
the loss of head by friction in the supply and exhaust pipes.
The mechanical losses remain practically constant at all loads.

The fact that the efficiency of turbines diminishes at part loads
must, therefore, in large measure be due to the losses by shock
being increased more than the friction losses are diminished.

By suitably designing the vanes, the greatest efficiency of
inward flow and mixed flow turbines can be obtained at some
fraction of full load.

204. Regulation of the flow by cylindrical gates.

When the speed of the turbine is adjusted by a gate between
the guides and the wheel, and the flow is less than the normal, the
velocity U with which the water leaves the guide is altered in
magnitude but not in direction.

Let ABC be the triangle of velocities, Fig. 244, when the flow is
normal.

Let the flow be diminished until the velocity with which the
water leaves the guides is U , equal to AD.

Then BD is the relative velocity of U and v, and u is the
radial velocity of flow into the wheel.

Draw DK parallel to AB. Then for the water to move along
the vane a sudden velocity equal to KD must be impressed on

& (KD) 2
the water, and there is a head lost equal to ^ --.

To keep the velocity U more nearly constant Mr Swain has
introduced the gate shown in Fig. 245. The gate g is rigidly
connected to the guide blades, and to adjust the flow the guide
blades as well as the gate are moved. The effective width of the
guides is thereby made approximately proportional to the quantity
of flow, and the velocity TJ remains more nearly constant. If the
gate is raised, the width b of the wheel opening will be greater
than bi the width of the gate opening, and the radial velocity u

Fig. 245. Swain Gate.

Fig. 246.

TURBINES

365

into the wheel will consequently be less than the radial velocity u
from the guides. If U is assumed constant the relative velocity of
the water and the vane will suddenly change from BC to BOi,
Fig. 246. Or it may be supposed that in the space between the
guide and the wheel the velocity U changes from AC to ACi.

The loss of head will now be

k (COO 2
29

205. The form of the wheel vanes between the inlet and
outlet of turbines.

The form of the vanes between inlet and outlet of turbines
should be such, that there is no sudden change in the relative
velocity of the water and the wheel.

Consider the case of an inward flow turbine. Having given
a form to the vane and fixed the width between the crowns of the
wheel the velocity relative to the wheel at any radius r can be
found as follows.

Take any circumferential section ef at radius r, Fig. 247. Let
b be the effective width between the crowns, and d the effective
width ef between the vanes, and let q be the flow in cubic feet
per second between the vanes Ae and B/.

lug. 247. Relative velocity of tlie water and the vanes.

Fig. 248.

366 HYDRAULICS

The radial velocity through e/is

Find by trial a point near the centre of ef such that a circle
drawn with as centre touches the vanes at M and MI.

Suppose the vanes near e and / to be struck with arcs of circles.
Join to the centres of these circles and draw a curve MCMi
touching the radii OM and OMi at M and MI respectively.

Then MCMi will be practically normal to the stream lines
through the wheel. The centre of MCMi may not exactly
coincide with the centre of ef, but a second trial will probably
make it do so.

If then, b is the effective width between the crowns at C,

6 . MMi . v r = q.
MMi can be scaled off the drawing and v r calculated.

The curve of relative velocities for varying radii can then be
plotted as shown in the figure.

Fig. 249.

It will be seen that in this case the curve of relative velocities
changes fairly suddenly between c and h. By trial, the vanes
should be made so that the variation of velocity is as uniform
as possible.

If the vanes could be made involutes of a circle of radius E ,

TURBINES 367

as in Fig. 249, and the crowns of the wheel parallel, the relative
velocity of the wheel and the water would remain constant.
This form of vane is however entirely unsuitable for inward
flow turbines and could only be used in very special cases for
outward flow turbines, as the angles < and which the involute
makes with the circumferences at A and B are not independent,
for from the figure it is seen that,

, . . Jtio

and sin $ - rf

sin0 R,

or -r r = - .

sin 9 r

The angle must clearly always be greater than <.

206. The limiting head for a single stage reaction
turbine.

Eeaction turbines have not yet been made to work under heads
higher than 430 feet, impulse turbines of the types to be presently
described being used for heads greater than this value.

From the triangle of velocities at inlet of a reaction turbine,
e.g. Fig. 226, it is seen that the whirling velocity V cannot be
greater than

v + u cot <.

Assuming the smallest value for <f> to be 30 degrees, and the
maximum value for u to be 0'25 V2grH, the general formula

S..B

becomes, for the limiting case,

If v is assumed to have a limiting value of 100 feet per second,
which is higher than generally allowed in practice, and e to
be 0'8, then the maximum head H which can be utilised in a one
stage reaction turbine, is given by the equation

25-6H- 346 VS = 10,000,
from which H = 530 feet.

207. Series or multiple stage reaction turbines.

Professor Osborne Eeynolds has suggested the use of two
or more turbines in series, the same water passing through them
successively, and a portion of the head being utilised in each.

For parallel flow turbines, Reynolds proposed that the wheels

3Go

HYDRAULICS

and fixed blades be arranged alternately as shown in Fig. 250*.
This arrangement, although not used in water turbines, is very
largely used in reaction steam turbines.

Fig. 250.

f^""^

mrnmm^

Toothed
quadrant:

Figs. 251, 252. Axial Flow Impulse Turbine.
* Taken from Prof. Reynolds' Scientific Papers, Vol. x.

TURBINES

369

208. Impulse turbines.

Girard turbine. To overcome the difficulty of diminution of
efficiency with diminution of flow,
Girard introduced, about 1850, the
free deviation or partial admission
turbine.

Instead of the water being
admitted to the wheel throughout
the whole circumference as in the
reaction turbines, in the Girard
turbine it is only allowed to enter
the wheel through guide passages
in two diametrically opposite
quadrants as shown in Figs. 252
254. In the first two, the flow is
axial, and in the last radial.

Fig. 253.

In Fig. 252 above the guide crown are two quadrant-shaped
plates or gates 2 and 4, which are made to rotate about a vertical
axis by means of a toothed wheel. When the gates are over the
quadrants 2 and 4, all the guide passages are open, and by turning
the gates in the direction of the arrow, any desired number of the
passages can be closed. In Fig. 254 the variation of flow is
effected by means of a cylindrical quadrant-shaped sluice, which,
as in the previous case, can be made to close any desired number
of the guide passages. Several other types of regulators for
impulse turbines were introduced by Girard and others.

Fig. 253 shows a regulator employed by Fontaine. Above the
guide blades, and fixed at the opposite ends of a diameter DD,
are two indiarubber bands, the other ends of the bands being
connected to two conical rollers. The conical rollers can rotate
on journals, formed on the end of the arms which are connected
to the toothed wheel TW. A pinion P gears with TW, and by
rotating the spindle carrying the pinion P, the rollers can be made
to unwrap, or wrap up, the indiarubber band, thus opening or
closing the guide passages.

As the Girard turbine is not kept full of water, the whole of
the available head is converted into velocity before the water
enters the wheel, and the turbine is a pure impulse turbine.

To prevent loss of head by broken water in the wheel, the air
should be freely admitted to the buckets as shown in Figs. 252
and 254.

For small heads the wheel must be horizontal but for large
heads it may be vertical.

This class of turbine has the disadvantage that it cannot
L. H. 24

370

HYDRAULICS

run drowned, and hence must always be placed above the tail
>vater. For low and variable heads the full head cannot therefore
be utilised, for if the wheel is to be clear of the tail water, an
amount of head equal to half the width of the wheel must of
necessity be lost.

Fig. 254. Girard Eadial flow Impulse Turbine.

To overcome this difficulty Grirard placed the wheel in an air-
tight tube, Fig. 254, the lower end of which is below the tail water
level, and into which air is pumped by a small auxiliary air-pump,
the pressure being maintained at the necessary value to keep the
surface of the water in the tube below the wheel.

TUKBINES 371

Let H be the total head above the tail water level of the supply
water, the pressure head due to the atmospheric pressure, H

the distance of the centre of the wheel below the surface of the
supply water, and h the distance of the surface of the water in
the tube below the tail water level. Then the air-pressure in
the tube must be

and the head causing velocity of flow into the wheel is, therefore,

W W

So that wherever the wheel is placed in the tube below the tail
water the full fall H is utilised.

This system, however, has not found favour in practice, owing
to the difficulty of preserving the pressure in the tube.

209. The form of the vanes for impulse turbines, neg-
lecting friction.

The receiving tip of the vane should be parallel to the relative
velocity Y r of the water and the edge of the vane, Fig. 255.

For the axial flow turbine Vi equals v and the relative velocity v r
at exit, Fig. 255, neglecting friction, is equal to the relative
velocity V r at inlet. The triangle of velocities at exit is AG-B.

For the radial flow turbine, Figs. 254 and 258, there is a

centrifugal head impressed on the water equal to ^ - - and,

2 ~\7 2 2 2

neglecting friction, ^j- = -- + ^- - |- . The triangle of velocities

at exit is then DEF, Fig. 256, and Ui equals DF.

If the velocity with which the water leaves the wheel is Ui,
the theoretical hydraulic efficiency is

and is independent of the direction of Ui .

It should be observed, however, that in the radial flow turbine
the area of the section of the stream by the circumference of the
wheel, for a given flow, will depend upon the radial component of
Ui, and in the axial flow turbine the area of the section of the
stream by a plane perpendicular to the axis will depend upon the
axial component of Ui . That is, in each case the area will depend
upon the component of Ui perpendicular to Vi .

242

372

HYDRAULICS

Now the section of the stream must not fill the outlet area of
the wheel, and the minimum area of this outlet so that it is just
not filled will clearly be obtained for a given value of Ui when Ui
is perpendicular to v*, or is radial in the outward flow and axial in
the parallel flow turbine.

For the parallel flow turbine since BC and BG-, Fig. 255, are
equal, Ui is clearly perpendicular to v l when

and the inclinations a. and </> of the tips of the vanes are equal.

Figs. 255, 256.

Fig. 257.

If H and r are the outer and inner radii of the radial flow
turbine respectively,

* It is often stated that this is the condition for maximum efficiency but it only
is so, as stated above, for maximum flow for the given machine. The efficiency
only depends upon the magnitude of T^ and not upon its direction.

TURBINES 373

For Ui to be radial

v r = Vi sec a

= -- sec a.
r

If for the parallel flow turbine v is made equal to -^ , Y r from

y
Fig. 255 is equal to -^sec^, and therefore,

r
sec a = ^ sec <f>.

210. Triangles of velocity for an axial flow impulse tur-
bine considering friction.

The velocity with which the water leaves the guide passages
may be taken as from 0'94 to 0*97 V20H, and the hydraulic losses
in the wheel are from 5 to 10 per cent.

If the angle between the jet and the direction of motion of the
vane is taken as 30 degrees, and U is assumed as 0'95 \/2#H, and v
as 0'45\/2#H, the triangle of velocities is ABC, Fig. 257.

Taking 10 per cent, of the head as being lost in the wheel, the
relative velocity v r at exit can be obtained from the expression

H now the velocity of exit Ui be taken as 0*22N/2#H, and
circles with A and B as centres, and Ui and v r as radii be
described, intersecting in D, ABD the triangle of velocities at exit
is obtained, and Ui is practically axial as shown in the figure.
On these assumptions the best velocity for the rim of the wheel is
therefore '45 \/20H instead of *5 x/2#H.

The head lost due to the water leaving the wheel with velocity
u is '048H, and the theoretical hydraulic efficiency is therefore
95'2 per cent.

The velocity head at entrance is 0*9025H and, therefore, "097H
has been lost when the water enters the wheel.
The efficiency, neglecting axle friction, will be
H - 01H - 0-048H - 0-097H

T-
= 76 per cent, nearly.

211. Impulse turbine for high heads.

For high heads Girard introduced a form of impulse turbine,
of which the turbine shown in Figs. 258 and 259, is the modern
development.

The water instead of being delivered through guides over an
arc of a circle, is delivered through one or more adjustable nozzles.

TURBINES 375

In the example shown, the wheel has a mean diameter of 6'9 feet
and makes 500 revolutions per minute; it develops 1600 horse-
power under a head of 1935 feet.

The supply pipe is of steel and is 1*312 feet diameter.

The form of the orifices has been developed by experience, and
is such that there is no sudden change in the form of the liquid
vein, and consequently no loss due to shock.

The supply of water to the wheel is regulated by the sluices
shown in Fig. 258, which, as also the axles carrying the same,
are external to the orifices, and can consequently be lubricated
while the turbine is at work. The sluices are under the control
of a sensitive governor and special form of regulator.

As the speed of the turbine tends to increase the regulator
moves over a bell crank lever and partially closes both the orifices.
Any decrease in speed of the turbine causes the reverse action to
take place.

The very high peripheral speed of the wheel, 205 feet per
second, produces a high stress in the wheel due to centrifugal
forces. Assuming the weight of a bar of the metal of which the
rim is made one square inch in section and one foot long as
3'36 Ibs., the stress per sq. inch in the hoop surrounding the
wheel is

3-36. tf

9
= 4400 Ibs. per sq. inch.

To avoid danger of fracture, steel laminated hoops are shrunk
on to the periphery of the wheel.

The crown carrying the blades is made independent of the disc
of the wheel, so that it may be replaced when the blades become
worn, without an entirely new wheel being provided.

The velocity of the vanes at the inner periphery is 171 feet per
second, and is, therefore, 0*484 \/2#H.

If the velocity U with which the water leaves the orifice is
taken as 0*97 \/2#H, and the angle the jet makes with the tangent
to the wheel is 30 degrees, the triangle of velocities at entrance is
ABC, Fig. 260, and the angle </> is 53*5 degrees.

The velocity Vi of the outer edges of the vanes is 205 feet per
second, and assuming there is a loss of head in the wheel, equal to
6 per cent, of H,

^p + 205* m*

2g 2g 2g 2g
and v r = 220 ft. per second,

376

HYDKAUL1CS

If then the angle a is 30 degrees the triangle of velocities at
exit is DEF, Fig. 261.

The velocity with which the water leaves the wheel is then
Ui = 111 feet per sec., and the head lost by this velocity is 191 feet
or '099H.

Fig. 260.

Fig. 261.

The head lost in the pipe and nozzle is, on the assumption
made above,

and the total percentage loss of head is, therefore,

6 + 9-9 + 6-20-5,
and the hydraulic efficiency is 78' 1 per cent.

Fig. 262. Pelton Wheel.

TURBINES

377

The actual efficiency of a similar turbine at full load was found
by experiment to be 78 per cent.; allowing for mechanical losses
the hydraulic losses were less than in the example.

212. Pelton wheel.

A form of impulse turbine now very largely used for high heads
is known as the Pelton wheel.

A number of cups, as shown in Figs. 262 and 266, is fixed to a
wheel which is generally mounted on a horizontal axis. The
water is delivered to the wheel through a rectangular shaped
nozzle, the opening of which is generally made adjustable, either
by means of a hand wheel as in Fig. 262, or automatically by a
regulator as in Fig. 266.

As shown on page 276, the theoretical efficiency of the wheel is
unity and the best velocity for the cups is one-half the velocity of
the jet. This is also the velocity generally given to the cups
in actual examples. The width of the cups is from 2J to
4 times the thickness of the jet, and the width of the jet is about
twice its thickness.

The actual efficiency is between 70 and 82 per cent.

Table XXXVIII gives the numbers of revolutions per minute,
the diameters of the wheels and the nett head at the nozzle in
a number of examples.

TABLE XXXVIII.

Particulars of some actual Pelton wheels.

Head
in feet

Diameter
of wheel
(two wheels)

Kevolutions
per minute

H. P.

262

39-4"

375

64-5

129

500

*233'

2100

125

*197

20"

650

56-5

112

722

39"

650

111

215

167

382

60"

300

156

144

*289

54"

310

136

400

508

90"

200

180

300

* Picard Pictet and Co., the remainder by Escher Wyss and Co.

213. Oil pressure governor or regulator.

The modern applications of turbines to the driving of electrical
machinery, has made it necessary for particular attention to be
paid to the regulation of the speed of the turbines.

The methods of regulating the flow by cylindrical speed gates
and moveable guide blades have been described in connection with

378

HYDRAULICS

various turbines but the means adopted for moving the gates and
guides have not been discussed.

Until recent years some form of differential governor was
almost entirely used, but these have been almost completely
superseded by hydraulic and oil governors.

Figs. 263 and 264 show an oil governor, as constructed by
Messrs Escher Wyss of Zurich.

Figs. 263, 264. Oil Pressure Kegulator for Turbines.

A piston P having a larger diameter at one end than at the
other, and fitted with leathers I and Zi, fits into a double cylinder
Ci . Oil under pressure is continuously supplied through a pipe S
into the annulus A between the pistons, while at the back of the
large piston the pressure of the oil is determined by the regulator.

TURBINES

379

Fig. 265.

Suppose the regulator to be in a definite position, the space
behind the large piston being full of oil, and the
turbine running at its normal speed. The valve V
(an enlarged diagrammatic section is shown in
Fig. 265) will be in such a position that oil cannot
enter or escape from the large cylinder, and the
pressure in the annular ring between the pistons
will keep the regulator mechanism locked.

If the wheel increases in speed, due to a
diminution of load, the balls of the spring loaded
governor Gr move outwards and the sleeve M
rises. For the moment, the point D on the lever
MD is fixed, and the lever turns about D as a
fulcrum, and thus raises the valve rod NY. This
allows oil under pressure to enter the large
cylinder and the piston in consequence moves to
the right, and moves the turbine gates in the manner described later.
As the piston moves to the right, the rod R, which rests on the
wedge W connected to the piston, falls, and the point D of the
lever MD consequently falls and brings the valve Y back to its
original position. The piston P thus takes up a new position
corresponding to the required gate opening. The speed of the
turbine and of the governor is a little higher than before, the
increase in speed depending upon the sensitiveness of the governor.
On the other hand, if the speed of the wheel diminishes, the
sleeve M and also the valve Y falls and the oil from behind the
large piston escapes through the exhaust E, the piston moving
to the left. The wedge W then lifts the fulcrum D, the valve Y
is automatically brought to its central position, and the piston P
takes up a new position, consistent with the gate opening being
sufficient to supply the necessary water required by the wheel.

A hand wheel and screw, Fig. 264, are also provided, so that
the gates can be moved by hand when necessary.

The piston P is connected by the connecting rod BE to a crank
EF, which rotates the vertical shaft T. A double crank KK is
connected by the two coupling rods shown to a rotating toothed
wheel R, Fig. 241, turning about the vertical shaft of the turbine,
-and the movement, as described on page 360, causes the adjust-
ment of the speed gates.

214. Water pressure regulators for impulse turbines.
Fig. 266 shows a water pressure regulator as applied to regulate
the flow to a Pel ton wheel.

The area of the supply nozzle is adjusted by a beak B which

380

HYDRAULICS

M A 1 N

Figs. 266, 267. Pelton Wheel and Water Pressure Regulator.

TURBINES

381

rotates about the centre O. The pressure of the water in the
supply pipe acting on this beak tends to lift it and thus to open
the orifice. The piston P, working in a cylinder C, is also acted
upon, on its under side, by the pressure of the water in the supply
pipe and is connected to the beak by the connecting rod DE.
The area of the piston is made sufficiently large so that when the
top of the piston is relieved of pressure the pull on the connecting
rod is sufficient to close the orifice.

The pipe p conveys water under the same pressure, to the
valve V, which maybe similar to that described in connection with
the oil pressure governor, Fig. 265.

A piston rod passes through the top of the cylinder, and carries
a nut, which screws on to the square thread cut on the rod. A
lever eg, Fig. 268, which is carried on the fixed fulcrum e, is made
to move with the piston. A link /A connects ef with the lever
MN, one end M of which moves with the governor sleeve and the
other end N is connected to the valve rod NV. The valve V is
shown in the neutral position.

Fig. 268.

Suppose now the speed of the turbine to increase. The
governor sleeve rises, and the lever MN turns about the fulcrum
A which is momentarily at rest. The valve V falls and opens the
top of the cylinder to the exhaust. The pressure on the piston
P now causes it to rise, and closes the nozzle, thus diminishing
the supply to the turbine. As the piston rises it lifts again the
lever MN by means of the link A/ ; and closes the valve V. A
now position of equilibrium is thus reached. If the speed of the

382

HYDRAULICS

governor decreases the governor sleeve falls, the valve Y rises,
and water pressure is admitted to the top of the piston, which is
then in equilibrium, and the pressure on the beak B causes it to
move upwards and thus open the nozzle.

Hydraulic valve for water regulator. Instead of the simple
piston valve controlled mechanically, Messrs Escher Wyss use, for
high heads, a hydraulic double-piston valve Pp, Fig. 269.

This piston valve has a small bore through its centre by means
of which high pressure water which is admitted below the valve
can pass to the top of the large piston P. Above the piston is a
small plug valve Y which is opened and closed by the governor.

Fig. 269. Hydraulic valve for automatic regulation.

If the speed of the governor decreases, the valve Y is opened,
thus allowing water to escape from above the piston valve, and the
pressure on the lower piston p raises the valve. Pressure water is
thus admitted above the regulator piston, and the pressure on the
beak opens the nozzle. As the governor falls the valve Y closes,
the exhaust is throttled, and the pressure above the piston P rises.
When the exhaust through Y is throttled to such a degree that
the pressure on P balances the pressure on the under face of the
piston p, the valve is in equilibrium and the regulator piston is
locked.

TURBINES 383

If the speed of the governor increases, the valve Y is closed,
and the excess pressure on the upper face of the piston valve
causes it to descend, thus connecting the regulator cylinder to
exhaust. The pressure on the under face of the regulator piston
then closes the nozzle.

Filter. Between the conduit pipe and the governor valve V,
is placed a filter, Figs. 270 and 271, to remove any sand or grit
contained in the water.

Within the cylinder, on a hexagonal frame, is stretched a
piece of canvas. The water enters the cylinder by the pipe E, and
after passing through the canvas, enters the central perforated
pipe and leaves by the pipe S.

Figs. 270, 271. Water Filter for Impulse Turbine Regulator.

To clean the filter while at work, the canvas frame is revolved
by means of the handle shown, and the cock R is opened. Each
side of the hexagonal frame is brought in turn opposite the
chamber A, and water flows outwards through the canvas and
through the cock E, carrying away any dirt that may have
collected outside the canvas.

Auxiliary valve to prevent hammer action. When the pipe line
is long an auxiliary valve is frequently fitted on the pipe near to
the nozzle, which is automatically opened by means of a cataract
motion* as the nozzle closes, and when the movement of the nozzle
beak is finished, the valve slowly closes again.

If no such provision is made a rapid closing of the nozzle
means that a large mass of water must have its momentum
quickly changed and very large pressures may be set up, or in
other words hammer action is produced, which may cause fracture
of the pipe.

When there is an abundant supply of water, the auxiliary
valve is connected to the piston rod of the regulator and opened
and closed as the piston rod moves, the valve being adjusted so
that the opening increases by the same amount that the area of
the orifice diminishes.

* See Engineer, Vol. xc., p. 255.

384 HYDRAULICS

If the load on the wheel does not vary through a large range
the quantity of water wasted is not large.

215. Hammer blow in a long turbine supply pipe.
Let L be the length of the pipe and d its diameter.
The weight of water in the pipe is

Let the velocity change by an amount dv in time dt. Then the

rate of change of momentum is TT-, an( l n & cross section of

got

the lower end of the column of water in the pipe a force P must
be applied equal to this.

mi P T\ ^ wljd* dv

Therefore P = 7 -- 57 .

4 g dt

Referring to Fig. 266, let b be the depth of the orifice and da its
width.

Then, if r is the distance of D from the centre about which the
beak turns, and n is the distance of the closing edge of the beak
from this centre, and if at any moment the velocity of the piston
is t? feet per second, the velocity of closing of the beak will be

In any small element of time dt the amount by which the
nozzle will close is

Let it be assumed that U, the velocity of flow through the
nozzle, remains constant. It will actually vary, due to the
resistances varying with the velocity, but unless the pipe is very
long the error is not great in neglecting the variation. If then v
is the velocity in the pipe at the commencement of this element of
time and v - dv at the end of it, and A the area of the pipe,

v.A=fe.<Z,.TJ .............................. (1)

and (v-dtOA^fc- dA.di.IJ . ...(2).

\ T /

Subtracting (2) from (1),

TURBINES 385

If W is the weight of water in the pipe, the force P in pounds
that will have to be applied to change the velocity of this water
by dv in time Ct is

g ot "
Therefore P = ~T- >

and the pressure per sq. inch produced in the pipe near the
nozzle is

~ g r A 2 "

Suppose the nozzle to be completely closed in a time t seconds,
and during the closing the piston P moves with simple harmonic
motion.

Then the distance moved by the piston to close the nozzle is

and the time taken to move this distance is t seconds.
The maximum velocity of the piston is then

and substituting in (3), the maximum value of r is, therefore,

dv
dt~
and the maximum pressure per square inch is

vWb.d 1 .'U *.W.Q ?r Wv
Pm 2^A 2 2g.t. A 2 2t' gA.'

where Q is the flow in cubic feet per second before the orifice
began to close, and v is the velocity in the pipe.

Example. A 500 horse-power Pelton Wheel of 75 per cent, efficiency, and working
under a head of 260 feet, is supplied with water by a pipe 1000 feet long and
2' 3" diameter. The load is suddenly taken off, and the time taken by the
regulator to close the nozzle completely is 5 seconds.

On the assumption that the nozzle is completely closed (1) at a uniform rate,
and (2) with simple harmonic motion, and that no relief valve is provided,
determine the pressure produced at the nozzle.

The quantity of water delivered to the wheel per second when working at full
power is

500x33.000

The weight of water in the pipe is

W=62-4x^. (2-25) 2 xlOOO

= 250,000 Ibs.
L. H. 25

6 HYDRAULICS

21*7

Tne velocity is -7^ = 5-25 ft. per sec.
D *y u

In case (1) the total pressure acting on the lower end of the column of water in
e pipe is

250,000x5-25

g x 5

= 8200 Ibs.
The

386 HYDRAULICS

Tne velc

In case I
the pipe is

= 8200 lbs.
pressure per sq. inch is

8200
p = - = 14-5 Ibs. per sq. inch.

TT W . v
In case (2) p m = ^ L -^=22-8 lbs. per sq. inch.

EXAMPLES.

(1) Find the theoretical horse-power of an overshot water-wheel 22 feet
diameter, using 20,000,000 gallons of water per 24 hours under a total head
of 25 feet.

(2) An overshot water-wheel has a diameter of 24 feet, and makes 3 '5
revolutions per minute. The velocity of the water as it enters the buckets
is to be twice that of the wheel's periphery.

If the angle which the water makes with the periphery is to be 15
degrees, find the direction of the tip of the bucket, and the relative velocity
of the water and the bucket.

(3) The sluice of an overshot water-wheel 12 feet radius is vertically
above the centre of the wheel. The surface of the water in the sluice
channel is 2 feet 3 inches above the top of the wheel and the centre of the
sluice opening is 8 inches above the top of the wheel. The velocity of the
wheel periphery is to be one-half that of the water as it enters the buckets.
Determine the number of rotations of the wheel, the point at which the
water enters the buckets, and the direction of the edge of the bucket.

(4) An overshot wheel 25 feet diameter having a width of 5 feet, and
depth of crowns 12 inches, receives 450 cubic feet of water per minute, and
makes 6 revolutions per minute. There are 64 buckets.

The water enters the wheel at 15 degrees from the crown of the wheel
with a velocity equal to twice that of the periphery, and at an angle of 20
degrees with the tangent to the wheel.

Assuming the buckets to be of the form shown in Fig. 180, the length
of the radial portion being one-half the length of the outer face of the
bucket, find how much water enters each bucket, and, allowing for centri-
fugal forces, the point at which the water begins to leave the buckets.

(5) An overshot wheel 32 feet diameter has shrouds 14 inches deep, and
is required to give 29 horse-power when making 5 revolutions per minute.

Assuming the buckets to be one -third filled with water and of the same
form as in the last question, find the width of the wheel, when the total
fall is 32 feet and the efficiency 60 per cent.

TURBINES 387

Assuming the velocity of the water in the penstock to be If times that
of the wheel's periphery, and the bottom of the penstock level with the top
of the wheel, find the point at which the water enters the wheel. Find also
where water begins to discharge from the buckets.

(6) A radial blade impulse wheel of the same width as the channel in
which it runs, is 15 feet diameter. The depth of the sluice opening is
12 inches and the head above the centre of the sluice is 3 feet.. Assuming
a coefficient of velocity of 0'8 and that the edge of the sluice is rounded so
that there is no contraction, and the velocity of the rim of the wheel is 0*4
the velocity of flow through the sluice, find the theoretical efficiency of
the wheel.

(7) An overshot wheel has a supply of 30 cubic feet per second on a fall
of 24 feet.

Determine the probable horse-power of the wheel, and a suitable
width for the wheel.

(8) The water impinges on a Poncelet float at 15 with the tangent to
the wheel, and the velocity of the water is double that of the wheel. Find,
by construction, the proper inclination of the tip of the float.

(9) In a Poncelet wheel, the direction of the jet impinging on the floats
makes an angle of 15 with the tangent to the circumference and the tip of
the floats makes an angle of 30 with the same tangent. Supposing the
velocity of the jet to be 20 feet per second, find, graphically or otherwise,
(1) the proper velocity of the edge of the wheel, (2) the height to which the
water will rise on the float above the point of admission, (3) the velocity
and direction of motion of the water leaving the float.

(10) Show that the efficiency of a simple reaction wheel increases
with the speed when frictional resistances are neglected, but is greatest
at a finite speed when they are taken into account.

If the speed of the orifices be that due to the head (1) find the efficiency,
neglecting friction ; (2) assuming it to be the speed of maximum efficiency,
show that f of the head is lost by friction, and ^ by final velocity of water.

(11) Explain why, in a vortex turbine, the inner ends of the vanes are
inclined backwards instead of being radial.

(12) An inward flow turbine wheel has radial blades at the outer

periphery, and at the inner periphery the blade makes an angle of 30 with

T>
the tangent. The total head is 70 feet and r=. Find the velocity of the

rim of the wheel if the water discharges radially. Friction neglected.

(13) The inner and outer diameters of an inward flow turbine wheel
are 1 foot and 2 feet respectively. The water enters the outer circumference
at 12 with the tangent, and leaves the inner circumference radially. The
radial velocity of flow is 6 feet at both circumferences. The wheel makes
3 -6 revolutions per second. Determine the angles of the vanes at both
circumferences, and the theoretical hydraulic efficiency of the turbine.

(14) Water is supplied to an inward flow turbine at 44 feet per second,
and at 10 degrees to the tangent to the wheel. The wheel makes 200

252

388 HYDRAULICS

revolutions per minute. The inner radius is 1 foot and the outer radius
2 feet. The radial velocity of flow through the wheel is constant.

Find the inclination of the vanes at inlet and outlet of the wheel.

Determine the ratio of the kinetic energy of the water entering the
wheel per pound to the work done on the wheel per pound.

(15) The supply of water for an inward flow reaction turbine is 500
cubic feet per minute and the available head is 40 feet. The vanes are
radial at the inlet, the outer radius is twice the inner, the constant
velocity of flow is 4 feet per second, and the revolutions are 350 per
minute. Find the velocity of the wheel, the guide and vane angles,
the inner and outer diameters, and the width of the bucket at inlet and
outlet. Lond. Un. 1906.

(16) An inward flow turbine on 15 feet fall has an inlet radius of 1 foot
and an outlet radius of 6 inches. Water enters at 15 with the tangent to
the circumference and is discharged radially with a velocity of 3 feet per
second. The actual velocity of water at inlet is 22 feet per second. The
circumferential velocity of the inlet surface of the wheel is 19^ feet per
second.

Construct the inlet and outlet angles of the turbine vanes.
Determine the theoretical hydraulic emciency of the turbine.
If the hydraulic emciency of the turbine is assumed 80 per cent, find the
vane angles.

(17) A quantity of water Q cubic feet per second flows through a
turbine, and the initial and final directions .and velocities are known.
Apply the principle of equality of angular impulse and moment of
momentum to find the couple exerted on the turbine.

(18) The wheel of an inward flow turbine has a peripheral velocity of
50 feet per second. The velocity of whirl of the incoming water is 40 feet
per second, and the radial velocity of flow 5 feet per second. Determine
the vane angle at inlet.

Taking the flow as 20 cubic feet per second and the total losses as
20 per cent, of the available energy, determine the horse-power of the
turbine, and the head H.

If 5 per cent, of the head is lost in friction in the supply pipe, and the
centre of the turbine is 15 feet above the tail race level, find the pressure
head at the inlet circumference of the wheel.

(19) An inward flow turbine is required to give 200 horse -power under
a head of 100 feet when running at 500 revolutions per minute. The
velocity with which the water leaves the wheel axially may be taken as
10 feet per second, and the wheel is to have a double outlet. The diameter
of the outer circumference may be taken as If times the inner. Determine
the dimensions of the turbine and the angles of the guide blades and
vanes of the turbine wheel. The actual efficiency is to be taken as 75 per
cent, and the hydraulic efficiency as 80 per cent.

(20) An outward flow turbine wheel has an internal diameter of 5*249
feet and an external diameter of 6'25 feet. The head above the turbine is
141-5 feet. The width of the wheel at inlet is 10 inches, and the quantity

TURBINES 389

of water supplied per second is 215 cubic feet. Assuming the hydraulic
losses are 20 per cent., determine the angles of tips of the vanes so that
the water shall leave the wheel radially. Determine the horse-power of
the turbine and verify the work done per pound from the triangles of

velocities.

(21) The total head available for an inward-flow turbine is 100 feet.

The turbine wheel is placed 15 feet above the tail water level.

When the flow is normal, there is a loss of head in the supply pipe of
3 per cent, of the head ; in the guide passages a loss of 5 per cent. ; in the
wheel 9 per cent. ; in the down pipe 1 per cent. ; and the velocity of flow
from the wheel and in the supply pipe, and also from the down pipe is
8 feet per second.

The diameter of the inner circumference of the wheel is 9^ inches and
of the outer 19 inches, and the water leaves the wheel vanes radially.
The wheel has radial vanes at inlet.

Determine the number of revolutions of the wheel, the pressure head in
the eye of the wheel, the pressure head at the circumference to the wheel,
the pressure head at the entrance to the guide chamber, and the velocity
which the water has when it enters the wheel. From the data given

(22) A horizontal inward flow turbine has an internal diameter of
5 feet 4 inches and an external diameter of 7 feet. The crowns of the
wheel are parallel and are 8 inches apart. The difference in level of the
head and tail water is 6 feet, and the upper crown of the wheel is just below
the tail water level. Find the angle the guide blade makes with the tangent
to the wheel, when the wheel makes 32 revolutions per minute, and the
flow is 45 cubic feet per second. Neglecting friction, determine the vane
angles, the horse-power of the wheel and the theoretical hydraulic efficiency.

(23) A parallel flow turbine has a mean diameter of 11 feet.

The number of revolutions per minute is 15, and the axial velocity of
flow is 3*5 feet per second. The velocity of the water along the tips of the
guides is 15 feet per second.

Determine the inclination of the guide blades and the vane angles that
the water shall enter without shock and leave the wheel axially.

Determine the work done per pound of water passing through the wheel.

(24) The diameter of the inner crown of a parallel flow pressure turbine
is 5 feet and the diameter of the outer crown is 8 feet. The head over the
wheel is 12 feet. The number of revolutions per minute is 52. The radial
velocity of flow through the wheel is 4 feet per second.

Assuming a hydraulic efficiency of 0'8, determine the guide blade angles
and vane angles at inlet for the three radii 2 feet 6 inches, 3 feet 3 inches
and 4 feet.

Assuming the depth of the wheel is 8 inches, draw suitable sections of
the vanes at the three radii.

Find also the width of the guide blade in plan, if the upper and lower
edges are parallel, and the lower edge makes a constant angle with the

390 HYDRAULICS

plane of the wheel, so that the stream lines at the inner and the outer
crown may have the correct inclinations.

(25) A parallel flow impulse turbine works under a head of 64 feet.
The water is discharged from the wheel in an axial direction with a
velocity due to a head of 4 feet. The circumferential speed of the wheel
at its mean diameter is 40 feet per second.

Neglecting all frictional losses, determine the mean vane and guide
angles. Lond. Un. 1905.

(26) An outward flow impulse turbine has an inner diameter of 5 feet,
an external diameter of 6 feet 3 inches, and makes 450 revolutions per
minute.

The velocity of the water as it leaves the nozzles is double the velocity
of the periphery of the wheel, and the direction of the water makes an
angle of 80 degrees with the circumference of the wheel.

Determine the vane angle at inlet, and the angle of the vane at outlet so
that the water shall leave the wheel radially.

Find the theoretical hydraulic efficiency. If 8 per cent, of the head
available at the nozzle is lost in the wheel, find the vane angle at exit that
the water shall leave radially.

What is now the hydraulic efficiency of the turbine ?

(27) In an axial flow Girard turbine, let V be the velocity due to the
effective head. Suppose the water issues from the guide blades with the
velocity 0'95V, and is discharged axially with a velocity '12 V. Let the
velocity of the receiving and discharging edges be 0'55 V.

Find the angle of the guide blades, receiving and discharging angles of
wheel vanes and hydraulic efficiency of the turbine.

(28) Water is supplied to an axial flow impulse turbine, having a mean
diameter of 6 feet, and making 144 revolutions per minute, under a head of
100 feet. The angle of the guide blade at entrance is 30, and the angle the
vane makes with the direction of motion at exit is 30. Eight per cent, of
the head is lost in the supply pipe and guide. Determine the relative
velocity of water and wheel at entrance, and on the assumption that 10 per
cent, of the total head is lost in friction and shock in the wheel, determine
the velocity with which the water leaves the wheel. Find the hydraulic
efficiency of the turbine.

(29) The guide blades of an inward flow turbine are inclined at 30
degrees, and the velocity U along the tip of the blade is 60 feet per second.
The velocity of the wheel periphery is 55 feet per second. The guide blades
are turned so that they are inclined at an angle of 15 degrees, the velocity
U remaining constant. Find the loss of head due to shock at entrance.

If the radius of the inner periphery is one-half that of the outer and the
radial velocity through the wheel is constant for any flow, and the water
left the wheel radially in the first case, find the direction in which it leaves
in the second case. The inlet radius is twice the outlet radius.

(30) The supply of water to a turbine is controlled by a speed gate
between the guides and the wheel. If when the gate is fully open the
velocity with which the water approaches the wheel is 70 feet per second

TURBINES 391

and it makes an angle of 15 degrees with the tangent to the wheel, find
the loss of head by shock when the gate is half closed. The velocity of
the inlet periphery of the wheel is 75 feet per second.

(31) A Pelton wheel, which may be assumed to have semi-cylindrical
buckets, is 2 feet diameter. The available pressure at the nozzle when it
is closed is 200 Ibs. per square inch, and the supply when the nozzle is
open is 100 cubic feet per minute. If the revolutions are 600 per minute,
estimate the horse -power of the wheel and its efficiency.

(32) Show that the efficiency of a Pelton wheel is a maximum
neglecting frictional and other losses when the velocity of the cups equals
half the velocity of the jet.

25 cubic feet of water are supplied per second to a Pelton wheel through
a nozzle, the area of which is 44 square inches. The velocity of the cups
is 41 feet per second. Determine the horse-power of the wheel assuming
an efficiency of 75 per cent.

CHAPTER X.

PUMPS.

Pumps are machines driven by some prime mover, and used
for raising fluids from a lower to a higher level, or for imparting
energy to fluids. For example, when a mine has to be drained
the water niay be simply raised from the mine to the surface, and
work done upon it against gravity. Instead of simply raising the
water through a height h, the same pumps might be used to
deliver water into pipes, the pressure in which is wh pounds per
square foot.

A pump can either be a suction pump, a pressure pump, or
both. If the pump is placed above the surface of the water in
the well or sump, the water has to be first raised by suction;
the maximum height through which a pump can draw water,
or in other words the maximum vertical distance the pump can
be placed above the water in the well, is theoretically 34 feet, but
practically the maximum is from 25 to 30 feet. If the pump
delivers the water to a height h above the pump, or against a
pressure-head h, it is called a force pump.

216. * Centrifugal and turbine pumps.

Theoretically any reaction turbine could be made to work as
a pump by rotating the wheel in the opposite direction to that in
which it rotates as a turbine, and supplying it with water at the
circumference, with the same velocity, but in the inverse direction
to that at which it was discharged when acting as a turbine. Up
to the present, only outward flow pumps have been constructed,
and, as will be shown later, difficulty would be experienced in
starting parallel flow or inward flow pumps.

Several types of centrifugal pumps (outward flow) are shown
in Figs. 272 to 276.

The principal difference between the several types is in the
form of the casing surrounding the wheel, and this form has con-
siderable influence upon the efficiency of the pump. The reason

* See Appendix.

CENTRIFUGAL PUMPS

393

for this can be easily seen in a general way from the following
consideration. The water approaches a turbine wheel with a
high velocity and in a direction making a small angle with the
direction of motion of the inlet circumference of the wheel, and

Fig. 272. Diagram of Centrifugal Pump.

thus it has a large velocity of whirl. When the water leaves the
wheel its velocity is small and the velocity of whirl should be zero.
In the centrifugal pump these conditions are entirely reversed;
the water enters the wheel with a small velocity, and leaves

394

HYDRAULICS

it with a high velocity. If the case surrounding the wheel
admits of this velocity being diminished gradually, the kinetic
energy of the water is converted into useful work, but if not, it is
destroyed by eddy motions in the casing, and the efficiency of the
pump is accordingly low.

In Fig. 272 a circular casing surrounds the wheel, and prac-
tically the whole of the kinetic energy of the water when it leaves
the wheel is destroyed ; the efficiency of such pumps is generally
much less than 50 per cent.

Fig. 273. *Centrifugal Pnmp with spiral casing.

The casing of Fig. 273 is made of spiral form, the sectional
area increasing uniformly towards the discharge pipe, and thus
being proportional to the quantity of water flowing through the
section. It may therefore be supposed that the mean velocity of
flow through any section is nearly constant, and that the stream
lines are continuous.

The wheel of Fig. 274 is surrounded by a large whirlpool
chamber in which, as shown later, the velocity with which the
water rotates round the wheel gradually diminishes, and the
velocity head with which the water leaves the wheel is partly
converted into pressure head.

The same result is achieved in the pump of Figs. 275 and 276

* See page 542.

CENTRIFUGAL PUMPS

395

by allowing the water as it leaves the wheel to enter guide
passages, similar to those used in a turbine to direct the water
to the wheel. The area of these passages gradually increases
and a considerable portion of the velocity head is thus converted
into pressure head and is available for lifting water.

This class of centrifugal pump is known as the turbine pump.

Fig. 274. Diagram of Centrifugal Pump with Whirlpool Chamber.

217. Starting centrifugal or turbine pumps.

A centrifugal pump cannot commence delivery unless the wheel,
casing, and suction pipe are full of water.

If the pump is below the water in the well there is no difficulty
in starting as the casing will be maintained full of water.

When the pump is above the water in the well, as in Fig. 272,
a non-return valve Y must be fitted in the suction pipe, to prevent
the pump when stopped from being drained. If the pump becomes
empty, or when the pump is first set to work, special means have
to be provided for filling the pump case. In large pumps the air
may be expelled by means of steam, which becomes condensed and
the water rises from the well, or they should be provided with

396

fiYDRAULICS

an air-pump or ejector as an auxiliary to the pump. Small pumps
can generally be easily filled by hand through a pipe such as
shown at P, Fig. 276.

With some classes of pumps, if the pump has to commence
delivery against full head, a stop valve on the rising main,
Fig. 296, is closed until the pump has attained the speed necessary
to commence delivery*, after which the stop valve is slowly
opened.

Fig. 275.

Turbine Pump.

Fig. 276.

It will be seen later that, under special circumstances, other
provisions will have to be made to enable the pump to commence
delivery.

218. Form of the vanes of centrifugal pumps.

The conditions to be satisfied by the vanes of a centrifugal
pump are exactly the same as for a turbine. At inlet the direction
of the vane should be parallel to the direction of the relative
velocity of the water and the tip of the vane, and the velocity
with which the water leaves the wheel, relative to the pump case,
is the vector sum of the velocity of the tip of the vane and the
velocity relative to the vane.

* See page 409-

CENTRIFUGAL PUMPS 397

Suppose the wheel and casing of Fig. 272 is full of water, and
the wheel is rotated in the direction of the arrow with such a
velocity that water enters the wheel in a known direction with a
velocity U, Fig. 277, not of necessity radial.

Let v be the velocity of the receiving edge of the vane or inlet
circumference of the wheel; Vi the velocity of the discharging
circumference of the wheel ; Ui the absolute velocity of the water
as it leaves the wheel ; Y and Vi the velocities of whirl at inlet
and outlet respectively; Y r and v r the relative velocities of the
water and the vane at inlet and outlet respectively ; u and u the
radial velocities at inlet and outlet respectively.

The triangle of velocities at inlet is ACD, Fig. 277, and if the
vane at A, Fig. 272, is made parallel to CD the water will enter
the wheel without shock.

A * C B * E

Triangle oC velocities Triangle of velocities

cut inlet. at exit.

Fig. 277. Fig. 278.

The wheel being full of water, there is continuity of flow, and
if A and AI are the circumferential areas of the inner and outer
circumferences, the radial component of the velocity of exit at the
outer circumference is

If the direction of the tip of the vane at the outer circum-
ference is known the triangle of velocities at exit, Fig. 278, can be
drawn as follows.

Set out BG radially and equally to HI, and BE equal to VL

Draw GF parallel to BE at a distance from BE equal to Ui,
and EF parallel to the tip of the vane to meet GF in F.

Then BF is the vector sum of BE and EF and is the velocity
with which the water leaves the wheel relative to the fixed casing.

219. Work done on the water by the wheel.

Let B and r be the radii of the discharging and receiving
circumferences respectively.

The change in angular momentum of the water as it passes
through the wheel is ViB/$rVr/0 per pound of flow, the plus
sign being used when V is in the opposite direction to Y J; as in
Figs. 277 and 278.

398 HYDRAULICS

Neglecting frictional and other losses, the work done by the
wheel on the water per pound (see page 275) is

9 ' 9 '

If U is radial, as in Fig. 272, Y is zero, and the work done on
the water by the wheel is

- foot Ibs. per Ib. flow.

If then H , Fig. 272, is the total height through which the water
is lifted from the sump or well, and u d is the velocity with which
the water is delivered from the delivery pipe, the work done on
each, pound of water is

and therefore,

9 ' 2#

Let (180 -* <) be the angle which the direction of the vane at
exit makes with the direction of motion, and (180 -,0) the angle
which the vane makes with the direction of motion at inlet. Then
ACD is and BEF is <f>.

In the triangle HEF, HE = HF cot <, and therefore,

Vi = Vi -MI cot <#>.
The theoretical lift, therefore, is

29 9

If Q is the discharge and AI the peripheral area of the dis-
charging circumference,

v\ Vi -- cot <f>
and H = - =1 - ........................ (1).

If, therefore, the water enters the wheel without shock and all

p
resistances are neglected, the lift is independent of the ratio , and

depends only on the velocity and inclination of the vane at the
discharging circumference.

220. Ratio of V x to v r

As in the case of the turbine, for any given head H, Vi and Vi
can theoretically have any values consistent with the product

CENTRIFUGAL PUMPS

399

being equal to #H, the ratio of V x to v l simply depending upon
the magnitude of the angle <j>.

The greater the angle <j> is made the less the velocity ^ of the
periphery must be for a given lift.

Fig. 279.

This is shown at once by equation (1), section 219, and is
illustrated in Fig. 279. The angle <j> is given three values,
30 degrees, 90 degrees and 150 degrees, and the product V^i and
also the radial velocity of flow % are kept constant. The theo-
retical head and also the discharge for the three cases are there-
fore the same. The diagrams are drawn to a common scale, and it
can therefore be seen that as < increases Vi diminishes, and Ui
the velocity with which the water leaves the wheel increases.

221. The kinetic energy of the water at exit from the
wheel.

Part of the head H impressed upon the water by the wheel
increases the pressure head between the inlet and outlet, and the
remainder appears as the kinetic energy of the water as it leaves

400 HYDRAULICS

U 2
the wheel. This kinetic energy is equal to 7^-, and can only be

utilised to lift the water if the velocity can be gradually diminished
so as to convert velocity head into pressure head. This however
is not very easily accomplished, without being accompanied by a
considerable loss by eddy motions. If it be assumed that the same

Ui 2
proportion of the head ~- in all cases is converted into useful

work, it is clear that the greater Ui, the greater the loss by eddy
motions, and the less efficient will be the pump. It is to be ex-
pected, therefore, that the less the angle </>, the greater will be
the efficiency, and experiment shows that for a given form of
casing, the efficiency does increase as < is diminished.

222. Gross lift of a centrifugal pump.

Let h a be the actual height through which water is lifted;
h s the head lost in the suction pipe ; Tid the head lost in the delivery
pipe ; and u d the velocity of flow along the delivery pipe.

Any other losses of head in the wheel and casing are incident

to the pump, but h s , hd, and the head ^ should be considered as

external losses.

The gross lift of a pump is then

and this is always less than H.

223. Efficiencies of a centrifugal pump.
Manometric efficiency. The ratio g , or

g .h

e ~ " Q~~~ """'

Ui 2 Vi -r- cot <}>
Ai

is the manometric efficiency of the pump at normal discharge.

The reason for specifically defining e as the manometric
efficiency at normal discharge is simply that the theoretical lift H
has been deduced from consideration of a definite discharge Q,
and only for this one discharge can the conditions at the inlet edge
be as assumed.

A more general definition is, however, generally given to e, and
for any discharge Q, therefore, the manometric efficiency may
be taken as the ratio of the gross lift at that discharge to the
theoretical head

tf-^-Scot*

CENTRIFUGAL PUMPS 401

This manometric efficiency of the pump must not be confused
with the efficiency obtained by dividing the work done by the
pump, by the energy required to do that work, as the latter in
many pumps is zero, when the former has its maximum value.

Hydraulic efficiency. The hydraulic efficiency of a pump is
the ratio of the gross work done by the pump to the work done
on the pump wheel.

Let W = the weight of water lifted per second.

Let h = the gross head

Let E == the work done on the pump wheel in foot pounds
per second.

Let Bh = the hydraulic efficiency. Then

W.h

e =~w

The work done on the pump wheel is less than the work done
on the pump shaft by the belt or motor which drives the pump,
by an amount equal to the energy lost by friction at the bearings
of the machine. This generally, in actual machines, can be
approximately determined by running the machine without load.

Actual efficiency. From a commercial point of view, what is
generally required is the ratio of the useful work done by the
pump, taking it as a whole, to the work done on the pump shaft.

Let E s be the energy given to the pump shaft per sec. and
e m the mechanical efficiency of the pump, then

E-E s .e OT ,
and the actual efficiency

W.h a

Gross efficiency of the pump. The gross efficiency of the pump
itself, including mechanical as well as fluid losses, is

_W.h
e g - Es

224. Experimental determination of the efficiency of a
centrifugal pump.

The actual and gross efficiencies of a pump can be determined
directly by experiment, but the hydraulic efficiency can only be
determined when at all loads the mechanical efficiency of the
pump is known.

To find the actual efficiency, it is only necessary to measure
the height through which water is lifted, the quantity of water
L. ii. 26

402 HYDRAULICS

discharged, and the energy E s given to the pump shaft in unit
time.

A very convenient method of determining E, with a fair
degree of accuracy is to drive the pump shaft direct by an electric
motor, the efficiency curve* for which at varying loads is known.
A better method is to use some form of transmission dynamo-
meter t.

225. Design of pump to give a discharge Q.

If a pump is required to give a discharge Q under a gross
lift h, and from previous experience the probable manometric
efficiency e at this discharge is known, the problem of determining
suitable dimensions for the wheel of the pump is not difficult.
The difficulty really arises in giving a correct value to e and in
making proper allowance for leakage.

This difficulty will be better appreciated after the losses in
various kinds of pumps have been considered. It will then be
seen that e depends upon the angle <, the velocity of the wheel,
the dimensions of the wheel, the form of the vanes of the wheel,
the discharge through the wheel, and upon the form of the casing
surrounding the wheel; the form of the casing being just as
important, or more important, than the form of the wheel in
determining the probable value of e.

Design of the wheel of a pump for a given discharge under a
given head. If a pump is required to give a discharge Q under an
effective head h aj the gross head h can only be determined if h sj

h d , and |^- , are known.

Any suitable value can be given to the velocity Ud. If the
pipes are long it should not be much greater than 5 feet per second
for reasons explained in the chapter on pipes, and the velocity u 8
in the suction pipe should be equal to or less than u*. The
velocities u s and u& having been settled, the losses h 8 and ha can be
approximated to and the gross head h found. In the suction pipe,
as explained on page 395, a foot valve is generally fitted, at which,
at high velocities, a loss of head of several feet may occur.
The angle < is generally made from 10 to 90 degrees. Theoreti-
cally, as already stated, it can be made much greater than
90 degrees, but the efficiency of ordinary centrifugal pumps might
be very considerably diminished as <f> is increased.

The manometric efficiency e varies very considerably ; with
radial blades and a circular casing, the efficiency is not generally

* See Electrical Engineering, Thomaleu-Howe, p. 195.
t See paper by Stanton, Proc. Inst. Mech. Engs. y 1903.

CENTRIFUGAL PUMPS 403

more than 0'3 to 0'4. With a vortex chamber, or a spiral casing,
and the vanes at inlet inclined so that the tip is parallel to the
relative velocity of the water and the vane, and <j> not greater than
90 degrees, the manometric efficiency e is from 0*5 to 0'75, being
greater the less the angle <, and with properly designed guide
blades external to the wheel, e is from 0'6 to '85.

The ratio of the diameter of the discharging circumference to
the inlet circumference is somewhat arbitrary and is generally
made from 2 to 3. Except for the difficulty of starting (see
section 226), the ratio might with advantage be made much
smaller, as by so doing the frictional losses might be considerably
reduced. The radial velocity Ui may be taken from 2 to 10 feet
per second.

Having given suitable values to u t and to any two of the three
quantities, e, v, and <, the third can be found from the equation

, e (vi Viiii cot <)
ri - .
9

The internal diameter d of the wheel will generally be settled from
consideration of the velocity of flow u% into the wheel. This may
be taken as equal to or about equal to u, but in special cases
it may be larger than u.

Then if the water is admitted to the wheel at both sides, as in
Fig. 273,

from which d can be calculated when ^ and Q are known.

Let b be the width of the vane at inlet and B at outlet, and D
the diameter of the outlet circumference.

Then & = --

and E

If the water moves toward the vanes at inlet radially, the
inclination of the vane that there shall be no shock is such that

a u
tan = - .
v 9

and if guide blades are to be provided external to the wheel, as in
Fig. 275, the inclination a of the tip of the guide blade with the
direction of v l is found from

tan a = y- .

, The guide passages should be so proportioned that the velocity
Ui is gradually diminished to the velocity in the delivery pipe.

262

404 HYDRAULICS

Limiting velocity of the rim of the wheel. Quite apart from
head lost by friction in the wheel due to the relative motion of
the water and the wheel, there is also considerable loss of energy
external to the wheel due to the relative motion of the water and
the wheel. Between the wheel and the casing there is in most
pumps a film of water, and between this film and the wheel,
frictional forces are set up which are practically proportional to
the square of the velocity of the wheel periphery and to the area
of the wheel crowns. An attempt is frequently made to diminish
this loss by fixing the vanes to a central diaphragm only, the
wheel thus being without crowns, the outer casing being so
formed that there is but a small clearance between it and the
outer edges of the vanes. At high velocities these frictional resist-
ances may be considerable. To keep them small the surface of
the wheel crowns and vanes must be made smooth, and to this
end many high speed wheels are carefully finished.

Until a few years ago the peripheral velocity of pump wheels
was generally less than 50 feet per second, and the best velocity
was supposed to be about 30 feet per second. They are now, how-
ever, run at much higher speeds, and the limiting velocities are
fixed from consideration of the stresses in the wheel due to centri-
fugal forces. Peripheral velocities of nearly 200 feet per second
are now frequently used, and Eateau has constructed small pumps
with a peripheral velocity of 250 feet per second*.

Example. To find the proportions of a pump with radial blades at outlet
(i.e. = 90) to lift 10 cubic feet of water per second against a head of 50 feet.

Assume there are two suction pipes and that the water enters the wheel from
both sides, as in Fig. 273, also that the velocity in the suction and delivery pipes
and the radial velocity through the wheel are 6 feet per second, and the manometric
efficiency is 75 per cent.

First to find Vj.

Since the blades are radial, *75 = 50,

from which 1^=46 feet per sec.

To find the diameter of the suction pipes.
The discharge is 10 cubic feet per second, therefore

from which <Z = l-03'=12f".

If the radius R of the external circumference be taken as 2r and r is taken equal
to the radius of the suction pipes, then B = 12f", and the number of revolutions
per second will be

The velocity of the inner edge of the vane is
v = 23 feet per sec.

* Engineer, 1902.

CENTRIFUGAL PUMPS 405

The inclination of the vane at inlet that the water may move on to the vane
without shock is

and the water when it leaves the wheel makes an angle a with v x such that

If there are guide blades surrounding the wheel, a gives the inclination of these
blades.

The width of the wheel at discharge is

M> = 7r.D.6 / = 7r.2-0
= 3 inches about.
The width of the wheel at inlet =6^ inches.

226. The centrifugal head impressed on the water by
the wheel.

Head against which a pump will commence to discharge. As
shown on page 335, the centrifugal head impressed on the water as
it passes through the wheel is

, _V v*

hc ~2g-W

but this is not the lift of the pump. Theoretically it is the head
which will be impressed on the water when there is no flow
through the wheel, and is accordingly the difference between the
pressure at inlet and outlet when the pump is first set in motion ;
or it is the statical head which the pump will maintain when

running at its normal speed. If this is less than , the pump

theoretically cannot start lifting against its full head without
being speeded up above its normal velocity.

The centrifugal head is, however, always greater than

as the water in the eye of the wheel and in the casing surrounding
the wheel is made to rotate by friction.

For a pump having a wheel seven inches diameter surrounded
by a circular casing 20 inches diameter, Stanton* found that, when
the discharge was zero and the vanes were radial at exit, h c was

Q , and with curved vanes, <f> being 30 degrees, h was ^ .

For a pump with a spiral case surrounding the wheel, the
centrifugal head h c when there is no discharge, cannot be much

greater than - , as the water surrounding the wheel is prevented

from rotating by the casing being brought near to the wheel at
one point.

* Proceedings Inst. M. E. t 1903.

406 HYDRAULICS

Parsons found for a pump having a wheel 14 inches diameter
with radial vanes at outlet, and running at 300 revolutions per

minute, that the head maintained without discharge was 9,
and with an Appold* wheel running at 320 revolutions per minute
the statical head was ~ . For a pump, with spiral casing,

having a rotor 1*54 feet diameter, the least velocity at which
it commenced to discharge against a head of 14*67 feet was

OK 2

392 revolutions per minute, and thus h c was ^ l , and the least
velocity against a head of 17*4 feet was 424 revolutions per
minute or h c was again ~ - . For a pump with circular casing

larger than the wheel, h c was ~-^- . For a pump having guide

passages surrounding the wheel, and outside the guide passages
a circular chamber as in Fig. 275, the centrifugal head may also

be larger than ~; the mean actual value for this pump was

found to be T087.

Stanton found, when the seven inches diameter wheels mentioned
above discharged into guide passages surrounded by a circular

chamber 20 inches diameter, that h c was - ; - when the vanes of

the wheel were radial, and -^ - when < was 30 degrees.

That the centrifugal head when the wheel has radial vanes is
likely to be greater than when the vanes of the wheel are set back
is to be seen by a consideration of the manner in which the water
in the chamber outside the guide passages is probably set in
motion, Fig. 280. Since there is no discharge, this rotation cannot
be caused by the water passing through the pump, but must be
due to internal motions set up in the wheel and casing. The
water in the guide chamber cannot obviously rotate about the
axis 0, but there is a tendency for it to do so, and consequently
stream line motions, as shown in the figure, are probably set
up. The layer of water nearest the outer circumference of the
wheel will no doubt be dragged along by friction in the direction
shown by the arrow, and water will flow from the outer casing to
take its place ; the stream lines will give motion to the water in
the outer casing.

* See page 415.

CENTRIFUGAL PUMPS

407

When the vanes in the wheel are radial and as long as a vane is
moving between any two guide vanes, the straight vane prevents
the friction between the water outside the wheel and that inside,
from dragging the water backwards along the vane, but when the
vane is set back and the angle < is greater than 90 degrees, there
will be a tendency for the water in the wheel to move backwards
while that in the guide chamber moves forward, and consequently
the velocity of the stream lines in the casing will be less in the
latter case than in the former. In either case, the general
direction of flow of the stream lines, in the guide chamber, will
be in the direction of rotation of the wheel, but due to friction
and eddy motions, even with radial vanes, the velocity of the stream

Fig. 280.

lines will be less than the velocity i\ of the periphery of the wheel.
Just outside the guide chambers the velocity of rotation will be
less than VL In the outer chamber it is to be expected that the
water will rotate as in a free vortex, or its velocity of whirl will
be inversely proportional to the distance from the centre of the
rotor, or will rotate in some manner approximating to this.

The head which a pump, with a vortex chamber, will theoreti-
cally maintain when the discharge is zero. In this case it is
probable that as the discharge approaches zero, in addition to the
water in the wheel rotating, the water in the vortex chamber will
also rotate because of friction,

408 HYDRAULICS

The centrifugal head due to the water in the wheel is

If K = 2r, this becomes j ^- .
4 Zg

The centrifugal head due to the water in the chamber is,
Fig. 281,

v<?dr

r and VQ being the radius and tangential velocity respectively of
any ring of water of thickness dr.

Fig. 281.

If it be assumed that v r is a constant, the centrifugal head
due to the vortex chamber is

tfrV P*dr = vfr? /_! J_\

g !r w r 3 2g W R.V'
The total centrifugal head is then

i, -^.^o-^LYi- _L\
~2^ 2^ + 2^ W R.V*

If r; is 2r and R, is

The conditions here assumed, however, give h c too high. In
Stanton's experiments h was only ~ . Decouer from experi-

CENTRIFUGAL PUMPS 409

ments on a small pump with a vortex chamber, the diameter being

I .q 2

about twice the diameter of the wheel, found h c to be ~ 1 - .

Let it be assumed that h c is -~-*- in any pump, and that the lift

*9
of the pump when working normally is

, _ eViVi _ e fa 2 - vMi cot <fr)

9 mv* g
Then if h is greater than -^, the pump will not commence to

discharge unless speeded up to some velocity v z such that
my* e (vi - ViUi cot <ft)
2g ' ~T

After the discharge has been commenced, however, the speed
may be diminished, and the pump will continue to deliver against
the given head*.

For any given values of m and e the velocity v z at which delivery
commences decreases with the angle <. If <f> is 90 or greater than
90 degrees, and m is unity, the pump will only commence to
discharge against the normal head when the velocity is v i9 if the
manometric efficiency e is less than 0*5. If < is 30 degrees and m
is unity, v z is equal to Vi when e is 0'6, but if </> is 150 degrees v 2
is equal to Vi when e is 0'428.

Nearly all actual pumps are run at such a speed that the
centrifugal head at that speed is greater than the gross head
against which the pump works, so that there is never any
difficulty in starting the pump. This is accounted for (1) by the
low manometric efficiencies of actual pumps, (2) by the angle <
never being greater than 90 degrees, and (3) by the wheels being
surrounded by casings which allow the centrifugal head to be

greater than .

It should be observed that it does not follow, because in many
cases the manometric efficiency is small, the actual efficiency of
the pump is of necessity low. (See Fig. 286.)

227. Head-velocity curve of a centrifugal pump at zero
discharge.

For any centrifugal pump a curve showing the head against
which it will start pumping at any given speed can easily be
determined as follows.

On the delivery pipe fit a pressure gauge, and at the top

* See pages 411, 419 and 542.

410

HYDRAULICS

of the suction pipe a vacuum gauge. Start the pump with
the delivery valve closed, and observe the pressure on the two
gauges for various speeds of the pump. Let p be the absolute
pressure per sq. foot in the delivery pipe and pi the absolute

pressure per sq. foot at the top of the suction pipe, then -
is the total centrifugal head h .

60-

,4V

WOO 1800 2000 2200

Revolutions per Minute. '
Fig. 282.

A curve may now be plotted similar to that shown in Fig. 282
which has been drawn from data obtained from the pump shown
in Fig. 275.

When the head is 44 feet, the speed at which delivery would
just start is 2000 revolutions per minute.

On reference to Fig. 293, which shows the discharge under
different heads at various speeds, the discharge at 2000 revolutions
per minute when the head is 44 feet is seen to be 12 cubic feet
per minute. This means, that if the pump is to discharge against
this head at this speed it cannot deliver less than 12 cubic feet
per minute.

228, Variation of the discharge of a centrifugal pump
with the head when the speed is kept constant*.

Head-discharge curve at constant velocity. If the speed of a
centrifugal pump is kept constant and the head varied, the dis-
charge varies as shown in Figs. 283, 285, 289, and 292.

* See also page 418.

CENTRIFUGAL PUMPS

411

The curve No. 2, of Fig. 283, shows the variation of the head
with discharge for the pump shown in Fig. 275 when running at
1950 revolutions per minute; and that of Fig. 285 was plotted
from experimental data obtained by M. Rateau on a pump having
a wheel 11*8 inches diameter.

The data for plotting the curve shown in Fig. 289* was
obtained from a large centrifugal pump having a spiral chamber.
In the case of the dotted curve the head is always less than the
centrifugal head when the flow is zero, and the discharge against
a given head has only one value.

701

3 4-

Radii Velocity dffiow from. Wheel.
Fig. 283. Head-discharge curve for Centrifugal Pump. Velocity Constant.

Fig. 284. Velocity-discharge curve for Centrifugal Pump. Head Constant.

In Fig. 285 the discharge when the head is 80 feet may be
either *9 or 3'5 cubic feet per minute. The work required to drive
the pump will be however very different at the two discharges,
and, as shown by the curves of efficiency, the actual efficiencies
for the two discharges are very different. At the given velocity
therefore and at 80 feet head, the flow is ambiguous and is
unstable, and may suddenly change from one value to the other,
or it may actually cease, in which case the pump would not start
again without the velocity ^ being increased to 707 feet per
second. This value is calculated from the equation

Proceedings last. Mech. Engs. t 1903.

412

HYDRAULICS

the coefficient m for this pump being 1'02. For the flow to be
stable when delivering against a head of 80 feet, the pump should
be run with a rim velocity greater than 70'7 feet per second, in
which case the discharge cannot be less than 4J cubic feet per
minute, as shown by the velocity-discharge curve of Fig. 287.
The method of determining this curve is discussed later.

Pump Wheel fl-Sctiam/.

* 10
%60

'* W

#20
*/>

^~ ;

~~-^

ad-Disc)

large Gar

r e

v,*=66'

oersec.

/</

Fig. 235.

3 4>

Discharge in c.fl. per mm/.

Fig. 286.

-pischarge
f jonstarub=

Carve

Fig. 287.

Example. A centrifugal pump, when discharging normally, has a peripheral
velocity of 50 feet per second.

The angle at exit is 30 degrees and the manometric efficiency is 60 per cent.
The radial velocity of flow at exit is 2 ^//i.

Determine the lift h and the velocity of the wheel at which it will start delivery
under full head.

V = 50 -(2 cos 130
* 60- 1-73 *.

CENTRIFUGAL PUMPS 413

Therefore

from which h = 37 feet.

Let ? 2 be the velocity of the rim of the wheel at which pumping commences.
Then assuming the centrifugal head, when there is no discharge, is

v 2 =48-6 ft. per sec.

229. Bernoulli's equations applied to centrifugal pumps.

Consider tlie motion of the water in any passage between two
consecutive vanes of a wheel. Let p be the pressure at inlet, pi at
outlet and p a the atmospheric pressure per sq. foot.

If the wheel is at rest and the water passes through it in
the same way as it does when the wheel is in motion, and all
losses are neglected, and the wheel is supposed to be horizontal, by
Bernoulli's equations (see Figs. 277 and 278),

w 2g w 2g

But since, due to the rotation, a centrifugal head

is impressed on the water between inlet and outlet, therefore,

_
w 2g w 2g 2g 2g

p, p , * V r 2 v?
w-w=2g-2- g + 2g:-2g
From (3) by substitution as on page 337,

w 2g w 2g g g
and when U is radial and therefore equal to u,

w 2g w 2g g

If now the velocity Ui is diminished gradually and without
shock, so that the water leaves the delivery pipe with a velocity
u d , and if frictional losses be neglected, the height to which the
water can be lifted above the centre of the pump is, by Bernoulli's
equation,

h = P 1+ W_P_uJ (?)>

w 2g w 2g

If the centre of the wheel is h feet above the level of the water
in the sump 01 well, and the water in the well is at rest,

& = *. + *+ ...(8).

w w 2g

414 HYDRAULICS

Substituting from (7) and (8) in (6)

9 2 2gr

= H,+ g = H ..................... (9).

This result verifies the fundamental equation given on page 398.
Further from equation (6)

^1^IZL_2_^L = TT +- d -
w 2g w 2g 2g*

Example. The centre of a centrifugal pump is 15 feet above the level of the
vater in the sump. The total lift is 60 feet and the velocity of discharge from the
delivery pipe is 5 feet per second. The angle <j> at discharge is 135 degrees, and
the radial velocity of flow through the wheel is 5 feet per second. Assuming there
are no losses, find the pressure head at the inlet and outlet circumferences.

At inlet *S4' -!?-.

w 64

= 18-6 feet.
The radial velocity at outlet is

! = 5 feet per second,

and = ll = 60 25

9 <J 64'

and therefore, v 1 2 + 5v 1 =1940 ....................................... (1),

from which Vj = 41 -6 feet per second,

and V = 46-6

The pressure head at outlet is then

w w

= 45 feet.
To find the velocity v when <f> is made 30 degrees.

cot 0=^/3,

therefore (1) becomes vf - 5 */3 . v l = 1940,

from which v 1 = 48 > 6 ft. per sec.

and V 1 = 40

Then ^L = 25-4 feet, and % = 53-6 feet.

2g w

230. Losses in centrifugal pumps.

The losses of head in a centrifugal pump are due to the same
causes as the losses in a turbine.

Loss of head at exit. The velocity Ui with which the water
leaves the wheel is, however, usually much larger than in the
case of the turbine, and as it is not an easy matter to diminish
this velocity gradually, there is generally a much larger loss of
velocity head at exit from the wheel in the pump than in the
turbine.

CENTRIFUGAL PUMPS 415

In many of the earlier pumps, which had radial vanes at exit,

U 2
the whole of the velocity head ^- was lost, no special precautions

being taken to diminish it gradually and the efficiency was
constantly very low, being less than 40 per cent.

The effect of the angle < on the efficiency of the pump. To
increase the efficiency Appold suggested that the blade should be
set back, the angle < being thus less than 90 degrees, Fig. 272.

Theoretically, the effect on the efficiency can be seen by
considering the three cases considered in section 220 and illustrated

TJ 2
in Fig. 279. When < is 90 degrees -~- is *54H, and when < is

U 2

30 degrees -^- is *36H. If, therefore, in these two cases this head

is lost, while the other losses remain constant, the efficiency in
the second case is 18 per cent, greater than in the first, and the
efficiencies cannot be greater than 46 per cent, and 64 per cent.
respectively.

In general when there is no precaution taken to utilise the
energy of motion at the outlet of the wheel, the theoretical lift is

and the maximum possible manometric efficiency is

Substituting for Vi, i - Ui cot <}>, and for Ui 2 , Vi 8 + u*,

Ht =lrS 2cosec '*'

, _- fa - U, COt <ft) 2 + U?

v Ui cosec <
~ 2vi (vi Ui cot <#>) "

When v l is 30 feet per second, Ui 5 feet per second and <
30 degrees, e is 62'5 per cent, and when < is 90 degrees e is
48'5 per cent.

Experiments also show that in ordinary pumps for a given lift
and discharge the efficiency is greater the smaller the angle <f>.

Parsons* found that when < was 90 degrees the efficiency of a
pump in which the wheel was surrounded by a circular casing
was nearly 10 per cent, less than when the angle < was made
about 15 degrees.

* Proceedings Inst. C. E. t Vol. XLVH. p. 272.

416 HYDKAULICS

Stanton found that a pump 7 inches diameter having radial
vanes at discharge had an efficiency of 8 per cent, less than when
the angle $ at delivery was 30 degrees. In the first case the
maximum actual efficiency was only 39'6 per cent., and in the
second case 50 per cent.

It has been suggested by Dr Stanton that a second reason for
the greater efficiency of the pump having vanes curved back at
outlet is to be found in the fact that with these vanes the variation
of the relative velocity of the water and the wheel is less than
when the vanes are radial at outlet. It has been shown experi-
mentally that when the section of a stream is diverging, that is
the velocity is diminishing and the pressure increasing, there is
a tendency for the stream lines to flow backwards towards the
sections of least pressure. These return stream lines cause a loss
of energy by eddy motions. Now in a pump, when the vanes are
radial, there is a greater difference between the relative velocity
of the water and the vane at inlet and outlet than when the angle
</> is less than 90 degrees (see Fig. 279), and it is probable there-
fore that there is more loss by eddy motions in the wheel in the
former case.

Loss of head at entry. To avoid loss of head at entry the vane
must be parallel to the relative velocity of the water and the
vane.

Unless guide blades are provided the exact direction in which
the water approaches the edge of the vane is not known. If there
were no friction between the water and the eye of the wheel it
would be expected that the stream lines, which in the suction pipe
are parallel to the sides of the pipe, would be simply turned to
approach the vanes radially.

It has already been seen that when there is no flow the water
in the eye of the wheel is made to rotate by friction, and it is
probable that at all flows the water has some rotation in the eye
of the wheel, but as the delivery increases the velocity of rotation
probably diminishes. If the water has rotation in the same
direction as the wheel, the angle of the vane at inlet will clearly
have to be larger for no shock than if the flow is radial. That
the water has rotation before it strikes the vanes seems to be
indicated by the experiments of Mr Livens on a pump, the vanes
of which were nearly radial at the inlet edge. (See section 236.)
The efficiencies claimed for this pump are so high, that there
could have been very little loss at inlet.

If the pump has to work under variable conditions and the
water be assumed to enter the wheel at all discharges in the same
direction, the relative velocity of the water and the edge of the

CENTRIFUGAL PUMPS 417

vane can only be parallel to the tip of the vane for one discharge,
and at other discharges in order to make the water move along
the vane a sudden velocity must be impressed upon it, which
causes a loss of energy.

Let u. 2) Fig. 288, be the velocity with which the water enters a
wheel, and and v the inclination
and velocity of the tip of the vane \*- us ->j
at inlet respectively.

The relative velocity of u 2 and v
is V/, the vector difference of u*
and v.

The radial component of flow
through the opening of the wheel
must be equal to the radial com-
ponent of u 2 , and therefore the
relative velocity of the water along the tip of the vane is V r .

If Uz is assumed to be radial, a sudden velocity

u 8 - v - u* cot
has thus to be given to the water.

If Us has a component in the direction of rotation u a will be
diminished.

It has been shown (page 67), on certain assumptions, that if
a body of water changes its velocity from v a to v* suddenly, the

head lost is ^-^ , or is the head due to the change of velocity.

*9
In this case the change of velocity is u s , and the head lost may

ku 2
reasonably be taken as -^- . If k is assumed to be unity, the

*9
effective work done on the water by the wheel is diminished by

Ug__ (V-Uy COt #) 2

2<r 2 3

If now this loss takes place in addition to the velocity head
being lost outside the wheel, and friction losses are neglected,
then

V? Q 2 z ,

- Si 2 cosec 9

- " 2 cot. a

t. H. 27

418

HYDRAULICS

Example. The radial velocity of flow through a pump ia 5 feet per second.
The angle is 80 degrees and the angle Q is 15 degrees. The velocity of the
outer circumference is 50 feet per sec. and the radius is twice that of the inner
circumference.

Find the theoretical lift on the assumption that the whole of the kinetic energy
is lost at exit.

v,* 5 2 (25 - 5 cot 15)*

h = cosec 2 30 - s

2g 2g 2g

= 37-0 feet.

The theoretical lift neglecting all losses is 64-2 feet, and the manometric
efficiency is therefore 58 per cent.

231. Variation of the head with discharge and with the
speed of a centrifugal pump.

It is of interest to study by means of equation (1), section 230,
the variation of the discharge Q with the velocity of the pump
when h is constant, and the variation of the head with the
discharge when the velocity of the pump is constant, and to
compare the results with the actual results obtained from
experiment.

The full curve of Fig. 289 shows the variations of the head
with the discharge when the velocity of a wheel is kept constant.
The data for which the curve has been plotted is indicated in
the figure.

v t -30Ft,.p<
v = 15 ' o * "

fi i 1

zrSet

. Normal radial veloct

\1 \Z \3 |4

f-"

RacbLaJL velocity of Fkw= H
A,

Fig. 289. Head-discharge curve at constant velocity.

When the discharge is zero

h = pr PT = 10*5 feet.
2g 2g

The velocity of flow -~ at outlet has been assumed equal to

-5r at inlet.
A.

Values of 1, 2, etc. were given to ~ and the corresponding
values of h found from equation (1).

CENTRIFUGAL PUMPS 419

When the discharge is normal, that is, the water enters the

wheel without shock, ~ is 4 feet and h is 14 feet. The theoretical

head assuming no losses is then 28 feet and the manometric
efficiency is thus 50 per cent. For less or greater values of -f

the head diminishes and also the efficiency.

The curve of Fig. 290 shows how the flow varies with the
velocity for a constant value of ft, which is taken as 12 feet.

Radial Velocity through, Wheei.
Fig. 290. Velocity-discharge curve at constant head for Centrifugal Pnrap.

It will be seen that when the velocity t?i is 31*9 feet per second
the velocity of discharge may be either zero or 8'2 feet per second.
This means that if the head is 12 feet, the pump, theoretically,
will only start when the velocity is 31*9 feet per second and the
velocity of discharge will suddenly become 8*2 feet per second.
If now the velocity Vi is diminished the pump still continues to
discharge, and will do so as long as Vj. is greater than 26*4 feet per
second. The flow is however unstable, as at any velocity v it may
suddenly change from CB to CD, or it may suddenly cease, and it
will not start again until ^ is increased to 31*9 feet per second.

232. The effect of the variation of the centrifugal head
and the loss by friction on the discharge of a pump.

If then the losses at inlet and outlet were as above and were
the only losses, and the centrifugal head in an actual pump was
equal to the theoretical centrifugal head, the pump could not be
made to deliver water against the normal head at a small velocity
of discharge. In the case of the pump considered in section 231,
it could not safely be run with a rim velocity less than 31*9 ft.
per sec., and at any greater velocity the radial velocity of flow
could not be less than 8 feet per second,

272

420 HYDRAULICS

In actual pumps, however, it has been seen that the centrifugal
head at commencement is greater than

There is also loss of head, which at high velocities and in small
pumps is considerable, due to friction. These two causes consider-
ably modify the head-discharge curve at constant velocity and the
velocity-discharge curve at constant head, and the centrifugal
head at the normal speed of the pump when the discharge is zero,
is generally greater than any head under which the pump works,
and many actual pumps can deliver variable quantities of water
against the head for which they are designed.

The centrifugal head when the flow is zero is

m being generally equal to, or greater than unity. As the flow
increases, the velocity of whirl in the eye of the wheel and in
the casing will diminish and the centrifugal head will therefore
diminish.

Let it be assumed that when the velocity of flow is u (supposed
constant) the centrifugal head is

7, _^L_^ -

flc ~'2g 20 20

and n being constants which must be determined by experiment.
When u is zero

and if m is known Jc can at once be found.

Let it further be assumed that the loss by friction* and eddy

cV
motions, apart from the loss at inlet and outlet is -~- .

* The loss of head by friction will no doubt depend not only upon u but also
upon the velocity v l of the wheel, and should be written as

Cu 2 o
or, as 27 + -!r +

If it be supposed it can be expressed by the latter, then the correction
fcV 2nku 1 v 1 ^
~2g'~ 2g 2g '

if proper values are given to &, n^ and k^ , takes into account the variation of the
centrifugal head and also the friction head v l .

CENTRIFUGAL PUMPS 421

The gross head h is then,

2vu cot 9 A
~^- ~*<*>**

nu* cV

2g -~2j ............... '

If now the head h and flow Q be determined experimentally,
the difference between h as determined from equation (1), page 4J 7,
and the experimental value of h, must be equal to

V 2nkuv l

2g
hi being equal to (c 2 -w 2 ).

The coefficient Jc being known from an experiment when u is
zero, for many pumps two other* experiments giving corresponding
values of h and u will determine the coefficients n and fa.

The head-discharge curve at constant velocity, for a pump such
as the one already considered, would approximate to the dotted
curve of Fig. 289. This curve has been plotted from equation (2),
by taking k as 0'5, n as 7*64 and fa as - 38.

Substituting values for fa n, fa, cosec < and cot <, equation (2)
becomes

C and Ci being new coefficients ; or it may be written

Q being the flow in any desired units, the coefficients C 2 and C 8
varying with the units. If * equation (4) is of the correct form,
three experiments will determine the constants m, C 2 and C 3
directly, and having given values to any two of the three
variables h, v, and Q the third can be found.

233. The effect of the diminution of the centrifugal head
and the increase of the friction head as the flow increases, on
the velocity-discharge curve at constant head.

Using the corrected equation (2), section 232, and the given
values of k, rh and fa the dotted curve of Fig. 290 has been plotted.

From the dotted curve of Fig. 289 it is seen that u cannot
be greater than 5 feet when the head is 12 feet, and therefore the
new curve of Fig. 290 is only drawn to the point where u is 5.

The pump starts delivering when v is 27*7 feet per second and
the discharge increases gradually as the velocity increases.

* See page 544.

422 HYDRAULICS

The pump will deliver, therefore, water under a head of
12 feet at any velocity of flow from zero to 5 feet per second.

In such a pump the manometric efficiency must have its
maximum value when the discharge is zero and it cannot be
greater than

COt '

This is the case with many existing pumps and it explains why,
when running at constant speed, they can be made to give any
discharge varying from zero to a maximum, as the head is
diminished.

234. Special arrangements for converting the velocity
head ^- with which the water leaves the wheel into pressure

head.

The methods for converting the velocity head with which the
water leaves the wheel into pressure head have been indicated on
page 394. They are now discussed in greater detail.

Thomson's vortex or whirlpool chamber. Professor James
Thomson first suggested that the wheel should be surrounded by
a chamber in which the velocity of the water should gradually
change from Ui to u d the velocity of flow in the pipe. Such a
chamber is shown in Fig. 274. In this chamber the water forms
a free vortex, so called because no impulse is given to the water
while moving in the chamber.

Any fluid particle ab, Fig. 281, may be considered as moving
in a circle of radius r with a velocity v and to have also a
radial velocity u outwards.

Let it be supposed the chamber is horizontal.

If W is the weight of the element in pounds, its momentum

perpendicular to the radius is - and the moment of mo-

~\/\ / H 7*

mentum or angular momentum about the centre C is - .

For the momentum of a body to change, a force must act upon
it, and for the moment of momentum to change, a couple must act
upon the body.

But since no turning effort, or couple, acts upon the element
after leaving the wheel its moment of momentum must be
constant.

CENTRIFUGAL PUMPS 423

Therefore,

is constant or V r = constant.

If the sides of the chamber are parallel the peripheral area of
the concentric rings is proportional to r , and the radial velocity of
flow u for any ring will be inversely proportional to r , and there-
fore, the ratio is constant, or the direction of motion of any

element with its radius r is constant, and the stream lines are
equiangular spirals.

If no energy is lost, by friction and eddies, Bernoulli's theorem
will hold, and, therefore, when the chamber is horizontal

2g + 2g + w

is constant for the stream lines.

This is a general property of the free vortex.
If u is constant

?r" + = constant.
2g w

Let the outer radius of the whirlpool chamber be R, and
the inner radius r w . Let v fw and v Rw be the whirling velocities
at the inner and outer radii respectively.

Then since v ^o is a constant,

and - ? + - = constant,

w 2g

w w 2g 2g

= ^ + w( l ~^'
When U w = 2r w ,

w w 4* 2g

If the velocity head which the water possesses when it leaves
the vortex chamber is supposed to be lost, and hi is the head of
water above the pump and p a the atmospheric pressure, then
neglecting friction

u d *

- = i ?i -- "

w 2g w

, PR W UA Pa

424 HYDRAULICS

If then h is the height of the pump above the well, the total
lift h% is hi + ho.
Therefore,

/, -7, + P + v '-
k-^ + +

But ^zfc.p *

to 10 z#
also Pr; = pi, ?*> = R, and v ru , = V lt

Therefore

, _pi_ p_^ V^/, R 2 \ _^
^ w 2g 2g V " R.V 2<? '

But from equation (6) page 413,

tt; w 2g g 2g
Therefore

Ui V^A R 2 \

^ ^A B2/'

This might have been written down at once from equation (1),
section 230. For clearly if there is a gain of pressure head

V 2 / R 2 \
in the vortex chamber of -^- fl ~-p~2J> ^ ne velocity head to

be lost will be less by this amount than when there is no vortex
chamber.

Substituting for Vi and Ui the theoretical lift h is now

, _V*-V 1 Ui COt<j> U? fa - Ui COt <^>) a R 2 ^ n ^

g -fy~ ~W -'&,

When the discharge or rim velocity is not normal, there is a
further loss of head at entrance equal to

and

.-. cot*

(2).

When there is no discharge v rtt , is equal to Vi and

J, = ^1_^

CENTRIFUGAL PUMPS 425

R = 2 RW and v = 2^1

Correcting equation (1) in order to allow for the variation of
the centrifugal head with the discharge, and the friction losses,

, _ Vi - ViUi COt < Ui (Vi Ui COt <ft) 2 R 2

~~ ""

(v u cot 0)* k?v* 2nkuvi

which reduces to h -

The experimental data on the value of the vortex chamber
per se, in increasing the efficiency is very limited.

Stanton* showed that for a pump having a rotor 7 inches
diameter surrounded by a parallel sided vortex chamber 18 inches
diameter, the efficiency of the chamber in converting velocity head
to pressure head was about 40 per cent. It is however questionable
whether the design of the pump was such as to give the best results
possible.

So far as the author is aware, centrifugal pumps with vortex
chambers are not now being manufactured in England, but it
seems very probable that by the addition of a well-designed
chamber small centrifugal pumps might have their efficiencies
considerably increased.

235. Turbine pumps.

Another method, first suggested by Professor Reynolds, and
now largely used, for diminishing the velocity of discharge Ui
gradually, is to discharge the water from the wheel into guide
passages the sectional area of which should gradually increase
from the wheel outwards, Figs. 275 and 276, and the tangents to the
tips of the guide blades should be made parallel to the direction
of Uj.

The number of guide passages in small pumps is generally four
or five.

If the guide blades are fixed as in Fig. 275, the direction of
the tips can only be correct for one discharge of the pump,
but except for large pumps, the very large increase in initial cost
of the pump, if adjustable guide blades were used, as well as
the mechanical difficulties, would militate against their adoption.

Single wheel pumps of this type can be used up to a head of
100 feet with excellent results, efficiencies as high as 85 per cent.
* Proceedings Inst. C.E., 1903. See also page 542.

426 HYDRAULICS

having been claimed. They are now being used to deliver water
against heads of over 350 feet, and M. Rateau has used a single
wheel 3'16 inches diameter running at 18,000 revolutions per
minute to deliver against a head of 936 feet.

Loss of head at the entrance to the guide passages. If the
guide blades are fixed, the direction of the tips can only be correct
for one discharge of the pump. For any other discharge than the
normal, the direction of the water as it leaves the wheel is not
parallel to the fixed guide and there is a loss of head due to
shock.

Let a be the inclination of the guide blade and < the vane
angle at exit.

Let Ui be the radial velocity of
flow. Then BE, Fig. 291, is the
velocity with which the water leaves
the wheel.

The radial velocity with which
the water enters the guide passages must be Ui and the velocity
along the guide is, therefore, BF.

There is a sudden change of velocity from BE to BF, and on
the assumption that the loss of head is equal to the head due to the
relative velocity FE, the head lost is

fa - -MI cot <ft - Uj cot cp a

At inlet the loss of head is

(v-u cot (9) 2

20
and the theoretical lift is

cot cfr (v-u cot 0) a fa Ui cot <j> - u\ cot a) a
~

= ~W 2g

_ v* v 9 2v 1 u l cot a 2vu cot
= 2g~2~g* ~~2g~ 20

Ui (cot <f> + cot a) 2 u? cot 2 m

~^~ %

To correct for the diminution of the centrifugal head and to

allow for friction,

tfv* _Zkvin^Ui _ -, u?
29 " 2^ Cl 2g>

must be added, and the lift is then
, _ Vi v 2 2viUi cot a 2vu cot U* (cot <ft + cot a) 9
h= 2g-*} + 20 2g 2g

u* cot 2 feV ^ ZJcnViUj TKU?
2g h "20 20 " 20 '

CENTRIFUGAL PUMPS

427

which, since u can always be written as a multiple of Ui, reduces
to the form

2gh = mv*+ CuiVi + du* (2).

Equations for the turbine pump shown in Fig. 275. Character-
istic curves. Taking the data

= 5 degrees, cot = 11 '43
= 1732

equation (2) above becomes
20fc = '

cot a =19*6

- 587

eo-

(3)

3)i<$cftarge, in/ Cubic Feet per J&nute. ,^

_l L_ i i_ __J i 1.1

f 23

Velocity cub Exit/ fronv the Wlieet/. Feet fer Second/.

Fig. 292. Head-discharge curves at constant speed for Turbine Pump.

From equation (3) taking Vi as 50 feet per second, the head-
discharge curve No. 1, of Fig. 283, has been drawn, and taking h
as 35 feet, the velocity-discharge curve No. 1, of Fig. 284, has been
plotted.

In Figs. 292 4 are shown a series of head-discharge curves at

428

HYDRAULICS

constant speed, velocity-discharge curves at constant head, and
head-velocity curves at constant discharge, respectively.

The points shown near to the curves were determined experi-
mentally, and the curves, it will be seen, are practically the mean
curves drawn through the experimental points. They were how-
ever plotted in all cases from the equation

2gh = l-087t>! a + 2'26tM>! - 62' W,

obtained by substituting for m, C and d in equation (2) the values
1*087, 2'26 and - 62*1 respectively. The value of m was obtained
by determining the head h, when the stop valve was closed, for
speeds between 1500 and 2500 revolutions per minute, Fig. 282.
The values of C and Ci were first obtained, approximately, by
taking two values of Ui and Vi respectively from one of the
actual velocity-discharge curves near the middle of the series, for
which h was known, and from the two quadratic equations thus
obtained C and Ci were calculated. By trial C and Ci were then
corrected to make the equation more nearly fit the remaining
curves.

ZOOO 2100

Speed* Revolutions per Alutute,.
Fig. 293. Velocity-Discharge curves at Constant Head.

No attempt has been made to draw the actual mean curves in
the figures, as in most cases the difference between them and the
calculated curves drawn, could hardly be distinguished. The
reader can observe for himself what discrepancies there are between
the mean curves through the points and the calculated curves. It

CENTRIFUGAL PUMPS

429

will be seen that for a very wide range of speed, head, and
discharge, the agreement between the curves and the observed
points is very close, and the equation can therefore be used with
confidence for this particular pump to determine its performance
under stated conditions.

It is interesting to note, that the experiments clearly indicated
the unstable condition of the discharge when the head was kept
constant and the velocity was diminished below that at which the
discharge commenced.

Fig. 294. Head-velocity curves at Constant Discharge.

236. Losses in the spiral casings of centrifugal pumps.

The spiral case allows the mean velocity of flow toward the
discharge pipe to be fairly constant and the results of experiment
seem to show that a large percentage of the velocity of the water
at the outlet of the wheel is converted into pressure head.
Mr Livens* obtained, for a purnp having a wheel 19 \ inches
diameter running at 550 revolutions per minute, an efficiency of
71 per cent, when delivering 1600 gallons per minute against a
head of 25 feet. The angle < was about 13 degrees and the mean
of the angle for the two sides of the vane 81 degrees.

For a similar pump 21| inches diameter an efficiency of 82 per
cent, was claimed.

* Proceedings Inst. Mech. Engs., 1903.

430 HYDRAULICS

The * author finds the equation to the head- discharge curve for
the 19 inches diameter pump from Mr Livens' data to be

118v 1 2 + 3^1-142^ = 2gh .................. (1),

and for the 21 inches diameter pump

I'18v l *-4,'5u l v 1 = 2gh ..................... (2).

The velocity of rotation of the water round the wheel will be
less than the velocity with which the water leaves the wheel and
there will be a loss of head due to the sudden change in velocity.

k U 2
Let this loss of head be written -75-^ . The head, when Ui is the

radial velocity of flow at exit and assuming the water enters the
wheel radially, is then

, tti 2 -j;i^icot< /CsTJi 8 (v-ucotOy

g ' 2g 2g

Taking friction and the diminution of centrifugal head into
account,

, _ v* ViUiCoi<j> _ fe 3 Ui a _ (v ucotOy Jcv* __ nJm } v l _ Jc^t?

g ' 2g ' 2g 2*7 ~ ~2<T "~2g'

which again may be written

7, = mv * + C ^i^i , Gi^i a
" 20 2g 2g '

The values of m, C and Ci are given for two pumps in equations
(1) and (2).

237. General equation for a centrifugal pump.
The equations for the gross head h at discharge Q as determined
for the several classes of pumps have been shown to be of the form

_
=

2g 2g >
or, if u is the velocity of flow from the wheel,

Cuv

in which m varies between 1 and 1'5. The coefficients C 2 and C 3
for any pump will depend upon the unit of discharge.

As a further example and illustrating the case in which at
certain speeds the flow may be unstable, the curves of Figs.
285287 may be now considered. When v l is 66 feet per second
the equation to the head discharge curve is

. 15-5Qt?i _ 236Q a

Q being in cubic feet per minute.

* See Appendix 11.

CENTRIFUGAL PUMPS 431

The velocity-discharge curve for a constant head of 80 feet as
calculated from this equation is shown in Fig. 287.

To start the pump against a head of 80 feet the peripheral
velocity has to be 70' 7 feet per second, at which velocity the
discharge Q suddenly rises to 4'3 cubic feet per minute.

The curves of actual and manometric efficiency are shown in
Fig. 286, the maximum for the two cases occurring at different
discharges.

238. The limiting height to which a single wheel centri-
fugal pump can be used to raise water.

The maximum height to which a centrifugal pump can raise
water, depends theoretically upon the maximum velocity at which
the rim of the wheel can be run.

It has already been stated that rim velocities up to 250 feet
per second have been used. Assuming radial vanes and a mano-
metric efficiency of 50 per cent., a pump running at this velocity
would lift against a head of 980 feet.

At these very high velocities, however, the wheel must be of
some material such as bronze or cast steel, having considerable
resistance to tensile stresses, and special precautions must be
taken to balance the wheel. The hydraulic losses are also
considerable, and manometric efficiencies greater than 50 per
cent, are hardly to be expected.

According to M. Eateau *, the limiting head against which it is
advisable to raise water by means of a single wheel is about
100 feet, and the maximum desirable velocity of the rim of the
wheel is about 100 feet per second.

Single wheel pumps to lift up to 350 feet are however being
used. At this velocity the stress in a hoop due to centrifugal forces
is about 7250 Ibs. per sq. incht.

239. The suction of a centrifugal pump.

The greatest height through which a centrifugal or other class
of pump will draw water is about 27 feet. Special precaution has
to be taken to ensure that all joints on the suction pipe are perfectly
air-tight, and especially is this so when the suction head is greater
than 15 feet; only under special circumstances is it therefore de-
sirable for the suction head to be greater than this amount, and it
is always advisable to keep the suction head as small as possible.

* "Pompes Centrifuges," etc., Bulletin de la Societe de I'Industrie minfrale,
1902 ; Engineer, p. 236, March, 1902.

t See Swing's Strength of Materials ; Wood's Strength of Structural Members ;
The Steam Turbine Stodola.

432

HYDRAULICS

CENTRIFUGAL PUMPS

433

240. Series or multi-stage turbine pumps.

It has been stated that the limiting economical head for a single
wheel pump is about 100 feet, and for high heads series pumps
are now generally used.

Fig. 296. General Arrangement of Worthington Multi-stage Turbine Pump.

By putting several wheels or rotors in series on one shaft, each
rotor giving a head varying from 100 to 200 feet, water can be
lifted to practically any height, and such pumps have been
L. n, 28

434

HYDRAULICS

constructed to work against a head of 2000 feet. The number
of rotors, on one shaft, may be from one to twelve according
to the total head. For a given head, the greater the number of
rotors used, the less the peripheral velocity, and within certain
limits the greater the efficiency.

Figs. 295 and 296 show a longitudinal section and general
arrangement, respectively, of a series, or multi-stage pump, as
constructed by the Worthington Pump Company. On the motor
shaft are fixed three phosphor-bronze rotors, alternating with fixed
guides, which are rigidly connected to the outer casing, and to
the bearings. The water is drawn in through the pipe at the left
of the pump and enters the first wheel axially. The water leaves
the first wheel at the outer circumference and passes along an
expanding passage in which the velocity is gradually diminished
and enters the second wheel axially. The vanes in the passage
are of hard phosphor-bronze made very smooth to reduce friction
losses to a minimum. The water passes through the remaining
rotors and guides in a similar manner and is finally discharged
into the casing and thence into the delivery pipe.

'///////////////////////////w
Fig. 297. Sulzer Multi-stage Turbine Pump.

The difference in pressure head at the entrances to any two
consecutive wheels is the head impressed on the water by one
wheel. If the head is h feet, and there are n wheels the total
lift is nearly nh feet. The vanes of each wheel and the directions
of the guide vanes are determined as explained for the single
wheel so that losses by shock are reduced to a minimum, and
the wheels and guide passages are made smooth so as to reduce
friction.

Through the back of each wheel, just above the boss, are
a number of holes which allow water to get behind part of the
wheel, under the pressure at which it enters the wheel, to balance
the end thrust which would otherwise be set up.

CENTRIFUGAL PUMPS 435

The pumps can be arranged to work either vertically or
horizontally, and to be driven by belt, or directly by any form
of motor.

Fig. 297 shows a multi-stage pump as made by Messrs Sulzer.
The rotors are arranged so that the water enters alternately
from the left and right and the end thrust is thus balanced.
Efficiencies as high as 84 per cent, have been claimed for multi-
stage pumps lifting against heads of 1200 feet and upwards.

The Worthington Pump Company state that the efficiency
diminishes as the ratio of the head to the quantity increases, the
best results being obtained when the number of gallons raised
per minute is about equal to the total head.

Example. A pump is to be driven by a motor at 1450 revolutions per minute, and
is required to lift 45 cubic feet of water per minute against a head of 320 feet.
Required the diameter of the suction, and delivery pipes, and the diameter and
number of the rotors, assuming a velocity of 5 '5 feet per second in the suction and
delivery pipes, and a manometric efficiency at the given delivery of 50 per cent.

Assume provisionally that the diameter of the boss of the wheel is 3 inches.

Let d be the external diameter of the annular opening, Fig. 295.

Then, f(^-3 2 ) ^

144 = 60 x 5-5 *
from which eZ=6 inches nearly.

Taking the external diameter D of the wheel as 2d, D is 1 foot.

1450
Then, t?i = -^- x v - 76 feet per sec.

Assuming radial blades at outlet the head lifted by each wheel is

=90 feet.
Four wheels would therefore be required.

241. Advantages of centrifugal pumps.

There are several advantages possessed by centrifugal pumps.

In the first place, as there are no sliding parts, such as occur in
reciprocating pumps, dirty water and even water containing com-
paratively large floating bodies can be pumped without greatly
endangering the pump.

Another advantage is that as delivery from the wheel is
constant, there is no fluctuation of speed of the water in the
suction or delivery pipes, and consequently there is no necessity
for air vessels such as are required on the suction and delivery
pipes of reciprocating pumps. There is also considerably less
danger of large stress being engendered in the pipe lines by
"water hammer*."

Another advantage is the impossibility of the pressure in the
See page 384.

282

436 HYDRAULICS

pump casing rising above that of the maximum head which the
rotor is capable of impressing upon the water. If the delivery
is closed the wheel will rotate without any danger of the pressure
in the casing becoming greater than the centrifugal head (page
335). This may be of use in those cases where a pump is de-
livering into a reservoir or pumping from a reservoir. In the first
case a float valve may be fitted, which, when the water rises to
a particular height in the reservoir, closes the delivery. The
pump wheel will continue to rotate but without delivering water,
and if the wheel is running at such a velocity that the centri-
fugal head is greater than the head in the pipe line it will start
delivery when the valve is opened. In the second case a similar
valve may be used to stop the flow when the water falls below a
certain level. This arrangement although convenient is uneco-
nomical, as although the pump is doing no effective work, the
power required to drive the pump may be more than 50 per cent,
of that required when the pump is giving maximum discharge.

It follows that a centrifugal pump may be made to deliver
water into a closed pipe system from which water may be taken
regularly, or at intervals, while the pump continues to rotate at a
constant velocity.

Pump delivering into a long pipe line. When a centrifugal
pump or air fan is delivering into a long pipe line the resistances
will vary approximately as the square of the quantity of water
delivered by the pump.

Let > 2 be the absolute pressure per square inch which has
to be maintained at the end of the pipe line, and let the
resistances vary as the square of the velocity v along the pipe.
Then if the resistances are equivalent to a head hs=kv*, the

pressure head at the pump end of the delivery pipe must be

ES-fc+W

w w

-p* + fcQ!

"w A 2 '

A being the sectional area of the pipe.

Let - be the pressure head at the top of the suction pipe, then

w
the gross lift of the pump is

h== Pl-P = P? + l_P f
www A. 2 w

If, therefore, a curve, Fig. 298, be plotted having
fej^p) W
w A 3

CENTRIFUGAL PUMPS

437

as ordinates, and Q as abscissae, it will be a parabola. If on
the same figure a curve having h as ordinates and Q as abscissae
be drawn for any given speed, the intersection of these two
curves at the point P will give the maximum discharge the pump
will deliver along the pipe at the given speed.

Discharge in/ C. Ft/, per Second/.
Fig. 298.

242. Parallel flow turbine pump.

By reversing the parallel flow turbine a pump is obtained
which is similar in some respects to the centrifugal pump, but
differs from it in an essential feature, that no head is impressed on
the water by centrifugal forces between inlet and outlet. It
therefore cannot be called a centrifugal pump.

The vanes of such a pump might be arranged as in Fig. 299,
the triangles of velocities for inlet and outlet being as shown.

The discharge may be allowed to take place into guide
passages above or below the wheel, where the velocity can be
gradually reduced.

Since there is no centrifugal head impressed on the water
between inlet and outlet, Bernoulli's equation is

From which, as in the centrifugal pump,

g w w 2g 2g

If the wheel has parallel sides as in Fig. 299, the axial velocity
of flow will be constant and if the angles < and are properly
chosen, V r and v r may be equal, in which case the pressure at
inlet and outlet of the wheel will be equal. This would have
the advantage of stopping the tendency for leakage through the
clearance between the wheel and casing.

438

HYDRAULICS

Such a pump is similar to a reversed impulse turbine, the
guide passages of which are kept full. The velocity with which
the water leaves the wheel would however be great and the lift
above the pump would depend upon the percentage of the velocity
head that could be converted into pressure head.

Fig. 299.

Since there is no centrifugal head impressed upon the water,
the parallel-flow pump cannot commence discharging unless the
water in the pump is first set in motion by some external means,
but as soon as the flow is commenced through the wheel, the full
discharge under full head can be obtained.

Fig. 300.

Fig. 301.

To commence the discharge, the pump would generally have to
be placed below the level of the water to be lifted, an auxiliary
discharge pipe being fitted with a discharging valve, and a non-
return valve in the discharge pipe, arranged as in Fig. 300.

CENTRIFUGAL PUMPS 439

The pump could be started when placed at a height h above
the water in the sump, by using an ejector or air pump to exhaust
the air from the discharge chamber, and thus start the flow
through the wheel.

243. Inward flow turbine pump.

Like the parallel flow pump, an inward flow pump if constructed
could not start pumping unless the water in the wheel were first
set in motion. If the wheel is started with the water at rest
the centrifugal head will tend to cause the flow to take place
outwards, but if flow can be commenced and the vanes are
properly designed, the wheel can be made to deliver water at its
inner periphery. As in the centrifugal and parallel flow pumps,
if the water enters the wheel radially, the total lift is

g w w g g
From the equation

p_ Vr 8 _ PI v* v* v\

w 2g w 2g 2g 2g'

it will be seen that unless V r 2 is greater than

?L. 4. ^- _ ^L
2g + 2g 2g>

U 2
Pi is less than p, and ^- will then be greater than the total

lift H.

Yery special precautions must therefore be made to diminish
the velocity U gradually, or otherwise the efficiency of the pump
will be very low.

The centrifugal head can be made small by making the
difference of the inner and outer radii small.

f-f ?L. + *. - ^

2g + 2g 2g

Y a

is made equal to 7p- , the pressure at inlet and outlet will be the

^9
same, and if the wheel passages are carefully designed, the

pressure throughout the wheel may be kept constant, and the
pump becomes practically an impulse pump.

There seems no advantage to be obtained by using either
a parallel flow pump or inward flow pump in place of the centri-
fugal pump, and as already suggested there are distinct dis-
advantages.

244. Reciprocating pumps.

A simple form of reciprocating force pump is shown dia-
grammatically in Fig. 301. It consists of a plunger P working in

440

HYDllAULICS

Fig. 301 a. Vertical Single-acting Keciprocating Pump.

RECIPROCATING PUMPS 441

a cylinder C and has two valves Y s and Y D , known as the suction
and delivery valves respectively. A section of an actual pump
is shown in Fig. 301 a.

Assume for simplicity the pump to be horizontal, with the
centre of the barrel at a distance h from the level of the water
in the well; h may be negative or positive according as the
pump is above or below the surface of the water in the well.

Let B be the height of the barometer in inches of mercury.
The equivalent head H, in feet of water, is

H13'596 . B -I.IQQ-R
~12~ 183B '

which may be called the barometric height in feet of water.
"When B is 30 inches H is 34 feet.

When, the plunger is at rest, the valve Y D is closed by the head
of water above it, and the water in the suction pipe is sustained by
the atmospheric pressure.

Let ho be the pressure head in the cylinder, then

ho = H h,
or the pressure in pounds per square inch in the cylinder is

p = '43(H-/i),

p cannot become less than the vapour tension of the water. At
ordinary temperatures this is nearly zero, and h cannot be greater
than 34 feet.

If now the plunger is moved outwards, very slowly, and there
is no air leakage the valve Y s opens, and the atmospheric pressure
causes water to rise up the suction pipe and into the cylinder,
h remaining practically constant.

On the motion of the plunger being reversed, the valve YS
closes, and the water is forced through Y D into the delivery
pipe.

In actual pumps if h Q is less than from 4 to 9 feet the
dissolved gases that are in the water are liberated, and it is there-
fore practically impossible to raise water more than from 25 to
30 feet.

Let A be the area of the plunger in square inches and L the
stroke in feet. The pressure on the end of the plunger outside the
cylinder is equal to the atmospheric pressure, and neglecting
the friction between the plunger and the cylinder, the force neces-
sary to move the plunger is

P = '43 {H - (H - h)} A = -43fc . A Ibs.,
and the work done by the plunger per stroke is
E = '43h. A. L ft. Ibs.

442 HYDRAULICS

If Y is the volume displacement per stroke of the plunger
in cubic feet

E = 62-4/i. Y ft. Ibs.

The weight of water lifted per stroke is *43AL Ibs., and the
work done per pound is, therefore, h foot pounds.

Let Z be the head in the delivery pipe above the centre of the
pump, and Ud the velocity with which the water leaves the delivery
pipe.

Neglecting friction, the work done by the plunger during the

delivery stroke is Z + ^ foot pounds per pound, and the total work

in the two strokes is therefore h + Z + ~ foot pounds per pound.

The actual work done on the plunger will be greater than this
due to mechanical friction in the pump, and the frictional and
other hydraulic losses in the suction and delivery pipes, and at the
valves; and the volume of water lifted per suction stroke will
generally be slightly less than the volume moved through by the
plunger.

Let W be the weight of water lifted per minute, and lit the
total height through which the water is lifted.

The effective work done by the pump is W . h t foot pounds per
minute, and the effective horse-power is

33,000'

245. Coefficient of discharge of the pump. Slip.

The theoretical discharge of a plunger pump is the volume
displaced by the plunger per stroke multiplied by the number of
delivery strokes per minute.

The actual discharge may be greater or less than this amount.
The ratio of the discharge per stroke to the volume displaced by
the plunger per stroke is the Coefficient of discharge, and the
difference between these quantities is called the Slip.

If the actual discharge is less than the theoretical the slip is
said to be positive, and if greater, negative.

Positive slip is due to leakage past the valves and plunger,
and in a steady working pump, with valves in proper condition,
should be less than five per cent.

The causes of negative slip and the conditions under which it
takes place will be discussed later*.

* See page 461.

RECIPROCATING PUMPS

443

246. Diagram of work done by the pump.

Theoretical Diagram. Let a diagram be drawn, Fig. 302, the
ordinates representing the pressure in the cylinder and the abscissae
the corresponding volume displacements of the plunger. The
volumes will clearly be proportional to the displacement of the
plunger from the end of its stroke. During the suction stroke,
on the assumption made above that the plunger moves very
slowly and that therefore all frictional resistances, and also the
inertia forces, may be neglected, the absolute pressure behind the
plunger is constant and equal to H - h feet of water, or 62'4 (H h)
pounds per square foot, and on the delivery stroke the pressure is

(2\
Z + H + 2^- J pounds per square foot.

The effective work done per suction stroke is ABCD which equals
62*4 . h . V, and during the delivery stroke is EADF which equals

iCtA 1 T7 'M'd \

62 4 Z + ^ ) ,
\ 2g / '

and EBCF is the work done per cycle, that is, during one suction
and one delivery stroke.

E F

* '

20-
AtTTL

A i B

Freesu

t |

J"f r *f

i Tie

SccuLe, of

Fig. 302. Theoretical diagram of pressure in a Eeciprocating Pump.

Strokes per

Fig. 303.

Actual diagram. Fig. 303 shows an actual diagram taken by
means of an indicator from a single acting pump, when running
at a slow speed.

The diagram approximates to the rectangular form and only

444

HYDRAULICS

differs from the above in that at any point p in the suction stroke,
pq in feet of water is equal to h plus the losses in the suction
pipe, including loss at the valve, plus the head required to
accelerate the water in the suction pipe, and qr is the head
required to lift the water and overcome all losses, and to accelerate
the water in the delivery pipe. The velocity of the plunger being
small, these correcting quantities are practically inappreciable.

The area of this diagram represents the actual work done on
the water per cycle, and is equal to W (Z + h), together with the
head due to velocity of discharge and all losses of energy in the
suction and delivery pipes.

It will be seen later that although at any instant the pressure
in the cylinder is effected by the inertia forces, the total work
done in accelerating the water is zero.

247. The accelerations of the pump plunger and of the
water in the suction pipe.

The theoretical diagram, Fig. 302, has been drawn on the
assumption that the velocity of the plunger is very small and
without reference to the variation of the velocity and of the
acceleration of the plunger, but it is now necessary to consider
this variation and its effect on the motion of the water in the suction
and delivery pipes. To realise how the velocity and acceleration
of the plunger vary, suppose it to be driven by a crank and
connecting rod, as in Fig. 304, and suppose the crank rotates with
a uniform angular velocity of w radians per second.

Fig. 304.

If r is the radius of the crank in feet, the velocity of the crank
pin is V = wr feet per second. For any crank position OC, it is
proved in books on mechanism, that the velocity of the point B is

By making BD equal to OK a diagram of velocities

Y.OK

00
EDF is found.

When OB is very long compared with CO, OK is equal to
00 sin 0, and the velocity v of the plunger is then Ysin#, and

RECIPROCATING PUMPS

445

EDF is a semicircle. The plunger then moves with simple
harmonic motion.

If now the suction pipe is as in Fig. 304, and there is to be
continuity in the column of water in the pipe and cylinder, the
velocity of the water in the pipe must vary with the velocity of
the plunger.

Let v be the velocity of the plunger at any instant, A and
a the cross-sectional areas of the plunger and of the pipe respect-

v A.
ively. Then the velocity in the pipe must be : .

As the velocity of the plunger is continuously changing, it is
continuously being accelerated, either positively or negatively.

Let I be the length of the connecting rod in feet. The
acceleration* F of the point B in Fig. 305, for any crank angle
0, is approximately

F = o>V (cos

j-cos 20) .

Plotting F as BG-, Fig. 305, a curve of accelerations MNQ is
obtained.

When the connecting rod is very long compared with the
length of the crank, the motion is simple harmonic, and the
acceleration becomes

F = wV cos 0,

and the diagram of accelerations is then a straight line.

Velocity and acceleration of the water in the suction pipe. The
velocity and acceleration of the plunger being v and F respectively,
for continuity, the velocity of the water in the pipe must be

v and the acceleration
a

F.A

* See Balancing of Engines, W. E. Dalby.

44C HYDRAULICS

248. The effect of acceleration of the plunger on the
pressure in the cylinder during the suction stroke.

When the velocity of the plunger is increasing, F is positive,
and to accelerate the water in the suction pipe a force P is
required. The atmospheric pressure has, therefore, not only to
lift the water and overcome the resistance in the suction pipe,
but it has also to provide the necessary force to accelerate the
water, and the pressure in the cylinder is consequently diminished.

On the other hand, as the velocity of the plunger decreases,
F is negative, and the piston has to exert a reaction upon the
water to diminish its velocity, or the pressure on the plunger is
increased.

Let L be the length of the suction pipe in feet, a its cross-
sectional area in square feet, f a the acceleration of the water in
the pipe at any instant in feet per second per second, and w the
weight of a cubic foot of water.

Then the mass of water in the pipe to be accelerated is w . a . L
pounds, and since by Newton's second law of motion
accelerating force = mass x acceleration,
the accelerating force required is

The pressure per unit area is

f-s^./.n.

and the equivalent head of water is

, _L

9 ' "'

, F.A

or since f a = ,

g.a
This may be large if any one of the three quantities, L, , or

B
F is large.

Neglecting friction and other losses the pressure in the
cylinder is now

H Ji h a ,

and the head resisting the motion of the piston is h + h a .

249. Pressure in the cylinder during the suction stroke
when the plunger moves with simple harmonic motion.

If the plunger be supposed driven by a crank and very long

RECIPROCATING PUMPS

447

connecting rod, the crank rotating uniformly with angular velocity
u> radians per second, for any crank displacement 0,

F = w 2 r cos 0,

and

, L.A.<o 2 r

tla = . COS

The pressure in the cylinder is

TT 7 L AwV cos

When is zero, cos is unity, and when is 90 degrees, cos
is zero. For values of between 90 and 180 degrees, cos# is
negative.

The variation of the pressure in the cylinder is seen in
Fig. 306, which has been drawn for the following data.

A
Ef

ALPr.

Fig. 306.

Diameter of suction pipe 3J inches, length 12 feet 6 inches.
Diameter of plunger 4 inches, length of stroke 7| inches.

Number of strokes per minute 136. Height of the centre of
the pump above the water in the sump, 8 feet. The plunger is
assumed to have simple harmonic motion.

The plunger, since its motion is simple harmonic, may be
supposed to be driven by a crank 3f inches long, making 68 revo-
lutions per minute, and a very long connecting rod.

The angular velocity of the crank is

27T.68 h -. 1 -,. ,

w = =71 radians per second.

The acceleration at the ends of the stroke is

E2 M I 7*12 v /VQ1O
= o> . r = / 1 x (j 6\.i

- 15*7 feet per sec. per sec.,

and

12-5. 15-7. 1-63
32

10 feet.

448 HYDRAULICS

The pressure in the cylinder neglecting the water in the
cylinder at the beginning of the stroke is, therefore,

34 -(10 + 8) =16 feet,

and at the end it is 34-8+ 10-36 feet. That is, it is greater
than the atmospheric pressure.

When is 90 degrees, cos is zero, and h a is therefore zero,
and when is greater than 90 degrees, cos is negative.

The area AEDF is clearly equal to GADH, and the work done
per suction stroke is, therefore, not altered by the accelerating
forces; but the rate at which the plunger is working at various
points in the stroke is affected by them, and the force required to
move the plunger may be very much increased.

In the above example, for instance, the force necessary to
move the piston at the commencement of the stroke has been
more than doubled by the accelerating force, and instead of
remaining constant and equal to '43. 8. A during the stroke, it
varies from

P = -43 (8 + 10) A
to P = '43 (8 -10) A.

Air vessels. In quick running pumps, or when the length
of the pipe is long, the effects of these accelerating forces tend to
become serious, not only in causing a very large increase in the
stresses in the parts of the pump, but as will be shown later, under
certain circumstances they may cause separation of the water in
the pipe, and violent hammer actions may be set up. To reduce
the effects of the accelerating forces, air vessels are put on the
suction and delivery pipes, Figs. 310 and 311.

250. Accelerating forces in the delivery pipe of a plunger
pump when there is no air vessel.

When the plunger commences its return stroke it has not only
to lift the water against the head in the delivery pipe, but, if no
Y/air vessel is provided, it has also to accelerate the water in the
cylinder and the delivery pipe. Let D be the diameter, a x the area,
and Li the length of the pipe. Neglecting the water in the
cylinder, the acceleration head when the acceleration of the piston
is F, is

, L!.A.F

ha == - *

and neglecting head lost by friction etc., and the water in the
cylinder, the head resisting motion is

U d *

If F is negative, h a is also negative.

RECIPROCATING PUMPS

449

When the plunger moves with simple harmonic motion the
diagram is as shown in Fig. 307, which is drawn for the same
data as for Fig. 306, taking Z as 20 feet, LI as 30 feet, and the
diameter D as 3J inches.

Fig. 307.

The total work done on the water in the cylinder is NJKM,
which is clearly equal to HJKL. If the atmospheric pressure is
acting on the outer end of the plunger, as in Fig. 301, the nett
work done on the plunger will be SNRMT, which equals HSTL.

251. Variation of pressure in the cylinder due to friction
when there is no air vessel.

Head lost by friction in the suction and delivery pipes. If v is
the velocity of the plunger at any instant during the suction
stroke, d the diameter, and a the area of the suction pipe, the
velocity of the water in the pipe, when there is no air vessel, is

, and the head lost by friction at that velocity is
a

2gda* '

Similarly, if Oi, D, and L x are the area, diameter and length
respectively of the delivery pipe, the head lost by friction, when
the plunger is making the delivery stroke and has a velocity v, is

When the plunger moves with simple harmonic motion,
v = wr sin 0,

and

L. H.

450

HYDRAULICS

If the pump makes n strokes per second, or the number of

revolutions of the crank is ~ per second, and I* is the length of

the stroke,

iD = 7rn,

and I, = 2r.

Substituting for <*> and r,

Plotting values of h f at various points along the stroke, the
parabolic curve EMF, Fig. 308, is obtained.

When is 90 degrees, sin# is unity, and h f is a maximum.
The mean ordinate of the parabola, which is the mean frictional
head, is then

2 /AVVLZ,'
'

3 2gda?
M~~"~

Fig. 308.

and since the mean frictional head is equal to the energy lost per
pound of water, the work done per stroke by friction is

all dimensions being in feet.

Fig. 309.
Let Do be the diameter of the plunger in feet. Then

and

RECIPROCATING PUMPS 451

Therefore, work done by friction per suction stroke, when
there is no air vessel on the suction pipe, is

The pressure in the cylinder for any position of the plunger
during the suction stroke is now, Fig. 309,

ho = H h h a h/.

At the ends of the stroke h/ is zero, and for simple harmonic
motion h a is zero at the middle of the stroke.

The work done per suction stroke is equal to the area
AEMFD, which equals

ARSD + EMF = 62-4W +
Similarly, during the delivery stroke the work done is

The friction diagram is HKGr, Fig. 309, and the resultant
diagram of total work done during the two strokes is EMFGrKH.

252. Air vessel on the suction pipe.

As remarked above, in quick running pumps, or when the
lengths of the pipes are long, the effects of the accelerating forces
become serious, and air vessels are put on the suction and delivery
pipes, as shown in Figs. 310 and 311. By this means the velocity ^
in the part of the suction pipe between the well and the air
vessel is practically kept constant, the water, which has its
velocity continually changing as the velocity of the piston \
changes, being practically confined to the water in the pipe
between the air vessel and the cylinder. The head required to
accelerate the water at any instant is consequently diminished,
and the friction head also remains nearly constant.

Let Z t be the length of the pipe between the air vessel and
the cylinder, I the length from the well to the air vessel, a the
cross-sectional area of each of the pipes and d the diameter of the
pipe.

Let h v be the pressure head in the air vessel and let the air
vessel be of such a size that the variation of the pressure may for
simplicity be assumed negligible.

Suppose now that water flows from the well up the pipe AB
continuously and at a uniform velocity. The pump being single
acting, while the crank makes one revolution, the quantity of
water which flows along AB must be equal to the volume the
plunger displaces per stroke.

292

452

HYDRAULICS

The time for the crank to make one revolution is

27T

t = sees.,

therefore, the mean velocity of flow is

_ A 2rw_ Awr

(For a double acting pump v m = . )

\ a TT /

During the delivery stroke, all the water is entering the air
vessel, the water in the pipe BC being at rest.

Fig. 310.

Then by Bernoulli's theorem, including friction and the velocity
head, other losses being neglected, the atmospheric head

H = x lh , A 2 coV t 4/AWZ a)

The third and fourth quantities of the right-hand part of the
equation will generally be very small and h v is practically equal
to H-OJ.

When the suction stroke is taking place, the water in the pipe
BC has to be accelerated.

Let H B be the pressure head at the point B, when the velocity
of the plunger is v feet per second, and the acceleration F feet per
second per second.

RECIPROCATING PUMPS 453

Let hf be the loss of head by friction in AB, and h/ the loss in
BC. The velocity of flow along BC is , and the velocity of
flow from the air vessel is, therefore,

v . A Awr

__ ^

a no,

Then considering the pipe AB,

H , AW

HB ~ ~ h ~2^~ hf >

and from consideration of the pressures above B,

/vA. A "~^ 2

7TO, J

Neglecting losses at the valve, the pressure in the cylinder is
then approximately

, AW
fl TV 5 s

Neglecting the small quantity

For a plunger moving with simple harmonic motion

By putting the air vessel near to the cylinder, thus making
Z t small, the acceleration head becomes very small and

h Q = TL h-hf nearly,
and for simple harmonic motion

j, -H 7, 4 > VA ' *
^- n - h --^NT^-

The mean velocity in the suction pipe can very readily be
determined as follows.

Let Q be the quantity of water lifted per second in cubic feet.
Then since the velocity along the suction pipe is practically

constant v m = and the friction head is

454

HYDRAULICS

Wlien the pump is single acting and there are n strokes per
second.

and therefore,

and

A . l s . n

If the pump is double acting,

h = /AWZ

ga?d

For the same length of suction pipe the mean friction head,
when there is no air vessel and the pump is single acting, is -7r 2
times the friction head when there is an air vessel.

253. Air vessel on the delivery pipe.

An air vessel on the delivery pipe serves the same purpose
as on the suction pipe, in diminishing the mass of water which
changes its velocity as the piston velocity changes.

Fig. 311.

*As the delivery pipe is generally much longer than the suction
pipe, the changes in pressure due to acceleration may be much
greater, and it accordingly becomes increasingly desirable to
provide an air vessel.

Assume the air vessel so large that the pressure head. in it
remains practically constant.

RECIPROCATING PUMPS 455

Let Z 2 , Fig. 311, be the length of the pipe between the pump and
the air vessel, Id be the length of the whole pipe, and i and D the
area and diameter respectively of the pipe.

Let hz be the height of the surface of the water in the air vessel
above the centre of the pipe at B, and let H be the pressure head
in the air vessel. On the assumption that Ht, remains constant,
the velocity in the part BC of the pipe is practically constant.

Let Q be the quantity of water delivered per second.

The mean velocity in the part BC of the delivery pipe will be

The friction head in this part of the pipe is constant and equal to

Considering then the part BC of the delivery pipe, the total
head at B required to force the water along the pipe will be

But the head at B must be equal to H w + 7i 2 nearly, therefore,

-W + H ............... (1).

In the part AB of the pipe the velocity of the water will vary
with the velocity of the plunger.

Let v and F be the velocity and acceleration of the plunger
respectively.

Neglecting the water in the cylinder, the head H r resisting the
motion of the plunger will be the head at B, plus the head
necessary to overcome friction in AB, and to accelerate the water
in AB.

rm, P TT TT L 4/^A 2 F.A.I,

Therefore, H r

For the same total length of the delivery pipe the acceleration
head is clearly much smaller than when there is no air vessel.
Substituting for H v + 7^ from (1),

If the pump is single acting and the plunger moves with simple
harmonic motion and makes n strokes per second,

and

456 HYDRAULICS

Therefore,

0,1 g

Neglecting the friction head in Z 2 and assuming 1 2 small com-
pared with Idj

4/rWA 2 Z d AZ 2 , /j

H r = Z + H + -r. + - w r cos ft

254. Separation during the suction stroke.

In reciprocating pumps it is of considerable importance that
during the stroke no discontinuity of flow shall take place, or
in other words, no part of the water in the pipe shall separate
from the remainder, or from the water in the cylinder of the pump.
Such separation causes excessive shocks in the working parts of
the pump and tends to broken joints and pipes, due to the hammer
action caused by the sudden change of momentum of a large mass
of moving water overtaking the part from which it has become
separated.

Consider a section AB of the pipe, Fig. 301, near to the inlet
valve. For simplicity, neglect the acceleration of the water in the
cylinder or suppose it to move with the plunger, and let the
acceleration of the plunger be F feet per second per second.

If now the water in the pipe is not to be separated from that in
the cylinder, the acceleration / of the water in the pipe must not

be less than - feet per second per second, or separation will not

FA
take place as long as - / a .

FA
If f a at any instant becomes equal to - , and f a is not to be-

come less than , the diminution 8/of / a , when F is diminished

by a small amount 9F, must not be less than dF, or in general

the differential of f a must not be less than times the differential

of F.

The general condition for no separation is, therefore,

fdF<3/ .............................. (1).

Perhaps a simpler way to look at the question is as follows.

Let it be supposed that for given data the curve of pressures
in the cylinder during the suction stroke has been drawn as in
Fig. 309. In this figure the pressure in the cylinder always remains
positive, but suppose some part of the curve of pressures EF to

RECIPROCATING PUMPS

457

come below the zero line BC as in Fig. 312*, The pressure in the
cylinder then becomes negative; but it is impossible for a fluid
to be in tension and therefore discontinuity in the flow must
occur t.

In actual pumps the discontinuity will occur, if the curve EFGr
falls below the pressure at which the dissolved gases are liberated,
or the pressure head becomes less than from 4 to 10 feet.

Fig. 312.

At the dead centre the pressure in the cylinder just becomes
zero when h + h a = H, and will become negative when h + h a > H.
Theoretically, therefore, for no separation at the dead centre,

- or

If separation takes place when the pressure head is less than
some head /&,,, for no separation,

li a 2i H h m h,

and

Neglecting the water in the cylinder, at any other point in the
stroke, the pressure is negative when

FAL + /i + ^A 2 >H
a g f 2g a?
And the condition for no separation, therefore, is

rm,' . T JD^V.U 7 V

That is, when h + + h f + ~-

(2).