r
REESE LIBRARY.
OK THK,
UNIVERSITY OF CALIFORNIA
loot
No,
THE
GRAPHICAL SOLUTION
OF
HYDRAULIC PROBLEMS
TREATING OP THE FLOW OF WATER
THROUGH PIPES, IN CHANNELS
AND SEWERS, OVER WEIRS,
ETC.
BY
FREEMAN C. COFFIN,
Member of the American Society of Civil Engineers.
FIRS T EDITION.
FIRST THOUSAND.
NEW YORK:
JOHN WILEY & SONS.
LONDON : CHAPMAN & HALL, LIMITED.
1897.
Copyright, 1897,
BY
FREEMAN C COFFIN.
ROBERT DRUMMOND, ELHCTROTYPHR AND PRINTER, NEW YORK.
CONTENTS.
CHAPTER PACK
I. INTRODUCTION i
II. UNITS AND SYMBOLS 4
III. FLOW OF WATER IN PIPES ... 6
Formulas 6
Small Pipes, ^-3 Inch 8
Explanation of Diagrams 8
Examples 9
Large Pipes, 4-60 Inch 10
Explanations and Examples 10
Problems Outside of Range of Diagrams n
Compound Mains 14
Complex Mains 16
Old or Tuberculated Pipe-lines 24
IV. FLOW OF WATER THROUGH SHORT TUBES 31
Formula and Coefficients 31
Examples 32
V. RECTANGULAR WEIRS 36
Formulas 36
Explanation of Diagrams 37
Examples 38
Wide Crests 40
Velocity of Approach 41
VI. FLOW OF WATER IN CHANNELS 44
Kutter's Formula 44
Values of n , 45
Explanations and Examples 47
iii
IV CONTENTS.
CHAPTER PAGE
VII. PIPE-SEWERS 50
Formula and Coefficient 50
Explanation of Diagram 50
Examples 50
VIII. FIRE-STREAMS AND DISCHARGE OF HOSE-NOZZLES. .. 52
Height of Streams and Head Required 52
Discharge of Hose-nozzles with Head or Pres-
sure Indicated at the Hydrant 53
Discharge of Hose-nozzles with Head or Pres-
sure Indicated at Base of Play-pipe 53
IX. MISCELLANEOUS PROBLEMS 55
Horse-power of Falling Water, also Horse-power
Required to Raise Water 55
Capacity and Size of Pumps. . ; 56
Coal Required for Pumping Water 58
TABLES.
Table No. i. Equivalents of U. S. Gallons 60
2. Discharge in Gallons per Minute for Different
Velocities in Circular Pipes 62
" 3. n/6 and 6/u Powers of Numbers 65
4. Coefficients of Friction in Old Pipes 66
5- " Discharge in Old Pipes 66
6. " Velocity of Approach for Weirs 67
7. Areas and Values of r in Circular Channels. 68
8. " " " " r in Rectangular Chan-
nels 69
9. With Side Slopes of i to i 70
" 10. " " " " 2 to i 71
" " IT. " " " " 3 to i 72
" 12. Equivalents of Pounds Pressure and Feet
Head 73
" " 13. Height and Discharge of Fire-streams 74
CONTENTS.
DIAGRAMS.
Diagram No. I. Flow of Water through Pipes -3 Inches
in Diameter.
*' " 2-17. Flow of Water through Pipes 4-60 Inches
in Diameter.
" " 18, 19. Flow through Short Tubes or Entry
Head.
" " 20. Value of c in V c(rs)\.
21. Weirs with Wide Crests.
" " 22-24. " " Thin " .
*' " 25,26. Flow of Water in Channels.
" 27. " in Pipe-sewers.
14 28. Horse-power of Water.
" " 29. Size and Capacity of Pumps.
" *' 30. Coal Required in Pumping.
" " 3i 32. Discharge of Hose-nozzles.
M " A and B. Experiments on Old Pipes.
GRAPHICAL SOLUTION
.OF
HYDRAULIC PROBLEMS,
CHAPTER I.
INTRODUCTION.
IT is not the purpose of this book to discuss the laws
governing the flow of water, the hydraulic experiments
that have been made, or the formulas derived from them.
The object of the book, as conceived by the author, is
to provide a convenient instrument or tool for the prac-
tising engineer (who is already familiar with hydraulic
laws and formulas) with which he can solve quickly
and correctly the commonly occurring hydraulic prob-
lems by means of diagrams and with a minimum of cal-
culation, either mental or written.
The diagrams are constructed upon well-known for-
mulas, using coefficients that are generally accepted as
safe for the conditions to which they are intended to
apply. While it was intended to have the computations
correctly made and carefully checked, no attempt has
been made to secure mathematical refinements, such
being in most cases entirely out of place in hydraulic
2 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
problems. The computations for these diagrams were
made with the slide-rule and logarithmic cross-section
paper.* Such computations are correct to two figures,
and sometimes three or four, and for these problems
this is a greater degree of accuracy than can usually be
secured in all of the data.
In regard to the possibility of obtaining sufficiently
accurate results from the diagrams themselves, this can
be said : That in all problems relating to the flow of
water through short tubes, long pipes, sewers, and chan-
nels, the result can be read from the diagram with a
greater degree of precision than the engineer is warranted
in relying upon in practice. The variations in the com-
mercial sizes of pipes will probably cause greater errors
than those that result from inability to read the diagram
closely. Differences that cannot be estimated in the
surfaces of pipes, sewers, or channels will produce differ-
ences in result beside which any errors from the above
sources will be insignificant. Details of actual construc-
tion, small variations from the design, differences in the
character of surfaces and many other elements, tend to
render a refinement in figures absurd.
Of the subjects relating to the flow of water treated
in this book, it is possible that in weirs alone a few cases
will arise where the diagrams will not give results suffi-
ciently close ; but such cases are only those in which the
weir has been constructed in perfect conformity with
the conditions from which the formulas were derived.f
* The paper used was that designed by Mr. John R. Freeman.
M. Am. Soc. C. E., with four complete squares of lo-inch base on
one sheet.
f Weirs are excepted from the general statement because the
conditions of their construction are more under the control of the
engineer than those of the other classes.
INTRODUCTION. 3
There are, however, probably few weirs used for ordinary
purposes that are so constructed in every detail. If the
diagrams are based upon proper formulas, the computa-
tions correctly made, and the drawing carefully done,
results can be read from them with sufficient precision
for all practical purposes.
The diagrams are offered to the profession in the hope
that they may prove useful, as similar ones have been a
valuable aid to the author for several years. Any of the
usually occurring hydraulic problems can be solved by
them, either in field or office, with no other tables, for-
mulas, or information than those contained in this book,
and in nearly every case without the use of pencil or
paper.
CHAPTER II.
UNITS AND SYMBOLS.
IN choosing the units to be used, preference is given
to those in most general use.
Measures of quantity of water are always given in
U. S. gallons per minute. Whenever possible, a second
scale gives cubic feet per second.
Table No. i, page 60, gives equivalents of U. S. gal-
lons per minute.
All linear dimensions are given in feet and decimals,
except in the case of diameters of pipes, which are given
in inches.
SYMBOLS.
The following characters will always have the meaning
here assigned, unless otherwise stated :
a = area in square feet ;
_ j coefficient in weir formula,
( coefficient in Chezy formula ;
D = diameter in feet ;
D' = depth of water in channels in feet ;
d = diameter in inches ;
g = acceleration of gravity ;
h = head in feed for pipes and weirs ;
/ = length in feet of pipes and weirs ;
UNITS AND SYMBOLS. 5
n Kutter's coefficient for character of surface of
pipes and channels ;
p = wetted perimeter in feet ;
Q = cubic feet of water per second ;
q = U. S. gallons of water per minute ;
R = radius = ;
r = hydraulic mean radius of channels and pipes = ;
for circular pipes running full r = ;
s = hydraulic slope for pipes and channels = ;
v = mean velocity in feet per second ;
for circular pipes running full v =
W '= width of channels.
Effective head and loss of head by friction in long
pipes is given on diagrams as " friction-head in feet."
CHAPTER III.
FLOW OF WATER IN PIPES.
PROBLEMS connected with the flow of water in pipes
are of constant occurrence in the practice of the hy-
draulic engineer.
They may be divided into three general classes :
(I) The size (Z>), length (/), and head (ti) being given
to find the quantity (Q) or discharge.
(II) The size, length, and quantity being given to find
the required head or loss by friction.
(III) The quantity, head, and length of pipe line
being given to find the size of pipe required.
There are many formulas, based upon numerous ex-
periments, in use for the solution of these problems.
Without discussing the respective merits of these, it may
be said that the Chezy formula is used as the basis of
computation for the diagrams relating to this subject.
This formula was selected on account of its simplicity*
and because it seemed possible with a varying coefficient
to meet the conditions of different diameters of the pipes
and of varying velocities more nearly than by any other
form. This formula is expressed as follows :
I. v = frs * ;
0$ WATER IN PIPES. 1
With these three forms of the equation, the value of
either factor of the problem can be found when the other
two are given.
Note. When r is known, the diameter of a circular
pipe running full can be obtained by multiplying r. by 4.
The values of the coefficient c used in these diagrams
are those given by Hamilton Smith, Jr., in his work on
Hydraulics. These values seem to have been derived
with great care from the most reliable hydraulic experi-
ments. The reader is referred to the above-named book
for further information on this subject.* These values
of c are given graphically on diagram No. 30. The full
curves are copied from Smith's Hydraulics ; the dotted
curves are interpolated by the author.
Although the Chezy formula is very simple, there is
this practical difficulty in using it in the calculation of
individual problems, in the second form in which it is
given, i.e., v = c(rs)* : that c cannot be taken correctly
until v is known, and an approximation must first be
made by taking a value of c according to some assumed
or estimated value of v. One or two approximations
will generally bring the result. This objection to the
formula does not hold, however, in constructing dia-
grams, as the first form, or s = , or, reduced to the
simpler expression, s = - f-J is the one best adapted for
such computations.
ACCURACY OF FORMULAS FOR THE FLOW OF
WATER IN PIPES.
It should be borne in mind that, while it is desirable
that formulas should give with accuracy values for cer-
* " Hydraulics," by Hamilton Smith, Jr. (John Wiley & Sons).
S GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
tain given conditions of pipes, exactly these conditions
rarely exist in practice. Therefore too much weight
must not be given to mathematical refinements in ordi-
nary practice, however desirable these refinements are in
experimentation. When, for instance, the value of the
coefficient c is in the vicinity of 100, a variation in the
coefficient in several formulas of 4 or 5 should not be
considered as invalidating either for practical purposes,
as perhaps the difference in the commercial sizes of pipes
of the same nominal diameter will cause a greater vari-
ation. Differences in the interior surfaces of the pipes,
that with our present knowledge are impossible of exact
estimation, will produce a greater variation than that
caused by such differences of the coefficient. Therefore,
while it is desirable to use the best formula for clean,
straight pipes as a basis, a sufficient margin should be
allowed for these differences of condition. The subject
of the effect of the age of pipe lines is discussed else-
where in this chapter.
EXPLANATION OF DIAGRAMS.
SMALL PIPES.
The diagrams for the flow of water in small pipes
(those from three eighths of an inch to three inches in
diameter) are constructed for a length of 100 feet. Re-
sults for other lengths can be taken proportionately.
These sizes are plotted upon one sheet, diagram No. i.
The diameter of the pipes is given by curves originat-
ing at the upper left-hand corner, the discharge in gal-
lons per minute by a vertical scale on the right of sheet,
the velocity by dotted curves, and the loss of head from
friction by a horizontal scale at the top of the sheet.
FLOW OF WATER IN PIPES, 9
Examples of Use.
CASE I. Size, length, and discharge given to find head.
Example: What is the loss of head in a i-inch pipe
75 ft. long, delivering 16 gallons per minute ?
On diagram No. i, at the intersection of the line repre-
senting 16 gallons of water with the curve representing
i-inch pipe, find the friction in feet, viz , 22 ft.; this is
in a length of 100 ft.; therefore 22 X = 16.5 ft., the
required answer.
CASE II. Size, length, and head given to find discharge.
Example: What is the capacity of a 2-inch pipe 350 ft.
long with effective head of 40 feet ?
To use this diagram the length must be reduced to 100
ft., and the head in like ratio ; thus 40 X =11.4 per
35
100 feet in length.
At intersection of 11.4 friction-head and 2-inch curve
find capacity in gallons = 70 gallons per minute, flowing
with velocity of 7 ft. per second.
CASE III. Length, head, and discharge given to find
size.
Example: What is the size of pipe necessary to dis-
charge 175 gallons per minute through 500 ft. in length
with a head of 80 ft. ?
80 X = 16 ft. per 100 ft. in length. The intersec-
500
tion of 175 gallons and 16 feet head is between curves of
z\ and 3 inches diameter, and of the usual sizes, the latter
must be used.
10 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
LARGE PIPES.
For pipes larger than 3 inches in diameter there is a
diagram for each size. The size is given in the upper
right-hand corner of diagram-sheet. On these diagrams
the length of pipe is given by a vertical scale on the right-
hand side ; the friction-head by a horizontal scale on top
and bottom of sheet ; the quantity in gallons per minute
by oblique lines radiating from a common origin at the
upper left-hand corner. Cubic feet per second are not
given on these diagrams, but may be taken from the
Table No. i of equivalents of U. S. gallons per minute
on page 60. The velocity in feet per second is given by
a heavy broken line parallel with the oblique line of
quantity. These diagrams are more convenient than
those for small pipes, because the solution of all problems
within a wide range can be read from them directly for
the proper length without multiplication or division.
Examples of Use.
CASE I. Size, length, and effective head given to find
discharge.
Example: What is the discharge of a 1 4-inch conduit-
line 7500 ft. long with an effective head of 30 ft.?
Ans. 1860 gallons per minute; velocity 3.9 ft. per sec-
ond. (From diagram No. 7 for 14-inch pipe.)
CASE II. Size and length of pipe and discharge given
to find loss of head or friction.
Example: What is the loss of head in 5500 ft. of 12-
inch pipe delivering uoo gallons per minute?
The answer will be found at intersection of lines repre-
senting 1 100 gallons and 5500 ft. of pipe on diagram No.
6 for i2-inch pipe. It is 18,2 ft. head. The velocity is
3.1 ft. per second.
FLOW OF WATER IN PIPES. Il
CASE III. Length, head, and required discharge given
to find size of pipe.
Example: What is the size of pipe necessary to dis-
charge 2500 gallons per minute through 7500 ft. of pipe
with 30 ft. head ?
The diagrams show that a 1 4-inch pipe will discharge
1860 gallons, and a i6-inch pipe will discharge 2675 gal-
lons; and of the usual commercial sizes, the latter is the
one that must be used.
These diagrams cover in their range nearly all of the
problems met in practice. If the solution of problems
that do not come within their limits is desired, it can be
found indirectly as follows.
Problems Outside of Limits of Diagrams.
When the length of line is greater than that given on the
diagram.
CASE IV. Size, length, and discharge given to find the
friction-head.
Divide the given length by some number that will give
a quotient within limits of the diagram; find head for
this length, and multiply by the number used to divide
length.
Example: What is the friction-head in 12,000 ft. of 12-
inch pumping main, with a delivery of 1000 gallons per
minute, or velocity = 2.85 ?
12,000 ~- 2 = 6000; friction in 6000 ft. = 16.7 ft.;
16.7 X 2 33.4 feet head.
CASE V. Size, length, and head given to find discharge.
Divide both length and head by the same number,
and find discharge for the quotient, which will be the
answer.
12 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
Example: What is the discharge through 15,000 ft. of
lo-inch pipe with head of 50 ft.?
15,000 -j- 2 = 7500 ft.; 50 -h 2 = 25 ft. head; discharge
from diagram for 25 ft. head through 7500 ft. of pipe
= 680 gallons; v 2.8.
When the length given is less than that for which it is
convenient to read results on the diagram.
CASE VI. Size, length, and discharge given to find
head.
Multiply length by any convenient number, and divide
the resultant head by same number.
Example: What is the loss of head in 36-inch penstock
400 ft. long, discharging 25 cu. ft. per second or 11,200
gallons per minute ?
400 X ioo = 40,000 ft.; 11,200 gallons through 40,000
ft. of 36-inch pipe = 39.2 ft. head; 39.2 -5- ioo = 0.392
head.
CASE VII. Size, length, and head given to find dis-
charge.
Multiply both head and length by any convenient
number, and find discharge corresponding to the prod-
ucts.
Example: What is the discharge of a 3o-inch pipe 200
ft. long with an effective head of 0.4 ft.?
200 X ioo = 20,000 ft. long; 0.4 X ioo = 40 ft. head;
the discharge of 3o-inch pipe through 20,000 ft. with 40
ft. head = 10,000 gallons; vel. = 4.5.
Note. The entry and velocity heads in the last two
examples, an important element, should be considered in
all cases of comparatively short lines. This subject is
treated in the next chapter.
OF WATER IN PIPES.
Where the discharge or head is greater than that given
on diagram.
CASE VIII. Size, length, and discharge given to find
head.
Divide the given discharge by any number that will
bring it within the limits of the diagram. Multiply the
friction due to the quotient by the n/6 power of the
number used.*
A table of n/6 and 6/n powers is given on page
65-
Example: What is the loss of head in 500 ft. of 6-inch
pipe discharging 1500 gallons per minute?
1500 -r- 2 = 750; 750 gallons through 500 ft. of 6-
inch pipe = 24 ft. head; 24 X (2 11 / 6 = 3.56) = 85.5 ft.,
answer.
CASE IX. Size, length, and head given to find dis-
charge.
Divide the given head by the u/6 power* of any
number that will bring head within limits of diagram.
On given length find discharge due to this quotient;
multiply this discharge by the number taken.
* Although the exponent of the formula v = c(rs$ is of the
second power, the variation in the coefficient c renders the results
practically identical with those of a formula of the n/6th power in
which c is constant for all velocities.
The reader is referred to a paper in the Journal of Associated
Engineering Societies for June, 1894, by Mr. W. E. Foss, in which
he shows that a formula of the n/6th power is in practical accord-
ance with the results of experiments. He proposes such a formula
for the flow of water in pipes and channels which is very simple,
and, with the tables given in the paper, renders the computation of
such problems much less tedious than by most formulas in use.
See page 22 for remarks on the use of the second power instead
of the n/6th.
14 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
Example: What is the discharge through 2000 ft. of
8-inch pipe with effective head of 125 ft.?
125 -T- (2 II/6 = 3.56) = 35.1; discharge due to 35.1 ft.
head = 925 gallons X 2 = 1850 gallons per minute.
Where the discharge or the head is less than can be
easily read from the diagram.
iL X. Size, length, and discharge given to find head.
Multiply the given discharge by a number that will
give a product large enough to be easily read from the
diagram; divide the head obtained by the n/6 power
of that number.
Example: What is the head in a i6-inch pipe 5000 ft.
long, delivering 100 gallons per minute ?
100 X 10 = 1000; head for 1000 gallons = 3.4ft; 3.4
-f. jo 11 / =0.05 ft.
CASE XI. Size, length, and head given to find dis-
charge.
Multiply the given head by the n/6 power of some
number; divide the discharge due to the product by the
number.
Example: What is the discharge of a i6-inch pipe
6500 ft. long with a head of .25 ft.?
0.25 X io II/6 = 17; discharge due to head of 17 ft.
= 2120; 2120 -5- 10 = 212 gallons per minute.
COMPOUND MAINS.
A compound main is a single line of pipe which is not
of uniform diameter for its entire length. It is com-
posed of two or more sizes in one line.
The friction in such a main can easily be computed
when the discharge is given.
FLOW OF WATER IN PIPES. lj
To find the discharge when the total head is given is
more difficult.
The finding of both the friction and discharge is much
facilitated by the use of these diagrams.
CASE I. Size, length, and discharge given to find head.
Find head for each size separately, and add results for
total head.
Example: What is the head in a compound main com-
posed of 3000 ft. of i2-inch pipe and 7000 ft. of 16-
inch, delivering 1500 gallons per minute?
Head in 3000 ft. of 1 2-inch pipe = 17.4 ft.
" " 7000 " " 1 6- " " = 9.6 "
Total head, 27.0 " Ans.
CASE II. Size, length, and head given to find discharge.
Assume some quantity for the discharge; find the head
due to such discharge in the given line, % the same as in
Case I. Divide the correct head by this head, and
multiply the assumed discharge by the 6/n power of the
quotient.
Example i: What is the discharge of a compound
main of 4500 ft. of lo-inch pipe and 7000 ft. of i6-inch
pipe with a total effective head of 40 ft.?
Assume a discharge of 1000 gallons.
Head in 4500 ft. of lo-inch pipe = 30.2 ft.
" " 7000 " " 16 " " = 4.6 "
Total, 34.8 '
\6/n
-) = 1000
correct discharge.
^O N6/II
Then 1000 X ( ~ 1 = 1000 X 1.075 = IO 75 gallons,
l6 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
Check on above calculation :
1075 gals, in 4500 ft. of lo-inch pipe = 34.7 ft. head.
" " 7000 " " 16- " " = 5.3 " "
Total, 40.0 " "
Example 2: What is the discharge of a compound
main of 10,000 ft. of i6-inch, 6000 ft. of 1 2-inch, and
2000 ft. of lo-inch pipe with total head of 50 ft.?
Assume discharge of 1500 gallons per minute.
Head in 10,000 ft. of i6-inch = 13.8
" " 6,000 " " 12 " =34-9 [VIII, page 13.
" " 2,000" " 10 " = 28. i, by method of Case
Total head for assumed discharge, 76. 8
/ c \ 6 /"
Then 1500 X Hrs) = Il8 4 gallons, correct dis-
charge.
*
COMPLEX MAINS.
The problems involved in the flow of water in complex
mains, or a system of mains in which the water flows to
the point of discharge through two or more lines of pipe
of different sizes and lengths, are of great importance
in practice and should receive careful attention. Owing
to the immense amount of work required to solve them
by the ordinary methods, they are undoubtedly left un-
solved in many cases.
These problems arise in nearly all cases where an
efficient fire system is to be designed for a town or city,
or where the efficiency of one already constructed is to
be ascertained. A valuable and suggestive paper on
this subject was read by Mr. John R. Freeman, M. Am.
Soc. C. E., before the New England Water-works Asso-
FLOW OF WATER IN PIPES. l^
ciation, and published in the Journal of that Association,
Vol. VII.
The diagrams for the flow of water in long pipes pro-
vide a ready means of solving such problems, and the
following examples are given of their use.
CASE I. System of piping and discharge given to find
head.
A
2500'12" B SOOO'lO" C_
Fig- 1
2000 '8" *
Example : With reservoir at A and arrangement of
piping as shown in Fig. i, what will be the loss of head
with a draft of 1000 gallons per minute at C?
First, find head in single line AB = 7 ft.
The flow will divide at B, part of it going through BC
and the remainder through BEDC.
It is self-evident that the total friction or loss of head
must be the same in one line as in the other, or the flow
from the two lines would meet at C under different pres-
sures, which is impossible.
Assume a loss of head for these lines, and on the dia-
grams find the discharge due to this assumed head.
Thus assume 20 ft. head; then
discharge in BC, 2000 ft. of 10 in. = 1240 gals.
" " BEDC, 4000 " " 8 " = 470 "
Total discharge for both lines, 1710 "
Divide the given discharge by the discharge of the as-
sumed head, and multiply the n/6 power of the quotient
by the assumed head, and the product will be the true
head.
I& GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
X 20 = 0.37 X 20 = 7.4, true head.
/loooN 11 / 6
\i7ioj
Add head in AB = 7
Total loss of head, 14.4 for 1000 gals.
(a) If, instead of assuming 20 ft. head, 8 ft. head had
been assumed, then
discharge in BC = 755 gals.
Total assumed discharge, 1035 "
1 / 6
x8 =7 ' 50feet -
Add head in AB = 7
Total loss of head, 14.50 "
Note. The slight variation in the results of the two
examples are due to slight errors in reading diagrams.
(b) After the correct head for one rate of discharge is
obtained, it is very easy to find it for any other by the
following formula :
/required discharge y/6 ( hea< ? f r , re '
r. r - 5 ~ X known head = < quired dis-
V known discharge / | charge.
Thus when draft of 1000 gallons through a given sys-
tem of pipes causes a loss of head of 14.50 ft. as above,
othe~ drafts would be as follows :
/5oo V I/6
- X 14.50= 4.06 ft.
iooo/
/I 200V 1 / 6
X 14.50 = 20.3 "
\iooo/
\ II/6
X 14.50 = 30.6 "
^
Viooo/
FLOW OF WATER IN PIPES. 10
CASE II. System of piping and head given to find dis-
charge.
Example: With same arrangement of piping as in Case
I, what will be the discharge with total head of 30 ft.?
First assume the head in the lines BC and BEDC ;
thus assume 15 ft.
BC = 2000 ft. of 10 in., discharge = 1060 gals.
BEDC = 4000 " " 8 ", " = 400 "
Total discharge with 15 ft. head, 1460 "
The loss of head in line AB, or 2500 ft. of 12 in.,
due to a discharge of 1460 gallons = 13.9 ft.
Add head in BC and BEDC =15 "
Total assumed head, 28.9 "
To find the discharge due to the given head use fol-
lowing formula:
l,
given head \ 6/ " j j- i (true dis-
-r-r 3 X assumed discharge = j .
ssumed head/ ( charge.
\ 6/I1
X 1460 =: 1.02 X 1460 = 1488 gallons,
'
^28. 9'
the true discharge for 30 ft. head.
The most complex problems can be solved by the
application of the foregoing principles.
Where there are a number of mains, with cross-mains
between, it will of course require the exercise of thought
and judgment to decide which lines are effective and how
they should be classified. The experienced engineer can
usually make a very close approximation in such cases
without going into the detail of every line, and thus avoid
much tedious work.
2O GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
The following case is given as an illustration of a pos-
sible treatment of a rather intricate problem.
__.500Q-10 : '
^.
A
Fig. 2
2000 '12" >/
f
4000'8"
E
1
^^
w 3000'S"
y
F
p^- -
X
z D
With the above arrangement of piping and reservoirs
at A, what will be the loss of head caused by a draft of
ten i-inch* hydrant streams or 2000 gallons per minute
from hydrants on line CD ?
As the hydrants will be fed both ways through short
lines, the friction in line CD can safely be ignored,
although if the location of the hydrants were fixed it
could be computed.
The cross-lines Bo, woe, yz can also be ignored. The
slight effect that they would have would be on the safe
side.
Assume the loss of head in lines BC, BE, and BmF\
it will be the same in the three.f One of these lines,
BmF, is a compound main, and the discharge will be
assumed here to start with. Assume 200 gallons; then
in line Bm, 2000' of 6", head 8.6 feet
" " mF, 3000' " 8", " = 3.2 "
Total, 1 1.8 "
* See Table No. 12 for definition of fire-steam.
f This would only be true in case the pressure in the line CEF
was the same at all points ; it is assumed here that it is practi-
cally so.
FLOW OF WATER IN PIPES. 21
Therefore the discharge with assumed head of n.8
feet will be :
in line BC, 5000' of 10", = 562 gals.
" " BE, 4000' " 8", - 353 "
" " BmF = 200 "
Total discharge through BC,
BE, and BmF, 1115 "
Loss of head in AB due to flow of 1115 gallons :
1115 gallons through 2000' of 12", = 6.8 ft.
Add for BC, BE, and BmF, 1 1.8 "
Total head for assumed discharge
in system ABCF, 18.6 "
It is evident that the loss of head in line APD must
be equal to that in ABCF. As APD is a compound
pipe, assume flow to be 1000 gallons. The head due to
this flow is as follows :
In AP, 1500' of 12", 4.2 head
" PD, 6000' " 14", 7.8 "
Total, 12 "
Then discharge due to 18.6 feet, or head in ABCF, is
/i8.6W
I I X 1000 = 1270 gallons per minute.
Therefore total discharge with loss of head of 18.6
ft. is :
In system ABCF, 1115 gallons per minute
" ' " APD, 1270 "
Total, 2385
22 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
Finally, to find loss of head due to the required dis-
charge of 2000 gallons per minute, use the formula
/required discharge V 1 / 6
, ,. , X assumed loss of head ;
Vassumed discharge/
discharge
or
/2000V 1 / 6
X 18.6 = .728 X 18.6 = 13.50 ft. head, ans.
If it were desired to find loss of head due to a draft
of ten i-J-inch fire-streams or 2500 gallons per minute,
or any other quantity, use the same formula.
If the discharge for a given loss of head is required in
the above system, use the formula
/ given head W" . [ required
. r -H X assumed discharge = < ,.
\assumed head / ( discharge.
In the above case, suppose that a head of but 15 feet
were available, or that it were not desirable to allow of a
greater loss of head ; then
(ic
^
6 /"
X 2385 = .86 X 2385 = 2050 gallons,
the discharge for a loss of head of 15 ft.
Note. In all of the cases given in this chapter the
2d power instead of the n/6, and the 1/2 power instead
of the 6/1 1, could be used ; and if the assumed quanti-
ties closely approximated the real ones, the difference in
results would be very small. The following table shows
the difference in a few cases.
When the friction caused by a draft of 1000 gals, is
14.50 :
FLOW OF WATER IN PIPES. 23
Gals. Head by n/6 power. Head by 2d power.
1000 14-5 J 4-5
500 4-06 3.62
1200 20.30 21.00
1500 30.60 32.60
2000 51-60 58.00
When a head of 10 feet would cause a discharge of
1000 gallons, the difference for different heads is shown
by the following table :
Head. Discharge by 6/n power. Discharge by 1/2 power.
10 1000 1000
5 68 S 707
12 noo I0 95
15 1245 1225
20 1458 1415
The advantage of using the 2d power instead of the
1 1/6 is that the formula can be computed on the slide-
rule with one setting, which would be as follows :
For formula
/required discharge \ a
, rr-. : X assumed head = true head
Vassumed discharge /
set assumed discharge on lower slide over required dis-
charge on lower scale, and over assumed head on upper
slide read true head on upper scale.
For formula
/ given head V/ 2 j- *_
X assumed discharge = true discharge
\assumed head/
set assumed head on upper slide under given head on
upper scale, and under assumed discharge on lower slide
find the true discharge on lower scale,
24 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
OLD OR TUBERCULATED PIPE-LINES.
It is now generally recognized that the loss of head
caused by the friction in pipes increases with the age of
the pipes. There is, however, no generally accepted
formula that takes into account the factor of age.*
Kutter gives a variable coefficient n for roughness of
the sides of channels. In old pipe-lines there are gen-
erally no data relating to the condition of these sur-
faces, and the age of the line is the only easily deter-
mined element. While this is not so satisfactory as the
known condition of the pipes in question would be,
which is affected by the character of the water and of
the coating as well as by the age of the line, there are
no other data relating to this condition given in the
recorded experiments upon old pipes. It is desirable to
have some ready means of approximately estimating the
effect of age upon pipe-lines, as the formulas given for
clean pipe cannot be safely used for old lines without
correction. Although good judgment should always be
used, some mathematical assistance is needed.
COEFFICIENTS OF LOSS OF HEAD.
Wishing to provide some systematic means of correct-
ing the results given by the diagrams for clean pipes, so
that they may be safely used for old lines, the author
made a study of all the experiments upon old pipes that
were available to him. As might be expected, the
results of these are conflicting and do not furnish a
satisfactory basis for a mathematical formula.
* Since the above was written a valuable book by Edmund B.
Weston, " The Friction of Water in Pipes " (D. Van Nostrand Co ),
giving the friction caused by the flow of water in pipes, has been
published, which gives coefficients for friction in old lines.
FLOW OF WATER IN PIPES. 25
The graphical method seemed to be the most satisfac-
tory one for arriving at some conclusion. Accordingly,
the results of the experiments expressed as a percentage
of the excess of loss of head by friction over the loss
under the same condition in new pipes, as computed by
the formula adopted in this book, was plotted as shown
on diagram A, using the horizontal scale for the percent-
age of excess of friction, and the vertical scale for the
velocity of flow in feet per second. The result of each
experiment of a series is represented by a small circle,
and all results of a series are connected by a broken
line. By graphical comparison, curves * that seemed to
the author to represent the average results of the ex-
periments were drawn for different ages of pipe. It may
be seen that with a few exceptions the general direction
of the broken lines is inclined from the lower left- to the
upper right-hand corner. This indicates that the per-
centage of increase of friction for any age of pipe is not
constant, but increases with the increase in velocity.
The rate of this increase (or the inclination of the
broken lines) differs greatly in the experiments. The
full lines drawn to represent this are inclined in the
ratio of one horizontal to two vertical, and represent an
increase in the percentage which varies with the square
root of the increase in velocity.
This rate almost coincides with that of one of the
most carefully made experiments, viz., that upon a 48-
inch pipe by Mr. Desmond Fitz-Gerald. It is also an
approximate average of all the experiments.
It is indicated by the experimental results that the
excess of friction is greater in small pipes than in large
* As the base of the diagram is logarithmic, these curves are
straight lines.
26 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
ones, other conditions being the same. There are not
sufficient data to make any classification based upon size.
In giving the value to the different ages, an endeavor
was made to use the approximate average for all sizes.
The curve for fifteen years nearly coincides with the
above-named experiment on a 48-inch pipe eighteen
years old.
Table No. 4, giving coefficients by which to multiply
the friction loss in new, clean pipe, was compiled from
diagram A by adding the percentage of excess of friction-
loss, as shown by the curves for several ages of pipe, to
unity.
For example, the excess of head in a pipe ten years
old with a velocity of flow of 2 ft. per second is 30 per
cent; adding this to TOO per cent (the friction-loss in
clean pipe) makes 130 per cent, or written as a coeffi-
cient, 1.30.
COEFFICIENTS OF DISCHARGE.
Diagram B was designed as a means of deriving
coefficients of velocity or discharge from diagram A.
This is also constructed upon a logarithmic base. The
horizontal scale represents the loss of head by friction ;
the vertical scale, velocity in feet per second. Line
CD represents the friction in new, clean pipe when a
velocity of one foot causes a loss of head of one foot
(it is immaterial what value is given to the loss of head
caused by a velocity of one foot) as calculated by the
formula used in this book. Lines CD', C"Z>", etc.,
represent the friction in pipes of different ages. The
small circles through which these lines are drawn were
obtained by multiplying the loss of head given at the
intersection of CD with line of velocity upon which
FLOW OF WATER IN PIPES. 27
the circles are plotted, by the proper coefficient frorh
diagram A. For example, on velocity = 2 the loss of
head given at CD is 3.56 ft.; this X (14 per cent from
diagram A at v = 2 and age 5 years -f- unity = coef.
1.14) = 4.06 ft., the loss of head in pipes five years old
with v = 2.
The coefficients of discharge are derived as follows:
On vertical lines EF drawn through the intersection of
velocity lines with CD, find the velocity at the inter-
section of CD', C"D",etc. t as for example: On vertical
EF drawn through intersection of v = 3 with CD find
the velocity at C" D" = 2.55. This is the velocity in a
pipe ten years old under the conditions and with the
head that produces a velocity of 3 ft. in new pipes.
Then = 0.85 = coefficient of discharge in a pipe ten
years old. The coefficients in Table No. 5 were derived
in this manner. The velocities in the first column are
those that would be produced in new, clean pipes under
the conditions of the problem, as given by the diagrams
for flow of water in clean pipe.
It is not claimed that there is mathematical exactness
in these coefficients as applied to any particular case.
It is hoped that they represent an approximate average
of the results of recorded experiments, and may, when
intelligently used, be of some assistance in the solution
of problems relating to old lines.
Following is a list of the experiments the results of
which were used, with references to the sources from
which they are taken.
28 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
]
Experi-
menter.
Place.
Diam-
eter.
Age, Years.
Source from which Data
were taken.
Fitz-Gerald
Boston
48"
18
Trans. Am. Soc. C. E.,
vol. xxxv.
Darcy
Paris
//
old
^1
Ehmann .. .
Stuttgart...
10"
6
Iben
Hamburg..
" n
12
Ganguillet & Kutter,
2
Flow of Water, translated
//
15
by Hering and Trautwine.
n
6"
25
j
Leslie"!!".
Edinburgh.
5"
3
I Hamilton Smith's Hydraul-
44 t ....
"
6"
8 or 9
ics.
Weston ...
Providence
6"
4
Simpson . .
2"
7
"
2"
Unknown
2"
4
Flow of Water in Pipes.
9"
13
Weston, Trans. Am. Soc.
30"
{Heavily
C. E., vol. xxn.
Green. .
Brooklyn. ..
36"
tuber-
culated
Gale
Glasgow.. .
48"
8
Forbes
Brookline..
*! and
18
[journal N. E. Water-works
(16"
[ Assn., vol. vi.
"
"
16"
18
j
Hastings...
Cambridge.
J 3 and
8
Same, vol. vm.
( 36"
Coffin
Randolph..
14"
8
See following table.
Note. All experiments were upon cast-iron pipe.
RESULTS OF EXPERIMENTS UPON A PIPE-LINE EIGHT
YEARS OLD AT RANDOLPH, MASS.
A 2,7do ft. 14" Pipe.
Loss of Head by Friction.
Actual.
Computed.
Per Cent of
Computed.
Per Cent in
Excess.
T
.80
T.50
2.85
53
2
1.40
6-95
7.90
88
3
1.82
13.80
12.70
1 08
8
4
2.09
18.30
16.40
in
II
5
2-34
23.50
20.30
116
16
6
2.50
28.00
22.70
123
23
7
2.80
34-00
28.
121
21
FLOW OP WATER IN PIPES. 29
Examples of Use of Tables Nos. 4 and 5.
CASE I. When the discharge is given and the head re-
quired in an old line.
Example : What is the friction-head in a 20-inch pipe
line 10,000 ft. long, 15 years old, with a discharge of
2000 gallons per minute ?
Find velocity and friction-head for clean pipe from
diagram No. 10: V = 2; friction-head = 7.6. On Table
No. 4 find opposite V = 2 and under 15 years :he co-
efficient = 1.47; then 7.6 X 1.47 = 1 1. 2 feet-head.
(a) What is the head in the above example with a dis-
charge of 5500 gallons per minute ?
Friction-head for clean pipes from diagram 10 = 49.5;
V = 5.6 ft., coefficient fiom Table No. 4 = 1.80; 49.5
X i. 80 = 89 ft.-head.
CASE II. When the head is given and the discharge
required.
Example : What is the discharge of a lo-inch pipe-line
one mile long, 20 years old, with head of 40 feet ?
From diagram No. 5 find discharge 1070 gallons;
v 4.4. On Table No. 5, opposite V = 4 to 5, and
under 20 years, find coefficient 0.72; then 1070 X 0.72 =
.770 gallons.
For compound and complex mains use the same meth-
ods, taking an approximate average of the velocities in
the different sizes.
As an approximate method and for preliminary work
the coefficients for an age of 20 years and velocity of
3 feet per second may be memorized and used. They
are 1.80 for loss of head and 0.75 for discharge.
An age of 20 years and velocity of 3 feet are very close
approximations to the values used in ordinary practice.
30 GRAPHICAL SOLUTION O? HYDRAULIC PROBLEMS.
Note. Nothing has been said in this chapter about
the loss of head caused by bends, branches, and gates in
pipe-lines. The author is of the opinion that in lines of
new pipes with the ordinary velocities the friction loss
as given on the diagrams will include all loss of head in
clean-coated pipe-lines as usually laid.* The coefficients
for old lines are intended to apply to lines of pipes with
the usual special castings and gates.
In designing all engineering works it is customary to
use a factor of safety. There should be no exception to
this custom in the design of pipe lines or systems. It
cannot be considered good practice to design a line that
will according to the formula used discharge exactly the
required amount of water. For instance, if it is abso-
lutely necessary that a line shall discharge ten million
gallons of water per day, safety would require that a di-
ameter be chosen which, when computing the discharge
as carefully as possible, will deliver at least eleven mil-
lions per day, or an increase of 10 per cent. This is a
small factor of safety as compared with that used in other
branches of engineering.
It is generally the case that the amount supposed to
be required is only an approximate estimate of that actu-
ally required. In such, as in all cases, the engineer must
use his judgment gained from study and experience.
No formula, table, or diagram, however correct within
their limitations, can alone assure successful design.
They are but useful tools for the skilful workman.
Note. For sizes of pipes not given on diagram, take
area and value of r from table No. 7 and find solution
on diagram No. 25 for channels, using n = .on.
* See paper on Friction in Several Pumping-mains in the Journal
of the N. E. W. W. Assoc., vol. x, No. 4.
CHAPTER IV.
FLOW OF WATER THROUGH SHORT TUBES.
THE flow of water through short tubes is treated in this
book on account of its relation to the entry-head of long
pipes. By a short tube is understood a pipe that has a
length equal to about three times its diameter. The loss
of head caused by the flow of water through short tubes
includes the loss due to friction at entrance and the head
required to generate velocity, and will be designated sim-
ply as entry-head hereafter in this book. This loss is
constant in any given pipe for any given velocity without
regard to the length of the pipe-line. Consequently it
must be determined separately from the loss of head
caused by friction in a long pipe which is in direct pro-
portion to its length.
In very long pipe-lines the entry-head is an insignificant
element in the problem, but in comparatively short lines,
with high velocities, it is an important factor, and should
not be ignored.
The diagrams for entry-head are constructed from the
formula
a
v = o(2gh)*, or h =
with the following value for o as given in Hamilton
Smith's Hydraulics:
32 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
Circular pipes:
Mouth flush with side of reservoir 0.825
" projecting into reservoir, with square
ends -7 I 5
Bell-shaped mouthpiece, small velocities 0.950
large velocities '995
2g was taken as 64.36.
There are two diagrams on this subject:
No. i for pipe from 4 to 20 inches in diameter ;
No, 2 " " " 20 " 60 "
The diameter of the pipe is shown by a curve from the
upper left-hand corner; the loss of head in feet by hori-
zontal scales; the one at the top of the diagram for pipes
with the mouth flush with the side of the reservoir; that
at the bottom for pipes projecting into the reservoir and
with square ends. These are the types most in use. If
it is required to find the loss for bell-shaped mouth-
pieces, multiply the head for flush pipes, or that given at
the top of the diagram, by 0.75. The scale of gallons per
minute is given on the right-hand side of the diagram,
and the scale of cubic feet per second on the left.
Examples.
CASE I. Diameter of pipe and discharge given, to find
the head.
Example : What is the loss of head in a lo-inch pipe dis-
charging 4500 gallons per minute or 10 cu. ft. per second
when pipe is flush
Ans. (from diagram No. 18, top scale), 7.7 ft.
OF THE '
UNIVERSITY
FLOW OF WATER THROUGH SHORT TUBES. 33
(a) What would be the loss of head in the above ex-
ample, if the pipe projected into reservoir ?
Ans. (from lower scale), 10.25 ft.
(b) If the mouth of the pipe were bell-shaped ?
Ans. 7.7 ft. loss for flush pipe X 0.75 = 5.77 ft.
CASE II. Diameter and head given, to find discharge
in a short tube or pipe.
Example : What will be the discharge through a short
tube 12 inches in diameter, flush ends, under head of
4.2 ft.?*
Ans. (using upper scale of diagram No. 18), 4850
gallons per minute.
If the end projects into reservoir, in above exam-
ple, take the head, 4.2 ft. on lower scale, and the dis-
charge will be 4190 gallons per minute.
CASE III. Available head and required discharge given,
to find diameter of pipe. x
Example : What size of pipe is required to discharge
2000 gallons per minute with flush ends and head of
2 ft.?
Using upper scale of diagram No. 18 for head, the
intersection of h = 2 and q = 2000 falls between the
curves of 8 and 10 inch pipes ; therefore of the usual
commercial sizes a lo-inch pipe would be required.
CASE IV. The diameter of a long pipe-line and its dis-
charge given, to find total loss of head.
Find the loss due to friction in the pipe by the methods
of Chapter III, and add to that amount the loss of head
caused by entry and velocity found on the diagrams for
entry-head, as in Case I : the sum will be the total head.
Example : What is the total loss of head in a pipe-line
* Head is measured from surface of water to centre of pipe,
34 GRAPHICAL SOLUTION OF HY.DRAULIC PROBLEMS.
14 inches in diameter, 1000 ft. long, flush end, discharg-
ing 4000 gallons per minute ?
The friction loss, taken from diagram No.
7 for flow of water in long pipes, will be. 16.35 ft-
The loss for entry, as shown on diagram
No. 18 for entry-head, will be 1.60 "
Total, 17.95 "
CASE V. The diameter and length of a long pipe and
the available head given, to find discharge.
This problem cannot be solved directly, as the loss of
head caused respectively by the entry and the friction is
not known.
The methods described in Chapter III, Compound
Mains, Case II, may be adopted, and a discharge as-
sumed.
Example: What will be the discharge of a i6-inch
pipe 1000 ft. long, with an effective head of 6 ft., with
flush end?
Assume a discharge of 3000 gallons per minute ; for
this quantity, by diagrams
No. 8, the friction-head will be 4.95 ft.
No. 18, the entry-head will be 0.5 "
Total, 5.45 "
Then, following method of Chapter III, Case II, of
Compound Main,
( given head ) T / 2 ,.
\ , -=-5 , ,. r - [ X assumed discharge
( head due to assumed discharge j
/ 6 \ I/2
= true discharge, or ( ) X 3000 = 3150* gallons.
\5'4S /
* See note on page 22.
FLOW OF WATER THROUGH SHORT TUBES. 35
The loss of head due to each cause can be taken from
the diagram after true discharge is found. In above ex-
ample, for a discharge of 3150 gallons,
from diagram No. 8, friction-head = 5.40 ft.
" " entry-head = .58 "
Total, 5.98 "*
Note. It will be found advantageous in such cases as
the above, where the pipe is comparatively short, to find
the friction for a longer length ; in this case the friction-
head can be found more closely by finding it for 5000 ft.
instead of 1000, and dividing result by 5.
The diagrams for entry-head are very convenient for
finding loss of head in circular penstocks and other short
pipe-lines. In such cases both the friction and the
entry-head must be considered.
* The entry-head increases as the second power of the increase
of velocity, and when it is as great as the friction-head the second
power will be as correct as the n/6.
CHAPTER V.
RECTANGULAR WEIRS.
THE diagrams for the discharge of weirs are computed
from the well-known Francis' formula, Q = clffl 2 . For
heads above 0.5 ft., c = 3.33.
For heads below .5, c = the values given in the follow-
ing table compiled from the paper of Fteley and Stearns,*
which describes their experiments on low heads:
VALUE OF C FOR LOW HEADS IN Q =
Head.
c.
Head.
c.
0-5'
3-33
0.2'
3.388
0.4
3-337
0.15
3-430
0-3
3-353
O.I
3.528
0.25
3.368
0.06
3-750
END CONTRACTIONS.
For each end contraction deduct .ih from the length,
or ,2h where there is contraction at both ends.
Note. Hamilton Smith, Jr., gives a tables of values
for the coefficient c in the formula Q = \c(2g)*lh. \
* Published in vol. xu, Transactions of the Am. Soc. Civil
Engineers.
f This formula can be reduced to the form of Francis' formula
by substituting values of c and 2g.
36
RECTANGULAR WEIRS. 37
Values are given for weirs both with and without end
contraction.
These coefficients give results slightly different from
those obtained by Francis' formula (from one to three
per cent), and it is probable that in cases where the
weirs are constructed entirely in accordance with the
given conditions that the results may be more accurate
than with a constant coefficient, even where a correction
is made for end contraction, as above.
See " Hydraulics," by Hamilton Smith, Jr., for full
discussion of the subject and description of experiments.
Smith says that no weir measurements of water should
be made with h less than 0.2' where accuracy is essential.
EXPLANATION OF WEIR DIAGRAMS.
Very little explanation is necessary for an understand-
ing of these diagrams. They are three in number:
No. i gives lengths from o to 4 ft.
" heads " o.oi " .5 "
discharge " o " 800 gals, per min.
" " " o " 1.77 cu. ft. per sec.
No. 2 gives lengths from o to 8 ft.
" heads " 0.05 " 1.50 ft.
" discharge " o " 8000 gals, per min.
" " o " 17.7 cu. ft. per sec.
No. 3 gives lengths from o to 16 ft.
" heads " o.i " 2.5 "*
". discharge " o " 40,000 gals, per min.
" o " 89 cu. ft. per sec.
* Caution : It is not safe to rely implicitly upon the results
with heads greater than 2 feet. Mr. Francis considered the limits
of his formula to be from 0.5 to 2 feet.
38 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
The lengths of the weirs are given in feet by a vertical
scale on the side of the diagrams ; the horizontal scale
at the top gives discharge in gallons per minute ; the
scale at the bottom in cubic feet per second, and the
oblique lines radiating from the upper left-hand corner
give the head in feet.
Examples.
Example i : What is the discharge of a weir 3 ft. long,
with head of 0.26 ft., with contraction at each end ?
First find corrected length for end contraction, or
3 .2h = 2.948 ft. On diagram No. 22, at the intersec-
tion of lines representing the above values, viz., / = 2.948
and h = 0.26, find discharge = 597 gallons per minute,
or 1.325 cu. ft. per second.
Example 2: In the above example, if the head is 0.45,
what will be the discharge ?
Corrected length = 3 (.2h = .09) = 2.91 ft.
It will be seen that the intersection of / = 2.91 and
h 0.45 does not fall on diagram No. 22; therefore look
for it on No. 23. It is 1315 gallons per min., or 2.9 cu.
ft. per second.
Diagram No. 23 is on a smaller scale than No. 22,
and if it is desired to determine the discharge more
closely by using No. 22, proceed as follows: Divide the
corrected length by 2 or any convenient number; thus,
2.91 -T- 2 = 1.455; find the discharge for this length
655 gallons; multiply this result by 2, or the number
used to divide length, 655 X 2 = 1310.
On one or other of the three diagrams results for any
head up to 1.40 ft. and any length up to 16 ft. can be
read directly; also any head up to 2 ft. with lengths
under 9.5 ft.
RECTANGULAR WEIRS. $$
If it is required to find results where the intersection
of / and h does not fall on the diagram that it is desired
to use, or if the length is greater than that given on any
of the diagrams, advantage can be taken of the principle
that the discharge is proportional to the length, and the
result found as in Example No. 2 (a).
N. B. If there is end contraction, the length must be
corrected for it before dividing.
Example 3: What is the flow over a weir 20 feet long
with contraction at each end and head of 0.2 ft?
Corrected length, 20 (.2/1 = .04) = 19.96 ft.
To find result on diagram No. 22, where the length is
but 4 feet: 19.96 -=- 5 = 3.992 ft. discharge with h =
.2 of 554 gallons per min. Multiply this by the divisor
used, 554 X 5 = 2770 gallons per min., total discharge.
It is often desirable to know how long to construct a
weir in order to discharge a certain maximum quantity
of water and not exceed a certain head.
Example 4: What should be the length of a weir to
discharge 2000 gallons per min., h not to exceed 0.6 ? On
diagram No. 23 at intersection of h = .6 and q = 2000,
find length = 2.96. A weir 3 feet long will meet the re-
quirements, including correction for end contraction.
All problems connected with rectangular weirs can be
solved by these diagrams, and with sufficient accuracy
for all except the most careful work.
CONDITIONS TO WHICH THE DIAGRAMS APPLY.
The conditions of weirs to which the formula and
diagrams apply are as follows:
The crest must be horizontal, the sides vertical, and
the plane of the face of the weir must be approximately
at right angles to the line of flow. The edges of the
40 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
weir must be so thin that the escaping water shall only
come in contact with the up-stream corner edges. For
accurate measurement stiff metallic plates with smooth,
straight edges should be used. For weirs where there is
no end contraction the sides of the canal should be pro-
longed down-stream beyond the crest, but should not ex-
tend below its level, as this would prevent the access of
air under the escaping vein. For full contraction the
water must discharge freely into the air.
The air must be given free entrance under the escaping
vein.
The head should be measured 6 feet up-stream from
the weir upon a stake or fixed point set level with its
crest. For weirs with end contractions the sides of the
weir openings should not be less than twice the maximum
head from the sides of the canal, and the same distance
from the bottom. The length of the weir should not be
less than three times the maximum head. When the
sheet falls on an apron, the latter should be not less than
the total head below the crest. If the discharge falls
into a body of water the latter may be level with the
crest.* The error will not exceed one per cent from this
cause if the weir is submerged to the extent of 15 per
cent of the head on the crest.f
WIDE CRESTS.
It often happens that measurements are made over
weirs with wide crests.
Messrs. Fteley and Stearns made a series of experi-
ments on the flow over wide crests, which are described
* Smith says that if air is admitted freely behind the falling
sheet the surface of water may be nearly to level of crest.
f See "Submerged Weirs" in the paper of Fteley and Stearns
referred to above.
RECTANGULAR WEIRS. 41
in the paper already referred to. As the result of their
work they give a table of corrections to be applied to the
depth or head on a wide crest in order to obtain the
depth on a sharp crest which will pass the same quantity
of water. The results of their table as worked out are
given in the form of a percentage of the discharge over a
thin-plate weir on diagram No. 21. To use this diagram
take the discharge due to the measured head from the
diagrams for thin crests in the usual manner, and multi-
ply it by the percentage given on diagram No. 21 for
given width of crest and head.
Example : What is the discharge of a weir 5 ft. long,
without end contraction, and with head of 0.55 over a
crest 6 inches wide ?
The discharge from the diagram No. 23 for this crest
35 gallons per minute. Multiply this by the per-
centage from diagram No. 21 for 6-inch crest with head
of 0.55 ft. or 3050 X 0.90 = 2745 gallons per minute.
The diagram gives crests from 2 to 24 inches wide,
and also gives the proper percentage for overflows of the
type of the Lawrence dam worked out from Mr. James
B. Francis' formula for that type, Q = 3.0 ^oS//* 1 ^ 3 .
This dam is 3 ft. wide on top, with a slope on the up-
stream side of about 3 horizontal to i vertical.
VELOCITY OF APPROACH.
When there is an appreciable current towards the weir
the velocity of approach must be considered if accuracy
is required.
This subject is thoroughly discussed in Smith's Hy-
draulics. In the paper of Fteley and Stearns, before re-
ferred to, is given a description of a series of experiments
42 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
and their conclusions from them. Any one wishing to
make extremely careful measurements is referred to the
above, and also to Francis* work on hydraulics, " The
Lowell Experiments."
For ordinarily careful work Table No. 6 gives correc-
tions to add to the measured head before taking results
from the diagram. This table is based upon the coeffi-
cients given by Fteley and Stearns for weirs with and
without end contractions, and for different heads and
depths below the crest.
The mean velocity of approach is found by dividing
the approximate discharge over weir by the area of cross-
section of the channel. The head due to this velocity is
found by the formula h = when v mean velocity of
2 g
approach and zg = 64.36.
The coefficients of Fteley and Stearns vary from 1.33,
to 1.87 for weirs without, and 1.88 to 2.42 for those with,
end contraction, and are used to multiply h as found
above, and the product is added to the measured head.
The method of using the table is as follows : Find the
discharge in cubic feet per second due to the measured
head from the diagram ; divide this discharge by the
area of the cross-section of the channel, to find mean
velocity of approach ;* find the nearest number to ve-
locity as found in column i of Table No. 6, and oppo-
site it, under the column giving the conditions nearest
those of the case, find the correction in feet to be added
to the measured head ; finally, find on the diagram the
result due to the corrected head.
Example : What will be the discharge over a weir 4 ft.
long with end contraction and measured head of 0.6 ft. ?
* The velocity may also be obtained by observations with floats
or a current-meter.
RECTANGULAR WEIRS. 43
The channel is 6 ft. wide and 1.5 ft. deep = 9 sq. ft.
area, depth below crest i foot.
Corrected length = 4 (.2h or .12) = 3.88 ft.; the ap-
proximate discharge, from diagram No. 23, due to .6
head is 6 cu. ft. per second. Velocity of approach
= 6 -T- 9 = 0.667 f eet P er second. ' On Table No. 6, for
weirs with end contraction, opposite v = 0.7, and in
column for depth below crest = i foot and head of
0.60, find the correction = 0.013 ; add this to the
measured head, = 0.6 + 0.013 0.613.
Then, with corrected length = 4 2h = 3.877, the
discharge for corrected head of 0.613 6.2 cu. ft. per
second or 2780 gallons per minute.
Note. Fteley and Stearns give a diagram in the paper
referred to from which the head to be added for velocity
of approach can be taken directly, without first finding
the mean velocity. This diagram only applies to weirs
without end contraction.
CHAPTER VI.
FLOW OF WATER IN CHANNELS.
THE engineer meets many problems connected with
the flow of water in channels of other than circular
sections. There is such variety in the shape of these
sections that it is impracticable to construct diagrams
for any special forms, ranging as they do from an
egg-shaped sewer to a canal of trapezoidal section in
earth. Instead of trying to treat these forms (except
that of circular pipe-sewers) on special diagrams, a gen-
eral diagram is constructed which gives the values of
v =i velocity per second, s = slope or fall per 1000, and
r = hydraulic mean radius. When any two of these
three elements of the problem are known, the third can
be read from the diagram. Thus all problems of flow
in channels of any section can be readily solved.
The basis of these diagrams (Nos. 25 and 26) is the
Chezy formula: v = c(rsY' z , as in the case of circular
pipes. For channels, however, the value of the coeffi-
cient c is found by Kutter's formula, which is
1.811 . . .00281 "1
+ 4..6+
\' |
.oo28i
-
all symbols having the meaning given in Chapter II.
44
FLOW OF WATER IN CHANNELS. 45
This formula seems better adapted to channels than
any other, on account of the different values that Kut-
ter gives to n corresponding with different degrees of
roughness of the surface of the channel. These values
are given in the following table :
VALUES OF IN KUTTER'S FORMULA.
Character of Surface. Value of .
Well-planed timber 0.009
Plastered with neat cement )
. . . , . > oio
Also glazed pipe j
Plastered with" mortar composed of one part sand
to three of cement on
Unplaned timber and uncoated C-I pipe 012
Ashlar and first-class brickwork .013
Second-class brickwork and well-dressed stone ... .015
Rubble in cement in good order 017
" inferior condition, and very firm and
regular sides in gravel 020
Rivers and canals in good order, free from vegeta-
tion 025
Same, having stones and weeds occasionally 030
" in bad condition, with much vegetation 035
There are two diagrams for channels : No. 25, for
those having surfaces corresponding to Kutter's n
.on and .013 ; No. 26, for surfaces corresponding to
n = .025 and .030. On these diagrams the slope or fall
per 1000 is given by curves from the upper left hand ;
the hydraulic mean radius, or r, by a vertical scale on
each side; and the velocity, or v, by a horizontal scale.
'On diagram No. 25 the upper scale is for ;/ = .on and
the lower scale for n = .013 ; on diagram No. 26 the
46 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
upper scale is for n = .025 and the lower one for
n = .030.
For other values of n the following table gives the
percentage to use. These percentages are only a mean of
those for different slopes and values of / ; they are usu-
ally not more than two or three per cent in error. For
n = .015 and .017 and ,035 they are about 10 per cent
in error for values of r = . i and 4 ; while for r = i they
are about 3 per cent in error for all slopes and values of
n. While it may not be considered that the table of
percentages gives a very close result in some cases, it
must be remembered that this is a subject that contains
a very uncertain factor in the value of n. It is probable
that the accuracy as given by the percentages is suffi-
cient for all practical purposes of design or computation.
When greater accuracy is desired, the reader is referred
to the table giving values of coefficient c in v c(rs)*/ 2
in Hering and Trautwine's translation of " The Flow of
Water," by Ganguillet and Kutter.* This gives the
results of Ganguillet and Kutter's formula for said co-
efficient, worked out for different slopes and values of r
and s.
TABLE GIVING PERCENTAGE OF v FOR DIFFERENT
VALUES OF .
For Diagram No. 25, upper scale
When n .009, multiply or divide v by 1.25
n .010,
v
I.IO
" tf=.OII,
" " "
1. 00
" n = .012, "
^ tt
.90
(t tt
" " 9 "
.80
" = - OI 5> "
" V "
.70
" n = .017, "
tt v tt
.60
John Wiley & Sons.
FLOW OF WATER IN CHANNELS. 47
For diagram No. 26, upper scale.
When n = .020, multiply or divide v by 1.28
" n = .025, ' ' " v " i.oo
" n = .030, " " " v " .80
* = .035, " " " ? " .68
For other values of n than those given on diagrams :
When v and s are given, multiply v as found on the
diagram by the percentage in the table. When r and v
are given, divide v by the percentage before reading
result from the diagram.
N. B. Always use upper scale of diagram for values
of v when using percentages.
Explanation and Examples.
In using the diagrams, the hydraulic mean radius of
the channel, or r, must be calculated. When two of the
factors r, v, and s are known, the third can be found by
inspection of the diagrams, s is given as the fall in
1000.
CASE I. When r and s are given, to find v.
Example i: What will be the velocity in a rectangular
channel 6 ft. wide, 2 ft. depth of water, fall of 0.25 in
1000, with surface corresponding to n = .on ?
Find the value of r :
_^__ 6X2 _ f or find area and value
p 6 + 2 + 2 ( of r on Table No. 8.
On diagram No. 25, at the intersection of r = 1.2 and
s = 0.25 per 1000, find the value of v on upper scale
= 2.50.
If the discharge is required, Q = va = 2.50 X 12 = 30
cu. ft. per second.
48 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
(a) If, in the above example, the surface corresponds
with n .013, the velocity will be found on the lower
scale, = 2.
(/;) If the surface corresponds with n = .015, multiply
v, as found in Example i, by the percentage taken from
table on page 46, or v = 2.5 X .70 = 1.75.
Example 2: What will be the velocity in a canal of
trapezoidal cross-section, 10 ft. wide on the bottom, side
slopes 3 to i, depth of water 5 ft., s = 0.125 per 1000,
with surface corresponding to n = .030 ?
Find area and value of r for this section on Table
No. ii. Area = 125 ; r = 3.01.
At intersection of r = 3.01 and s = 0.125, ^ n( ^ velocity
on lower scale, diagram No. 26, = 1.12 per sec.
CASE II. When r and v are given, to find s.
Example: What is the slope required in a rectangular
channel 6 ft. wide, with depth of water of i ft., surface
n = .013, to produce a velocity of 3 ft. per second ?
On Table No. 8 find area = 6 ft, r 0.75.
On diagram No. 25, at the intersection of r 0.75 and
v = 3, measured by lower scale, find the value of s = 1.02
per 1000.
If in the above example the surface corresponds
with n .020, what will be the required slope ?
First divide v by the percentage from the table on
page 47, for n = .020 = = 2.34.
1.28
At the intersection of r = 0.75 and v = 2.34, on upper
scale of diagram No. 26, find value of s = 2.80.
CASE III. When the quantity of water to be delivered
and the slope is given, to find the size and shape of the
channel.
The problem is met in this form in designing channels.
As v is involved in the quantity and area of the chan-
FLOW OF WATER IN CHANNELS. 49
nel, this form of the problem cannot be directly solved
by the diagram. Tables Nos. 7, 8, 9, 10, and n give the
width and depth of rectangular and trapezoidal channels
and diameters of circular ones, with areas and values of
r t and will, with the diagrams 25 and 26, facilitate the
solution of the above problem.
Example : What will be the width, depth, and velocity
of a rectangular channel, with slope of 0.5 in 1000 and
surfaces corresponding to n = .013, to deliver 30 cu. ft.
per second ?
It is evident that there are many combinations of
width and depth that will meet the conditions. A prefer-
ence may be had, however, for a certain approximate
width. Assume that to be 6 ft.; there will also probably
be a limit to the allowable velocity, say 3 ft.
This, with Q = 30, will set the minimum limit of area
at IQ ft. Therefore _ = 1.66 deep ; now, with
W = 6 and D = 1.66, r = approx. i.i from Table No. 8.
With r i.i and s = 0.5, v = 2.68, but with the above
area v must = 3. D must be increased ; try D = 1.80 ;
then r approx. 1.13, area 10.8, v from diagram
= 2.70 ; 2.70 X 10.8 29.2 cu. ft. per second. If a
close design is required, a trifling change in D will give
the correct result.
In this way, by the use of the tables and diagrams, the
suitable size and shape of a channel is easily found.
CHAPTER VII.
PIPE-SEWERS.
THE discharge and velocity of pipe-sewers are given
on diagram No. 27 for sizes from 6 to 24 inches, when
running full ; when running half full, velocity is the
same, and the discharge is one-half. This diagram was
constructed from Kutter's formula, and gives results for
two values of the coefficient n. The scale at the top of
the sheet gives the fall or slope in feet per 100 when
'=.oii. The scale at the bottom gives slope when
n = .013. The diameter of the sewer is given by curves
originating at the upper left-hand corner. The discharge
in gallons per minute is given by a vertical scale on the
right, and cubic feet per second by a vertical scale on
the left. The velocity in feet per second is given by
the broken curved lines.
EXPLANATION OF DIAGRAM.
CASE I. Size of pipe and slope per 100 ft. given, to
find velocity and discharge.
Example : What will be the discharge of a 10-inch
pipe-sewer, running full, with a slope of 0.48 per 100, with
surface corresponding to Kutter's n = .on ?
At the intersection of diameter = 10 inch and slope
= 0.48 per 100 taken on the upper scale, find the dis-
charge = 790 gallons per minute, or 1.78 cu. ft. per sec-
ond, with velocity = 3.25.
PIPE-SEWERS. 51
If in the above example the surface corresponds
with n = .013, use the lower scale for slope, and the
answer would be 650 gallons per minute ; velocity
= 2.60.
CASE II. The slope per 100 and required velocity or.
discharge being given, to find the size of the pipe.
Example: What size of pipe is required to secure a
velocity of 3 ft. per minute when the slope is 0.32 per-
100 and the surface corresponds with n .on ?
At the intersection of velocity = 3, and slope on the
upper scale =- 0.32, find next larger size of pipe = 12
inch.
If the required discharge is given instead of the ve-
locity, the method of solution is the same.
CASE III. Size of sewer and required velocity or dis-
charge given, to find fall per 100.
Example: What is the necessary fall in a 15-inch
sewer to produce a velocity of 2.5 ft. per second = dis-
charge 1375 gallons per minute, with surface correspond-
ing to n = .013 ?
At intersection of v 2.5 and diameter = 15 inch,
find value of fall read from the lower scale = 0.24.
Note. When it is required to find velocity more
closely than can be read from velocity-curves, it may be
done by reading the discharge instead and taking veloc-
ity from Table No. 2, which gives the number of gallons
for different velocities in the different diameters.
Note. For sizes of sewers, either pipe or brick, not
given on diagram No. 27 use table No. 7 to find area and
value of /, and solve problem by diagram No. 25 or 26,
using such value of n as the case requires.
CHAPTER VIII.
FIRE-STREAMS AND DISCHARGE OF NOZZLES.
IN designing pipe systems for fire-protection, it is
desirable to know the pressure or head required at the
hydrant to throw a stream to a given height, and the
quantity of water required for such stream.
Mr. John R. Freeman, M. Am. Soc. C. E., in a valua-
ble paper published in the Transactions of the Am. Soc.
of Civil Engineers, vol. xxi, describes and gives the
results of a great number of careful experiments made
by him upon the pressure and discharge of fire-streams.*
In this paper are tables giving the height and discharge
of fires-treams under various conditions of pressure,
length of hose, size and shape of nozzle, etc. Table
No. 13 in this book is compiled from Mr. Freeman's
tables and gives the head required at the hydrant to
deliver fire-streams through lengths of hose varying
from 50 to 500 ft.; the height to which 'the streams
reach, and the number of gallons per minute required for
each. The table gives the results for i-, i-J-, and i^-inch
smooth nozzles with the ordinary best quality of 2^-
inch rubber-lined hose.
* A similar paper by the same author with the same tables was
also published in the Journal of the N. E. Water-works Associa-
tion, March 1890.
52
FIRE-STREAMS AND DISCHARGE OF NOZZLES. 53
This table is intended for ready reference, and com-
prizes those sizes of nozzles in most common use. The
reader is referred to the paper mentioned above for more
extended tables and a discussion of the subject.
Discharge of Hose-nozzles. Diagram No. 31 is con-
structed from tables given in the paper referred to. It
gives the discharge of hose-nozzles through 50 and 100
ft. of 2^-inch ordinary best quality of rubber-lined hose,
and the head or pressure indicated by a gauge at the
hydrant.
The diagram may be used when the approximate
quantity of water delivered by a pump or flowing through
a pipe is to be found. Mr. Freeman says that when
the hose corresponds with the conditions of his tables
the results are close enough for practical purposes, and
when careful judgment is used to interpolate between
the values of rough and smooth hose, results within 5 per
cent can be obtained. The diagram only gives results
for ordinary best quality rubber-lined hose with the
inside smooth. When testing pumps or lines it is advis-
able to use that class of hose. Fifty feet of hose should
be used in preference to one hundred. Full curves on
diagram represent the size of the nozzles with 50 ft. of
hose; the dotted curves with 100 ft.
Diagram No. 32 gives the discharge of nozzles and
the head or pressure as indicated by a gauge at the base
of the play-pipe. In this case the effect of friction in
the hose is eliminated, and with smooth, true, carefully-
calibred nozzles very accurate results can be obtained.*
This diagram is constructed from the formula
v o(2gh) T / 2 , using o = 0.99.
* See paper on The Nozzle as an Accurate Water-meter, by
J. R. Freeman. Trans. Am. Soc. Civil Engineers, vol. xxiv, 1891.
54 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
Mr. Freeman found this coefficient to be 0.995 as tne
result of his experiments.
On the diagram one curve represents two sizes of
nozzles, the larger just double the diameter of the smaller,
as f" and i*".
The scale of gallons is given for the smaller sizes; it
must be multiplied by four for the larger ones.
CHAPTER IX.
MISCELLANEOUS DIAGRAMS.
UNDER this head there are several diagrams that the
author has found useful in his practice, and that may be
called hydraulic diagrams.
HORSE-POWER.
Diagram No. 28. This diagram gives the horse-power
for falling water, and also the horse-power required to
raise water. The vertical scale at the right of the diagram
gives the quantity of water in cubic feet per second ;
that at the left gives the same in gallons per minute.
The horizontal scale at the top gives the horse-power as
represented by the total energy or 100$ efficiency; that
at the bottom gives the same with an efficiency of 75$,
or the amount usually taken as the efficiency of a good
water-wheel.
CASE I. When the quantity of water and effective
head are given to find the power.
Example : What is the total horse-power of 8.5 cu. ft.
per second with an effective head on wheel of 13 ft. ? At
intersection of cu. ft. per sec. = 8.5 and head = 13, find
value of horse-power as read on upper scale = 12.6.
(a) In the above example what would be the effective
power from a wheel with an efficiency of 75$ ? The
answer, read from the lower scale, = 9.4 horse-power.
55
56 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
CASE II. Power required to pump water.
When the quantity to be pumped and the total head
to be pumped against are given to find the effective
horse-power required.
Example : What is the power required to pump 2000
gallons per minute against a total head on the purnp of
150 ft. ?
Answer taken from upper scale = 76 horse-power.
DISCHARGE OF SINGLE PUMPS
DIAGRAM No. 29.
This diagram is designed to give the capacity of recip-
rocating pumps or pumping-engines, either single, duplex,
or triplex. The horizontal scale at the top and bottom
of the sheet gives the capacity in gallons per minute;
the horizontal scale between the two parts of the diagram
gives the revolutions per minute. The vertical scale on
the sides gives the piston-speed in feet per minute. The
oblique lines radiating from the upper left-hand corner
give the diameters of pistons and plungers in inches.
Those from the upper right-hand corner give the length
of stroke in inches.
The upper part of the diagram is for the smaller sizes
of pumps, and the lower part for the larger ones. The
diagrams are designed for single pumps with double-
acting pistons or plungers. For duplex double-acting
pumps multiply the discharge, as taken from the diagram,
by 2. For triplex single-acting pumps multiply by i.
CASE I. Given the diameter of plungers and piston-
speed to find the discharge.
Example i: W T hat is the capacity of a single-acting
pump with a piston 10" in diameter and a piston-speed
of 80 feet per minute, diameter of piston-rod 2 inches ?
MISCELLANEOUS DIAGRAMS. 57
At the intersection of lines representing the above
values of speed and diameter find the discharge as fol-
lows:
Discharge of 10" plunger, 80' piston-speed = 326 gallons
Less 1/2 discharge of 2" rod at same
piston-speed, from diagram, = ^ = 7 "
Total, 319 "
(a) If the above were a duplex pump, multiply by 2:
319 X 2 = 638 gallons per minute.
When the length of stroke and revolutions per minute
are given instead of piston-speed, first find, at the inter-
section of line representing the number of revolutions
with that of length of stroke, the line representing piston-
speed, and follow this line to its intersection with that of
the given diameter, and read the discharge at that point
as before.
Example 2: What is the discharge of a duplex pump
with i2-inch plungers and 1 8-inch stroke at 35 revolu-
tions per minute, diameter of piston-rod 3 inches ?
At the intersection of 35 revolutions and i8-inch
stroke find piston-speed (107 ft.) ; follow this to
intersection of 1 2-inch diameter and there read gallons
= 620 gals, per min.
Less one half rod same speed = - 4 / = 20 " " "
Total, 600 " " "
As it is a duplex pump, multiply by 2 = 1200 gallons.
CASE II. When the required discharge and piston-
speed are known, to find the diameter of plunger.
Example i: What size of plunger is necessary in a
duplex pump to discharge 4000 gallons per minute; the
58 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
piston-speed not to exceed 130 ft., diameter of rod 3?
inches ?
As it is a duplex pump, first divide 4000 by 2
= 2000 -f- loss caused by 1/2 rod at given speed, or ^
= 2033 gallons. At the intersection of 2033 gallons and
130 ft. piston-speed find the nearest diameter given on
the diagram, which is 20 inches. Then 20-inch diameter
delivering 2033 gallons =127 ft. piston-speed. On the
line representing this speed find the intersection of the
length of stroke desired, and read the number of revolu-
tions from the proper scale.
In the above case, if the stroke is 30 inches, the number
of revolutions will be 25.5.
COAL KEQUIRED IN PUMPING ONE MILLION GALLONS OF
WATER.
DIAGRAM No. 30.
This diagram is designed to give the weight of coal
consumed in pumping-plants of different duties to raise
one million gallons of water to various heads from o to
250. The head in feet is given on the horizontal scale.
The duty is given by oblique lines from the upper left-
hand corner. The weight of coal in net tons of 2000 Ibs.
is given on the vertical scale on the left, and the same in
gross tons of 2240 Ibs. on the right.
The following examples will be sufficient explanation
of the use of the diagram.
Example i: What weight of coal will be consumed in
a pumping-plant capable of a duty of 60 million ft.-lbs.
per 100 Ibs. of coal to raise one million gallons of water
against 150 ft. head?
At the intersection of 60 millions duty and 150 ft. head
find weight of coal = 1.04 net tons or o 92 gross tons.
MISCELLANEOUS DIAGRAMS. 59
Example 2: What is the duty of a pumping-plant that
raises one million gallons of water for 2.3 net tons of coal
consumed against a total head of 200 ft. ? At intersection
of 2.3 tons and 200 ft. head find duty = 36 millions
approximately.
If problems of higher duties than 100 million are to be
solved, divide the duty by some number, find answer for
the quotient, and divide weight of coal found by same
number.
Example : How much coal is required to pump one
million gallons against a head of 175 feet with machinery
capable of a duty of 125 million? Divide 125 million
by 5 = 25 million; this with head of 175 will require 2.60
gross tons; this divided by 5 = .52 tons, answer for 125
million duty.
Reverse the process to find duty when coal is given.
60 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
TABLE No. 1.
TABLE OF U. S. GALLONS PER MINUTE AND THEIR
EQUIVALENTS.
Gallons
per
Minute.
Gallons
per 24
Hours.
Cubic
Feet per
Second.
Gallons
per
Minute.
Gallons
per 24
Hours.
Cubic
Feet per
Second.
I
1440
O.OO2
350
504000
0,780
10
14400
.022
360
518400
.802
20
28800
.044
370
532800
.825
30
43200
.06 7
380
547200
.847
40
57600
.089
390
561600
.869
50
72000
.121
4OO
576000
.892
60
86400
.134
4IO
590400
.914
70
lOOSoo
.156
42O
604800
.936
80
II52OO
.178
430
619200
.958
9
129600
.200
440
633600
.98!
100
I44OOO
.223
450
648000
1.003
no
158400
245
460
662400
-025
120
172800
.268
470
676800
.048
130
187200
.290
480
691200
.069
I4O
2OI6OO
312
490
705600
.091
150
2I6OOO
335
500
72OOOO
.112
1 6O
230400
357
510
734400
.136
170
244800
.380
520
748800
159
180
259200
.401
530
763200
.I8l
190
273600
.424
540
777600
.202
200
288000
.446
550
792000
.222
210
302400
.468
560
806400
.248
22Q
316800
.490
570
820800
.269
230
331200
.513
580
835200
.291
240
345600
535
590
849600
312
250
360000
557
600
864000
337
260
374400
579
610
878400
359
27O
388800
.601
620
892800
.381
280
403200
.624
630
907200
.402
290
417600
.647
640
921600
.426
300
432OOO
.669
650
936000
449
310
446400
.691
660
950400
.470
320
460800
.713
670
964800
492
330
475200
736
680
979200
515
340
489600
.758
690
993600
538
TABLES.
U. S. GALLONS AND THEIR EQUIVALENTS.
61
Gallons
per
Minute.
Gallons
per 24
Hours.
Cubic
Feet per
Second.
Gallons
per
Minute.
Gallons
per 24
Hours.
Cubic
Feet per
Second.
700
IOO8OOO
559
1250
1800000
2.785
710
1022400
.58!
I3CO
1872000
2.893
720
1036800
.602
1350
1944000
3-009
730
IO5I2OO
.627
1400
2OI6OOO
3.U9
740
1065600
.649
M50
2088000
3.230
750
loSoooo
.671
1500
2160000
3-341
7 60
1094400
.692
1550
2232OOO
3-453
770
1108800
715
1600
2304000
3.562
780
1123200
733
1650
2376000
3.676
790
1137600
.760
1700
2448000
3.785
800
1152000
.782
1750
2520000
3-899
810
1166400
.802
1800
2592000
4.010
820
1180800
.827
1850
2664000
4.121
830
1195200
.849
IOXX)
2736000
4-233
840
1209600
.8 7 I
1950
2808000
4-344
850
1224000
.892
2000
2880000
4.456
860
1238400
.918
2050
2952000
4-567
870
1252800
.936
2IOO
3024000
4-683
880
1267200
.960
2150
3096000
4.790
890
1281600
.982
22OO
3168000
4.901
900
1296000
2.005
2250
3240000
5-013
910
1310400
2.027
2300
33I2OOO
5-125
920
1324800
2.048
2350
3384000
5.235
930
1339200
2.073
24OO
3456000
5.347
940
1353600
2.093
2450
3528000
5.458
950
1368000
2.II4
2500
3600000
5-570
960
1382400
2.138
2550
3672000
5.681
970
1396800
2.161
2600
3744000
5-792
980
1411200
2.181
2650
3816000
5-904
990
1425600
2.202
2700
3888000
6.015
IOOO
1440000
2.228
2750
3960000
6.127
1050
1512000
2-339
2800
4032000
6.245
IIOO
1584000
2 450
2850
4104000
6-349
1150
1656000
2 562
2900
4176000
6.464
1200
1728000
2.672
2950
4248000
6-573
3000
4320000
6.684
62 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
TABLE No. 2.
GIVING DISCHARGE OF CIRCULAR PIPES RUNNING FULL IN
GALLONS PER MINUTE FOR DIFFERENT VELOCITIES.
Velocity in
Feet per
Second.
Diameter of Pipe in Inches.
4"
6"
8"
10"
12"
14"
.2
8
.18
31
49
70
96
4
16
35
63
98
141
192
.6
24
53
94
147
211
288
.8
3i
70
125
196
282
384
I.O
39
88
157
245
352
480
.2
47
106
188
294
422
576
4
55
123
219
342
493
672
.6
63
141
250
391
563
768
.8
70
159
282
441
634
864
2.0
78
176
313
489
704
960
.2
86
194
345
538
774
1056
4
94
211
376
588
845
1152
.6
IO2
229
407
637
915
1248
.8
ICQ
247
438
686
986
1344
3-o
117
264
470
734
1056
1440
.2
125
282
502
783
1126
1536
4
133
300
532
832
1197
1632
.6
141
317
564
882
1267
1728
.8
149
335
596
930
1338
1824
4.0
156
352
627
979
1408
1920
.2
164
370
658
1027
1478
2016
4
172
388
689
1076
1549
2112
.6
1 80
405
721
1125
1619
2208
.8
188
423
752
1174
1689
2304
5-0
196
441
783
1223
1760
24OO
5
215
485
861
1345
1936
2640
6.0
235
529
940
1468
2112
2880
5
254
573
1017
1590
2288
3120
7.0
274
617
1096
1712
2464
3360
5
293
66 1
H73
1835
2640
3600
8.0
313
705
1252
1957
28l6
3840
5
332
749
1330
2079
2992
4080
9.0
352
793
1410
22OI
3168
4320
5
372
837
1489
2324
3344
4560
IO.O
39i
881
1567
2446
3520
4800
TABLES.
DISCHARGE OF CIRCULAR PIPES FOR DIFFERENT
VELOCITIES. Continued.
Velocity in
Feet per
Second.
Diameter of Pipe in Inches.
15" 16"
18"
20"
24"
30"
.2
110
125
159
196
282
440
4
22O
251
318
391
564
880
.6
330
376
476
587
846
1320
.8
441
501
635
782
1128
1760
I.O
551
627
794
978
1410
2 2OO
.2
66 1
752
953
1174
1692
2640
4
771
877
III2
1369
1974
3080
.6
882
1003
1270
1565
2256
3520
.8
993
1128
1429
1760
2538
3960
2.0
1103
1253
1588
1956
2820
4400
.2
1213
1379
1747
2152
3102
4840
4
1323
1504
1906
2347
3384
5280
.6
1434
1629
2064
2543
3666
5720
.8
1544
1755
2223
2738
3948
6160
3-o
1653
1880
2382
2934
4230
6600
.2
1763
2005
2541
3130
4512
7040
4
1873
2131 .
2700
3325
4794
748o
.6
1984
2256
2858
3521
5076
7920
.8
2094
2381
3017
37i6
5358
8360
4.0
2204
2507
3176
3912
5640
8800
.2
2314
2632
3335
4108
5922
9240
4
2424
2757
3494
4302
6204
9680
.6
2535
2883
3652
4499
6486
IOI20
.8
2645
3008
3811
4694
6768
10560
5-o
2755
3133
3970
4890
7050
IIOOO
5
3031
3447
4367
5379
7755
I2IOO
6.0
3306
3760
4764
5863
8460
13200
5
3582
4073
5161
6357
9165
14300
7.0
3857
4387
5558
6846
9870
15400
5
4133
4700
5955
7335
10575
16500
8.0
4408
5013
6352
7824
11280
17600
5
4684
5326
6749
8313
11985
18700
9.0
4959
5641
7146
8802
12690
19800
5
5235
5954
75-13
9291
13395
2O9OO
IO.O
55io
6267
7940
9780
14100
22OOO
64 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
DISCHARGE OF CIRCULAR PIPES FOR DIFFERENT
VELOCITIES. Continued
Velocity in
Feet per
Second.
Diameter of Pipe in Inches.
36"
42"
4 8"
54"
60"
.2
6 3 I
863
1128
1427
1756
4
1262
1727
2256
2854
3512
.6
1893
2590
3384
4281
5269
.8
2524
3454
4512
57o8
7025
I.O
3155
4317
5639
7136
8781
.2
3786
5180
6767
8564
10537
4
4417
6044
7895
9990
12293
6
5048
6907
9023
11417
14049
.8
5679
7771
10151
12844
15807
2.0
6310
8634
11279
14272
17562
.2
6941
9497
12407
15699
I93I9
4
7572
10361
13535
17126
21075
.6
8203
11224
14663
18553
22831
.8
8834
12088
I579I
19981
24588
3-o
9465
12951
16919
21408
26344
.2
10096
13814
18046
22835
28100
4
10727
14678
I9I75
24262
29856
.6
H358
I554I
20303
25689
31613
.8
11989
16405
21431
27116
33369
4.0
12620
17268
22558
28544
35125
.2
I325I
18131
23686
29971
36881
4
13882
18995
24814
31398
38637
.6
T45I3
19858
25942
32825
40394
.8
I5M4
20722
27070
34252
42150
5-o
15775
21585
28198
35679
43906
5
17353
23744
31018
39247
48297
6.0
18930
25902
33838
42815
52687
5
20507
28060
36657
46383
57078
7.0
22085
30219
39474
49951
61468
5
23662
32378
42297
53519
65858
8.0
25240
34536
45H7
57087
70249
5
26817
36694
47937
60655
74639
9.0
28395
38853
50756
64223
79030
5
29973
41012
53576
67791
83420
IO.O
31550
43170
56396
71359
87810
TABLES.
TABLE No. 3.
TABLE OF 11/6 AND 6/11 POWERS.
Num-
ber.
1 1/6
Power.
6/1 1
Power.
Num
her.
1 1/6
Power.
6/1 1
Power.
Num-
ber.
1 1/6
Power.
6/1 1
Power.
0.50
0.28
0.685
.12
23
.06
74
2.77
35
52
30
.70
.14
27
.07
./6
2.8 3
36
54
32
.715
.16
31
.08
.78
2.89
37
.56
35
73
.18
.36
.09
.80
2-95
38
08
37
744
.20
.40
.IO
.82
3-01
.38
.60
39
.758
.22
44
.11
.84
3-07
39
.62
.42
77
.24
49
.12
.86
3-13
.40
64
.44
.785
.26
53
-13
.88
3-19
41
.66
47
.798
.28
58
.14
.90
3.26
42
.68
49
.81
i -SO
.62
15
92
3-32
43
.70
^52
.824
32
67
.16
94
3.38
43
72
55
.835
i -34
71
17
1.96
3-45
.44
74
.58
.848
.36
76
.18
1.98
3.52
45
76
.61
.86
38
.Si
.19
2.
3.56
.46
.78
64
873
.40
.86
.20
2.5O
5-40
.65
.80
.67
855
.42
.91
.21
3-00
7-55
.82
.82
.70
.899
44
.96
.22
3.50
10.
98
.84
73
.91
.46
2.OI
23
4.00
12.75
2.13
.86
76
.92
48
2.06
.24
4.50
15.85
2.27
.88
79
932
50
2. II
25
5-oo
19.20
2.40
.90
.82
944
.52
2.16
25
5 50
22.90
2-53
.92
.86
955
54
2.21
.26
6.00
26.90
2.65
.94
-89.
.966
56
2.27
27
6.50
31-20
2.77
.96
93
977
.58
2.32
.28
7. co
35-70
2.88
.98
97
.988
.60
2-37
.29
7 50
40.70
2.99
.00
i.
i.
.62
2.43
30
8.00
45-6o
3.10
.02
.04
1. 01
.64
2.49
31
8.50
51.00
3.20
.04
.07
1.02
.66
2-54
32
9.00
56.70
3-30
.06
.11
1.0 3
.68
2.60
33
9-50
62.70
3-40
.08
15
1.04
.70
2.65
33
10.00
68.00
3.50
.IO
.19
1.05
.72
2.71
34
66 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
TABLE No. 4.
COEFFICIENTS FOR LOSS OF HEAD BY FRICTION
IN OLD PIPE-LINES.
Velocity,
Age of Pipe in Years.
Feet
per
Second. /
5
10
15
20
25
30
40
5
I
.IO
.21
1-33
1.46
1.58
I.7I
1.96
2 23
2
.14
30
47
165
1.82
2.00
2 37
2-74
3
17
37
- -57
i. 80
2.OO
2.22
2.69
3-13
4
.20
43
.66
i-93
2.16
2.41
2.96
3-45
5
.22
.48
74
2.04
2.30
2-57
3-20
3 73
6
.24
52
.81
2.13
2.42
2.72
3.41
3-99
7
.26
.56
.88
2.22
2-53
2.86
3.61
4.22
8
.28
.60
93
2.31
2.63
2-99
3-79
4.44
9
30
.64
99
2-39
2-73
3-n
3-97
4-65
10
32
1.67
2.04
2.46
2.82
3-22
4-13
4-85
TABLE No. 5
COEFFICIENTS OF DISCHARGE IN OLD PIPE-LINES.
Velocity,
Feet
per
Second.
I
2
3
4
6
8
9
10
Age of Pipe in Years.
5
IO
'5
20
25
30
40
50
0-95
0.91
0.86
0.82
0.79
0.77
0.72
0.68
93
.87
.82
.78
75
.72
.66
.62
92
.85
79
75
71
.68
63
59
.91
.84
.78
73
69
.66
.60
56
.90
.83
77
7i
.67
.64
.58
54
.89
.82
75
.69
.66
.62
57
53
.89
.80
74
.68
.65
.61
.56
52
.88
79
.72
.67
.63
.60
54
51
.88
79
72
.66
.63
59
53
50
87
78
71
.65
.62
58
53
49
TABLES.
6 7
TABLE No. 6.
GIVING CORRECTIONS IN FEET TO BE ADDED TO HEAD
ON THIN-CREST WEIRS FOR VELOCITY OF APPROACH.
WITHOUT END CONTRACTIONS.
Depth of Bottom of Channel below Crest.
Veloc-
Depth below
Crest = .50.
Depth below
Crest = i.
Depth below Crest = 2.50.
Ap-
proach.
Head on Crest.
Head on Crest.
Head on Crest.
.20
.30t0.50
.4010.60
.80 to i oo
50
I.OO
1.50
2.OO
.2
.001
.001
.001
.001
.001
.00:
.001
.001
3
.OO2
.002
.002
.002
.002
.002
.002
.002
4
.004
.004
.004
.004
.004
.004
.003
.003
5
.007
.006
.007
.006
.006
.006
.005
.005
.6
.OI
.009
.01
.009
.008
.008
.008
.007
7
.012
.013
.OI2
.Oil
.OI I
.on
.01
.8
.cts
.017
.016
015
.014
.014
.013
9
.022
.O2I
.019
.018
.017
.017
.0
....
.027
.026
.023
.022
.021
.021
.1
....
'033
.031
.028
.02 7
.026
.025
.1
...
039
037
.033
.032
o->8
.031
.030
4
.051
045
.044
.042
035
.041
5
....
....
.058
.052
05
.048
.047
064.
'062*
'o6 3
.8
060
067
9
.08
.077
.075
....
....
....
.083
WITH END CONTRACTIONS.
.2
.001
.001
.001
.001
.OOI
.OOI
.001
.001
3
.003
.003
.003
.003
.003
.003
.003
.003
-4
.006
.005
.006
.005
-005
.005
.005
.005
5
.009
.008
.009
.008
.008
.008
.008
.007
.6
.013
.012
.013
.012
.Oil
.Oil
.Oil
.on
7
.017
.016
.017
.017
.015
.015
.015
.014
.8
....
.021
.023
.022
.02
.02
.019
.019
9
.029
.027
.026
02 S
.024
.024
.0
....
.036
0^4
.032
.031
3
.029
.1
.043
.041
.038
037
.036
.036
.2
.051
.048
.046
.044
.043
.042
3
....
....
057
053
.052
.051
.049
4
.066
.062
.06
59
.058
5
...
.076
.071
.069
.068
.065
.6
.o8l
7
O79
.089
.077
.087
75
.085
.8
9
....
.099
.III
.097
.108
.0^4
.105
....
.117
68 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
TABLE No. 7.
GIVING AREA AND VALUE OF r OF CIRCULAR CHANNELS
RUNNING FULL.
Diameter in
Ft. and In.
e
?
<
Values of r.
Diameter in
Ft. and In.
.n
rt CT
#
Values of r.
~
ai
<u a
|
c
& o*
Hc/5
<
Values of r.
l"
0.005
O.O2
3' 5"
9.17
-85
6' 9"
35.78
.69
2
.022
.04
6
9.62
.88
IO
36.67
71
3
.049
.06
7
10.08
.90
ii
37-57
73
4
.087
.08
8
10.56
.92
j'
38.48
75
5
.136
.IO
9
II.O4
-94
i
39-41
-77
6
.196
13
IO
H-54
.96
2
40.34
.79
7
.267
15
ii
12.05
.98
3
41.28
.81
8
349
17
4'
12.57
1. 00
4
42.24
83
9
.442
.19
i
13.10
1.02
5
43-20
.85
10
545
.21
2
13.64
1.04
6
44.18
.88
ii
.660
23
3
14.19
1. 06
7
45-17
.90
i'
.785
25
4
14-75
1. 08
8
46.16
-92
i
.921
27
5
I5-32
I.IO
9
47-17
-94
2
.07
30
6
15.90
I-I3
10
48.19
.96
3
23
31
7
16.50
I-I5
ii
49.22
97
4
.40
33
8
17.10
I.I7
8'
50.27
2.OO
5
.58
35
9
17.72
I.I9
i
51-32
2.02
6
77
38
10
iS-35
1. 21
2
52.38
2.04
7
97
.40
ii
18.99
1.23
3
53.46
2.O6
8
2.18
.42
5'
19.63
1.25
4
54-54
2.08
9
2.41
44
i
20.29
1.27
5
55.64
2.10
10
2.64
.46
2
20.97
1.29
6
56.75
2.13
ii
2.89
.48
3
21.65
I-3I
7
57-86
2.15
2'
3-14
50
4
22.34
1-33
8
58.99
2.17
i
3-41
52
5
23.04
1-35
9
60.13
2.19
2
3-69
54
6
23-76
1.38
IO
61.28
2.21
3
3-98
56
i 7
24.48
1.40
ii
62.44
2.23
4
4.28
58
8
25.22
1.42
9'
63.62
2.25
5
4-59
.61
9
25-97
1.44
i
64.80
2.27
6
4.91
.63
IO
26.73
I. 4 6
2
66.
2.29
7
5-24
.65
ii
27.49
I. 4 8
3
67.20
2.31
8
5-59
.67
6'
28.27
1.50
4
68.42
2-33
9
5-94
.69
i
29.07
1-52
5
69.64
2-35
10
6.31
7i
2
29.87
1.54
6
70.88
2.38
ii
6.68
73
3
30 68
1.56
7
72.13
2.40
3'
7.07
75*
4
31.50
1.58
8
73-39
2.42
i
7-47
77
5
32.34
i. 60
9
74.66
2.44
2
7.88
79
6
33-iS
1.63
IO
75-94
2.46
3
8.30
.81
7
34-04
1.65
ii
77.24
2.48
4
8-73
83
8
34-91
1.67
10'
7854
2.50
TABLES.
6 9
to Tt ^ ^- MOOO w Ttmeo Ot M o O o
oi-' cnuir^oo N -rr^O v~> \r> o u^o^cs in
Nor->. in ir>
O c^oo ir>r^.o N
xr>\o O r^oo O'-i
WCM
....?
cococOTtTtinoo r^ O O*
co Tt mo t^oo OMcoinr^c>
CM TtO ooOCMin
CMcoOOco
O N 00 Tt
HHCOTto'oOOd
-toco O N Ttco NO Oooo Tt N O
nhH^-NNNNcocOTtTtinor^oo
N TtO oo m
M ci co TtO r^.
co coo
6
w N CM M CO CO
*
M M coTt-min
cor^O^i-i comr^O N coo co O^ O >-<
ooo r^i^.f^r^cococooococo o O^
N TtO CO
6
Tt mo t^co O* O
70 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
TABLE No. 9.
CHANNELS OF TRAPEZOIDAL SECTION. SIDE SLOPES 1 TO 1.
GIVING AREA AND VALUES OF r.
W-
i Ft.
W -
2 Ft.
W =
3 Ft.
arm
4 Ft.
Area.
r
Area.
r
Area.
r
Area.
r
0-5
0-75
0.31
1-25
0.37
i-75
0.40
2.25
0.42
75
1.32
.42
2.O7
-50
2.82
55
3-57
58
i
2
52
3
.62
4
.69
5
73
1-25
2.82
.62
4.07
-74
5-32
.81
6-57
.87
1.50
3-75
72
5-25
.84
6-75
93
8.25
1.75
4.81
.81
6.56
94
8.31
05
10.06
.12
2
6
.90
8
.04
IO
15
12
.24
2.25
7-35
99
9.60
.14
11.85
.26
14.10
36
2.50
8.75
.08
11.25
.24
13.75
37
16.25
47
2-75
10.30
17
13-05
33
15.80
47
18.55
57
3
12
.27
15
43
18
57
21
.63
3-50
15-75
45
19.25
.62
22.75
77
26.25
.89
4
20
63
24
.80
28
.96
32
2.09
4-50
24-75
.80
29.25
99
33-75
2.14
38.25
2.29
5
30
.98
35
2.17
40
2-33
45
2.48
W =
5 Ft.
w-
6 Ft.
W =
8 Ft.
IT.
10 Ft.
Area.
r
Area.
r
Area.
r
Area.
r
0-5
2-75
0-43
3-25
0-44
4-25
0.45
5-25
0.46
-75
4-32
.61
5-07
63
6.57
65
8.07
67
i
6
77
7
79
9
-83
II
.86
1-25
7.82
.92
9.07
95
n-57
14.07
.04
1.50
9-75
1.05
11-25
i. ii
14-25
.16
I7.25
.21
i-75
ii. 81
1.18
I3-56
1.24
17.06
32
20.56
38
2
14
.31
16
37
20
47
24
53
2.25
16.35
43
18.60
50
23.10
.61
27.60
.68
2-50
18.75
55
21.25
63
26.25
74
31-25
.83
2-75
21.30
.67
24.05
75
29-55
87
35.05
97
3
24
78
27
.88
33
2
39
2. II
29-75
2
33-25
2.09
40-25
2.25
47-25
2.38
4
36
2.21
40
2.31
48
2.48
56
2.6 3
4.50
42-75
2.41
47-25
2.52
56.25
2.71
65-25
2.87
5
50
2.61
55
2-73
65
2-94
75
3-io
TABLES.
TABLE No. 10.
CHANNELS OF TRAPEZOIDAL SECTION. SIDE SLOPES 2 TO
GIVING AREA AND VALUES OF r.
w =
i Ft.
w =
2 Ft.
W =
3 Ft.
w =
4 Ft.
Depth.
Area.
T
Area.
r
Area.
r
Area.
r
0.50
I
0.31
1-50
0-35
2
0.38
2.50
0.40
75
1.88
43
2.6 3
49
3-38
52
4-13
56
i
3
55
4
.62
5
.67
6
71
1.25
4-37
.66
5-62
74
6.87
.80
8.12
.84
1.50
6
78
7-50
.86
9
93
10.50
.98
1-75
7-87
.90
9.62
.98
n-37
-05
13-12
I. II
2
10
.01
12
- 10
14
17
16
1.24
2.25
12.40
.12
14.65
.21
16.90
.29
I9- I 5
.36
2-50
15
23
I7-50
33
20
.41
22.50
I. 4 8
2 75
17-85
34
20.60
44
23-35
53
26.10
.60
3
21
45
24
56
27
.64
30
72
3-50
28
.68
31-50
79
35
.87
38-50
.96
4
36
.90
40
2.OI
44
2. II
48
2.19
4-50
45
2.13
49.50
2.24
54
2-34
58.50
2.42
5
55
2.36
60
2.46
65
2.56
70
2.66
w =
5 Ft.
w =
6 Ft.
W ' =
8 Ft.
w =
10 Ft.
Depth.
Area.
r
Area.
r
Area.
T
Area.
r
0.50
3
0.41
3-50
0.42
4-50
0-44
5-50
o-45
75
4.88
58
5-6 3
.60
7-13
63
8.63
-65
i
7
74
8
. 7 6
IO
.80
12
-83
1.25
9-37
.88
10.62
.92
13-12
.96
15.62
1.50
12
1.02
I3-50
.06
16.50
.12
19.50
17
i-75
14.87
I- 15
16.62
.20
20.12
.27
23.62
33
2
18
1.29
20
34
24
-42
28
1.48
2.25
21.40
1.42
23.65
47
28.15
56
32.65
1.63
2.50
25
1-55
27.50
.60
32.50
.69
37-50
-77
2-75
28.85
1.67
31.60
1-73
37.10
83
42.60
.91
3
33
1-79
3 6
1.85
42
.96
48
2.05
3-50
42
2-03
45-50
2.10
52.50
2.22
59-50
2.32
4
52
2.27
56
2-34
6 4
2-47
72
2.58
4-50
63
2.51
67.50
2-59
76.50
2.72
85-50
2.84
5
75
2.74
80
2.82
90
2-97
100
3-09
72 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
TABLE No. 11.
CHANNELS OF TRAPEZOIDAL SECTION. SIDE SLOPES 3 TO 1.
GIVING AREA AND VALUES OF r.
W = i Ft.
W=-z Ft.
W = 3 Ft.
W = 4 Ft.
Depth.
Area.
r
Area.
r
Area.
r
Area.
r
0.50
1.25
0.30
1-75
0-34
2.25
0-37
2-75
0-39
75
2.44
43
3-19
47
3-94
57
4.69 .54
i
4
-55
5
.60
6
-65
7
.68
1.25
5-95
.67
7.20
73
8-45
.78
9.70
.82
1.50
8.25
79
9-75
85
11-25
.90
12-75
95
i-75
10.92
.91
12.67
97
14.42
.02
16. 17
.07
2
14
1.03
16
. 10
18
15
20
20
2.25
17-50
15
19-75
.21
22
.27
24.25
33
2.50
21.25
.26
23-75
.33! 26.25
39
28.75
45
2-75
25-50
-38
28.25
45
31
52
33-75
-57
3
30
50
33
57
36
.64
39
.70
3-50
40.25
74
43-75
1. 8l
47-25
.88
50.75
94
4
52
97
56
2.05
60
2. 12
64
2.18
4-50
65.30
2. 2O
69.80
2.28
74-30
2-35
78.80
2.42
5
30
2-45
85
2-53
90
2.60
95
2.67
W = 5 Ft.
W = 6 Ft.
W = 8 Ft.
W= 10 Ft.
Depth.
Area.
r
Area.
r
Area.
r
Area.
r
0.50
3-25
0.40
3-75
0.41
4-75
0-43
5-75
0.44
75
5-44
55
6. 19
58
7.69
.60
9.19
.62
i
8
70
9
73
n
77
13
.80
1.25
10.95
85
12.20
.88
14.70
93
17.20
.96
1.50
14.25
99
15-75
.02
18-75
07
21-75
1. 12
i-75
17.92
.12
19.67
15
23.17
.21
26.67
1.27
2
22
25
24
.29
28
.36
32
42
2.25
26.50
38
28.75
.42
33-25
49
37-75
-56
2.50
31-25
.50
33-75
55
38.75
63
43-75
70
2.75
36.50
63
39- 2 5
.68
44-75
.76
50-25
83
3
42
i-75
45
.80
5i
.89
57
97
3-50
54-25
2
57-75
2.05
64-75
2.15
71-75
2.24
4
68
2.24
72
2.30
80
2.40
88
2-49
4-50
83-30
2. 4 8
87.80
2-59
96.80
2.64
105.80
2-74
5
100
2-73
105
2.80 115
2.91
125
3-oi
TABLES.
TABLE No. 12.
CONVERSION TABLE OF PRESSURE IN POUNDS AND HEAD
IN FEET.
Pressure in
Pounds.
a
o w
F
Pressure in
Pounds.
l
^
Pressure in
Pounds.
c
T3 U
rt v
&
Pressure in
Pounds.
Head in
Feet.
Pressure in
Pounds.
jl
I
2.3
41
94.6
66
152.3
91
2IO
Il6
267.7
5
ii. 5 i
42
96.9
67
154 6
92
212.3
117
270
10
23-1 i
43
99.2
68
156.9
93
214.6
Il8
272.3
15
34-6
44
101.5
69
159.2
94
2l6.9
119
274.6
20.
46.2
45
103.9
70
161.6
95
219.2
120
276.9
21
48.5
46
106.2
71
163.9
96
221.5
121
279 2
22
50.8
47
108.5
72
166.2
97
223.9
122
281.6
23
53-i
48
no. 8
73
168.5
98
226.2
123
283.9
24
55-4
49
113-1
74
170.8
99
228.5
124
286.2
25
57-7
50
II5-4
75
I73-I
100
230.8
125
288.5
26
60
5i
117.7
76
175.4
101
233-1
130
300
27
62.3
52 .
1 20
77
177-7
102
235-4
135
3H.6
28
64.6
53
122.3
78
1 80
103
237-7
140
323.1
29
66.9
54
124.6
79
182.3
104
240
145
334-6
30
69.2
55
126.9
So
184.6
105
242.3
150
346.2
31
71-5
56V
129.2
81
186.9
106
244.6
155
357-7
32
73-9
57
I3I-5
82
189.2
107
2469
160
369.3
33
76.2
58
133.9
83
191.6
108
249.2
165
380.8
34
78-5
59
136.2
84
193-9
109
251.6
170
392.3
35
80.8
60
138.5
85
196.2
no
253-9
175
403.9
36
83.1
61
140.8
86
198.5
in
256.2
1 80
4I5.4
37
85.4
62
I43.I
87
200.8
112
258.5
185
427
38
87.7
63
145-4
88
203. r
H3
260.8
190
438.5
39
90
64
147.7
89
205.4
114
263.1
195
450.1
40
92.3
65
150
90
207.7
US
265.4
200
461.6
74 GRAPHICAL SOLUTION OF HYDRAULIC PROBLEMS.
c
'
1
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;
fe 1
5.
33 U
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1000 M
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sgs
8
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er/5 <u
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m ajjs
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sa
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lit
o
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t-s * O
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in
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3
5
TABLES.
75
I I
^U&? ! ?Q *
00 M I $
mOO
OT ! ~ M M
ID t^
O 00
N W T ft
~
O M I ro 4-
t^ O O M
in O
qSiH 1033 jaj 01
JIB j B SB jitnn
30 O O ^ tT% 10VO
as
00 ft I 00 10
oo r^ ! >o m
M o I *
^ O I -^
m io\o i oo o N
aH*
Hff
N ro
-3JIJ JIE,J B SB IJUi
M *
oo o o
10 t^OO
FRICTION HEAD
10 20
IN" FEET IN CLEAN PIPES
30 40
50
\
r
art
10
20
30
40
50
FLOW OF WATER IN LONG PIPES.
DIAGRAM No.
LIB
OF THF
UNIVERSITY
FRICTION HEAD IN FEET IN CLEAN PIPES
10 20 30 40
50
600
fx4
jgsTN
&U
\ \ \
m
<i,s
v->
1000
1200
z
1400
^
1600
10
20
30
40
50
FLOW OF WATER IN LONG PIPES.
DIAGRAM No. 3
FRICTION HEAD IN FEET IN CLEAN PIPES
10 20 30 40
50
8
vv>
XI
1000
XVv^
&\\
n ,i\
\>
2000
z
3500
4000
10
20
30
40
50
FLOW OF WATER IN LONG PIPES.
DIAGRAM No. 4
FRICTION HEAD IN FEET IN
10 20 30
CLEAN PIPES
40
50
1000
2000
3000
4000
5000
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FRICTION HEAD IN FEET IN CLEAN PIPES
10 20 30 40
50
10000
20000
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Is
LOOOO
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50000
60000
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10
30
50
FLOW OP WATER IN LONG PIPES.
DIAGRAM No. 16
FRICTION HEAD IN FEET IN CLEAN PIPES
10 20 30 40 50
1[- : 10000
20000
30000
40000
50000
60000
70000
80000
FLOW OF WATER
IN LONG PIPES.
DIAGRAM No. 17
-*p*^
UNIVERSITY
Of
HEAD
FLOW OF WATER IN SHOE
D IN FEET.
8 10
FLUSH ENDS.
12 14
2000
16000
2
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14
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HEAD
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ELUSH ENDS.
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es
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UNIVERSITY
PER. CENT. OF DISCHARGE.
70 80 ^90 100 110
BEE
60
70
80
90
100
110
DISCHARGE OF WIDE CREST WEIRS.
DIAGRAM No. 21
GALLONS
200
300
DISCHARGE
NS PER MIN.
400 500
600
700
800
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1.5
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3.5
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ECTANGULAR WEIRS.
DIAGRAM No. 22
2000
GALLONS
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6 8
CUBIC FEET
DISCHARGE OF EEC
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DIAGRAM No. 23
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FALL IN FE
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FEET PER. 1 00
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4000
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IARGE OF PIPE SEWERS.
DIAGRAM No. 27
10.
OF THK
UNIVERSITY
10
20
SCALE OF FP. |(
30
1000
6000
7000
20 30
SCALE OF FP. 75
HORSE i
GIVING HORSE POWER OF FALLING
REQUIRED TO PUMP WATER TO DIPl
WEIGHT OF WATER TAKEN AT 62
1 00 X- EFFICIENCY.
40 50
3() 40
.75% EFFICIENCY.
: POWER
ING WATER ALSO OF HORSE POWER
DIFFERENT HEIGHTS
I 62.4 LBS. PER. CUBIC FOOT.
DIAGRAM No. 28
5Q
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60
Q
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S 80
100
120
140
CO
160
180
200
300
70
70
500 100
60
60
50 4
REVOLUTIOI
50 4(
1500 2000 2500
GALLONS P
DISCHARGE OF S
FOR DUPLEX MULTIPLY BY 2.
LONS PER. MIN
400
500 600
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DIAGRAM No. 29
50
HEAD IN FEET.
100 ISO
200
250
7
50
150
200
250
COAL REQUIRED IN PUMPING.
1000000 GALLONS OF WATER. DIAGRAM No.30
50
HEAD IN FEET.
100 150
200
250
20
40 60 80
PRESSURE IN LBS.
DISCHARGE OF NOZZLES.
100
HLL CURVE FOR 50 HOSE.
OTTED 100'
PRESSURE INDICATED AT HYDRANT.
DIAGRAM No. 31
OF THK
UNIVERSITY
50
HEAD IN FEET.
100 150
200
250
20
PRESSURE IN LBS.
100
DISCHARGE OF NOZZLES.
PRESSURE INDICATED AT BASE OF PLAY PIPE. DIAGRAM
FOR LARGE NOZZLES MULTIPLY GALLONS BY 4. No. 32
VELOCITY IN FEET.
10 CD . 01 Ot CO
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UNIVEKSITY OF CALIFORNIA LIBRARY,
BERKELEY
THIS BOOK IS DUE ON THE LAST DATE
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expiration of loan period.
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