GIFT OF MICHAEL REE&E MECHANICS OF ENGINEERING. 12] [FLUIDS.] A TREATISE ON HYDRAULICS AND PNEUMATICS. FOR USE IN TECHNICAL SCHOOLS. BY IRVING P. CHURCH, C.E., ASSISTANT PROFESSOR OF CIVIL ENGINEERING, CORNELL UNIVERSITY. (IN CHARGE OF APPLIED MECHANICS.) NEW YORK: JOHN WILEY & SONS, 15 ASTOR PLACE. 1889. , Copyright, 1889, BY IRVING P. CHURCH. PnuMMOND & NBU, FERRIS BROS., Klectrotypera, Printers, to 7 Hague Street, 828 Pearl Street, New York. New Y ork. G. f. PREFACE. THE same general design has been kept in view in the prep- aration of the following work as in the preceding pages on Solids, viz. : to combine clearness and consistency in the setting forth and illustration of theoretical principles ; to provide nu- merous and fully-lettered diagrams, in which in most cases the notation of the accompanying text can be apprehended at a glance ; and to invite close attention to the proper use of systems of units in numerical examples, the latter being introduced very copiously and with detailed explanations. Advantage has been taken of the results of the most reeent experimental investigations in Hydraulics in assigning values of the numerous coefficients necessary in this science. The re- searches of Messrs. Fteley and Stearns in 1880 and of M. Bazin in 1887 on the flow of water over weirs, and of Mr. Clemens Herschel in testing his invention the "Venturi Water-meter/' are instances in point ; as also some late experiments on the transmission of compressed air and of natural gas. Though space has forbidden dealing at any great length with the action of fluid motors, 'sufficient matter is given in treating of the mode of working of steam, gas, and hot-air engines, air- compressors, and pumping-engines, together with numerical ex- amples, to be of considerable advantage, it is thought, to students not making a specialty of mechanical engineering. Special acknowledgment is due to Col. J. T. Fanning, the well-known author of " Hydraulic and Water-supply Engineer- ing," for his consent to the use of an abridgment of the table of coefficients for friction of water in pipes, given in that work ; and to Prof. C. L. Crandall, of this university, for permission to incorporate the chapter on Retaining-walls. References to original research in the Hydraulic Laboratory of the Civil Engineering Department at this institution will be found on pp. 694 and 729. CORNELL UNIVERSITY, ITHACA, N. Y., May 1889. i Hi CONTENTS. PAKT IV. HYDKAULICS. CHAPTER I. DEFINITIONS. FLUID PRESSURE. HYDRO- STATICS BEGUN. PAGE 406-417. Perfect fluids. Liquids and Gases. Principle of ' ' Equal Transmission of Pressure." Non-planar Pistons. . . 515 418-427. Hydraulic Press. Free Surface of Liquid. Barometers and Manometers. Safety-valves. Strength of Thin Hollow Cylinders against Bursting and Collapse. . . 526 CHAPTER II. HYDROSTATICS CONTINUED. PRESSURE OF LIQUIDS IN TANKS AND RESERVOIRS. 428-434. Liquid in Motion, but in "Relative Equilibrium." Pressure on Bottom and Sides of Vessels. Centre of Pressure of Rectangles, Triangles, etc 540 435-444. Stability of Rectangular and Trapezoidal Walls against Water Pressure. High Masonry Dams. Proposed Quaker Bridge Dam. Earthwork Dam. Water Pressure on both Sides of a Gate 554 CHAPTER III.-EARTH PRESSURE AND RETAINING WALLS. 445-455. Angle of Repose. Wedge of Maximum Thrust. Geo- metrical Constructions. Resistance of Retaining Walls. Results of Experience 572 CHAPTER IV. HYDROSTATICS CONTINUED. IMMERSION AND FLOTATION. 456-460. Buoyant Effort. Examples of Immersion. Specific Gravity. Equilibrium of Flotation. Hydrometer.. 586 461-465. Depth of Flotation. Draught and Angular Stability of Ships. The Metacentre 592 VI CONTENTS. CHAPTER V. HYDROSTATICS CONTINUED. FLUIDS. GASEOUS 466-478. Thermometers. Absolute Temperature. Gases and Va- pors. Critical Temperature. Law of Charles. Closed Air-manometer. Mariotte's Law. Mixture of Gases. Barometric Levelling. Adiabatic Change.. 604 479-489. Work Done in Steam-engine Cylinders. Expanding Steam. -Graphic Representation of Change of State of Gas. Compressed-air Engine. Air-compressor. Hot-air Engines. Gas-engines. Heat- efficiency. Duty of Pumpiug-engines.' Buoyant Effort of the Atmosphere 624 CHAPTER VI. HYDRODYNAMICS BEGUN. STEADY FLOW OF LIQUIDS THROUGH PIPES AND ORIFICES. 489a-495. \ 496-500. 501-508. 509-513. 513o-518. 519-526. S 527-536. Phenomena of a " Steady Flow." Bernoulli's Theorem for Steady Flow without Friction, and Applications. Orifice in " Thin Plate" 646 Rounded Orifice. Various Problems involving Flow through Orifices. Jet from Force-pump. Velocity and Density; Relation. Efflux under Water. Efflux from Vessel in Motion. Barker's Mill 663 Efflux from Rectangular and Triangular Orifices. Pon- celet's Experiments. Perfect and Complete Con- traction, etc. Overfall Weirs. Experiments of Francis, Fteley and Stearns, and Bazin. Short Pipes or Tubes : . . 672 Conical Tubes. Venturi's Tube. Fluid Friction. Froude's Experiments. Bernoulli's Theorem with Friction. Hydraulic Radius. Loss of Head. Prob- lems involving Friction Heads in Pipes. Accumu- lator 692 Loss of Head in Orifices and Short Pipes. Coefficient of Friction of Water in Pipes. Fanning 's Table. Petroleum Pumping. Flow through Long Pipes. . 70& Chezy's Formula. Fire-engine Hose. Pressure-energy. Losses of Head due to Sudden Enlargement of Sec- tion; Borda's Formula. Diaphragm in Pipe. Ven- turi Water-meter 714 Sudden Diminution of Section. Losses of Head due to Elbows, Bends, Valve-gates, and Throttle-valves. Examples, Prof. Bellinger's Experiments on Elbows. Siphons. Branching Pipes. Time of Emptying Vessels of Various Forms; Prisms, Wedges. Pyra- mids, Cones, Paraboloids, Spheres, Obelisks, and Volumes of Irregular Form using Simpson's Rule. . 727 CONTENTS. Vil PAGE CHAPTER VII. HYDRODYNAMICS, CONTINUED; STEADY FLOW OF WATER IN OPEN CHANNELS. : 538-542a. Nomenclature. Velocity Measurements and Instru- ments for the same. Ritchie-Haskell Direction Current-meter. Change of Velocity with Depth. Pitot's Tube. Hydrometric Pendulum. Wolt- mann's Mill. Gauging Streams. Chezy's Formula for Uniform Motion in Open Channel. Experi- ments 749 . 5425-547. Kutter's Formula. Sections of Least Resistance. Trape- zoidal Section of Given' Side Slope and Minimum Friction. Variable Motion in Open Channel. Bends. Formula introducing Depths at End Sec- tions. Backwater 759 CHAPTER VIII. DYNAMICS OF GASEOUS FLUIDS. 548-556. Theorem for Steady Flow of Gases without Friction. Flow through Orifices by Water-formula; with Isothermal Expansion; with Adiabatic Expansion. Maximum Flow of Weight. Experimental Co- efficients for Orifices and Short Pipes. Flow con- sidering Velocity of Approach 773 557-561&. Transmission of Compressed Air through Long Pipes. Experiments in St. Gothard Tunnel. Pipes of Vari- able Diameter. The Piping of Natural Gas 786 CHAPTER IX. IMPULSE AND RESISTANCE OF FLUIDS. 563-569. Reaction of a Jet of Liquid. Impulse of Jet on Curved Vanes, Fixed and in Motion. Pitot's Tube. The California "Hurdy-gurdy." Impulse on Plates. Plates Moving in a Fluid. Plates in Currents of Fluid 798 570-575. Wind-pressure. Smithsonian Scale. Mechanics of the Sail-boat. Resistance of Still Water to Immersed Solids in Motion. Spinning Ball, Deviation from Vertical Plane. Robinson's Cup anemometer. Re- sistance of Ships. Transporting Power of a Cur- rent.., . 818 PART IV. HYDRAULICS. CHAPTEK I. DEFINITIONS FLUID PRESSURE HYDROSTATICS BEGUN. 406. A Perfect Fluid is a substance the particles of which are capable of moving upon each other with the greatest free dom, absolutely without friction, and are destitute of mutual attraction. In other words, the stress between any two con- tiguous portions of a perfect fluid is always one of compression and normal to the dividing surface at every point ; i.e., no shear or tangential action can exist on any imaginary cutting plane. Hence if a perfect fluid is contained in a vessel of rigid ma- terial the pressure experienced by the walls of the vessel is normal to the surface of contact at all points. For the practical purposes of Engineering, water, alcohol, mercury, air, steam, and all gases may be treated as perfect fluids within certain limits of temperature. 407. Liquids and Gases. A fluid a definite mass of which occupies a definite volume at a given temperature, and is in- capable both of expanding into a larger volume and of being compressed into a smaller -volume at that temperature, is called a Liquid, of which water, mercury, etc., are common examples ; whereas a Gas is a fluid a mass of which is capable of almost indefinite expansion or compression, according as the space within the confining vessel is made larger or smaller, and al- ways tends to fill the vessel, which must therefore be closed in every direction to prevent its escape. 515 516 MECHANICS OF ENGINEERING. Liquids are sometimes called inelastic fluids, and gases elastic fluids. 408. Remarks. Though practically we may treat all liquids as incompressible, experiment shows them to be compressible to a slight extent. Thus, a cubic inch of water under a pres- sure of 15 Ibs. on each of its six faces loses only fifty millionths (0.000050) of its original volume, while remaining at the same temperature; if the temperature be sufficiently raised, how- ever, its bulk will remain unchanged (provided the initial tem- perature is over 40 Fahr.). Conversely, by heating a liquid in a rigid vessel completely filled by it, a great bursting pressure may be produced. The slight cohesion existing between the particles of most liquids is too insignificant to be considered in the present connection. The property of indefinite expansion, on the part of gases, by which a confined mass of gas can continue to fill a confined space which is progressively enlarging, and exert pressure against its walls, is satisfactorily explained by the " Kinetic Theory of Gases," according to which the gaseous particles are perfectly elastic and in continual motion, impinging against each other and the confining walls. Nevertheless, for prac- tical purposes, we may consider a gas as a continuous sub- stance. Although by the abstraction of heat, or the application of great pressure, or both, all known gases may be reduced to liquids (some being even solidified); and although by con- verse processes (imparting heat and diminishing the pressure) liquids may be transformed into gases, the range of tempera- ture and pressure in all problems to be considered in this work is supposed kept within such limits that no extreme changes of state, of this character, take place. A gas approaching the point of liquefaction is called a Vapor. Between the solid and the liquid state we find all grades of intermediate conditions of matter. For example, some sub- stances are described as soft and plastic solids, as soft putty, moist earth, pitch, fresh mortar, etc.; and others as viscous and sluggish liquids, as molasses and glycerine. In sufficient bulk, DEFINITIONS FLUID PRESSURE HYDROSTATICS. 517 however, the latter may still be considered as perfect fluids. Even water is slightly viscous. 409, Heaviness of Fluids. The weight of a cubic unit of a homogeneous fluid will be called its heaviness, or rate of weight (see 7), and is a measure of its density. Denoting it by y, and the volume of a definite portion of the fluid by V, we have, for the weight of that portion, G= Vy. (1) This, like the great majority of equations used or derived in this work, is of homogeneous form ( 6), i.e., admits of any sys- tem of units. E.g., in the metre-kilogram-second system, if y is given in kilos, per cubic metre, V must be expressed in cubic metres, and G will be obtained in kilos.; and similarly in any other system. The quality of y, = G -f- F, is evidently one dimension of force divided by three dimensions of length. In the following table, in the case of gases, the temperature and pressure are mentioned at which they have the given heaviness, since under other conditions the heaviness would be different ; in the case of liquids, however, for ordinary pur- poses the effect of a change of temperature may be neglected (within certain limits). HEAVINESS OF VARIOUS FLUIDS.* [In ft. Ib. sec. system; y = weight in Ibs. of a cubic foot.] Liquids. J At temp, of melting ice; and 14.7 1 Ibs. per sq. in. tension. Freshwater, y= 62.5 Sea water 64.0 Mercury 848. 7 Alcohol 49.3 Crude Petroleum, about 55.0 (N.B. A cubic inch of water weighs 0.036034 Ibs.; and a cubic foot 1000 av. oz.) Atmospheric Air 0. 08076 Oxygen 0.0892 Nitrogen 0.07'86 Hydrogen 0.0056 Illuminating ) from 0.0300 Gas, fto 0.0400 Natural Gas, about 0.0500 * See Trautwine's Civ. Engineer's Pocket Book for an extended table p. 380, edition of 1885. 518 MECHANICS OF ENGINEERING. For use in problems where needed, values for the heaviness of pure fresh water are given in the following table (from Rossetti) for temperatures ranging from freezing to boiling ; as also the relative density, that at the temperature of maxi- mum density, 39. 3 Fahr. being taken as unity. The temper- atures are Fahr., and y is in Ibs. per cubic foot. Temp. Rel. Dens. Y. Temp. Rel. Dens. y. Temp. Rel. Dens. V- 32 .99987 62.416 60 .99907 62.366 140 .98338 61.386 35 .99996 62.421 70 .99802 62.300 150 .98043 61.203 39. 3 1.00000 62.424 80 .99669 62.217 160 .97729 61.006 40 .99999 62.423J 90 .99510 62.118 170 .97397 60.799 43 .99997 62.422 100 .99318 61.998 180 .97056 60.586 45 .99992 62.419 110 .99105 61.865 190 .96701 60.365 50 .99975 62.408 120 .98870 61.719 200 .96333 60.135 55 .99946 62.390 130 .98608 61.555 212 .95865 59.843 EXAMPLE 1. What is the heaviness of a gas, 432 cub. in. of which weigh 0.368 ounces? Use ft.-lb.-sec. system. 432 cub. in. = \ cub. ft. and 0.368 oz. = 0.023 Ibs. = 0.092 Ibs. per cub. foot. EXAMPLE 2. Required the weight of a right prism of mer- cury of 1 sq. inch section and 30 inches altitude. OA V =30 X 1 = 30 cub. in. = -^-- cub. feet ; while from the 1728 table, y for mercury = 848.7 Ibs. per cub. ft. OA .-. its weight = G = Vy = -%- X 848.7 = 14.73 Ibs. 410. Definitions. By Hydraulics we understand the me- chanics of fluids as utilized in Engineering. It may be divided into Hydrostatics, treating of fluids at rest ; and Hydrodynamics (or Hydrokinetics), which deals with fluids in motion. (The name Pneumatics is sometimes used to cover both the statics and dynamics of gaseous fluids.) DEFINITIONS FLUID PRESSURE HYDROSTATICS. 519 [Rankine's nomenclature has been adopted in the present work. Some recent writers use the term Hydromechanics for mechanics of fluids, subdividing it into Hydrostatics and Hydrokinetics, as above ; they also use the term Dynamics to embrace both of the two divisions called Statics and Dynamics by Rankine, which by them are called Statics and Kinetics re- spectively. Though unusual, perhaps, the term Hydraulics is here used to cover the applied Mechanics of Fluids as well as of Liquids.] Before treating separately of liquids and gases, a few para- graphs will be presented applicable to both kinds of fluids. 411. Pressure per Unit Area, or Intensity of Pressure. As in 180 in dealing with solids, so here with fluids we indicate the pressure per unit area between two contiguous portions of fluid, or between a fluid and the wall of the containing vessel, by p, so that if dP is the total pressure on a small area dF, we have >- ........ as the pressure per unit area, or intensity of pressure (often, though ambiguously, called the tension in speaking of a gas) on the small surface dF. If pressure of the same intensity exists over a finite plane surface of area = F, the total pres- sure on that surface is P = fpdF=pfdF= Fp, ] P [ . (2)" = >. (!N".B. For brevity the single word " pressure" will some- times be used, instead of intensity of pressure, where no am- biguity can arise.) Thus, it is found that, under ordinary con- ditions at the sea level, the atmosphere exerts a normal pressure (normal, because fluid pressure) on all surfaces, of an intensity of about p 14.7 Ibs. per sq. inch (= 2116. Ibs. per sq. ft.). This intensity of pressure is called one atmosphere. For ex- 520 MECHANICS OF ENGINEERING. ample, the total atmospheric pressure on a surface of 100 sq. in. is [inch, lb., sec.] P = Fp = 100 X 14.7 = 1470 Ibs. (= 0.735 tons.) The quality of p is evidently one dimension of force divid- ed by two dimensions of length. 412. Hydrostatic Pressure; per Unit Area, in the Interior of a Fluid at Rest. In a body of fluid of uniform heaviness, at rest, it is required to find the mutual pressure per unit area be- tween the portions of fluid on opposite sides of any imaginary cutting plane. As customary, we shall consider portions of the fluid as free bodies, by supplying the forces exerted on them by all contiguous portions (of fluid or vessel wall), also- those of the earth (their weights), and then apply the condi- tions of equilibrium. First) cutting plane horizontal. Fig. 451 shows a body of homogeneous fluid confined in a rigid vessel closed at the top with a small air- tight but frictionless piston (a horizontal disk) of weight = G and exposed to at- mospheric pressure ( ^? a per unit area) on its upper face. Let the area of piston- face be = F. Then for the equilibrium of the piston the total pressure between its under surface and the fluid at must be FIG. 451. and hence the intensity of this pressure is (1) It is now required to find the intensity,^, of fluid pressure between the portions of fluid contiguous to the horizontal cut- ting plane BC&i a vertical distance = h vertically below the pis- ton O. In Fig. 452 we have as a free body the right parallelo- FLUID PRESSURE. olliil piped OBC oi Fig. 451 with vertical sides (two || to paper and four ~] to it). The pressures acting on its six faces are normal to them respectively, and the weight of the prism is vol. Xy = Fhy, supposing y to have the same value at all parts of the column (which is practically true for any height of liquid and for a small height of gas). Since the ^-^ ^ n prism is in equilibrium under the forces shown in the figure, and would still be so were it to become rigid, we may put ( 36) .2 (vert, compons.) = and hence obtain Fp-Fp.- Fhy = 0. . . (2) J (In the figure the pressures on the ver- tical faces || to paper have no vertical com- ponents, and hence are not drawn.) From FIG. 453. (2) we have P =P + (3) (hy, being the weight of a column of homogeneous fluid of unity cross-section and height A, would be the total pressure on the base of such a column, if at rest and with no pressure on the upper base, and hence might be called intensity due to iveigfit.) Secondly, cutting plane oblique. Fig. 453. Consider free an infinitely small right triangular prism bed, whose bases are || to the. paper, while the three side faces (rectangles), having areas = dF, dF^ , and dF^ , are respectively hori- zontal, vertical, and oblique ; let angle cbd = a. The surface be is a portion of the plane BC of Fig. 452. Given p (= intensity of pressure on dF) and a, required^, the intensity of pressure on the oblique face bd, of area dJF 1 ^ [N. B. The prism is taken very small in order that the intensity of pressure may be considered con- stant over any one face ; and also that the weight of the prism may be neglected, since it involves the volume (three dimen- 522 MECHANICS OF ENGINEERING. sions) of the prism, while the total face pressures involve only two, and is hence a differential of a higher order.] From 2 (vert, compons.) = we shall have pdF = ; but dF '--f- dF t = cos a ; which is independent of the angle a. Hence, the intensity of fluid pressure at a given point is the same on all imaginary cutting planes containing the point. This is the most important property of a fluid, and is true whether the liquid is at rest or has any kind of motion ; for, in case of rectilinear accelerated motion, e.g., although the sum of the force-components in the direction of the accelera- tion does not in general = 0, but mass X ace., still, the mass of the body in question is = weight -v- 100 feet (for, gas being compressible, the lower strata are generally more dense than the upper), but in (3) the pistons must be fixed, and P e and P Q refer solely to the in- terior pressures. 524 MECHANICS OF ENGINEERING. Again, if A is small or p very great, the term hy may be omitted altogether in eqs. (2) and (3) (especially with gases, since for them y (heaviness) is usually small), and we then have, from (2), P=P.\ (4) being the algebraic form of the statement: A lody of fluid at rest transmits pressure with equal intensity in every direc- tion and to all of its parts. [Principle of "Equal Transmis- sion of Pressure."] 414. Moving Pistons. If the fluid in Fig. 454 is inelastic and the vessel walls rigid, the motion of one piston (o) through a distance s causes the other to move through a distance s e de- termined by the relation F s F e s e (since the volumes de- scribed by them must be equal, as liquids are incompressible) ; but on account of the inertia of the liquid, and friction on the vessel walls, equations (2) and (3) no longer hold exactly, still are approximately true if the motion is very slow and the vessel short, as with the cylinder of a water-pressure engine. But if the fluid is compressible and elastic (gases and vapors ; steam, or air) and hence of small density, the effect of inertia and friction is not appreciable in short wide vessels like the cylinders of steam- and air-engines, and those of air-compres- sors ; and eqs. (2) and (3) still hold, practically, even with high piston-speeds. For exam pie, in the space AB, Fig. 455, between the piston and cylinder-head of a steam-engine (piston moving toward the right) the intensity of pressure, J9, of the steam against the moving piston B is prac- FIG. 455. tically equal to that against the cylinder-head A at the same instant. 415. An Important Distinction between gases and liquids (i.e., between elastic and inelastic fluids) consists in this : A liquid can exert pressure against the walls of the contain- ing vessel only by its weight, or (when contined on all sides) by transmitted pressure coming from without (due to piston pressure, atmospheric pressure, etc.); whereas FLUID PRESSURE. 525 A gas, confined, as it must be, on all sides to prevent dif- fusion, exerts pressure on the vessel not only by its weight, but by its elasticity or tendency to expand. If pressure from without is also applied, the gas is compressed and Exerts a still greater pressure on the vessel walls. \ I 416. Component, of Pressure, in a Given Direction, Let ABCD, whose area = dF, be a small element of \a surface, plane or curved, and j9 the intensity of fluid pressure upon this element, then the total pressure upon it \$>pdF, and is of course normal to it. Let A' B' CD be the projection of the element dF upon c \^ /\\fe--''* a plane CDM. making an angle a with the element, and let it be required to find the value of the component of pdF in a direction normal to this last plane (the other component being understood to be || to the same plane). We shall have < Compon. ofpdF ~\ to CDM = pdFcos a = p(dFiOQ). (1)"' But dF . cos a = area A'B'CD, the projection of the plane CDM. .*. Compon. "1 to plane CDM =p X (project. ofdFon CDM]\ i.e., the component of fluid pressure (on an element of a sur- face) in a given direction (the other component being "I to the first) is found ly multiplying the intensity of the pressure ty the area of the projection of the element upon a plane ~\ to the given direction, v It is seen, as an example of this, that if the fluid pressures on the elements of the inner surface of one hemisphere of a hollow sphere containing a gas are resolved into components 1 and || to the plane of the circular base of the hemisphere, the sum of the former components simply 7rr*p, where r is the radius of the sphere, and^> the intensity of the fluid pressure ; for, from the foregoing, the sum of these components is just the same as the total pressure would be, having an intensity p > 526 MECHANICS OF ENGINEERING. on a great circle of the sphere, the area, m?, of this circle being the sum of the areas of the projections, upon this circle as a base, of all the elements of the hemispherical surface. (Weight of fluid neglected.) A similar statement may be made as to the pressures on e inner curved surface of a right cylinder. 417. Non-planar Pistons. From the foregoing it follows that I/ >fehe sum of the components || to the piston-rod, of the fluid i/ pressures upon the piston at A, Fig. 457, is just the same as at B, if the cylinders are of equal size and the steam, or air, is at the same tension. For the sum of the projections of all the elements of the curved surface of A upon a plane ~| to the piston-rod is always =: nr* = area of section of cylinder-bore. SP |^! : .v5 S If the surface of A is symmetrical about the axis of the cylin- der the other components (i.e., those ~| to the piston-rod) will neutralize each other. If the line of intersection of that sur- face with the surface of the cylinder is not symmetrical about the axis of the cylinder, the piston may be pressed laterally against the cylinder-wall, but the thrust along the rod or " working force" 1 ( 128) is the same (except for friction in- duced by the lateral pressure), in all instances, as if the surface were plane and ~] to piston-rod. 418. Bramah, or Hydraulic, Press, This is a familiar instance of the principle of transmission of fluid pressure. Fig. 458. Let the small piston at have a diameter d = 1 inch = -^ ft., while the plunger E, or large piston, has a diameter d' = AE = CD = 15 in. = ft. The lever MJV weighs G, = 3 IDS., and a weight G = 40 Ibs. is hung at M. The lever-arms of these forces about the fulcrum N are given in the figure. The apparatus being full of water (oil is often used), the fluid pressure P against the small piston is found by putting FLUID PRESSURE. 52T ^(moms. about 2V) = for the equilibrium of the lever; whence [ft., lb., sec.] p o x i _ 40 X 3 - 3 X 2 = 0. /. P = 126 Ibs. * FIG. 458. But, denoting atmospheric pressure by p a , and that of the water against the piston by p (per unit area), we may also write Solving for p , we have, putting p a 14.7 X 144 Ibs. per eq. ft, r 1 = 126 -f- (^M + 14.7 X 144 = 25236 Ibs. per sq. ft. Hence at e the press, per unit area, from 409, and (2), 413, is p e =p n -f- hy = 25236 + 3 X 62.5 = 25423 Ibs. per sq. ft. = 175.6 Ibs. per sq. inch or 11.9 atmospheres, and the total upward pressure at e on base of plunger is n ^p e = 7t(*-y x 25423 = 31194 Ibs, or almost 16 tons (of 2000 Ibs. each). The co repressive force upon the block or bale, C, = P less the weight of the plunger and total atmos. pressure on a circle of 15 in. diameter. 628 MECHANICS OF ENGINEEKING. 419. The Dividing Surface of Two Fluids (which do not mix) in Contact, and at Rest, is a Horizontal Plane. For, Fig. 459, sup- posing any two points e and O of this sur- face to be at different levels (the pressure at being _> , that at ep e , and the heavi- nesses of the two fluids y l and y^ respec- tively), we would have, from a considera- tion of the two elementary prisms eb and bO (vertical and horizontal), the relation . FIG. 459.- while from the prisms ec and cO, the relation These equations are conflicting, hence the aoove supposition is absurd. Therefore the proposition is true. For stable equilibrium, evidently, the heavier fluid must oc- cupy the lowest position in the vessel, and if there are several fluids (which do not mix), they will arrange themselves vertically, in the order of their den- sities, the heaviest at the bottom, Fig. 460. On account of the property called diffusion the par- ticles of two gases placed in contact soon inter- mingle and form a uniform mixture. This fact gives strong support to the " Kinetic Theory of Gases" ( 408). FIG. 460. 420. Free Surface of a Liquid at Rest. The surface (of a liquid) not in contact with the walls of the containing vessel is called a free surface, and is necessarily horizontal (from 419) when the liquid is at rest. Fig. 461. (A gas, from its tendency to indefinite expansion, is incapable of hav- ing a free surface.) This is true even if the space above the liquid is vacuous, for if the surface were inclined or curved, points in the body of the liquid and in the same horizon- tal plane would have different heights (or " heads") of liquid FIG. 461. TWO LIQUIDS IN BENT TUBE. 529 between them and the surface, producing different intensities of pressure in the plane, which is contrary to 413. When large bodies of liquid like the ocean are considered, gravity can no longer be regarded as acting in parallel lines ; consequently the free surface of the liquid is curved, being ~| to the direction of (apparent) gravity at all points. For ordi- nary engineering purposes (except in Geodesy) the free surface of water at rest is a horizontal plane. 421. Two Liquids (which do not mix) at Rest in a Bent Tube open at Both Ends to the Air, Fig. 460 ; water and mercury, for instance. Let their heavinesses be Yl and YI respectively. The pressure at e may be written ( 413) either or according as we refer it to the water column or the mercury column and their respective free surfaces where the pressure JPO, =^o a = p a atmos. press. e is the surface of contact of the two liquids. Hence we have i.e., the heights of the free surfaces of the two liquids above the surface of contact are inversely proportional to their respec- tive heavinesses. EXAMPLE. If the pressure at e = 2 atmospheres ( 396) we shall have from (1) (inch-lb.-sec. system of units) h^ =p e p a = 2x 14.7 14.7 14.7 Ibs. per sq. inch. .-. A 2 must = 14.7 -r- [848.7 + 1728] = 30 inches (since, for mercury, y z = 848.7 Ibs. per cub. ft.). Hence, from (3), KY* 30 X [848.7 -*- 1728] 530 MECHANICS OF ENGINEERING. i.e., for equilibrium, and that p e may = 2 atmospheres, h t and A a (of mercury and water) must be 30 in. and 34 feet respec- tively. 422. City Water-pipes. If h = vertical distance of a point B of a water-pipe below the free surface of reservoir, and the water be at rest, the pressure on the inner surface of the pipe at B (per unit of area) is p =p + hy ; and here j? =p a = atmos. press. EXAMPLE. If h = 200 ft. (using the inch, lb., and second) p = 14.7 + [200 X 12] [62.5 + 1728] = 101.5 Ibs. per sq. in. The term hy, alone, = 86.8 Ibs. per sq. inch, is spoken of as the hydrostatic pressure due to 200 feet height, or *\Head," of water. (See Trautwine's Pocket Book for a table of hydro- static pressures for various depths.) If, however, the water \& flowing through the pipe, the pres- sure against the interior wall becomes less (a problem of Hy- drodynamics to be treated subsequently), while if that motion is suddenly checked, the pressure becomes momentarily much greater than the hydrostatic. This shock is called "water- ram" and " water-hammer," and may be as great as 200 to 300 Ibs. per sq. inch. 423. Barometers and Manometers for Fluid Pressure. If a tube, closed at one end, is filled with water, and the other ex- tremity is temporarily stopped and afterwards opened under water, the closed end being then a (vertical) height = h above the surface of the water, it is required to find the intensity, p , of fluid pressure at the top of the tube, sup- posing it to remain filled with water. Fig. 463. At E inside the tube the pressure is 14.7 Ibs. per sq. inch, the same as that outside at the same level ( 413) ; hence, from p E = p BAROMETERS. 531 EXAMPLE. Let h = 10 feet (with inch-lb.-sec. system) ; then p = 14.7 120 X [62.5 -j- 1728] = 10.4 Ibs. per sq. inch, or about of an atmosphere. If now we inquire the value of h to make p = zero, we putp E hy = and obtain h = 408 inches, = 34 ft., which is called the height of the water- barometer. Hence, Fig. 463$, ordinary atmospheric pressure will not sustain a column of water higher than 34 feet. If mercury is used instead of water the height supported by one atmosphere is I = 14.7 -=- [848.7 -f- 1728] = 30 inches, = 76 centims. (about), and the tube is of more manageable proportions than with water, aside from the ad- vantage that no vapor of mercury forms above the liquid at ordinary temperatures [In fact, the water-barometer height b = 34 feet has only a theoretical existence since at ordinary tempera- tures (40 to 80 Fahr.) vapor of water would form above the column and depress it by from 0.30 to 1.09 ft.]. Such an apparatus is called a Barometer ', and is used not only for measuring the varying tension of the atmosphere (from 14.5 to 15 Ibs. per sq. inch, according to the weather and height above sea-level), but also that of any body of gas. Thus, Fig. 464, the gas in D is put in communication with the space above the mercury in the cistern at (7; and we have p hy, where y = heav. of mercury, and JP is the pressure on the liquid in the cistern. For delicate measurements an at- tached thermometer is also used, as the heavi- ness Y varies slightly with the temperature. If the vertical distance CD is small, the ten- sion in C is considered the same as in D. For gas-tensions greater than one atmosphere, the tube may be left open at the top, forming an open ma- FIG. 463a. FIG. 464. 532 MECHANICS OF ENGINEERING. nometer, Fig. 465. In this case, the tension of the gas above the mercury in the cistern is (i) FIG. 465. in which ~b is the height of mercury (about 30 in.) to which the tension of the atmosphere above the mercury column is equivalent. EXAMPLE. If h 51 inches, Fig. 465, we have (ft., lb., sec.) p = [4.25 ft. + 2.5 ft.] 848.7 = 5728 Ibs. per sq. foot = 39.7 Ibs. per sq. inch = 2.7 atmospheres. Another form of the open manometer consists of a U tube, Fig. 464, the atmosphere having access to one branch, the gas to be examined, to the other, while the mercury lies in the curve. As before, we have (2) tn where p a = atmos. tension, and b as above. The tension of a gas is sometimes spoken of as measured by so many inches of mer- cury. For example, a tension of 22.05 FIG. 466. Ibs. .per sq. inch (1J atmos.) is measured by 45 inches of mer- cury -in a vacuum manometer (i.e., a common barometer), Fig. 464. "With the open manometer this tension (1 atmos.) would be indicated by 15 inches of actual mercury, Figs. 465 and 466. An ordinary steam-gauge indicates the 'excess of tension over one atmosphere ; thus " 40 Ibs. of steam" implies a tension of 40 + 14.7 = 54.7 Ibs. per sq. in. The Bourdon steam-gauge in common use consists of a curved elastic metal tube of flattened or elliptical section (with the long axis ~] to the plane of the tube), and has one end fixed. The movement of the other end, which is free and TENSION OF GASES. 533 closed, by proper mechanical connection gives motion to the pointer of a dial. This movement is caused by any change of tension in the steam or gas admitted, through the fixed end, to O * O the interior of the tube. As the tension increases the ellip- tical section becomes less flat, i.e., more nearly circular, caus- ing the two ends of the tube to separate more, widely, i.e., the free end moves away from the fixed end ; and vice versa. Such gauges, however, are not always reliable. They are graduated by comparison with mercury manometers; and should be tested from time to time in the same way. 424. Tension of Illuminating Gas. This is often spoken of as measured by inches of water (from 1 to 3 inches usually). Strictly it should be stated that this water-height measures the excess of its tension over that of the atmos- phere. Thus, in Fig. 466, water being used instead of mercury, h = say 2 inches, while b = 408 inches. This difference of tension may be largely affected by a change in the barometer due to the weather/ or by a difference in altitude, as the follow- ing example will illustrate : EXAMPLE. Supposing the gas at rest, and the tension at the gasometer A, Fig. 467, to be "two inches of water," required the water-column h" (in open tube) that the gas will support in the pipe at B, 120 feet (vertically) above the gasometer. Let the temperature be freezing (nearly), and the outside air at a tension of 14.7 Ibs. per sq. inch ; the heaviness of the gas at this temperature being 0.036 Ibs. per cubic foot. For the small difference of 120 ft. we may treat both the atmosphere and the gas as liquids, that is, of constant density throughout the vertical column, and therefore apply the principles of 413 ; with the following result : The tension of the outside air at JS y supposed to be at the same temperature as at A, will sustain a water-column less than the 408 inches at A by an amount corresponding to the FIG. 467. 534 MECHANICS OF ENGINEERING. 120 feet of air between, of the heaviness .0807 Ibs. per cub. ft. 120 feet of air weighing .0807 Ibs. per cub. ft. will balance 0.154 ft. of water weighing 62.5 Ibs. per cubic ft., i.e., 1.85 inches of water. Now the tension of the gas at B is also less than its tension at J., but the difference is not so great as with the outside air, for the 120 ft. of gas is lighter than the 120 ft. of air. Since 120 ft. of gas weighing 0.036 Ibs. per cubic ft. will balance 0.0691 ft., or 0.83 inches, of water, therefore the difference between the tensions of the two fluids at B is greater than at A by (1.85 0.83 = ) 1.02. inches; or, at B the total difference is 2.00 + 1.02 = 3.02 inches. v Hence if a small aperture is made in the pipe at B the gas will flow out with greater velocity than at A. At Ithaca, !N. Y., where the University buildings are 400 ft. above the gas-works, this phenomenon is very marked. "When the difference of level is great the decrease of tension as we proceed upward in the atmosphere, even with constant temperature, does not follow the simple law of 413; see 477. For velocity of flow of gases through orifices, see 548, etc. 425. Safety-valves. Fig. 468. Kequired the proper weight G to be hung at the extremity of the horizontal lever AB Y with fulcrum at B, that the flat disk- valve E shall not be forced upward by the steam pressure, p' , until the latter reaches a given value p. Let the weight of the arm be 6r, , its centre of grav- ity being at 7, a distance = o from B ; the other horizontal distances are marked in the figure. 1 Suppose the valve on the point of rising ; then the forces acting on the lever are the fulcrum-reaction at B, the weights G and G l , and the two fluid-pressures on the disk, viz. : Fp a (atmospheric) downward, and Fp (steam) upward. Hence, from ^(moms. B ) = 0, Gb + G,c + Fp a a - Fpa = 0. ... (1) BURSTING OF PIPES. 535 Solving, we have (2) EXAMPLE. With a = 2 inches, = 2 feet, c = 1 foot G l = 4 Ibs.,j9 = 6 atmos., and diam. of disk = 1 inch; with the foot and pound, X m X 144 - 1 X 14.7 X 144] -4 .-. G = 2.81 Ibs. [Notice the cancelling of the 144; for F(p p^) is pounds, being one dimension of force, if the pound is selected as the unit of force, whether the inch or foot is used in both fac- tors.] Hence when the steam pressure has risen to 6 atmos. (= 88.2 Ibs. per square inch) (corresponding to 73.5 Ibs. persq. in. by steam-gauge) the valve will open if G = 2.81 Ibs., or be on the point of opening. 426. Proper Thickness of Thin Hollow Cylinders (i.e,, Pipes and Tubes) to Resist Bursting by Fluid Pressure. CASE I. /Stresses in the cross-section due to End Pressure; Fig. 469. Let AB be the circular cap clos- ing the end of a cylindrical tube containing fluid at a tension = p. Let r = internal radius of the tube or pipe. Then considering the cap free, neglecting its weight, we have three sets of || forces in equilibrium (see II in figure), viz.: the iuternal fluid pres- sure = 7tr*p-, the external fluid pressure = nr*p a ; while the total stress (tensile) on the small ring, whose area now exposed is 2 nrt (nearly), is = < ^nrtp l , where t is the thickness of the pipe, and j^ the tensile stress per unit area induced by the end-pres- sures (fluid). 636 MECHANICS OF ENGINEERING. For equilibrium, therefore, we may put ^(hor. comps.) = ; i = 0; ..... (1) _ r (p -p a ) Pi ^ (Strictly, the two circular areas sustaining the fluid pressures are different in area, but to consider them equal occasions but a small error.) Eq. (1) also gives the tension in the central section of a thin hollow sphere, under bursting pressure. CASE II. Stresses in the longitudinal section of pipe, due to radial fluid pressures.* Consider free the half (semi-circular) of any length I of the pipe, be- tween two cross-sections. Take an axis X (as in Fig. 470) "| to the longitudinal section which has been made. Let p z denote the tensile stress (per unit area) produced in the narrow rectangles exposed at A and B (those in the half-ring edges, having no X components, are not drawn in the figure). On the in- ternal curved surface the fluid pres- sure is considered of equal intensity = ^> at all points (practically true even with liquids, if 2r is small compared with the head of water producing p). The fluid pressure on any dF or elementary area of the internal curved surface is = pdF. Its X component (see 416) is obtained by multiplying^? by the projection of dF on the ver- tical plane ABC, and since p is the same for all the dF'& of the curved surface, the sum of all the X components of the in- ternal fluid pressures must = p multiplied by the area of rect- angle ABCD, = %rlp ; and similarly the X components of the FIG. 470. * Analytically this problem is identical with that of the smooth cord on a smooth cylinder, 169, and is seen to give the same result. BURSTING OF PIPES. v external atmos. pressures = %rlp a (nearly). The tensile stresses ( || to X) are equal to 2% 3 ; hence for equilibrium, ^2X gives 3 %rlp + 2rlp a = ; This tensile stress, called hoop tension, p^, opposing rupture by longitudinal tearing, is seen to be double the tensile stress^ induced, under the same circumstances, on the annular cross' section in Case I. Hence eq. (2), and not eq. (1), should be used to determine a safe value for the thickness of metal, , or any other one unknown quantity involved in the equation. For safety against rupture, we must put p z T' ', a safe tensile stress per unit area for the material of the pipe or tube (see 195 and 203) ; (For a thin hollow sphere, t may be computed from eq. (1) ; that is, need be only half as great as with the cylinder, other things being equal.) EXAMPLE. A pipe of twenty inches internal diameter is to contain water at rest under a head of 340 feet ; required the proper thickness, if of cast-iron. 340 feet of water measures 10 atmospheres, so that the in ternal fluid pressure is 11 atmospheres ; but the external pres- sure p a being one atmos., we must write (inch, lb., sec.) (pp a ) = 10 X 14. Y = 147.0 Ibs. per sq. in., and r = 10 in., while ( 203) we may put T' = J of 9000 = 4500 Ibs. per sq. in. ; whence t = , = 0.326 inches. 4500 538 MECHANICS OF ENGINEERING. But to insure safety in handling pipes and imperviousness to the .water, a somewhat greater thickness is adopted in practice than given by the above theory. Thus, Weisbach recommends (as proved experimentally also) for Pipes of sheet iron, t = [0.00172 rA + 0.12" "0.00476 rA + 0.34T U00296 rA + 0.16 = "0.01014 rA + 0.21 "0.00484 rA + 0.16 = u .. cast t = copper t = " " lead t = I " " zinc t = inches ; in which t = thickness in inches, r = radius in inches, and A = excess of internal over external fluid pressure (i.e., p p a ) expressed in atmospheres. For instance, for the example just given, we should have (cast-iron) t .00476 X 10 X 10 + 0.34 = 0.816 inches. If the pipe is subject to " water-ram" ( 422) the strength should be much greater. To provide against " water-ram," Mr. J. T. Fanning, on p. 453 of his " Hydraulic and "Water- supply Engineering," advises adding 230 feet to the static head in computing the thickness of cast-iron pipes. For thick hollow cylinders see Rankine's Applied Mechan- ics, p. 290, and Cotterill's Applied Mechanics, p. 403. 427. Collapsing of Tubes under Fluid Pressure. (Cylindrical boiler-flues, for example.) If the external exceeds the internal fluid pressure, and the thickness of metal is small compared with the diameter, the slightest deformation of the tube or pipe gives the external pressure greater capability to produce a further change of form, and hence possibly a final collapse ; just as with long columns ( 303) a slight bending gives great advantage to the terminal forces. Hence the theory of 426 is inapplicable. According to Sir Wm. Fairbairn's experi- ments (1858) a thin wrought-iron cylindrical (circular) tube will not collapse until the excess of external over internal pressure is COLLAPSE OF TUBES. 539 j?(in Ibs. per sq. in.) = 9672000 1~. . . (1) . . (not homog.) (t, I, and d must all be expressed in the same linear unit.) Here t = thickness of the wall of the tube, d its diameter, and Z its length ; the ends being understood to be so supported aa to preclude a local collapse. EXAMPLE. With I = 10 ft. = 120 inches, d = 4 in., and t = J$ inch, we have p = 9672000 P-j^- -5- (120 X 4)1 = 201.5 Ibs. per sq. inch. For safety, % of this, viz. 40 Ibs. per sq. inch, should not be exceeded ; e.g., with 14.7 Ibs. internal and 54.7 Ibs. external. 540 MECHANICS OF ENGINEERING. CHAPTEE II. HYDROSTATICS (Continued) PRESSURE OF LIQUIDS IN TANKt AND RESERVOIRS. 428. Body of Liquid in Motion, but in Relative Equilibrium. By relative equilibrium it is meant that the particles are not changing their relative positions, i.e., are not moving among each other. On account of this relative equilibrium the fol- lowing problems are placed in the present chapter, instead of under the head of Hydrodynamics, where they strictly belong. As relative equilibrium is an essential property of rigid bodies, we may apply the equations of motion of rigid bodies to bodies of liquid in relative equilibrium. CASE I. All the particles moving in parallel right lines with equal velocities ; at any given instant (i.e., a motion of translation.) If the common velocity is constant we have a uniform translation, and all the forces acting on any one par- ticle are balanced, as if it were not moving at all (according to Newton's Laws, 54); hence the relations of internal pressure, free surface, etc., are the same as if the liquid were at rest. Thus, Fig. 471, if the liquid in the moving tank is at rest rel- v atively to the tank at a given instant, with its free surface horizontal, and the motion of the tank be one of translation with a uni- form velocity, the liquid will remain in this condition of relative rest, as the motion FIQ. 471. proceeds. But if the velocity of the tank is accelerated with a constant acceleration =p (this symbol must not be confused with p for pressure), the free surface will begin to oscillate, and finally come to relative equilibrium at some angle a with the horizon- tal, which is thus found, when the motion is horizontal. See Fig. 472, in which the position and value of of are the same, whether the motion is uniformly accelerated from left to right RELATIVE EQUILIBRIUM OF LIQUIDS. 541 FIG. 472. or uniformly retarded from right to left. Let be the lowest point of the free surface, and Ob a _v n *.? - small prism of the liquid with its axis horizontal, and of length = % ; nb is a vertical prism of length = *~T~ 3, and extending from the extremity of Ob to the free surface. The pressure at both and n is p a = atmos. pres. Let the area of cross- section of both prisms be = dF. Now since Ob is being accelerated in direction ^(horizont.), the difference between the forces on its two ends, i.e., its ^X y must = its mass X accel, ( 109). .\p b dF p a dF = [xdF. y-?-g]p. . . . (1)' (y = heaviness of liquid ; p b = press, at b) ; and since the ver- tical prism nb has no vertical acceleration, the -^(vert. com- pons.) for it must = 0. s.p b dF-p a dF-zdF.y=Q (2) From (1) and (2), ffiy ~ . p frf\ V "T Hence On is a right line, and therefore tan a. or , =^ g (4) [Another, and perhaps more direct, method of deriving this result is to consider free a small particle of the liquid lying in. the surface. The forces acting on this particle are two : the first its weight = dG ; and the second the resultant action of its immediate neighbor-particles. Now this latter force (point- ing obliquely upward) must be normal to the free surface of the liquid, and therefore must make the unknown angle a with the vertical. Since the particle has at this instant a rectilinear accelerated motion in a horizontal direction, the resultant of the two forces mentioned must be horizontal and have a value = mass X acceleration. That is, the diagonal formed on the two 542 MECHANICS OF ENGINEERING. forces must be horizontal and have the value mentioned, = (dG -r- g)p ; while from the nature of the figure (let the stu- dent make the diagram for himself) it must also = dG tan a. -, , dG tan a = . p ; or, tan a = . g 9 Q. E. D.I ' If the translation were vertical, and the acceleration upward [i.e., if the vessel had a uniformly accelerated upward motion or a uniformly retarded downward motion], the free surface would be horizontal, but the pressure at a depth = h below the surface instead of p =p a -\-hy would be obtained as follows : Considering free a small vertical prism of height = h with upper base in the free surface, and putting ^(vert. compons.) = mass X acceleration, we have dF.p - dF.p a - hdF. y = MF ' y . p; (5) If the acceleration is downward (not the velocity necessarily) we make p negative in (5). If the vessel falls freely, p = g and .'.p =p a , in all parts of the liquid. Query : Suppose p downward and > g. CASE II. Uniform Rotation about a Vertical Axis. If the narrow vessel in Fig. 473, open at top and containing a liquid, be kept rotating at a uniform angu- lar velocity GO (see 110) about a vertical axis Z, the liquid after some oscillations will be brought (by fric- tion) to relative equilibrium (rotat- ing about Z, as if rigid). Required the form of the free surface (evi- dently a surface of revolution) at each point of which we know P=Pa- Let be the intersection of the axis Zwith the surface, and n any point in the surface ; 1) being Fio. 473. UNIFORM ROTATION OF LIQUID IN VESSEL. 543 a point vertically under n and in same horizontal plane as 0. Every point of the small right prism nb (of altitude z and sectional area dF) is describing a horizontal circle about 2, and has therefore no vertiacl acceleration. Hence for this prism, free, we have 2Z = 0; i.e., dF. p b - dF.p a - zdF. y = ..... (1) Now the horizontal right prism Ob (call the direction ... 5, X) is rotating uniformly about a vertical axis through one ex- tremity, as if it were a rigid body. Hence the forces acting on it must be equivalent to a single horizontal force, (*?Mp, (1220*) coinciding in direction with X. [M= mass of prism = its weight -=- r / x c = |A a . Similarly, for a triangle with _v*i base horizontal and vertex down, Fig. 481, we find that (3) If the base is in the surface, A 1 = and FIG. 48i. (3) reduces to x c = |7i 2 . It is to be noticed that in the case of the triangle the value of x c is the same whatever be its shape, so long as h l and A 2 remain unchanged and the base is horizontal. If the base is not horizontal, we may easily, by one horizontal line, divide the triangle into two triangles whose bases are horizontal and whose combined areas make up the area of the first. The re- sultant pressure on each of the component triangles is easily found by the foregoing principles, as also its point of applica- tion. The resultant of the two parallel forces so determined will act at some point on the line joining the centres of pres- sures of the component triangles, this point being easily found by the method of moments, while the amount of this final re- sultant pressure is the sum of its two components, since the latter are parallel. An instance of this procedure will be given in Example 3 of 433. Similarly, the rectangle of Fig. 479 may be distorted into an oblique parallelogram with hori- zontal bases without affecting the value of x c , nor the amount of resultant pressure, so long as h l and A 2 remain unchanged. 432. Centre of Pressure of Circle. Fig. 482. It will lie on the vertical diameter. Let r = radius. From eq. (3), Fx F (See eq. (4), 88, and also 91.) F:o. 482. x CENTRE OF PRESSURE. 551 FIG. 483. 433. Examples. It will be noticed that although the total pressure on the plane figure depends for its value upon the head, a, of the centre of gravity, its point of application is al- ways lower than the centre of gravity. EXAMPLE 1. If 6 ft. of a vertical sluice-gate, 4 ft. wide, Fig. 483, is below the water-surface, the total water pressure against it is (ft, lb., sec. ; eq. (1),430) P = Fzy = 6 X 4 X 3 X 62.5 = 4500 Ibs., and (so far as the pressures on the vertical posts on which the gate slides are concerned) is equivalent to a single horizontal force of that value applied at a distance x c = f of 6 = 4 ft. below the surface ( 431). EXAMPLE 2. To (begin to) lift the gate in Fig. 483, the gate itself weighing 200 Ibs., and the coefficient of friction between the gate and posts being/" = 0.40 (abstract numb.) (see 156), we must employ an upward vertical force at least = P' = 200 + 0.40 X 4500 = 2000 Ibs. EXAMPLE 3. It is required to find the resultant hydro- static pressure on the trapezoid in Fig. 483& with the dimen- sions there given and its bases horizontal ; also its point of ap- plication, i.e., the centre of pressure of r~; the plane figure in the position there A B c D shown. From symmetry the C. of P. will be in the middle vertical of the figure, as also that of the rectangle B CFE, and that of the two triangles ABE and CDF taken together (conceived to be shifted horizontally so that CF and BE coincide on the middle vertical, thus forming a single triangle of 5 ft. base, and having the same total pressure and C. of P. as the two actual triangles taken together). Let P l = the total pressure, and a?/ refer to the C, of P., for the rectangle ; P 9 and a?/, for the 5 ft. tri- EF- 5' FIG. 4S3a. 552 MECHANICS OF ENGINEERING. angle; h t = 4 ft. and A 2 = 10 ft. being the same for both. Then from eq. (1), 430, we have (with the ft., lb., and sec.) P l = 30 X 1y = and f = i X 6 X 5 X 6y = 90 r ; while from eqs. (1) and (3) of 431 we have also (respectively) 7.438 feet; 2 1000 64 3 100 - 16 2 936 3'~84~ -V, 9 * ~ 2" 80 100 228 8+10 2X18 = 6.333 feet. The total pressure on the trapezoid, being the resultant of P l and P 2 , has an amount = P 1 + P 2 (since they are parallel), and has a lever-arm sc c about the axis OY to be found by the principle of moments, as follows : P V + />/ _ (210 X 7.438 + 90 X 6.33) x \ + P, ~(210 + "^-- = T *. = The total hydrostatic pressure on the trapezoid is (for fresh water) P = P, + P 2 = [210 + 90] 62.5 = 18T50 Ibs. EXAMPLE 4. Required the horizontal force P', Fig. 484, to be applied at N (with a leverage of a' = 30 inches about the fulcrum M) necessary to (begin to) lift the circular disk AB of radius r = 10 in., covering an opening of equal size. NMAB is a single rigid lever weighing # / = 210 Ibs. The centre of gravity, G, of disk, being a ver- tical distance ~z = O'G 40 inches from the surface, is 50 inches (viz., the sum of OM = k = 20" and MG = 30") from axis Y ; i.e., x = 50 inches. The centre of gravity of the whole lever is a horizontal distance J 7 , = 12 inches, from M. FIG. 484. EXAMPLES CENTRE OF PRESSURE. 553 For impending lifting we must have, for equilibrium of the lever, -k); . . . . (1) where P = total water pressure on circular disk, and x c = OC. From eq. (1), 430, (using inch, lb., and sec.,) P = fzy = 7tr*zy = arlOO X 40 X -ss" = 454.6 Ibs. 1728 From 432, x c = OC = x + ~ = 50 + 1 . = 50.5 in. 4 2; 4: = ~ [210 X 12 + 454.6 x 30.5] = 546 Ibs. 30 434. Example of Flood-gate. Fig. 485. Supposing the rigid double gate AD, 8 ft. in total width, to have four hinges ; two at 0, and two at/ 1 ft. from top and bottom of water chan- "^=j-'=i-='1 , nel ; required the pressures upon them, FT taking dimensions from the figure (ft., \ lb., sec.). Wat. press. = P Fzy = 72 X 4J- X 62.5 = 20250 pounds, and its point of application (cent, of press.) is a dis- tance x c = f of 9' 6' from O ( 431). Considering the whole gate free and taking moments about 0, we shall have (press, at f)xT = 20250 x 5 ; .-. press, at / = 14464 Ibs. (half on each hinge at/), and /. press, at e = P press, at/ = 5875 Ibs. (half coming on eacli hinge). 554 MECHANICS OF ENGINEERING. If the two gates do not form a single rigid body, and hence are not in the same plane when closed, a wedge-like or toggle- joint action is induced, producing much greater thrusts against the hinges, and each of these thrusts is not ~| to the plane of the corresponding gate. Such a case forms a good exercise for the student. 435. Stability of a Vertical Rectangular Wall against Water Pressure on One Side. Fig. 486. All dimensions are shown in the figure, except I, which is the length of wall ~] to paper. Supposing the wall to be a single rigid block, its weight G' Vh'ly' (y' being its heaviness ( 7), and I its length). Given the water depth A, required the proper width b f for stability. For proper security : First, the resultant of G' and the Fl - 4g 6. water-pressure P must fall within the base BD (or, which amounts to the same thing), the moment of G about D, the outer toe of the wall, must be numerically greater than that of P ; and Secondly, P must be less than the sliding friction/Vjr' (see 156) on the base BD. Thirdly, the maximum pressure per unit of area on the base must not exceed a safe value (compare 348). NowJ 3 = Fzy = Til y = h*ly (y heaviness of water) ; 2 2 and x c -f A. Hence for stability against tipping about D, P^h must oe < G'^b' ; i.e., \tfly < ^b'*h'ly' ; . (1) while, as to sliding on the base, i.e., $h*ly " " brickwork, 0.60 For common bricks on common bricks, 0.64 To satisfactorily investigate the third condition requires the detail of the next paragraph. 436, Parallelopipedical Reservoir Walls. More Detailed and Exact Solution. If (1) in the last paragraph were an exact equality, instead of an inequality, the resultant R of P and G f would pass through the corner J9, tipping would be impending, and the pressure per unit area at D would be theoretically infinite. To avoid this we wish the wall to be wide enough that the re- sultant 7?, Fig. 487, may cut BD in such a point, E' , as to cause the pressure per unit area, p m , at D to have a definite safe value (for the pressure p m at D, or quite near _Z>, will evidently be greater than elsewhere on BD ; i.e., it is the maximum pressure to be found on BD). This may be done by the principles of 346 and 362. First, assume that R cuts BD outside of the 'middle third; i.e., that VE ! , = nb', > #'(<" n > |) ; FIG. 487. where n denotes the ratio of the distance of E' from the mid- dle of the base to the whole width, >', of base. Then the pres- sure (per unit area) on small equal elements of the base BD (see 346) may be considered to vary as the ordinates of a triangle MND (the vertex M being within the distance BD), and E~D will = \MD ; i.e., 556 MECHANICS OF ENGINEERING. The mean pressure per unit area, on and hence the maximum pressure (viz., at D\ being double the mean, is p m = iff' -=- [8J7(i - n)] ; . . . . (0) and if p m is to equal ^'(see 201 and 203), a safe value for the crushing resistance, per unit area, of the material, we shall have &7(i -n)C f = \G' = ty'k'ly', i 2 hy ' V s it fan (1) To find &', knowing w,, we put the ^(moms.) of the G r and P at E, about E',=- zero (for the only other forces acting on the wall are the pressures of the foundation against it, along MD ; and since the resultant of these latter passes through E\ the sum of their moments about E' is already zero) ; i.e., QW - pk = o or nb n h'l= Having obtained 5 r , we must also ascertain if P is 0, the angle of friction, with the ver- tical. If n, computed from (1), should prove to be < , our first assumption is wrong, and we therefore assume n < ^, and pro- ceed thus : Secondly, n being < (see 346 and 362), we have a STABILITY OF KESfcRVOIK WALLS. 557 trapezoid of pressures, instead of a triangle, on BD. Let the pressure per unit area at D be p m (the maximum on base). The whole base now receives pressure, the mean pressure (per unit area) being = G-' -5- [5'Z] ; and therefore, from 362, Case I, we have ; ..... (Oa) and since, here, G' = Vh'ly' ', we may write p m = (Qn + 1)A y. For safety as to crushing resistance we put 6(n + l)hy = C'; whence n = | ["^-7 - ll . . (la) Having found n from eq. (la), we determine the proper width of base I' from eq. (2), in case the assumption n < -J- is verified. EXAMPLE. In Fig. 486, let h f = 12 ft., h = 10 ft., while the masonry weighs (y 1 =) 150 Ibs. per cub. ft. Supposing it desirable to bring no greater compressive stress than 100 Ibs. per sq. inch (= 14400 Ibs. per sq. ft.) on the cement of the joints, we put C' 14400, using the ft.-lb.-sec. system of units. Assuming n > -J-, we use eq. (1), and obtain _ 1 _ 2 12 X 150 _ _5_ ~ 2 " 3 ' 14:400 " ~ 12' which is > -J- ; hence the assumption is confirmed, also the propriety of using eq. (1) rather than (la). Passing to eq. (2), we have But, as regards frictional stability, we find that, wiibf = 0.30,, a low value, and V = 3.7ft. (ft., lb., sec.), 558 MECHANICS OF ENGINEERING. P_ h*ly _ _ 100 X 62.5 _ = 15. f ~fbWl? "" 2X0.3 X 3.7 X 12 X 150 ~ which is greater than unity, showing the friction to be insuf- ficient to prevent sliding (with /=0.30); a greater width must therefore be chosen, for frictional stability. If we make n = -J-, i.e., make R cut the base at the outer edge of middle third ( 362), we have, from eq. (2), / = iox 62 ' 5X] X 12 X 150 and the pressure at D is now of course well within the safe limit ; while as regards friction we find P -r-fG' = 0.92, < unity, and therefore the wall is safe in this respect also. With a width of base = 3.7 feet first obtained, the portion MD, Fig. 487, of the base which receives pressure [according to Navier's theory ( 346)] would be only 0.92 feet in length, or about one sixth of the base, the portion BM tending to open, and perhaps actually suffering tension, if capable (i.e., if cemented to a rock foundation), in which case these tensions should properly be taken into account, as with beams ( 295), thus modifying the results. It has been considered safe by some designers of high masonry dams, to neglect these possible tensile resistances, as has just been done in deriving ~b' =. 3.7 feet ; but others, in view of the more or less uncertain and speculative character of Navier's theory, when applied to the very wide bases of such structures, prefer, in using the theory (as the best available), to keep the resultant pressure within the middle third at the base (and also at all horizontal beds above the base), and thus avoid the chances of tensile stresses. This latter plan is supported by Messrs. Church and Fteley, as engineers of the proposed Quaker Bridge Dam in connec- tion with the New Croton Aqueduct of New York City, in iheir report of 1887. See 438. RESEEVOIR WALLS. 437. Wall of Trapezoidal Profile. Water-face Vertical- Economy of material is favored by using a trapezoidal profile,, Fig. 488. With this form the stability may be investigated in a corresponding manner. The portion of wall above each horizontal bed should be ex- amined similarly. The weight G' acts through the centre of gravity of the whole mass. Detail. Let Fig. 488 show the vertical cross-section of a trapezoidal wall, with notation for dimensions as indicated ; the portion considered having a length = I, ~] to the paper. Let y = heaviness of water, y' that of the masonry (assumed homo- geneous), with n as in 436. Fora triangle of pressure, MD, on the base, i.e., withn > -J, or resultant falling outside the middle third (neglecting pos- sibility of tensile stresses on left of M\ if the intensity of pressure j9 TO at D is to C' ( 203), we put, as in 436, FIG. 488. - n\ C' = .e. = whence For a trapezoid of pressure, i.e. with n < -J-, or the resultant of P and G' falling within the middle third, we have, as be- fore ( 362, Case I), ~ whence n = - i n = 560 MECHANICS OF ENGINEERING. From the geometry of the figure, having joined the middles of the two bases, we have { 26, Prob. 6), and, by similar triangles, OT : KV ::gO:h r , whence The lines of action of ' and P meet at E, and their result- ant cuts the base in some point E'. The sum of their moments about E' should be zero, i.e., P . h G f . 'OE'\ that is, (see eq. (a) above, and eq. (1), 430,) i.e., cancelling, + V') (2)' Hence we have two equations for finding two unknowns viz.: (l)'and (2)' when n > -J- ; and (la) 7 and (2) r when /i < -J-. For dams of small height (less than 40 ft., say), if we im- mediately put n = -J-, thus restricting the resultant pressure to the edge of middle third, and solve (2) 7 for J 7 , 1)" being as- sumed of some proper value for a coping, foot- walk, or road- way, while h' may be taken enough greater than h to provide against the greatest height of waves, from 2.5 to 6 ft., the value of p m at D will probably be < C'. In any case, for a value of n =, or <, we put p m for <7'in equation (la) 7 and solve for p m , to determine if it is no greater than C' . Mr. Fanning recommends the following values for C' (in Us. KESEKVOIR WALLS. 561 per sq.foof) with coursed rubble masonry laid in strong mor- tar: For Limestone. Sandstone. Granite. Brick. O = 50,000 50,000 60,000 35,000 of ) in > ft. f 152 132 154 120 Ibs. per cub." As tofrictional resistance, P must be " = in the preceding article, the trapezoid becomes a right triangle, and the equations reduce to the following : and l-}forn" = from 6 to 22 feet at the top, and a sufficient h" (see figure) to ex- ceed the maximum height of waves, the up-stream outline A CM is made nearly vertical and perhaps somewhat concave, while the down-stream profile BDN, by computation or graphical trial, or both, is so formed that when the reservoir is full the resultant R, of the weight G of the portion AECD of ma- sonry above each horizontal bed, as 1 ^j (7Z>, and the hydrostatic pressure P on the corresponding up-stream face AC, shall cut the bed CD in such a point E' as not to cause too great compression p m at the outer edge D (not over 85 Ibs. per sq. inch accord- ing to M. Krantz in " Keservoir FIG. 490. Walls"), p m being computed by one of the equations [(0) and (Oa) of 1 436] 5b*4 MECHANICS OF ENGINEERING. For E' outside the middle third ) 26r \ CD. I where I = length of wall 1 to paper, usually taken = one foot, or one inch, according to the unit of length adopted ; for n, see 436. Nor, when the reservoir is empty and the water pressure lacking, must the weight G resting on each bed, as CD, cut the bed in a point E" so near the edge C as to produce exces- sive pressure there (computed as above). The figure shows the general form of profile resulting from these conditions. The masonry should be of such a character, by irregular bond- ing in every direction, as to make the wall if possible a mono- lith. For more detail see next paragraph. 440. Quaker Bridge Dam (on the New Croton Aqueduct). Attempts, by strict analysis, to determine the equation of the curve BN, AM being assumed straight, so as to bring the point E' at the outer edge of the middle third of its joint, or to make the pressure at D constant below a definite joint, have failed, up to the present time; but approximate and tentative methods are in use which serve all practical purposes. As an illustration the method set forth in the report on the Quaker Bridge Dam will be briefly outlined ; this method confines E' to the middle third. The width AB I" is taken 22' for a roadway, and h" = 7 ft. The profile is made a vertical rectangle from A down to a depth of 33 ft. below the water surface (reservoir full). Combining the weight of this rectangle of masonry with the corresponding water pressure (for a length of wall = one foot), we find the resultant pressure conies a little within the outer edge of the middle third of the base of the rectangle, while p m is of course small. The rectangular form of profile might be continued below this horizontal joint, as far as complying with the middle QUAKER BRIDGE DAM. 565 third requirement, and the limitation of pressure-intensity, are concerned ; but, not to make the widening of the joints too abrupt in a lower position where it would be absolutely re- quired, a beginning is made at the joint just mentioned by forming a trapezoid between it and a joint 11 ft. farther down, making the lower base of the latter of some trial width, which can be altered when the results to which it gives rise become evident. Having computed the weight of this trapezoid and constructed its line of action through the centre of gravity of the trapezoid, the value of the resultant G of this weight and that of the rectangle is found (by principle of moments or by an equilibrium polygon) in amount and position, and combined with the water pressure of the corresponding 44 ft. of water to form the force _Z?, whose point of intersection with the new joint or bed (lower base of trapezoid) is noted and the value of p m computed. These should both be somewhat nearer their limits than in the preceding joint. If not, a different width should be chosen, and changed again, if necessary, until satis- factory. Similarly, another layer, 11 ft. in height and of trapezoidal form, is added below and treated in the same way ; and so on until in the joint at a depth of 66 ft. from the water surface a width is found where the point E' is very close upon its limiting position, while p m is quite a little under the limit set for the upper joints of the dam, 8 tons per square foot. For the next three 11 ft. trapezoidal layers the chief governing element is the middle-third requirement, E' being kept quite close to the limit, while the increase of p m to 7.95 tons per sq. ft. is unobjectionable; also, we begin to move the left-hand edge to the left of the vertical, so that when the reservoir is empty the point E" shall not be too near the up- stream edge C. Down to a depth of about 200 ft. the value of p m is allowed to increase to 10.48 tons per sq. ft., while the position of E r gradually retreats from the edge of its limit. Beyond 200 ft. depth, to prevent a rapid increase of width and consequent extreme flattening of the down-stream curve, p m is allowed to mount rapidly to 16.63 tons per sq. ft. (=231 Ibs. per sq. in.), which value it reaches at the point ^Tof the base of 566 MECHANICS OF ENGINEEEING. the dam, which has a width = 216 ft., and is 258 feet below the water surface when the reservoir is full. The heaviness of the masonry is taken as y f = 156.25 Ibs. per cubic foot, just f of y = 62.5 Ibs. per cub. foot, the heavi- ness taken for water. *, When the reservoir is empty, we have the weight G of the superincumbent mass resting on any bed CD, and applied through the point E" ; the pressure per unit area at C can then be computed by eq. (lfl)'", 439, n being the quotient of (faCD CE") -r- Cl) for this purpose. In the present case we find E" to be within middle third at all joints, and the pressures at C to be under the limit. For further details the reader is referred to the report itself (reprinted in Engineering News, January, 1888, p. 20). The graphic results were checked by computation, Wegmann's method, applied to each trapezoid in turn. 441. Earthwork Dam, of Trapezoidal Section. Fig. 491. It is . & ^ required to find the conditions of sta- bility of the straight earthwork dam ABDE, whose length I, L to paper, as regards sliding horizontally on the plane AE\ i.e., its frictional stability. With the dimensions of the figure, y and y' being the heavi- FIG. 49i. nesses of the water and earth respec- tively (see 7), we have Weight of dam = G, = vol. X / = lh$ + ^(a, + c)]y'. (1) Eesultant water press. = P = Fzy = OA X I X %hy. . (2) Horiz. comp. of P = II = P sin a = [OA sin a] \Uy - %tfly. .* . . (3) From (3) it is evident that the horizontal component of P is just the same, viz., = hi . %hy, as the water pressure would be on a vertical rectangle equal to the vertical projection of OA EAKTHWORK DAM. 567 and with its centre of gravity at the same depth (JA). Com- pare 416. Also, Vert. comp. of P = V = P cos a = [OA cos a]%kly =*%akly, ... (4) and is the same as the water pressure on the horizontal projec- tion of OA if placed at a depth = O'G = %h. For stability against sliding, the horizontal component of P must be less than the friction due to the total vertical pressure on the plane AE, viz., G 1 + F; hence if /"is the coefficient of friction on AE, we must have H )j_-sin(/3-(!>) - G cos (|3 - )~| -fsin(|3 - ) cos (j3 + 8 - <) dtfT ~ sln2 [j8 -f 6 - ] For P to be a maximum we must put numerator of above (#} To find a geometrical equivalent of , denote A C by Z, and draw AE, making an angle = d

= %Z?d(j) . . . (neglecting infinitesimal of 2d order). Now dG = y X area J. CI X unity ; /. -=-- = i/Z 2 ; /. (a) becomes sin(/3 + d 0)^Z 3 sin(/? 0) sin (/? + (J 0)^cos(^ 0) + # sin (/? 0) cos (/? + d 0) = ; i.e., G = sin (ft 0) sin (ft + 0) _ sin (/? -f d 0) cos (/? 0) cos (yff + d 0) sin (/? 0) - ./ 576 MECHANICS OF ENGINEERING. when P is a maximum ; and hence, calling G-' and 0' and Z' the values of 6r, 0, and Z, for max. P, we have in (ft - , (2) and therefore from (1) P max. itself is 447. Geometric Interpretation and Construction. If in Fig. 496 we draw CF^ making angle d with AD, C being any point on the ground surface JBD, we have sn Drop a perpendicular FH from F to A C, and we shall have = CF. sin (ft + S - '\ Describe a circle on ED as a diameter, and draw 580 MECHANICS OF ENGINEERING. BX ~l to BD, thus fixing X in the curve. With centre E describe a circular arc through X, cutting BD in C', required. Having AC' (i.e., L'\ 0' is known ; hence from eq. (3) we obtain the earth thrust or pressure P'\ or, with F 1 as centre and radius = C'F', describe arc C'J'\ then the triangle C'F'J' is the base of a prism of unity height whose weight = P' (as in 447). Centre of Pressure. Applying the method of 448, Fig. 498, to this case, we find that the successive L' 's are propor- tional to the depths ab, ac, ad, etc., and that the successive P'B are proportional [see (3)] to the squares of the depths ; hence the area in Fig. 498 must be triangular in this case, and the point of application of the resultant pressure on AB is one third of AB from A : just as with liquid pressure. 451. Resistance of Retaining Walls. (Fig. 500.) Knowing the height of the wall we can find its weight, = G l , for an as- sumed thickness, and unity width ~| to paper. The resultant of Cr 1 , acting through the centre of gravity of wall, and P\ the thrust of the embankment, in its proper line of action, should cut the base A V within the middle third and make an angle with the normal (to the base) less than the angle of friction. For the straight wall and straight earth-profile of Fig. 499 and 450, the FIG. 500. length Z', = A C ', can be expressed in terms of the (vertical) height, A, of wall, thus : ' = AC' = B{n (C ~ g) sin (C ~ a - . sin (C ) cos a sin ( ) .-. eq. (3) becomes P' = l r Jt- Bin(/?-0Qsin*(C-) K ^ r cos 2 a ' sin d sin 2 (C - 00 cos a a [A representing the large fraction for brevity.] RETAINING WALLS. 581 This equation will require, for a wall of rectangular section, that the thickness, d, increase as A, in order that its weight may increase as A 3 (i.e., as P f ) and that its resisting moment may- increase with the overturning moment. By this equality of moments is meant that P'a = Gfi ; where a and b are the respective lever-arms of the two forces about the front edge of the middle third. (AB is the back of the wall.) In other words, their resultant will pass through this point. The following table is computed on the basis just mentioned, viz., that the resultant of P' and G- shall pass through the front edge of the midde third. The wall is vertical, i.e., a is = 0, and is of rectangular sec- tion ; and we further suppose that the heaviness of the earth is two thirds that of the masonry of the wall ; d is tne proper safe thickness to be given to the wall on the basis spoken of, h being its altitude. Whether the wall is safe against sliding on its base, and whether a safe compression per unit area is ex- ceeded on the front edge of the base, are matters for separate consideration. See Figs. 499 and 500, and the foregoing text, for the meaning of all symbols employed. The above assump- tion as to the relative densities of wall and earth is realized if the wall is of first-quality masonry weighing 150 Ibs. per cubic foot, supporting earth of 100 Ibs. per cubic foot. Note that d a + 8 ; i.e., d 8 f or this table. a = 0; i.e., wall is vertical; also density of wall = f that of the earth. I. II. III. ' A d f A d *' A d 1.0 45 22^ .17 Mh 26 .18 .22* 45 .71 .33* 1.5 56i 28 .29 Mli 33 .26 .30* 56 .83 ,43* 2.0 63| 31| .38 .51* 38 .33 .36* 63 .89 .51* 4.0 76 38 .61 .64* 45 .54 .50* 76 .97 .65* Infinity 90 45 1.00 .82* 90 1.00 .82* 90 1.00 .82* In Case I of table, since a = 0, = 90 and C = 90 ; d = 90, and hence C'F' of Fig. 499 is 1 to AD, so that 582 MECHANICS OF ENGINEEKING. (since the area of &ABC' = &AC f F f ) 0' must = fyff. These values, in (5), give P' = %yh* tan 2 %ft ; i.e., A = tan 2 \ft. . . . (6) In Case II, since C = 90, a = and 8 = ft, .-. 3 = ft ; and (5) reduces to ;in 2 (ft (f>'\ . . sin 2 (ft r ) ,^. In Case III, C = /? and will be || to J.Z>, Z> being at infinity. See Fig. 501. Through FIG. SOL BI1 1 to u4Z>, and BF" e d with ^LI>. C' is to be located on BD, so as tQ make (area of) &ABC' = ( area of ) A AC'f" (according to 447), the angle C 'F'A being = <$ = a+ Q\ = 0, in this case, and hence also = ft. Conceive B and F ' to be joined. Now 'F' = &ABF" + But &ABC' = kBF'F" (equal bases and altitudes). Hence A ABC' cannot = i^AC'F' unless C f is moved out to infinity ; and then 0' becomes ytf, and eq. (5) reduces to P'= ^ytf sin ft ; i.e., A = sin ft. (8) [Increasing from zero will decrease the thickness d ; i.e., inclining the wall inwards will decrease the required thickness, but diminish the frictional stability at the base, unless the lat- ter be "1 to AB. The back of the wall is frequently inclined outwards, making the section a trapezoid, to increase the fric- tional stability at the base when necessary, as with timber walls supporting water.] KETAINLNG WALLS. 583 452. Practical Considerations. An examination of the values of A and d in the table of 451 will show that in sup- porting quicksand and many kinds of clay which are almost fluid under the influence of water, it is important to know what kind of drainage can be secured, for on that will depend the thickness of the wall. With well compacted material free from water-bearing strata, an assumed natural slope of If to 1 (i.e., If hor. to 1 vert.) will be safe ; the actual pressure below the effect of frost and surface water will be that due to a much steeper slope on account of cohesion (neglected in this theory). The thrust from freshly placed material can be reduced by depositing it in layers sloping back from the wall. If it is not so placed, however, the natural slope will seldom be flatter than If to 1 unless reduced by water. In supporting material which contains water-bearing strata sloping toward the wall and overlain by strata which are liable to become semi-fluid and slippery, the thrust may exceed that due to semi-fluid ma- terial on account of the surcharge. If these strata are under the wall and cannot be reached by the foundation, or if resist- ance to sliding cannot be obtained from the material in front by sheet-piling, no amount of masonry can give security. Water at the back of the wall will, by freezing, cause the material to exert an indefinitely great pressure, besides disinte- grating the wall itself. If there is danger of its accumulation, drainage should be provided by a layer of loose stone at the back leading to "weep-holes" through the wall. A friction-angle at the back of the wall equal to that of the filling should always be realized by making the back rough by steps, or projecting stones or bricks. Its effect on the required thickness is too great to be economically ignored. The resistance to slipping at the base can be increased, when necessary, by inclining the foundation inwards; by stepping or sloping the back of the wall so as to add to its effective weight or incline the thrust more nearly to the vertical; by sheet-piling in front of the foundation, thus gaining the resist- ance offered by the piles to lateral motion ; by deeper founda- tions, gaining the resistance of the earth in front of the wall. 584 MECHANICS OF ENGINEERING. The coefficient of friction on the base ranges, according to Trautwine, from 0.20 to 0.30 on wet clay ; " .50 to .66 " dry earth; " .66 to .75 " sand or gravel ; " .60 on a dry wooden platform ; to .75 on a wet one. If the wall is partially submerged, the buoyant effort should be subtracted from G 1 , the weight of wall. 453. Eesults of Experience. (Trautwine.) In railroad prac- tice, a vertical wall of rectangular section, sustaining sand, gravel, or earth, level with the top [p. 682 of Civ. Eng. Pocket Book] and loosely deposited, as when dumped from carts, cars r etc., should have a thickness d, as follows : If of cut stone, or of first-class large ranged rubble, in mortar. . . . d = . 35& " good common scabbled mortar-rubble, or brick d AOh " well scabbled dry rubble d = .5Qh Where h includes the total height, or about 3 ft. of foundations. (a) For the best masonry of its class h may be taken from the top of the foundation in front. (5) A mixture of sand or earth, with a large proportion of round boulders or cobbles, will weigh more than the backing assumed above ; requiring d to be increased from one eighth to one sixth part. ( G', and FIG. 502.] 588 MECHANICS OF ENGINEERING. to preserve equilibrium the body is attached by a cord to the bottom of the vessel. The tension in this cord is %>= V f y-G f (1) At (c) V'y is < G\ and the cord must be attached to a support above, and its tension is S C =G'- V'y. . ...... v (2) If in eq. (2) [( = (}' + V = 6 -=- 0.032 = 187.5 Ibs. per cub. ft., and the ratio of y' to y is 187.5 : 62.5 = 3.0 (abstract num- ber) ; i.e., the substance of this solid is three times as dense, or three times as heavy, as water. [The buoyant effort of the air has been neglected in giving the true weight as 6 Ibs.] 458. Specific Gravity. By specific gravity is meant the ratio of the heaviness of a given homogeneous substance to that of a standard homogeneous substance ; in other words, the ratio of the weight of a certain volume of the substance to the weight of an equal volume of the standard substance. Dis- tilled water at the temperature of maximum density (4 Centi- grade) under a pressure of 147 Ibs. per sq. inch is sometimes taken as the standard substance, more frequently, however, at 62 Fahrenheit (16.6 Centigrade). Water, then, being the standard substance, the numerical example last given illustrates a common method of determining experimentally the specific gravity of a homogeneous solid substance, the value there ob- tained being 3. The symbol cr will be used to denote specific gravity, which is evidently an abstract number. The standard substance should always be mentioned, and its heaviness y ; then the heaviness of a substance whose specific gravity is cr is (i) and the weight G' of any volume V of the substance may be written G 1 = Vy = V' V'y, i.e., if V'y' is > V'y, or a- > 1, it sinks ; while if G-' is < V'y, ...... or tr < 1, it rises ; i.e., according as the weight G' is > or < than the buoyant effort. Other methods of determining the specific gravity of solids, liquids, and gases are given in works on Physics. 45"9. Equilibrium of Flotation. In case the weight G' of an immersed solid is less than the buoyant effort V'y (where V is the volume of displacement, and /the heaviness of liquid) the body rises to the surface, and after a series of oscillations comes to rest in such a position, Fig. 503, that its centre of gravity C and the centre of buoyancy B (the new B, belonging to the new volume of displacement, which is limited above by the horizontal plane of the free surface of the liquid) are in the same vertical (called the axis of flotation, or line of support), and that the volume of displacement has diminished to such a new value V, that Vy=G' ........ (1) In the figure, V vol. AND, below the horizontal plane AN, and the slightest motion of the body will change the form of this volume, in general (whereas with complete immersion the volume of dis- placement remains constant). For stable equilibrium it is not essential in every case that C (centre of gravity of body) should be below B (the centre of buoy- ancy) as with complete immersion, since if FIG. 503. " the solid is turned, B may change its posi- tion in the body, as the form of the volume AND changes. There is now no definite relation between the volume of displacement Fand that of the body, V, unless the latter is homogeneous, and then for G' we may write V'y', i.e. V'y' = Vy (for a homogeneous solid) ; . . (2) or, the volumes are inversely proportional to the heavinesses. FLOTATION. 591 The buoyant effort of the air on the portion ANE may be neglected in most practical cases, as being insignificant. If the solid is hollow, the position of its centre of gravity C may be easily varied (by shifting ballast, e.g.) within certain limits, but that of the centre of buoyancy J3 depends only on the geometrical form of the volume of displacement AND, below the horizontal plane AN. EXAMPLE. (Ft., lb., sec.) Will a solid weighing G 400 Ibs., and having a volume V = 8 cub. feet, without hollows or recesses, float in water? To obtain a buoyant effort of 400 Ibs., we need a volume of displacement, see eq. (1), of G' 400 ^ = = ^ * = only 6.4 cub. ft. !-* WATFR- Hence the solid will float with 8 6.4, or 1.6, cub. ft. pro- jecting above the water level. Query : A vessel contains water, reaching to its brim, and also a piece of ice which floats without touching the vessel. When the ice melts will the water overflow 2 460. The Hydrometer is a floating instrument for determin- ing the relative heavinesses of liquids. Fig. 504 shows a sim- ple form, consisting of a bulb and a cylin- drical stem of glass, so designed and weighted as to float upright in all liquids whose heavinesses it is to compare. Let F denote the uniform sectional area of the stem (a circle), and suppose that when float- ing in water (whose heaviness = y) the water surface marks a point A on the stem ; and that when floating in another liquid, say petroleum, whose heaviness, = y p , we wish to determine, it floats at a greater depth, the liquid surface now marking A on the stem, a height x above A. G' is the same in both experiments ; but while the volume of dis- placement in water is F, in petroleum it is V -f- Fx. There- fore from eq. (1), 459, FIG. 504. 592 MECHANICS OF ENGINEERING. in the water G Vy, '. .... (1) and in the petroleum G = ( V+ Fx)y p ; . . (2) from which, knowing G', F, x, and y, we find Fand y p , i.e., (3) [N.B. jPis best determined by noting the additional dis- tance, = I, through which the instrument sinks in water under an additional load P, not immersed ; for then EXAMPLE. [Using the inch, ounce, and second, in which system y = 1000 -=- 1728 = 0.578 ( 409).] With ' = 3 ounces, and F= 0.10 sq. inch, x being observed, on the graduated stem, to be 5 inches, we have for the petroleum 3 X 0.578 3 + 0.10X5X0.578 = 56.7 Ibs. per cub. foot. Temperature influences the heaviness of most liquids to some extent. In another kind of instrument a scale-pan is fixed to the top of the stem, and the specific gravity computed from the weight necessary to be placed on this pan to cause the hydrometer to sink to the same point in all liquids for which it is used. 461, Depth of Flotation. If the weight and external shape of the floating body are known, and the centre of gravity so situated that the position of flotation is known, the depth of the lowest point below the surface may be determined. FLOTATION. 593 CASE I. Eight prism or cylinder with its axis vertical. Fig. 505. (For stability in this position, see 464#.) Let G' = weight of cylin- der, j^the area of its cross-section (full circle), h f its altitude, and h the un- known depth of flotation (or draught) ; then from eq. (1), 426, 9L V in which y heaviness of the liquid. If the prism (or cylinder) is homo- geneous (and then (7, at the middle of h ', is higher than and y its heaviness, we then have FIG. 505. . . . . (2) in which cr = specific gravity of solid referred to the liquid as standard. (See 458.) CASE II. Pyramid or cone with axis vertical and vertex down. Fig. 506. Let V = volume of whole pyramid (or cone), and V = vol- ume of displacement. From similar ~~f" pyramids, /Tr But O' = Vy\ or, V= ; whence FIG. 506. (3) 594 MECHANICS OF ENGINEERING. CASE III. Ditto, ~but vertex up. Fig. 50Y. Let the nota- tion be as before, for V and V. The part out of water is a pyramid of volume = V" = V V, and is similar to the whole pyramid ; V- Y: Y' :: A" 3 : h' 9 . FIG. 507. ===^ Also, .-., finally, h = A'pL - j/1 - [#' -f- F>]~|. ... (4) CASE IY. /Sphere. Fig. 508. The volume immersed is Y = and hence, since Yy = G f = weight of sphere, nW G f Ttrh = . 3 y (5) From which cubic equation h may be ^Ji*Hz: obtained by successive trials and ap- FlG - m - [An exact solution of (5) for the unknown h is impossible, as it falls under the irreducible case of Cardan's Rule.] CASE Y. Right cylinder with axis horizontal. Fig. 509. Ier, J of seg. Aj)B\ X Fia.509. hence, since F= , r - sin 2a = . EXAMPLES OF FLOTATION. 595 Prom this transcendental equation we can obtain <*, by trial, in radians (see example in 428), and finally A, since h = r(l cos a). ' (7) EXAMPLE 1. A sphere of 40 inches diameter is observed to have a depth of flotation h = 9 in. in water. Required its weight G' . From eq. (5) (inch, lb., sec.) we have G' = [62.5 + 1728]7r9 2 [20 - -J X 9] = 156.5 Ibs. The sphere may be hollow, e.g., of sheet metal loaded with shot ; constructed in any way, so long as G' and the volume T^of displacement remain unchanged. But if the sphere is homogeneous, its heaviness ( 7) y' must be = G' -f- V = G 1 -f- f-Trr 3 = (156.5) -f- f tf20 3 = .00466 Ibs. per cubic inch, and hence, referred to water, its specific gravity is OK, M is nega- ~^~" -w\j tive, indicating an upsetting couple. FlG - 514 - That is, for 0=0 the equilibrium is stable, but for = OK, unstable ; and M = in both positions. From eq. (4) we see why, if C is above JE>, instability does not necessarily follow. 464a. Metacentre of a Ship. Eeferring again to Fig. 512, we note that the entire couple [#', Vy] will be a righting couple, or an upsetting couple, according as the point m (the 600 MECHANICS OF ENGINEERING. intersection of the vertical through B l , the new centre of buoyancy, with BC prolonged) is above or below the centre of gravity C of the ship. The location of this point m changes with ; but as becomes very small (and ultimately zero) m approaches a definite position on the line DE, though not oc- cupying it exactly till = 0. This limiting position of m is called the metacentre, and accordingly the following may be stated : A ship floating iipright is in stable equilibrium if its metacentre is above its centre of gravity ; and vice versa. In other words, for a slight inclination from the vertical a righting, and not an upsetting, couple is called into action, if m is above C. To find the metacentre, by means of the dis- tance Bin, we have, from eq. (3), -, Vy sm (5) and wish ultimately to make = 0. Now the moment ( V w y)db = the sum of the moments about the horizontal fore- and-aft water-line axis OL, Fig. 515, of the buoyant efforts due to the immersion of the separate vertical elementary prisms of the wedge OLN^N, plus the moments of those lost, from emersion, in the wedge .,. ___ OLA.A. Let OA.LN, be the new water-line section of the ship when inclined a small FIG. sis. angle from the vertical (0 = NO,N^ and OALN the old water-line. Let 2 = the "1 distance of any elementary area dF oi the water-line section from OL (which is the intersection of the two water-line planes). Each dF is the base of an elementary prism, with altitude = (f>z, of the wedge N^OLN (or of wedge A^OLA when z is negative). The buoyant effort of this prism = (its vol.) X y = yzdF, and its moment about OL is Hence the total moment, = Qa, or V w ya, of Fig. 505, = 70 X I OL THE METACENTRE. 601 of water-line section, in which I OL denotes the " moment of inertia" ( 85) of the plane figure OALNO about the axis OL. Hence from (5), putting = sin (true when = 0), we have mB = I OL -T- V\ and therefore the distance mC\ of the meta- centre m above (7, the centre of gravity of the ship, Fig. 512, is ~~n -L IOL (of water-line sec.) . f ~ mC, = A m , = Z* -- _ -- 1 e , ... (6) in which e = BC = distance from the centre of gravity to the centre of buoyancy, the negative sign being used when C is above B\ while V= whole volume of water displaced by the ship. We may also write, from eqs. (6) and (1), for small values Mom. of righting couple = M = Vy sin -~ e , . (7) or M=ysm 0[/ 0i Ve] ..... (7)' Eqs. (7) and (7)' will give close approximations for < 10 or 15 with ships of ordinary forms. EXAMPLE 1. A homogeneous right parallelepiped, of heaviness y r , floats upright as in Fig. 516. Find the distance mC = h m for its metacentre in this A ^ \ M f\ position, and whether the equilibrium -^ is stable. Here the centre of gravity, = C, being the centre of figure, is of ; course above B, the centre of buoy- i>z ancy ; hence e is negative. B is the =^ centre of gravity of the displacement, and is therefore a distance ^h below the water-line. We here assume that I is greater than From eq. (2), 461, B--"- DJ, 602 MECHANICS OF ENGINEERING. and since CD = %h', and BD %h, .', e = i(A'' A); .e. = while ( 90) I OL , of the water-line section AN, = -% Also, and hence, from eq. (6), we have Hence if V is > 6A /a fl V the position in Fig. 516 is r v 7 y one of stable equilibrium, and vice versa. E.g., if y f = Jy, #' = 12 inches and A 7 = 6 inches, we have (inch, pound, sec.) h m = ^C = jf [144 - 6 X = 2.5 in. The equilibrium will be unstable if, with y' =. %y, b' is made less than 1.225 ti '; for, putting mC = Q, we obtain J r = 1.225 A'. EXAMPLE 2. (Ft., lb., sec.) Let Fig. 517 represent the half water-line section of a loaded ship of G' = Vy = 1010 tons FIG. 517. displacement; required the height of the metacentre above the centre of buoyancy, i.e., mB 2 (See equation just before eq. (6).) Now the quantity / OL , of the water-line section, may, from symmetry, (see 93,) be written IL = (1) METACENTRE. 603 in which y = the ordinate 1 to the axis OL at any point; and this, again, by Simpson's Rule for approximate integration, OL being divided into an even number, n, of equal parts, and ordinates erected (see figure), may be written / -?. fr- L 3 * 3n From which, by numerical substitution (see figure for dimen- sions ; n = 8), 125 1728 729 or ' 2197 2744 343 1331 IOL = W>.125 +4 X 4393 + 2 X 4804+ 0.125] . = 120801 biquad. ft,, = = = 3.8 feet. That is, the metacentre is 3.8 feet above the centre of buoyancy, and hence, if J3C=2 feet, is 1.90 ft. above the centre of gravity. [See Johnson's Cyclopaedia, article Naval Architec- ture.'} 465. Metacentre for Longitudinal Stability. If we consider the stability of a vessel with respect to pitching, in a manner similar to that just pursued for rolling, we derive the position of the metacentre for pitching or for longitudinal stability and this of course occupies a much higher position than that for rolling, involving as it does the moment of inertia of the water-line section about a horizontal gravity axis ~| to the keel. "With this one change, eq. (6) holds for this case also. In large ships the height of this metacentre above the centre of gravity of the ship may be as groat as 90 feet. CHAPTER Y. HYDROSTATICS (Continued) GASEOUS FLUIDS. 466. Thermometers. The temperature, or " liotness" of liquids has, within certain limits, but little influence on their statical behavior, but with gases must always be taken into account, since the three quantities, tension, temperature, and volume, of a given mass of gas are connected by a nearly in- variable law, as will be seen. An air-thermometer, Fig. 518, consists of a large glass bulb filled with air, from which projects a fine straight tube of even bore (so that equal lengths represent equal volumes). A small drop of liquid, A, sepa- rates the internal from the ex- ternal air, both of which are at a tension of (say) one at- mosphere (14.7 Ibs. per sq. inch). When the bulb is placed in melting ice (freezing-point) the drop stands at some point F in the tube ; when in boiling water (boiling under a pressure of one atmosphere), the drop is found at B, on account of the expansion of the internal air under the influence of the heat imparted to it. (The glass also expands, but only about -3-^5- as much ; this will be neglected.) The distance FB along the tube may now be divided into a convenient number of equal parts called degrees. If into one hundred degrees, it is found that each degree represents a volume equal to the -n^VW (.00367) part of the total volume occupied by the air at freez- ing-point ; i.e., the increase of volume from the temperature of freezing-point to that of the boiling-point of water = 0.367 of the volume at freezing, the pressure being the same, and even having any value whatever (as well as one atmosphere), within ordi- nary limits, so long as it is the same both at freezing and boil- 604 THERMOMETERS. 605 ing. It must be understood, however, that by temjjerature of boiling is always meant that of water boiling under one at- mosphere pressure. Another way of stating the above, if one hundred degrees are used between freezing and boiling, is as follows : That for each degree increase of temperature the in- crease of volume is ^fj of the total volume at freezing ; 273 being the reciprocal of .00367. As it is not always practicable to preserve the pressure con- stant under all circumstances with an air-thermometer, we use the common mercurial thermometer for most practical pur- poses. In this, the tube is sealed at the outer extremity, with a vacuum above the column of mercury, and its indications agree very closely with those of the air-thermometer. That equal absolute increments of volume should imply equal incre- ments of heat imparted to these thermometric fluids (under constant pressure) could not reasonably be asserted without satisfactory experimental evidence. This, however, is not al- together wanting, so that we are enabled to say that within a moderate range of temperature equal increments of heat pro- duce equal increments of volume in a given mass not only of atmospheric air, but of the so-called "perfect" or "permanent" gases, oxygen, nitrogen, hydrogen, etc. (so named before it was found that they could be liquefied). This is nearly true for mercury also, and for alcohol, but not for water. Alcohol has never been frozen, and hence is used instead of mercury as a thermometric substance to measure temperatures below the freezing-point of the latter. The scale of a mercurial thermometer is fixed ; but with an air-thermometer we should have to use a new scale, and in a new position on the tube, for each value of the pressure. 467, Thermometric Scales. In the Fahrenheit scale the tube between freezing and boiling is marked off into 180 equal parts, and the zero placed at 32 of these parts below the freez- ing point, which is hence -f- 32, and the boiling-point + 212. The Centigrade, or Celsius, scale, which is the one chiefly used in scientific practice, places its zero at freezing, and 100 at boiling-point. Hence to reduce f306 MECHANICS OF ENGINEERING. Fahr. readings to Centigrade, subtract 32 and multiply by -J ; Cent. " " Fahrenheit, multiply by and add 32. 468. Absolute Temperature. Experiment also shows that if a mass of air or other perfect gas is confined in a vessel whose volume is but slightly affected by changes of temperature, equal increments of temperature (and therefore equal incre- ments of heat imparted to the gas, according to the preceding paragraph) produce equal increments of tension (i.e., pressure per unit area) ; or, as to the amount of the increase, that when the temperature is raised by an amount 1 Centigrade, the ten- sion is increased ^-|- of its value at freezing-point. Hence, theoretically, an ideal barometer (containing a liquid unaffected by changes of temperature) communicating with the confined gas (whose volume practically remains constant) would by its indications serve as a thermometer, Fig. 519, and the attached scale could be graduated accordingly. Thus, if the col- umn stood at A when the temperature was freezing, A would be marked on the Centigrade system, and the degree spaces above and below A would each FIG. 519. _ ^i_^ Q fo e height AB, and therefore the point B (cistern level) to which the column would sink if the gas-tension were zero would be marked 273 Centi- grade. But a zero-pressure, in the Kinetic Theory of Gases ( 408), signifies that the gaseous molecules, no longer impinging against the vessel walls (so that the press. = 0), have become motionless; and this, in the Mechanical Theory of Heat, or Thermodynamics, implies that the gas is totally destitute of heat. Hence this ideal temperature of 273 Centigrade, or 460 Fahrenheit, is called the Absolute Zero of Temperature, and by reckoning temperatures from it as a starting-point, our formulas will be rendered much more simple and compact. Tempera- ture so reckoned is called absolute temperature, and will be denoted by the letter T. Hence the following rules for re- duction : GASES AND VAPORS. 607 Absol. temp. T in Cent, degrees = Ordinary Cent. + 273 ; Absol. temp. T in Fahr. degrees = Ordinary Fahr. -f- 460. For example, for 20 Cent., T = 293 Abs. Cent. 469. Distinction Between Gases and Vapors. All known gases can be converted into liquids by a sufficient reduction of temperature or increase of pressure, or both ; some, however, with great difficulty, such as atmospheric air, oxygen, hydro- gen, nitrogen, etc., these having been but recently (1878) re- duced to the liquid form. A vapor is a gas near the point of liquefaction, and does not show that regularity of behavior under changes of temperature and pressure characteristic of a gas when at a temperature much above the point of liquefac- tion. All gases treated in this chapter (except steam) are sup- posed in a condition far removed from this stage. The fol- lowing will illustrate the properties of vapors. See Fig. 520. Let a quantity of liquid, say water, be intro- THERM . duced into a closed space, previously vacuous, of considerably larger volume than the water, and furnished with a manometer and ther- mometer. Vapor of water immediately be- gins to form in the space above the liquid, and continues to do so until its pressure attains a definite value dependent on the temperature, and not on the ratio of the volume of the vessel and the origi- nal volume of water ; e.g., if the temperature is 70 Fahren- heit, the vapor ceases to form when the tension reaches a value of 0.36 Ibs. per sq. inch. If heat be gradually applied to raise the temperature, more vapor will form (with ebullition ; i.e., from the body of the liquid, unless the heat is applied very slowly), but the tension will not rise above a fixed value for each temperature (independent of size of vessel) so long as there is any liquid left. Some of these corresponding values, for water, are as follows : For a Fahr. temp. = 70 100 150 212 220 287 300 Tension (Ibs ) = Q 36 Q 93 3 6g m n 2 5 persq. in.) j = one atm. At any such stage the vapor is said to be saturated. 608 MECHANICS OF ENGINEERING. Finally, at some temperature, dependent on the ratio of the original volume of water to that of the vessel, all of the water will have been converted into vapor (i.e., steam); and if the temperature be still further increased, the tension also increases and no longer depends on the temperature alone, but also on the heaviness of the vapor when the water disappeared. The vapor is now said to be superheated, and conforms more in its properties to perfect gases. 470. Critical Temperature. From certain experiments there seems to be reason to believe that at a certain temperature, called the critical temperature, different for different liquids, all of the liquid in the vessel (if any remains, and supposing the vessel strong enough to resist the pressure) is converted into vapor, whatever be the size of the vessel. That is, above the critical temperature the substance is necessarily gaseous, in the most exclusive sense, incapable of liquefaction by pres- sure alone ; while below this temperature it is a vapor, and lique- faction will begin if, by compression in a cylinder and conse- quent increase of pressure, the tension can be raised to a value corresponding, for a state of saturation, to the temperature (in such a table as that just given for water). For example, if vapor of water at 220 Fahrenheit and tension of 10 Ibs. per sq. inch (this is superheated steam, since 220 is higher than the temperature which for saturation corresponds to p 10 Ibs. per sq. inch) is compressed slowly (slowly, to avoid change of temperature) till the tension rises to 17.2 Ibs. per sq. in., which (see above table) is the pressure of saturation for a tem- perature of 220 Fahrenheit for water- vapor, the vapor is satu- rated, i.e., liquefaction is ready to begin, and during any fur- ther slow reduction of volume the pressure remains constant and some of the vapor is liquefied. By " perfect gases," or gases proper, we may understand, therefore, those which cannot be liquefied by pressure unac- companied by great reduction of temperature; i.e., whose " critical temperatures" are very low. The critical temperature of NO 2 , or nitrous oxide gas, is between 11 and + 8 Cen- tigrade, while that of oxygen is said to be at 118 Centi- LAW OF CHARLES. 609 grade. [See p. 471, vol. 122 of the Journal of the franklin Institute. For an account of the liquefaction of oxygen, etc., see the same periodical, January to June, 1878.] 471. Law of Charles (and of Gay Lussac). The mode of gradu- ation of the air-thermometer may be expressed in the follow- ing formula, which holds good (for practical purposes) within the ordinary limits of experiment for a given mass of any perfect gas, the tension remaining constant : y=Y,+ 0.00367 V t = F (l + .00367*) ;..(!) in which ~F denotes the volume occupied by the given mass at freezing-point under the given pressure, Y its volume at any other temperature t Centigrade under the same pressure. Now, 273 being the reciprocal of .00367, we may write (273+Q. F.. T < press. ). (2) y 273 ' 'T;~r o ' 1 const, f' (see 468 ;) in which T = the absolute temperature of freezing- point, = 273 absolute Centigrade, and T the absolute tem- perature corresponding to t Centigrade. Eq. (2) is also true when T and T are both expressed in Fahrenheit degrees (from absolute zero, of course). Accordingly, we may say that, the pressure remaining the same, the volume of a given mass of gas varies directly as the absolute temperature. Since the weight of the given mass of gas is invariable at a given place on the earth's surface, we may always use the equation Yy = Y y , (3) pressure constant or not, and hence (2) may be rewritten ^ = -... (press, const.) ; . (4) Y ^o i.e., if the pressure is constant, the heaviness (and therefore the specific gravity} varies inversely as the absolute tempera- ture. 610 MECHANICS OF ENGINEERING. Experiment also shows ( 468) that if the volume [and there- fore the heaviness, eq. (3)] remains constant, while the tem- perature varies, the tension p will change according to the following relation, in which p Q = the tension when the tem- perature is freezing : (5) t denoting the Centigrade temperature. as before, we have Hence transforming, J> _ T^ j vol., ' ' and ( heav., const. (6) or, the volume and heaviness remaining constant, the tension of a given mass of gas varies directly as the absolute tempera- ture. This is called the Law of Charles (or of Gay Lussac). 472. General Formulae for any Change of State of a Perfect Gas. If any two of the three quantities, viz., volume (or heavi- ness), tension, and temperature, are changed, the new value of the third is determinate from those of the other two, according to a relation proved as follows (remember- ing that henceforth the absolute temperature only will be used, T, 468) : Fig. 521. At A a certain mass of gas at a tension of j? , one atmosphere, and absolute tempera- ture T (freezing), occupies a volume V . Let it now be heated to an absolute temp. = T', without change of tension (expanding behind a piston, for instance). Its volume will increase to a value "F which from (2) of 471 will satisfy the relation FIG. 521. Z - r] L V ~~ T * v o * (7) (See B in figure.) Let it now be heated without change of volume to an abso- lute temperature T (C in figure). Its volume is still V, but LAWS OF PERFECT GASES. 611 the tension has risen to a value p, such that, on comparing B and C by eq. (6), we have Combining (7) and (8), we obtain for any state in which the tension is _p, volume V, and absolute temperature T, in (General) . . . %-= - = ^7fr - 5 or ^=- = a constant ; . (9) or (General) .... -*^L w ^ n n , (10) which, since (General). . Vy =-- V,y, = V m y m = V nYn , . . . (11) is true for any change of state, we may also write (General) . . . J . -= = -=%r, ( 12 ) or P P* n Q\ y m T m -y n T n - These equations (9) to (13), inclusive, hold good for any state of a mass of any perfect gas (most accurately for air). The subscript refers to the state of one-atmosphere tension and freezing-point temperature, in and n to any two states what- ever (within practical limits) ; y is the heaviness, 7 and 409, and 7 7 the absolute temperature, 468. If j?, T 7 ", and T oi equation (9) be treated as variables, and laid off to scale as co-ordinates parallel to three axes in space, respectively, the surface so formed of which (9) is the equation is a hyperbolic paraboloid. 473. Examples. EXAMPLE 1. What cubic space will be occupied by 2 Ibs. of hydrogen gas at a tension of two atmos- pheres and a temperature of 27 Centigrade? 612 MECHANICS OF ENGINEERING. With the inch-lb.-sec. system we have p = 14.7 lbs. per sq. inch, y. = [.0056 -f- 1728] Ibs. per cubic inch, and T = 273 absolute Centigrade, when the gas is at freezing-point at one atmosphere (i.e., in state sub-zero). In the state mentioned in the problem, we havej? = 2 X 14.7 Ibs. per sq. in., T 273 + 27 = 300 absolute Centigrade, while y is required. Hence, from eq. (12), 2 X 14.7 = 14.7 Y 300 ~ (.0056 -r- 1728)273 ' /. Y = ' Ibs. per cub. in. = .0102 Ibs. per cub. foot ; and if 17-28 the total weight, = 6r, = Vy, is to be 2 Ibs., we have (ft., lb., sec.) V 2 -f-.0102 = 196 cubic feet. Ans. EXAMPLE 2. A mass of air originally at 24 Centigrade and a tension indicated by a barometric column of 40 inches of mercury has been simultaneously reduced to half its former volume and heated to 100 Centigrade ; required its tension in this new state, which we call the state n, in being the original state. Use the inch, lb., sec. "We have given, there- fore, p m = nx 14 - 7 lbs - P er s q- incll > T = 2 ? 3 + 24 = 29 ? absolute Centigrade, the ratio V m : V n = 2 : 1, and T n = 273 + 100 =373 Abs. Cent.; while p n is the unknown quantity. From eq. (10), hence, p n = ^ . ^- .p m = 2 X m f * X 14.7 = 49.22 Ibs. per sq. in., * n -*-m which an ordinary steam-gauge would indicate as (49.22 - 14.7) = 34.52 lbs. per sq. inch. (That is, if the weather barometer indicated exactly 14.7 lbs. per sq. inch.) EXAMPLES. PERFECT GASES. 613 EXAMPLE 3. A mass of air, Fig. 522, occupies a rigid closed vessel at a temperature of 15 Centigrade (equal to that of sur- rounding objects) and a tension of four atmospheres [state m~\. By opening a stop-cock a few seconds, thus allowing a portion of the gas to escape quickly, and then shutting it, the remainder FIG. 522. of the air [now in state ri\ is found to have a tension of only 2.5 atmospheres (measured immediately) ; its temperature can- not be measured immediately (so much time being necessary to affect a thermometer), and is less than before. To compute this temperature, T n , we allow the air now in the vessel to come again to the same temperature as surrounding objects (15 Centigrade) ; find then the tension to be 2.92 atmospheres. Call the last state, state r (inch, lb., sec.). The problem then stands thus : p m = 4 x 14.7 rm = ? T m = 288 Abs. Cent. p n = 2.5 X 14.7 princip unknown n _ y j principal Pr = 2.92 X 14.7 y r = y n (since V r = Vn) T r = T m = 288 Abs. Cent. In states n and r the heaviness is the same ; hence an equa* tion like (6) of 471 is applicable, whence or 27 Centigrade ; considerably below freezing, as a result of allowing the sudden escape of a portion of the air, and the con- sequent sudden expansion, and reduction of tension, of the re- mainder. In this sudden passage from state m to state n, the remainder altered its heaviness (and its volume in inverse ratio) in the ratio (see eqs. (11) and (10) of 472) * m ~T n 2.5 X 14.7 288 4 X 14.7 * 246 ISTow the heaviness in state m (see eq. (12), 472) was 614 MECHANICS OF ENGINEERING. V - 7m 14.7 .0807 273 .306 p. 288 1728 14.7 1728 Ibs. per cub. in. = .306 Ibs. per cub. ft. .-. y n = 0.73 X y m = 0.223 Ibs. per cub. ft., and also, since V m = 0.73 T^, about T 2 T 7 F of the original quan- tity of air in vessel has escaped. [JSToxE. By numerous experiments like this, the law of cooling, when a mass of gas is allowed to expand suddenly (as, e.g., behind a piston, doing work) has been determined ; and vice versa, the law of heating under sudden compression ; see T' it] aL 474. The Closed Air-manometer. If a manometer be formed of a straight tube of glass, of uniform cylindrical bore, which is partially filled with mercury and then inverted in a cistern of mercury, a quantity of air having been left between the mercury and the upper end of the tube, which is closed, the tension of this confined air (to be computed from its observed volume and tem- perature) must be added to that due to the mercury column, in order to obtain the tension^' to be measured. See Fig. 523. The advantage of this kind of instrument is, that to meas- ure great tensions the tube need not be very long. Let the temperature T 7 , of whole instrument, and the tension p 1 of the air or gas in the cistern, be known when the mercury in the tube stands at the same level as that in the cistern. The tension of the air in the tube must now be p, also, its temperature T 7 , , and its. volume is V 1 = Fh^ , F being the sectional area of the bore of the tube ; see on left of figure. "When the instrument is used, gas of unknown tension p' is admitted to the cistern, the tem- perature of the whole instrument being noted (= T\ and the heights h and h" are observed (h -f- h" cannot be put = //., FIG. 523. CLOSED AIR-MANOMETER. 615 unless the cistern is very large), p' is then computed as fol- lows (eq. (2), 413) : (i) in which p = the tension of the air in the tube, and y m the heaviness of mercury. But from eq. (10), 472, putting 7, T h, T f^T'-T^-ST Hence finally, from (1) and (2), (3) Since T 7 , , p l , and h l are fixed constants for each instrument, we may, from (3), compute p 1 for any observed values of h and T(N".B. T and T t are absolute temperatures), and construct a series of tables each of which shall give values of p' for a range of values of A, and one special value of T. EXAMPLE. Supposing the fixed constants of a closed air- manometer to be (in inch-lb.-sec. system) p^ 14.7 (or one atmosphere), T, = 285 Abs. Cent, (i.e., 12 Centigrade), and Aj 3' 4" = 40 inches ; required the tension in the cistern indicated by h" 25 inches and k 15 inches, when the temperature is 3 Centigrade, or T = 270 Abs. Cent. For mercury, y m = [848.7 -=- 1728] ( 409) (though strictly it should be specially computed for the temperature, since it varies about .00002 of itself for each Centigrade degree). Hence, eq. (3), Ibs. per sq. inch, or nearly 3J atmospheres [steam-gauge would read 34.7 Ibs. per sq. in.]. 475. Marietta's Law, (or Boyle's,) Temperature Constant ; i.e., Isothermal Change. If a mass of gas be compressed, or al- 616 MECHANICS OF ENGINEERING. lowed to expand, isothermally, i.e., without change of tern, perature (practically this cannot be done unless the walls of the vessel are conductors of heat, and then the motion must be slow), eq. (10) of 472 now becomes (since T m = T n ) ( MariottJs Law, ) TT - v p n " V m ' i.e., the temperature remaining unchanged, the tensions are inversely proportional to the volumes, of a given mass of a perfect gas ; or, the product of volume by tension is a constant quantity. Again, since V m y m V n y n for any change of state, ( .MariottJs Law, ) Pm_Ym Qr Pm __ Pn . /2\ [ Temp, constant j p n ~ y n ' y m y n ' i.e., the pressures (or tensions are directly proportional to the (first power of the) heavinesses, if the temperature is the same. This law, which is very closely followed by all the perfect gases, was discovered by Boyle in England and Mariotte in France more than two hundred years ago, but of course is only a particular case of the general formula, for any change of state, in 472. It may be verified experimen- p ' ' tally in several ways. E.g., in Fig. 524, the tube M being closed at the top, while PN is open, let mercury be poured in at P until it reaches the level A ' B '. The air in OA is now at a tension of one atmosphere. Let more mer- cury be slowly poured in at P, until the air confined in has been compressed to a volume OA" = i of OA, and the height B"E" then measured ; it will be found to be 30 inches ; i.e., the tension of the air in O is now two atmos- pheres (corresponding to 60 inches of mercury). FIG. 524. Again, compress the air in O to \ its original volume (when at one atmosphere), i.e., to volume OA'" = \OA', and the mercury height B'"E"' will be 60 inches, show- ing a tension of three atmospheres in the confined air at (90 MAKIOTTES LAW. 617 inches of mercury in a barometer). It is understood that the temperature is the same, i.e., that time is given the compressed air to acquire the temperature of surrounding objects after being heated by the compression, if sudden. [NOTE. The law of decrease of steam-pressure in a steam- engine cylinder, after the piston has passed the point of " cut- off " and the confined steam is expanding, does not materially differ from Mariotte's law, which is often applied to the case of expanding steam ; see 479.] While Mariotte's law may be considered exact for practical purposes, it is only approximately true, the amount of the deviations being different at different temperatures. Thus, for decreasing temperatures the product Vp of volume by tension becomes smaller, with most gases. EXAMPLE 1. If a mass of compressed air expands in a cylinder behind a piston, having a tension of 60 Ibs. per sq. inch (45.3 by steam-gauge) at the beginning of the expansion, which is supposed slow (that the temperature may not fall) ; then when it has doubled in volume its tension will be only 30 Ibs. per sq. inch ; when it has tripled in volume its tension will be only 20 Ibs. per sq. inch, and so on. EXAMPLE 2. Diving-bell. Fig. 525. If the cylindrical diving-bell AB is 10 ft. in height, in what depth, h = ?, of salt water, can it be let down to the bottom, without allowing the water to rise in the bell more than a distance a = 4 ft. ? Call the horizontal sectional area, F. The mass of air in the bell is constant, at a constant temperature. First, algebraically ; at the surface this mass of air occupied a volume V m = FTi" at a tension p m = 14. Y X 144 Ibs. per sq. ft., while at the depth mentioned it is compressed to a volume V n = F(h" a\ and is at a tension p n =p m + (h a)y w , in which y w = heaviness of salt water. Hence, from eq. (1), XT-- Vm - -d); . . (3) 618 MECHANICS OF ENGINEERING. hence, numerically, (ft.. lb., sec.,) 476. Mixture of Gases. It is sometimes stated that if a vessel is occupied by a mixture of gases (between which there is no chemical action), the tension of the mixture is equal to the sum of the pressures of each of the component gases present ; or, more definitely, is equal to the sum of the pressures which the separate masses of gas would exert on the vessel if each in turn occupied it alone at the same temperature. This is a direct consequence of Mariotte's law, and may be demonstrated as follows : Let the actual tension be p, and the capacity of the vessel V. Also let Fj , F 2 , etc., be the volumes actually occupied by the separate masses of gas, so that V t + V,+ . . . = V; ..... (1) and Pupt, etc., the pressures they would individually exert when occupying the volume V alone at the same tempera- ture. Then, by Mariotte's law, Vp t =V lP -, Vp t =V#; etc.; ... (2) whence, by addition, we have ...) = (Vi + F 2 +...f; i.e., 1> =?,+!>, -Jr.--.. . .... (3) Of course, the same statement applies to any number of separate parts into which we may imagine a mass of homo- geneous gas to be divided. For numerical examples and practical questions in the solu- tion of which this principle is useful, see p. 239, etc., Ran- kine's Steam-engine. (Rankine uses 0.365, where 0.367 has been used here.) BAROMETRIC LEVELLING. 619 477. Barometric Levelling. By measuring with a barometer the tension of the atmosphere at two different levels, simul- taneously, and on a still day, the two localities not being widely separated horizontally, we may compute their vertical distance apart if the temperature of the stratum of air between them is known, being the same, or nearly so, at both m stations. Since the heaviness of the air is different in different layers of the vertical : column between the two elevations J^and M, ': Fig. 526, we cannot immediately regard the whole of such a column as a free body (as was .. ILr:y; -f- dp. Let the area of the base of lamina be F; then the vertical forces acting on the lamina are Fp, down- ward ; its weight yFdz downward ; and F(p + dp) upward. For its equilibrium ^(vert. compons.) must = ; /. F(p + dp) - Fp - Fydz = 0; i.e., dp = ydz, (1) which contains three variables. But from Mariotte's law, 475, eq. (2), if p n and y n refer to the air at N, we may substitute y=^p and obtain, after dividing by p, to separate Pn the variables/* and 2, 620 MECHANICS OF ENGINEERING. :*;*. (2) Yn P Summing equations like (2), one for each lamina between M (where p =p m and s = 0) and N( where p =j} n &ndz = h\ we have = /V which gives A, the difference of level, or altitude, between M and N, in terms of the observed tensions p n and^? TO , and of y n , the heaviness of the air at ^V, which may be computed from eq. (12), 472, substituting from which we have finally in which the subscript refers to freezing-point and one at- mosphere tension ; T n and T are absolute temperatures. For the ratio p n : p m we may put the equal ratio h n : h m of the actual barometric heights which measure the tensions. The log. e (or Naperian, or natural, or hyperbolic, log.) (common log. to base 10) X 2.30258. From 394, y of air 0.0807ft Ibs. per cub. ft., and^ = 14.701 Ibs. per sq. inch ; T = 273 Abs. Cent. If the temperature of the two stations (both in the shade) are not equal, a mean temp. = %(T m -{- T n ) may be used for T n in eq. (4), for approximate results. Eq. (4) may then be written ... (5) The quantity ^ = 26213 ft., just substituted, is called the Yo height of the homogeneous atmosphere, i.e., the ideal height which the atmosphere would have, if incompressible and non- BAROMETRIC LEVELLING ADIABATIC LAW. 621 expansive like a liquid, in order to exert a pressure of 14.701 Ibs. per sq. inch upon its base, being throughout of a constant heaviness = .08076 Ibs. per cub. foot. By inversion of eq. (4) we may also write .r. ' rj (6) where e = 2.71828 = the Naperian Base, which is to be raised rri to the power whose index is the abstract number . - . A Po T n a,nd the result multiplied by j? OT to obtain p n . EXAMPLE. Having observed as follows (simultaneously) : At lower station jV, h n = 30.05 in. mercury ; temp. = 77.6 F. ; "upper M,h m = 23.26" "' = 70.4F.; required the altitude h. From these figures we have a mean absolute temperature of 460 + (77.6 + 70.4) = 534 Abs. Fahr. ; hence, from (5), h = 26213 X Hi X 2.30258 X log. 10 = 6787.9 ft. (Mt. Guanaxuato, in Mexico, by Baron von Humboldt.) Strictly, we should take into account the latitude of the place, since y varies with g (see 76), and also the decrease in the intensity of gravitation as we proceed farther from the earth's centre, for the mercury in the barometer weighs less per cubic inch at the upper station than at the lower. Tables for use in barometric levelling can be found in Traut- wine's Pocket-book, and in Searles's Field-book for Railroad Engineers, as also tables of boiling-points of water under dif- ferent atmospheric pressures, forming the basis of another method of determining heights. 478. Adiabatic Change Poisson's Law. By an adiabatic change of state, on the part of a gas, is meant a compression or expansion in which work is done upon the gas (in compress- 622 MECHANICS OF ENGINEERING. ing it) or by the gas (in expanding against a resistance) when there is no transmission of heat between the gas and enclosing vessel, or surrounding objects, by conduction or radiation. This occurs when the volume changes in a vessel of non-con- ducting material, or when the compression or expansion takes place so quickly that there is no time for transmission of heat to or from the gas. The experimental facts are, that if a mass of gas in a cylinder be suddenly compressed to a smaller volume its temperature is raised, and its tension increased more than the change of vol- ume would call for by Mariotte's law ; and vice versa, if a gas at high tension is allowed to expand in a cylinder and drive a piston against a resistance, its temperature falls, and its tension diminishes more rapidly than by Mariotte's law. Again (see Example 3, 473), if -$fc of the gas in a rigid vessel, originally at 4 atmos. tension and temperature of 15 Cent., is allowed to escape suddenly through a stop-cock into the outer air, the remainder, while increasing its volume in the ratio 100 : 73, is found to have cooled to 27 Cent., and its tension to have fallen to 2.5 atmospheres; whereas, by Mariotte's law, if the temperature had been kept at 288 Abs. Cent., the tension would have been lowered to y 7 ^ of 4, i.e., to 2.92 atmospheres only. The reason for this cooling during sudden expansion is, ac- cording to the Kinetic Theory of Gases, that since the " sensi- ble heat" (i.e., that perceived by the thermometer), or " hot- ness" of a, gas depends on the velocity of its incessantly moving molecules, and that each molecule after impact with a receding piston has a less velocity than before, the temperature neces- sarily falls; and vice versa, when an advancing piston com- presses the gas into a smaller volume. If, however, a mass of gas expands without doing work, as when, in a vessel of two chambers, one a vacuum, the other full of gas, communication is opened between them, and the gas allowed to fill both chambers, no cooling is noted in the mass as a whole (though parts may have been cooled tem- porarily). By experiments similar to that in Example 3, 473, it has ADIABATIC CHANGE EXAMPLES. 623 been found that for air and the " perfect gases," in an adiabatic change of volume [and therefore of heaviness], the tension varies inversely with the 1.41th power of the volume. This. is called Poissorfs Law. For ordinary purposes (as Weisbach suggests) we may use f instead of 1.41, and hence write Adiabat. \ Pm _ f7 m \* _ p m _ ( V n \* Change] ~ (*/' ' T = W' ' ' and combining this relation with the general eqs. (10) and (13), 472, we have also Adiabat. ) Pm _ I T^ Change ] ' p n " \~T n ) > i.e., the tension varies directly as the cube of the absolute tem- perature; also, Adiabat. ) / KA _ (T n \ a y n _ iT n \\ Change] fe$^W' \Ej' i.e., the volume is inversely, and the heaviness directly, as the square of the absolute temperature. Here m and n refer to any two adiabatically related states. T is the absolute temperature. EXAMPLE 1. Air in a cylinder at 20 Cent, is suddenly compressed to -J- its original volume (and therefore is six times as dense, i.e., has six times the heaviness, as before). To what temperature is it heated ? Let m be the initial state, and n the final. From eq. (3) we have = 718 Abe. Cent, or nearly double the absolute temperature of boiling water. EXAMPLE 2. After the air in Example 1 has been given time to cool again to 20 Cent, (temperature of surrounding objects) it is allowed to resume, suddenly, its first volume, i.e., 624 MECHANICS OF ENGINEERING. to increase its volume sixfold by expanding behind a piston. To what temperature has it cooled ? Here T m 293 Abs. Cent., the ratio V m I V n = , and T n is required. Hence, from (3), ' = 293 ^ v ~*~ = 119<5 Abs - or = 154 Cent., indicating extreme cold. From these two examples the principle of one kind of ice- making apparatus is very evident. As to the work necessary to compress the air in Example 1, see 483. It is also evident why motors using compressed air expansively have to encoun- ter the difficulty of frozen watery vapor (present in the air to some extent). EXAMPLE 3. What is the tension of the air in Example 1 (suddenly compressed to -J- its original volume) immediately after the compression, if the original tension was one atmos- phere? That is, with V n : V m = 1 : 6, and j? w = 14.7 Ibs. per sq. inch, p n = 2 From eq. (1), (in., lb., sec.,) p n = 14.7 X 6* = 14.7 V2l6 = 216 Ibs. per sq. inch ; whereas, if, after compression and without change of volume, it cools again to 20 Cent., the tension is only 14.7 X 6 = 88.2 Ibs. per sq. inch (now using Mariotte's law). 479. Work of Expanding Steam following Mariotte's Law. Although gases do not in general follow Mariotte's law in ex- panding behind a piston (without special provision for sup- plying heat), it is found that the tension of saturated steam (i.e., saturated at the beginning of the expansion) in a steam- engine cylinder, when left to expand after the piston has passed the point of " cut-af" diminishes very nearly in accordance with Mariotte's law, which may therefore be ap- plied in this case to find the work done per stroke, and thence the power. In Fig. 527 a horizontal steam-cylinder is EXPANDING STEAM. 625 I- shown in which the piston is making its left-to-right stroke. The " back-pressure" is con- stant and = Fq, F being the area of the piston and q the intensity (i.e., per unit area) of the back or exhaust pres- sure on the right side of the piston ; while the forward pressure on the left face of the piston = Fp, in which p is the steam-pressure per unit area, and is different at different points of the stroke. While the piston is passing from O" to D" ,p is constant, being p b = the boiler-pressure, since the steam-port is still open. Between D" and C" , however, the steam being cut off (i.e., the steam-port is closed) at D" , a dis- tance a from 0' ', p decreases with Mariotte's law (nearly), and its value is (Fa -r- Fx)p b at any point on C"D fl ', x being the distance of the point from 0" . Above the cylinder, conceive to be drawn a diagram in which an axis OX\ \\ to the cylinder-axis, OY an axis 1 to the same, while is vertically above the left-hand end of the cylinder. As the piston moves, let the value of p correspond- ing to each of its positions be laid off, to scale, in the vertical immediately above the piston, as an ordinate from the axis X. Make OD' = q by the same scale, and draw the horizontal D'C'. Then the effective work done on the piston-rod while it moves through any small distance dx is d W = force X distance = F( p q)dx, and is proportional to the area of the strip RS, whose width is dx and length = p q ; so that the effective work of one stroke is --q)dx, . . . (1) 626 MECHANICS OF ENGINEERING. and is represented graphically by the area A'ARBC'D'A'* From 0" to D" p is constant and = p b (while q is constant at all points), and x varies from to a ; q )a, . . (2) which may be called the work of entrance, and is represented by the area of the rectangle A' ADD'. From D" to Q " p is variable and, by Mariotte's law, = p b ; .,- q (l-a) ... (3) = the work of expansion, adding which to that of entrance, we have for the total effective work of one stroke By effective work we mean that done upon the piston-rod and thus transmitted to outside machinery. Suppose the engine to be " double-acting" ; then at the end of the stroke a communication is made, by motion of the proper valves, be- tween the space on the left of the piston and the condenser of the engine ; and also between the right of the piston and the boiler (that to the condenser now being closed). On the return stroke, therefore, the conditions are the same as in the forward stroke, except that the two sides of the piston have changed places as regards the pressures acting on them, and thus the same amount of effective work is done as before. If n revolutions of the fly-wheel are made per unit of time (two strokes to each revolution), the effective work done per unit of time, i.e., the power of the engine, is Z = 2n W= ZnF [ap^l + log.. (1)1 - ql^. . (5) WORK OF STEAM-ENGINES. 6 '27 For simplicity the above theory has omitted the considera- tion of " clearance" that is, the fact that at the point of " cut- off " the mass of steam which is to expand occupies not only the cylindrical volume Fa, but also the " clearance" or small space in the steam-passages between the valve and the entrance of the cylinder, the space between piston and valve which is never encroached upon by the piston. " Wire-drawing" has also been disregarded, i.e., the fact that during communication with the boiler the steam-pressure on the piston is a little less than boiler-pressure. For these the student should consult special works, and also for the consideration of water mixed with the steam, etc. Again, a strict analysis should take into account the difference in the areas which receive fluid-pressure on the two sides of the piston. EXAMPLE 1. A reciprocating steam-engine makes 120 revo- lutions per minute, the boiler-pressure is 40 Ibs. by the gauge (i.e.,^> & = 40 -f- 14.7= 54.7 Ibs. per sq. inch), the piston area is F= 120 sq. in., the length of stroke I = 16 in., the steam being "cut off" at J stroke (/. a = 4 in., and I : a = 4.00), and the exhaust pressure corresponds to a " vacuum of 25 inches" (by which is meant that the pressure of the exhaust steam will balance 5 inches of mercury), whence q -f$ of 14.7 = 2.45 Ibs. per sq. inch. Required the work per stroke, W, and the corresponding power Z. Since I : a = 4, we have log. e 4 = 2.302 X .60206 = 1.386, and from eq. (4), (foot, lb., sec.,) W= | (54.7 X 144) . j. . [2.386] - -fjf (2.45 X 144) . f = 5165.86 - 392.0 = 4773.868 ft. Ibs. of work per stroke, and therefore the power at 2 rev. per sec. (eq. 5) is Z = 2 X 2 X 4773.87 = 19095.5 ft. Ibs. per second. Hence in horse-powers, which, in ft.,-lb.-sec. system, = Z-i-550, Power = 19095.5 -5- 550 = 34.7 H. P. EXAMPLE 2. Required the weight of steam consumed per 628 MECHANICS OF ENGINEERING. second by the above engine with given data ; assuming with Weisbach that the heaviness of saturated steam at a definite pressure (and a corresponding temperature, 469) is about f of that of air at the same pressure and temperature. The heaviness of air at 54.7 Ibs. per sq. in. tension and temperature 287 Fahr. (see table, 469) would be, from eq. (12) of 472 (see also 409), _yj\ j> 0807X493 64.7 T ' p o 460 + 287 '14.7 Ibs. per cub. foot, f of which is 0.1237 Ibs. per cub. ft. Now the volume of steam, of this heaviness, admitted from the boiler at each stroke is V = Fa = |f . = 0.2777 cub. ft. ? and therefore the weight of steam used per second is 4 X .2777 X 0.1237 = 0.1374 Ibs. Hence, per hour, 0.1374 X 3600 = 494.6 Ibs. of feed-water are needed for the boiler. If, with this same engine, the steam is used at full boiler pressure throughout the whole stroke, the power will be greater, viz. = %nFl(p b q) 33440 ft. Ibs. per sec., but the consumption of steam will be four times as great; and hence in economy of operation it will be only 0.44 as efficient (nearly). 480. Graphic Representation of any Change of State of a Con- fined Mass of Gas. The curve of expansion AB in Fig. 527 is an equilateral hyperbola, the axes JTand Y being its asymp- totes. If compressed air were used instead of steam its ex- pansion curve would also be an equilateral hyperbola if its temperature could be kept from falling during the expansion (by injecting hot-water spray, e.g.), and then, following Mariotte's law, we would have, as for steam, ( 475,) p V= con- stant, i.z.,pFx = constant, and therefore px = constant, which is the equation of a hyperbola, p being the ordinate and x the abscissa. This curve (dealing with a perfect gas) is also called an isothermal, the x and y co-ordinates of its points being pro- GRAPHICS OF CHANGE OF STATE OF GAS. 629 X portional to the volume and tension, respectively, of a mass of air (or perfect gas) whose temperature is maintained constant. Hence, in general, if a mass of gas be confined in a rigid cylinder of cross-sec- tion F (area), provided with an air-tight pis- ton, Fig. 528, its vol- ume, Fx, is propor- tional to the distance OD =-- x (of the piston from the closed end of the cylinder) taken as an abscissa, while its o tension p at the same instant may be laid off as an ordinate from D. Thus a point A is fixed. Describe an equilateral hyperbola through A, asymptotic to X and I 7 , and mark it with the ob- served temperature (absolute) of the air at this instant. In a similar way the diagram can be filled up with a great number of equilateral hyperbolas, or isothermal curves, each for its own temperature. Any point whatever (i.e., above the critical temperature) in the plane angular space YOX will indicate by its co-ordinates a volume and a tension, while the correspond- ing absolute temperature T will be shown by the hyperbola passing through the point, since these three variables always satisfy the relation ( 472) 528. pV . pFx -Zj,- = const. ; i.e., ^- . . . . (1)' Any change of state of the gas in the cylinder may now be represented by a line in the diagram connecting the two points corresponding to its initial and final states. Thus, a point moving along the line AB, a portion of the isothermal marked 293 Abs. Cent, represents a motion of the piston from D to E, and a consequent increase of volume, accompanied by just sufficient absorption of heat by the gas (from other bodies) to maintain its temperature at that figure (viz., its temperature at 630 MECHANICS OF ENGINEERING. A). If the piston move from D to E, without transmission of heat, i.e., adiabatically ( 478), the tension falls more rapidly, and a point moving along the line AB 1 represents the corresponding continuous change of state. AB' is a portion of an adiabatic curve, whose equation, from 478, is iL = [llfj ' r 1** =#***? = <**&.; 00 in which p K and X K refer to the point K where this particular adiabatic curve cuts the isothermal of freezing-point. Evi- dently an adiabatic may be passed through any point of the diagram. The mass of gas in the cylinder may change its state from A to B' by an infinite number of routes, or lines on the diagram, the adiabatic route, however, being that most likely to occur for a rapid motion of the piston. For example, we may cool it without allowing the piston to move (and hence without altering its volume Xor the abscissa x) until the pres- sure falls to a value p B > = DL = EB\ and this change is rep- resented by the vertical path from A to L ; and then allow it to expand, and push the piston from D to E (i.e., do external work), during which expansion heat is to be supplied at just such a rate as to keep the tension constant, ^=pp> =PLI this latter change corresponding to the horizontal path LB' from L to B f . It is further noticeable that the work done by the expanding gas upon the near face of the piston (or done upon the gas when compressed) when the space dx is described by the piston, is = Fpdx, and therefore is proportional to the area pdx of the small vertical strip lying between the axis X and the line or route showing the change of state ; whence the total work done on the near piston-face, being = Ffpdx, is represented by the area fpdx of the plane figure between the initial and final ordinates, the axis X and the particular route followed be- tween the initial and final states (K.B. We take no account here of the pressure on the other side of the piston, the latter depending on the style of engine). For example, the work done on the near face of the piston during adiabatic expansion WORK OF ADIABATIC EXPANSION. 631 from D to E is represented by the plane figure AB'EDA, and is measured by its area. The mathematical relations between the quantities of heat imparted or rejected by conduction and radiation, and trans- formed into work, in the various changes of which the con- fined gas is capable, belong to the subject of Thermodynamics, which cannot be entered upon here. It is now evident how the cycle of changes which a definite mass of air or gas experiences when used in a hot-air engine, compressed-air engine, or air-compressor, is rendered more in- telligible by the aid of such a diagram as Fig. 528 ; but it must be remembered that during the entrance into, or exit from, the cylinder, of the mass of gas used in one stroke, the distance x does not represent its volume, and hence the locus of the points in the diagram determined by the co-ordinates p and x during entrance and exit does not indicate changes of state in the way just explained for the mass when confined in the cylinder. However, the work done by or upon the gas during entrance and exit will still be represented by the plane figure included by that locus (usually a straight horizontal line, pressure constant) and the axis of X and the terminal ordinates. 481. Adiabatic Expansion in an Engine using Compressed Air. Fig. 529. Let the compressed air at a tension p m and an absolute temperature T m be supplied from a reservoir (in which the loss is continually made good by an air-corn- pressor). Neglecting the resistance of the port, its tension and temperature when behind the piston are still p m and T m . Let x n = length of stroke, and o let the cut-off (or closing of communi- cation with the reservoir) be made at some point D where x x m , the posi- tion of D being so chosen (i.e., the ratios? : x n so computed) that after adia- , , . . - T\ j. TV A* FIG. 529. batic expansion from D to E the pres- sure shall have fallen fromjp m at M (state m) to a value p n = 632 MECHANICS OF ENGINEERING. = one atmosphere at JV(state n\ at the end of stroke ; so that when the piston returns the air will be expelled ("exhausted") at a tension equal to that of the external atmosphere (though at a low temperature). Hence the back-pressure at all points either way will be = p n per unit area of piston, and hence the total back-pressure = Fp n , F being the piston area. From to D the forward pressure is constant and = Fp n . and the effective work, therefore, or work on piston-rod from O to D, is W = F[p m Pn\m) * (1). represented by the rectangle M'MLN'. The cut-off being made at D, the volume of gas now in the cylinder, viz., Y m = Fx m , is left to expand. Assuming no device adopted (such as injecting hot-water spray) for preventing the cooling and rapid decrease of tension during expansion, the latter is- adiabatic, and hence the tension at any point P between M and N will be P=p m * . [see 478; V=JM]; . . (ay .: Work of expansion -p n }dx = Ff*pdx - Fp n (x n - O, (2) m I and is represented by the area MPNL. i /I \t Now substitute (3) in (2) and then add (2) to (1), noting that COMPEESSED-AIR ENGINE. 633 which furthermore, since n and m are adiabatically related [see (0)], can be reduced to and we have finally : Total work on piston- \ ^ Q ^ I", /aj w \*~| rod per stroke \ ~~ = 8J ^ [ ^ J' * But Fx m = V m , and the adiabatic relation holds good, e therefore we may also write TT= 3 F MJ t>l -?; .... (5) in which V m = the volume which the mass of air used per stroke occupies in the state m, i.e., in the reservoir, where the tension is^? m and the absolute temperature = T m . To find the work done per pound of air used (or other unit of weight), we must divide W by the weight G = V m y m of the air used per stroke, remembering (eq. (13), 472) that V m y m = [ Vp m y.T.] * (T m p.). Work per unit of weight of\_ T p, H _(Pn^ r \ /g\ air used in adiabatic working \ ' m y o T\_ \p m ) J The back-pressure p n =p a = one atmosphere. In (6) y = .0807 Ibs. per cub. foot, p 14.7 Ibs. per sq inch, and T = 273 Abs. Cent, or 492 Abs. Fahr. 634 MECHANICS OF ENGINEERING. It is noticeable in (6) that for given tensions p m and p n) the work per unit of weight of air used is proportional to the ab- solute temperature T m of the reservoir. The temperature T m to which the air has cooled at the end of the stroke is obtained as in Example 2, 478, and may be far below freezing-point unless T m is very high or the ratio of expansion, x m : x n , large. EXAMPLE. Let the cylinder of a compressed-air engine have a section of F = 108 sq. in. and a stroke x n = 15 inches. The compressed air entering the cylinder is at a tension of 2 atmos. (i.e., p m = 29.4 Ibs. per sq. in., and p n --p m = -J), and at a temperature of 27 Cent, (i.e., T m = 300 Abs. Cent.). Ke- quired the proper point of cut-off, or a? m = ? , in order that the tension may fall to one atmosphere at the end of the stroke ; also the work per stroke, and the work per pound of air. Use the/002, pound, and second. From eq. (a), above, we have m =x n f^ = 1*354 / T = 0.7875 ft. = 9.45 inches, W V 4 and hence the volume of air in state m, used per stroke [eq. (5)] is V m = Fx m = |f| X 0.7875 = 0.5906 cubic feet ; while the work per stroke is W = 3 X 0.5906 X 29.4 X 144 X [1 - (i)* ] = 1545 ft. Ibs., and the work obtained from each pound of air, eq. (6), ft. Ibs. per pound of air used. The temperature to which the air has cooled at the end of stroke [eq. (2), 478] is COMPKESSED-AIR ENGINE. 635 T n = T m : &- = 300 X V i = 300 X .794 = 238 Abs. C. ; i.e., 35 Centigrade. 482. Remarks on the Preceding. This low temperature is objectionable, causing, as it does, the formation and gradual accumulation of snow, from the watery vapor usually found in small quantities in the air, and the ultimate blocking of the ports. By giving a high value to T m , however, i.e., by heat- ing the reservoir, T n will be correspondingly higher, and also the work per pound of air, eq. (6). If the cylinder be encased in a " jacket" of hot water, or if spray of hot water be injected behind the piston during expansion, the temperature may be maintained nearly constant, in which event Mariotte's law will hold for the expansion, and more work will be obtained per pound of air; but the point of cut-off must be differently placed. Thus if, in eq. (4), 479, we make the back-pressure, which = (Fa -s- Fl)p b , equal to the value to which the air- pressure has fallen at the end of the stroke by Mariotte's law, we have Work per stroke with 1 __ c* i (l\_ Y 1 (^\ C[\ isotherm, expans. f ~~ a ^ b &' \a/~ b ^ b ' \^"/> v and hence Work per unit of weight of air, \ _ j, p i /_ \ ,~ with isothermal expansion ) m y T ^'* v/ Applying these equations to the data of the example, we obtain Work per unit of weight of air with iso- \ _ Q go y 1 P . thermal expansion ) " m y,T\ ' whereas, with adidbatic expansion, work I _ Q gg T 7 P per unit of weight of air is only } " "V ^V 636 MECHANICS OF ENGINEERING. 483. Double-acting Air-compressor, with Adiabatic Compres- sion. This is the converse of 481. In Fig. 530 we have the piston moving from right to left, compressing a mass of air which at the beginning of the stroke fills the cylinder. This is brought about by means of an external motor (steam-engine or turbine, e.g.) which exerts a thrust or pull along the piston-rod, enabling it with the help of the atmospheric pressure of the fresh supply of air flowing in behind x it, to first compress a cylinder-full of air to the tension of the compressed p air in the reservoir, and then, the Lj port or valve opening at this stage, to force or deliver it into the reservoir. Let the temperature and tension of the cylinder-full of fresh air be T Ul and p ni , and the tension in the reservoir be p mi . Suppose the compression adiabatic. As the piston passes from E toward the left, the air on the left has no escape and is compressed, its tension and temperature increasing adiabatically until it reaches a value p mi = that in reservoir, at which instant, the piston being at some point D, a valve opens and the further progress of the piston simply transfers the compressed air into the re- servoir without further increasing its tension. Throughout the whole stroke the piston-rod has the help of one atmosphere pressure on the right face, since a new supply of air is entering on the right to be compressed in its turn on the return stroke. The work done from E to D may be called the work of com- pression that from D to 0, the work of delivery. [Since, here, dx and dW(or increment of work) have con- trary signs, we introduce the negative sign as shown.] r D The work of compression = J E F(p p n ^)dx. . . . (Ic) The work of delivery = Pm.pn^dx. . . (Id) AIR-COMPKESSOK. 637 In these equations only^> and x are variables. In the sum- mation indicated in (1 TOI |~1 ( Y~| ' ( 2 ) \prnj J and Work per unit of weight ) _. 077 P O F-, _(Pn L \*~] /o\ o/" o*> compressed j " mi ^ j 7 [_ \p m I The value of T 7 ^ , at the immediate end of the sudden com- pression, by eq. (2) of (478), is (4) The temperature of the reservoir being T m , as in 481 (usually much less than T mi ), the compressed air entering it cools down gradually to that temperature, T m , contracting in volume correspondingly since it remains at the same tension p mi . The mechanical equivalent of this heat is lost. Let us now inquire what is the efficiency of the combination of air-compressor and compressed-air engine, the former sup- plying air for the latter, both working adiabatically, assuming that no tension is lost by the compressed air in passing along the reservoir between, i.e., that p mi = p m . Also assume (as already implied, in fact) that j? Wl =p n = one atmos., and that the temperature, T n ^ of the air entering the compressor cyl- inder is equal to that, T m , of the reservoir and transmission- pipe. To do this we need only find the ratio of the amount of work obtained from one pound (or other unit of weight) in the compressed-air engine to the amount spent in compressing one pound of air in the compressor. Calling this ratio t?, the 638 - MECHANICS OF ENGINEERING. efficiency, and dividing eq. (6) of 481 by eq. (3) of this para- graph, we have, with substitutions just mentioned, _ T m _ Abs. temp, of outer free air ,_ _ _ of sudden compression, T mi ( Abs. temp, of air at end ( of sudd or, substituting from eq. (4), and remembering that T ni = T we have also (6) also, since T ^ ... -* m we may write _ T n __ Ab. tern, air leaving eng. cyl. ,^ T m Ab. tern, outer free air. For practical details of the construction and working of engines and compressors, and the actual efficiency realized, the student may consult special works, as they lie somewhat be- yond the scope of the present work. EXAMPLE 1. In the example of 445, the ratio of p m to p n was . Hence, if compressed air is supplied to the reser- voir under above conditions, the efficiency of the system is, from eq. (6), 77 = V^ 0.794, about 80 per cent. 7} 1 EXAMPLE 2. If the ratio of the tensions is as small as = , Pm 6 the efficiency would be only (-J-)* = 0.55 ; i.e., 45 per cent of the energy spent in the compressor is lost in heat. EXAMPLE 3. What horse-power is required in a blowing engine to furnish 10 Ibs. of air per minute at a pressure of 4 atmos., with adiabatic compression, the air being received by the compressor at one atmosphere tension and 27 Cent, (ft.-lb.-sec. system). Since 27 C. = 300 Abs. C. = T ni , we have, from eq. (4), T mi = 300 (f ) = 477 Abs. Cent. ; and hence, eq. (3), HOT-AIB ENGINES. 630 1 4 7 v 1 44 F /I Yin TteworJcper pound of air = 3 X 477^^ ^[l - g) = 50870 ft. Ibs. per pound of air. Hence 10 Ibs. of air will require 508700 ft. Ibs. of work ; and if this is done every min- ute we have the req. H. P. = *ff- = 15A H ' R NOTE. If the compression could be made isothermal, an approximation to which is obtained by injecting a spray of cold water, we would have, from eqs. (1) and (2) of 482 : Workper \_ T p , / #, \ 300 X 14.7 X 144 II. air \ - T ^f 10g ' e fc) = .0807 X 273 X l ^ = 39950 ft. Ibs. per lb., and the corresponding H. P. = 12.1 ; a saving of about 25 per cent, compared with the former. The difference was employed in heating the air in the air-com- pressor with adiabatic compression, and was lost when that extra heat was dissipated in the reservoir as the air cooled again. This difference is easily shown graphically by compar- ing in the same diagram the areas representing the work done in the two cases. 484. Hot-air Engines. Since we have seen that the tension of air and other gases can be increased by heating, if the vol- ume be kept the same, a mass of air thus treated can after- wards be allowed to expand in a working cylinder, and thus become a means of converting heat into work. In Stirling^ hot-air engine a definite confined mass of air is used indefinitely without loss (except that occasional small supplies are needed to make up for leakage), and is alternately heated and cooled. A displacement-plunger, or piston, fitting loosely in a bell-like chamber, is so connected with the piston of the working cylinder and the fly-wheel, that its forward stroke is made while the other piston waits at the beginning of its stroke. In this motion the plunger causes the confined air to pass in a thin sheet over the top and sides of the furnace dome, thus greatly increasing its tension. The air then expands behind the working piston with falling tension and temperature, and, 640 MECHANICS OF ENGINEERING. while that piston pauses at the end of its forward stroke, is again shifted in position, though without change of volume, by the return stroke of the plunger, in such a way as to pass through a coil of pipes in which cold water is flowing. This reduces both its temperature and tension, and hence its resist- ance to the piston on the return stroke is at first less than at- mospheric, but is gradually increased by the compression. This cycle of changes is repeated indefinitely, and is easily traced on a diagram like that in Fig. 528, and computations made accordingly. A special invention of Stirling's is the " regenerator" or box filled with numerous sheets of wire gauze, in its passage through which the working air, after expansion, deposits some of its heat, which it re-absorbs to some extent when, after further cooling in the " refrigerator" or pipe coil and com- pression by the return stroke of the piston, it is made to pass backward through the regenerator to be further heated by the furnace in readiness for a forward stroke. This feature, how- ever, has not realized all the expectations of its inventor and improvers, as to economy of heat and fuel. In Ericsson? s hot-air engine, of more recent date, the dis- placement-plunger fits its cylinder air-tight, but valves can be opened through its edges when moving in one direction, thus causing it to act temporarily as a loose plunger, or shifter. The two pistons move simultaneously in the same direction in the same cylinder, but through different lengths of stroke, so that the space between them is alternately enlarged and con- tracted. The working piston also has valves opening through it for receiving a fresh supply of air into the space between the two pistons. During the forward stroke a fresh instal- ment from the outer air enters through the working piston into the space between it and the other, whose valves are now closed and which is now expelling from its further face, through proper valves, the air used in the preceding stroke ; no work is done in this stroke. On the return stroke this fresh supply of air is free to expand behind the now retreating working piston, while its tension is greatly increased by its being shifted (at least a large portion of it) over the furnace GAS-ENGINES. 641 dome through the valves (now open) of the plunger piston, by the motion of the latter, which now acts as a loose plunger. The engine is therefore only single-acting, no work being done in each forward stroke. For further details, see Goodeve's and Rankine's works on the steam-engine; also the article " Hot-air Engine" in Johnson's Cyclopedia by Fres. Barnard, and Rontgen's Thermodynamics. 485. Brayton's Petroleum-engine. Although a more recent invention than the gas-engines to be mentioned presently, this motor is more closely related to hot-air engines than the latter. By a slow combustion of petroleum vapor the gaseous products of combustion, while under considerable tension, are enabled to follow up a piston with a sustained pressure, being left to expand through the latter part of the stroke. Thus we have the furnace and working cylinder combined in one. The gradual combustion is accomplished by making use of the principle of the Davy safety-lamp that flame will not spread through layers of wire gauze of proper fineness. 486. Gas-engines. We again have the furnace and working cylinder in one in a " gas-engine" where illuminating gas and atmospheric air are introduced into the working cylinder in proper proportions (about ten parts of air to one of gas, by weight) to form an explosive mixture of more or less violence and exploded at a certain point of the stroke, causing a very sudden rise of temperature and tension', after which the mass expands behind the piston with falling pressure. On the return stroke the products of combustion are expelled, and no work done, these engines being single-acting. In some forms the mixture is compressed before explosion, since it has been found that under this treatment a mixture containing a larger proportion of air to gas can be made to ignite, and that then the resulting pressure is more gradual and sustained, like that of steam or of the mixture in the Brayton engine. That is, the effect is analogous to that of " slow-burning powder" in a gun. In the " Otto Silent Gas-engine" the explosion occurs only every fourth stroke, and one side of the piston is always open 642 MECHANICS OF ENGINEERING. to the air. The action on the other side of the piston is as follows : (1) In the forward stroke a fresh supply of explosive mixture is drawn into the cylinder at one atmosphere tension. (2) The next (backward) stroke compresses the mixture into about one fourth of its original bulk, this operation occurring at the expense of the kinetic energy of the fly-wheel. (3) The mixture is ignited, the pressure rises to 6 or 7 atmospheres, and work is done on the piston through the next (forward) stroke, the tension of the products of combustion having fallen to one atmosphere (nearly) at the end of the stroke. (4) In the next (backward) stroke the products of combus- tion are expelled arid no work is done. The Atkinson " Cycle Gas-engine," an English invention of recent .date (see the London Engineer for May 1887 ; pp. 361 and 380) also makes an explosion every fourth stroke, but the link work connecting the piston and fly-wheel is of such de- sign that the latter makes but one revolution during the four strokes. Also the length of the expansion or working stroke is greater than that of the compression stroke and the products of combustion are completely expelled. Consequently the effi- ciency of this motor is at present greater than that of any other gas-engine. See 487. One of the most simple gas-engines is made by the Economic Motor Company of New York. The piston has no packing, being a long plunger ground to fit the cylinder accurately and kept well lubricated. As with most gas-engines the cylinder is encased in a water-jacket to prevent excessive heating of the working parts and consequent decomposition of the lubricant. For further details on these motors, see Kankine's " Steam- engine," Clark's "Gas-engines" in Yan Nostrand's Science Series, and article " Gas-engine" in Johnson's Cyclopaedia ; also Prof. Thurston's report on Mechanical Engineering at the Vienna Exhibition of 1873, and proceedings of the " Society of Engineers" (England) for 1881. 487. Efficiency of Heat-engines. According to the mechan- ical theory of heat, the combustion of one pound of coal, pro- GAS-ENGINES. 643 ducing, as it does, about 14,000 heat-units (British Thermal units ; see 149, Mechanics) should furnish 14,000 X 772 = 10,808,000 ft. Ibs. of work, if entirely converted into work. Let us see how nearly this is accomplished in the performance of the most recent and economical marine engines of the present day, viz., the triple expansion engines of some Atlantic steamers, which are claimed to have consumed per hour only 1.25 Ibs. of coal for each measured ( u indicated ") horse-power of effective work done in their cylinders. The work-equivalent of 1.25 Ibs. of coal per hour is 1.25 X 14,000 X 772 = 13,510,000 ft. Ibs. per hour; while the actual work per hour implied in " one H. P. per hour" is 33000 X 60 = 1,980,000 ft. Ibs. per hour. That is, the engines utilize only one seventh of the heat of com- bustion of the fuel. According to Prof. Thurston, this is a rather extravagant claim (1.25), the actual consumption having probably been 1.4 Ibs. of coal per H. P. per hour. The ordinary compound marine engine is stated to use as little as 2.00 Ibs. per hour for each H. P. Most of the heat not utilized is dissipated in the condenser. Similarly, the water-jacket, a necessary evil in the operation of the gas-engine, is a source of great loss of heat and work. Still, Mr. Wm. Anderson in his recent work, " Conversion of Heat into Work" (London, 1887), mentions a motor of this class as having converted into work -J- of the heat of combus- tion [an Otto "Silent Gas-engine," tested at the Stevens Institute, Hoboken, 1ST. J.,in 1883] ; while Prof. Unwin found the Atkinson engine (see last paragraph) capable of returning (in the cylinder) fully -^ of the heat-equivalent of the gas con- sumed. [This latter result was confirmed in Philadelphia in Jan. 1889 by Prof. Barr, under direction of Prof. Thurston.] 644 MECHANICS OF ENGINEERING. 488. Duty of Pumping-engines. Another way (often used in speaking of the performance of pumping-enginesj of ex- pressing the degree of economy attained in the use of fuel by the combined furnace, boiler, and engine is to give the num- ber of foot-pounds of work obtained from each 100 Ibs. of coal consumed in the furnace, calling it the " duty" of the engine. For example, by a duty of 99,000,000 ft. Ibs. it is meant that from each pound of coal 990,000 ft. Ibs. of work are obtained. From this we gather that, since one horse-power consists of 33,000 X 60 = 1,980,000 ft. Ibs. per hour, the engine men- tioned must use each hour 1,980,000 -f- 990,000 = 2 Ibs. of coal for each H. P. developed ; which is as low a figure as that attained by the marine engines last quoted. 489. Buoyant Effort of the Atmosphere, In the case of a body of large bulk but of small specific gravity the buoyant effort of the air (due to the same cause as that of water, see 456) becomes quite appreciable, and may sometimes be greater than the weight of the body. This buoyant effort is equal to the weight of air displaced, i.e., = Yy, where Y is the volume of air displaced, and y its heaviness. If G l total weight of the body producing the displace- ment, the resultant vertical force is (i) and for equilibrium, or suspension in the air, we must Lave P = 0, i.e., G,= Yy ........ (2) We may therefore find approximately the elevation where a given balloon will cease to ascend, by determining the heavi- ness y of the air at that elevation from eq. (2) ; then, know- ing approximately the temperature of the air at that elevation, we may compute its tension p [eq. (13), 472], and finally, from eqs. (3), (4), or (5) of 477, obtain the altitude required. EXAMPLE. The car and other solid parts of a balloon weigh BALLOONS. 645 400 Ibs., and the bag contains 12,000 cub. feet of illuminating gas weighing 0.030 Ib. per cub. foot at a tension of one at- mosphere and temperature of 15 Cent., so that its total weight =12,000 X 0.030 = 360 Ibs. Hence G- l = 760 Ibs. We may also write with sufficient accuracy : Whole volume of displacement =V= 12,000 cub. ft. As the balloon ascends the exterior pressure diminishes, and the confined gas tends to expand and so in- -M-O_ ' ^ crease the volume of displacement V\ but --- '' this we shall suppose prevented by the strength of the envelope. At the surface of the ground (station n of Fig. 531 ; see also Fig. 526) let the barometer read 29.6 inches and the temperature be 15 Cent. Then T n = 288 Abs. Cent., and the heavi- FIG. 531. ness of the air at n is .0807 X 273 2.M X 14.7 14.7 288 273 29.6 'm-ir-* 1 At the unknown height 7i, where the balloon is to come to rest, i.e., at J/~, G l must = Vy [eq. (2)] ; therefore G l 760 Ibs. y m = y- = p^r- r-sr- = - 0633 lbs - P er cub. and if the temperature at M be estimated to be 5 Cent, (or T m 278 Abs. Cent.) (on a calm day the temperature de- creases about 1 Cent, for each 500 ft. of ascent), we shall _. 388 m . -0633 ' 278 ~ and hence, from eq. (5), 477, with |(7 T m + T n ) put for T K , h = 26213 X Iff X 2.30258 X log.,, 1.206 = 5088 ft. & CHAPTER Vt HYDRODYNAMICS BEGUN STEADY FLOW OF LIQUIDS THROUGH PIPES AND ORIFICES. 489a. The subject of Water in Motion presents one of the most unsatisfactory branches of Applied Mechanics, from a mathematical stand-point. The internal eddies, cross-currents, and general intricacy of motion of the particles among each other, occurring in a pipe transmitting a fluid, are almost en- tirely defiant of mathematical expression, though the flow of water through a circular orifice in a thin plate into the air pre- sents a simpler case, where the conception of " stream lines" is probably quite close to the truth. In most practical cases we are forced to adopt as a basis for mathematical investigation the simple assumption that the particles move side by side in such a way that those which at any instant form a lamina or thin sheet, ~| to the axis of the pipe or orifice, remain together as a lamina during the further stages of the flow. This is the Hypothesis of Flow in Plane Layers, or Laminated Flow. Experiment is then relied on to make good the discre- pancies between the indications of the formulae resulting from this theory and the actual results of practice ; so that the science of Hydrodynamics is largely one of coefficients determined by experiment. " 7 490. Experimental Phenomena of a " Steady Flow." As pre- liminary to the analysis on which the formulae of this chapter are based, and to acquire familiarity with the quantities involved, it will be advantageous to study the phenomena of the appara- tus represented in Fig. 532. A large tank or reservoir I>O is connected with another, DE, at a lower level, by means of a rigid pipe opening under the water-level in each tank. This 64G STEADY FLOW OF A LIQUID. 647 pipe lias no sharp curves or bends, is of various sectional areas at different parts, the changes of section being very gradual, and the highest point N^ not being more than 30 ft. higher than EC) the surface-level of the upper tank. Let both tanks FIG. 532. be filled with water (or other liquid), which will also rise to H and to ./Tin the pipe. Stop the ends L and N of the pipe, and through J/, a stop-cock in the highest curve, pour in water to fill the remainder of the pipe ; then, closing M, unstop L and NI . I If the dimensions are not extreme (and subsequent formulae will furnish the means of testing/ such points) the water will now begin to flow from the upper tank into the lower, and all parts of the pipe will continue full of water as the flow goes on. Further, suppose the upper tank so large that its surface- level sinks very slowly / or that an influx at A continually makes good the efflux at E; then the flow is said to be a Steady Flow ; or, a state of permanency is said to exist ; i.e., the cir- cumstances of the flow at each section of the pipe are per- manent, or steady. 648 MECHANICS OF ENGINEERING. By measuring the volume, V, of water discharged at E in a time t y we obtain the volume of flow per unit of time, viz., while the weight of flow per unit of time is G = QY, : ..... (2) where y = heaviness ( 7) of the liquid concerned. Water being incompressible and the pipe rigid, it follows that the same volume of water per unit of time must be pass- ing at each cross-section of the pipe. But this is equal to the volume of a prism of water having F, the area of the section^ as a base, and, as an altitude, the mean velocity = v with which the liquid particles pass through the section. Hence for any section we have , . (3) in which the subscripts refer to different sections. If the flow were unsteady, e.g., if the level BC were sinking, this would be true for a definite instant of time ; but when steady, we see that it is permanently true; e.g., F l v l at any instant = Fjp^ at the same or any other instant, subsequent or previous. In other words, in a steady flow the velocity at a given section remains unchanged with lapse of time. [N.B. "We here assume for simplicity that the different particles of water passing simultaneously through a given sec- tion (i.e., abreast of each other) have equal velocities, viz., the velocity which all other particles will assume on reaching this section. Strictly, however, the particles at the sides are some- what retarded by friction on the surface of the pipe. This as- sumption is called the Assumption of Parallel Flow, or Flow in Plane Layers, or Laminated FlowJ] Let us suppose Q to have been found as already prescribed. We may then, knowing the internal sectional areas at different parts of the pipe, -ZT, , JV a , etc., compute the velocities STEADY FLOW OF LIQUIDS. 649 Vi=Q + F>, v,= Q + F*> etc., ( a? which the water must have in passing those sections, respec- tively. It is thus seen that the velocity at any section has no direct connection with the height or depth of the section from the plane, EC, of the upper reservoir surface. The fraction - will be called the height due to the velocity, v, or simply *y the velocity-head, for convenience. Next, as to the value of the internal fluid pressure', p, per unit-area (in the water itself and against the side or wall of pipe) at different sections of the pipe. If the end N of the pipe were stopped, the problem would be one in Hydrostatics, and the pressure against the side of the pipe at JVj (also at N z on same level) would be simply measured by a water column of height in which p a = one atmosphere, and Z> = 34 ft. = height of an ideal water barometer, and y = 62.5 Ibs. per cubic foot ; and this would be shown experimentally by screwing into the side of the pipe at j~, a small tube open / at both ends ; the water would rise in it to the level EC. That is, a column of water of height h l would be sustained in it, which indicates that the internal pressure at N l corresponds to an ideal water col- umn of a height But when a steady flow is proceeding, the case being now one of Hydrodynamics, we find the column of water sustained at rest in the small tube (called an open piezometer) Nfi has a height f/ 1 , less than 7^ , and hence the internal fluid pressure is 650 MECHANICS OF ENGINEERING. less than it was when there was no flow. This pressure being called^, , the ideal water column measuring it has a height at N l , and will be called the pressure-head at the section re- ferred to. We also find experteeTfCa11j~iffiat while the flow is steady the piezometer-height y (and therefore the pressure- head b -\- y) at any section remains unchanged with lapse of time, as a characteristic of a steady flow. [For correct indications, the extremity of the piezometer should have its edges flush with the inner face of the pipe wall, where it is inserted.] At ^3, although at the same level as JV 19 we find, on in- serting an open piezometer, TF, that with F z = F l (and there- fore with v 3 = v^ y z is a little less than ?/, ; while if F 9 < F v (so that v a > ^j), y 3 is not only less than y, , but the dif- ference is greater than before. We have therefore found experimentally that, in a general way, when water is flowing in a pipe it presses less against the side of the pipe than it did before the flow was permitted, or (what amounts to the same thing) the pressure between the transverse laminae is less than the hydrostatic pressure would be. In the portion HN Z of the pipe we find the pressure less than one atmosphere, and consequently a manometer register- ing pressures from zero upward (and not simply the excess over one atmosphere, like the Bourdon steam-gauge and the open piezometer just mentioned) must be employed. At N^ , e.g., we find the pressure = J atmos., i.e., ^ = 17 ft. Even below the level BC, by making the sections quite nar- row (and consequently the velocities great) the pressure may be made less than one atmosphere. At the surface BC the pres- sure is of course just one atmosphere, while that in the jet at N^ , entering the right-hand tank under water, is necessarily p 4 = 1 atmos. -|- press, due to col. h f of water practically at rest ; STEADY FLOW OF LIQUIDS. 651 i.e., pressure-bead at J\\ = I -f- A'; (whereas if N^ were stopped by a diaphragm, tbe pressure- bead just on tbe right of tbe diaphragm would be ~b -f- A', and that on the left + A 4 .) Similarly, when a jet enters the atmosphere in parallel fila- ments its particles are under a pressure of one atmosphere, i.e., their pressure-head = 1) = 34 ft. (for water) ; for the air im- mediately around the jet may be considered as a pipe between which and the water is exerted a pressure of one atmosphere. 491. Recapitulation and Examples. We have found experi- mentally, then, that in a steady flow of liquid through a rigid pipe there is at each section of the pipe a definite velocity and pressure which all the liquid particles assume on reaching that section ; in other words, at each section of the pipe the liquid velocity and pressure remain constant with progress of time. EXAMPLE 1. If in Fig. 532, the flow having become steady, the volume of water flowing in 3 minutes is found on meas- urement to be 134 cub. feet, the volume per second is, from <3q. (1), 490, Q ||4. = 0.744 cub. ft. per second. EXAMPLE 2. If the flow in 2 min. 20 sec. is 386.4 Ibs., the volume of flow per second is [ft., lb., sec. ; eqs. (1) and (2)] V G 386.4 1 Q =; -- r- 1 = - . --- = 0.0441 cub. ft. per sec. t y 62.5 140 EXAMPLE 3. In Fig. 532 the height of the open piezometer at N l is y l = 9 feet ; what is the internal fluid pressure ? [Use the inch, lb., and sec.] The internal pressure is Pi =p a + y^ 14. Y + 108 X -&& = 18.6 Ibs. per sq. inch. The pressure on the outside of the pipe is, of course, one at- mosphere, so that the resultant bursting pressure at that point is 3.9 Ibs. per sq. in. EXAMPLE 4. The volume of flow per second being .0441 652 MECHANICS OF ENGINEERING. cub. ft. per sec., as in Example 1, required the velocity at a section of the (circular) pipe where the diameter is 2 inches. [Use ft, lb., and sec.] O 0. ^ 2 persec>; while at another section of the pipe where the diameter is four inches (double the former) and the sectional area, F, is there- fore four times as great, the velocity is \ of 2.02 = 0.505 ft. per sec. 492. Bernoulli's Theorem for Steady Flow; without Friction. If the pipe is comparatively short, without sudden bends, elbows, or abrupt changes of cross-section, the effect of friction of the liquid particles against the sides of the pipe and against each other (as when eddies are produced, disturbing the paral- lelism of flow) is small, and will be neglected in the present analysis, whose chief object is to establish a formula for steady flow through a short pipe and through orifices. An assumption, now to be made, oiflow in plane layers, or laminated flow, i.e., flow in laminae ~| to the axis of the pipe at every point, may be thus stated : (see Fig. 533, which shows a steady flow proceeding, through a pipe CD of indefinite extent.) All the liquid particles which at any instant form a small lamina, or sheet, as AB y ~1 to axis of pipe, keep company as a lamina throughout the whole flow. FIG. 533. The thickness, ds', of this lamina re- mains constant so long as the pipe is of constant cross-section, but shortens up (as at C) on passing through a larger section, and lengthens out (as at D) in a part of the pipe where the section is smaller (i.e., the sectional area, F, is smaller). The mass of such a lamina is Fds'y -r- g [ 55], its velocity at any section will be called v (pertaining to that point of the pipe's axis), the pressure of the lamina just behind it is Fp, upon the rear face, while the resistance (at the same instant) offered by its neighbor just ahead is F( p -\- dp) on the front face ; also BERNOULLI S 'THEOREM. 653 its weight is the vertical force Fds'y. Fig. 534 shows, as a free body, the lamina which at any instant is passing a point A of the pipe's axis, where the velocity is v and pressure jp. Note well the forces acting ; the pressures of the pipe wall on the edges of the lamina have no components in the direction of v, for the wall is considered smooth, i.e., those pressures are 1 to wall; in other words, -no FIG. 534. friction is considered. To this free body apply eq. (7) of 74, for any instant of any curvilinear motion of a material point vdv (tang, acceleration) X ds, . . . . (1)' in which ds = a small portion of the path, and is described in the time dt. Now the tang, accel. = J(tang. compons. of the acting forces) -f- mass of lamina, i.e., tang. ace. = * -=- g (2) Now, Fig. 535, at a definite instant of time, conceive the volume of water in the pipe to be subdivided into a great number of laminae of equal mass (which implies equal volumes FIG. 535. in the case of a liquid, but not with gaseous fluids), and let the ds just mentioned for any one lamina be the distance from its centre to that of the one next ahead ; this mode of subdivision 6,)4 MECHANICS OF ENGINEERING. makes the ds of any one lamina identical in value with its thickness ds' , i.e., have also ds cos = dz] or ds' cos = da ; . . (4) z being the height of the centre of a lamina above any con- venient horizontal datum plane. Substituting from (2), (3), and (4) in (1), we derive finally vdv + ^dp + dz = 0. . (5V 7 9 > The flow being steady, and the subdivision into laminae being of the nature just stated, each lamina in some small time dt moves into the position which at the beginning of dt was filled by the lamina next ahead, and acquires the same velocity, the same pressures on its faces, and the same value of z, that the front lamina had at the beginning of dt. Hence, considering the simultaneous advance made by all the laminae in this same dt, we may write out an equation like (5) for each of the laminae between any two cross-sections n and m of the pipe, thus obtaining an infinite number of equations, from which by adding corresponding terms, i.e., by integra- tion, we obtain whence, performing the integrations and transposing, v m * Pm\, _ v n , Pn , , ( Bernoulli s \ m ~2g~'~y~ m ~"ty "y~"~ n * '( Theorem \ ' Denoting by Potential Head the vertical height of any section of the pipe above a convenient datum level, we may state Bernoulli's Theorem as follows : In steady flow without friction, the sum of the velocity- head, pressure-head, and potential head at any section of the pipe is a constant quantity, being equal to the sum of the cor- responding heads at any other section. APPLICATIONS OF BERNOULLI'S THEOREM. 655 It is noticeable that in eq. (7) each of the terms is a linear quantity, viz., a height, or head, either actual, such as z n and z m , or ideal (all the others), and does not bring into account the absolute size of the pipe, nor even its relative dimensions (v m and v n , however, are connected by the equation of continuity F m v m F n v n \ and contains no reference to the volume of water flowing per unit of time \_Q] or the shape of the pipe's axis. "When the pipe is of considerable length compared with its diameter the friction of the water on the sides of the pipe cannot be neglected ( 512). It must be remembered that Bernoulli's Theorem does not hold unless the flow is steady^ i.e., unless each lamina, in com- ing into the position just vacated by the one next ahead (of equal mass), conies also into the exact conditions of velocity and pressure in which the other was when in that position. [N.B. This theorem can also be proved by applying to all the water particles between n and m, as a collection of small rigid bodies (water being incompressible) the theorem of Work and Energy for a collection of Rigid Bodies in 142, eq. (xvi), taking the respective paths which they describe simultaneously in a single dt.~\ 493. First Application of Bernoulli's Theorem without Friction. Fig. 536 shows a large tank from which a vertical pipe of uniform section leads to another tank and dips below the sur- face of the water in the latter. Both surfaces are open to the air. The vessels and pipe being filled with water, and the lower end m of the pipe un- stopped, a steady flow is established almost immediately, the surface BC being very large compared with F, the area of the (uni- form) section of the pipe. , Given F, and the heights h Q and A, re- quired the velocity v m of the jet at m and also the pressure, p n , at n (in pipe near en- trance of same), m is in the jet, just clear of the pipe, and practically in the water- level, AD. The velocity v m is unknown, but the pressure p m is practically =p a = one atmosphere, since FIG. 536. 656 MECHANICS OF ENGINEERING. the pressure on the sides of the jet is necessarily the hydro- static pressure due to a slight depth below the surface AD. .*. Press.-head at m is = = I = 34 feet. . . '( 423) y y Now apply Bernoulli's Theorem to sections m and n, taking a horizontal plane through m as a datum plane for potential heads, so that z n = h and z m = 0, and we have (1) But, assuming that the section of the pipe is filled at every point, we must have for, in the equation of continuity j7 v __ 2? v if we put F m = F n , the pipe being of uniform section, we ob- tain v m = v n Hence eq. (1) reduces to 2 = b - h = 34 f t. - h. y (2) Hence the pressure at n is less than one atmosphere, and if a small tube communicating with an air-tight receiver full of air were screwed into a small hole at n, the air in the receiver would gradually be drawn off until its tension had fallen to a value p n . [This is the principle of SprengeVs air-pump, mercury, how- ever, being used instead of water, as for this heavy liquid & = only 30 inches.] If h is made > & for water, i.e. > 34 feet (or > 30 inches for mercury), p n would be negative from eq. (2), which is impossible, showing that the assumption of full pipe-sections is not borne out. In this case, h > &, only a portion, mri ', (in length somewhat less than &,) of the tube will be kept full FIG. 537. APPLICATIONS OF BEENOULLl'S THEOREM. 657 during the flow (Fig. 537); while in the part AV vapor of water, of low tension corresponding to the temperature { 469), will surround an internal jet which does not fill the pipe. As for the value of v m , Bernoulli's Theorem, applied to BC and w, in Fig. 536, gives finally v m = V%gh . EXAMPLE. If h = 20 feet, Fig. 536, and the liquid is water, the pressure-head at n is (ft., lb., sec.) * = b - h = 34' - 20' = 14 ft., r and therefore p n =14x 62.5 = 875 Ibs. per sq. ft. = 6.07 Ibs. per sq. in. 494. Second Application of Bernoulli's Theorem without Fric- tion. Knowing by actual measurement the open piezometer height y n at the section n in Fig. 538 (so that the pressure- B ^- head, = I -f- y n , at n is r known) ; knowing also the vertical distance h n from n to m, and the respective cross-sections F n and ^ m (F m being the sectional area of the jet, flowing into the air, so that = &), required the volume of flow per sec.; i.e., required Q, which = F n v n = F m v m . (1} 4 n n m m \ / The pipe is short, with smooth curves, -if any, and friction will therefore be neglected. From Bernoulli's Theorem [eq. (7), 492J, taking m as a datum plane for potential heads, we have But from (1) we have MECHANICS OF ENGINEERING. substituting which in (2) we obtain, solving for v r and hence the volume per unit of time becomes known, viz.> Q = F m v m . ...... ''. (4) . If the cross-section F m of the nozzle, or jet, is > F n / v m becomes imaginary (unless y n is negative (i.e., p n < one at- mos.), and numerically > h n ) ; in other words, the assigned cross-sections are not filled by the flow. EXAMPLE. If y n 17 ft. (thus showing the internal fluid pressure at n to be p n = y(y n + ft) = 1-J- atmos.), A n = 10 ft., and the (round) pipe is 4 inches in diameter at n and 3 inches at the nozzle m, we have from (3) (using ft.-lb.-sec. system of units in which g 32.2) 4/2x32.2(17- L0)_ = ^ [N.B. Since F m -r- 7^ is a ratio and therefore an abstract number, the use of the inch in the ratio will give the same result as that of the foot.] Hence, from (4), Q = F m v m = %7c(-fzJ X 50.4 = 2.474 cub. ft. per sec. 495. Orifices in Thin Plate, Fig. 539. When efflux takes place through an orifice in a thin plate, i.e., a sharp-edged orifice in the plane wall of a tank, a contracted vein (or " vena ORIFICE IN THIN PLATE. 659 contracta") is formed, the filaments of water not becoming parallel until reaching a plane, w, parallel to the plane of vessel wall, which for circular orifices is at a dis- tance from the interior plane of vessel wall equal .to the radius of the circular aperture ; and not until reaching this plane does the internal fluid pres- sure become equal to that of the sur- rounding medium (atmosphere, here), FlG - 539 - i.e., surrounding the jet. We assume that the" width of the orifice is small compared with A, unless the vessel wall is horizontal. The area of the cross-section of the jet at m, called the con- tracted section, is found on measurement to be from .60 to .64 of the area of the aperture with most orifices of ordinary shapes, even with widely different values of the area of aper- ture and of the height, or head, A, producing the flow. Call- ing this abstract number [.60 to .64] the Coefficient of Con- traction, and denoting it by C, we may write F m =GF, in which. F = area of the orifice, and F m = that of the con- tracted section. C ranges from .60 to .64 with circular*orifices, but may have lower values with some rectangular forms. (See table in 503.) A lamina of particles of water is under atmospheric pressure at n (the free surface of the water in tank or reser- voir), while its velocity at n is practically zero, i.e. v n = (the surface at B being very large compared with the area of orifice). It experiences increasing pressure as it slowly de- scends until in the immediate neighborhood of the orifice, when its velocity is rapidly accelerated and pressure decreased, in accordance with Bernoulli's Theorem, and its shape length- ened out, until finally at m it forms a portion of a filament of a jet, its pressure is one atmosphere, and its velocity, = v m , we wish to determine. The course of this lamina we call a 660 MECHANICS OF ENGINEERING. " stream-line" and Bernoulli's Theorem is applicable to it, just as if it were enclosed in a frictionless pipe of the same form. Taking then a datum plane through the centre of m, we have while = 0, and i> w =?; - also =b, z n = h, . and v n = 0. Hence Bernoulli's Theorem gives (1) and That is, ^0 velocity of the jet at m is theoretically the same as that acquired ~by a body falling freely in vacuo through a height =h= the " head of water." We should therefore ex- pect that if the jet were directly ver- tically upward, as at m, Fig. 540, a height j! _ would be actually attained. [See 52 and 53.] Experiment shows that the height of the jet (at m) does not materially differ from h if h is not > 6 or 8 feet. For h > 8 ft., however, the actual height reached is < A, the difference being not only absolutely but relatively greater as h is taken greater, since the resistance of the air is then more and more effective in depressing and breaking up the stream. At m', Fig. 540, we have a jet, under a head = A', directed FIG. 540. OKIFICE IN THIN PLATE. at an angle <* with the horizontal. Its form is a parabola ( 81), and the theoretical height reached is h" = ti sin a a ( 80)- _ The jet from an orifice in thin plate is very limpid and clear. From eq. (1), we have theoretically v m = (an equation we shall always use for efflux into the air through orifices and short pipes in the plane wall of a large tank whose water-surface is very large compared with the orifice, and is open to the air), but experiment shows that for an " orifice in thin plate" this value is reduced about 3$ by friction at the edges, so that for ordinary practical purposes we may write v m = (f>V~fyh = Q.WV'fyh, .... (2) in which is called the coefficient of velocity. Hence the volume of flow, Q, per time-unit will be ~ Q = F m v m = CF(f> Vfyh, on the average = Q.^FVltyh. (3) " e It is to be understood that the flow is steady, and that the reservoir surface (very large) and the jet are both under at- mospheric pressure. 0(7 is called the coefficient of efffax* EXAMPLE 1. Fig. 539. Required the velocity of efflux, v m , at m, and the volume of the flow per second, Q, into the air, if h = 21 ft. 6 inches, the circular orifice being 2 in. in diam. ; take C = 0.64. [Ft., lb., and sec.] From eq. (2), v m = 0.97 1/2 X 32.3 X 21.5 = 36.1 ft. per sec. ; hence the discharge is Q = F m v m = 0.64 X fu|j X 36.1 = 0.504 cub. ft. per second. EXAMPLE 2. [Weisbach.] Under a head of 3.396 metres the velocity v m in the contracted section is found by measure- 662 MECHANICS OF ENGINEERING. ments of the jet-curve to be 7.98 metres per sec., and the dis- charge proves to be 0.01825 cub. metres per sec. Required the coefficient of velocity (0) and that of contraction ((7), if the area of the orifice is 36.3 sq. centimetres. Use the metre-Jcilogram-second system of units, in which g 9.81 met. per sq. second. From eq. (2), V 2g/i V2 X 9.81 + 3.396 while from (3) we have xv vis vi^ v/-LO^O s^ s*t*.^ O= %_ = -~- = = = 0.631. F V %y/i ^ v m ttfiftr X 7.98 and (7, being abstract numbers, are independent of the sys- tem of concrete units adopted. NOTE. To find the velocity v m of the jet at the orifice by measurements of the jet-curve, as mentioned in Example 2, we may proceed as follows : Since we cannot very readily as- sure ourselves that the direction of the jet at the orifice is horizontal, we consider the angle a a of the parabola (see Fig. 93 and 80) as unknown, and therefore have two unknowns to deal with, and obtain the necessary two equations by meas- uring the x and y (see page 84) of two points of the jet, re- membering that if we use the equation (3) of page 84 in its present form points of the jet below the orifice will have nega- tive 2/'s. The substitution of these values x l , o? 2 , y l , and ?/ 2 in equation (3) furnishes two equations between constants, in which only a and A are unknown. To eliminate or , for we write 1 4- tan 3 a , and taking a? 2 = %x l for COIL- COS a venience, we finally obtain h^-.^-^- ^V~^ ' aTld "'* Vm ~\/ Q~ > in which y l and y 2 are the vertical distances of the two points HOUNDED ORIFICE. 663 chosen "below the orifice ; that is, we have already made them negative in eq. (3) of page 84. The h of the preceding equa- tion simply denotes v m * -r- 2^, and must not be confused with that of the last two figures. For accuracy the second point should be as far from the orifice along the jet as possible. 496, Orifice with Bounded Approach. Fig. 541 shows the general form and proportions of an orifice or mouth-piece in the use of which contraction does not take place beyond the edges, the inner surface being one u of revolution," and so shaped that the liquid filaments are parallel on passing the outer edge m; hence the pressure-head at m is = b (= 34 ft. for water and 30 inches for mercury) in Bernoulli's Theorem, if efflux takes place into the air. We have also the sectional area F m F = that of final edge of orifice, i.e., the coefficient of contraction, or (7, = unity = 1.00, so that the discharge per time-unit has a volume \ ^*1 Q _ TT v _ jf v *& * m u m <* "m* \4f *' n ' The tank being large, as in Fig. 540, Bernoulli's Theorem applied to m and n will give, as before, FlG - 541 - v= as a theoretical result, while practically we write and (1) As an average is found to differ little from 0.97 with this orifice, the same value as for an orifice in thin plate ( 495). 497. Problems in Efflux Solved by Applying Bernoulli's Theorem. In the two preceding paragraphs the pressure- heads at sections m and n were each =jp a -f- y = height of 664 MECHANICS OF ENGINEEKINO. the liquid barometer = b ; but in the following problems this will not be the case necessarily. However, efflux is to take place through a simple orifice in the side of a large reservoir, whose upper surface (n) is very large, so that v n may be put = zero. Problem I. Fig. 542. What is the velocity of efflux, v m , at the orifice m (i.e., at the contracted sec- tion, if it is an orifice in thin plate) of a jet of water from a steam-boiler, if the free surface at n is at a height = h above m, and the pressure of the steam over the water is p n , the discharge tak- ing place into the air? Applying Bernoulli's Theorem to sec- Fio.648. tion m at the orifice [where the pres- sure-head is ~b and velocity-head v m * -f- 2g (unknown)] and to section n at water-surface (where velocity-head = and pres- sure-head = p n - y\ we have, taking m as a datum for poten- tial heads so that z m and z n k< (i> v EXAMPLE. Let the steam-gauge read 40 Ibs. (and hence p n = 54. 7 Ibs. per sq. inch) and A = 2 ft. 4 in. ; required v m . Also if J^= 2 sq. in., in thin plate, the volume of discharge per sec. = Q = ? For variety, use the inch-lb.-second system of units, in which g 386.4 inches per sq. second, while I = 408 inches, and the heaviness of water, y, [62.5 -f- 1728] Ibs. per cubic inch. Hence, from eq. (1), v m = A /2 X 386.4F- n^-j:^ - 408 + 28~] = J 935 ' 3 i y L62.5 H- 1728 J ( per sec. in. theoretically ; but practically PROBLEMS OF EFFLUX ORIFICES. v m = 935.3 X 0.97 = 907 in. per sec., so that the rate of discharge (volume) is Q = 0.64 F V and as theoretical results. But practically we must write and in which JP = area of orifice in thin plate, and C= coefficient of contraction about 0.62 approximately [see 495]. 666 MECHANICS OF ENGINEERING. EXAMPLE. If in the condenser there is a "vacuum" of 2TJ inches (meaning that the tension of the vapor would support 2^- inches of mercury, in a barometer), so that Pm = [|f X 14.7] Ibs. per sq. inch, and Ji 12 feet, while the orifice is inch in diameter ; we have, using the ft., lb., and sec., X 32.2 X = 51.1 ft. per sec. v (We might also have written, for brevity, ^ = [2J- : 30] X 34 = 2.833, since the pressure-head for one atmos. = 34 feet, for water. Hence, for a circular orifice in thin plate, we have the volume discharged per unit of time, Q = CFv = 0.62 X ?f JsV-X 51.1 = 0.0431 cub. ft. per sec. 4: \1^' 497. Efflux through an Orifice in Terms of the Internal and External Pressures. Fig. 544. Let efflux take place through a small orifice from the plane side of a large tank, in which at the level of the orifice the hydrostatic pressure was = p' be- fore the opening of the orifice, that of the medium surround- ing the jet being =p". "When a steady flow is established, after opening the orifice, the pressure in the water on a level with the ori- fice will not be materially changed, except in tE.||::f the immediate neighborhood of the orifice [see ^Srlj 495] ; hence, applying Bernoulli's Theorem ' to m in the jet, where the filaments are parallel, and a point ^, in the body of the liquid and at the same level as wi, and where the particles FIG. 544. are practically at rest [i.e., v n = 0] (hence not too near the FORCE-PUMP. 667 orifice), we shall have, cancelling out the potential heads which are equal, T. . . . (I/ (In Fig. 544 p' would be equal to p a -\- hy.) Eq. (1) is con- veniently applied to the jet produced by a force-pump, supposing, for simplicity, the orifice to be in the head of the pump- cylinder, as shown in Fig. 545. Let the thrust (force) exerted along the piston- rod be = P, and the area of the piston be = F' . Then the intensity of internal pressure produced in the chamber AB (when the piston moves uniformly) is P FIG. 545. while the external pressure in the air around the jet is simply p a (one atmos.). Tto (1)' (KB. Of course, at points near the orifice the internal pressure is < p'\ read 495.) EXAMPLE. Let the force, or thrust, P, [due to steam-pres- sure on a piston not shown in figure,] be 2000 Ibs., and the diameter of pump-cylinder be d = 9 inches, the liquid being salt water (so that y 64 Ibs. per cubic foot). Then F r = **(&)' = 0.442 sq. ft., and [ft., lb., sec.] 668 MECHANICS OF ENGINEERING. v m = 0.97^/2 X 32.2 x gjj^ - 65.4 ft. per sec. If the orifice is well rounded, with a diameter of one inch, the volume discharged per second is Q = F m v m = Fv m = (i)*X 65.4 = 0.353 cub. ft. per sec. To maintain steadily this rate of discharge, the piston must move at the rate [veloc. = v'~\ of v'=Q + F' = .353 -T- ~ = 0.806 ft. per sec., and the force P must exert a, power ( 130) of L = Pv f = 2000 X 0.816 = 1632 ft. Ibs. per sec. = about 3 horse-power (or 3 H. P.). If the water must be forced from the cylinder through a pipe or hose before passing out of a nozzle into the air, the velocity of efflux will be smaller, on account of "fluid fric- tion" in the hose, for the same P\ such a problem will be treated later [ 513]. Of course, in a pumping-engine, by the iise of several pump-cylinders, and of air-chambers, a practically steady flow is kept up, notwithstanding the fact that the mo- tion of each piston is not uniform, and must be reversed at the end of each stroke. 498. Influence of Density on the Velocity of Efflux in the Last Problem. From the equation of the preceding paragraph, where p" is the external pressure around the jet, and p' the internal pressure at the same level as the orifice but well back of it, where the liquid is sensibly RELATION OF DENSITY TO VELOCITY OF EFFLUX. 669 at rest, we notice that for the same difference of pressure [p'p rr \ the velocity of efflux is inversely proportional to the square root of the heaviness of the liquid. Hence, for the same (p f p"\ mercury would flow out of the orifice with a velocity only 0.272 of that of water ; for :5 = / 1 . 272 :8 " V 13.5 1000' Again, assuming that the equation holds good for the flow of gases (as it does approximately when^' does not greatly exceed p"; e.g., by 6 or 8 per cent), the velocity of efflux of atmospheric air, when at a heaviness of 0.807 Ibs. per cub. foot, would be times as great as for water, with the same p' p". (See 548, etc.) rl :ViR- 499. Efflux under Water. Simple Orifice. Fig. 546. Let and A 2 be the depths of the (small) ori- fice below the levels of the " head " and " tail " waters respectively. Then, using the formula of 457, we have for the pressure at n (at same level as m, the jet) FIG. 546. and for the external pressure, around the jet at m, whence, theoretically, where h = difference of level between the surfaces of the two bodies of water. 670 MECHANICS OF ENGINEERING. Practically, v m = (2) but the value of for efflux under water is somewhat uncer- tain ; as also that of C, the coefficient of contraction. Weis- bach says that /*, = 06 Y , is -^-g- part less than for efflux into the air ; others, that there is no difference (Trautwine). See also p. 389 of vol. 6, Jour, of Engin. Associations, where it is stated that with a circular mouth-piece of 0.37 in. diam., and of " nearly the form of the vena contracta" p was found to be .952 for discharge into the air, and .945 for submerged dis- charge. 500. Efflux from a Small Orifice in a Vessel in Motion. CASE I. When the motion is a vertical translation and uni- formly accelerated. Fig. 547. Suppose the vessel to move up- ward with a constant acceleration p. (See 49a.) Taking m and n as in the two preceding paragraphs, we know that p m =p" = external pressure one at- FIG. 547. mos. =p a (and /. & J). As to the internal pressure at n (same level as m, but well back of orifice), p n , this is not equal to (b + h)y, because of the acceler- ated motion, but we may determine it by considering free the vertical column or prism On of liquid, of cross-section = dF, the vertical forces acting on which are p a dF, downward at 0, p n dF upward at n, and its weight, downward, lidFy. All other pressures are horizontal. For a vertical upward acceler- ation =p, the algebraic sum of the vertical components of all the forces must = mass X vert, accel., i.e., whence (1) Putting p n and p a equal to the p' and j9 r/ , respectively, of the equation, we have ORIFICE IN MOTION". 671 of 497, (2) It must be remembered that v m is the velocity of the jet rel- atively to the orifice, which is itself in motion with a variable velocity. The absolute velocity w m of the particles of the jet is found by the construction in 83, being represented graph- ically by the diagonal of a parallelogram one of whose sides is v m , and the other the velocity c with which the orifice itself is moving at the instant, as part of the vessel. The jet may make any angle with the side of the vessel. On account of the flow the internal pressures of the water against the vessel are no longer balanced horizontally, and the latter will swing out of the vertical unless properly constrained. If p g = acc. of gravity, v m = V 2 V 2gh. v If p is nega- tive and = g, v m ; i.e., there is no flow, but both the vessel and its contents fall freely, without mutual action. CASE II. When the liquid and the vessel both have a uni- form rotary motion about a vertical axis with an angular veloc- ity = GO ( 110). Orifice small, so that we may consider the liquid inside (except near the orifice) to be in relative equilibrium. Suppose the jet horizontal at m, Fig. 548, and the radial distance of the orifice from the axis to be = x. The external pressure p m = p a , and the internal [see 428, eqs. (3) and (4)] is G0V FIG. 548. hence the velocity of the jet, relatively to the orifice, is (from 497, since p n and p m correspond to the p f and p" of that article), Apn~P m }_ <*)', 672 MECHANICS OF ENGINEEKING. (3) in which w, = cox, = the (constant) linear velocity of the ori- fice in its circular path. The absolute velocity w m of the par- ticles in the jet close to the orifice is the diagonal formed on w and v m ( 83) / Hence by properly placing the orifice in the casing, w m may be made small or large', and thus the kinetic energy carried away in the effluent water be regulated, within certain limits. Equation (3) will be utilized subsequently in the theory of Barker's Mill. EXAMPLE. Let the casing make 100 revol. per min. (whence G? = [2?rlOO -r- 60] radians per sec.), A = 12 feet, and x = 2 ft. ; then (ft., lb., sec.) Vm = % X 32.2 X 12 + = 34.8 ft per sec. (while, if the casing is not revolving, v m = V2gh = only 27. 8 ft. per sec.). If the jet is now directed horizontally and backward, and also tangentially to the circular path of the centre of the orifice, its absolute velocity (i.e., relatively to the earth) is w m = v m cox= 34.8 20.9 = 13.9 ft. per sec., and is also horizontal and backwards. If the volume of flow is Q = 0.25 cub. feet per sec., the kinetic energy earned away with the water per second ( 133) is ft. Ibs. per second = 0.085 horse-power. 501. Theoretical Efflux through Rectangular Orifices of Con- siderable Vertical Depth, in a Vertical Plate. If the orifice is so large vertically that the velocities of the different filaments in a vertical plane of the stream are theoretically different, hav- ing different " heads of water," we proceed as follows, taking into account, also, the velocity of approach, c, or mean velocity EECTANGULAS OKIFICES. 673 (if any appreciable), of the water in the channel approaching the orifice. Fig. 549 gives a section of the side of the tank and orifice. Let b = width of the rectangle, the sills of the latter being horizontal^ and a = A 2 A, , its height. Disregarding con- traction for the present, the theoretical volume of discharge per unit of time is equal to the sum of the volumes like v^dF ( vjbdx), in which v m = the velocity of any filament, as m, in the jet, and bdx = cross-sec- tion of the small prism which passes through any horizontal strip of the area of orifice, in a unit of time, its altitude being v m . For each strip there is a FIG. 549. different x or " head of water," and hence a different velocity. Now the theoretical discharge (volume) per unit of time is />*=/- Q = sum of the volumes of the elem. prisms =J v m c *dx. But from Bernoulli's Theorem, if "k c* -^- 2g = the velocity- head at 7i, the surface of the channel of approach nC,b being the pressure-head of n, and x its potential head referred to m as datum (!N~.B. This 5 = 34 ft. for water, and must not be con- fused with the width 5 of orifice), we have [see 492, eq. (7)] (2)' and since dx = d(x + Tc\ Jc being a constant, we have, from (1)' and (2)', Theoret. O = bV~ 674 MECHANICS OF ENGINEERING. or Theoret. Q = &VTg [(A, + ty - & + *)] - (I/ (b now denotes the width of orifice.) If c is small, the chan- nel of approach being large, we have Theoret. Q = $> Vty (hf hf) ...... (2)*' (c being = Q -f- area of section of nC). v If A x = 0, i.e., if the orifice becomes a notch in the side, or an overfall [s.ee Fig. 550, which shows the contraction which actually occurs in all these cases], we have for an overfall Theoret. Q = & 2g\_(h, + &)1 - #J (3) NOTE. Both in (1) and (2) h, and A 2 are the vertical depths. ... . .. . .-. -.v . ^ the respective sills of the orifice from the surface of the water three or four feet ~back of the plane of the orifice, where the surface is comparatively level. This must be specially attended to in deriv- Fl - 55 - ing the actual discharge from the theoretical (see 503). If Q were the unknown quantity in eqs. (1) and (3) it would be necessary to proceed by successive assumptions and ap- proximations, since Q is really involved in & ; for and ~ (where F^ is the sectional area of the channel of approach nC\ With Tc = (or c very small, i.e., F n very large), eq. (3) re- duces (for an overfall) to Theoret. Q = |5A 2 Vfyh, , ...... (3 J) ' or $ as much as if all parts of the orifice had the same head of water = A 2 (as for instance if the orifice were in the horizontal bottom of a tank in which the water was A 2 deep, the orifice having a width = b and length = A a ). TRIANGULAR ORIFICE. 675 502, Theoretical Efflux through a Triangular Orifice in a Thin Vertical Plate or Wall. Base Horizontal. Fig. 551. Let the channel of approach be so large that the velocity of approach may be neglected, h^ and A 2 = depths of sill and vertex, which is downward. The analysis- differs from that of the preceding article only in having k = and the length u, of a horizontal strip of the orifice, variable ; 5 being the length of the base of the triangle. From similar triangles we have x -(A,-*). /. Theoret. Q = fv m dF=fv m udx= and finally, substituting from eq. (2)' of 501, with Jc = 0, FIG. 551. Theoret. Q = FIG. 552. For a triangular notch as in Fig. 552, this reduces to Thewet. = 15 2 (5)' i.e., ^ of the volume that would be discharged per unit of 676 MECHANICS OF ENGINEERING. time if the triangular orifice witli base b and altitude A 2 were cut in the horizontal bottom of a tank under a head of h t The measurements of ^ 2 and b are made with reference to the level surface back of the orifice (see figure) ; for the water- surface in the plane of the orifice is curved below the level surface in the tank. Prof. Thomson has found by experiment that with 1) = 2A 2 , the actual discharge theoret. disch. X 0.595 ; and with I = 4A 2 , actual = theoret. disch. X 0.620. 503. Actual Discharge through Sharp-edged Rectangular Ori- fices (sills horizontal) in the vertical side of a tank or reservoir. CASE I. Complete and Perfect Contraction. The actual volume of water discharged per unit of time is much less than the theoretical values derived in 501, chiefly on account of contraction. By complete contraction we mean that no edge of the orifice is flush with the side or bottom of the reservoir ; and by perfect contraction, that the channel of approach, to whose surface the heads h l and 7* 2 are measured, is so large that the contraction is practically the same if the channel were of infi- nite extent sideways and downward from the orifice. For this case (A, not zero) it is found most convenient to use the following practical formula (b = width) : Actual Q = / + in which (see Fig. 553) a = the height of orifice, h l = the ver- tical depth of the upper edge of the orifice below the level of the reservoir surface, measured a few feet back of the plane of the orifice, and /* is a coefficient of efflux (an abstract number), dependent on experiment. With /i = 0.62 approximate results (within 3 or 4 per cent) may be obtained from eq. (6) with openings not more than (I Y i l , ( . << u^^i s \r RECTANGULAR ORIFICES. 677 18 inches, or less than 1 inch, high ; and not less than 1 inch wide; with heads (h, + -\ from 1 ft. to 20 or 30 feet. EXAMPLE. What is the actual discharge (volume) per min- ute through the orifice in Fig. 553, 14 inches wide and 1 foot high, the upper sill being 8 ft. 6 in. below the surface of still water ? Use eq. (6) with the ft., lb., and sec. as units, and /* = 0.62. Solution : Q = 0.62 X 1 X H X V2 X 32.2[8J+^]m 17.41 cub. ft. per. sec. while the flow of weight is G = Qy = 17.41 X 62.5 = 1088 Ibs. per second. Poncelet and Lesbros* Experiments. For comparatively ac- curate results, values of // taken from the following table (computed from the careful experiments of Poncelet and Les- bros) may be used for the sizes there given, and, where prac- ticable, for other sizes by interpolation. To use the table, the values of A t , #, and b must be reduced to metres, which can be done by the reduction-table below ; but in substituting in eq. (6), if the metre-kilogram-second system of units be used g must be put = 9.81 metres per square second (see 51), and Q will be obtained in cubic metres per second. Since /* is an abstract number, once obtained as indicated above, it does not necessitate any particular system of units in making substitutions in eq. (6). The ft., lb., and sec. will be used in subsequent examples. TABLE FOR REDUCING FEET AND INCHES TO METRES. 1 foot = 0.30479 metre. 1 inch 0.0253 metre. 2 feet = 0.60959 u 2 inches 3= 0.0507 u 3 u = 0.91438 u 3 u 0.0761 it 4 a = 1.21918 metres. 4 u 0.1015 tt 5 u = 1.52397 u 5 u 0.1268 tt 6 a = 1.82877 u 6 u 0.1522 tt 7 a = 2.13356 tt 7 0.1776 tt 8 u = 2.43836 u 8 tt 0.2030 tt 9 u = 2.74315 a 9 u 0.2283 tt 10 u = 3.04794 a 10 u 0.2536 tt 11 u 0.2790 tt 678 MECHANICS OF ENGINEERING. TABLE, FROM PONCELET AND LESBROS. VALUES OF //<>, FOR EQ. (6), FOR RECTANGULAR ORIFICES IN THIN PLATE. (Complete and perfect contraction.) Value of HH Fig. 553 (in metres). 6 = .20*. a = .20*- b = .20 m - a = .10>- 6 = .20"" a = .05 ra - b = .20">- a = .03 m - b = .20- a = .fej- 6 = .20- a = .Ol m - b = .80 a =.20" 6 = .60 m - a = 02 m - Mo Mo Mo Mo Mo Mo Mo Mo 0.005 0.705 .010 0.607 0.630 0.660 .701 0.644 .015 0.593 .612 .632 .660 .697 .644 .020 0.572 .596 .615 .634 .659 .694 .043 .030 .578 .600 .620 .638 ' .659 .688 0.593 .642 .040 .582 .603 .623 .640 .658 .683 .595 .642 .050 .585 .605 .625 .640 .658 .679 .597 .641 .060 .587 .607 .627 .640 .657 .676 .599 .641 .070 .588 .609 .628 .639 .656 .673 .600 .640 .080 .589 .610 .629 .638 ! .656 .670 .601 .640 .090 .591 .610 .629 .637 .655 .668 .601 .639 .100 .592 .611 .630 .637 .654 .666 .602 .639 .120 .593 .612 .630 .636 .653 .663 .603 .638 .140 .595 .613 .630 .635 .651 .660 .603 .637 I .160 .596 .614 .631 .634 .650 .658 .604 .637 .180 .597 .615 .630 .634 .649 .657 .605 .636 .200 .598 .615 .630 .633 .648 .655 .605 .635 .250 .599 .616 .630 .632 .646 .653 .606 .634 .300 .600 .616 .629 .632 .644 .650 .607 .633 .400 .602 .617 .628 .631 .642 .647 .607 .631 .500 .603 .617 .628 .630 .640 .644 .607 .630 .600 .604 .617 .627 .630 .638 .642 .607 .629 .700 .604 .616 .627 .629 .637 .640 .607 .628 .800 .605 .616 .627 .629 .636 .637 .606 .62$ .900 .605 .615 .626 .628 .634 .635 .606 .627 1.000 .605 .615 .626 .628 .633 .632 .605 .626 1.100 .604 .614 .625 .627 .631 .629 .604 .626 1.200 .604 .614 .624 .626 .628 .626 .604 .625 1.300 .603 .613 .622 .624 .625 .622 .603 .624 1.400 .603 .612 .621 .622 .622 .618 .603 .624 1.500 .602 .611 .620 .620 .619 .615 .602 .623 1.600 .602 .611 .618 .618 .617 .613 .602 .623 1.700 .602 .610 .617 .616 .615 .612 .602 .622 1.800 .601 .609 .615 .615 .614 .612 .602 .621 1.900 .601 .608 .614 .013 .612 .611 .602 .621 2.000 .601 .607 .613 .612 .612 .611 .602 .620 3.000 .601 .603 .606 .608 .610 .609 .601 .615 i i EXAMPLE. With h, = 4 in. [=0.10 met.], .a = 8 in. [= 0.20 met] , Z> = 1 f t. 8 in. [= 0.51 met] , required the (actual) volume discharged per second. See Fig. 553. RECTANGULAR ORIFICES. 679 From the foregoing table, for h, = 0.10 m -, I = 0.60 m - and a = 0.20 m -, we find /* = .602 " h, = 0.10-, I = 0.20 m - " a = 0.20 m -, " /*. = .592 diff. = ."ulO Hence, by interpolation, for h, = 0.10 m - J = 0.51 m -, and a = 0.20 m -, we have V = 0.602 - A [-602 - 0.592] = 0.600. v Hence [ft., lb., sec.], remembering that /* is an abstract num- ber, from eq. (6), Q = 0.600 X X - 4.36 cub. ft. per second. CASE II. Incomplete Contraction. This name is given to the cases, like those shown in Fig. 554, where one or more sides of the orifice have an interior border flush with the sides or bottom of the (square-cornered) tank. Not only is the general direction of the stream altered, but the discharge is greater, on account of the larger size of the contracted section, since contraction is prevented on those sides which have a border. It is assumed that the contraction which does occur (on the other edges) is perfect / i.e., the cross-sec- tion of the tank is large, compared with the orifice. According to the experiments of Bidone and Weisbach with Poncelet's ori- fices (i.e., orifices in thin plate mentioned in the preceding table), the actual volume discharged per unit of time is (7) FIG. 554. (differing from eq. (6) only in the coefficient of efflux /*), in which the abstract number >w is found thus : Determine a coefficient of efflux ja as if eq. (6) were to be used in Case I ; i.e., as if contraction were complete and perfect ; then write 680 MECHANICS OF ENGINEERING, / = /![! + 0.155 ], (7)' where n = the ratio of the length of periphery of the orifice with a border to the whole periphery. E.g., if the lower sill, only, has a border, while if the lower sill and both sides have a border, w = (20 + &)-*- [2(0 + &)]. EXAMPLE. If h, = 8 ft. (= 2.43 m -), I = 2 ft. (= 0.60 m -), a = 4 in. (= 0.10 m -), and one side is even with the side of the tank, and the lower sill even with the bottom, required the volume discharged per second. (Sharp-edged orifice, in ver- tical plane, etc.) Here for complete and perfect contraction we have, from Poncelet's tables (Case I), /i = 0.608. Now n = i; hence,, from eq. (7)', = 0.608 [1 + 0.155 X i]= 0.6551 ; hence, eq. (7), Q = 0.655 X 2 X A V2x32.2(8+i.A) = 10.23 cub. ft. per sec. CASE III. Imperfect Contraction. If there is a submerged channel of approach, symmetrically placed as regards the orifice, and of an area (cross-section), = Gr 9 not much larger than that, = F, of the orifice (see Fig. 555), the contraction is less than in Case I, and is called imperfect contraction. Upon his experiments with Poncelet's orifices, with imperfect contraction, Weisbach. bases the following formula for the discharge (volume) per unit of time, viz., (8) Fl - 555 - EECTANGULAK OKIFICES. 681 (see Fig. 553 for notation), with the understanding that the co- efficient ...... (8)' where /< IB the coefficient obtained from the tables of Case I (as if the contraction were perfect and complete), and ft an ab- stract number depending on the ratio F : G = ra, as follows : = 0.0760 [9-- 1.00]. TABLE A. To shorten computation Weisbach gives the following table for/?: EXAMPLE. Let h, = V 9J-" (= 1.46 met.), the dimensions of the orifice being width = b = 8 in. (= 0.20 m -); height = a = 5 in. (= 0.126 m -) ; while the channel of approach (CD, Fig. 555) is one foot square. From Case I, we have, for the given di- mensions and head, m. /3. m. 0. .05 .009 .55 .178 .10 .019 .60 .208 .15 .030 .65 .241 .20 .042 .70 .278 .25 .056 .75 .319 .30 .071 .80 .365 .35 .088 .85 .416 .40 .107 .90 .473 .45 .128 .95 .537 .50 .152 1.00 .608 = 0.610; We find [Table A] 144 sq. in. p = 0.062 ; and hence ;* = /* (1.062), from eq. (8)'. Therefore, from eq. (8), with ft., lb., and sec., Q = 0.610 X 1.062 X A A V2 X 32.2 X 5 = 3.22 cub. ft. per sec. CASE IV. Head measured in Moving Water. See Fig. 556. If the head h^ , of the upper sill, cannot be measured to the level of still water, but must be taken to the surface of a channel of approach, where the velocity of approach is quite 682 MECHANICS OF ENGINEERING. appreciable, not only is the contraction imperfect, but strictly we should use eq. (1) of 501, in which the velocity of approach is considered. Let F = area of orifice, and G that of the cross-section of the channel of approach ; then the velocity of approach is c = Q -f- G, and ~k (of above eq.) = c 2 -f- 2^ = Q' -f- 2^ 2 ; but Q itself being unknown, a substitution of ~k in terms of Q in eq. (1), 501, leads to an equation of high degree with respect to Q. Practically, therefore, it is better to write (9) and determine fit by experiment for different values of the ratio F -- G. Accordingly, "Weisbach found, for Poncelet's orifices, that if j* is the coefficient for complete and perfect contraction from Case I, we have l = / i.(l + /8') ....... (9)' ft' being an abstract number, and being thus related tof-T- G, = 0.641 (9)" h^ was measured to the surface one metre back of the plane of the orifice, and F : G did not exceed 0.50. Weisbach gives the following table computed from eq. (9)" : TABLE B. EXAMPLE. A rectangular water-trough 4 ft. wide is dammed up with a vertical board in which is a rectangular orifice, as in Fig. 556, of width b = 2 ft. (= 0.60 met.), and height a = 6 in. (= 0.15 met.) ; and when the water-level be- hind the board has ceased rising (i.e., when the flow has become steady), we find that h 1 = 2 ft., and the depth behind in the trough to be 3 ft. Kequired Q. Since F: G = l sq. ft. -r- 12 sq. ft. = .0833, we find (Table B) ft' 0.005 ; and ^ being = 0.612 from Pon- celet's tables, Case I, we have finally, from eq. (9), F-t-G. /3'. 0.05 .002 .10 .006 .15 .014 .20 .026 .25 .040 .30 .058 .35 .079 .40 .103 .45 .130 .50 .160 DISCHARGE OF OVERFALL-WEIRS. 683 Q = 0.612(1.005) 2 xi V2 X 32.2 X 2.25 = 7.41 cub. f-t. per second. 504. Actual Discharge of Sharp-edged Overfalls (Overfall- weirs; or Rectangular Notches in a Thin Vertical Plate), CASE I. Complete and Perfect Contraction (the normal case), Fig. 557 ; i.e., no edge is flush with the side or bottom of the reservoir, whose sectional area is very large compared with that, &A 3 , of the notch. By depth, A 2 , of the notch, we are to understand the depth of the sill Mow the surface a few feet back of the notch where it is level. In the plane of the notch the vertical thickness of the stream is only from J to -^j- of A 2 . Putting, therefore, the velocity of approach = zero, and hence k = 0, in eq. (3) of 501, we have for the FIG. 557. Actual Q = ^ |JA 2 1/2 = 0.586. But, the ends being flush with the sides of the reservoir or channel, and G being 6 sq. ft. (see figure), which is not excessively large compared with J^= l/i t = 2 sq. ft., we have, from Table E, with F : G = 0.333, FIG. 559. and hence [eq. (13) and (13)'], /* being .586 as in last example, Q = | x 0.586 X (1 + .081) X 2 X 1 X V64.4 X 1.0 = 6.Y8 cu.b. ft. per sec. 505. Francis' Formula for Overfalls (i.e., rectangular notches). From extensive experiments at Lowell, Mass., in 1851, with rectangular overfall-weirs, Mr. J. B. Francis deduced the fol- lowing formula for the volume, Q, of flow per second over such weirs 10 feet in width, and with A 2 varying from 0.6 to 1.6 feet (from sill of notch to level surface of water a few feet back) : OVERFALL-WEIRS. 687 . . (14) in which l> = width. This provides for incomplete contraction, as well as for com- plete and perfect contraction, by making n = 2 for perfect and complete contraction (Fig. 557) ; n = 1 when one end only is flush with side of channel ; n = when both ends are flush with sides of channel. The contraction was considered complete and perfect when the channel of approach was made as wide as practicable, = 13.96 feet, the depth being about 5 feet. Mr. Francis also experimented with submerged or " drowned" weirs in 1883 ; such a weir being one in which the sill is be- low the level of the tail-water (i.e., of receiving channel). 506. Fteley and Stearns' s Experiments at Boston, Mass., in 1877 and 1880. These may be found in the Transactions of the American Society of Civil Engineers, vol. xii, and gave rise to formulae differing slightly from those of Mr. Francis in some particulars. In the case of suppressed end-contractions^ like that in Fig. 559, they propose formulae as follows : When depth of notch is not large, Q (in cub. ft. per sec.) = 3.31 IhJ + 0.007 ~b . (15) (b and A 2 both infeet\ "A 2 , the depth on the weir, should be measured from the sur- face of the water above the curvature of the sheet." " Air should have free access to the space under the sheet." The crest must be horizontal. The formula does not apply to depths on the weir less than 0.07 feet. When the depth of notch is quite large, a correction must be made for velocity of approach, =[0.731 + -j-V (.016)]|/|y 1/64.4 X 12 =1.79 cub.ft.per sec. The tube must be at least 3 times as long as wide, to be filled. 509. Conical Diverging, and Converging, Short Tubes. With conical convergent tubes, as at A, Fig. 565, with inner edges not rounded, D'Aubuisson and Castel found by experiment values of the coefficient of velocity, 0, and of that of efflux, //, [from which the coefficient of contraction, C = /* -5- 0, may be FIG. 565. computed,] for tubes 1.55 centimeters wide at the narrow end, and 4.0 centimeters long, under a head of h = 3 metres, and with different angles of convergence. By angle of converg- ence is meant the angle between the sides CE and DB, Fig. 565. In the following table will be found some values of /* and founded on these experiments, for use in the formulae v m = Vtyh and Q = in which T^ 7 denotes the area of the outlet orifice EB. CONICAL SHOKT PIPES. 693 TABLE G (CONICAL CONVERGING TUBES). Angle of ) _ go 1Q , convergence f 8 10 20' 13 30' 19 30' 30 49 // = .895 .930 .938 .946 .924 .895 .847 0= .894 .932 .951 .963 .970 .975 .984 Evidently JJL is a maximum for With a conically divergent tube as at MN, having the in- ^ternal diameter MO = .025 metre, the internal diam. NP " .032 metre, and the angle between J/TTand PO = 4 50' ? Weisbach found that in the equation Q = ^F V%gh (where F area of outlet section NP) j* should be = 0.553; the great loss of velocity as compared with V%gh being due to the eddying in the re-expansion from the contracted section at M (corners not rounded), as occurs also in Fig. 549. The jet was much troubled and pulsated violently. "When the angle of divergence is too great, or the head h too large, or if the tube is not wet by the water, efflux with the tube filled cannot be maintained, the flow then taking place as in Fig. 563. Yenturi and Eytelwein experimented with a conically di- vergent tube (called now " Ven- turis tube"), with rounded en- trance to conform to the shape of the contracted vein, as in Fig. 566, having a diameter of one inch at m r (narrowest part), where the sectional area = F' = O.T854 sq. in., and of 1.80 inches at m (outlet), where area = F\ the length being 8 ins., and the angle of convergence 5 9'. With Q = pFV^yh they found p = 0.483. Hence 2J- times as much water was discharged as would have flowed out under the same head through an orifice in thin plate with area = F' = the smallest section of the divergent tube, and 1.9 times as much as through a short pipe of sec- tion = F' . A similar calculation shows that the velocity at m f must have been v m > = 1.55 V%gh, and hence that the pres- sure at in! was much less than one atmosphere. FIG. 566. 694 MECHANICS OF ENGINEERING. Mr. J. B. Francis also experimented with Yenturi's tube (see " Lowell Hydraulic Experiments"). See also p. 389 of vol. 6 of the Journal of Engineering Societies, for experi- ments with diverging short tubes discharging under water. The highest coefficient (yw) obtained by Mr. Francis was 0.782. 509a. New Forms of the Venturi Tube. The statement made in 507, in connection with Fig. 562, was based on purely theoretic grounds, but has recently (Dec. 1888) been com- pletely verified by experiments* conducted in the hydraulic laboratory of the Civil Engineering Department at Cornell University. Three short tubes of circular section, each 3 in. in length and 1 in. in internal diameter at both ends, were ex- perimented with, under heads of 2 ft. and 4 ft. Call them A, B, and C. A was an ordinary straight tube as in Fig. 561 ; the longitudinal section of B was like that in Fig. 562, the narrowest diameter being 0.80 in. [see 495 ; (0.8) 2 = 0.64] ; while C was somewhat like that in Fig. 566, being formed like B up to the narrowest part (diameter 0.80 in.), and then made conically divergent to the discharging end. The results of the experiments are given in the following table : Name of Tube. A A Head. Number of Experi- ments. Range of Values of /.. Average Values of /A. h = 3 ft. ft = 4 ft. 4 3 From 0.804 to 0.823 " 0.819 to 0.823 0.814 0.821 B B li = 2 ft. h = 4 ft. 5 4 " 0.875 to 0.886 " 0.881 to 0.902 0.882 0.892 C C h = 2 f t. k = 4 ft. 5 4 " 0.890 to 0.919 " 0. 902 to 0.923 0.901 0.914 The fact that B discharges more than A is very noticeable, while the superiority of C to B, though evident, is not nearly so great as that of B to A, showing that in order to increase the discharge of an (originally) straight tube (by encroaching on the passage-way) it is of more importance to fill up with solid substance the space around the contracted vein than to make the transition from the narrow section to the discharg- ing end very gradual. * See Journal of the Franklin Inst., for April, 1869. " FLUID FRICTION." 510. " Fluid Friction." By experimenting with the flow of water in glass pipes inserted in the side of a tank, Prof. Rey- nolds of England has found that the flow goes on in parallel filaments for only a few feet from the entrance of the tube, and that then the liquid particles begin to intermingle and cross each other's paths in the most intricate manner. To render this phenomenon visible, he injected a fine stream of colored liquid at the inlet of the pipe and observed its further motion, and found that the greater the velocity the nearer to the inlet was the point where the breaking up of the parallel- ism of flow began. The hypothesis of laminated flow is, nevertheless, the simplest theoretical basis for establishing practical formulae, and the resistance offered by pipes to the flow of liquids in them will therefore be attributed to the fric- tion of the edges of the laminse against the inner surface of the pipe. The amount of this resistance (often called skin-friction) for a given extent of rubbing surface is by experiment found 1. To be independent of the pressure between the liquid and the solid ; 2. To vary nearly with the square of the relative velocity ; 3. To vary directly with the amount of rubbing surface; 4. To vary directly with the heaviness [y, 409] of the liquid. Hence for a given velocity v, a given rubbing surface of area = S 9 and a liquid of heaviness y, we may write v* Amount of friction (force) = fSy ^ , (1) *&> in which/ is an abstract number called the coefficient of fluid friction, to be determined by experiment. For a given liquid, given character (roughness) of surface, and small range of velocities it is approximately constant. The object of intro- ducing the 2^ is not only because is a familiar and useful y function of v, but that v* -f- 2^ is a height, or distance, and there- fore the product of S (an area) by v* -r- 2g is a volume, and this volume multiplied by y gives the weight of an ideal prism of MECHANICS OF ENGINEERING. the liquid ; hence S - - y is & force and f must be an abstract Zg number and therefore the same in all systems of units, in any given case or experiment. In his experiments at Torquay, England, the late Mr. Froude found the following values for/", the liquid being salt water, while the rigid surfaces were the two sides of a thin straight wooden board -f$ of an inch thick and 19 inches high, coated or prepared in various ways, and drawn edgewise through the water at a constant velocity, the total resistance being measured by a dynamometer. 511. Mr. Froude' s Results. (Condensed.) [The velocity was the same = 10 ft. per sec. in each of the following cases. For other velocities the resistance was found to vary nearly as the square of the velocity, the index of the power varying from 1.8 to 2.16.] TABLE H. Character of Surface. Value of / [from eq. (1), 510]. Varnish / 2 ft. long. When the 8 ft. long. board was 20 ft. long. 50 ft. long. 0.0041 .0038 .0030 .0087 .0081 .0090 .0110 0.0032 .0031 0028 .0063 .0058 .0062 .0071 0.0028 .0027 .0026 .0053 .0048 .0053 .0059 0.0025 Paraffin e . Tinfoil .0025 .0047 .0040 .0049 Calico Fine Sand Medium Sand Coarse Sand N.B. These numbers multiplied by 100 also give the mean f fictional resistance in Ibs. per sq. foot of area of surface in each case (v = 10' per sq. sec.), considering the heaviness of sea water, 64 Ibs. per cubic foot, to cancel the 2g = 64.4 ft. per sq. sec. of eq. (1) of the preceding paragraph. For use in formulae bearing on flow in pipes, f is best deter- mined directly by experiments of that very nature, the results of which will be given as soon as the proper formulae have been established. 512. Bernoulli's Theorem for Steady Flow, with Friction. [The student will now re-read the first part of 492, as far as eq. (1).] Considering free any lamina of fluid, Fig. 567, (according to the subdivision of the stream agreed upon in 492 referred BERNOULLI'S THEOREM WITH FRICTION. 697 to,) the frictions on the edges are the only additional forces as compared with the system in Fig. 534. Let w denote the length of the wetted perimeter of the base of this lamina (in case of a pipe running full, as we here postulate, the wetted perimeter is of course the whole perimeter^ but in the case of an open chan" nel or canal, w is only a portion of the whole perimeter of the cross-section). Then, since the area of rubbing surface at the edge is S wds f , the total fric- tion for the lamina is [by eq. (1), 510] =fwy (v* -r- %g)ds' . Hence from vdv = (tan. accel.) X ds, and from (tan. accel.) [^"(tang. compons. of acting forces)] -f- (mass of lamina), we have v* Fp F(p -\- dp) -f- Fyds' cos fwy ds f vdv = ^ j-. .ds...(a) Fyds' -^ g As. in 492, so here, considering the simultaneous advance of all the laminae lying between any two sections m and n during the small time dt, putting ds' = ds, and ds f cos = dz (see Fig. 568), we have, for any one lamina, vdv, -\ dp -\-dz- g Y (1)'' fi 1 * 9, -/. ^J Now conceive an infinite number of equations to be formed like eq. (1), one for each la- mina between n and m, for the same dt, viz., a dt of such length that each lamina at the end of dt will occupy the same position, and acquire the same values of v, z, and p, that the lamina next in front had at the beginning of the FIG. 568. dt (this is the characteristic of a steady flow). Adding up 698 MECHANICS OF ENGINEERING. the corresponding terms of all these equations, we have (re- membering that for a liquid y is the same in all laminae), 1 f* m 1 f* m (* m -f f* m _ rib FRICTION IN PIPES. 699 -':.-.'.. : :.:? ~ i ^ r i z^r-^^ i TI T!H ^'. "- y ">?: -\ =>^ _ ^ ^ 3 r -j } . --_ # : * =-==( "~ .?& =r~ r FIG. 569. (513. Problems involving Friction-heads; and Examples of Bernoulli's Theorem with Friction. PROBLEM I. Let the portion of pipe between n and m be level, and of uniform cir- ^..~... cular section and diameter = d. The jet at m dis- charges into the air, and has the same sectional area, F= ^-7r^ 2 ,as the pipe; then the pressure-head at m is 34 feet (for ^ = = r water), and the velocity- head at m is = that at n, since v m = v n . The height of the water column in the open piezometer at n is noted, and = y n (so that the pressure-head at n is S^L = y n + b) ; while the r length of pipe from n to m is I. Knowing I, d, y n , and having measured the volume Q, of flow, per unit of time, it is required to find the form of the friction-head and the value off. From F m v m = Q, or %7fd?v m = Q, '. . . . (1) v m becomes known. Also, v m is known to be = v n , and the velocity at each ds is v ~ v m , since F (sectional area) is con- stant along the pipe, and Fv Q. The hydraulic radius is (2) the same for all the ds's between n and m. Substituting in eq. (3) of 512, with the horizontal axis of the pipe as a datum for potential heads, we have 2 o / 2 /m 4- 5 -I- = ^- 4- y n -4- b -4- == . I ds: . (3) zg %g ^d 2 O T-l OS GO g 3 OS GO Ci O i> O OS GO t- IO 10 ^ TH T* ^ 1 10 10 10 Ttf -rtf CO T-l OS GO 10 10 O CO ^ O O TH IO ^ CO T I O OS IO IO IO IO -^ ^ CO b- OS O OS t- JO '. CO CO CO 8 (M -* C5 CO CO ^ CQ T-H IO 1O 1O IO % 8 IO IO IO O T-I IO *> O C5 OS IO CO O GO CO IO "tfl CO CO CO IO O IO IO ^ t- CO GO CO Tt< t- t- t- / IO CO ^ 8 TH O GO O -I C~ -* T-I OS GO CO CO CO IO O OS CO ^ TH TH O TH T-l T-l o o o J> CO CO CO o t> -* o O CO CO TH CO CO O O O O TH IO 01 o o o o o o o 00 ^ CO GO C<1 CO O 710 MECHANICS OF ENGINEERING. be put = ; the jet at m being of the same size as the pipe r the velocity in the pipe is =v m , and therefore v m = 4.4 ft. per sec. Notice that m, the down-stream section, is at a higher level than n. From Bernoulli's Theorem, 516, we have, with n as a datum level, Using the ft., lb., and sec., we have h = 700 ft., v m 2 + 2g = 0.30 ft., while 1000 and . = _ 55 y 55 Hence, in eq. (1), 30 X 5280 (4.4) -0.30 + 38.5 + 700 = 2618 - 4/. 64.4 Solving for/*, we have f = .00485 (whereas for water, with v = 4A ft. per sec. and d=% ft., the table, p. 146, gives /=. 00601. If the y of the petroleum had been 50 Ibs. per cubic foot, instead of 55, we would have obtained ? = 2880 feet and f Y = .0056. 518. Flow through a Long Straight Cylindrical Pipe, including both friction-head and entrance loss of head (corners not rounded); reservoir large. Fig. 573. The jet issues directly I from the end of the pipe, ! in parallel filaments, into * | A 7 therefore .-"..v Ijas same section as pipe; hence, also, v m of the jet E FIG. 573. = v in the pipe (which is assumed to be running full), and is COEFFICIENT OF LIQUID FRICTION. 711 therefore the velocity to be used in the loss of head C# - at the entrance ^( 515). Taking m and n as in figure and applying Bernoulli's Theorem ( 474), with m as datum level for the potential heads z m and z n , we have -< . (1) Three different problems may now be solved : First, required the head li to keep up a flow of given volume = Q per unit of time in a pipe of given length I and diameter = d. From the equation of continuity we have Q = F m v m = %nd*v m ; 4:0 .'. veloc. of jet, which = veloc. in pipe, = v m = ~. . . (2) Having found v m = v, from (2), we obtain from (1) the re- quired A, thus : Now G, = 0.505 if a = 90 (see 515), while / may be taken from the table, 517, for the given diameter and com- puted velocity [v m = i>, found in (2)], if the pipe is clean ; if not clean, see end of 517, for slightly tuberculated and for foul pipes. Secondly. Given the head A, and the length I and diameter d of pipe, required the velocity in the pipe, viz., v, = v m , that of jet ; also the volume delivered per unit of time, Q. Solv- ing eq. (1) for v m , we have d 712 MECHANICS OF ENGINEERING. whence Q becomes known, since (5) [NOTE. The first radical in (4) might for brevity be called a coefficient of velocity, 0, for this case. Since the jet has the same diameter as the pipe, this radical may also be called a coefficient of efflux.'} Since in (4) f depends on the unknown v as well as on the known d, we must first put/" .006 for a first approximation for v m ; then take a corresponding value for f and substitute again ; and so on. Thirdly, knowing the length of pipe and the head h, we wish to find the proper diameter d for the pipe to deliver a given volume Q of water per unit of time. Now which substituted in (1) gives that is, As the radical contains d, we first assume a value for d, with/* .006, and substitute in (7). With the approximate value of d thus obtained, we substitute again with a new value for f based on an approximate v from eq. (6) (with d its first approximation), and thus a still closer value for d is de- rived ; and so on. (Trautwine's Pocket-book contains a table of fifth roots and powers.) If I is quite large, we may put d = for a first approximation. In connection with these examples, see last figure. LONG PIPES. 713 EXAMPLE 1. What head h is necessary to deliver 120 cub. ft. of water per minute through a clean straight iron pipe 140 ft. long and 6 in. in diameter ? From eq. (2), with ft., lb., and sec., we have v = v m = [4 X W] * *(*)'= 10 - 18 ft - Per sec. Now for v = 10 ft. per sec. and d = % ft., we find (in table, 517) /= .00549 ; and hence, from eq. (3), + = 12-23 ft, of which total head, as we may call it, 1.60 ft. is used in pro- ducing the velocity 10.18 ft. per sec. (i.e., v m * -- 2g = 1.60 ft.), while 0.808 ft. ( = dz^M is lost at the entrance ^(with a = 90), and 9.82 ft. (friction-head) is lost in skin-friction. EXAMPLE 2. [Data from Weisbach.] Required the de- livery, Q, through a straight clean iron pipe 48 ft. long and 2 in. in diameter, with 5 ft. head (= fi). V, = v m , being un- known, we first takey= .006 and obtain [eq. (4)] Vm= I . . _ . 4x. 006X48^ 1 _L. .505 -f = 6.18 ft. per sec. From the table, 5 IT, f or v = 6.2 ft. per sec. and d = 2 in., /= .00638, whence / Vi+^5+* x -T x4sV ^ 7T = 6.04 ft. per sec., which is sufficiently close. Then, for the volume per second, Q = - d*v m J7r(^) 2 6.04 = 0.1307 cub. ft. per sec. 4 714 MECHANICS OF ENGINEERING. [Weisbach's results in this example are v m = 6.52 ft. per sec. and Q 0.1420 cub. ft. per sec., but his values for/" are slightly different.] EXAMPLE 3. [Data from Weisbach.] What must be the diameter of a straight clean iron pipe 100 ft. in length, which is to deliver Q = \ of a cubic foot of water per second under 5ft. head (= A) I With/^ .006 (approximately), we have from eq. (7), put- ting d under the radical for a first trial (ft., lb., sec.), 4:0 whence v = -~ = 7 ft. per sec. net* For d = 3.6 in. and v = 7 ft. per sec., we find/= .00601 ; whence, again, -, 5 /1.505 x .30 + 4 X .00601 X 100 /4 X JV n OCM* . = V~ 2 x 32.2 X 5 ";rTv = and the corresponding v = 6.06 ft. For this d and v we find/= .00609, whence, finally, d= . 7005 x .30 + 4x .00609 x 100 /2V ft _ y 2 X 32.2 X 5 \nl [Weisbach's result is d .318 ft] 519. Ch6zy's Formula. If, in the problem of the preceding paragraph, the pipe is so long, and therefore I : d so great, that 4/1 -r- d in eq. (3) is very large compared with 1 + C#, we may neglect the latter term without appreciable error; whence eq. (3) reduces to ig. 573), . . (8) CHEZY'S FOBMULA. 715 which is known as Chezy's Formula. For example, if Z = 100 ft. and d 2 in. = \ ft, and/approx. = .006, we have A/ -7- = 144, while 1 + C# for square corners = 1.505 only. If in (8) we substitute (8) reduces to .A = _- /.|L . . . (very long pipe):, .... (9) 7t 01 Ag so that for a very long pipe, considering f as approximately constant, we may say that to deliver a volume Q per unit of time through a pipe of given length = I, the necessary head, h, is inversely proportional to the ffth power of the diameter. And again, solving (9) for Q, we find that the volume con- veyed per unit of time is directly proportional to the jift/i power of the square root of the diameter directly proportional to the square root of the head / and inversely proportional to the square root of the length. (Not true for short pipe ; see above example.) If we conceive of the insertion of a great number of piezom- eters along the long straight pipe, of uniform section, now under consideration, the summits of the respective water columns maintained in them will lie in a straight line joining the discharging (into the air) end of the pipe with a point in the reservoir surface vertically over the inlet extremity (prac- tically so), and the " slope" of this line (called the Hydraulic Grade Line or Gradient), i.e., the tangent (or sine ; the angle is so small, generally) of the angle which it makes with the horizontal is = , and may be denoted by s. Putting also ^d R = the hydraulic radius of the section of the pipe, and v m = v = velocity in pipe, we may transform eq. (8) into or, v = A(By, . . . (9) 716 MECHANICS OF ENGINEERING. which is the form by which Mr. Hamilton Smith (see 506) interprets all the experiments quoted by him on long pipes. As to notation, however, he uses n for A, and r for H. With the foot and second as units, the quantity A (not an abstract number) varies approximately between 60 and 140. For a given A we easily find the corresponding f from the relation 2# y = -^ . If the pipe discharges under water, h the differ- ence of elevation of the two reservoirs. If the pipe is not horizontal, the use of the length of its horizontal projection instead of its actual length in the relation s = occasions an i error, but it is in most cases insignificant. Similarly, if a steady flow is going on in a long pipe of uni- form section, at the extremities of any portion of which we have measured the piezometer heights (or computed them from the readings of steam or pressure gauges), we may apply eq. (9), putting for A the difference of level of the piezometer summits, and for I the length of the pipe between them. 520. Coefficient f in Fire-engine Hose. Mr. Geo. A. Ellis, C.E., in his little book on " Fire-streams," describing experi- ments made in Springfield, Mass., gives a graphic comparison (p. 45 of his book) of the friction-heads occurring in rubber hose, in leather hose, and in clean iron pipe, each of 2j- in. diameter, with various velocities; on which the following state- merits may be based : That for the given size of hose and pipe (d = 2j- in.) the coefficient f for the leather and rubber hose respectively may be obtained approximately by adding to ./for clean iron pipe (and a given velocity) the per cent of itself shown in the accompanying table. EXAMPLE. For a clean iron pipe 2^ in. diam., for a velocity = 10 ft. per sec., we have, from 517, f = .00593. Hence for a leather hose of the same diameter, we have, for v = 10 ft. per sec., Velocity Rubber Leather ft. per sec. hose 2 in. diam. hose 2 in. diam. 3.0 50# 300# 6.5 20 80 10 16 43 13 12.5 32 16 12 30 /= .00593 + .43 X .00593 = .00848. PKESSURE-ENERGY. 717 521. Bernoulli's Theorem as an Expression of the Conservation of Energy for the Liquid Particles. In any kind of flow with- out friction^ steady or not, in rigid immovable vessels, the aggregate potential and kinetic energy of the whole mass of liquid concerned is necessarily a constant quantity (see 148 and 149), but individual particles (as the particles in the sink- ing free surface of water in a vessel which is rapidly being emptied) may be continually losing potential energy, i.e., reaching lower and lower levels, without any compensating in- crease of kinetic energy or of any other kind ; but in a steady flow without friction in rigid motionless vessels, we may state that the stock of energy of a given particle, or small collection of particles, is constant during the flow, provided we recognize a third kind of energy which may be called Pressure-energy, or capacity for doing work by virtue of internal fluid pressure ; as may be thus explained : In Fig. 574 let water, with a very slow motion and under a pressure p (due to the reservoir-head -f- atmosphere-head be- hind it), be admitted behind a pis- ton the space beyond which is vacuous. Let s = length of I stroke, and F= the area of pis- ton. At the end of the stroke, f ^*s^ by motion of proper valves, com- 7/ VAC - munication with the reservoir is FIG - 574 ' cut off on the left of the piston and opened on the right, while the water in the cylinder now on the left of the piston is put in 'communication with the vacu- ous exhaust-chamber. As a consequence the internal pressure of this water falls to zero (height of cylinder small), and on the return stroke is simply conveyed out of the cylinder, neither helping nor hindering the motion. That is, in doing the work of one stroke, viz., W = force X distance = Fp X s Fps, a volume of water V =. Fs, weighing Fsy (Ibs. or other unit) r has been used, and, in passing through the motor, has experi- enced* no appreciable change in velocity (motion slow), and 718 MECHANICS OF ENGINEERING. therefore no change in kinetic energy, nor any change of level, and hence no change in potential energy, but it has given up all its pressure. (See 409 for y.) Now TF, the work obtained by the consumption of a weight G = Vy of water, may be written . . (1) Hence a weight of water = G is capable of doing the work G x = G X head due to pressure p, i.e., G x pressure- head, in giving up all its pressure p or otherwise, while still having a pressure^, a weight G of water possesses an amount of energy, which we may call pressure-energy, of an amount =# , where y = the heaviness ( 7) of water, and - a height, or head, measuring the pressure p ; i.e., it equals the pressure-head. We may now state Bernoulli's Theorem without friction in a new form, as follows : Multiplying each term of eq. (7), 451, by Qy, the weight of water flowing per second (or other time-unit) in the steady flow, we have But Qy = -~^-Vm = i X mass flowing per time unit X %g 2 9 square of the velocity = the kinetic energy inherent in the volume Q of water on passing the section m, due to the veloc- ity at m. Also, Qy = the pressure-energy of the volume Q at m, due to the pressure at m ; while Qyz m = t\iQ potential energy of the volume Q at m due to its height z m above the arbitrary datum plane. Corresponding statements may be made for the terms on the right-hand side of (2) referring to the other section, ??, of the pipe. Hence (2) may be thus read : The aggregate amount of energy (of the three kinds mentioned) resident in the particles of liquid when passing section in is LOSS OF ENERGY. 719- equal to that when passing any other section, as n ; in steady flow without friction in rigid motionless vessels ; that is, the store of energy is constant. 522. Bernoulli's Theorem with Friction, from the Standpoint of Energy. Multiply each term in the equation of 516 by Qy,, as before, and denote a loss of head or height of resistance due to any cause by h r , and we have = Qrlr+Qy- n 2g y Each term Qyh r (Q.g., Qy 4/ -, due to skin-friction in a ff long pipe, and Qy C^^ due to loss of head at the reservoir entrance of a pipe) represents a loss of energy, occurring between any position n and any other position m down-stream from % but is really still in existence in the form of heat generated by the friction of the liquid particles against each other or the sides of the pipes. As illustrative of several points in this connection, consider two short lengths of pipe in Fig. 575, A and B, one offering a gradual, the other a sudden, enlargement of section, but otherwise identical in dimen- sions. We suppose them to occupy places in separate lines of pipe in each of which a steady flow with full cross-sec- tions is proceeding, and so reg- ulated that the velocity and in- ternal pressure at n, in A, are equal respectively to those at n FIG. 575. in B. Hence, if vacuum piezometers be inserted at n, the- 720 MECHANICS OF ENGINEERING. smaller section, the water columns maintained in them by the T) internal pressure will be of the same height, , for both A and B. Since at m, the larger section, the sectional area is the same for both A and B, and since F n in A =* F n in J5, so that QA = QB> hence v m in A = v m in B and is less than v n . Now in B a loss of head occurs (and hence a loss of energy) between n and ra, but none in A (except slight friction-head); hence in A we should find as much energy present at m as at n y only differently distributed among the three kinds, while at m in B the aggregate energy is less than that at n in B. As regards kinetic energy, there has been a loss between n and m in both A and B (and equal losses), for v m is less than v n . As to potential energy, there is no change between n and m either in A or B, since n and m are on a level. Hence if the loss of kinetic energy in B is not compensated for by an equal gain of pressure-energy (as it is in A\ the pressure-head (~\ at m in B should be less than that (S=\ at rain A. Ex- \ri* \y I A periment shows this to be true, the loss of head being due to the internal friction in the eddy occasioned by the sudden en- largement ; the water column at m in B is found to be of a less height than that at m in A, whereas at n they are equal. (See p. 467 of article " Hydromechanics" in the Ency. Bri- tannica for Mr. Froude's experiments.) In brief, in A the loss of kinetic energy has been made up in pressure-energy, with no change of potential energy, but in B there is an actual absolute loss of energy = Qyh r , or v a = Qy -^-, suffered by the weight Qy of liquid. The value ? of C in this case and others will be considered in subsequent paragraphs. Similarly, losses of head, and therefore losses of energy, occur at elbows, sharp bends, and obstructions, causing eddies and internal friction, the amount of each loss for a given weight, 6r, of water being = Gh r = GC, ~ ; A r = C being / t/ the loss of head occasioned by the obstruction ( 474). It is SUDDEN ENLARGEMENTS IN PIPES. 721 therefore very important in transmitting water through pipes for purposes of power to use all possible means of preventing disturbance and eddying among the liquid particles. E.g., sharp corners, turns, elbows, abrupt changes of section, should be avoided in the design of the supply-pipe. The amount of the losses of head, or heights of resistance, due to these various causes will now be considered (except skin-friction, already treated). Each such loss of head will be written in the form / y 2 C , and we are principally concerned with the value of the o abstract number , or coefficient of resistance , in each case. The velocity v is the velocity, known or unknown, where the resistance occurs y or if the section of pipe changes at this place, then v = velocity on the down-stream section. The late Professor "Weisbach, of the mining-school of Freiberg, Saxony, was one of the most noted experimenters in this respect, and will be frequently quoted. 523. Loss of Head Due to Sudden (i.e., Square-edged) Enlarge- ment. Borda's Formula. Fig. 576. An eddy is formed in the t angle with consequent loss of energy. Since eacn particle of water of weight = G l , arriving with the velocity v l in the small pipe, may be t considered to have an impact against the base . 576. of the infinitely great and more slowly moving column of water in the large pipe, and, after the impact, moves on with the same velocity, -y a , as that column, just as occurs in inelastic direct central impact ( 60), we may find the energy lost by this particle on account of the impact by eq. (1) of 138, in which, putting M l = 6r, -f- But since Fp^ = F^v^ , we have < = (*)v - a )x, and hence the pressure-head at F, (substituting from equations above) is and /. pi = ||- of 14.7 = 11.6 Ibs. per sq. inch, which is greater than zero ; hence efflux with the tube full in both parts can be maintained under 9 ft. head. If, with F l and F^ as before (and .'. C), we put p, = 0, and solve for 7i, we obtain h = 42.5 ft. as the maximum head under which efflux with the large portion full can be secured. 524. Short Pipe, Square-edged Internally. This case, already 724 MECHANICS OF ENGINEERING. treated in 507 and 515 (see Fig. 578 ; a repetition of 560), presents a loss of head due to the sudden enlargement from the contracted section at m f (whose sec- * tional area may be put CF, C being i an unknown coefficient, or ratio, of j contraction) to the full section F of :'_l:r the pipe. From 515 we know that ^j=^=. the loss of head due to the short pipe is h r = CJT^ (for a = 90), in which FIG. 578. CE = 0.505 ; while from Borda's For- mula, 523, we have also E = ^ 1 . Equating these, we find the coefficient of internal contraction at m f to be C= - -= = - - -- = 0.584, or about 0.60 (compare with C= .64 for thin-plate contrac- tion, 497). It is probably somewhat larger than this (.584), since a small part of the loss of head, A r , is due to friction at the corners and against the sides of the pipe. By a method similar to that pursued in the example of 523, we may show that unless h is less than 40 feet, about, the tube cannot be kept full, the discharge being as in Fig. 551. If the efflux takes place into a " partial vacuum," this limiting value of h is still smaller. "Weisbach's experiments confirm these statements (but those in the C. U. Hyd. Lab. seem to indicate that the limiting value for h in the first case is about 50 ft.). 525. Diaphragm in a Cylindrical Pipe. Fig. 579. The dia- phragm being of "thin plate," , let the circular opening in it (concentric with the pipe) have an area = F, while the sectional area of pipe = F 9 . Beyond F, the FIG. 579. stream contracts to a section of area = CF F l , in enlarging SHOKT PIPE. DIAPHRAGM PIPE. 725 -from which to fill the section F^ , of pipe, a loss of head occurs which by Borda's Formula, 523, is F 1 -- * 1 -- where v 2 is the velocity in the pipe (supposed full). Of course F l (or OF) depends on F\ but since experiments are necessary in any event, it is just as well to give the values of itself, as determined by Weisbach's experiments, viz. : For~- = .10 #1 .20 .30 .40 .50 .60 .70 .80 .90 1.00 C = 226. 48. 17.5 7.8 3.7 1.8 .8 .3 .06 0.00 By internal lateral filling, Fig. 580, the change of section may be made gradual and eddying prevented ; and then but little loss of head (and therefore little loss of energy) occurs, besides the slight amount due to skin-friction along this small surface. FIG. 580. On p. 467 of the article Hydromechanics in the Encyclopaedia Britannica may be found an account of experiments by Mr. Froude, illustrating this fact. 526. "The Venturi Water-meter." The invention bearing this name was made by Mr. Clemens Herschel (see Trans. Am- Soc. Civ. Engineers, for November 1887), and may be de- scribed as a portion of pipe in which a gradual narrowing of section is immediately succeeded by a more gradual enlarge- ment, as in Fig. 580 ; but the dimensions are more extreme. During the flow the piezometer-heights are observed at the three positions r, n, and m (see below), and the rate of dis- charge may be computed as follows : Referring to Fig. 580, let us denote by r the (up-stream) position where the narrow- ing of the pipe begins, and by m that where the enlargement ends, while n refers to the narrowest section. F m = F r . Applying Bernoulli's Theorem to sections r and n, assuming 726 MECHANICS OF ENGINEERING. no loss of head between, we have, as the principle of the ap- paratus, . whence, since F r v r = F n v n , in which represents the first radical factor. should differ F but little from unity with - small (and such was found to be /V the case by experiment). Its theoretical value is constant and greater than unity. In the actual use of the instrument the and & are inferred from the observed piezometer-heights y r and y n (since & = y r + 1, and ^ = y n + &, 5 being = 34 ft.), and then the quantity flowing per time-unit computed, from Q F n v n , v n having been obtained from eq. (2). This pro- cess gives a value of Q about four per cent in excess of the truth, according to the second set of experiments mentioned below, if v n =35 ft. per sec. ; but only one per cent excess with v n = 5 or 6 ft. per sec. Experiments were made by Mr. Herschel on two meters of this kind, in each of which F n was only one ninth of F r , a ratio so extreme that the loss of head due to passage through the instrument is considerable. E.g., with the smaller appara- tus, in which the diameter at n was 4 in., the loss of head be- tween r and in was 10 or 11 ft., when the velocity through n was 50 ft. per sec., those at other velocities being roughly pro- portional to the square of the velocity. In the larger instru- ment d n was 3 ft., and the loss of head between r and m was much more nearly proportional to the square of the velocity than in the smaller. (E.g., with v n = 34.56 ft. per sec. the loss of head was 2.07 ft., while with v n = 16.96 ft. per sec. it SUDDEN DIMINUTION OF SECTION OF PIPE. 727 was 0.49 ft.) The angle of divergence was much smaller in these meters than that in Fig. 580. 527. Sudden Diminution of Cross-section, Square Edges. Fig. 581. Here, again, the resistance is ^ due to the sudden enlargement from =^ f^^\^_ _____ ^ the contracted section to the full sec- -^.-J^&j tion F^ of the small pipe, so that in ^~~~-^3 the loss of head, by Borda's formula, FIG. 581. the coefficient depends on the coefficient of contraction C\ but this latter is influenced by the ratio of F^ to F , the sectional area of the larger pipe, C being about .60 when F is very large (i.e., when the small pipe issues directly from a large reservoir so that F z : F practically = 0). For other values of this ratio TVeisbach gives the following table for (7, from his own ex- periments : For F* : F = .10 .20 .30 .40 .50 .60 .70 .80 .90 1.00 C= .624 .632 .643 .659 .681 .712 .755 .813 .892 1.00 C being found, we compute C from eq. (2) for use in eq. (1). 528. Elbows. The internal disturbance caused by an elbow, Fig. 582 (pipe full, both sides of elbow), occasions a loss of head FIG. 582. in which, according to Weisbach's experi- ments with tubes 3 centims., i.e. 1.2 in., in diameter, we may put 728 MECHANICS OF ENGINEERING. For a - 20 40 60 8(T 90 100 110 120 130 140 C= .046 .139 .364 .740 .984 1.26 1.556 1.86 2.16 2.43 computed from the empirical formula ; C = .9457 sin' %a + 2.047 sin 4 ; v is the velocity in pipe ; a as in figure. For larger pipes C would probably be somewhat smaller ; and vice versa. If the elbow is immediately succeeded by another in the same plane and turning the same way, Fig. 583, the loss of head is not materially increased, since the eddying takes place chiefly in the further branch of the second elbow ; but if it turns in the reverse direction, Fig. 584, but still in the same plane, the total loss of head is double that of one elbow ; while if the plane of the second is ~| to that of the first, the total loss of head is 1-J- times that of one alone. (Weisbach.) FIG. 583. FIG. 584. 529. Bends in Pipes of Circular Section. Fig. 585. Weis- bach bases the following empirical formula for C, the coefficient of resist- ange of a quadrant bend in a pipe of circular section, on his own experi- FIG. 585. ments and some of Dubuat's, viz. : C = 0.131 + 1.847 for use in (1> (2) where a = radius of pipe, r radius of bend (to centre of pipe), and v velocity in pipe ; h r = loss of head due to bend. ELBOWS AND BENDS IN PIPES. 729 It is understood that the portion EC of the pipe is kept full by the flow ; which, however, may not be practicable unless BG\ more than three or four times as long as wide, and is full at the outset. A semicircular bend occasions about the same loss of head as a quadrant bend, but two quadrants forming a re- verse curve in the same plane, Fig. 586, occasion a double loss. By enlarging the pipe at the bend, or providing internal thin partitions parallel to the sides, the loss of head may be considerably dimin- ished. Weisbach gives the following table com- FIG. 586. puted from eq. (1), but does not state the absolute size of the pipes. For -=.10 r .20 .30 .40 .50 .60 .70 .80 .90 1.0 C=.131 .138 .158 .206 .294 .440 .661 .977 1.40 1.98 Accounts of many of Weisbach's hydraulic experiments are contained in the Civilingenieur, vols. ix, x, and xi. 529a. Common Pipe-elbows. Prof. L. F. Bellinger of Nor- wich University, Vermont, conducted a set of experiments in 1887, when a student at Cornell, on the loss of head occasioned by a common el- bow (for wrought-iron pipe), whose longi- tudinal section is shown in Fig. 586a. The elbow served to connect at right angles two wrought-iron pipes having an internal diameter of 0.482 in. The internal diameter of the short bend or elbow was f in., and the radius of its curved circular axis (a quadrant) was in. Its internal surface was that of an ordinary rough casting. A straight pipe of the same character and size and 14 feet long was first used, and the loss of head due to skin-friction (the only loss of head in that case) carefully determined for a range of velocities from 2 to 20 ft. per sec. Two lengths of similar pipe were then joined by the elbow FIG. 586a. 730 MECHANICS OF ENGINEERING. mentioned, forming a total length of 14 feet, and the total loss of head again determined through the same range of velocities. By subtraction, the loss of head due to the elbow was then easily found for each velocity, and assuming the form (1) for the loss of head, C was computed in each case. From Fig. 586a it is seen that the stream meets with a sud- den enlargement and a sudden diminution, of section, as well as with the short bend ; so that the disturbance is of a rather complex nature. The principal results of Prof. Bellinger's experiments, after the adjustment of the observed quantities by " least squares," were found capable of being represented fairly well by the formula C = 0.621 + [2 W - 1] X 0.0376, .. . ' . (2) where n = [veloc. in pipe in ft. per sec.] -r- 5. The following table was computed from eq. (2) (where v is in ft. per second) : V = 2 4 6 8 10 12 14 16 18 20 &= .633 .649 .670 .697 .734 .782 .845 .929 1.039 1.185 530. Valve-gates and Throttle-valves in Cylindrical Pipes. Adopting, as usual, the form *-'t|p ...... a) for the loss of head due to a valve-gate, Fig. 587, or for a find ,^r^ r_ i____J 1 nm ^Sf^jt ~\ ~ ^>> WX7<^ FIG. 587. FIG. 588. throttle-valve, Fig. 588, each in a definite position, Weisbach's VALVE-GATE. THROTTLE-VALVE. 731 experiments furnish us with a range of values of C in the case of these obstacles in a cylindrical pipe 1.6 inches in diameter, :as follows (for meaning of s, d, and a, see figures, v is the velocity in the full section of pipe, running full on both .sides.) Valve-gate. Throttle-valve. S d a 9 5 .24 1.0 .00 10 .52 1 .07 15 20 .90 1.54 1 .26 25 30 2.51 3.91 * .81 35 6.22 40 10.8 1 2.06 45 18.7 1 5.52 50 55 32.6 58.8 I 17.00 60 65 118.0 256.0 * 97.8 70 751. 531. Examples involving Divers Losses of Head. We here suppose, as before, that the pipes are full during the flow. Practically, provision must be made for the escape of the air which collects at the high points. If this air is at a tension greater than one atmosphere, automatic air-valves will serve to provide for its escape ; if less than one atmosphere, an air- pump can be used, as in the case of a siphon used at the Kansas City Water Works. (See p. 346 of the Engineering News for November 1887.) :'";..',;. ; .." m X 90 ... -,-- -^ 7 n = 50 _rr-r^ ~-\vA A f 7 on ' ^E=^='- :) n . S I 90 K fei'V l 1 To " & FIG. 589. EXAMPLE 1. Fig. 589. What head, = A, will be required to deliver \ U. S. gallon (i.e. 231 cubic inches) per second 732 MECHANICS OF ENGINEEKING. through the continuous line of pipe in the figure, containing two sizes of cylindrical pipe (d = 3 in., and d 2 = 1 in.), and two 90 elbows in the larger. The flow is into the air at ra, the jet being 1 in. in diameter, like the pipe. At E, a = 90, and the corners are not rounded ; at JT, also, corners not rounded. Use the ft.-lb.-sec. system of units in which g = 32.2. Since Q = % gal. = j- T 2 T y B = .0668 cub. ft. per sec., and therefore the velocity of the jet v m = v, = Q -f- i?^) 2 = 12.25 ft. per sec.; hence the velocity in the large pipe is to be v = (-i-)X = 1.36 ft. per sec. From Bernoulli's Theorem, we have, with m as datum plane, involving six separate losses of head, for each of which there is no difficulty in finding the proper or/*, since the velocities and dimensions are all known, by consulting preceding para- graphs. (Clean iron pipe.) From 515, table, for a = 90 we have . ./' ,. C^ = 0.505 " 51Vor<=3in.,and^ =1.36ft.persec.,/ =: -00725 " " " d, = l in., and v 9 -=12.25 " " / 2 = .00613 * 528 (elbows), for = 90 .... C el =0.984 " 527, for sudden diminution at K we have [since F^ + F = I 2 ~ 3 2 = 0.111, /. (7 = 0.625] Solving the above equation for A, then, and substituting above numerical values (in ft.-lb.-scc.-system), we have (noting that v m '= v and v = ^ 2 ) (.606 + 9 ^ A , 4 X. 00613X201. .360 -| -- - U EXAMPLES; WITH LOSSES OF HEAD. 733 i.e., h = ^1^ [ 1 +(.00623 +.07160 + .0243)+(.360+5.8848~]; t)4:.4: L_ .-. h = 2.323 X 7.3469 = 17.09 ft.Ans. It is here noticeable how small are the losses of head in the large pipe, the principal reason of this being that the velocity in it is so small (v = only 1.36 ft. per sec.), and that in gen- eral losses of head depend on the square of the velocity (nearly). In other words, the large pipe approximates to being a reser- voir in itself. With no resistances a head h = OT a -~ 2g = 2.32 ft. would be sufficient. EXAMPLE 2. Fig. 590. "With the valve-gate Thalf raised (i.e., 8 = %d in Fig. 587), required the volume delivered per second through the clean pipe here shown. The jet issues FIG. 590. from a short straight pipe, or nozzle (of diameter d^ = 1 in.) inserted in the end of the larger pipe, with the inner corners not rounded. Dimensions as in figure. Hadius of each bend = r = 2 in. The velocity v m of the jet in the air = velocity i> 2 in the small pipe ; hence the loss of head at Now v m is unknown, as yet ; but V Q , the velocity in the large pipe, is = v m ntF]> i - e -J v = A 1 "*- From Bernoulli's The- 734 MECHANICS OF ENGINEERING. orem (m as datum level) we obtain, after transposition, Of the coefficients concerned, f alone depends on the un- known velocity v . For the present [first approximation], put / = .006 From 515, with a = 90, . ....... C E = .505 From 517, valve-gate with s = %d, C F = 2.06 From 529, with a:r = 0.5, C B = 0.294 While at K, from 527, having (F, : F ) = (f) 2 : 2 2 = -& = 0.562; we find from table, . . . , . . ' . . . . C = 0.700 and .-. c*=(J_-l) 2 =(0.428) 2 . . . . i.e., *= 0.183 Substituting in eq. (1) above, with v* = ( T V) 2 ^m 2 > we have ' -, in which the first radical, an abstract number, might be called a coefficient of velocity, 0, for the whole delivery pipe ; and also, since in this case Q, = F m v m = Fjv z , may be written Q = }JiF^ V^g/iy it may be named a coefficient of efflux, p. Hence V2 x 32.2 x 25: A v m = 0.421 i/2^A = 0.421 ^2 X 32.2 X 25 = 16.89 ft. per sec, (The .421 might be called a coefficient of velocity.) The volume delivered per second is Q = \nd; v m = i?r(A) 3 16 - 89 = - 2 7 cu ^. ft. per sec. (As the section of the jet F m F^ , that of the short pipe or nozzle, we might also say that .421 = /* = coefficient of efflux, for we may write Q = pF^ V'2gh, whence /* .421.) FLOW THROUGH SIPHONS. 735 532. Siphons. In Fig. 532, 490, the portion HNJ) is above the level, BC> of the surface of the water in the head reservoir BL, and yet under proper conditions a steady flow can be maintained with all parts of the pipe full of water, in- cluding HJV^O. If the atmosphere exerted no pressure, this would be impossible ; but its average tension of 14.7 Ibs. per sq. inch is equivalent to an additional depth of nearly 34 feet of water placed upon BC. With no flow, or a very small velocity, the pipe may be kept full if JV Z is not more than 33 or 34 feet above BC\ but the greater -y a , the velocity of flow at NI, and the greater and more numerous the losses of head between L and N^ the less must be the height of N^ above BC for a steady flow. The analytical criterion as to whether a flow can be main- tained or not, supposing the pipe completely filled at the out- set, is that the internal pressure must be > at all parts of the pipe. If on the supposition of a flow through a pipe of given design the pressure^? is found < 0, i.e. negative, at any point \N^ being the important section for test] the supposition is inadmissible, and the design must be altered. For example, Fig. 532, suppose LN^N^ to be a long pipe of uniform section (diameter = d, and length = Z), and that under the assumption of filled sections we have computed i> 4 , the velocity of the jet at ^ 4 ; i.e., To test the supposition, apply Bernoulli's Theorem to the surface BC and the point N^ where the pressure is ^> 2 , velocity y a (= -y since we have supposed a uniform section for whole pipe), and height above BC=h^. Also, let length of pipe = Z a . Whence we have [BC being datum, plane.] 736 MECHANICS OF ENGINEERING. Solving for , we have !' = 34 feet - [A, + g- (l + Cji + 4/|]. . . (3) We note, then, that for j? a to be < 0, ---^f 1 -.. (4) In the practical working of a siphon it is found that atmos- pheric air, previously dissolved in the water, gradually collects at N^ , the highest point, during the flow and finally, if not re- moved, causes the latter to cease. See reference below. One device for removing the air consists in first allowing it to collect in a chamber in communication with the pipe be- neath. This communication is closed by a stop-cock after the water in it has been completely displaced by air. Another stop-cock, above, being now opened, water is poured in to re- place the air, which now escapes. Then the upper stop-cock is shut and the lower one opened. The same operation is again necessary, after some hours. On p. 346 of the Engineering News of November 1887 may be found an account of a siphon which has been employed since 1875 in connection with the water-works at Kansas City. It is 1350 ft. long, and transmits water from the river to the artificial " well " from which the pumping engines draw their supply. At the highest point, which is 16 ft. above low- water level of the river, is placed a " vacuum chamber " in which the air collects under a low tension corresponding to the height, and a pump is kept constantly at work to remove the air and prevent the " breaking" of the (partial) vacuum. The diam- eter of the pipe is 24 in., and the extremity in the " well " dips 5 ft. below the level of low water. See Trautwine's Pocket- book, for an account of Maj. Crozet's Siphon. 532a. Branching Pipes. If the flow of water in a pipe is caused to divide and pass into two others having a common BRANCHING PIPES. 737 junction with the first, or vice versa ; or if lateral pipes lead out of a main pipe, the problem presented may be very com- plicated. As a comparatively simple instance, let us suppose that a pipe of diameter d and length I leads out of a reservoir, and at its extremity is joined to two others of diameters d l and d^ and lengths l t and Z 2 respectively, and that the further extrem- ities of the latter discharge into the air without nozzles under heads A x and A 2 below the reservoir surface. Call these two pipes Nos. 1 and 2. That is, the system forms a Y in plan. Assuming that all entrances and junctions are smoothly rounded, so that all loss of head is due to skin-friction, it is re- quired to iind the three velocities of flow, v, v l9 and v a , in the respective pipes. First applying Bernoulli's Theorem to a stream-line from the reservoir surface through the main pipe to the jet at the discharging end of pipe No. 1, we have and similarly, dealing with a stream-line through pipe and No. 2, while the equation of continuity for this case is . .... (3) From these three equations, assuming/' the same in all pipes as a first approximation, we can find the three velocities (best by numerical trial, perhaps) ; and then the volume of discharge of the system per unit of time (4) 533. Time of Emptying Vertical Prismatic Vessels (or Inclined Prisms if Bottom is Horizontal) under Variable Head. CASE I. Through an orifice or short pipe in the bottom and opening into the air. Fig. 591. As the upper free surface, 738 MECHANICS OF ENGINEERING. of area = F', sinks, F' remains constant. Let z = head of water at any stage of the emptying ; it = 2 at the outset, and = when the vessel is empty. At any FIG. 593. a? = -^r , and rr /ox '' For the pyramid } f _2 (I ^) @ &,) /7 - (Case III, 534) f ' ^-5* Hence to empty the whole reservoir we have .e., . . (4) EXAMPLE. Let a reservoir of above form, and with & = 50 ft., I = 60 ft., 5j = 10 ft., I, = 20 ft., and depth of water h = 16 ft., be emptied through a straight iron pipe, horizontal, and leaving the side of the reservoir close to the bottom, at an angle a =. 36 with the inner plane of side. The pipe is 80 ft. long and 4 inches in internal diameter ; and of clean surface. The jet issues directly from this pipe into the air, and hence F=\n($f sq. feet To find /<, the "coefficient of efflux" (= 0, the coefficient of velocity in this case, since there is no contraction at discharge orifice), we use eq. (4) (the first radical) of 518, with f approx. = .006, and obtain V (N.B. Since the velocity in the pipe diminishes from a value v = .361 4/20r X 16 = 11.6 ft. per sec. at the beginning of the flow to v = zero at the close, f = .006 is a reasonably approximate average with which to compute the average above ; see 517. 746 MECHANICS OF ENGINEERING. Hence from eq. (4) of this paragraph (ft.-lb.-sec. system) _ [3 X 50 X 60 + 8 X 10 X 20 + 2(50 X 20+20 X 60)]2 15 X 0.361 xVax 32.2 = 31630 sec. = 8 hrs. 47 min. 10 sec. i * ro > l Z J 1 ^** r ( 3$ of the truth. 536. Time of Emptying Reservoirs of Irregular Shape, Simp- son's Rule. From eq. (11), 534, we have, for the time in which the free surface of water in a vessel of any shape what- ever sinks through a vertical distance =d& 9 dt = - -^--, whence \ time = -- / Sz~*dz, . . (1) ^ F ^g ' U = , o lifVSgJ^ where S is the variable area of the free surface at any in- stant, and z the head of water at the same instant, efflux proceeding through a small orifice (or extremity of pipe) of area =. F. If. S can be expressed in terms of z, we can in- tegrate eq. (1) (i.e., provided that Sz~l has a known anti- derivative); but if not, the vessel or reservoir being irregular in form, as in Fig. 600 (which shows a pond whose bottom has been accurately surveyed, so that we know the value of S for any stage of the emptying), we can still get an approximate solution by using Simpson's Rule for approximate inte- gration. Accordingly, if we inquire the time in which the surface will sink from to the entrance Eoi the pipe in Fig. 600 (any point n ; at E. or short of that), we divide the vertical distance from to n (4 in this figure) into an even number of equal parts, and from the known form of the pond compute the area $ corresponding to each point of division, calling them $, $,, etc. Then the required time is approximately TIME OF EMPTYING POND. 747 * i 4 r +- 600. Suppose we have a pipe Em of the same design as in the example of 535, and an initial head of 2 = 16 ft., so that the same value of /*, = .361, may be used. Let z n z = 8 feet, and divide this interval (of 8 ft.) into four equal vertical spaces of 2 ft. each. If at the respective points of division we find from a previous survey that $ Q = 400000 sq. ft., S, = 320000 sq. ft., #, = 270000 sq. ft., &, = 210000 sq. ft., and 4 = 180000 sq. ft. ; while n = 4, p .361, and the area F= i^) 2 = .0873 sq. ft., we obtain (ft.,lb., sec.) G_ _ 16-8 ptOOOOO 4 X 320000 ~ 0.361 X. 0873 I/2^<~32^X 3X4 L ylg" " 2 X 270000 4 X 210000 ISOOOOn = 2444000 sec. " ^8" J= 28d - 6h ' 53m ' volume discharged, T 7 ', may also be found by Simpson's Rule, thus : Since each infinitely small horizontal lamina has a volume or, approximately, 4S $S IS Hence with n = 4 we have (ft., lb., sec.) + 180000~| = 2,160,000 cub. ft. 748 MECHANICS OF ENGINEEBING. 537. Volume of Irregular Reservoir Determined by Observing Progress of Emptying. Transforming eq. (11), 534, we have But Sdz is the infinitely small volume d V oi water lost by the reservoir in the time di, so that the volume of the reser- voir between the initial and final (0 and n) positions of the horizontal free surface (at beginning and end of the time t n ) may be written (1) \= This can be integrated approximately by Simpson's Eule, if the whole time of emptying, = t n , be divided into an even number of equal n parts, and the values ^o 5 s i > z* 5 e t c -> f the head of water noted at these equal inter- vals of time (not of vertical height). The corresponding sur- face planes will not Whence for the particular case FIG. be equidistant, in general, when n = 4 (see Fig. 601) J 4*]' ' ' ( 2 > CHAPTEE VII. HYDRODYNAMICS (Continued) STEADY FLOW OF WATER IN OPEN CHANNELS. 538. Nomenclature, Fig. 602. When water flows in an open channel, as in rivers, canals, mill-races, water-courses, ditches, etc., the bed and banks being rigid, the upper surface is free to conform in shape to the dynamic conditions of each case, which therefore regu- late to that extent the shape of the cross-sec- tion. In the vertical trans- FlG - 602> verse section A C in figure, the line AC is called the air-profile (usually to be considered horizontal and straight), while the line ABC, or profile of the bed and banks, is called the ivetted perimeter. It is evident that the ratio of the wetted perimeter to the whole perimeter, though never < -J, varies with the form of the transverse section. In a longitudinal section of the stream, EFGH, the angle made by a surface filament EF with the horizontal is called the slope, and is measured by the ratio s = h : I, where I is the length of a portion of the filament and h = the fall, or vertical distance between the two ends of that length. The angle be- tween the horizontal and the line HG along the bottom is not necessarily equal to that of the surface, unless the portion of the stream forms a prism ; i.e., the slope of the bed does not necessarily = s = that of surface. EXAMPLES. The old Croton Aqueduct has a slope of 1.10 ft. per mile; i.e., s .000208. The new aqueduct (for New 749 750 MECHANICS OF ENGINEERING. York) has a slope * = .000132, with a larger transverse section. For large sluggish rivers s is much smaller. 539. Velocity Measurements. Various instruments and methods may be employed for this object, some of which are the following : Surface-floats are small balls, or pieces of wood, etc., so colored and weighted as to be readily seen, and still but little affected by the wind. These are allowed to float with the cur- rent in different parts of the width of the stream, and the sur- face velocity c in each experiment computed from c =. I -=- #, where I is the distance described between parallel transverse alignments (or actual ropes where possible), whose distance -apart is measured on the bank, and t = the time occupied. Double-floats. Two balls (or small kegs) of same bulk and condition of surface, one lighter, the other heavier than water, are united by a slender chain, their weights being so adjusted that the light ball, without projecting notably above the surface, buoys the other ball at any assigned depth. Fig. 603. It is assumed that the combination ^^^^w/wm^^m moves with a velocity c', equal to the FIG. 603. arithmetic mean of the surface veloc- ity c of the stream and that, c, of the water filaments at the depth of the lower ball, which latter, 606< for equilibrium. Now P (see 572) for a ball of given size and character of surface varies (nearly) as the square of the velocity ; i.e., if P f is the impulse on a given stationary ball, when the velocity of the current == c 1 ', then for any other velocity c we have P impulse c* (2) c P From this and the relation tan = ^ we derive Cr ^^tantf (3) With a given instrument and a specified system of units, the numerical value of the first radical may be determined as a single quantity, by experimenting with a known velocity and the value of 6 then indicated, and may then, as a constant fac- tor, be employed in (3) for finding the value of c for any ob- served value of ; but the same units must be used as before. 540. Velocities in Different Parts of a Transverse Section. The results of velocity-measurements made by many experi- menters do not agree in supporting any very definite relation between the greatest surface velocity (c max .) of a transverse 754 MECHANICS OF ENGINEERING. section and the velocities at other points of the section, but establish a few general propositions : 1st. In any vertical line the velocity is a maximum quite near the surface, and diminishes from that point both toward the bottom and toward the surface. 2d. In any transverse horizontal line the velocity is a maxi- mum near the middle of the stream, diminishing toward the banks. 3d. The mean velocity v, of the whole transverse section, i.e., the velocity which must be multiplied by the area, F, of the section, to obtain the volume delivered per unit of time, Q = Fv, . . . .' .~r. ,- (1) is about 83 per cent of the maximum surface velocity (c max.) observed when the air is still ; i.e., v = 0.83 X (0. Of eight experimenters cited by Prof. Bowser, only one gives a value (= 0.62) differing more than .05 from .83, while others obtained the values .82, .78, .82, .80, .82, .83. In the survey of the Mississippi River by Humphreys and Abbot, 1861, it was found that the law of variation of the velocity in any given vertical line could be fairly well repre- sented by the ordinates of a parabola (Fig. 607) with its axis .*'.: ".^.:\ :/>. v .;';: ::. : *, horizontal and its vertex at a distance d l ~'i -=V- -t^'-" -i== below the surface according to the follow- X. I i _\ , ' I -I I ,^~^\~ E ing relation, f being a number dependent ^- . T ' ,. i 1 # f" j 1 T / (* r\. (* _i_ Y ~ >\ c dr-~ on the force of the wind (from for no ' f=L wind to 10 for a hurricane) : = [0.31T0.06/"]rf; . . (3) =- where d is the total depth, and the double sign is to be taken -j- for an up-stream, FIG. 607. for a down-stream, wind. The following relations were also based on the results of the survey : (putting, for brevity, B = 1.69 -r- Yd + 1.5,) . (4) VELOCITIES IN OPEtf CHANNEL. 755 /- .7x2 -vm?-=^\, (5) and c^ = c m + -fy^v. .-..:*' . . (7) (These equations are not of homogeneous form, but call for the foot and second as units.) In (4), (5), and (6), c = velocity at any depth & below the surface ; c m = mean velocity in the vertical curve ; c dl = max. " " " Cfa = " at mid-depth ; c d = velocity at bottom ; v = mean velocity of the whole transverse section. It was also found that the parameter of the parabola varied inversely as the square root of the mean velocity c m of curve. In general the bottom velocity ( which is of the same form as Chezy's formula in 519 for a very long straight pipe (the slope s of the actual surface in this case corresponding to the slope along piezometer-summits in that of a closed pipe). In (4) the coefficient A = V%g -s-f is not, like/", an abstract number, but its numerical value depends on the system of units employed. 542a. Experiments on the Flow of Water in Open Channels. Those of Darcy and Bazin, begun in 1855 and published in 1865 (" Recherches Hydrauliques"), were very carefully con- ducted with open conduits of a variety of shapes, sizes, slopes, and character of surface. In most of these a uniform flow was secured before the taking of measurements. The velocities ranged between from about 0.5 to 8 or 10 ft. per second, the hydraulic radii from 0.03 to 3.0 ft., with deliveries as high as 182 cub. ft. per second. For example, the following results were obtained in the canals of Marseilles and Craponne, the quantity A being for the foot and second. The sections were nearly all rectangular. See eq. (4) above. No. (cub.' ft. per sec.) R. (ft.) s. abs. numb. V. (ft. per sec.) A. (foot and sec.) Character of the masonry surface. 1 182.73 1.504 .0037 10.26 137.1 Very smooth. 2 143.74 1.774 .00084 5.55 125. Quite 3 43.93 .708 .029 11.23 78.4 4 43.93 .615 .060 13.93 72.5 Hammered stone. 5 43.93 .881 .0121 7.58 73.5 Rather rough. 6 43.93 .835 .014 8.36 77.3 7 167.68 2.871 . 00043 2.54 72.2 Mud and vegetation. [In Experiment No. 7 the flow had not fully reached a state of permanency.] Fteley and Stearns's experiments on the Sudbury conduit at Boston, Mass. (Trans. A. 8. C. E., '83), from 1878 to 1880, are also valuable. This open channel was of brick masonry with UNIFORM MOTION. good mortar joints, and about 9 ft. wide ; the depths of water ranging from 1.5 to 4.5 ft. With plaster of pure cement on the bed in one of the experiments the high value of A = 153.6 was reached (foot and second), with v = 2.805 ft. per second, R = 2.111 ft., s = .0001580, and Q = 87.17 cu. ft. per second. Captain Cunningham, in his experiments on the Ganges Canal at Roorkee, India, in 1881, found A to range from 48 to 130 (foot and second). Humphreys and Abbot's experiments on the Mississippi River and branches (see 540), with values of R = from 2 or 3 ft. to 72 ft., furnish values of A = from 53 to 167 (foot and second). 542b. Kutter's Formula. The experiments upon which Weisbach based his deductions for/*, the coefficient of fluid friction, were scanty and on too small a scale to warrant gener- al conclusions. That author considered that/* depended only on the velocity, disregarding altogether the degree of rough- ness of the bed, and gave a table of values in accordance with that view, these values ranging from .0075 for 15 ft. per sec. to .0109 for 0.4 ft. per sec. ; but in 1869 Messrs. Kutter and Ganguillet, having a much wider range of experimental data at command, including those of Darcy and Bazin, and those obtained on the Mississippi River, evolved a formula, known as Gutter's Formula, for the uniform motion of water in open channels, which is claimed to harmonize in a fairly satisfactory manner the chief results of the best experiments in that direc- tion. They make the coefficient A in eq. (4) (or rather the factor contained in A) a function of JR, $, and also n an J abstract number, or coefficient of roughness, depending on the nature of the surface of the bed and banks ; viz., *un 41.6 1.811 .00281 1 N, r per f sec. J 1+^41.6 .00281 J n 1 VI ?(in feet) which is Kutter^ s Formula. 760 MECHANICS OF ENGINEERING. That is, comparing (5) with (4), we have /a function of It, and s, as follows : ^ n . . . (6) From (6) it appears that f decreases with an increasing j??, as has been also noted in the case of closed pipes ( -517) ; that it increases with increasing roughness of surface ; and that it is somewhat dependent on the slope. The makers of the formula give the following values for n. Values of n. n = .009 for well-planed timber bed ; .010 for plaster in pure cement ; .011 for plaster in cement with \ sand ; .012 for unplaned timber ; .013 for ashlar and brickwork ; * .015 for canvas lining on frames ; .017 for rubble ; .020 for canals in very firm gravel ; .025 for rivers and canals in perfect order and regimen, and perfectly free from stones and weeds ; .030 for rivers and canals in moderately good order and regi- men, having stones and weeds occasionally ; .035 for rivers and canals in bad order and regimen, overgrown with, vegetation and strewn with stones or detritus of any sort. Kutter's Formula is claimed to apply to all kinds and sizes of watercourses, from large rivers to sewers and ditches ; for uni- form motion. If 4/7? is the unknown quantity, Kutter's For- mula leads to a quadratic equation ; if s the slope, to a cubic. Hence, to save computation, tables have been prepared, some of which will be found in vol. 28 of Yan Nostrand's Magazine * For ordinary brick sewers Mr. R. F. Hartford claims that n = .014 gives good results. See Jour. Eng. Societies for '84- '85, p. 220. UNIFORM MOTION. OPEN CHANNEL. 761 (pp. 135 and 393) (sewers), and in Jackson's works on Hydrau- lics (rivers). The following table will give the student an idea of the variation of the coefficient A, = --, of eq. (4), or large t/ bracket of eq. (5), with different hydraulic radii, slopes, and values of ?&, according to Kutter's Formula ; from R = ft., for a small ditch or sluice-way (or a wide and shallow stream), to R 15 ft., for a river or canal of considerable size. Under each value of R are given two values of A ; one for a slope of s' .001, and the other for s" = .00005. All these values of A imply the use of the foot and second. These values of A have been scaled by the writer from a diagram given in Jackson's translation of Kutter's " Hydraulic Tables^ and are therefore only approximate. The corre- sponding values of /*, the coefficient of fluid friction, can be computed from f = ^ . n 0.010 0.015 0.020 0.025 0.030 0.035 R = * f t. R = 1 f t. R = 3 f t. R = 6 f t. R = 15 ft. for for s' s" for for s' s" for for s' s" for f el- s' s" for for s' s" 133 104 83 68 58 49 45 38 36 31 28 25 149 137 96 87 70 63 54 48 43 40 38 34 174 174 118 118 87 87 70 70 58 58 50 50 187 196 128 137 98 106 80 85 66 72 58 64 199 222 138 158 110 126 90 106 77 90 68 81 The formula used in designing the New Aqueduct for New York City, in 1885, by Mr. Fteley, consulting engineer, was [see (4)] " ' v (ft. per sec.} = 142 VE(irift.) X s, ... (7) whereas Kutter's Formula gives for the same case (a circular section of 14 ft. diameter, and slope of 0.7 ft. to the mile), with n = 0.013, v(ft per xec.) = 140.7 V E(ln ft.} X *. . . . (8) 762 MECHANICS OF ENGINEERING. To quote from a letter of Mr. I. A. Shaler of the Aqueduct Corps of Engineers, " Mr. Fteley states that the cleanliness of the conduit (Sudbury) had much to do in affecting the flow. He found the flow to be increased by 7 or 8 per cent in a por- tion which had been washed with a thin wash of Portland cement." EXAMPLE 1. A canal 1000 ft. long of the trapezoidal sec- tion in Fig. 611 is required to deliver 300 cubic ft. of water per second with the water 8 ft. deep at all sections (i.e., with uniform motion), the slope of the bank being such that for a depth of 8 ft. the width of the water surface (or length of air-profile) will be 20 ft.; and the coefficient for roughness being n .020. What is the neces- sary slope to be given to the bed (slope of bed = that of sur- face, here) (ft., lb., sec.) \ The mean velocity v = Q + F= 300 -f- i (20 + 8) 8 = 2.6T ft. per sec. [So that the surface velocity of mid-channel in any section would probably be ( Equations (4), (5), ... (8) hold good, then, for the trapezoi- dal section of least f rictional resistance for a given angle 6. PKOBLEM. Required the dimensions of the trapezoidal sec- tion of minimum f rictional resistance for 6 = 45, which with k = 6 inches fall in every 1200 feet ( Z) is required to de- liver Q = 360 cub. ft. of water per minute with uniform motion. Here we have given, with uniform motion, A, Z, and Q r with the requirement that the section shall be trapezoidal, with 6 = 45, and of minimum frictional resistance. The following equations are available : Eq. of continuity . . . Q = Fv, ...... (1 ; ) Eq. (8) preceding, for con- ) 7?' A / s ^ dition of least resistance j 2 y 2 cos 6 . (2'Y There are three unknown quantities, v, F, and R' . Solve 768 MECHANICS OF ENGINEERING. (I/) for v ; solve (2') for R'\ substitute their values in (3') ; whence h I/sin ! cos# .(4') Since /"cannot be exactly computed in advance, for want of knowing the value of R, we calculate it approximately [eq. (6)> 542b] for an assumed value of R, insert it in the above equation (4/), and thus find an approximate value of F\ and then, from (8), a corresponding value of R, from which a new value of /"can be computed. Thus after one or two trials a satisfactory adjustment of dimensions can be secured. 545. Variable Motion. If a steady flow of water of a de- livery Q, = Fv, = constant, takes place in a straight open channel the slope of whose bed has not the proper value to maintain a " uniform motion" then " variable motion" ensues (the flow is still steady, however); i.e., although the mean velocity in any one transverse section remains fixed (with lapse of time), this velocity has different values for different sections ; but as the eq. of continuity, etc., still holds (since the flow is steady), the different sections have different areas. If, Fig. 616, a stream of water flows down an inclined trough without friction, the relation between the velocities v and v l at any two FIG. 616. sections and 1 will be the same as for a material point sliding down a guide without friction (see 79, latter part), viz. : (1) VARIABLE MOTION. OPEN CHANNEL. 769 an equation of heads (really a case of Bernoulli's Theorem, 492). But, considering friction on the bed, we must sub- tract the mean friction-head f -= --- [see eqs. (3) and (3'), H 2<7 542] lost between and 1 ; this friction-head may also be 7 2 written thus : f - ~- ; and therefore eq. (1) becomes which is the formula for variable motion ; and in it I is the length of the section considered, which should be taken short enough to consider the surface straight between the end-sec- tions, and the latter should differ but slightly in area. The subscript m may be taken as referring to the section midway between the ends, so that v m 2 = j-(i> 2 -f- v*). The wetted pe- ri meter w m = i(w. + w,), and F m = %(F Q + F^. Hence eq. (2) becomes * _ , ,. ,o\ B SF ""%" ~^+*r ~^~ and again, by putting v Q -=- F Q , v, Q -=- F, , we may write F whence , * n From eq. (4), having given the desired shapes, areas, etc., of the end-sections and the volume of water, Q, to be carried per unit of time, we may compute the necessary fall, A, of the sur- face, in length = I ; while from eq. (5), having observed in an actual water-course the values of the sectional areas F Q and JF 1 l , the wetted perimeters w and w l , the length, = I, of the por- 770 MECHANICS OF ENGINEERING. tion considered, we may calculate Q and thus gauge the stream approximately, without making any velocity measurements. As to the value of f, we compute it from eq. (6), 542b, using for R a mean between the values of the hydraulic radii of the end-sections. 546. Bends in an Open Channel. According to Humphreys and Abbot's researches on the Mississippi Kiver the loss of head due to a bend may be put , ^"536 IT' (1) in which v must be in ft. per sec., and #, the angle ABC, Fig. 617, must be in 7r-measure, i.e. in radians. The section ^must be greater than 100 sq. ft., and the slope s less than .0008. v is the mean velocity of the water. Hence if a bend occurred in a portion of a stream of length I, q. (3) of 542 be- comes FIG. 617. 6 while eq. (2) of 545 for variable motion would then become = -^- -\-h --^rr- -^TT-I - (ft- and sec.). . (3) 2# 2# J 1 ' m Qg Ooo 7t (v and d as above.) (For " radian" see p. 544.) 547. Equations for Variable Motion, introducing the Depths. Fig. 618. The slope of the bed being sin a (or simply ^, Trmeas.), while that of the surface is different, viz., sin fi = s = h -4- /, we may write h = d -f- 1 sin a d l , FIG. 618. VARIABLE MOTION. OPEN CHANNEL. 771 in which d and d l are the depths at the end-sections of the portion considered (steady flow with variable motion). With these substitutions in eq. (4), 545, we have, solving for Z, From which, knowing the slope of the bed and the shape and size of the end-sections, also the discharge Q, we may compute the length or distance, Z, between two sections whose depths differ by an assigned amount (d d^. But we can- not compute the change of depth for an assigned length Z from (6). However, if the width b of the stream is constant, and the same at all depths ; i.e., if all sections are rectangles hav- ing a common width ; eq. (6) may be much simplified by intro- ducing some approximations, as follows : We may put > ' F? ._ ' and, similarly, w m ^ 1 m n which approx. = ~- d b Zg Hence by substitution in eq. (6) we have 547a. Backwater. Let us suppose that a steady flow las been proceeding with uniform motion (i.e., the surface parallel 772 MECHANICS OF ENGINEERING. to the bed) in an open channel of indefinite extent, and that a vertical wall is now set up across the stream. The water rises and flows over the edge of the wall, or weir, and after a time a steady flow is again established. The depth, y , of the water close to the weir on the up-stream side is greater than d , the original depth. "We now have " variable motion " above the weir, and at any distance x up-stream from the weir the new depth y is greater than d . This increase of depth is called backwater, and, though decreasing up-stream, may be percep- tible several miles above the weir. Let s be the slope of the original uniform motion (and also of present bed), and v the v* velocity of the original uniform motion, and let ~k = . y Then, if the section of the stream is a shallow rectangle of constant width, we have the following relation (Rankine) : 0.)], where is a function of -, as per following table : For = 1 . d

a "With this table and eq. (1), therefore, we can find a?, the dis- tance ( u amplitude of backwater") from the weir of the point where any assigned depth y (or " height of backwater," y d ) will be found. For example, Prof. Bowser cites the case from D'Aubuis- son's Hydraulics of the river Weser in Germany, where the erection of a weir increased the depth at the weir from 2.5 ft. to 10 ft., the flow having been originally " uniform" for 10 miles. Three miles above the dam the increase (y d ) of depth was 1.25 ft., and even at four miles it was 0.75 ft. CHAPTER YIII. DYNAMICS OF GASEOUS FLUIDS. 548. Steady Flow of a Gas. [KB. The student should now review 492 up to eq. (5).] The differential equation from which Bernoulli's Theorem was derived for any liquid, with- out friction, was [eq. (5), 492] l r vdv + ds + -d j p = 0, ..... (A) / / and is equally applicable to the steady flow of a gaseous fluid, but with this difference in subsequent work, that the heaviness, y ( A f the gas passing different sections of the pipe or stream-line is, or may be, different (though always the same at a given point or section, since the flow is steady). For the present we neglect friction and consider the flow from a large receiver, where the great body of the gas is practically at rest, through an orifice in a thin plate, or a short nozzle with a rounded entrance. In the steady flow of a gas, since y is different at different points, the equation of continuity takes the form Flow of weight per time-unit = F l v 1 y 1 = F^v^y^ = etc. ; . (a) i.e., the weight of gas passing any section, of area F, per unit of time, is the same as for any other section, or Fvy = con- stant, y being the heaviness at the section, and v the velocity. 549. Flow through an Orifice Remarks. In Fig. 619 we have a large rigid receiver containing gas at some tension, p n , higher than that, p m , of the (still) outside air (or gas), and at some absolute temperature T n , and of some heaviness y n \ that , is, in a state n. The small orifice of area F being opened, the gas begins to escape, and if the receiver is very large, or if the supply is continually kept up (by a blowing-engine, e.g.), after 773 774 MECHANICS OF ENGINEERING. STILL. . AIR. . FIG. 619. a very short time the flow becomes steady. Let nm represent any stream-line ( 495) of the flow. According to the ideal subdivision of this stream-line into J T ^ ; _ .,, laminae of equal mass or weight (not equal volume, necessarily) in estab- lishing eq. (A) for any one lamina, each lamina in the lapse of time dt moves into the position just vacated by the lamina next in front, and assumes precisely the same velocity, pressure, and volume (and there- fore heaviness] as that front one had at the beginning of the dt. In its progress toward the orifice it expands in volume, its tension diminishes, while its velocity, insensible at n, is gradually accelerated on account of the pressure from behind always being greater than that in front, until at m, in the " throat" of the jet, the velocity has become v m , the pressure (i.e., tension) has fallen to a value p m , and the heaviness has changed to y m . The temperature T m (absolute) is less than T n , since the expansion has been rapid, and does not depend on the temperature of the outside air or gas into which efflux takes place, though, of course, after the effluent gas is once free from the orifice it may change its temperature in time. We assume the pressure ^> m (in throat of jet) to be equal to that of the outside medium (as was done with flow of water), so long as that outside tension is greater than .527^ TC ; but if it is less than .527 p n and is even zero (a vacuum), experiment seems to show that p m remains equal to 0.527 of the interior tension p n \ probably on account of the expansion of the effluent gas beyond the throat, Fig. 620, so ,///, that although the tension in the outer edge, ^" at a, of the jet is equal to that of the outside medium, the tension at m is greater because of the centripetal and centrifugal forces devel- FIO. 620. oped in the curved filaments between a and m. (See 553.) 550. Flow through an Orifice; Heaviness assumed Constant during Flow, The Water Formula. If the inner tension p n ex- STEADY FLOW OF GASES. 775 ceeds the outer, p m , but slightly, we may assume that, like water, the gas remains of the same heaviness during flow. Then, for the simultaneous advance made by all the laminae of a stream-line, Fig. 619, in the time dt, we may conceive an equation like eq. (A) written out for each lamina between n and w, and corresponding terms added ; i.e., {For orifices) . . - fvdv + fdz + /**& = <>. . (B) qjn ' t//i ' Jn y In general, y is different in the. different laminae, but in the present case it is assumed to be the same in all ; hence, with m as datum level and h vertical distance from n to w, we have, from. eq. (B), ^ _ V^_ , Q _ h , Pm _ Pn = Q . .. fy *g r r But we may put v n ; while A, even if several feet, is small compared with . E.g., with p m ^ 15 Ibs. per sq. in. and p n = 16 Ibs. per sq. in., we have for atmospheric air at freezing temperature y y Hence, putting v n and h = in eq. (1), we have ^m _ Pn Pm ( Water formula ; for small \ /^ 2g y n \ difference of pressures, only. \ The interior absolute temperature T n being known, the y n (interior heaviness) may be obtained from y n =p n y*T Q -r- T n p ( 472), and the volume of flow per unit of time then obtained (first solving (2) for v m ) is (3) where F m is the sectional area of the jet at m. If the mouth- piece or orifice has well-rounded interior edges, as in Fig. 541, 776 MECHANICS OF ENGINEERING. its sectional area F may be taken as the area F m . But if it is an orifice in "thin plate," putting the coefficient of contraction = C 0.60, we have F m = CF = 0.60 F-, and Q m = 0.60 Fv m . . (4) This volume, Q m , is that occupied by the flow per time-unit when in state m, and we have assumed that y m = y n ; hence the weight of flow per time-unit is O = Q m y m = F m v mY ^ = F m v m7n . . . . (5) EXAMPLE. In the testing of a blowing-engine it is found capable of maintaining a pressure of 18 Ibs. per sq. inch in a large receiver, from whose side a blast is steadily escaping through a " thin plate" orifice (circular) having an area F = 4 sq. inches. The interior temperature is 20 Cent, and the out- side tension 15 Ibs. per sq. in. Required the discharge of air per second, both volume and weight. The data are: p n = lS Ibs. per sq. in., T n = %93 Abs. Cent., F= 4 sq. inches, and p m = 15 Ibs. per sq. in. Use ft.-lb.-sec. system. First, the heaviness in the receiver is Then, from eq. (2), /o Pnp m _ /2X32.2[144X18144X15]_J 555.3 v m \ / ^9 '~ \ A AQn ) feet V Yn V 0<089 I per sec. (97 per cent of this would be more correct on account of fric- tion.) .-. Q m =F' m v m =.6JFv m = TViirX 555.3 = 9.24 cub. ft. per sec. at a tension of 15 Ibs. per sq. in., and of heaviness (by hypoth- esis) = .089 Ibs. per cub. ft. Hence weight = G = 9.24 X .089 = .82 Ibs. per sec. FLOW OF GASES BY MARIOTTE'S LAW. 777 The theoretical power of the air-compressor or blowing-en- gine to maintain this steady flow can be computed as in Exam- ple 3, 483. 551. Flow through an Orifice on the Basis of Mariotte's Law ; or Isothermal Efflux. Since in reality the gas expands during flow through an orifice, and hence changes its heaviness (Fig. 619), we approximate more nearly to the truth in assuming this change of density to follow Mariotte's law, i.e., that the heaviness varies directly as the pressure, and thus imply that the temperature remains unchanged during the flow. We again integrate the terms of eq. (B\ but take care to note that, now, y is variable (i.e., different in different laminae at the same instant), and hence express it in terms of the variable p (from eq. (2), 475), thus : Therefore the termy of eq. (_Z?) becomes and, integrating all the terms of eq. (B\ neglecting A, and call- ing v n zero, we have *V* _ Pn i Pn ( efflux ly Mariotte's ) /2\ %g ~ Yn Pm ' ' \ Law through orifice ) ' T 7) As before, y n ^ f y , and the flow of volume per time- *n Po unit at m is Q m = F m v m ; ....... (3) while if the orifice is in thin plate, F m may be put = .60 F^ and the weight of the flow per time-unit G = F m VmY.' '(*) If the mouth-piece is rounded, F m F= area of exit orifice of mouth-piece. 778 MECHANICS OF ENGINEERING. EXAMPLE. Applying eq. (2) to the data of the example in 550, where y n was found to be .089 Ibs. per cub. ft., we have [ft., lb., sec.] P % Q m = F m v m = 0.60 X T f X 584.7 = 9.745 cub. ft. per sec. Since the heaviness at m is, from Mariotte's law, y m = SOL Yn = ff o f .089, i.e., y m = .0741 Ibs. per cub. ft., Pn hence the weight of the discharge is G = QmYm = 9.745 X .0741 = 0.722 Ibs. per sec., or about 12 per cent less than that given by the " water for- mula." If the difference between the inner and outer tensions had been less, the discrepancy between the results of the two methods would not have been so marked. 552. Adiabatic Efflux from an Orifice. It is most logical to assume that the expansion of the gas approaching the orifice, being rapid, is adiabatic ( 478). Hence (especially when the difference between the inner and outer tensions is considerable) it is more accurate to assume y as varying according to Pois- son's Law, eq. (1), 478 ; i.e., y = [y n +p n *]p*, in integrat- ing eq. (B). Then the term . PJ ' ADIABATIC FLOW OF GASES THROUGH ORIFICES. 779 and eq. (^), neglecting h as before, and with v n = 0, becomes (See Fig. 619) L = l - " - (Adiabaticflow; orifice.) . (1) "9 y^ L- \Pn' -J Having observed ^> n and 7^ in the reservoir, we compute /y~7 y n P^ Q (from 472). The gas at m, jnst leaving the *nj^ orifice, having expanded adiabatically from the state n to the state m, has cooled to a temperature T m (absolute) found thus ( and is of a heaviness and the flow per second occupies a volume (immediately on exit) and weighs G = Fj> mYm ....... (5) EXAMPLE 1. Let the interior conditions in the large reser- voir of Fig. 619 be as follows (state n) : p n 22 -J- Ibs. per sq. in., and T n = 294 Abs. Cent, (i.e., 21 Cent.) ; while ex- ternally the tension is 15 Ibs. per sq. inch, which may be taken as being = p m tension at m, the throat of jet. The opening is a circular orifice in " thin plate" and of one inch diameter. Required the weight of the discharge per second [ft., lb., sec.; g = 32.2]. First, r . = . X .0807 = 0.114 Ibs. per cub. ft. Then, from (1), Vm = /^n-MH A V r* L w - t _ f f] = 8M ft . per sec . 780 MECHANICS OF ENGINEERING. Now F= \n(-^= .00546 sq. ft. /. Q m = CFv m = 60 Fv n = 0.60 X .00546 X 844 = 2.765 cub. ft. per sec., at a temperature of T m = 294 Vf = 257 Abs. Cent. = - 16 Cent, and of a heaviness y m = 0.114 V(f7 = 0.085 Ibs. per cub. ft., so that the weight of flow per sec. = O = Q m y m = 2.765 X .085 = .235 Ibs. per sec. EXAMPLE 2. Let us treat the example already solved by the two preceding approximate methods ( 550 and 551) by the present more accurate equation of adiabatic flow, eq. (1). The data were (Fig. 619) : p n = 18 Ibs. per sq. in. ; T n = 293 Abs. Cent. ; p m = 15 " " " ; and F 4 sq. inches [/'"being the area of orifice]. y n was found = .089 Ibs. per cub. ft. in 550 ; hence, from eq. (1), 18X144 From (4), Q m =F m v m =.6Fv m =.eXr^X^^ = 9.603 cub. ft. per sec.; and since at m it is of a heaviness Ym = .089 V(3y = .0788 Ibs. per cub. ft, we have weight of flow per sec. = G = QmYm = 9.603 X .0788 = 0.756 Ibs. per sec. THEORETICAL MAXIMUM FLOW OF WEIGHT OF GAS. 781 Comparing the three methods for this problem, we see that By the " water formula? . . . 0.82 Ibs. per sec. " isothermal formula, . . G = 0.722 " " " adiabatic formula, . . G = 0.756 u u 553. Practical Notes. Theoretical Maximum Flow of Weight. If in the equations of 552 we write for brevity p m --p n = x we derive, by substitution from (1) and (3) in (5), Weight of flow) n n T? */~5 r-i m i per unit of time \ = G = Qr=F Vfyp n y n [l-afl-a*. . (1) This function of x is of such a form as to be a maximum for = (j?-^ n )=(|.) s =.512; .... (2) i.e., theoretically, if the state n inside the reservoir remains the same, while the outside tension (considered =p m of jet, Fig. 619) is made to assume lower and lower values (so that a?, p m -p n , diminishes in the same ratio), the maximum flow of weight per unit of time will occur when p m = .512 p n , a little more than half the inside tension. (With the more ac- curate value 1.41 (1.408), instead of f, see 478, we should obtain .527 instead of .512 for dry air; see 549.) Prof. Cotterill says (p. 544 of his " Applied Mechanics") : " The diminution of the theoretical discharge on diminution of the external pressure below the limit just now given is an anomaly which had always been considered as requiring ex- planation, and M. St. Tenant had already suggested that it could not actually occur. In 1866 Mr. R. D. Napier showed by experiment that the weight of steam of given pressure dis- charged from an orifice really is independent of the pressure of the medium into which efflux takes place * ; and in 1872 Mr. Wilson confirmed this result by experiments on the reac- tion of steam issuing from an orifice." " The explanation lies in the fact that the pressure in the * When the difference between internal and external pressures is great, should be added. 782 MECHANICS OF ENGINEERING. centre of the contracted jet is not the same as that of the sur- rounding medium. The jet after passing the contracted sec- tion suddenly expands, and the change of direction of the fluid particles gives rise to centrifugal forces" which cause the pres- sures to be greater in the centre of the contracted section than at the circumference ; see Fig. 620. Prof. Cotter-ill then advises the assumption ttmt^ w =.527p n (for air and perfect gases) as the mean tension in the jet at in (Fig. 619), whenever the outside medium is at a tension less than .527j? n . He also says, u Contraction and friction must be allowed for by the use of a coefficient of discharge the value of which, however, is more variable than that of the corresponding coefficient for an incompressible fluid. Little is certainly known on this point." See 549 and 554. For air the velocity of this maximum flow of weight is r /7M Vd. of max. G = 997 A / ~ ft. per sec., . v -* o ' (3) where T n abs. temp, in reservoir, and T Q = that of freezing point. Rankine's Applied Mechanics ( p. 584) mentions ex- periments of Drs. Joule and Thomson, in which the circular orifices were in a thin plate of copper and of diameters 0.029 in., 0.053 in., and 0.084 in., while the outside tension was about one half of that inside. The results were 84 per cent of those demanded by theory, a discrepancy due mainly, as Rankine says, to the fact that the actual area of the orifice was used in computation instead of the contracted section; i.e., con- traction was neglected. 554. Coefficients of Efflux by Experiment. For Orifices and Short Pipes. Small Difference of Tensions. Since the discharge through an orifice or short pipe from a reservoir is affected not only by contraction, but by slight friction at the edges, oven with a rounded entrance, the theoretical results for the volume and weight of flow per unit of time in preceding para- graphs should be multiplied both by a coefficient of velocity and one for contraction (7, as in the case of water ; i.e., by a coefficient of efflux /*, = 0(7. (Of course, when there is no COEFFICIENTS OF EFFLUX. GAS. 783 contraction, C= 1.00, and then ju = as with a well-rounded mouth-piece, for instance, Fig. 541, and with short pipes.) Hence for practical results, with orifices and short pipes, we should write for the weight of flow per unit of time Pnl V \Pn (from the equations of 552 for adiabatic flow, as most accu- rate ; p m -T-p n may range from \ to 1.00). F= area of orifice,. or of discharging end of mouth-piece or short pipe. y n = heaviness of air in reservoir and T p n y -=- T n p , eq. (13) of 437 ; and /* = the experimental coefficient of efflux. From his own experiments and those of Koch, D'Aubuis- son, and others, "Weisbach recommends the following mean values of ^ for various mouthpieces, when p n is not more than J larger than p m (i.e., about 17 % larger), for use in eq, (1) : 1. For an orifice in a thin plate, ....... /*=0.56 2. Forashortcylindricalpipe(innercornersnotrounded),/f==0.75 3. For a well-rounded mouth-piece (like that in Fig. 541), /*=0.98 4. For a short conical convergent pipe (angle about 6), /^=0.92 EXAMPLE. (Data from Weisbach's Mechanics.) "If the sum of the areas of two conical tuyeres of a blowing-machine is F =. 3 sq. inches, the temperature in the reservoir 15 Cent., the height of the attached (open) mercury manometer (see Fig. 464) 3 inches, and the height of the barometer in the ex- ternal air 29 inches," we have (ft., lb., sec.) Ab, Cent.; Pn - (H) 14.7 X 144 Ibs. per sq. ft. ; y n = |||.ff X 0.0807 = 0.0816 Ibs. per cub. ft, while F= yf^ sq. ft. and (see above) // = 0.92 ; hence O = 0.92 X T| T (ff)* V2 X 32.2 X 3X If X 14.7 X 144 X .0816 [1 - f f "fl; 784 MECHANICS OF ENGINEEEING. i.e.. G .6076 Ibs. per second ; which will oc'jupy a volume y % = a -T- y = & -5-. -0807 = 7.59 cub. ft. at one atmosphere tension and freezing-point temperature; while at a temperature of T n = 288 Abs. Cent, and tension of p m J-9. of one atmosphere (i.e., in the state in which it was on entering the blowing-engine) it occupied a volume v = iff -H X 7.59 = 8.24 cub. ft. (This last is Weisbach's result, obtained by an approximate formula.) 555. Coefficients of Efflux for Orifices and Short Pipes for a Large Difference of Tension. For values > -J and < 2, of the ratio p n : p m , of internal to external tension, Weisbach's ex- periments with circular orifices in thin plate, of diameters (= d) from 0.4 inches to 0.8 inches, gave the following results : Pn : Pm 1.05 1.09 1.40 1.65 1.90 2.00 . or d = .4 in -; jj. = .55 .59 .69 .72 .76 .78 d=.8-,ju = .56 .57 .64 .68 .72 Whence it appears that ju increases somewhat with the ratio of p n to p m , and decreases slightly for increasing size of orifice. With short cylindrical pipes, internal edges not rounded, and three times as long as wide, Weisbach obtained p, as follows : Pn : Pm = 1.05 1.10 1.30 1.40 1.70 1.74 diam. = .4 in - ; /* .73 .77 .33 " = ,6 in -; n = .81 .82 " =1.0 in - ; // = .83 When the inner edges of the 0.4 in. pipe were slightly rounded, /* was found = 0.93 ; while a well-rounded mouth- piece of the form shown in Fig. 541 gave a value /* = from .965 to .968, for p n : p m ranging from 1.25 to 2.00. These values of p are for use in eq. (1), above. 556. To find the Discharge when the Internal Pressure is measured in a Small Reservoir or Pipe, not much larger than the VELOCITY OF APPROACH. OASES. 785 Orifice. Fig. 621. If the internal pressure p n , and tempera- ture jT n , must be measured in a small reservoir or pipe, n, whose sectional area F n is not very large \ = compared with that of the orifice, - F, (or of the jet, F m ,) the velocity v n at n (velocity of approach) can- FIG. 621. not be put = zero. Hence, in applying eq. (E\ 550, to the successive laminae between n and m, and integrating, we shall have, for adiabatic steady flow, instead of eq. (1) of 552. But from the equation of continuity for steady flow of gases [eq. (a) of 548], F n v n y n F m v m y m ; F V 2 hence v^ ^ v m \ while for an adiabatic change from n -CnYn to m, = \~\ ; whence by substitution in (1), solving for v m , we have \ As before, from 472 and 478, i __ f ^V^ 1 w w (2) *.= frr-y. ( 3 ) j^o * n * rm=(*r]r. W \Pnl Having observed p n , p m , and T n , tnjn, and knowing the area F of the orifice, we may compute y n , y m , and v m , and finally the Weight of flow per time-unit = G = f^Fv m y m ^ . . (5) 786 MECHANICS OF ENGINEERING. taking /* from 554 or 555. In eq. (2) it must be remembered that for an orifice in " thin plate," F m is the sectional area of the contracted vein, and = CF\ where C may be put = . CM EXAMPLE. If the diameter of AB, Fig. 621, is 3 inches^ and that of the orifice, well rounded, = 2 in. ; if p n = 1^ at- mospheres i= i J- x 14.7 X 144 Ibs. per sq. ft., while p m = -^ of an atmos., so that ^ = |, and T n = 283 Abs. Cent., re- Pn quired the discharge per second, using the ft., lb., and sec. From eq. (3), Yn = If -fit X 0.0807 = .08433 Ibs. per cub. ft. ; 'whence (eq. (4)) Ym (f i-)i/ n = .07544 Ibs. per cub. ft. Then, from eq. (2), = 558.1 ft. per sec. ; /. G = 0.98 558.1 X .07544 = .9003 Ibs. per sec. 4V6/ 557. Transmission of Compressed Air; through very Long Level Pipes. Steady Flow. CASE I. When the difference between the tensions in the reservoirs at the ends of the pipe is small. Fig. 622. Under FIG. 622. these circumstances it is simpler to employ the form of formula that would be obtained for a liquid by applying Bernoulli's Theorem, taking into account the " loss of head " occasioned TRANSMISSION OF COMPKESSED AIR. 787 by the friction on the sides of the pipe. Since the pipe is very long, and the change of pressure small, the mean velocity in the pipe, v', assumed to be nearly the same at all points along the pipe, will not be large ; hence the difference be- tween the velocity-heads at n and m will be neglected ; a cer- tain mean heaviness y f will be assigned to all the gas in the pipe, as if a liquid. Applying Bernoulli's Theorem, with friction, 516, to the ends of the pipe, n and m, we have (as for a liquid) ^.^, = ^ + ^ + 0- 4/1^-. (1) W^? fy r f J <*fy Putting (as above mentioned) v m * v n * 0, we have, more simply, The value of f as coefficient of friction for air in long pipes is found to be somewhat smaller than for water ; see next paragraph. 558. Transmission of Compressed Air. Experiments in the St. Gothard Tunnel, 1878. [See p. 96 of Yol. 24 (Feb. '81), Yan Nostrand's Engineering Magazine.] In these experiments, the temperature and pressure of the flowing gas (air) were ob- served at each end of a long portion of the pipe which delivered the compressed air to the boring-machines three miles distant from the tunnel's mouth. The portion considered was selected at a distance from the entrance of the tunnel, to eliminate the fluctuating influence of the weather on the temperature of the flowing air. A steady flow being secured by proper regulation of the compressors and distributing tubes, observations were made of the internal pressure (p\ internal temperature (T), as well as the external, at each end of the portion of pipe con- sidered, and also at intermediate points ; also of the weight of flow per second G = Q y , measured at the compressors under standard conditions (0 Cent, and one atmos. tension). Then knowing the p and T at any section of the pipe, tha 788 MECHANICS OF ENGINEERING. heaviness y of the air passing that section can be computed r y T> T ~\ from - = - . - and the velocity v = G -=- Fy, F being L Yo Po -L ' the sectional area at that point. Hence the mean velocity v', and the mean heaviness 7', can be computed for this portion of the pipe whose diameter = d and length I. In the ex- periments cited it was found that at points not too near the tunnel-mouth the temperature inside the pipe was always about 3 Cent, lower than that of the tunnel. The values of /in the different experiments were then computed from eq. (2) of the last paragraph ; i.e., >n Pm _ Y' 7 ./a all the other quantities having been either directly observed, or computed from observed quantities. THE ST. GOTHARD EXPERIMENTS. [Concrete quantities reduced to English units.] No. l (feet.) d (ft.) (Ibs. cub. ft.) Atmospheres. Pn ~ Pm Ibs. sq. in. v' ft. per sec. mean temp. Cent. Pn Pm 1 15092 | 0.4058 5.60 5.24 5.29 19.32 21 .0035 2 15092 } 0.3209 4.35 4.13 3.23 16.30 21 .0038 8 15092 .2803 3.84 3.65 2.79 15.55 21 .0041 4 1712 * .3765 5.24 5.00 3.52 37.13 26.5 .0045 1712 . 3009 4.13 4.06 1.03 30.82 26.5 .0024(?) 6 1712 * .2641 3.65 3.54 1.54 29.34 26.5 .0045 In the article referred to (Yan Nostrand's Mag.) f is not computed. The writer contents himself with showing that "Weisbach's values (based on experiments with small pipes and high velocities) are much too great for the pipes in use in the tunnel. With small tubes an inch or less in diameter "Weisbach found, for a velocity of about 80 ft. per second,/* =.0060; for still higher velocities/ was smaller, approximately, in ac- cordance with the relation f .0542 yV(in ft. per sec.) TRANSMISSION OF COMPRESSED AIR. 789 On p. 370, vol. xxiv, Yan Nostrand's Mag., Prof. Kobinson of Ohio mentions other experiments with large long pipes. From the St. Gothard experiments a value off= .004 may be inferred for approximate results with pipes from 3 to 8 in. in diameter. EXAMPLE. It is required to transmit, in steady flow, a supply of G 6.456 Ibs. of atmospheric air per second through a pipe 30000 ft. in length (nearly six miles) from a reservoir where the tension is 6.0 atmos. to another where it is 5.8 atmos., the mean temperature in the pipe being 80 Fahr., = 24 Cent. (i.e. 297 Abs. Cent.). Kequired the proper diameter of pipe ; d = ? The value . /= .00425 will be used, and the ft- Ib.-sec. system of units. The mean volume passing per second in the pipe is Q> The mean velocity may thus be written : v f = -~ = - ; . (4) The mean heaviness of the flowing air, computed for a mean tension of 5.9 atmospheres, is, by 472, r'= -~ X .0807 = 0.431 Ibs. per cub. ft. ; and hence, see eq. (3), at tension of 5.9 atmos., and temperature 297 Abs. Cent. Now, from eq. (2), whence J/ y'l ~ 790 MECHANICS OF ENGINEERING. and hence, numerically, 5 / 4 X .00425 X 0.431 X 30000 X (14.74) a ~ Y (.7854)*[14.7 X 144(6.00 - 5.80)]2 X 32.2 ~ 559. (Case II of g 557) Long Pipe, with Considerable Differ- ence of Pressure at Extremities of the Pipe. Flow Steady. Fig. 623. If the difference between the end-tensions is compara- tively great, we can no longer deal with the whole of the air 2 p~ --^ *-fl /*> :::A| i B V:i :-./.> n . ' . ""' "? *<--> ds FIG. 623. in the pipe at once, as regards ascribing to it a mean velocity and mean tension, but must consider the separate laminae, such as AE (a short length of the air-stream) to which we may apply eq. (2) of 556 ; A and E corresponding to the n and m of Fig. 622. Since t\\Q p n p m , I, y f , and v' of 559 correspond to the dp, ds, y, and v of the present case (short section or lamina), we may write r (I'D , which substituted in eq. (2) enables us to Pn' write: T -pdp = ; Performing the integration, noting that at n r p =p n >, s 0, and at m f p = p m > and s = I, we have ir. _ > ,n - ^f l & Pn> j isothermal flow \ ,.. *UV p m \ -tyd'jr*"^,' ( in long pipes ] It is here assumed that the tension at the entrance of the pipe is practically equal to that in the head reservoir, and that at the end (ra') to that of the receiving reservoir; which is not strictly true, especially when the corners are not rounded. It will be remembered also that in establishing eq. (2) of 556 (the basis of the present paragraph), the "inertia" of the gas was neglected ; i.e., the change of velocity in passing along the pipe. Hence eq. (4) should not be applied to cases where the pipe is so short, or the difference of end-tensions so great, as to create a considerable difference between the velocities at the two ends of the pipe. EXAMPLE. A well or reservoir supplies natural gas at a ten- sion of p n > = 30 Ibs. per sq. inch. Its heaviness at Cent, and one atmosphere tension is .0484 Ibs. per cub. foot. In piping this gas along a level to a town two miles distant, a single four-inch pipe is to be employed, and the tension in the receiving reservoir (by proper regulation of the gas distributed from it) is to be kept equal to 16 Ibs. per sq. in. (which would sustain a column of water about 2 ft. in height in an open water manometer, Fig. 465). 792 MECHANICS OF ENGINEERING. The mean temperature in the pipe being 17 Cent., required the amount (weight) of gas delivered per second, supposing leakage to be prevented (formerly a difficult matter in practice). Solve (4) for G, and we have First, from 472, with T n , = T m , = 290 Abs. Cent., we compute 273 Hence with/ .005, x -30 x 4X.005 X 10560 x 46454: = 0.337 Ibs. per sec. (For compressed atmospheric air, under like conditions, we would have G = 0.430 Ibs. per second.) Of course the proper choice of the coefficient/" has an im- portant influence on the result. From the above result (G = 0.337 Ibs. per second) we can compute the volume occupied by this quantity of gas in the fi- receiving reservoir, using the relation Q m = . Ym! The heaviness y m > of the gas in the receiving reservoir is most easily found from the relation -^ = -^ .which holds Ym' Yn> good since the flow is isothermal. I.e., &2*'= 46454 ft.; Ym' whence y m > = 0.049 Ibs. per cubic foot, p m > being 16 X 144 Ibs. per sq. ft. Hence FLOW OF GAS IK PIPES. 793 It should be said that the pressure at the up-stream end of the pipe depends upon the rate of flow allowed to take place. With no flow permitted, the pressure in the tube of a gas- well has in some cases reached the high figure of 500 or 600 Ibs. per sq. in. . 560. Rate of Decrease of Pressure along a Long Pipe, Con- sidering further the case of the last paragraph, that of a straight, long, level pipe of uniform diameter, delivering gas from a storage reservoir into a receiving reservoir, we note that if in eq. (4) we retain p m > to indicate the tension in the receiving reservoir, but let p n > denote in turn the tension at points in the pipe successively further and further (a distance x] from the receiving reservoir w', we may write x for I and obtain the equation (between two variables, p n > and x) Pn? Pn/ = Const. X . (6) This can be used to bring out an interesting relation men- tioned by a writer in the Engineering News of July 1887 (p. 71), viz., the fact that in the parts of the pipe more distant from the receiving end, m', the distance along the pipe in which a given loss of pressure occurs is much greater than near the receiving end. To make a numerical illustration, let us suppose that the pipe is of such size, in connection with other circumstances, that the tension p n > at A, a distance x = six miles from m', is two atmospheres, the tension in the receiving reservoir being one atmosphere ; that is, that the loss of tension between A and m! is one atmosphere. If we express tensions in atmos- pheres and distances in miles, we have for the value of the constant in eq. (6), for this case, Const. = (4 1) -4- 6 f ; (for assumed units.} . . (7) Now let p n > = the tension at B, a point 18 miles from m', and we have, from eqs. (6) and (7), the tension at B 3.1 f> atmospheres. Proceeding in this manner, the following set of values is obtained : 794 MECHANICS OF ENGINEERING. Point. Total distance from m'. Distance be- tween consecu- tive points. Tension at point. Loss of ten- sion in each in- terval. F E D C B A m' 126 mi 90 60 36 18 6 les. 36 mi 30 24 "18 12 6 les. 8.00 at 6.78 5.56 4.35 3.16 2.00 1.00 m. 1.22 at 1.22 1.21 1.19 1.16 1.00 m. If the distances and tensions in the second and fourth columns be plotted as abscissae and ordinates of a curve, the latter is a parabola with its axis following the axis of the pipe ; its vertex is not at m', however. 561. Long Pipe of Variable Diameter, Another way of stat- ing the fact mentioned in the last paragraph is as follows : At the up-strearn end of the pipe of uniform diameter the gas is of much greater density than at the other extremity (the heaviness is directly as the tension, the temperature being as- sumed the same throughout the pipe), and the velocity of its motion is smaller than at the discharging end (in the same ratio). It is true that the frictional resistance per unit of length of pipe varies directly as the heaviness [eq. (1), 510], but also true that it varies as the square of the velocity ; so that, for instance, if the pressure at a point A is double that at B in the pipe of constant diameter, it implies that the heaviness and velocity at A are double and half, respectively, those at B, and thus the gas at A is subjected to only half the frictional resisting force per foot of length as compared with that at B. Hence the relatively small diminution, per unit of length, in the tension at the up-stream end in the example of the last paragraph. In the pipe of uniform diameter, as we have seen, the greater part of the length is subjected to a comparatively high ten- sion, and is thus under a greater liability to loss by leakage than if the decrease of tension were more uniform. The total "hoop-tension" ( 426) in a unit length of pipe, also, is proportional to the gas tension, and thinner walls might be employed for the down-stream portions of the pipe if the gas FLOW OF GAS IN PIPES. 795 tension in those portions could be made smaller than as shown in the preceding example. To secure a more rapid fall of pressure at the up-stream end of the pipe, and at the same time provide for the same delivery of gas as with a pipe of uniform diameter throughout, a pipe of variable diameter may be employed, that diameter being considerably smaller at the inlet than that of the uniform pipe but progressively enlarging down-stream. This will require the diameters of portions near the discharging end to be larger than in the uniform pipe, and if the same thickness of metal were necessary throughout, there would be no saving of metal, but rather the reverse, as will be seen ; but the diminished thickness made practicable in those parts from a less total hoop tension than in the corresponding parts of the uniform pipe more than compensates for the extra metal due to increased circumference, aside from the diminished liability to leakage, which is of equal importance. A simple numerical example will illustrate the foregoing. The pipe being circular, we may replace F by ^nd* in equation {4), and finally derive, G being given, i 7 li - (8) Let A be the head reservoir, and m f the receiving reservoir, and B a point half-way between. At A the tension is 10 at- mospheres ; at m', 2 atmospheres. For transmitting a given weight of gas per unit-time, through a pipe of constant diam- ter throughout, that diameter must be (tensions in atmospheres ; 2/ being the length), by eq. (8), d = 67! * = - 0208 * = - 46 - 8 If we substitute for the pipe mentioned, another having a con- stant diameter d l from A to B, where we wish the tension to be 5 atmospheres, and a different constant diameter d % from B to m', we derive similarly 100 1- 796 MECHANICS OF ENGINEERING. and It is now to be noted that the sum of d 1 and d^ is slightly greater than the double of d ; so that if the same thickness of metal were used in both designs the 'compound pipe would require a little more material than the uniform pipe; but, from the reasoning given at the beginning of this paragraph, that thickness may be made considerably less in the down- stream part of the compound pipe, and thus economy secured. [In case of a cessation of the flow, the gas tension in the whole pipe might rise to an equality with that of the head- reservoir were it not for the insertion, at intervals, of auto- matic regulators, each of which prevents the decrease of ten- sion on its down-stream side below a fixed value. To provide for changes of length due to rise and fall of temperature, the pipe is laid with slight undulations.] It is a noteworthy theoretical deduction that a given pipe of variable diameter connecting two reservoirs of gas at specified pressures will deliver the same weight of gas as before, if turned end for end. This follows from equation (3)', 559. With d variable, (3)' becomes (with F \ittf) S* , ***' , -. . (9) ''**-?' (C" is a constant.) f* m ' ds But J n , -'- is evidently the same in value if the pipe be turned end for end. In commenting on this circumstance, we should remember (see 559) that the loss of pressure along the pipe is ascribed entirely to frictional resistance, and in no de- gree to changes of velocity (inertia). On p. 73 of the Engineering News of July 1887 are given the following dimensions of a compound pipe in actual use, and delivering natural gas. The pressure in the head-reservoir is 319 Ibs. per sq. in.; that in the receiving reservoir, 65. For 2.84 miles from the head-reservoir the diameter of the pipe is NATURAL GAS. COEFFICIENT OF FLUID FRICTION. 797 8 in. ; throughout the next 2.75 miles, 10 in.*, while in the remaining 3.84 miles the diameter is 12 in. At the two points of junction the pressures are stated to be 185 and 132 Ibs. per sq. in., respectively, during the flow of gas under the conditions mentioned. 561a. Values of the Coefficient of Fluid Friction for Natural Gas. In the Ohio Keport on Economic Geology for 1888 may be found an article by Prof. S. W. Robinson of the University of that State describing a series of interesting experiments made by him on the flow of natural gas from orifices and through pipes. By the insertion of Pitot tubes approximate measurements were made of the velocity of the stream of gas / C5 in a pipe. The following are some of the results of these ex- periments, p 1 p^ representing the loss of pressure (in Ibs. per sq. inch) per mile of pipe-length, and /the coefficient of fluid friction, in experiments with a six-inch pipe : Pi -Pi 1.00 1.50 2.25 2.50 5.75 6.25 f 0.0025 0.0037 0.0052 0.0059 0.0070 0.0060 In the flow under observation Prof. Robinson concluded that f could be taken as approximately proportional to the fourth root of the cube of the velocity of flow ; though calling attention to the fact that very reliable results could hardly be expected under the circumstances. CHAPTEK IX. IMPULSE AND RESISTANCE OF FLUIDS. 562. The so-called "Reaction" of a Jet of Water flowing from a Vessel. In Fig. 624, if a frictionless but water-tight plug B be inserted in an orifice in the vertical side of a vessel mounted on wheels, the resultant action of the water on the rigid vessel (as a whole) consists of its weight 6r, and a force P' Fhy (in which F= the area of orifice) which is the excess of the horizontal hydro- static pressures on the vessel wall toward the right ( || to paper) over those toward the left, since the pressure P, = Fhy, exerted on the plug is felt by the post (7, and not by the vessel. Hence the post D receives a pressure P' = Fhy. (1) Let the plug B be removed. A steady flow is then set up through the orifice, and now the pressure against the post D is 2Fhy (as will be proved in the next paragraph) ; for not only is the pressure Fhy lacking on the left, because of the orifice, but the sura of all the horizontal components ( || to paper) of. the pressures of the liquid filaments against the vessel wall around the orifice is less than its value before the flow began, by an amount Fhy. A resistance R = %Fhy being provided, and the post removed, a slow uniform motion may be main- tained toward the right, the working force being %Fhy = P" 798 KE ACTION" OF A JET. FIG. 625. (see Fig. 625 ; R is not shown). If an insufficient resistance be furnished before removing the post D, .... .... ^ the vessel will begin to move toward the right with an acceleration, which will disturb the surface of the water and change the value of the horizontal force. This force is called the " reaction'* of the water-jet ; y is the heaviness of the liquid ( 7). Of course, as the flow goes on, the water level sinks and the ' reaction" diminishes accordingly. Looked upon as a motor, the vessel may be considered to be a piston-less and valve-less water-pressure engine, carrying its own reservoir with it. In Case II of 500 we have already had a treatment of the " Reaction-wheel " or " Barker's mill," which is a practical machine operating on this principle, an.d will be again con- sidered in "Notes on Hydraulic Motors." 563. " Reaction" of a Liquid Jet on the Vessel from which it Issues. Instead of showing that the pressures on the vessel close to the orifice are less than they were when there was no flow by an amount Fhy (a rather lengthy demonstration), another method will be given, of greater simplicity but some- what fanciful. If a man standing on the rear platform of a car is to take up in succession, from a basket on the car, a number of balls of equal mass M, and project each one in turn horizontally backward with an acceleration p, he can accomplish this only by exerting against each ball a pressure = Mp, and in the opposite direction against the car an equal pressure = -Mp. If this action is kept up continuously the car is subjected to a constant and continuous forward force of P" Mp. Similarly, the backward projection of the jet of water in the case of the vessel at rest must occasion a forward force against the vessel of a value dependent on the fact that in each small interval of time At a small mass AM of liquid has its velocity changed from zero to a backward velocity of v = V%gh ; that 800 MECHANICS OF ENGINEERING. is, has been projected with a mean acceleration of p = so that the forward force against the vessel is P" = mass X ace. = -- ..... (3) At If Q = the volume of water discharged per unit time, then Ov AM. = - At, and since also Q Fv FV%gh, eq. (3) be- / &/i^ comes "Reaction" of jet = P" %Fhy. . . . (4) (A similar proof, resulting in the same value for P" , is easily made if the vessel has a uniform motion with water sur- face horizontal.) If the orifice is in " thin plate," we understand by F the area of the contracted section. Practically, we have v = V%g"h> ( 495), and hence (3) reduces to P" = % = .94 for the same orifice, so that by eq. (4) we should have P" = ^(MJFhy = \MFhy. With an orifice in thin plate Mr. Ewart found P" "L.^Fhy. As for a result from eq. (4), we must put, for F\ the area of the contracted section .647^ ( 495), which, with = .96, gives P" = 2(.96) 2 . Fhy = LlSFhy. ... . (6) Evidently both results agree well with experiment. Experiments made by Prof. J. B. Webb at the Stevens Institute (see Journal of the Franklin Inst., Jan. '88, p. 35) also confirm the foregoing results. In these experiments the vessel was suspended on springs and the jet directly down- ward, so that the "reaction" consisted of a diminution of the tension of the springs during the flow. 564. Impulse of a Jet of Water on a Fixed Curved Vane (with Borders). The jet passes tangentially upon the vane. Fig. IMPULSE OF JET. 801 626. B is the stationary nozzle from which a jet of water of cross-section F (area) and velocity = c impinges tangentially upon the vane, which has plane borders, parallel to paper, to prevent the lat- eral escape of the jet. The curve of the vane is not circular necessarily. The vane being smooth, the velocity of the water in its curved path remains x < FIG. 626. c at ail points a^ong the curve. Conceive the curve divided into a great number of small lengths dP each = ds, and subtending some angle = d

and makes such an angle <*', Fig. 627, with the direction that , Y sin a tan &' = ---= - . X 1 cos a FIG. 62 FIG. 628. For example, if OL = 90, then a? = 45; while if a = 180, Fig. 628, we have a' = ; i.e., P" is parallel to the jet and its value is IMPULSE OF JET ON VANE. 803 565. Impulse of a Jet on a Fixed Solid of Revolution whose Axis is Parallel to the Jet. If the curved vane, with borders, of the pre- ceding paragraph be replaced by a solid of revolution, Fig. 629, with its axis in line of the jet, the resultant pressure of the jet upon it will simply be the sum of the X-components (i.e., = to BA) of the pressures on all ele- ments of the surface at a given instant ; i.e., FIG. 629. (5) while the components 1 to X, all directed toward the axis of the solid, neutralize each other. For a fixed plate, then, Fig. 630, at right angles to the jet, we have for the force, or " im- pulse" (with a = 90), The experiments of Bidone, made in 1838, confirm the truth of eq. (6) quite closely, as do also those of two students of FIG. 630. - the University of Pennsylvania at Phila- delphia (see Jour, of the Frank. Inst. for Oct. '87, p. 258). Eq.(6) is applicable to the theo- ry of Pitot's Tube (see 539), Fig. 631, if we consider the edge of the tube plane and quite wide. '.'.:'.' :-'' : ' : .'-.'-. : -:': V.B'- The water in the tube is at rest, and its section at A (of area F) may be treated as a flat vertical plate receiving not only the hydrostatic pressure Fxy, due to the depth x below the sur- - FIG. 63i. face, but a continuous impulse P" Fc*y -f- g [see eq. (6)]. 804 MECHANICS OF ENGINEERING. For the equilibrium of the end A, of the stationary column AD, we most have, therefore, Fxy + ' n i.e, A' = (2.0). .(7) The relation in equation (7) corresponds reasonably well with the results of Weisbaeh's experiments with the instru- ment mentioned in 539. Pitot himself, on trial of an in- strument in which the edges of the tube at A were made flar- ing or conically divergent, like a funnel, found (7)' while Darcy, desirous that the end of the tube should occasion as little disturbance as possible in the surrounding stream, made the extremity small and conically convergent. The latter obtained the relation h' = almost exactly (1.0) . . (7)" i/ (See 539.) If the solid of revolution is made cup- shaped, as in Fig. 632, we have (as in Fig. 628) a 180, and therefore, from eq. (5), ' FIG. 632. EXAMPLE. Fig. 632. If c = 30 ft. per sec. and the jet (cylindrical) has a diameter of 1 inch, the liquid being water, so that y = 62.5 Ibs. per cub. ft., we have [ft., lb., sec.] the impulse (force) = P" = 32.2 = 19.05 Ibs. Experiment would probably show a smaller result. IMPULSE OF JET ON MOVING VANE. 803 o - FIG. 633. 566. Impulse of a Liquid Jet upon a Moving Vane having Lateral Borders and Moving in the Direction of the Jet. Fig. 633. The vane has a motion of translation ( 108) in the same direction as the jet. Call this the axis X. It is moving with a velocity v away from the jet (or, if toward the jet, v is negative). "We con- sider v constant, its ac- celeration being prevented by a proper resistance (such as a weight = G) to balance the ^-com- ponents of the arc-pres- sures. Before coming in contact with the vane, which it does tangentially (to avoid sudden devia- tion), the absolute velocity ( 83) of the water in the jet = or < \c this efficiency will not be attained. 567. The California " Hurdy-gurdy ;" or Pelton Wheel, The efficiency of unity in the series of cups just mentioned is in practice reduced to 80 or 85 per cent from friction and lateral escape of water. The Pelton wheel or Cali- fornia " Hurdy-gur- dy," shown (in prin- ciple only) in Fig. 636, is designed to utilize the mechanical rela- tion just presented, and yields results con- TQis^Ms firming the above the- ' ^ ory, viz., that with the linear velocity of the FIG. 636. s* cup-centres regulated to equal , and with a = 180, the effi- 2 ciency approaches unity or 100 per cent. Each cup has a pro- jecting sharp edge or rib along the middle, to split the jet ; see Fig. 635. This wheel was invented to utilize small jets of very great velocities (c) in regions just deserted by " hydraulic mining" operators. Although c is great, still, by giving a large value s* to r, the radius of the wheel, the making of v = - does not 2 necessitate an inconveniently great speed of rotation (i.e., revolutions per unit of time). The plane of the wheel may be in any convenient position. In the London Engineer of May '84, p. 397, is given an ac- count of a test made of a " Hurdy-gurdy," in which the motor 810 MECHANICS OF ENGINEERING. showed an efficiency of 87 per cent. The diameter of the wheel was only 6 ft., that of the jet 1.89 in., and the head of the supply reservoir 386 ft., the water being transmitted through a pipe of 22 inches diameter and 6900 ft. in length. 107 H. P. was developed by the wheel. EXAMPLE.- If the jet in Fig. 636 has a velocity c = 60 ft. per second, and is delivered through a 2-inch nozzle, the total power due to the kinetic energy of the water is (ft., lb., sec.) t/ and if, by making the velocity of the cups = - = 30 ft. per sec., 85 per cent of this power can be utilized, the DO^ver of the wheel at this most advantageous velocity is L = .85 X 4566.9 = 3881 ft. Ibs. per sec. = 7.05 horse-power [since 3881 -r- 550 = 7.05] ( 132). For a cup-velocity of 30 ft. per sec., if we make the radius, ?, = 10 feet, the angular velocity of the wheel will be GO = v -z- r = 3.0 radians per sec. (for radian see Example in 428 ; for angular velocity, 110), which nearly = ?r, thus implying nearly a half-revolu- tion per sec. 568. Oblique Impact of a Jet on a Moving Plate having no Border. The plate has a motion of trans- lation with a uniform D veloc. v in a direc- tion parallel to jet, whose velocity is = c. At the filaments of liquid are deviated, so that in leaving the plate their particles are all found in the moving plane BB' of the plate surface, but the respective absolute velocities of these particles FIG. 637. OBLIQUE JET AND PLATE. 811 depend on the location of the point of the plate where they leave it, being found by forming a diagonal on the relative veloc. c v and the velocity v of the plate. For example, at B the absolute velocity of a liquid particle is w = BE = Vv* + (o v)* + 2v(c v) cos a, while at B' it is w' = B'E' 1/V + (c v)'* Zv(c v) cos a ; but evidently the component 1 to plate (the other component being parallel) of the absolute velocities of all particles leaving the plate, is the same and = v sin a. The skin-friction of the liquid on the plate being neglected, the resultant impulse of the jet against the plate must be normal to its surface, and its amount, jP, is most readily found as follows : Denoting by AM the mass of the liquid passing over the plate in a short time At, resolve the absolute velocities of all the liquid particles, before and after deviation, into com- ponents 1 to the plate (call this direction Y) and || to the plate. Before meeting the plate the particles composing AM have a velocity in the direction of Y of c v = c sin a ; on leav- ing the plate a velocity in direction of Y of v sin a : they have therefore lost an amount of velocity in direction of Y = (c v) sin a in time At ; i.e., they have suffered an average retardation (or negative acceleration) in a Indirection of ( neg. accelera- ) _ (c v) sin a ,_ *V? : jtiou || to Y f : ~~ZT Hence the resistance in direction of Y(i.e., the equal and op- posite of P in figure) must be P Y = mass X I^-accel. = (c v) sin a ; . . (2) At and therefore, since = M = - = mass of liquid passing At g 812 MECHANICS OF ENGINEEKING. over the plate per unit of time (not that issuing from nozzle), we have Impulse orpres- ) =p= Qy (c _ v )^ a = ^r (c _ v y sin (3) sure on plate j g ^ g ^ in which F = sectional area of jet before meeting plate. [N.B. Since eq. (3) can also be written P =. Me sin a Mv sin or, and Me sin a may be called the I^-momentum before contact, while Mv sin a is the JT-momentum after contact (of the mass passing over plate per unit of time), this method may be said to be founded on the principle of momentum which is nothing more than the relation that the accelerating force in any direction = mass X acceleration in that direction ; e.g.,, P x = Mp x ; P y = Mp y ', see74.] If we resolve P, Fig. 637, into two components, one, P', \\ to the direction of motion (v and _ ^p ^ \ v normal ) ,^ to moving plate } ~ ~ ^ ? ' %g ' ' ( to plate j * * " W And similarly for the impulse of an indefinite stream of fluid against a fixed plate ( ~| to velocity of stream), v being the velocity of the current, Impulse of current } __ p r'// 7 ^ j ^ norma l ) /o\ upon fixed plate ) ~ ^ ^ 2^~ ' ' ' ( to plate f * V* * The 2g is introduced simply for convenience ; since, having v given, we may easily find v* -=- %g from a table of velocity- heads ; and also (a ground of greater importance) since the co- efficients and C' which depend on experiment are evidently abstract numbers in the present form of these equations (for It and P are forces, and Fyv* -r- %g is the weight (force) of an ideal prism of fluid ; hence C and C,' must be abstract numbers.) From Col. Beaufoy's experiments (see above), we have for sea-water [ft., lb., sec.], putting R = 112 Ibs., F = 1 sq. ft.,, y = 64 Ibs. per cub. ft., and v = 10 ft. per second, 2 X 32.2 X 112 = 3 " 1.0 X 64 X 10 2 Hence in eq. (1) for sea-water, we may put C = 1.13 (with y = 64 Ibs. per cub. ft.). From the experiments of Dubuat and Thibault, "Weisbach computes that for the plate of Fig. 639, moving through either water or air, C = 1-25 for eq. (1), in which the y for air must be computed from 437 ; while for the impulse of water or air on fixed plates he obtains C' 1.86 for use in eq. (2). It is hardly reasonable to suppose that C and C' should not be identical in value, and Prof. Unwin thinks that the difference in the numbers just given must be due to errors of experi- ment. The latter value, C' = 1-86, agrees well with equation (6) below. For great velocities C and C' are greater for air than for water, since air, being compressible, is of greater heaviness in front of the plate than would be computed for 816 MECHANICS OF ENGINEERING. the given temperature and barometric height for use in eqs. (1) and (2) The experiments of Borda in 1763 led to the formula P= [0.0031 + 0.00035tf]#y a .... (3) for the total pressure upon a plate moving through the air in a direction l to its own surface. P is the pressure in pounds, c the length of the contour of the plate in feet, and S its surface in square, feet, while v is the velocity in miles per hour. Adopting the same form of formula, Hagen found, from experiments in 18T3, the relation P= [0.002894 -t-0.00014]xSV ... (4) for the same case of fluid resistance. Hagen's experiments were conducted with great care, but like Borda's were made with a "whirling machine," in which the plate was caused to revolve in a horizontal circle of only 7 or 8 feet radius at the end of a. horizontal bar rotating about a vertical axis. Hagen's plates ranged from 4 to 40 sq. in. in area, and the velocities from 1 to 4 miles per hour. The last result was quite closely confirmed by Mr. H. Allen Hazen at Washington in November 1886, the experiments being made with a whirling machine and plates of from 16 to 576 sq. in. area. (See the American Journal of Science, Oct. 1887, p. 245.) In Thibault's experiments plates of areas 1.16 and 1.531 sq. ft. were exposed to direct wind-pressure, giving the formula P = 0.00475#y 3 (5) Recent experiments in France (see R. R. and Eng. Journal, Feb. J 87), where flat boards were hung from the side of a rail- way train run at different velocities, gave the formula P = 0.00535/V (6) The highest velocity was 44 miles per hour. The magnitude of the area did not seemingly affect the relation given. More PLATES IN FLUIDS. 817 extended and elaborate experiments are needed in this field, those involving a motion of translation being considered the better, rather than with whirling machines, in which "centrif- ugal action" must have a disturbing influence. The notation and units for eqs. (4), (5), and (6) are the same as those given for (3). It may be of interest to note that if equation (3) of 568 be considered applicable to this case of the pressure of. an un- limited stream of fluid against a plate placed at right-angles to the current, with T^put equal to the area of the plate, we ob- tain, after reduction to the units prescribed above for the pre- ceding equations and putting a = 90, P = 0.0053$y 2 ....... (T) The value y = 0.0807 Ibs. per cub. ft. has been used in the substitution, corresponding to a temperature of freezing and a barometric height of 30 inches. At higher temperatures, of course, y would be less, unless with very high barometer. 569a. Example. Supposing each blade of the paddle-wheel of a steamer to have an area of 6 sq. ft., and that when in the lowest position its velocity [relatively to the water, not to the vessel] is 5 ft. per second ; what resistance is it overcoming in salt water ? From eq. (1) of 569, with C = 1.13 and y = 64 Ibs. per cubic foot, we have (ft., lb., sec.) If on the average there may be considered to be three pad- dles always overcoming this resistance on each side of the boat, then the work lost (work of " slip"} in overcoming these resistances per second (i.e., power lost) is Z, = [6 X 169.4] Ibs. X 5 ft. per sec. = 5082 ft.-lbs. per sec. or 9.24 Horse Power (since 5082 -5- 550 = 9.24). 818 MECHANICS OF ENGINEERING. If, further, the velocity of the boat is uniform and = 20 ft per sec., the resistance of the water to the progress of the boat at this speed being 6 X 169.4, i.e. 1016.4 Ibs., the power ex- pended in actual propulsion is Z 2 = 1016.4 X 20 = 20328 ft.-lbs. per sec. Hence the power expended in both ways (usefully in propul- sion, uselessly in "slip") is t = 25410 ft-lbs. per sec. = 46.2 H. P. Of this, 9.24 H. P., or about 20 per cent, is lost in " slip." 570. Wind-pressure on the surface of a roof inclined at an angle = a with the horizontal, i.e., with the direction of the wind, is usually esti- mated according to the empirical formula FIG. 640. (Button's) pp' [sin a] [1-84 cos a - 1] ? (1) in which p' = pressure of wind per unit area against a vertical surface ( ~| to wind), and p that against the inclined plane (and normal to it) at the same velocity. For a value of p' = 40 Ibs. per square foot (as a maximum), we have the following values for JP, computed from (1) : For a = 5 10 15 20 25 30 35 40 45 50 55 60 p=(lbs. sq. ft.) 5. 2 9.6 14 18.3 22.5 26.5 30.1 33.4 36.1 38.1 39.6 40. Duchemin's formula for the normal pressure per unit-area is , 2 sin 2 a p=p ? . 2 , (2) WIND AND SAIL. 819 with the same notation as above. Some experimenters in London tested this latter formula by measuring the pressure on a metal plate supported in front of the blast-pipe of a blow- ing engine ; the results were as follows : a = 15 20 60 90 p by experiment = (in Ibs. per sq. ft.) 1.65 1.60 2.05 2.02 3.01 3.27 3.31 3.31 By Duchemin's formula p = The scale of the Smithsonian Institution at Washington for the estimation and description of the velocity and pressure of the wind is as follows : Grade. Velocity iu miles per hour. Pressure per sq. foot in Ibs. Name. 0.00 Calm. 1 2 2 4 0.02 0.08 Very light breeze. Gentle breeze. 3 12 0.75 Fresh wind. 4 25 3.00 Strong wind. 5 35 6 High wind. 6 45 10 Gale. 7 60 18 Strong gale. 8 73 Violent gale. 9 90 Hurricane. 10 100 Most violent hurricane. 571. Mechanics of the Sail-boat. The action of the wind on a sail will be understood from the following. Let Fig. 641 represent the boat in horizontal projection and OS the sail, O being the mast. For , simplicity we consider the sail to be a and to remain At this instant the boat is moving in the direc- tion MB of its fore-and- aft line with a velocity = c, the wind having a velocity of the direction and magni- tude represented by w (purposely taken at an angle < 90 with the direction of motion of the boat). We are now to inquire the nature of the action of the wind on the boat, and whether vertical. ..i \ p ot FIG. 641. 820 MECHANICS OF ENGINEERING. in the present petition its tendency is to accelerate, or retard, the motion of the boat. If we form a parallelogram of which w is the diagonal and c one side, then the other -side OK, mak- ing some angle a with BM, will be the velocity v of the wind relatively to the boat (and sail), and upon this (and not upon w) depends the action on the sail. The sail, being so placed that the angle is smaller than a, will experience pressure from the wind ; that is, from the impact of the particles of air which strike the surface and glance along it. This pressure, P, is normal to the sail (considered smooth), and evidently, for the position of the parts in the figure, the component of P along MB points in the same direction as w, with 6 < a and > 0). 572. Resistance of Still Water to Moving Bodies, Completely Immersed. This resistance depends on the shape, position, and velocity of the moving body, and also upon the roughness of its surface. If it is pointed at both ends (Fig. 642) with its MOVING BODIES, IMMERSED. 821 FIG. 642. axis parallel to the velocity, v, of its uniform motion, the stream lines on closing tor gether smoothly at the hinder extremity, or stern, B^ exert normal pressures against the surface of the portion CD...B whose longitudinal compo- nents approximately balance the corresponding components of the normal pressures on CD . . . A ; so that the resistance R, which must be overcome to maintain the uniform velocity ^, is mainly due to the " skin-friction" alone, distributed along the external surface of the body ; the resultant of these resist- ances is a force R acting in the line AB of symmetry (sup- posing the body symmetrical about the direction of motion). If, however, Fig. 643, the stern, E..B..F\* too bluff, . eddies are formed round the corners ^and Fj and the pressure on the surface E . . . F is much less than in Fig. 642; i.e., the water pres- sure from behind is less than the backward (longitudinal) pressures from in front, and thus the resultant resistance R is due partly to skin- friction and partly to " eddy-making" ( 569). [NOTE. The diminished pressure on EF\& analogous to the loss of pressure of water (flowing in a pipe) after passing a nar- row section the enlargement from which to the original section is sudden. E.g., Fig. 644, supposing the velocity v and pressure p (per unit-area) to be the same respectively at A and A\ in the two pipes shown, with diameter AL = WK=A f L f = WK' ; then the pressure at M is FrG . 6 44. equal to that at A (disregarding skin-friction), whereas that at FIG. 643. 822 MECHANICS OF ENGINEERING. ' M ' is considerably less than that at A f on account of the head lost in the sudden enlargement. . (See also Fig. 575.)] It is therefore evident that Ihiffness of stern increases the resistance much more than Huff ness of bow. In any case experiment shows that for a given body sym- metrical about an axis and moving through a fluid (not only water, but any fluid) in the direction of its axis with a uni- form velocity = v, we may write approximately the resistance = (resistance at vel. v) = As in preceding paragraphs, F = area of the greatest section, "I to axis, of the external surface of body (not of the sub- stance) ; i.e., the sectional area of the circumscribing cylinder (cylinder in the most general sense) with elements parallel to the axis of the body, y ==. the heaviness ( 409) of the fluid, and v = velocity of motion ; while is an abstract number dependent on experiment. According to Weisbach, who cites different experimenters, we can put for spheres, moving in water, = about 0.55 ; for cannon-balls moving in water, = .467. According to Robins and Hutton, for spheres in air, we have For in mets. ) .. per sec. j" 5 25 100 200 300 400 rnn j metres U 1 per sec. C=.59 .63 .67 .71 .77 .88 .99 1.04 For musket-balls in the air, Piobert found = 0.451 (1 4- 0.0023 x veloc. in metres per sec.). From Dubuat's experiments, for the resistance of water to a right prism moving endwise and of length = Z, For (I: V'f)= 1 2 3 5 = 1.25 1.26 1.31 1.33 For a circular cylinder moving perpendicularly to its axis Borda claimed that is one-half as much as for the circura- MOVING BODIES, IMMERSED. 823 scribing right parallelepiped moving with four faces parallel to direction of motion. EXAMPLE. The resistance of the air at a temperature of freezing and tension of one atmosphere to a musket-ball \ inch in diameter when moving with a velocity of 328 ft. per sec. is thus determined by Piobert's formula, above : C = 0.451(1 + .0023 X 100) = 0.554 ; hence, from eq. (1), R = 0.554 X X - 0807 X = al018 lbs ' 572a. Deviation of a Spinning Ball from a Vertical Plane in Still Air. It is a well-known fact in base-ball playing that if a rapid spinning motion is given to the ball about a vertical axis as well as a forward motion of translation, its path will not remain in its initial vertical plane, but curve out of that plane toward the side on which the absolute velocity of an external point of the ball's surface is least. Thus, if the ball is thrown from North to South, with a spin of such character as to ap- pear " clock-wise" seen from above, the ball will curve toward the West, out of the vertical plane in which it started. This could not occur if the surface of the ball were perfectly smooth (there being also no adhesion between that surface and the air particles), and is due to the fact that the cushion of com- pressed air which the ball piles up in front during its progress, and which would occupy a symmetrical position with respect to the direction of motion of the centre of the ball if there were no motion of rotation of the kind indicated, is now piled up somewhat on the East of the centre (in example above), creating constantly more obstruction on that side than on the right ; the cause of this is that the absolute velocity of the sur- face-points, at the same level as the centre of ball, is greatest, and the friction greatest, at the instant when they are passing through their extreme Easterly positions; since then that velocity is the sum of the linear velocity of translation and that of rotation ; whereas, in the position diametrically oppo- 824 MECHANICS OF ENGINEERING. site, on the West side, the absolute velocity is the difference ; hence the greater accumulation of compressed air on the left (in the case above imagined, ball thrown from North to South, etc.). 573. Robinson's Cup-anemometer. This instrument, named after Dr. T. R. Robinson of Armagh, Ireland, consists of four hemispherical cups set at equal intervals in a circle, all facing in the same direction round the circle, and so mounted on a light but rigid framework as to be capable of rotating with but little friction about a vertical axis. When in a current of air (or other fluid) the apparatus begins to rotate with an ac- celerated velocity on account of the pressure against the open mouth of a cup on one side being greater than the resistance met by the back of the cup diametrically opposite. Yery soon,, however, the motion becomes practically uniform, the cnp- centre having a constant linear velocity v" the ratio of which to the velocity, v', of the wind at the same instant must be found in some way, in order to deduce the value of the latter from the observed amount of the former in the practical use of the instrument. After sixteen experiments made by Dr. Robinson on stationary cups exposed to winds of varying in- tensities, from a gentle breeze to a hard gale, the conclusion was reached by him that with a given wind- velocity the total pressure on a cup with concave surface presented to the wind was very nearly four times as great as that exerted when the convex side was presented, whatever the velocity (see vol. xxu of Transac. Irish Royal Acad., Pa?^t /, p. 163). Assuming this ratio to be exactly 4.0 and neglecting axle- friction, we have the data for obtaining an approximate value of m, the ratio of v f to the observed v", when the instrument is in use. The influence of the wind on those cups the planes of whose mouths are for the instant || to its direction will also be neglected. If, then, Fig. 645, we write the impulse on a cup when the hollow is presented to the wind [ 572, eq. (1)] '" ROBINSON'S CUP-ANEMOMETER. and the resistance when the convex side is presented 825 we may also put (3) In (1) and (2) v, and v 9 are relative velocities. ^Regarding only the two cups A and B, whose centres at a definite instant are mov- ing in lines parallel to the direction of the wind, it is evident that the motion of the cups does not become uniform until the rel- > v ative velocity v' v" of the wind and cup A (retreating before the wind) has become so small, and the relative velocity v' 4- v" with which B advances to meet the air- particles has become so great, that the im- pulse of the wind on A equals the resist- ance encountered by B; i.e., these forces, P h and P c , must be equal, having equal lever-arms about the axis. Hence, for uniform rotary motion, i.e. [see eq. (3)], 4 ~ * '= or Solving the quadratic for m, we obtain TW = 3.00 ........ (6) That is, the velocity of the wind is about three times that of the cup-centre. 574. Experiments with Robinson's Cup-anemometer. The ratio 3.00 just obtained is the one in common use in connec- tion with this instrument in America. Experiments by Mr. 826 MECHANICS OF ENGINEERING. Hazen at Washington in 1886 (Am. Jour. Science, Oct. '87, p. 248) were made on a special type devised by Lieut. Gibbon. The anemometer was mounted on a whirling machine at the end of a 16-ft. horizontal arm, and values for m obtained, with velocities up to 12 miles per hour, from 2.84 to 3.06 ; average 2.94. The cups were 4 in. in diameter and the distance of their centres from the axis 6.72 in., these dimensions being those usually adopted in America. This instrument was nearly new and was well lubricated. Dr. Robinson himself made an extensive series of experi- ments, with instruments of various sizes, of which an account may be found in the Philos. Transac. for 1878, p. 797 (see also the volume for 1880, p. 1055). Cups of 4 in. and also of 9 in. were employed, placed first at 24 and then at 12 in. from the axis. The cup-centres revolved in a (moving) vertical plane perpendicular to the horizontal arm of a whirling- machine ; this arm, however, was only 9 ft. long. A friction- brake was attached to the axis of the instrument for testing the effect of increased friction on the value of in. At high speeds of 30 to 40 miles per hour (i.e., the speed of the centre of the instrument in its horizontal circle, representing an equal speed of wind for an instrument in actual use with axis stationary) the effect of friction was relatively less than at low velocities. That is, at high speeds with considerable friction the value of m was nearly the same as with little friction at low speeds. With the large 9 in. cups at a distance of either 24 or 12 in. from the axis the value of m at 30 miles per hour ranged generally from 2.3 to 2.6, with little or much friction ; while with the minimum friction m rose slowly to about 2.9 as the velocity diminished to 10 miles per hour. At 5 miles per hour with minimum friction m was 3.5 for the 24 in. instru- ment and about 5.0 for the 12 in. The effect of considerable friction at low speeds was to increase m, making it as high as 8 or 10 in some cases. With the 4 in.-cups no value was ob- tained for m less than 3.3. On the whole, Dr. Robinson con- cluded that m is more likely to have a constant value at all velocities the larger the cups, the longer the arms, and the less the friction, of the anemometer. But few straight-line experi- VARIOUS ANEMOMETERS. 827 ments have been made with the cup-anemometer, the most noteworthy being mentioned on p. 308 of the Engineering News for October 188T. The instrument was placed on the front of the locomotive of a train running between- Baltimore and Washington on a calm day. The actual distance is 40 miles between the two cities, while from the indications of the anemometer, assuming m = 3.00, it would have been in one trip 46 miles and in another 47. The velocity of the train was 20 miles per hour in one case and 40 in the other. 575 Other Anemometers. Both Biram's and Castello's ane- mometers consist of a wheel furnished with radiating vanes set obliquely to the axis of the wheel, forming a small "wind- mill," somewhat resembling the current-meter for water shown in Fig. 604 ; having six or eight blades, however. The wheel revolves with but little friction, and is held in the current of air with its axis parallel to the direction of the latter, and very quickly assumes a steady motion of rotation. The number of revolutions in an observed time is read from a dial. The in- struments must be rated by experiment, and are used chiefly in measuring the velocity of the currents of air in the galleries of mines, of draughts of air in flues and ventilating shafts, etc. To quote from vol. v of the Keport of the Geological Sur- vey of Ohio, p. 370: "Approximate measurements (of the velocity of air) are made by miners by flashing gunpowder, and noting with a watch the speed with which the smoke moves along the air-way of the mine. A lighted lamp is sometimes used, the miner moving along the air-gallery, and keeping the light in a perfectly perpendicular position, noting the time required to pass to a given point." Another kind makes use of the principle of Pitot's Tube (p. 751), and consists of a U -tube partially filled with water, one end of the tube being vertical and open, while the other turns horizontally, and is enlarged into a wide funnel, whose mouth receives the impulse of the current of air ; the differ- ence of level of the water in the two parts of the U is a meas- ure of the velocity. 828 MECHANICS OF ENGINEERING. 576. Resistance of Ships. We shall first suppose the ship to be towed at a uniform speed ; i.e., to be without means of self- propulsion (under water). This being the case, it is found that at moderate velocities (under six miles per hour), the ship being of "fair" form (i.e., the hull tapering both at bow and stern, under water) the resistance in still water is almost wholly due to skin-friction, "eddy-making" (see 569) being done away with largely by avoiding a bluff stern. When the velocity is greater than about six miles an hour the resistance is much larger than would be accounted for by skin-friction alone, and is found to be connected with the sur- face-disturbance or waves produced by the motion of the hull in (originally) still water. The recent experiments of Mr. Froude and his son at Torquay, England, with models, in a tank 300 feet long, have led to important rules (see Mr. White's Naval Architecture and "Hydromechanics" in the Ency. Britann.} of so proportioning not only the total length of a ship of given displacement, but the length of the entrance (for- ward tapering part of hull) and length of run (hinder tapering part of hull), as to secure a minimum "wave-making resist- ance" as this source of resistance is called. To quote from Mr. White (p. 460 of his Naval Architecture, London, 1882): "Summing up the foregoing remarks, it appears : " (1) That / rictional resistance, depending upon the area of the immersed surface of a ship, its degree of roughness, its length, and (about) the square of its speed, is not sensibly affected by the forms and proportions of ships ; unless there be some unwonted singularity of form, or want of fairness. For moderate speeds this element of resistance is by far the most important ; for high speeds it also occupies an important position from 50 to 60 per cent of the whole resistance, probably, in a very large number of classes, when the bottoms are clean and a larger percentage when the bottoms become foul. " (2) That eddy-maldng resistance is usually small, except in special cases, and amounts to 8 or 10 per cent of the frictional RESISTANCE OF SHIPS. 829 resistance. A defective form of stern causes largely increased eddy-making. " (3) That wave-making resistance is the element of the total resistance which is most influenced by the forms and pro- portions of ships. Its ratio to the frictional resistance, as well as its absolute magnitude, depend on many circumstances ; the most important being the forms and lengths of the entrance and run, in relation to the intended full speed of the ship. For every ship there is a limit of speed beyond which each small increase in speed is attended by a disproportionate in- crease in resistance ; and this limit is fixed by the lengths of the entrance and run the 4 wave-making features ' of a ship. " The sum of these three elements constitutes the total re- sistance offered by the water to the motion of a ship towed through it, or propelled by sails ; in a steamship there is an ' augment ' of resistance due to the action of the propel- lers." In the case of a steamship driven by a screw propeller, this augment to the resistance varies from 20 to 45 per cent of the " tow-rope resistance," on account of the presence and action of the propeller itself ; since its action relieves the stern of some of the forward hydrostatic pressure of the water closing in around it. Still, if the screw is placed far back of the stern, the augment is very much diminished ; but such a position in- volves risks of various kinds and is rarely adopted. We may compute approximately the resistance of the water to a ship propelled by steam at a uniform velocity -y, in the following manner : Let L denote the power developed in the engine cylinder ; whence, allowing 10 per cent of L for engine friction, and 15 per cent for " work of slip" of the propeller- blade, we have remaining 0.75Z, as expended in overcoming the resistance R through a distance = v each unit of time ; i.e., (approx.) 0.75Z = 7fo (1) EXAMPLE. If 3000 indicated H. P. ( 132) is exerted by the engines of a steamer at a uniform speed of 15 miles per hour 330 MECHANICS OF ENGINEERING. (= 22 ft. per sec.), we have (with above allowances for slip and engine friction) [foot-lb.-sec.] | X 3000 X 550 = E X 22 ; .-. R = 56250 Ibs. Further, since J2 varies (roughly) as the square of the veloc- ity, and can therefore be written It = (Const.) X 'y 2 , we have from (1) L = a constant X v s (2) as a roughly approximate relation between the speed and the power necessary to maintain it uniformly. In view of eq. (3) involving the cube of the velocity as it does, we can understand why a large increase of power is necessary to secure a propor- tionally small increase of speed. 577. " Transporting Power," or Scouring Action, of a Current. The capacity or power of a current of water in an open channel to carry along with it loose particles, sand, gravel, pebbles, etc., lying upon its bed was investigated experimen- tally by Dubuat about a century ago, though on a rather small scale. His results are as follows : The velocity of current must be at least 0.25 ft. per sec., to transport silt ; 0.50 " loam; 1.00 " " sand; 2.00 " " gravel; 3.5 " pebbles 1 in. in diam.; 4.0 " " " broken stone : 5.0 " chalk, soft shale. However, more modern writers call attention to the fact that in some instances beds of sand are left undisturbed by currents of greater velocity than that above indicated for sand, and explain this fact on the theory that the water-particles may not move parallel to the bed, but in cycloids, approxi- mately, like the points in the rim of a rolling wheel, so as to have little or no scouring action on the bed in those cases. In case the particles move in filaments or stream-lines parallel to the axis of the stream the statement is sometimes made that the " transporting power" varies as the sixth power " TRANSPORTING POWER" OF CURRENTS. 831 FIG. 646. of the velocity of the current, by which is meant, more defi- nitely, the following : Fig. 646. Conceive a row of cubes (or other solids geometri- cally similar to each other) of many sizes, all of the same heavi- ness ( 7), and simi- larly situated, to be placed on the horizon- tal bottom of a trough and there exposed to a current of water, *** being completely im- mersed. Suppose the coefficient of friction between the cubes and the trough-bottom to be the same for all. Then, as the current is given greater and greater velocity v, the impulse P m (corresponding to a particular velocity v m ) against some one, m, of the cubes, will be just sufficient to move it, and at some higher velocity v n the impulse P n against some larger cube, n, will be just sufficient to move it, in turn. "We are to prove that P m : P n :: v m 6 : v n \ Since, when a cube barely begins to move, the impulse is equal to the friction on its base, and the frictions under the cubes (when motion is impending) are proportional to their volumes (see above), we have therefore (i) Also, the impulses on the cubes, whatever the velocity, are pro- portional to the face areas and to the squares of the velocities (nearly ; see 572) ; hence Pn ft.X' .... (2) From (1) and (2) we have = ^; (3) 'w "** 832 MECHANICS OF ENGINEERING. while from (3) and (2) we have, finally, P m :P n :: Vjn 6 :^ c (4) Thus we see in a general way why it is that if the velocity of a stream is doubled its transporting power is increased about sixty-four-fold ; i.e., it can now impel along the bottom pebbles that are sixty-four times as heavy as the heaviest which it could move before (of same shape and heaviness). Though rocks are generally from two to three times as heavy as water, their loss of weight under water causes them to encounter less friction on the bottom than if not immersed. INDEX TO HYDRAULICS. Absolute Temperature 606 Absolute Zero 606 Accumulator, Hydraulic 700 Adiabatic Change, 621, 629, 631, 636 Adiabatic Expansion in Com- pressed-air Engine 631 Adiabatic Flow of Gases from an Orifice 778 Air Collecting in Water-pipes, 731, 736 Air, Compressed, Transmission of 786-790 Air-compressor 636 Air-profile 749 Air-pump, Sprengel's 656 Air-thermometer 604 Amplitude of Backwater 772 Anemometer, Robinson's. . 824, 826 Anemometer, Biram's 827 Anemometer, Castello's 827 Angle of Repose 572 Angular Stability of Ships 597 Atkinson Gas-engine 642, 643 Atmosphere as a Unit Pressure, 519 Augment of Resistance of Screw Propeller 829 Backwater 771 Ball, Spinning, Deviation from Vertical Plane 823 Balloon 644 Barker's Mill 672 Barometers 530 Barometric Levelling 619 Baziu, Experiments."" 688 Beaufoy's Experiments 814 Bellinger, Prof., Experiments with Elbows 729 Bends, Loss of Head due to 728 Bends in Open Channels 770 3ent Tube, Liquids in 529 Bernoulli's Theorem and the Conservation of Energy 717 Bernoulli's Theorem for Gases, 773 Bernoulli's Theorem. General Form 706 Bernoulli's Theorem, Steady Flow without Friction. . .652, 654 j Bernoulli's Theorem with Fric- tion 696 Bidone, Experiments on Jets. . . 803 Blowing-engine, Test 776 Borda's Formula 722 Bourdon Steam-gauge 532 Boyle's Law , . . 615 Bramah Press 526 Branching Pipes . . 736 Bray ton's Petroleum-engine . . . 641 Buoyant Effort 586 Buoyant Effort of the Atmos- phere 644 Canal Lock, Time of Filling, 739, 740 Centigrade Scale 605 Centre of Buoyancy 586 Centre of Pressure 546 Change of State of Gas 610 Chezy's Formula 714 Chezy's Formula for Open Chan- nels 758 Church and Fteley, Report on Quaker Bridge Darn, 558, 563, 564 Clearance 627 Closed Air-manometer 614 Coal Consumption 643 Coal, Heat of Combustion 643 Coefficient of Contraction 659 Coefficients of Efflux, 661, 712, 676, 734, 738, 784 Coefficient of Fluid Friction, 707, 797 Coefficient of Resistance 704 Coefficient of Roughness. . .759, 760 Coefficient of Velocity, 661, 689, 704, 712, 723, 734 Collapse of Tubes 538 Communicating Prismatic Ves- sels 739 Complete and Perfect Contrac- tion 676 Component of Fluid Pressure.. 525 Compressed-air Engine 631 Compressed Air, Transmission of. 786 Compressibility of Water 516 Conical Short Tubes 692 Conservation of Energy and INDEX TO HYDRAULICS. PAGE Bernoulli's Theorem 717 Contraction 659 Contraction, Perfect 676 Cooling in Sudden Expansion of Gas 622 Critical Temperature of a Vapor, 608 Croton Aqueduct, Slope 749 Cunningham, Experiments on the River Ganges 759 Cup-anemometer 824, 826 Cups, Impulse of Jet on ... 804, 808 Current, Transporting Power of, 830 Current-meters 750 Curved Dams 562 Cylinders, Thin Hollow, Strength 538 Dams, High Masonry 562 Darcy, Experiments with Pitot's Tube 804 Darcy and Bazin, Experiments with Open Channels 758 Decrease of Tension of Gas along a Pipe 793 Depth of Flotation 592 Deviation of Spinning Ball from Vertical Plane 823 Diaphragm in a Pipe, Loss of Head 724 Displacement of a Ship 596 Divergent Tubes, Flow through, 692 Dividing Surface of Two Fl uids, 528 Diving-bell 617 Double Floats 750 Draught of Ships 596 Dubuat's Experiments, 707, 81 5, 822, 830 Duchemin's Formula for Wind- pressure 818 Duty of Pumping-engines. 644 Dynamics of Gaseous Fluids, 773-797 Earth Pressure 5T2 Earthwork Dam 566 Eddy-making Resistance . . 814, 828 Efficiency 637, 642 Efflux of Gases 773-797 Efflux from Steam-boiler 664 Efflux from Vessel in Motion. . 670 Efflux into Condenser 665 Efflux under Water 669 Egg-shaped Section for Sewers, 765 Elastic Fluids 516 Elbows, Loss of Head due to, 727, 729 Ellis, Book on Fire-streams . . . 716 Emptying Vessels, Time of. . 737-746 Engine, Gas- 641 Engine, Hot-air 639 Engine, Petroleum- 641 Engine, Steam- 624 Enlargement, Sudden, in Pipe, 721 Equal Transmission of Fluid Pressure 524 Equation of Continuity, 648, 737, 756 Equation of Continuity for Gases 773 Equation of Continuity for Open Channels 756 Equilibrium of Flotation 590 Ericsson's Hot-air Engine 640 Ewart's Experiments on Jets . . . ^00 Expanding Steam 625 Fahrenheit Scale 605 Fairbairn's Experiments on Col- lapse of Tubes 538 Fanning, Table of Coefficients of Fluid Friction 709 Feet and Meters, Table ... 677 Fire-engine Hose, Friction in. .. 716 Fire-streams, by Ellis 71 & Floating Staff 750 Flood-gate 553 Flotation 590 Flow in Plane Layers 648, 6o2 Flow in Open Channels 749 Flow of Gas in Pipes 786, 790 Fluid Friction 695, 797, 828 Fluid Friction, Coefficient for Natural Gas 797 Fluid Pressure, Equal Trans- mission of 524 Force-pump 667 Francis' Formula for Overfalls, 687 Free Surface of Liquid at Rest, 528 Free Surface a Paraboloid 544 Fresh Water, Heaviness, Table, 518 Friction-head in Open Channels, 757 Friction-head in Pipes 699 Friction, Fluid 695, 797, 828 Fronde's Experiments on Fluid Friction 696 Fronde's Experiments on Grad- ual Enlargement in Pipes . . . 725 Fronde's Experiments with Piezometers 720 Fteley and Stearns's Experi- ments on Overfalls 687 Fteley and Stearns's Experi- ments with Open Channels. . 758 Gas and Vapor 607 Gas-engines 641 Gaseous Fluids 604-645 Gas, Flow through Short Pipes, 784 Gas, Flow through Orifices. .773-784 Gas, Illuminating 517, 533 Gas, Natural, Flow in Pipes, 786, 79g Gas, Steady Flow 773 Gas, Velocity of Approach 784 Gases, Definition 515 Gauging of Streams 755- INDEX TO HYDRAULICS. XI PAGE Gay-Lussac's Law 609 Gradual Enlargement in Pipe.. 725 Granular Materials 572 Graphic Representation of Change of State of Gas 628 Head of Water 530 Heat-engines, Efficiency 642 Heaviness of Fluids 517 Heaviness of Fresh Water at Different Temperatures 518 Height Due to Velocity 649 Height of the Homogeneous Atmosphere 620 Herschel'sVenturi Water-meter, 726 High Masonry Dams 562 Hoop-tension 537 Hose, Rubber and Leather, Fric- tion in 716 Hot-air Engines 639 Humphreys and Abbot's Survey of the Mississippi River... 754, 759 Hurdy-gurdy, California 809 Button's Formula for Wind- pressure . . 818 Hydraulic Grade-line 715 Hydraulic Mean Depth, 698,757, 764 Hydraulic Press 526 Hydraulic Radius 698 Hydraulic Radius for Minimum Friction 764 Hydraulics, Definition 518 Hydrodynamics 646-832 Hydrodynamics or Hydrokinet- ics. Definition 518 Hydromechanics, Definition... 519 Hydrometers 591 Hydronietric Pendulum 753 Hydrostatic Pressure 522 Hydrostatics, Definition 518 Ice-making Machine 624 Illuminating Gas .517, 533 Immersion of Rigid Bodies. . . . 586 Imperfect Contraction 680, 684 Impulse and Resistance of Fluids 797-832 Impulse of Jet on Vanes 801, 805 Inclin^i Short Tubes, Efliux through 691 Incomplete Contraction 679, 684 Inelastic Fluids 516 Irres'ilar Shape, Emptying of Vesselsof 746 Isothermal Change 615, 629, 639 Isothermal Expansion 624, 635 Isothermal Flow of Gas in Pipes 790 Isothermal Flow of Gases through Orifices 777 Jacket of Hot Water 635 PAGE Jackson's Works on Hydraulics, 761 Jet from Force-pump 667 Jets, Impulse of 800, 803, 810 Jets of Water 660, 662 Joule, Experiment on Flow of Gas 782 Kansas City Water-works; Si- phon 731, 736 Kinetic Energy 672, 718 Kinetic Energy of Jet 672, 808 Kinetic Theory of Gases 516, 606, 622 Kutter's Diagram 761 Kutter's Formula 759 Kutter's Hydraulic Tables 761 Laminated Flow 648, 652 Land-ties 585 Law of Charles 609 Levelling, Barometric 619 Liquefaction of Oxygen 609 Liquid, Definition 515 Long Pipes, Flow of Water through 710-716 Loss of Head 698, 703, 721, etc. Loss of Head Due to Bends 728 Loss of Head Due to Elbows, 727, 729 Loss of Head Due to Throttle- valves 730 Loss of Head Due to Valve- gates 730 Manometers 530 Mariotte's Law 615, 777 Mechanics of the Sail-boat 819 Mendeleieff's Device for Spec- ula 544 Metacentre of a Ship 599 Metres and Feet, Table 677 Mill, Barker's 672 Minimum Frictional Resistance in Open Channel 764, 766 Mississippi River, Hydraulic Survey 754, 759, 770 Mixture of Gases 618 Momentum, Principle of 812 Moving Pistons 524 Moving Vane, Impulse of Jet on 805 Napier on Flow of Steam 781 Natural Gas, Flow in Pipes, 791, 797 Natural Slope, of Earth 573 Non-planar Pistons 526 Notch, Rectangular, Efflux from, 683, 741 Obelisk-shaped Vessel; Time of Emptying 744 Oblique Impact of Jet on Plate, 810 Open Channels, Flow in 749-797 Orifices in Thin Plate 658, 773 Otto Gas-engine 641, 643 Xll INDEX TO HYDRAULICS. PAGE Overfall, Emptying Reservoir through 741 Overfall Weirs. . . .677, 683, 688, 75G Overfall Weirs, Actual Dis- charge 683 Paddle-wheel of a Steamer 817 Paraboloid as Free Surface 544 Paraboloidal Vessel 742 Parallelopipedical Reservoir Walls 555 Pelton Wheel or Hurdy-gurdy, 809 Pendulum, Hydrometric 753 Perfect Fluid, Definition. 515 Permanent Gases 605, 608 Petroleum-engine 641 Petroleum Pumping 708 Piezometer 649, 657, 700 Pipes, Clean 708 Pipes, Foul 708 Pipes, Thickness of, for Strength 538 Pipes, Tuberculated 708 Piston Pressures 523 Pistons, Moving . 524 Pistons, Non-planar 526 Pilot's Tube 751, 804, 827 Plate between Two Levels of Water 568,569 Plates, Impulse of Jets on.. 801, 805, 810 Plates Moving in a Fluid 813 Plates, Resistance in Sea-water. 814 Pneumatics, Definition 518 Poisson's Law 621 Poncelet's Experiments with 830 Overfalls 677 Power Required to Propel Ships 830 Pressure, Centre of 546 Pressure on Bottom of a Vessel, 545 Pressure on Curved Surfaces. . . 569 Pressure on Sluice-gate 551 Pressure per Unit-area 519 Pressure-energy 717 Pressure-head 650 Principle of Momentum 812 Pyramidal Vessel, Emptying of, 743 Quaker Bridge Dam, Proposed, 564 Radian, Definition 544 Reaction of a Water- jet 798 Rectangular Orifices 672, 676 Refrigerator of Hot-air Engine, 640 Regenerator of Hot-air Engine, 640 Relative Equilibrium of a Liquid 540 Reservoir of Irregular Shape, Emptying of 746 Reservoir Walls 554-567 .Resistance, Eddy-making. .814, 828 PA.GE Resistance, Wave-making 828 Resistance of Fluid to Moving Bodies 820 Resistance of Fluid to Moving Plates.. 814 Resistance of Ships 828 Resistance of Still Water to Moving Solids 820 Retaining Walls 572-580 Righting Couple, of Floating Body." * 597 Ritchie-Haskell Direction Cur- rent-meter 750 Rivers, Flow in 749-797 Robinson, Prof., Experiments on Flow of Natural Gas 797 Robinson's Anemometer. .. .824, 826 Rounded Orifice 663 Safety-valves 534 Sail-boat, Mechanics of the 819 Sail-boatjMoving Faster than the Wind 820 Saturated Steam, Heaviness .. . 628 Saturated Vapor 607 Scouring Action of a Current. . 830 Screw Propeller, Augment of. . 829 Sewers, Flow in 761, 765 Ships, Resistance of 828 Ships, Stability of 597, 599 Short Cylindrical Pipes, Efflux through 689, 704 Short Pipes 722, 723 Short Pipe, Minimum Head for Full Discharge 724 Simpson's Rule 603, 747, 748 Siphons 735 Skin-friction 695, 828 "Slip" 829 Slope, in Open Channels 749 Smith, Mr. Hamilton; Hydrau- lics 689 Smithsonian Scale of Wind- pressures 819 Solid of Revolution, Impulse of Jet on 803 Specific Gravity 589 Spren gel's Air-pump 656 St. Gothard Tunnel, Experi- ments in 787 Stability of Rectangular Wall. . 554 Stability of Ships 597 State of Permanency of Flow. . . 647 Steady Flow, Definition 647 Steady Flow, Experimental Phenomena 646 Steady Flow of a Gas 773 Steam, Expanding 624 Steam. Flow of 781 Steam, Saturated, Heaviness of. . 628 INDEX TO HYDRAULICS. X11I Steam-engine, Examples 627 Steam-gauge, Bourdon 532 Stirling's Hot-air Engine 639 Stream-line 658 Sudbury Conduit, Experiments, 758, 762 Sudden Diminution of Section in a Pipe 727 Sudden Enlargement of Section in a Pipe 721 Surface Floats 750 Survey, Hydraulic, of Missis- sippi River 754, 759 Tachometer 750 Temperature, Absolute 606 Temperature, Influence on Flow of Water 703 Tension of Gas 519 Tension of Illuminating Gas. . . 533 Thermodynamics 606 Thermometers 604 Thickness of Pipes 538 Thickness of Pipe for Natural Gas 796 Thin Hollow Cylinders 535 Thin Plate, Orifices in 658 Throttle-valves, Loss of Head Due to 730 Time of Emptying Vessels of Various Forms 737-746 Transmission of Compressed Air 786, 790 Transporting Power of a Cur- rent 830 Trapezoidal Section for Open Channel 765 Trapezoidal Wall, Stability of.. 559 Triangular Orifices 675 Triangular Wall, ^ Stability 561 Uniform Motion in Open Chan- nel 756 Uniform Rotation of Liquid 542 Uniform Translation of Liquid.. 540 Upsetting Couple. . . 599 Vacuum- chamber, in Siphon. . . 736 Valve-gates, Loss of Head due to 730 Vanes, Impulse of Jets on. .801, 805 I PAGE ' Vapors 516, 607 Variable Diameter, Long Pipe of, 794 Variable Motion in Open Chan- nels 768 Velocities in Section of River. . 754 Velocity of Efflux as related to Density 668 Velocity-head 649 Velocity Measurements in Open Channel 750 Vena Contracta 659 Venturi's Tube 693 Venturi Tube, New Forms 694 Venturi Water-meter 725 Volume of Reservoir Found by Observing Time of Emptying, 748, Water, Compressibility of 519* Water-formula for Flow of , Gases 774 !Water, Heaviness, at Different i Temperatures 518 Water in Motion 646 Water-meter, Venturi 725 Water-ram 530, 538 Wave-making Resistance 828 Webb, Prof., Experiments on the Reaction of Jets 800 Wedge of Maximum Thrust . . . 573 Wedge-shaped Vessel, Time of Emptying 742 Weirs, Overfall. . .677, 683, 688, 756, 772- Weisbach's Experiments. . . .682, 685, 686, 691, 707, 721, 804 Weisbach's Experiments with Elbows and Bends 727, 728 Weser, River, Backwater in 772 Wetted Perimeter 697, 749 Wex, von, Hydrodynamik 688 Whirling Machine 816, 826 Wind-pressure 818 Wind -pressure, Smithsonian Scale of 819 Woltmann's Mill 750 Work of Compressed-air En- gine 631 Work of Expanding Steam 624 Work of Jet on a Vane 807 DATE YC I UNIVERSITY OF CALIFORNIA LIBRARY