HYDRAULICS BOOKS BY R. L. DAUGHERTY HYDRAULIC TURBINES Second Edition 190 pages, 6X9, Illustrated $2.00 CENTRIFUGAL PUMPS 192 pages, 6X9, Illustrated $2.00 HYDRAULICS 287 pages, 6X9, Illustrated $2.50 HYDRAULICS BY R. L. ^AUGHERTY, A. B., M. E. PROFESSOR OF HYDRAULIC ENGINEERING, RENS8ELAER POLYTECHNIC INSTITUTE; FORMERLY ASSISTANT PROFESSOR OF HYDRAULICS, 8IBLEY COLLEGE, CORNELL UNIVERSITY; AUTHOR OF "HYDRAULIC TURBINES" AND "CENTRIFUGAL PUMPS" SECOND EDITION REVISED AND ENLARGED McGRAW-HILL BOOK COMPANY, INC, 239 WEST 39TH STREET. NEW YORK LONDON: HILL PUBLISHING CO., LTD. 6 & 8 BOUVERIE ST., B.C. 1919 COPYRIGHT, 1916, 1919, BY THE MCGRAW-HILL. BOOK COMPANY, INC. THE. MAPLE- PRESS. YORK- PA PREFACE TO THE SECOND EDITION In this edition it has been possible to add new material which the author has found necessary in his present courses. This consists principally of graphical methods of solving certain practical problems, the determination of the economic size of pipe, and problems of flow through compound pipes, branching pipes, pipes with laterals, and through rotating channels. The treatment of various topics has also been extended in many sases, and certain other portions have been rewritten where sxperience has indicated that this was desirable for greater sffectiveness. The author has been aided in this revision by the helpful suggestions of many individuals and in particular of Professor F. G. Switzer of Cornell University. R. L. D. TROY, N. Y., May, 1919. PREFACE TO FIRST EDITION This book has been prepared as a text for students who are required to cover a wide field in hydraulics in a limited amount of time. Therefore the treatment has been made as brief and con- cise as is consistent with clearness. Attention has been given mostly to matters which are of fundamental importance and but little space has been devoted to those things which are of small practical value, except where necessary to illustrate basic principles. As a step in saving the student's time a liberal use has been made of diagrams, curves, and half-tones. These not only save words but often give a clearer idea at a glance than can be obtained in any other way. The treatment throughout has been made as consistent as is possible. The solution of all problems involving the flow of water is made to depend upon applications of Bernoulli's theorem, which is the key to a rational treatment of the subject. The student is not told in the very beginning that V = V^gh and then compelled to unlearn it later. Experience in the class room has shown that many students will persistently apply that formula whether it fits the case or not. By deriving it at a later time by an application of Bernoulli's theorem, they will more readily see that it is a very special case and thus realize more fully its limitations. An effort has been made to avoid special cases so far as is possible. The treatment in the text and the equations are for the most part perfectly general. Special cases are given only when necessary to illustrate the application of some general principle, or where a special case makes some proposition clearer, and when the general treatment is too complex. But the at- tention of the reader is called to the fact that the equations there given are not universally applicable. Class-room experience has shown that very few students ob- tain a true physical conception of the subject of hydraulics. To most of them, even some of the best, it is very largely an abstract subject. This is partly due to the fact that, with their limited experience and observation, they have actually seen but few of vii viii PREFACE the things with which the book deals and hence they can form no adequate mental picture of the physical facts. In order to overcome this, so far as possible, a large number of illustrations from photographs have been employed. As a further step in implanting a true physical idea in the mind of the student, a great deal of care has been exercised in the arrangement and pre- sentation of the subject and a constant attempt has been made to connect one part with another. In many cases the problems have been taken from actual practice and have also been ar- ranged so as to be instructive in themselves. In considering turbines and centrifugal pumps the first es- sential is to convey a fair idea as to the general appearance, construction, and arrangement of such machines and possibly some simple features of their operation, since it is useless to plunge directly into a mass of equations which are no more than mathematical gymnastics to most students. The second step should be the presentation of the principles of operation together with a general idea as to actual characteristics. These facts could then be explained by as much theory as one had time to go into. In this text but very little theory has been given and that of the simplest kind, though it is believed that what is given is both general and rational. By the aid of this theory the nature of the characteristics of these machines can be accounted for. After that one is ready to take up certain very useful and prac- tical commercial factors by the aid of which one can classify turbines or pumps, can compare one type with another, and can make an intelligent selection of the best type for certain conditions. The simple theory of hydraulic machinery that has been given here covers about all that is really useful in a text of this scope. The design of turbines and pumps is too empirical, and requires too much judgment and experience backed up by a good supply of test data, to be expressed by a few equations. Any brief treatment of this phase of the subject would be false and mis- leading, hence it has been omitted. For any more extended treatment of these subjects the reader is referred to other publi- cations of the author. The main idea underlying the entire text has been to present fundamental principles. After this ground has once been cov- ered, those who desire to specialize in hydraulics are prepared to PREFACE ix study certain topics more intensively. The devotion of con- siderable space to an account of experiments and test data is unwarranted here, though the student should not lose sight of the fact that the study of such is desirable when important work is undertaken. However, a sufficient amount of information on experimental coefficients and empirical factors has been given so that a correct idea may be formed both as to the range of values arid the considerations that enter into the choice of a suitable value for a given case. Very naturally some very important topics in practical hydrau- lics have been omitted, altogether or treated very briefly and superficially because they did not involve fundamental principles and hence were not within the scope of this text, or else were of such a nature as to belong to advanced treatises. The final apology which the author makes for this work is that it has been prepared primarily to meet the needs of his own classes. The author wishes to acknowledge his indebtedness to the various parties whose names are attached to certain of the illustrations for their kindness in furnishing the same. He is also indebted to E. H. Wood, Professor of Mechanics of Engineering in Sibley College, and to D. R. Francis, Instructor in Sibley College, for valuable assistance in the criticism of the manu- script and the reading of the proof. R. L. D. ITHACA, N. Y., April, 1916. CONTENTS PAGE PREFACE v NOTATION ^ . .,. xv CHAPTER I INTRODUCTION 1 Definition of subject Distinction between a solid and a fluid Distinction between a gas and a liquid Compressibility of water Density of water Accuracy of computations Notation Units Problems. CHAPTER II INTENSITY OF PRESSURE 7 Definition of intensity of pressure Variation of pressure in a liquid Surface of equal pressure Pressure the same in all direc- tions Pressure expressed in height of liquid Barometer Vacuum Absolute and relative or gage pressure Instruments for measuring pressure The hydraulic press Problems. CHAPTER III HYDROSTATIC PRESSURE ON AREAS 19 Total pressure on plane area Depth of the center of pressure Lateral location of center of pressure Graphical solution for pressure on plane area Resultant thrust on plane areas Horizontal pressure on curved surface Vertical pressure on curved surface Component of pressure in any direction Resultant pressure on curved surface Pipes under pressure Buoyant force of the water and flotation Metacenter. CHAPTER IV APPLICATIONS OF HYDROSTATICS 32 The gravity dam The framed dam The arch dam The earth dam Additional notes on dams Flashboards Problems. CHAPTER V HYDROKINETICS 44 Actual and ideal conditions Critical velocity Steady flow Rate of discharge Equation of continuity General equation for steady flow Use of the word "head" Energy and power mean- ing of head Correct and incorrect applications of Bernoulli's theorem Applications of general equation Problems. xi xii CONTENTS CHAPTER VI PAGE APPLICATIONS OF HYDROKINETICS 62 Definition of a jet Jet coefficients Flow through orifice Orifice in case of unequal pressures Submerged orifice Values of orifice coefficients Flow through short tubes Flow through nozzles Efficiency of nozzle Venturi meter Large- vertical orifice Weir The triangular weir The rectangular weir The Francis weir formula The Bazin weir formula Comments on weirs The Cippoletti weir Special weirs The Pitot tube The current meter Comments on Measurement of Water Discharge under varying head Problems. CHAPTER VII FLOW THROUGH PlPES 96 Loss of head in pipe friction Loss of head at entrance Loss of head at discharge Loss of head in nozzle Other minor losses of head Flow through long pipe Hydraulic gradient Effect of air at summit Hydraulic slope Chezy's formula Other formulas for pipe friction Values of friction factors Size of pipe for given discharge Power delivered by a pipe Pipe line with pump Pipe line with turbine Equation of energy with turbine or pump Economic size of pipe Compound pipes Branching pipes Pipe with laterals Construction of pipe lines Problems. CHAPTER VIII UNIFORM FLOW IN OPEN CHANNELS 125 Open channels Uniform flow Hydraulic gradient Equation for uniform flow Kutter's formula for c Manning's formula for c Bazin's formula for c Construction of open channels Non- uniform flow in open channels Stream gaging Rating curve Problems. CHAPTER IX HYDRODYNAMICS 141 Dynamic force exerted by a stream Dynamic force (second method) Dynamic action upon a stationary body Force exerted upon pipe Theory of Pitot tube Water hammer and surges in unsteady flow Relation between absolute and relative velocities Dynamic action upon moving body Im- pulse and reaction of a jet Distinction between an impulse and a reaction turbine Theorem of angular momentum Torque ex- erted upon turbine by water Torque exerted upon water by cen- trifugal pump Power Definitions of heads Definitions of tur- bine efficiencies Definitions of pump efficiencies Centrifugal action or forced vortex Free vortex Flow through rotating channel Problems. CONTENTS xiii CHAPTER X PAGE DESCRIPTION OF THE IMPULSE WHEEL 165 The impulse wheel Buckets Nozzles and governing Condi- tions of service. CHAPTER XI DESCRIPTION OF THE REACTION TURBINE 177 The reaction turbine Runners Gates and governing The draft tube Cases and settings Conditions of service. CHAPTER XII WATER POWER PLANTS 198 Elements of a water power plant High head plants Low head plants. CHAPTER XIII THEORY OF THE IMPULSE WHEEL 209 Action of the water Force exerted by jet Power of wheel Speed Head on impulse wheel Problems. CHAPTER XIV THEORY OF THE REACTION TURBINE 221 Introductory illustration Torque exerted Power Speed Values of c e and e for maximum efficiency Theory of the draft tube Head on reaction turbine Problems. CHAPTER XV TURBINE LAWS AND FACTORS 231 Operation under different heads Different sizes of runner Specific speed Uses of specific speed Factors affecting effi- ciency Problems. CHAPTER XVI THE CENTRIFUGAL PUMP 239 Definition Classification Description of the centrifugal pump Conditions of service Head developed Measurement of head Head imparted by impeller Centrifugal pump factors Specific speed Operation at different speeds Factors affecting efficiency Problems. APPENDIX 262 Table of areas of circles Table of standard pipe sizes Table of numbers to the % power Table of numbers to the % power Fundamental trigonometry. INDEX. 265 NOTATION A = angle between V and u (Fig. 142) a = angle between v and u (Fig. 142) c = coefficient of discharge or coefficient of flow c e = coefficient of contraction c v = coefficient of velocity D = diameter of turbine runner or pump impeller in inches d = diameter of pipe in feet d" = diameter of pipe in inches e = efficiency 6h = hydraulic efficiency e m = mechanical efficiency e v = volumetric efficiency F = area in square feet; in turbines and pumps it is the total area of the streams measured normal to the absolute velocity of the water / = friction factor in pipes = area in square feet in turbines or pumps measured normal to the relative velocity of the water G = any weight in pounds g = acceleration of gravity in feet per second per second H = total effective head in feet, = p + z + V*/2g H' = any loss of head in feet h = head in feet / = moment of inertia k = any coefficient of loss I = any length in feet m = hydraulic mean depth (or hydraulic radius) in feet N = revolutions per minute N, = specific speed, = N X VB.hp./h* 4 n = factor in Kutter's formula = any abstract number P = total pressure or force in pounds p = intensity of pressure in feet of water p' = intensity of pressure in pounds per square foot p" = intensity of pressure in pounds per square inch q = rate of discharge in cubic feet per second r = radius to any point in feet 8 = slope of hydraulic gradient, = H'/l = tangential component of absolute velocity, = V cos A T = torque or moment of a force in foot pounds u = linear velocity of a point in feet per second V = absolute velocity of water (or relative to earth) in feet per second XV xvi ABBREVIATIONS v = velocity of water relative to some moving point in feet per second W = pounds of water per second, = wq w = density of water in pounds per cubic foot z = any vertical distance in feet; in measuring "head" it is a vertical elevation above any arbitrary datum plane = ratio of peripheral speed of turbine runner or pump impeller to \/2gh e = value of for which the maximum efficiency is obtained w = angular velocity in radians per second, = 2-jrN/QQ = u/r Values of quantities at specific points will be indicated by subscripts. In the use of subscripts (1) and (2) the water is always assumed to flow from (1) to (2). ABBREVIATIONS G.P.M. = gallons per minute Sec. ft. = cubic feet per second R.p.m. = revolutions per minute Hp. = horsepower B.Hp. = brake horsepower = D.Hp. W.Hp. = water horsepower HYDRAULICS CHAPTER 1 INTRODUCTION 1. Definition of Subject. Hydromechanics is the science of the mechanics of fluids. It may be subdivided into three branches: Hydrostatics is the study of the mechanics of fluids at rest, hydrokinetics deals with the flow of fluids, while hydro- dynamics is concerned with the forces exerted by or upon fluids in motion. Hydraulics is practical hydromechanics, that is, it is the study of the applications of hydromechanics to engineering problems. 1 While it might deal with any fluid it is generally restricted to liquids and especially to water. By idealizing conditions and ignoring phenomena that are known to exist, it is possible to study hydromechanics as a subject in pure mathematics. But naturally the results of such studies, though interesting, are often of little practical value. The determination of actual results by rigorous mathematics is often impossible because of the fact that the exact nature of certain hydraulic phenomena are either unknown or if known are so complex that it is not feasible to express them as mathematical functions. We must, therefore, resort to a combination of rigid mathematics, empirical expressions, and experimental coeffi- cients. The science that results, based partly upon pure reason- ing and partly upon experimental evidence, is called hydraulics. It is seen that hydraulics is not an exact science. In its actual applications much depends upon the judgment and the ex- perience of the engineer. In many cases it is necessary to compute or estimate results for which satisfactory experimental data is lacking. And in applying any experimental factors or empirical formulas it is well to have some familiarity with the 1 The derivation of the word "hydraulics" means "flow of water in a pipe" but usage has given the word a much broader significance. 1 work upon which they were based in order to judge as to their application to the case in hand. 2. Distinction between a Solid and a Fluid. The distinction between a solid and a fluid is ordinarily quite clear but there are plastic solids which flow under the proper circumstances and even metals may flow under high pressures. On the other hand, there are certain very viscous liquids which do not flow readily and it is easy to confuse them with the plastic solids. The definition of a fluid as a substance which flows must be extended therefore. The distinction is that any fluid, no matter how vis- cous, will yield in time to the slightest stress. But a solid, no matter how plastic, requires a certain magnitude of stress to be exerted before it will flow. Also when the shape of a solid is altered by external forces the tangential stresses between adjacent particles tend to restore the body to its original figure. With a fluid these tangential stresses, which are proportional to the viscosity, can act only while the change is taking place. When motion ceases the tangential stresses disappear and the fluid does not tend to regain its original shape. 3. Distinction between a Gas and a Liquid. A fluid may be either a gas or a liquid. A gas is quite compressible and when all external pressure is removed it tends to expand indefinitely. A gas is, therefore, in equilibrium only when it is completely en- closed. A liquid, on the other hand, is relatively incompressible and if all pressure, except that of its own vapor, be removed the cohesion between adjacent particles holds them together so that the liquid does not expand indefinitely. Therefore, a liquid may have a free surface, that is, a surface from which all pressure is removed, except that of its own vapor. The volume of a gas is greatly affected by changes in either pressure or temperature or both. It is usually necessary, there- fore, to take account of changes in volume and temperature when dealing with gases. Since the mechanics of gases is largely one of heat phenomena it is called thermodynamics. The volume of a liquid is affected to a very small extent by changes in pressure or temperature and for most purposes the changes in volume or temperature may be ignored. 4. Compressibility of Water. Water is usually said to be incompressible and as compared with gases it is relatively so. But it is much more compressible than many solids such as steel or even wood where the elastic limit is not passed. Its bulk or INTRODUCTION 3 volume modulus of elasticity, the ratio of the change of pressure per unit area to the change per unit of volume, is E v = 294,000 Ib. per sq. in. This value holds only for pressures below 1,000 Ib. per sq. in. and for temperatures near the freezing point. For higher tempera- tures it increases slightly. -Thus at 77F. it is about 327,000 Ib. per sq. in. and at 212F. it is 360,000 Ib. per sq. in. Also for higher pressures than the above the modulus is materially larger. Thus at a pressure of 65,000 Ib. per sq. in. Kite found a value of E v = 650,000 Ib. per sq. in. Temperature - Degrees Centigrade 10 20 30 40 50 60 70 80 90 100 62.6 62.4 62.2 .62.0 C1 Q ' i ' ' ' ' ' ' ' ' ' "^ ^ ^ s^ \ X. Weight of Pure Water -Lb. per Cu gggg-gg22S2 bo b k> k. o bo o Vo '.&. b> c s s^ \ \ \ \ \ \ N * \ \ ^ 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 Temperature -Degrees Fahrenheit FIG. 1. Density of pure water. It may be seen that increasing the pressure from atmospheric to 1,000 Ib. per sq. in. will reduce the volume of a body of water by 985 -f- 294,000 or about ^ of 1-0 P er cent - Therefore, it is seen that the usual assumption regarding water as incompressible is justified. 5. Density of Water. The density of water varies somewhat with the temperature as well as with the pressure. In Fig. 1 can be seen the values of the density of pure water at atmospheric 4 HYDRAULICS pressure for the range of temperature from freezing to boiling. The presence of impurities increases these values somewhat. Thus ocean water may ordinarily be taken as weighing 64.0 Ib. per cu. ft. In the computations in this book it will be sufficient to take w = 62.4 Ib. per cu. ft. 6. Accuracy of Computations. No computed result can be more accurate than the data upon which it is based and it is therefore not only useless but also misleading to carry out results to more significant figures than the data warrant. It should be noted that the number of significant figures has no relation to the location of the decimal point. Thus 347,000, 34.7, 0.0000347 are all values given to three significant figures and are of the same degree of accuracy. It is incorrect in such a figure as the first to preserve any more figures such as 347,129 if three figure work is all that is warranted. And if it is warranted it is likewise incorrect in such a value as the last to abbreviate it to 0.00003 for that is equivalent to saying that its value is 0.0000300 to three significant figures. There are some quantities that may be known with a high degree of accuracy but in hydraulic work most experimental factors are uncertain in the third significant figure and there are some coefficients or values which are uncertain even in the second significant figure. Thus slide rule work is all that is usually justified. Suppose for example that the product is desired of two quanti- ties whose values are 34.7 and 125. Multiplying these two numbers together we obtain 4,337.5 but the answer that should be given is 4,340. If our values were known to be 34.700 and 125.00 then the exact product would be permissible. But if our values are experimental they may range for example from 34.6 to 34.8 and 124 to 126 respectively. The products of the minimum and maximum values in each case are 4,290 and 4,380, thus showing that our result of 4,340 is uncertain in the third significant figure as we should expect when the given data are not correct in the third figure. 7. Notation. The use of a systematic and consistent notation is highly desirable and familiarity with the notation will save time and trouble. A table of the notation employed in this book is given on page xiii. INTRODUCTION 5 So far as possible an attempt has been made to employ the same notation that the majority of other writers use in this and in related subjects. This will result in a few cases of the same letter being used for different quantities but in such instances the quantities are so unlike that it is believed no real confusion can result. Unfortunately, the necessity of avoiding real conflicts in notation prevents the mse of certain letters that most naturally suggest themselves for several different quantities, the quantities not being sufficiently removed from each other to permit the duplication. 8. Units. The standard system of units employed in this book is based upon the foot, pound, and second. With few exceptions all formulas are to be used with such units. There are some few exceptions that commercial practice makes necessary or desirable. For instance the diameter of a pipe is customarily given in inches rather than in feet. Any exceptions to the general rule will be clearly indicated. It should be noted that the units of the answer in any computa- tion can be determined from the units involved in the separate items. Thus the product of velocity and area is the product of (ft. /sec.) X sq. ft. = cu. ft. /sec. The familiar quantity v 2 /2g is (ft. /sec.) 2 /(ft. /sec. 2 ) = ft. The product of the depth of water by the density of water is ft. X lb. per cu. ft. = Ib. per sq. ft., etc. It will frequently be necessary to use the value of g, the ac- celeration of gravity. Its units are feet per second per second, often written ft. /sec. 2 The value of g varies with latitude and elevation. Its value for any locality may be computed by the following formula according to Pierce, g = 32.0894 (1 + 0.0052375 sin 2 Z)(l - 0.0000000957 e), where I is the latitude in degree and e is the altitude in feet. For ordinary purposes g may be taken as 32.2 ft. per sec. 2 9. PROBLEMS 1. If a body of water is subjected to a pressure of 65,000 lb. per sq. in. how much less will its volume be than in a perfect vacuum? . Ans. 10 per cent. 2. What pressure will be required to reduce the volume of a body of water by 0.1 of 1.0 per cent, if the temperature is 32F. and the initial pressure 10 lb. per sq. in. Ans. 304 lb. per sq. in. 6 HYDRAULICS 3. If the temperature is 77F. what would be the result in problem 2? Ans. 337 Ib. per sq. in. 4. A cubic foot of ocean water at the surface and at ordinary temperature weighs 64.0 Ib. At the surface it is under a pressure of 14.7 Ib. per sq. in. What will be the weight of a cubic foot at a depth such that the pressure is 2,000 Ib. per sq. in.? Assume E v = 310,000 Ib. per sq. in. (Density is inversely proportional to volume.) 6. The radiator of an automobile holds 2.0 cu. ft. of water. It is filled with water at 50F. After the engine has been running the temperature of the water is 180F. Assuming no loss by evaporation or otherwise and neglecting expansion of radiator, how much water will have run out the overflow? 6. If we multiply cubic feet of water by the density of water in pounds per cubic foot and by feet, in what units will the answer be? 7. If we multiply torque, which is the product of a force in pounds and a distance in feet, by angular velocity in radians per second, what units will be involved in the answer? 8. If we multiply pounds per second by feet per second and divide by g, in what units is the answer? 9. If we multiply a force in pounds by velocity in feet per second in what units is the answer? CHAPTER II INTENSITY OF PRESSURE 10. Definition of Intensity of Pressure. By intensity of pressure is meant pressure per unit area. It may be expressed in various units such as pounds per square inch, pounds per square foot, or, as will be seen later, in feet of water, inches of mercury, etc. If P represents the total pressure on some finite area, F, while dP represents the total pressure on an infinitesimal portion of area, dF, the intensity of pressure is |i " ' ?'= ." w If the pressure is uniformly distributed over the area in question the intensity of pressure would then be p f = P/F. If the pres- sure is not uniformly distrib- uted the latter expression will give the average value only. The word "pressure " is usu- ally used for "intensity of pressure" though the latter term should be employed where there is any possibility of misunderstanding. The word " pressure" is also used to designate the resultant force exerted on an area. In order, to clearly distinguish this usage from intensity it would be well to employ the term "re- sultant pressure" or "total pressure." 11. Variation of Pressure in a Liquid. Let us consider a slender prism of the liquid in Fig. 2 as a free body in equilibrium. The forces acting upon it will be the pressures on its various faces and the pull of gravity. If the intensity of pressure at A be denoted by p' i} the total pressure on the end at A will be p'idF, where dF is the cross-section area. In similar manner the total pressure on the end at B will be p' 2 dF. The weight of the volume 7 HYDRAULICS of liquid is evidently wdFl. Since the prism of water is in equilibrium the algebraic sum of the components in any direction of all the forces acting on it will be zero. If the forces be resolved along the axis A B the three forces mentioned will be the only ones that will appear since the forces acting on the sides of the prism are all normal to the axis. Hence we may write 2/1 dF - 2/2 dF + wdFl cos a = 0. Since I cos a = z% - z\ it follows that p'z - p'i = w(z 2 - zi) (2) This equation shows that the difference in the intensity of pressure at two different points varies directly as the difference in the depths of the two points. Also if point A be taken at the level where p'\ is zero and if z be the elevation of such level above any other point, then in general p f = wz (3) From this equation it can be seen that the intensity of pressure varies directly as the depth of the point in question below the level where p f is zero. The results of equations (2) and (3) are strictly true only for an incompressible fluid in which the density, w, is constant at all depths. For practical purposes water is an incompressible fluid and hence these equations may be applied. But, owing to the high degree of compressibility of gases, they should not be used for a gas except where there are relatively small differences in pressure. 12. Surface of Equal Pressure. It may be seen from equation (3) that all points in a connected body of water at rest are under the same intensity of pressure if they are at the same depth. This indicates that a surface of equal pressure is a horizontal plane. Strictly speaking it is a surface everywhere normal to the direc- tion of gravity and it is, therefore, approximately a spherical surface concentric with the earth. But a limited portion of such a spherical surface is practically a plane area. A free surface is strictly one on which there is no pressure. Usually however the surface of a liquid exposed only to the pres- sure of the atmosphere is said to be a free surface. 13. Pressure the Same in all Directions. In a solid, owing to the existence of tangential stresses between adjacent particles, INTENSITY OF PRESSURE 9 the stresses at a given point may be different in different direc- tions. But in a fluid at rest no tangential stresses can exist and the only forces exerted between adjacent surfaces are normal to the surfaces. Therefore, the intensity of pressure at a given point is the same in every direction. This can be proven by reference to Fig. 3 where we have a small triangular element of volume whose thickness perpendicular to the plane of the paper is constant and equal to dz. Let a be any angle, p' the intensity of pressure in any direction, and p' x the intensity of pressure on a vertical plane. The following forces act upon this volume of fluid: The pressure on the vertical face is p' x dydz, the pressure on the slanting face is p'dldz, then there are the pressures on the horizontal face and on the two faces parallel to the plane of the paper, and the weight of the volume. Their values are not required. Since this volume is a fluid body at rest there are no other forces besides these normal to the surfaces, and, since it is a condition of equilibrium, the sum of the components in any direction is equal to zero. Writing such an equation for components in a horizontal direction we have only p'dldz cos a p'x dydz = 0. Since dy = dl cos a, it is seen that This result is independent of the angle a, and, therefore, it follows that the intensity of pressure is the same upon any plane passing through 0. 14. Pressure Expressed in Height of Liquid. In Fig. 4 imagine a body of liquid upon whose surface there is no pressure. Then by equation (3) the intensity of pressure at any depth z is p f = wz. For a given liquid w is constant and thus there is a definite relation between p' and z. That is any pressure per unit area is equivalent to a corresponding height of liquid. In hydraulic work it is often more convenient to express intensity of pressure in terms of height of a column of water rather than in pressure per unit area. Even if the surface of the liquid in Fig. 4 is under some pressure 10 HYDRAULICS the relation stated is still true. For this pressure on the surface could be expressed in terms of height of the liquid and such value added to z. The resulting value of p f would thus be increased by the amount of this surface pressure. The intensity of pressure expressed in terms of the height of a column of liquid will be denoted by the letter p. It will thus be seen that we have the relation This equation is true for any con- sistent system of units. Thus if w is density in pounds per cubic foot, p must be expressed in feet, and FlG 4 p' will then be in pounds per square foot. For pure water at ordinary temperatures we have the relation p' = 62. 4p. It is quite com- mon to express intensity of pressures in pounds per square inch, but p is rarely found expressed in other units besides feet of water. Since p' = 144p" = 62.4p we have p" = 0.4333p and p = 2.308p" EXAMPLES 1. Neglecting the pressure of the atmosphere upon the surface, what is the pressure in pounds per square inch at a depth of 3,000 ft. in fresh water? At a depth of 3,000 ft. in the ocean? 2. A certain pump for a hydraulic press delivers water at a pressure of 5,000 Ib. per sq. in. To what height of pure water would that be equivalent? To what height of liquid having a density of 100 Ib. per cu. ft.? 3. The specific gravity of mercury is 13.57, that is, its density is 13.57 times that of pure water. How many feet of mercury is equivalent to a pressure of 100 Ib. per sq. in.? 4. If the specific gravity of mercury is 13.57 how many feet of water is equivalent to a pressure of 10 ft. of mercury? 6. The pressure of the atmosphere is about 14.7 Ib. per sq. in. To what height in feet of water is this equivalent? What is the equivalent height in inches of mercury? 15. Barometer. If a tube such as that in Fig. 5 has its lower end immersed in a liquid and the air is exhausted from the tube the liquid will rise in the latter. If the air is completely ex- hausted we shall have zero pressure on 'the surface of the liquid in the tube, and the liquid will have reached its maximum height. INTENSITY OF PRESSURE 11 This device is called the barometer and is used for measuring the pressure of the atmosphere. By Art. 12 it may be seen that the intensities of pressure at o (within the tube) and at a (at the surface of the liquid outside) are the same. That is p = p a - And, since the pressure on the surface of the liquid in the tube is zero, the intensity of pressure at o is by equation (3) p' = wy. And by equation (4) p' wp . Thus the pressure of the air in terms of height of the liquid column is Pa = y (5) The liquid employed is usually mercury because of the fact that its density is sufficiently great to enable a reasonably short tube to be used, and because its vapor pressure is negligible at ordinary tempera- tures. If water were employed the height of tube would be inconvenient and also its vapor pressure at ordinary temperatures is appreci- able so that instead of having a perfect vac- uum at the top of the tube we should have a space filled with water vapor. The height attained by the liquid would consequently be less than what would otherwise be the case, the tube should be at least 0.5 in. in order to eliminate errors due to capillarity. The "pressure of the atmosphere is different in different locali- ties, depending upon elevation, and at a given point it varies from time to time according to the temperature and other factors. In round numbers the pressure of the atmosphere may be taken as 14.7 Ib. per sq. in., 30 in. of mercury, and 34 ft. of water. (These values are not exact equivalents.) EXAMPLES 1. If the barometer reads 29.92 in. of mercury, what is the pressure in pounds per square inch? 2. If the pressure of water vapor at 80F. is 0.505 Ib. per sq. in. what would be the height of the water barometer if the atmospheric pressure were 14.7 Ib. per sq. in.? (Use correct density of water for this temperature.) 3. Assuming the density of air to be 0.0807 Ib. per cu. ft., what would be the height of the air surrounding the earth and producing a pressure of 14.7 Ib. per sq. in., if air were incompressible? FIG. 5. Barometer. The diameter of 12 HYDRAULICS 16. Vacuum. Pressures less than that of the atmosphere are usually called vacuums, a perfect vacuum meaning an entire absence of all pressure. Vacuum is usually measured from the pressure of the atmosphere as a base and is commonly, though not necessarily, measured in inches of mercury. If the atmospheric pressure is 30 in. of mercury, a perfect vacuum would then be a vacuum of 30 in. And a vacuum of 10 in. of mercury would mean that there was a real pressure of 20 in. of mercury. . EXAMPLES 1. If the barometer reads 28.5 in. of mercury and the absolute pressure in the condenser for a steam turbine is 1.5 in. of mercury, what is the value of the vacuum? 2. The barometer reads 30 in. of mercury and within a certain vessel there is a vacuum of 22 in. of mercury. What is the real pressure within that vessel in pounds per square inch? What is the excess external pressure on the walls of the vessel in pounds per square inch? 17. Absolute and Relative or Gage Pressures. If the pres- sure is measured above absolute zero pressure it is called absolute pressure. If it is measured from the atmospheric pressure as a base it is called relative or gage pressure, since it is only relative pressure that a gage meas- ures. Thus Fig. 6 shows a compound gage which will measure pressures either above or below that of the atmosphere. When the gage is open to the atmosphere the hand points to zero. If the gage is connected to any vessel in which there is a pressure above FIG. 6. Compound gage. that of the surrounding air the hand will turn in a clockwise direction from zero. If the pres- sure is a vacuum the hand will move in the opposite direction. Thus the gage measures only the difference between the pressure on the inside of the gage tube and that of the air surrounding the gage. In Fig. 7 let indicate entire absence of all pressure or absolute zero and the ordinate OA represent the pressure of the atmos- phere. Then suppose we have any pressure such as at B. The gage will read the value AB and this is the gage pressure. The absolute pressure is OB. Also if we have a vacuum of AC, the gage pressure is ACj the minus sign merely indicating a value below atmospheric just as a plus sign indicates a pressure above that of the atmosphere. But the absolute pressure is OC. Absolute Zero INTENSITY OF PRESSURE 13 When dealing with absolute pressures all values &re positive. In the case of gage pressures only values above that of the atmos- pheric pressure are positive, but the minus sign for pressures below that of the atmosphere serves only to indicate a vacuum. There may still be a real pressure between adjacent particles of water. A true negative pressure would mean that, the water was in a state of tension and as water can sustain only a very slight tensile stress it is impossible to have a pressure below absolute zero. Abso- lute zero is the point where the stress in the liquid would change from compression to tension. In most problems in hydraulics we are A Atraospheric Press e not interested in absolute pressures. We are c concerned with the differences in pressure in- side a vessel and that outside for example o and in general that would be the gage pres- FlG sure. And in many other cases the atmos- pheric pressure acts alike at all points and balances out. EXAMPLES 1. A gage reads 20 Ib. per sq. in. when the gage itself is surrounded by the atmosphere. If the air surrounding the gage be exhausted to a vacuum of 20 in. of mercury while the real pressure of the fluid on the inside of the gage tube remains the same, what will be the reading of the gage? 2. If the barometer reads 30 in. of mercury and a vacuum gage reads 5 in. of mercury, what is the absolute pressure? 3. A gage pressure of 25 ft. of water is how much less than a gage pres- sure of 10 ft. of water? 18. Instruments for Measuring Pressure. Gage. The famil- iar pressure or vacuum gages have already been referred to and the combination of the two is shown in Fig. 6. In this type of instrument a curved tube is caused to change its curvature by changes of pressure within the interior of the tube. The moving end of the tube then rotates a hand by means of some intermediate links. It is usually assumed that the pressure indicated by the gage is that existing at the center of the gage. Thus the location of the center of the gage should always be taken into consideration. For instance, referring to Fig. 8, the 14 HYDRAULICS pressure at A is the gage reading plus the distance z. If the gage reads pounds per square inch, as is customary, p A = 2.308 p" + z. Piezometer Tube. A piezometer tube is a simple device for measuring moderate pressures. It consists of a tube in which the liquid can freely rise, without overflowing, un- til equilibrium is established. To prevent error due to capillarity the diameter of the tube should be at least 0.5 in. The height of the surface of liquid in the tube will give the pressure de- sired directly. It should be noted that if the water, whose pressure is desired, is flowing the true pressure can be obtained only by having the axis of the tube at the point of connection perpendicular to the stream flow and further- more the interior opening should be smooth and free from all projections. If the end of the pipe projects into the stream, as in the case of the fourth tube in Fig. 94, the pressure read will be too low. Mercury U Tube. For high pressure the water piezometer is not suited and some modification must be adopted. The mercury FIG. 8. FIG. 9. Piezometer. FIG. 10. U tube shown in Fig. 10 may then be used. If s is the specific gravity of the mercury (or whatever liquid may be employed) the pressure at the point C is sy. This is also the pressure at B but the pressure at A is greater than this by the amount z, if the INTENSITY OF PRESSURE 15 tube from A' to B is filled with water. If it were filled with air, then, neglecting the slight weight of the air within the tube, the pressures at B and at the surface of the water at A' would be equal. In practice it would be difficult to insure the absence of air and if the tube were partially filled with air and partially with water it would be troublesome to make correction and accuracy would be impossible unless at were known just what proportion of the tube was filled with water and what with air. It is therefore desirable to provide some means of permitting all the air to escape and its place to be taken by water. If the con- necting tube in Fig. 10 is filled with water the pressure at A is p A = z + sy. FIG. 11. FIG. 12. In measuring a vacuum we must interpret y as a negative quantity in Fig. 11 so that we have, if the tube is filled with water, PA = z - sy. If this connecting tube from A' to B were filled with air then the correction for the height above A' would be negligible but it is difficult to insure this being filled with air and error will be intro- duced if it is not. Thus the arrangement in Fig. 12 is much better as that permits no air to collect in the tube and introduce errors in the readings. In this case z is negative so that PA = ~ z - 16 HYDRAULICS Differential Gage. The differential gage is used for measuring differences of pressure only. One form of this is shown in Fig. 13. Assuming the entire connecting tubing to be filled with water except that portion of the U that is filled with the denser liquid, such as mercury for instance, the pressure at A' will exceed that at B' by the amount sy. That is But and FIG. 13. Substituting these values we have PA ~ PB = sy + Z A - Z B = sy - y = (s - l)y. In the differential manometer the left-hand column of mercury, or whatever it is, has a column of water of height y resting upon it that is not balanced by a corresponding amount on the right- hand column, hence the pressure difference is not sy alone. EXAMPLES 1. Two pressure gages are connected to the same vessel containing water under pressure. One of these gages is 10 ft. below the point where the pres- INTENSITY OF PRESSURE 17 sure is measured and it reads 40 Ib. per sq. in. The other gage is located 15 ft. above the point in question. What will it read ? What is the pressure in the vessel? (It is assumed that the connecting pipes are filled with water.) 2. Two vessels are connected to a differential manometer using mercury (specific gravity = 13.57). When the mercury reading is 36 in. what is the difference in pressure in feet of water between the two vessels? 19. The Hydraulic Press. The most important device operat- ing upon the principle of equal transmission of intensity of pres- sure in all directions is the hydraulic press. If in Fig. 14 a force PI be applied to the small piston whose area is FI the intensity of pressure throughout the whole volume of liquid will be increased by the amount p' = Pi/Fi. Then the total additional force exerted upon the face of the large piston will be p'F 2 . = (Pi/Fi)F 2 = Pi(F 2 /Fi). It is thus seen that a small force m r FIG. 14. Hydraulic press. exerted on the smaller piston is enabled to oppose a much greater load on the large piston. If G\ and G% denote the weights of the pistons while z is the difference in elevation of their faces, we have for equilibrium P 2 + 2 -- = -- ff -- : r i ft Since the volume of liquid in the vessel must remain constant, it follows that the distance moved by the larger piston must be much less than that moved through by the smaller piston. EXAMPLES 1. In Fig. 15 the diameter of the small piston is % in. and that of the large one is 20 in. The big plunger weighs 1,000 Ib. and sustains an external load 2 18 HYDRAULICS of 6,000 Ib. The liquid used is water. What total force P applied to the small piston will secure equilibrium? Ans. 4.12 Ib. 2. When the small piston has descended 20 ft. how far will the plunger have been raised? 20. PROBLEMS 1. In Fig. 16 what are the values of absolute pressure at A, B, C, and D, assuming the liquid to be water? What are the values of gage pressure? FIG. FIG. 16. FIG. What is the value of the vacuum at C? (Give answers in feet of water, pounds per square inch, and inches of mercury in every case.) 2. In Fig. 17 the cylinder is 2 ft. in diameter and the weight of the piston and load is 4,000 Ib. What will be the gage reading in pounds per square inch? 3. If the mercury manometer in Fig. 17 reads 35 in., how far is the top of the lower mercury column below the piston? If the manometer remained at Open Tube 2000* FIG. 18. this same place but the connection were made to the tank at a different level, would the mercury reading change? 4. The small piston in Fig. 18 has a diameter of 3 in. Neglecting friction, when a force PI of 20,000 Ib. is applied to it, what will be the force P 2 that can be exerted by the plunger with a diameter of 24 in. ? To what height would water rise in the piezometer tube shown? CHAPTER III Free Surface HYDROSTATIC PRESSURE ON AREAS 21. Total Pressure on Plane Area. Since we are dealing with fluids at rest, no tangential forces can be exerted and hence all pressures are normal to the surfaces in question. If the pressure were uniformly distributed over an area, the total or resultant pressure would be the product of the area and the intensity of pressure and the point of application of the force would be the center of gravity of the area. In general the in- tensity of pressure is not uniform, hence further analysis is necessary. In Fig. 19 consider a vertical plane whose upper edge lies in the free sur- face. Let this plane be perpendicular to the plane of the paper so that AB is merely its trace. The intensity of pressure will vary from zero at A to BC at B. It will thus be seen that the total pressure P will be the sum- mation of the products of the ele- mentary areas and the intensities of pressure upon them. It is also ap- parent that the resultant of this sys- tem of parallel forces must be applied at a point below the center of gravity of the area, since the center of gravity of an area is the point of application of the re- sultant of a system of uniform parallel forces. If the plane be immersed to A'B' the intensity of pressure varies from A'D at A' to B'E at B'. Since the proportionate change of intensity of pressure from A' to B' is less than before, it is clear that the center of pressure will approach nearer the center of gravity. In Fig. 20 let AB be the trace of a plane making any angle B with the horizontal. The view to the right is the projection of this area upon a vertical plane which is also normal to the plane containing AB. Let z be the depth of any point and y be the 19 FIG. 19. 20 HYDRAULICS distance of any point from OX, the axis of intersection of the plane, produced if necessary, and the free surface. The coordin- ates of the center of gravity of the area may be denoted by z and y respectively. Take an element of area dF such that all portions of said ele- ment are at the same depth z. Then the total pressure on dF is dP = p'dF = wzdF] Hence P = wfzdF. But fzdF = zF, hence P = wzF (6) 22. Depth of the Center of Pressure. The point of applica- tion of the resultant force on an area is called the center of pres- sure. We usually locate the line of action of a force by taking moments. In this case it is convenient to take OX in Fig. 20 as the axis of moments. On any element of area dF the total pressure is dP = wzdF = wy sin BdF and its moment is ydP = wy* sin 6dF. If the distance of the center of pressure from OX be denoted by y' , ' fydP w sin 6fy 2 dF '' fdP " w sin BfydF But fy z dF is the moment of inertia of the area F about the axis OX, and fydF is the statical moment of the area with respect to the same axis, hence To y ~~ (8) This may be put in a more convenient form by noting that, if I a is the moment of inertia of the plane area about its gravity HYDROSTATIC PRESSURE ON AREAS 21 axis and k g is the radius of gyration about the gravity 'axis, we have < ''' From these equations it may be" seen that the location of the center of pressure is independent of the angle 0, that is, the plane area may be rotated about the axis OX without affecting the loca- tion of the center of pressure. However, this will not hold for 6 = zero since the value of P would also be zero. From equation (10) it may also be seen that the center of pressure is always below the center of gravity. Also as the depth of immersion is increased for a given value of 0, the distance y increases. But as k g remains constant in value it may be seen that the last term in equation (10) becomes relatively small, hence y f approaches y in value. The same thing would be true if the depth of the center of gravity z remained constant while the plane was rotated so as to approach a horizontal direc- tion. (This is entirely different from rotation about the axis OX, since y no longer remains constant.) EXAMPLES 1. A rectangular plane area is 5 ft. by 6 ft., the 5-ft. side is horizontal, and the 6-ft. side vertical. Determine the resultant pressure and the location of the center of pressure when: (a) the top edge is in the water surface; (6) the top edge is 1 ft. below the water surface; (c) the top edge is 100 ft. below the water surface. Ans. (a) P = 5,620 lb., y' = 4 ft.; (6) P = 7,500 lb., y' = 4.75 ft.; (c) P = 193,000 lb., y' = 103.03 ft. 2. Suppose in Fig. 20 that we have a rectangular area 5 ft. by 6 ft., that AB = 6 ft., the 5-ft. edge being normal to the plane of the paper, and that y 4 ft. Find the magnitude of the total pressure and the location of the center of pressure when 6 has values of 90, 60, 30, and 10. Ans. (a) P = 7,488 lb., y f = 4.75 ft.; (6) P = 6,490 lb.; (c) P = 3,744 lb.; (d) P = 1,302 lb. 3. Suppose that in problem (2) y was variable but that z = 4 ft. Solve with values of 6 of 90, 60, 30, and 0. Ans., (a] P = 7,488 lb., y' = 4.75 ft.; (6) y' = 5.265 ft.; (c) y 1 = 8.375 ft.; (d) y' = infinity, z' = 4 ft. 4. Find the depth of the center of pressure on a vertical triangular area whose altitude is h and whose base is 6 if : (a) its vertex lie's in the water surface and base is horizontal; (ft) its base lies in the water surface. Ans. (a) if = %h\ (6) ?/ = %h. * 22 HYDRAULICS 23. Lateral Location of Center of Pressure. For most prac- tical problems the depth of the center of pressure is all that re- quires solution since the areas with which we deal are usually such that a straight line can be drawn through the centers of all horizontal lines. In such cases the center of pressure is seen to lie on this line. But in case this is not so we should have to com- pute x' as in Fig. 20, x' being measured from any axis parallel to trace AB. Again we employ moments as in the preceding article. If x is the distance of an element from the axis in question the moment of dP is xdP = wxy sin0 dF Hence the value of x' is fxdP w sin BfxydF fdP ' w sin BfydF fxydF yF (11) This equation differs from (7) simply in the fact that we have fxydF instead of ftfdF. The latter quantity is more frequently met with, it is given a name, symbolized by the letter /, and values of I for different areas can usually be found in tables. The former expression is called " product of inertia," is symbolized by the letter J, but owing to the infrequent use that is made of it values of J cannot usually be obtained save by integration. Lacking the knowledge of the value of J for any area, we should simply proceed to evaluate fxydF just as we should evaluate fy*dF in case we did not know the value of / for the area in question. It will be found that reduction formulas can be used here as with moments of inertia. If J indicates the product of inertia with respect to the intersection of any two axes, while a and b are the coordinates of the center of gravity of an area about which the product of inertia is J g , it will be found that J = J + Fab. In using equation (11) it must be noted that y is to be measured as in Fig. 20, while x may be measured from any axis in the plane of the figure and perpendicular to OX. HYDROSTATIC PRESSURE ON AREAS 23 EXAMPLES 1. Given a right triangle with height, h, and base, b, with its vertex in the water surface and its plane vertical. Find the value of y' and then determine x'\ (a) by inspection; (6) by calculus. Ans. y' = %h; x' = %6. 2. Find the center of pressure on an area which is a quadrant of a circle. It is placed in a vertical plane and one edge lies in the water surface. Ans. y = 4r/37r; y' = 3*T/16^' = 3r/8. 23a. Graphical Solution for Pressure on Plane Area. It is not always feasible to apply equations (6), (8) and (11) directly, especially if the plane area in question is irregular in outline so that its center of gravity and moment of inertia cannot be readily determined. The problem may then be solved as follows. In Fig. 20a is shown a plane area at I which makes any angle B with the horizontal. As in Art. 21 let us taken an element of area such that every portion is at the same depth below the water surface. Then dF = xdy. Note also that z = y sin B. The intensity of pressure, p' = wz, is shown in //, values of wz being plotted perpendicular to the inclined base line. The total pressure on an element of area is dP = wzdF = wzxdy. The total pressure on the entire area is P = wfzxdy = w sin Bfxydy. In III xy is plotted as a function of y. The elementary rectangle there shown is of length xy and width dy. The sum of all such areas is the total area shown in III. But the sum of all values of xydy is the value of the definite integral. Hence to find the total pressure on the area shown at I, it is only necessary to 23a HYDRAULICS multiply values of x by the corresponding values of y and plot the product against the values of y. The area then represents to some scale the value of the integral between the limits used, and when multiplied by w sin 6 will give the value of the total force. 1 Since the ordinates of III, when multiplied by the proper constant, represent values of forces on elementary areas, the resultant force will act through the center of gravity of III. This then serves to locate the y co-ordinate of the center of pres- sure of the actual area in I. It will also be perceived that, in general, the centers of gravity of areas I and III do not coincide. Hence the center of gravity of the actual area I and its center of pressure never coincide, except in the special case where the in- tensity of pressure is uniform. To determine the location of the center of gravity of III or in other words to locate the center of pressure in I we take moments. Thus Py f = w sin8j*xy 2 dy The area shown in IV is seen to represent the value of this inte- gral. Since y' = Py'/P, , _ f*y*dy fxydy or y' is obtained by dividing the area of IV by the area of III after reducing each to its proper scale value. A similar procedure could be employed for finding the value of x' if desired. It may be noted that the area of III represents fxydy = fydF = yF or is the moment of the actual area in I. And the area repre- sented by IV is f xy*dy = fy*dF = 1 or is the moment of the area III or the moment of inertia of the actual area I. 1 In general when one has an integral of the form J'udv it may be impossible to integrate it either because the calculus solution cannot be discovered or because u cannot be expressed as .a mathematical function of v. In either case a numerical value of the definite integral may be obtained by plotting values of u against values of v and determining the area between the curve and the v axis. If u is plotted to such a scale that 1 in. = a units and v to such a scale that 1 in. =6 units, then the scale for the area is 1 sq. in. = ab units. If it is not convenient to plot the curve and measure the area with a planimeter, the ordinates can be computed and the area determined without actually plotting it by some method of approximation, such as Simpson's rule. HYDROSTATIC PRESSURE ON AREAS 236 EXAMPLES 1. A vertical plane area, whose upper edge coincides with the water sur- face, has the following widths starting with the surface and at 1 ft. intervals below it. 4.90 ft., 4.48, 4.00, 3.46, 2.82, 2.00, and 0. Plot values of zx and z 2 x and determine the magnitude of the resultant pressure and the depth of the center of pressure. Ana. P = 2930 lb., y' = z' = 3.43 ft. 2. Find the area of the plane and the depth of the center of gravity. Ans. F = 19.6 sq. ft., y = $"= 2.4 ft. 3. Solve the above problems by Simpson's rule. 24. Resultant Thrust on Plane Areas. So far we have dealt with the total pressures on one side of a plane area alone. Of course, when the area is completely immersed in a fluid as shown in some of the previous illustrations, the total pressure on one side is balanced by that on the other and the net effect is zero. But when the two sides are not subjected to the same pressure, there is a resultant thrust whose value we desire. AlVater Surface So far we have considered the surface of the liquid as being free from all pressure. Thus in Fig. 21 we should consider the intensity of pressure as varying from zero at A to BC at B. But in reality there is some pressure, in general, from the atmosphere acting upon the water surface equivalent to a height of about 34 ft. of water, and thus the true free surface might really be at point, O, the distance AO being equal to the height of the water baro- meter. The absolute intensity of pressure upon the left-hand side of the plane, AB, therefore varies from AD to BE. But in practical applications we desire the difference between the pressure on the left-hand side and that on the right-hand side. But the pressure on the right-hand side is that due to the atmosphere and its intensity is uniform from A to B being equal ioAD'. But AD' = AD = CE. Hence atmospheric pressure is added alike to both 24 HYDRAULICS sides, and it is useless to consider it. Therefore, we neglect atmospheric pressure altogether and treat the water surface as a true free surface in most calculations. Suppose we have an area such as AB in Fig. 22 with a fluid pressure on both sides but of different intensities. Of course, we could compute the magnitudes of the total pressures on both sides of the area and the difference would be the resultant desired. But we should also have to find the centers of pressure on both sides and then locate the line of action of the resultant of these two forces. The following analysis will indicate a much easier solution. At A the intensities of pressure on the two sides are A I and AK. If IJ be laid off equal to AK the net difference in the intensity of pressure will be AJ. In similar manner at B the net intensity of pressure is BF. And it is readily seen that, since CDE and HKG make the same angle with the vertical, the values of HDj AJ f and BF are equal. Thus the resultant intensity of pressure on the area, AB, is uniform and equal to HD in value. But HD is the intensity of pressure at the depth, h. Hence the resultant thrust on any area with both sides com- pletely covered by the same liquid is R = whF (12) where h is the difference in level of the two liquids. And since the net intensity of pressure is uniform, the resultant thrust will act through the center of gravity of the plane area. EXAMPLES 1. Suppose that a rectangular area is 2 ft. wide by 3 ft. high and that its upper edge lies in a water surface. What twisting moment will be necessary in a shaft through A (Fig. 21), perpendicular to the plane of the paper, to withstand the water pressure? It will be assumed that the gate received no support save what the shaft affords, and that atmospheric pressure acts alike on the water surface and the right-hand side of the gate. Ans. 1,123 ft.-lb. 2. Suppose that the right-hand side of the gate in problem (1) is under a vacuum of 30 in. of mercury, and that the barometer reading is 30 in. of mercury. What twisting moment would be required? Ans. 20,200 ft.-lb. 3. Suppose that the barometer reads 30 in. of mercury and that the right- hand side of the gate in problem (1) is under a vacuum of 20 in. of mercury. What twisting moment would be required? Ans. 13,800 ft.-lb. 4. Suppose the barometer reads 30 in. of mercury, that the right-hand HYDROSTATIC PRESSURE ON AREAS 25 side of the gate in problem (1) is under atmospheric pressure/ while the sur- face of the water is under a gage pressure of 50 Ib. per sq. in. What twisting moment would be required? Ans. 65,900 ft.-lb. 6. Suppose in Fig. 22 that A B is a circular gate of 3-ft. diameter, that BC = 10 ft. and BH = 4 ft. Find: (a) magnitude and line of action of total pressure on left-hand side only; (6) magnitude and line of action of total pressure on right-hand side only; (c) resultant thrust on gate. Ans. (a) 3,747 Ib., 1.566 ft: below top of gate; (6) 1,102 Ib., 1.725 ft. below top of gate; (c) 2,645 Ib., 1.500 ft. below top of gate. 25. Horizontal Pressure on Curved Surface. On any curved or irregular area in general, such as that whose trace is AB in Fig. 23, the pressures upon different elements are different in direction and an algebraic or calculus summation is impossible. FIG. 23. Hence equation (6) can be applied only to a plane area. But we may find the component of pressure in certain directions. Thus if we multiplied each dP by cos 0, being a variable angle which each elementary force makes with the horizontal, the total horizontal force would be p x = fdP cos 0. (13) In general it will be tedious to integrate the latter and often practically impossible. Hence the following procedure may be employed Project the irregular area in question upon a vertical plane, the trace of the latter being A'B'. The projecting elements are A A', BB r , etc. It is seen that these projecting elements, which are all horizontal, enclose a volume whose ends are the vertical 26 HYDRAULICS plane A'B r and the irregular area whose trace is A B. This volume of liquid is in equilibrium under the action of the following forces. Upon the vertical plane at the left there is a force P', gravity G f acts upon the volume and is vertical, the pressures on the projecting elements are all normal to these elements, hence normal to P f . Then there are the pressures upon the area in question at the right-hand end, the horizontal component of pressure being represented by P x and the vertical component by P y . Since we have a condition of equilibrium the sum of all the forces in any direction must be equal to zero. But in a hori- zontal direction the only forces are P f and P x . Hence P x = P'. (14) That is the component, in any given horizontal direction, of the pressure upon any area whatever is equal to the pressure upon the projection of the area upon a vertical plane which is per- pendicular to the given horizontal direction. The lines of action must also be the same. 26. Vertical Pressure on Curved Surface. The vertical com- ponent of pressure on an irregular surface can be found by a method similar to that for the horizontal pressure. Thus in Fig. 23 if we take a volume of liquid of which the area in question forms the base and vertical elements such as AD and BC form the sides, we find the following forces are acting. Considering CD a free surface the pressure on the upper face is zero. The pressure on the lower face is composed of the two components P x and P y . Gravity, G, is the only other vertical force, the pressures on the sides all being horizontal. Summing up the vertical forces and equating to zero we have P v = G. (15) Hence the vertical component of pressure on any area whatever is equal to the weight of that volume of liquid which would extend vertically from the area to the free surface. 27. Component of Pressure in any Dkection. In general the component of pressure in any direction aside from horizontal and vertical cannot be found, since the weight of the volume of liquid, such as AA'B'B in Fig. 23 would have to enter the equation. But if the depth of immersion is great so that the pressures on AB and A 'B' are great compared with the weight G' the latter may be neglected. Hence in such cases only, the component of HYDROSTATIC PRESSURE ON AREAS 27 pressure in any direction may be taken as the pressure upon an area projected in that direction upon a plane which is perpen- dicular to the given direction. Of course with a plane area the component of pressure in any direction may be found by multiplying P by the proper function of some angle. Or it may be convenient to find it by the methods of Arts. 25 and 26. Also-for a plane area, since P cos 6 = (wzF) cos 8, it may be seen that the component of pressure is the same as the pressure upon an area of value F cos 6 provided the center of gravity of such area be the same depth as the center of gravity of the given plane. 28. Resultant Pressure on Curved Surface. In general there is no single resultant pressure on an irregular surface, for a system of non-parallel and non-coplanar forces does not usually reduce to anything simpler than two single forces. Thus in general P x and P y are not in the same plane and hence cannot be com- bined. But in some special cases of symmetrical surfaces, these two components will lie in the same plane and hence can be combined into a single force. EXAMPLES 1. In Fig. 24 is shown a quadrant of a circular cylinder, AB, whose length perpendicular to the plane of the paper is 4 ft. (a) Find the horizontal com- ponent of pressure. (6) Find the vertical com- , , , , _,. , ., _, ' i i Water Surface ponent of pressure, (c) Find the magnitude and direction of the resultant water pressure, locates its line of action? 29. Pipes under Pressure. If the in- ternal pressure in a cylindrical pipe is great enough to be considered in determining the thickness of pipe wall necessary, it will be large enough so that the weight of the water may be disregarded. Hence according to Art. 27 we may compute the resultant pressure in any direction. Suppose that in Fig. 25, we pass a plane XY through a diameter of the pipe as shown. The total pressure on one-half of the pipe in any direction, such as that normal to XY, will evidently be p r X 2r X I, I being any length of pipe. This follows directly from Art. 27 or may be seen from the fact that the thrust of the water on the wall of the pipe normal to XY must be balanced by the thrust of the water on the plane XY. This pressure will tend to rupture the pipe across the plane XY and is resisted by the (d) What 28 HYDRAULICS tensions in the walls of the pipe, such as T. Evidently 2T = 2p'rl. If the thickness of the pipe wall be denoted by t, and the stress induced in it by S h , then T = S h tl. Hence S h t = p'r (16) From (16) the thickness of wall necessary may be computed for any allowable unit tensile stress. However, it is well to note that p' should be the maximum intensity of pressure that may occur and in case of water hammer these intensities are much greater than the static pressures alone. Also it may often be found that (16) gives entirely too thin a wall to stand ordinary handling and to allow for a certain amount of corrosion. In practice p r is, therefore, increased to allow for possible water hammer and the thickness determined by (16) is then increased FIG. Y FIG. 26. to a value necessary for these other reasons. The tension in the case shown is called hoop tension. Referring to Fig. 26 it may be seen that a cylindrical pipe may also be ruptured by forces parallel to the axis. Thus the pressure on the blank end is balanced by the tension in any section such as XY. The total pressure, assuming it to be of uniform intensity, is p' X TiT 2 . And the tension across a section XY is T Si X 2irrt. Hence, equating these two, 2S t t = p'r (17) This stress is called longitudinal tension and it may be seen that it is one-half the hoop tension. For cylinders with thin walls these formulas will hold, since they assume uniform intensity of stress across the metal. But with thick walls they do not hold. In the case of hoop tension in a cylinder with thick walls it is usually assumed that the intensity of stress is a maximum at the inner face and decreases to zero at the outside of the wall. Also the elasticity of the material enters HYDROSTATIC PRESSURE ON AREAS 29 into the hypothesis. John Sharp 1 gives the following empirical formula for hoop tension in a cast-iron cylinder with thick walls S log e ^ = p' (18) where r 2 = external radius and r internal radius. For wrought iron and steel cylinders he gives the empirical expression [(7-1)+ log. 7] = V: - (19) Equation (16) with S understood as compressive stress would also hold for external pressure provided the pipe remained truly cylindrical. But actually it may become slightly distorted from the cylindrical form and then there is a possibility of sudden collapse. A large thin tube which can stand a high internal pressure can withstand only a small external pressure. All formulas for determining the strength of pipes against external pressure are purely empirical. So far no satisfactory expression has been deduced, an.d sufficient data is lacking. 30. Buoyant Force of the Water and Flotation. Considering the body EHDK immersed in a fluid in Fig. 27, we see that it is acted upon by gravity and the pressures from the surrounding fluid at least. In addition there may be other forces applied. On the upper surface of the body the vertical component of the pressure, P v , will be equal to the weight of the volume of fluid AEHDC. In similar manner the vertical component of the pressure on the under surface, P' y , will be equal to the weight of the volume of fluid AEKDC. It is evident that P' v is greater than Py and that the total vertical force exerted by the fluid is upward and is equal in magnitude to P'v - PV = weight of volume AEKDC - weight of volume AEHDC. But the difference between these two volumes is the volume of the body EHDK. Hence for any body immersed in a fluid such as water the buoyant force of the water is equal to the weight of the water displaced. If the body remains in equilibrium in the position shown in Fig. 27, when no other forces are acting, it is seen that G = P' y P v . Hence the body must be of the same density as the fluid in which it is immersed. If it is lighter than the fluid, a downward 1 "Some Considerations Regarding Cast Iron and Steel Pipe." 30 HYDRAULICS force will have to be applied whose value is B G, B being the buoyant force of the fluid. If the body is denser than the fluid, it will have to be supported by a force whose value is G B. But if the body rests on the bottom of a body of fluid (Fig. 28) in such away that the fluid does not have access to the under side, there will be no buoyant effect for then P' y = zero. Thus in the case of a ship, for example, sunk in the mud at the bottom of a body of water, the pull T necessary to raise the ship is not only FIG. 27. FIG. 28. the weight of the ship but also the weight of the entire volume of water resting on top of it. Thus in Fig. 28, T = G + P y . If no external forces are applied to a body which is lighter than the fluid, it will float on the surface, such portion of its volume being immersed as is necessary to displace an amount of fluid equal in weight to the weight of the body. If the body is slightly heavier than the fluid, it will sink. If it is less compressible than the fluid and there is sufficient depth, it will sink until such a depth is reached that the density of the fluid is equal to its own density. If it is more compressible than ^^g the fluid its own density will be increased more rapidly than that of the water and it ^^= will sink to the bottom. i EXAMPLES FIG. 29. 1. A body whose volume is 2 cu. ft. weighs 200 Ib. What will be the force necessary to sustain it when it is immersed in fresh water? In ocean water? 2. In Fig. 29 the cube A is 12 in. along each edge and weighs 100 Ib. It is attached to the square prism B which is 6 in. by 6 in. by 8 ft. and weighs 30 Ib. per cu. ft. What length of B will project above the water surface? Ans. 1.76ft. HYDROSTATIC PRESSURE ON AREAS 31 3. A balloon weighs 250 Ib. and has a volume of 10,000 cu. ft. When it is filled with hydrogen which weighs 0.0056 Ib. per cu. ft. what load will it support in air which weighs 0.08 Ib. per cu. ft.? Ans. 494 Ib. 4. The specific gravity of a solid is 0.8. What portion of its volume will be above the surface of the water upon which it floats? 5. A body weighs 50 Ib. and has a volume of 4 cu. ft. What vertical force is necessary to sink it beneath the surface of the water? 31. Metacenter. For a body floating on the surface of the water, such as in Fig. 30, there are only the two vertical forces, its weight G and the buoyant force of the water B. The latter acts through the center of gravity of the water displaced. This point is called the center of buoyancy. If the body is in equilib- rium, these two forces must be in the same straight line. FIG. 30. Suppose that by some external agency the body is rolled or dis- placed through some angle 6. The center of gravity is naturally unchanged in its position in the section but the center of buoy- ancy, in general, will change. Thus G and B constitute a couple. In Fig. 30 (6) this is a righting couple since it tends to restore the body to the upright position. It may be seen that the line of action of B cuts the axis at point M. This point is called a metacenter. As the angle varies, the amount of this couple will vary and the point M will also change its location. The position which M approaches as 6 approaches zero is the true metacenter. It may be seen that if the couple is a righting couple the point M must always be above C the center of gravity. It is necessary in ship design to insure that M will be above the center of gravity for all angles of heel. Thus not only is it necessary to locate the true metacenter but also to compute the moment of the righting couple for all values of 6 which are likely to be encountered. Further consideration of this topic properly belongs to the subject of ship design. CHAPTER IV APPLICATIONS OF HYDROSTATICS 32. The Gravity Dam. One of the most important of the many applications of hydrostatics is the design of dams, of which there are several types. The gravity dam is one which depends for its stability upon its weight. A typical cross-section of such a dam is shown in Fig. 31. If the face AB is curved it will be necessary to compute the two components of the water pressure, H being equal to the pressure on a plane whose trace is A'B while V is the weight of the volume of water represented by FIG. 31. Cross-section of gravity dam. ABA'. In all computations it is customary to consider a length of dam (perpendicular to the plane of the figure) of 1 ft. Evi- dently the stability of a gravity dam is independent of the total length of the dam. The total water pressure P combined with the weight of the section G gives a resultant pressure on the base whose value is R. This pressure is distributed all over the base BO but may be considered to have a single point of application C. If R be resolved into two components at the point C, evidently the value of the horizontal component must be equal to H while that 32 APPLICATIONS OF HYDROSTATICS 33 of the vertical component will equal G + V . By taking moments of all the forces about it will be easy to locate the point C. If the dam rests solidly upon impervious rock and there is no leakage of water along any plane, or if a cutoff wall at B runs down deep enough to stop percolation, and the base of the dam is well drained, the above forces are all that act upon the struc- ture, excepting of course tKe support of the earth which is equal and opposite to R. But if water does have access to the under side of the dam there will be exerted upon BO a vertical upward pressure due to this. How much this may amount to depends upon conditions. Thus if water saturates the foundation but does not have an opportunity of escaping past the whole base of the dam will be subjected to a water pressure equal to BA' in intensity. But if the water can escape past there will be a flow of water under the dam and consequently the pressure must decrease from BA' at B to a very much smaller value at 0. It is often reasonable to assume the pressure as zero at 0. But in any event the admission of water to the base of the dam tends to decrease the safety of the structure. It may be seen that the horizontal thrust of the water H is opposed solely by the friction between the dam and the founda- tion upon which it rests. If the coefficient of friction here be denoted by /* then it is clear that if the dam is safe against sliding the value of H must be less than n(G + V). The factor of safety against sliding is the ratio of the latter quantity to H. Any leakage of water under the base of the dam decreases the pressure between the dam and the material upon which it rests and thus tends to decrease the frictional resistance. The frictional resistance can be increased by sinking portions of the dam into trenches such as in the case of the cutoff wall at B. If it were possible for the dam to act as a rigid body under all circumstances it could then fail by overturning about as an axis. It is seen that with as a center of moments, H tends to overturn the dam but is resisted by G and V. If water pressure acted upon the base it would also tend to overturn the dam. The factor of safety against overturning is the ratio of the moment of G + V to the moment of H and the water pressure on the base, if any is allowed for. However, before a masonry dam of any size would overturn, the material along the base near would be crushed due to the high intensity of pressure it would be under. Thus although the point C might be to the left of in Fig. 31 so that 34 HYDRAULICS the structure is safe against overturning, the base would still not be safe against crushing. Hence, the second consideration of the stability of the dam is not as to whether it will or will not overturn but is concerning the distribution of stresses along the base BO. Referring to Fig. 32, a uniform intensity of stress p' distributed over an area represented by A B gives a resultant pressure P applied midway between A and B. If, however, the stress varies uni- formly from P'A at A to zero at B, the resultant P will pass through a point one-third the distance from A to B. If the total pressure From a photograph by the author. FIG. 33. Concrete dam at Crystal Springs Lake, California. 145 feet high. P has the same value in both cases, it is clear that the intensity of pressure at A is greater in the latter case than in the former. And if P is applied at a point less than one-third the distance from A to B, the intensity of stress will be still greater at A and at B the intensity of stress P'B will be opposite in sign to that at APPLICATIONS OF HYDROSTATICS 35 A. It is thus clear that it is desirable to have the resultant pres- sure pass as nearly through the midpoint as possible. And if tensile stresses are to be avoided the resultant pressure must be kept within the middle third. As masonry is not supposed to en- dure tensile stresses, it is customary to so design the dam that the resultant pressure falls within the middle third of any section. It is not only necessary~to undertake such an analysis of the dam as a whole but also to investigate the stability of all portions , of the dam with respect to any horizontal plane. In all such studies the maximum height of water should be assumed. But also the pressures should be determined when the reservoir is empty as the inner face of the dam might then be subjected to excessive vertical stresses. 33. The Framed Dam. Contrasted with the gravity dam we have the framed dam shown in Fig. 34 which depends for its FIG. 34. Framed dam. stability upon the strength of its members. It consists of a water-tight deck AB supported by struts, trusswork, or buttresses at certain intervals along the length of the dam (perpendicular to the plane of the figure). The deck is always inclined so that the weight of the water upon it may hold the structure down and increase the factor of safety against sliding. 34. The Arch Dam. In the, case of a short high dam in a situation where firm support can be had from the walls on either side the arch dam is desirable. It is designed to withstand the water pressure by pure arch action and to transmit the pressures to the abutments at either end. The material in an arch dam is usually much less than in a pure gravity dam but any arch dam 36 HYDRAULICS acts to some extent as a gravity dam. Its analysis is not within the scope of this text. From a photograph by the author. FIG. 35. Lake Spaulding, Cal., variable radius arch dam. Ultimate height will be 325 ft. Bpillwaj Crest, EL 790 Btoney Gate8-0=: SIT ,1 s 3 ^^>W 7 .^s^%%! , \ ^^^?^:^Bagv\\x^:: *^^^-'*- -^ -/ csasaTOfiarV- " X ^.Downstream **C Approximate Line of Bed Bock 2, "West Side oUTewer East Side Spillway Ground Une Concrete Culrert FIG. 36. Section of Calaveras earth dam. 35. The Earth Dam. Under favorable circumstances the earth dam is a very economical type. A typical section of such APPLICATIONS OF HYDROSTATICS 37 a dam may be seen in Fig. 36. The slopes on both tne upstream and downstream faces are less than the angle of repose of the From a photograph by the author. FIG. 37. Upstream face of San Andreas earth dam. 90 ft. high. From a photograph by the author. FIG. 38. Incompleted Calaveras earth dam. Ultimate crest will be at dotted line making it the highest earth dam in the world. material used. In order to make such a dam water-tight it is provided with an impermeable core which may be a thin vertical wall of concrete or other material, or, as in Fig. 36, it may be 38 HYDRAULICS obtained by depositing fine earth under water. Fig. 38 shows the pool of water in the center of the dam where such a core is being formed. There is little mathematical analysis to be made for such a dam. The main problems are those of construction and careful selection of the materials employed. 36. Additional Notes on Dams. In most cases there are times when there is an excessive quantity of water that must be disposed of, usually by allowing it to flow over a spillway that is provided for that purpose. The spillway may be located at a different place from the dam so that no water ever overtops the latter as will be the case in Fig. 35. Again the spillway may From a photograph by the author. FIG. 39. Low dam at Ithaca, N. Y. occupy a portion of the crest of the dam as in Fig. 33 where the spillway can be seen in the middle. In other cases, such as in Figs. 36, 37, and 38, the spillways are located at one end of the dam and consist of rectangular canals through which the flood waters are discharged. But in Fig. 39 it may be seen that the entire crest of the dam is used for a spillway. This dam also shows the curved face that is provided to minimize the scouring effect of the waterfall upon the bed of the stream at the toe of the dam. For it must be recognized that water in falling over a dam acquires kinetic energy that must be expended in some way and unless suitable provision is made for this it may be expended in undermining the dam itself. APPLICATIONS OF HYDROSTATICS 39 37. Flashboards. In storing water by means of a' dam it is desirable to keep the water level as high as possible without flooding any lands upstream. If, therefore, the crest of the dam were located at the elevation allowable under normal conditions it would be excessively high in times of flood. In order to over- come this difficulty movable devices are employed called flash- boards, movable crests, amd various other names (Fig. 40). These are all schemes for increasing the height of the dam by equipment which can be removed when necessary. In some cases they work automatically, being either washed away when the water reaches a certain stage or caused to drop to a horizontal position. Other types require removal by hand in such emer- gencies. After the flood is past and the drier season comes on T \ FIG. 40. Flashboard. they may be replaced again. Some of these are entirely auto- matic in their action as in the case of the Stickney automatic crest outlined in Fig. 41. We have here two planes AB and BC rigidly connected and rotating about B. The water pressure on AB together with the weight of the shutters and the additional weight added at C tend to rotate the device in one direction but that is opposed by the pressure of the water on BC. By a suitable adjustment of area and weights it is possible to keep this crest in the position shown until the water reaches the level of A. Then the pressure on AB may be sufficient to cause it to drop to the position A'BC'. Hence the crest of the dam will then be reduced to the height of B, and the flood water will pour over the shutter B A f and hold it down. But when the excess waters have passed and the water level drops to B, or thereabouts, the pres- 40 HYDRAULICS sure on BC', no longer opposed by that on BA', will raise the crest to the initial position. ^^^^S^^^^//g^'siS?ij &^;4&jfc$$^ FIG. 41. Automatic dam crest. 38. PROBLEMS 1. The intake tower in Fig. 42 will be surrounded by water when the reservoir is filled and the outflow of water will take place through the openings provided in the tower. Assume one of these gates to be 3 ft. wide and 4 ft. high and to weigh 1,000 Ib. When the inside of the tower is sub- jected to the pressure of the air only, what vertical pull on a gate rod will be necessary to open the gate when the water stands 10 ft. above its top, if the coefficient of friction between the gate and its guides is 0.3? 2. The valve in Fig. 43 is 34 in. in diameter. If it is closed and under a pressure of 1,000 ft. of water on one side and atmospheric pressure on the other, what pull will have to be exerted on the valve stem to open it if the coefficient of friction is 0.4? APPLICATIONS OF HYDROSTATICS 41 From a photograph by F. H. Fowler. FIG. 42. Intake tower at Elizabeth Lake Reservoir on Los Angeles Aqueduct. From a photograph by the author. FIG. 43. A 34-in. high-pressure gate valve in the shop of the Pelton Water Wheel Co. 42 HYDRAULICS 3. Find the magnitude and point of application of the resultant pressure on the 2-ft. circular gate shown in Fig. 44. 4. The gate AB in Fig. 45 rotates about an axis through B. If the width is 4 ft., what torque applied to the shaft through B is required to keep the gate shut? 6. What value of 6 in Fig. 46 is necessary to keep the masonry wall from sliding? Masonry weighs 150 Ib. per cu. ft. and the coefficient of friction equals 0.4. Will it also be safe from overturning? If it has a factor of safety against sliding of 2, where will the resultant of the water pressure and its weight cut the base? FIG. 44. B FIG. 45. L FIG. 46. 6. In the framed dam shown in Fig. 47, the struts CD are placed 5 ft. apart along the dam (perpendicular to the plane of the figure). WTiat will be the load on each strut? What will be the value of the reaction at A? If the length BE is 4 ft. and the depth of the water flowing over the crest at E is 3 ft. what will be the load on the strut? 7. Assume the weight of the dam in Fig. 48 to be 150 Ib. per cu. ft., that there is no seepage of water under its base, and that the coefficient of friction between the dam and the material upon which it rests is 0.6. For 1 ft. T FIG. 47. '-H 6'r* 20^ H. FIG. 48. length compute : (a) Horizontal component of water pressure. (6) Vertical component of water pressure, (c) Weight of dam. (d) Is it safe against slid- ing? (e) Is it safe against overturning? (/) Where does the resultant of the water pressure and the weight of the dam cut the base? 8. In Fig. 40 the flashboard AB rests against a solid block at B but there is a pin at either end at A which is breakable. If the length of a section of flashboard is 6 ft., what must be the shearing strength of the pins if they give way when the water level reaches A ? APPLICATIONS OF HYDROSTATICS 43 9. In Fig. 41 what weight must be added at C per foot of len'gth in order that the crest may drop when the water level reaches A ? Neglect the weight of the rest of the movable crest, and assume EC =7.5 ft. 10. Figure 49 shows a cylindrical tank. What is the total pressure on the bottom? What is the total pressure on the annular surf ace A-A ? Find the maximum intensity of longitudinal tensile stress in side walls B-B: (a) If the tank is suspended from the top. (6) If it is supported on the bottom. Ans. 392 lb., 147 lb., (a) 20.8 lb. per sq. in., (6) 7.8 Ib. per sq. in. U-12Diam.-> T 12" M- 12" I N 24 Diam. H FIG. 49. 11. A pipe line 3 ft. in diameter is to carry water under a pressure of 1,000 ft. If the allowable tensile stress is 20,000 lb. per sq. in., what should be the thickness of steel used? 12. With the thickness of metal computed in the preceding problem what would be the tensile stress across a circumferential section if a valve was closed, the pressure on the other side of it being atmospheric? CHAPTER V HYDROKINETICS 39. Actual and Ideal Conditions. From the standpoint of pure mechanics the subject of hydrokinetics is rather unsatis- factory. This is due to the fact that so many assumptions are necessary, many of which are known not to be true. Thus in the greater portion of the work all particles of water in any cross- section of a flowing stream are assumed to move in parallel paths and with equal velocities. This is shown in Fig. 50, a o f A o ^^> "1 \ \ B 3 7 t 7 >7 *" O' C FIG. 51. particle of water at point moving along the axis of a pipe with a velocity OB. But every other particle of water across section 0'00 r is assumed to move with the same velocity giving us the velocity curve ABC, in this case a straight line. But it is well known that in a pipe the actual velocity curve is similar to ABC in Fig. 51, the velocity of a particle of water at O in the center tuwwwwwvu^w^^uu^w* of the pipe being OB while _^^ ,. ^s ^ na ^ f a particle near the /' ^""\ ^*L^_._^ '<>;* wall of the pipe is O'A. Ex- ^^_^,} ^^^-Jzj periment shows that in gen- ffffffffttffff era j 0j5 i s about twice the value of O'A and that the mean velocity of all the particles is about 0.84 OB. It is this mean velocity that is actually used in our computations. Hence our results are based upon a velocity which is possessed by only a few of the particles of water, the greater portion of them moving with either higher or lower velocities. We usually do not attempt to deal with the actual velocity curve ABC of Fig. 51 because we 44 FIG. 52. HYDROKINETICS 45 have no assurance as to its exact nature in every case 'and, if we did, our equations would be too complicated for practical use. But in Fig. 51 all particles of water have been assumed to be moving in straight lines parallel to the axis of the pipe, which we know is very seldom the case. In fact the path of a given par- ticle is very irregular as is shown in Fig. 52 and at the instant From a photograph by the author. FIG. 53. Showing vortices on surface of canal. in question a particle at point may be moving with some velocity OD. But in most practical problems we are concerned with OB which is the axial component of the true velocity. Thus not only do our equations ordinarily deal with a mean velocity, but they deal with a component of the true velocity. Instead of water flowing in parallel threads the true phenomena has been 46 HYDRAULICS very aptly compared to the motion of a cloud of feathers blown along by the wind. Water tends to travel in vortices as may often be observed upon the surface of an open stream such as the canal shown in Fig. 53. In this particular scene the water was flow- ing with a moderate velocity (about 3 miles an hour) over a reasonably smooth bed but the surface was covered with little vortices. Since actual conditions depart so widely from the ideal con- ditions assumed by our imperfect theory we can expect our theory to provide little more than a framework upon which may be hung the results of experimental investigation. The mean velocity at any section (strictly the mean axial component of velocity) is obtained by dividing the total rate of discharge by the total area of the section. That is V = q/F. EXAMPLES 1. Experiment indicates that the velocity curve ABC of Fig. 51 is approxi- mately a semi-ellipse and that OB is about twice O'A. Assuming this to be so, find the ratio between the mean velocity and the maximum velocity. (The total rate of discharge is f VdF and the value of this integral is the volume of the solid O'ABCO'. Dividing the solid by the area of the base, ?rr 2 , we should have the mean ordinate or in this case the mean velocity. The volume of an ellipsoid is two-thirds that of the circumscribing cylinder.) Ans. 0.833. 2. A stream is divided into five equal areas and the mean velocity of each portion is found by some method. These velocities are 3, 3, 4, 4, and 5 ft. per sec. What is the mean velocity of the entire stream? Ans. 3.80 ft. per sec. 3. Suppose that the areas are not equal but have values of 2.5, 2.5, 2.0, 2.0, and 1.0 sq. ft. while the velocities are 3, 3, 4, 4, and 5 ft. per sec., respect- ively. What is the total rate of discharge? What is the mean velocity? Ans. 36 cu. ft. per sec.; 3.60 ft. per sec. 40. Critical Velocity. The path followed or assumed to be followed by a single particle of fluid is called a stream line. It has been found that for very low velocities the stream lines are straight parallel lines as shown in Figs. 50 and 51, but that as soon as a certain velocity is exceeded the flow becomes turbulent or sinuous as in Fig. 52. The velocity at which the change occurs is called the critical velocity. The value of the critical velocity is affected by the temperature and also the size of the tube or pipe; the larger the latter the lower the critical velocity. For ordinary size pipes with which the engineer has to deal the critical velocity is so low that its value is of no interest. HYDROKINETICS 47 41. Steady Flow. By steady flow is meant that at any point in a stream all conditions remain constant with respect to time. This does not mean that the conditions at any one point are necessarily like those at some other point. Unsteady flow is met with in cases where change is taking place. Thus suppose a pipe line is flowing full of water and a From a photograph by the author. FIG. 54. The Los Angeles Aqueduct. valve is closed suddenly at its lower end. The velocity of the water would be brought to zero and in so doing there would be certain pulsations of pressure, which if violent enough would be recognized as water hammer. While such changes are in progress we should have unsteady flow. Again suppose that a gate is opened so as to admit water into an open canal originally empty. As the canal filled with water the level at any point would stead- 48 HYDRAULICS ily rise and also the velocity would in general be changing at all points. While such changes were under way the flow would be unsteady. But when equilibrium is finally established, the water level at any point and the velocity of flow across any section no longer vary from time to time and we then have steady flow. In the strictest sense of the word steady flow is seldom met with in ordinary engineering work as it would be found only with velocities below the critical velocity. For with all veloci- ties above the critical we have continual fluctuations of flow at any point due to the irregular motion of the individual par- ticles. It is for this reason that manometers or pressure gages attached to pipes, in which water is flowing, continually pulsate. Another evidence may be seen in Fig. 54, the dark band on either side of the water being where the latter has wet the concrete by wave action. For all practical purposes we disregard these slight fluctuations at individual points. If the average conditions over the entire section are reasonably constant with respect to time, we consider the flow as steady. While problems of unsteady flow are often problems of great practical value, especially in connection with the speed regulation of water power plants, they are rather difficult of mathematical treatment. Fortunately they are not as common as the more simple problems of steady flow. For the most part this text will be devoted to the latter. 42. Rate of Discharge. The volume of water flowing across any section per unit time is called the rate of discharge. It must not be confused with velocity, since it is the product of the cross- section area of the stream and the velocity of flow across the section. It may be expressed in various units such 'as cubic feet per minute, gallons per day, etc., depending upon the custom in that particular class of work. In the foot-pound second sys- tem of units such as are employed in this text it would naturally be in cubic feet per second. This is often called " second foot" for brevity and written as "sec. ft." 1 43. Equation of Continuity. In Fig. 56 it is apparent that the volume of water between any two sections such as (1) and (2) must remain constant if the flow is steady. Hence it follows that the rate at which water flows in at (1) must be equal to the rate at which it flows out at (2) , otherwise there would be a change 1 In irrigation work in India the term "cusec" has gained acceptance for this rate of discharge. HYDROKINETICS 49 in the volume contained between the two sections. Thus we may say that for steady flow, qi = g 3 . If the flow is unsteady this is not necessarily so. For suppose that the closure of a gate above (1) shut off the flow of water at (1), we would still find water flowing for a time past (2) though at the expense of the volume stored between the two sections. Hence in case of unsteady flow, where the volume in any distance is changing, the equation of continuity no longer applies. The equation of continuity states that for steady flow q = FiVi = F Z V 2 = . . . . . = FV = constant (20) This equation justifies the use of the term " rate of discharge " for the rate of flow across any section even though it be in the middle of a length of pipe or at some point in a river. For at some ultimate point the pipe or stream actually discharges in the usual sense of the word. And the rate of discharge at this point is equal, if the flow be steady, to the rate of volume flow at all sections throughout the stream. EXAMPLES 1. In Fig. 55 the portion of pipe between A and C is the frustum of a right circular cone with vertex at 0. If the rate of discharge is 10 cu. ft. per sec., what are the velocities at A, B, and C? Between A and C the velocity will vary as what function of the distance from 0? What shape of tube should be between A and C in order that the velocity may decrease uniformly with respect to distance? (Take origin at point where velocity would become zero.) FIG. 55. 2. The canal shown in Fig. 53 is 14.5 ft. wide and 4.2 ft. deep. If the velocity of the water is 3.5 miles per hour, what will be the rate of dis- charge in cubic feet per second? 3. The water in the canal of problem (2) finally flows down a steel pen- stock (Fig. Ill) which is 52 in. in diameter. What is the velocity of flow? 4. At the end of the pipe line in problem (3) the water is discharged through four nozzles the jets from which are approximately 7 in. in diameter. What is the jet velocity? 44. General Equation for Steady Flow. In the case of steady flow we may derive a very useful equation commonly known as 50 HYDRAULICS Bernoulli's theorem in honor of Daniel Bernoulli who proposed it in 1738. We shall make use of the principle of work and kinetic energy, and the following conditions will be assumed: (a) Flow is steady. (6) Fluid is incompressible. (c) Velocity across any cross-section is uniform. In Fig. 56 let A and B be any two cross-sections of a filament of a stream in steady flow. Suppose that during an infinitesimal time interval particles passing A and B move to A' and B f respectively. The pressure, elevation, velocity, and cross-section area between A and A' will be denoted by pi, Zi, Vi, and Fi respectively, while between sections B and B' these will be p 2 , 22, V 2 and F 2 respectively. Since the flow is steady and the fluid is incompressible, the volumes of water passing A and B during any time interval must be equal so that Fidsi FIG. 56. From the principle of work and kinetic energy, the net work done on the volume between A and B while it moves to the posi- tion between A' and B r is equal to the corresponding change in its kinetic energy. The net work done on the volume under consideration is the sum of three parts: (1) The work done by pressures normal to the external surface of the filament; (2) the work done by gravity; (3) the work done by frictional forces. In the first item it is necessary to consider only the work done by the pressures on the end cross-sections, since the side pressures do no work. These forces at A and B are wiFi and re- HYDROKINETICS 51 spectively and the displacements of their application points in the directions in which they act are dsi and ds z respectively. Hence the net work done by these forces is wp\F\dsi wpzFzds*. Since the location of the center of gravity of the portion of the filament between A' and B remains unchanged, the net work done by gravity during the time interval is equal to that due to the change of elevation of trie volume of water Fidsi from z\ to Zz. The net work done by gravity is thus wFidsi(zi z z ). The work done by friction will be neglected for the present. Since the kinetic energy of the portion between A' and B remains unchanged if the flow is steady, the whole change of kinetic energy is the difference between the kinetic energies of the parts between B and B r and between A and A'; that is 9 2 y z Combining all the work and energy terms in an equation and noting that Fidsi = F 2 dsz y 2 _ Y 2 Dividing both terms by wFidsi and rearranging, we have Zl This is Bernoulli's theorem, but, since all real fluids are viscous, it is impossible for flow to take place without fluid friction and hence a term should always be included to account for the energy converted into heat and hence lost. Experiment indicates that the friction loss is some function of the velocity and for the present we shall represent it as H' = kV n /2g, realizing that this is purely an empirical expression. Hence we may write the general equation as In case the velocity varies between (1) and (2) the V for the friction term might be taken as the average velocity, or by using 52 HYDRAULICS a suitable value of k it may be written as V\ or Fa- In practical work the difficulty of using equation (21) lies largely in estimat- ing proper values of k and n, and it is necessary to rely entirely upon experimental evidence. 45. Use of the Word "Head." Examining each term of equation (21) in detail we find: The term p indicates intensity of pressure expressed in feet of water, hence it is a linear quantity and indicates the height of a column of water necessary to pro- duce the given pressure. There may be no such real height of water in the problem, as in the case of a small volume of water enclosed within a cylinder and subjected to pressure by a piston. The quantity p is called pressure head. The elevation of a point above any arbitrary datum plane is indicated by z. It is a linear quantity and in our system of units it should be expressed in feet. It is called elevation head or potential head. The third ternvy 2 /2# may also be seen to reduce to a linear quantity when we analyze the units involved in V and g. The linear quantity equivalent to V 2 /2g is the height through which a body might fall in a vacuum from rest and acquire the velocity V. In many cases it is a purely artificial quantity in that there is no actual height in the figure illustrating the problem that gives any indication of its value. It may be called velocity head. Since all the other quantities in equation (21) are in linear dimensions, or feet in our system of units, it follows that kV n /2g must also be in feet. It may be called the lost head } and is represented by the letter H '. The sum of the pressure, elevation, and velocity heads at any section is called the total or the effective head at that section. However, the effective head, or any of the individual terms composing it, may be called "head" without any qualifying adjective. It is often convenient to let a single letter stand for the effective head, hence we may write H = P + z + |f (22) Using this brief notation we may rewrite equation (21) as H, - H' = H,' (23) We see that the effective head must decrease in the direction of flow by an amount H'. Hence, although either pressure, eleva- HYDROKINETICS 53 tion, or velocity may increase in the direction of flow, the sum of all three of them must continually decrease. Therefore an increase in one of these items must always be accompanied by a corresponding decrease in one or both of the other heads. EXAMPLES 1. Assuming a body of water at rest in Fig. 57, so that there is no loss of head, what are the values of the pressure head at A, B, C, and D? What are the values of the elevation head? What are the values of the effective head at these four points? 2. In Fig. 58 the point A is 30 ft. higher than B. Assuming the pipe to be of uniform diameter, in which FIG. 57. FIG. 58. direction will the water be flowing if the pressure at A is 20 Ib. per sq. in. and that at B is 40 Ib. per sq. in*? What is the head lost between the two points? What would the pressure be at A if the flow were to be in the opposite direction, the rate of discharge remaining the same? FIG. 59. 3. In Fig. 59 suppose 8 cu. ft. of water per sec. to be flowing from A to C. Assume a loss of head from A to B to be equal to 0.0017^ and an equal loss to occur between B and C. If the pressure head at B is 2 ft., how high will the water stand in piezometer tubes at A and C? 4. Neglecting all loss of head in Fig. 55, what kind of a curve would ex- press the variation of pressure from A to C? 54 HYDRAULICS 46. Energy and Power Meaning of Head. Suppose we multiply the elevation head z by the weight G of a definite volume of water. The product Gz being pounds times feet represents foot-pounds and we recognize it as potential energy. That is the body of weight G possesses Gz foot-pounds of potential energy by reason of its elevation z. 1 In like manner if we multiply the velocity head by G we have GV 2 /2g, which represents the kinetic energy of G pounds of water due to its mass and velocity. By analogy we might expect that if we multiplied the pressure head p by G we should also have Gp foot-pounds of energy which we could call pressure energy. But we here face the difficulty that we recognize energy in only two fundamental forms which we call potential energy or kinetic energy as the case might be. All other forms of energy may be reduced to one of these two. It is not clear that pressure energy can be reduced to either of these and so we have to seek further for an explanation of this term. Energy is ability to do work and we feel that water under pressure is capable of doing work. But if a particle of water should in some manner suddenly be disconnected from its fellows it would still have its initial elevation and velocity; these are qualities that it possesses in itself . But its pressure would be lost, since that is derived from contact with other particles. Thus water can do work due to pressure only so long as it is still connected with other particles. Hence we might conclude that pressure energy, if the term is permissible, is not something that a particle possesses but is merely energy that is transmitted from one to the other by virtue of the pressure and the motion. A good analogy to the way that energy may be transmitted past any point is offered by a belt connecting two pulleys. The belt possesses kinetic energy due to its mass and velocity and this energy is carried past any stationary reference point. But in addition to this the belt is under tension and is in motion and hence transmits energy from one pulley to the other that it does not in itself possess. In like manner in the case of a flowing stream of water, the water carries across any transverse section a certain amount of energy which it possesses in either the potential or the kinetic form or both. In addition to this it 1 Strictly speaking this energy is possessed by the system consisting of the earth and the water. HYDROKINETICS 55 may transmit energy across the section due to its pressure and motion. Suppose we consider a particle of water flowing from (1) to (2) in Fig. 56 and assume that there is no loss of head so that H = p + z + V 2 /2g = constant. In the case of a freely falling body acted upon by no other forces save gravity we should find that the loss of potential energy^ was compensated for by an equal increase in kinetic energy so that its energy remains constant. But if we assume that the stream of water is confined in a channel of uniform area the velocity is constant according to the equation of continuity, hence the kinetic energy cannot change. If the particle loses potential energy without any increase in its kinetic energy, it follows that its total energy must decrease. This is true for the particle. But the total energy in the system does not change for we are assuming no loss. The reason the velocity of the particle of water cannot increase is that negative work is being done upon.it by the pressure dp'dF (Fig. 56). But this negative work, although it reduces the energy of the particle, is not lost from the system. If conditions permit the particle of water to again ascend to a higher elevation or to increase in velocity the pressure acting on it will then do positive work and restore the potential or kinetic energy to it. The discussion may be concluded by stating that head repre- sents energy per unit weight of water. In the case of lost head H' represents the energy lost and dissipated in the form of heat per unit weight of water. Thus head may represent foot- pounds per pound of water. If we multiply the effective head H by the weight of water W flowing across any section per unit time, the product of the two will be energy per unit time. But energy per unit time, or the rate at which energy is transmitted, is power. In our system of units the product WH will be in foot-pounds per second. Thus head is equal to the energy per unit weight of water or it is equal to the power per unit rate of discharge. EXAMPLES 1. The surface of a lake is 500 ft. above a certain arbitrary datum plane with respect to which energy is to be measured, (a) What is the energy per pound of water? (6) If the lake is capable of furnishing 200 cu. ft. of water per sec. what power is available at the datum plane? 2. In a pipe line which is 24 in. in diameter we have water flowing with a velocity of 15 ft. per sec. under a pressure of 10 Ib. per sq. in. What power 56 HYDRAULICS is being transmitted through the pipe due to pressure? What is the total power delivered? Ans. 123.2 hp.; 141.8 hp. 3. A jet of water free from all pressure is 7 in. in diameter and has a veloc- ity of 250 ft. per sec. What is the horsepower? Ans. 7,390 hp. 4. A pipe line draws water from a lake and delivers it to a power house at a point 500 ft. below the level of the surface of the lake. The water is delivered at a velocity of 170 ft. per sec. by a jet 6 in. in diameter, and free from all pressure (save that of the atmosphere). What horsepower has been lost in the pipe line? 47. Correct and Incorrect Applications of Bernoulli's Theorem. Bernoulli's theorem states that along any stream line the effective head remains constant. But in a real fluid which is viscous there can be no flow whatever without some loss due to friction. Hence the correct statement is that along any stream line the effective head always decreases in the direction of flow. It should be emphasized that the general equation should be written only between two points in the same stream line so that a particle of water may be assumed to flow from one point to the other. If there were no loss of energy it would follow that the effective head is constant at all points throughout a connected body of fluid and in that event only we might apply the equation to any two points whatever. But in reality this would lead us to incorrect conclusions as the following will show. Suppose that we have no loss due to friction; it would then follow that we should have all particles of water at a section moving with equal velocities in parallel paths as shown in Fig. 50. All particles of water would have the same amount of kinetic energy and it is clear that all particles of water through the section would have the same amount of energy. But in reality the velocity across any section of a circular pipe, for example, is like that in Fig. 51. This is due to the fact that the greater frictional resistance near the walls of the pipe has retarded the water near them. Certain persons have incorrectly applied Bernoulli's theorem between a point near the wall and a point at the center of the pipe and reasoning that there could be no loss between two such points come to the conclusion that the pressure is less at the center of the pipe than it is near the wall because the velocity head is higher. This would lead to an excessive pressure difference if true. This reasoning has even been bol- HYDROKINETICS 57 stered up by claims of experimental evidence, but in reality the data were inaccurately determined. It is no more permissible to apply the general equation be- tween two points in adjacent stream lines than between two separate streams in different channels. It is a mistake to assume that the effective head is constant across any section. Correct experimental evidence shows that the pressure head across any section varies only according to the depth, the same as in the case of water at rest. Hence if the sum of pressure head and elevation head is constant across any section while the velocity head varies, it follows that the total head varies at different points in the section. This is in harmony with a correct appli- cation of Bernoulli's theorem along the different assumed stream lines. If all particles started with the same store of energy it is clear that those near the pipe wall would have lost more by friction than those near the center. Hence the energy of those particles near the pipe wall should be less than that of those in the center. In practical application of the general equation we do not deal with stream lines but with entire streams. Hence we have for H the average head across the section and for V the average velocity across the section. But we do equate the average head at some section of a stream to the average head at some other section of the same stream. In considering an entire stream, rather than a single stream line, we assume the kinetic energy per unit time to be WV 2 /2g, where V is the average velocity. This is not strictly true for, if the velocity varies from point to point over the section, the kinetic energy is the sum of the kinetic energies of all the in- dividual particles. Considering an elementary area dF, the flow through it will be wV'dF, where V is the actual velocity at the point in question. The kinetic energy of the elementary stream would be wV' 3 dF/2g. Hence the total kinetic energy for the entire stream is -' (v 20 J l W. (24) If the velocity is constant this will become wV*F/2g = WV 2 /2g since the true velocity at every point is the average velocity for the section. But in reality the velocity does vary to some extent over the section and hence (24) gives the true kinetic 58 HYDRAULICS energy. If the law of variation of V throughout the section is known this integral can be evaluated, but in any event it can be shown that the kinetic energy so obtained is greater than that computed by using the average velocity. 1 Thus making the assumption that the velocity in the center of a circular pipe is twice that near the walls and that the velocity curve is a semi- ellipse it will be found that the true kinetic energy is 1.06 times that based upon the mean velocity. Fortunately the difference is not great in important cases met with in practice. Thus in the case of a jet of water from a good nozzle, where there is little variation in velocity, the difference may be a matter of about 1 per cent, only. 2 A correct application of equation (21) would require us to insert some factor before the velocity head, based upon the average velocity, to give a correct value. But if the velocity curves at sections (1) and (2) are similar and the velocities nearly the same in value the error in one may nearly balance that in the other. Hence it is not customary to allow for this discrepancy between the true kinetic energy and that computed by using the average velocity. EXAMPLE 1. Assume that in a rectangular stream the velocity of the water is uni- form from side to side at any depth but that it varies from the top to the bottom inversely as the depth. If the velocity at the top is twice that at the bottom find the ratio between the true kinetic energy passing a section per unit time and that based upon the mean velocity. Ans. 1.11. 48. Applications of General Equation. For the solution of problems in hydrokinetics we have two fundamental equations, the equation of continuity (20) and the general equation for steady flow (21), usually known as Bernoulli's theorem. In most cases the following procedure may be employed: 1. Choose a datum plane through any convenient point. 2. Note at what sections the velocity is known or assumed. If at any point the cross-section is great as compared with its value elsewhere, the velocity will be so small that the velocity head may be disregarded. 3. Note at what points the pressure is known or assumed. In a body of water at rest with a free surface the pressure is !L. M. Hoskins, "Hydraulics," page 119. 2 W. R. Eckart, Jr., Inst. of Mech. Eng., Jan. 7, 1910. HYDROKINETICS 59 known at every point. The pressure in a jet is the Bame as that in the medium surrounding the jet. 4. Note if there is any point where the three items of pressure, elevation, and velocity are known. 5. Note if there is any point where there is only one unknown quantity. It is generally possible Jo write equation (21) between two points such that they fulfill conditions (4) and (5) respectively. Then the equation may be solved for the one unknown. If it is necessary to have two unknowns then equation (21) must be solved simultaneously with equation (20). The procedure is best shown by applications such as the following: In Fig. 60 we have a pipe BCD which is 6 in. in diameter through which water flows from reservoir A. The diameter of the stream discharging freely into the air at E is 3 in. Let us assume that n = 2 in equation (21) so that the loss of head due to friction is proportional to the square of the velocity. Then H' = kV 2 /2g, where V is the velocity in the pipe. Suppose that the roughness of the pipe and the lengths between the various points are such that the values of k from the reservoir to B, from B to C, from C to D, and from D to E are 2, 4, 4, and 1 respectively. Let it be required to find the pressure at C when flow takes place. At C there is both an unknown pressure and an unknown velocity, hence we cannot immediately apply equation (21) as one equation is capable of determining only one unknown. Let us then follow the procedure outlined. The location of a datum plane is immaterial in the solution of the problem but it is usually convenient to take it through the lowest point in the figure and thus avoid negative values of z. Therefore let us assume a datum plane through E. In the reservoir we find that the velocity is negligible because of the large area as compared with 60 HYDRAULICS the area of the pipe. At a point A on the surface of the water we find the pressure to be atmospheric, which is also the case with the stream at E. Thus whatever the pressure of the atmos- phere may be its effects can easily be shown to balance out and therefore we neglect it altogether. Hence at A we find that everything is known while at E the velocity head is the only unknown. We shall apply equation (21), or its equivalent (23), between points A and E. We find that # A =0 + 40 + # E = + + F /2 /2<7 #' A _E = 11 F 2 /2 the area of the nozzle opening being 43.02 sq. in. Com- pute the coefficients of velocity, contraction, and discharge using the values of Hi and Vz given. What is the efficiency of the nozzle? What is the horsepower in the jet? 2. What is the value of the head lost in hydraulic friction in the nozzle of Fig. 63? What is the value of kl 3. The velocity of water in a 6-in. pipe is 12 ft. per sec. At the end of the pipe is a nozzle whose velocity coefficient is 0.98. If the pressure in the pipe at the base of the nozzle is 10 Ib. per sq. in., what is the velocity of the jet? What is the diameter of the jet? What is the rate of discharge? 4. A jet 2 in. in diameter is discharged through a nozzle whose velocity coefficient is 0.98. In the pipe at the base of the nozzle there is a pressure of lOlb. persq. in., the diameter of the pipe at that point being 6 in. What is the velocity of the jet? What is the rate of discharge? 6. If the diameter of the jet in problem (4) were 1.0 in., all other data remaining the same, find the jet velocity and the rate of discharge. 69. Venturi Meter. If water is caused to flow through the device shown in Fig. 76, the increased velocity through the "throat" will produce a corresponding pressure drop. This drop in pressure may be made to serve as a measure of the rate of discharge. Such an instrument is called a Venturi meter. FIG. 76. Venturi meter. It may be seen that the Venturi meter is very similar in prin- ciple to the nozzle. In both there is an increase in velocity of the water accompanied by a corresponding pressure drop. And in both the rate of discharge may be found to be a function of the pressure drop. The only difference is that the pressure at the throat of the Venturi meter may be either somewhat greater or less than atmospheric, and the stream at that point is not a free jet but is expanded again to fill the pipe below the meter. Hence the APPLICATIONS OF HYDROKINETICS 75 equations for the nozzle would seem to apply directly to the Venturi meter, with pi equal to the pressure drop in both cases. Actually the coefficients for the Venturi meter are based upon a formula derived by a slightly different procedure. Thus we equate HI to H z assuming that H' is zero, and then intrpduce a coefficient as the very last step. Assuming the meter to be hori- zontal so that Zi = z Z) we have Pi -h K From this, - = -h By the equation of continuity Vi = (F 2 /Fi) V 2 , and hence V 2 - Since there is some slight loss of head between (1) and (2) the true velocity will be less than this and so we multiply it by a velocity coefficient. We then have (35) This may be seen to differ from equation (33) in that the term (Fz/Ftfis not multiplied by c v 2 , but both (33) and (35) could be made to yield the same numerical value by using a slightly different value of c v for the two. Custom has based values of c v upon (35) for the Venturi meter and upon (33) for the nozzle. With the Venturi meter we desire q, not F 2 , and hence, multi- plying (35) by F 2 and replacing c v by c, we have For a given meter F\ and F 2 are known quantities and, if K r = F 2 V^g/Vl - (F z /Fi)\ this may be reduced to q = cK'Vh (37) The coefficient c may be assumed to be 0.985 for a new meter and 0.980 for an old one, the interior of which will be slightly rougher and perhaps reduced in area through incrustation. These factors will give a result that is very accurate. The coefficient is practically a constant, though there is some slight reason to 76 HYDRAULICS believe that it increases slightly with higher rates of discharge. If c is assumed constant for any given meter, it is convenient to replace cK' by K and we then have q = K\/h (38) The Venturi meter, invented by Clemens Herschel in 1886, affords a most valuable and accurate means of measuring water, especially in large quantities. 1 By a suitable recording device Courtesy of Builder's Iron Foundry. FIG. 77. Venturi meter in wood pipe line. it is possible to make a continuous record of the flow of water through any pipe line in which a Venturi meter is installed. The sole objection to its permanent use in a pipe is that it must neces- sarily cause some slight friction loss or resistance to flow. If this loss be expressed as H r = kV z 2 /2g, we find that values of k range from about 0.1 to 0.2. The higher values of k naturally go with smaller values of F 2 /Fi. The usual ratio of the diameter of the throat to the diameter of 1 Trans. A. S. C. E., vol. 17, page 228 (1887). APPLICATIONS OF HYDROKINETICS 77 the pipe is about 1 to 3, making the ratio F z /Fi = 1/9. But in order to reduce the resistance as much as possible and also to avoid producing -pressures at the throat below atmospheric, it is quite common to make the diameters in the ratio of 1 to 2, making Fz/Fi = 1/4. Of course this reduces the magnitude of h for a given rate of discharge and hence makes the readings less accu- rate, especially for very low Discharges. Courtesy of Builder's Iron Foundry. FIG. 78. Venturi meter of riveted steel. A diverging stream is always less stable than a converging stream, that is it is more readily broken up into whirlpools and eddies, and hence more loss of energy takes place in the portion of the meter on the downstream side of the throat. In order to minimize this the down stream portion is made to taper much more gradually than the upstream side. EXAMPLES 1. A Venturi meter with a 4-in. throat is to be used in a 12-in. pipe line. Assuming a value of c = 0.985, determine the value of K for this meter. 2. If a differential manometer employing mercury (sp. gr. = 13.57) were to be used, determine the value of K for the Venturi meter in problem (1), replacing h by y (Fig. 13) in inches of mercury. 3. Suppose the throat of the meter in problem (1) were to be 6 in. the pipe remaining 12 in. Compute the value of K. 4. Suppose that 5 cu. ft. per sec. is flowing tnrough the Venturi meter. What are the values of h in problems (1) and (3)? 78 HYDRAULICS 5. Water flows through a pipe line 6 ft. in diameter with a velocity of 7 ft. per sec. In this pipe line is installed a Venturi meter with a throat diameter of 2 ft. Assuming the value of k to be 0.12, what will be the loss of head caused by the meter? What will be the power lost? 60. Large Vertical Orifice. In the case of an orifice whose vertical dimensions are large as compared with its depth below the free surface it is necessary to proceed as follows : Choose an elementary area dF such that all portions are at the same depth z below the free surface. ,Now by Art. 52 the rate of discharge through this strip may be expressed as dq = cV2gz dF (39) The rate of discharge through the entire orifice may be obtained by integrating equation (39) . Thus *dF (40) - t i h f i f i 1 ~7^ 1 1 f t A 2 . i , I " -e 2H > 1 i 1 2H \ FIG. 92. Limiting proportions of standard contracted weirs. made greater than the values given, but never less. The height of the crest above the bottom should preferably be at least 3H. Francis estimated that a height of 2H would increase the dis- charge about one per cent. If sufficient space cannot be had to secure perfect end contraction, the end contractions should be entirely suppressed, or one of them suppressed and all the space given to the other end. There is no known coefficient or method for dealing with imperfect contraction. Also the value of H should not exceed one-third of the width b. For the suppressed weir there are no standard dimensions to be observed. Experiments at Cornell University have indicated that the Francis coefficient in equation (47) is applicable for heads up to at least 5 ft. and for heads down to 0.3 ft. For H = 0.2 ft. it should be increased 3 per cent., and for H = 0.1 ft. it should be increased 7 per cent. If the area of the channel of approach exceeds QbH it can be shown that the velocity of approach is negligible. Hence it is APPLICATIONS OF HYDROKINETICS 89 seen that velocity of approach need not be considered in a con- tracted rectangular weir. But with a suppressed weir the depth of the channel would have to be QH, and that is not often the case. Hence most suppressed weirs have a velocity of approach that needs to be considered. The Francis coefficient was based upon work with weirs having a velocity of approach less than 1 ft. per sec. When the velocity of approach is high, the formula of Bazin should be applied. The most accurate type of weir is a suppressed weir with such a deep channel of approach that the velocity of approach is neg- ligible. A contracted weir for which the velocity of approach is negligible is about in the same class with a suppressed weir with a moderate velocity of approach. End contractions have been held to be a source of error and there appears to be no truly rational way to correct for them. The least desirable type of weir is the one with a high velocity of approach because of the difficulty not only of reading H accurately but also of allowing for the effect of this velocity in a scientific manner. It almost goes without saying that a weir should be set with its crest level and its plane vertical. An inclination upstream decreases and an inclination downstream increases the dis- charge for a given H. The crest should be sharp and in good condition. In using weirs for accurate work it is desirable to study the original experiments upon which the formulas are based and use the formula that has been derived under circumstances most nearly like those in hand. And it is likewise desirable to dupli- cate the original investigator's methods. The hook gage, for instance, should be located in the same way and at the same distance from the weir. Unless these precautions are followed one has no assurance that the coefficients given fit his own case. 67. The Cippoletti Weir. In order to avoid the trouble of correcting for end contractions, the sides of the Cippoletti weir are given such a batter (1:4) that they add enough to the effect- ive width of the stream to offset the contraction 0.2H of the con- 1 For information on weirs see: "Weir Experiments, Coefficients, and Formulas," by R. E. Horton, U. S. G. S. Water Supply and Irrigation Paper, No. 150, Revised No. 200. Hughes and Safford, "Hydraulics." Parker, "The Control of Water." 90 HYDRAULICS traded Francis weir. Thus computations may be made upon the basis of the width b at the crest by the following formula q = 3.367&#* (50) 68. Special Weirs. There are other types of weirs for special purposes. Thus we have a floating circular intake weir in Fig. 93. The purpose of this is not for accurate water measurement, but rather to float in such a manner in the reservoir that a certain From a photograph by F. H. Fowler. FIG. 93. Floating circular intake weir for Los Angeles aqueduct. depth of water continually flows through or over its passageways and down the intake in the center. Thus no matter how the water level in the reservoir rises and falls a uniform quantity will automatically be delivered to the intake. 69. The Pitot Tube. Among other water-measuring devices is the Pitot tube. This is an instrument which indicates the velocity of water at a point. From the velocity the rate of dis- charge may be obtained. The principle of the Pitot tube is illustrated in Fig. 94 and its theory will be discussed in a subsequent chapter. For an open stream only a single tube is necessary, but in a stream of water under pressure a second tube is necessary to record the pressure APPLICATIONS OF HYDROKINETICS 91 alone. The quantity desired is the difference between the two readings, which we shall call h. It can be shown by correct theory that if h is the value in feet of water of the dynamic pres- sure exerted by the impact of the stream against the opening of the tube, the velocity of the water is given by V = V2gh (51) T _i FIG. 94. This has been found by experiment to be true when there is smooth stream line flow, but in case of turbulent flow we should introduce a coefficient whose value is about 0.977, so that we should write V = 0.977V2gh (52) The fact that this coefficient is anything less than unity is not because our theory is at fault nor because of any defect in the in- strument itself, but is due to the fact that the instrument records the true velocity at the point while we desire, for practical pur- poses, the axial component of velocity. Hence the factor is designed to give us the axial component of velocity (OB in Fig. 52 rather than OD). 1 In using the Pitot tube it is often convenient to divide a cross- section of the stream up into parts of equal area and to determine the velocity in the center of each area. The average velocity of 1 L. F. Moody, " Measurement of the Velocity of Flowing Water." Proc. of the Engineer's Soc. of W. Penn., vol. 30, page 319 (1914). 92 HYDRAULICS the stream will be the average of the observed velocities. But if the areas are not equal the average of the velocities will have no significance. It will then be necessary either to plot a curve from which velocities at other points may be taken or to multiply each observed velocity by the area which it may be assumed to represent. The total rate of discharge of the entire stream is the sum of all such partial discharges. Thus q = ZF'V (53) where F f is a portion of the total area and V is the velocity through that area. If the average velocity is desired, it can be obtained by dividing the rate of discharge by the total area. 70. The Current Meter. For moderate velocities such as are found in canals and natural streams the current meter is well adapted. It consists of a wheel, -as in Fig. 95, or in other types a screw, which is rotated by the action of the water. By calibra- tion the relation is determined between the velocity of the water and the rate of rotation of the meter. In many current meters each revolution is recorded by a click in a telephone receiver at the ear of the observer, the click being produced by the wheel making an electric contact every revolu- tion. In most meters the contact is not made so frequently, every ten revolutions being the number commonly recorded. Other types of meter have some form of mechanical recording device. It is generally better to determine the time necessary for a given number of revolutions rather than to attempt to find the number of revolutions made in some specified time, owing to the difficulty of estimating fractions of a revolution or frac- tions of the number of revolutions that may be recorded as a unit. Current meters may be roughly divided into two classes, those with the axis vertical, as in Fig. 95, and those with the axis horizontal. In comparatively shallow water the meter may be rigidly fastened to a rod, and in this case the weight and tail, as shown in Fig. 95, are unnecessary. But for deeper water where the meter is suspended by a cable the latter are required to hold the meter in the proper position. Generally it is desired to find the velocity of the water flowing across some sectional area. If the stream lines are not perpen- dicular to the area in question, it is the normal component of the APPLICATIONS OF HYDROKINETICS 93 velocity that is desired rather than the value of the Actual veloc- ity. It may be seen that the type of meter shown in Fig. 95 will rotate with equal velocity no matter from which horizontal direction the water may come. It will also be rotated by a cur- rent that is vertical, or parallel to its axis. And in any case the Courtesy of W. & L. E. Curley. FIG. 95. Current meter. rotation is always in the same direction. Thus this meter tends to record the value of the velocity regardless of its direction. In other types, generally with the axis horizontal and the wheel made in some form similar to a screw propeller, the meter records only the component of the velocity parallel to its axis. And in case the meter is located in a portion of the stream where an 94 HYDRAULICS eddy causes a reverse current the meter will then give a negative reading, since it will be rotated in the opposite direction. Such a type of meter is more accurate in all cases where the flow is irregular or turbulent. However, the type shown in the figure is of excellent mechanical construction and is widely used. For many cases where the stream flow is fairly regular and extreme accuracy is not required, it is quite satisfactory. In using the current meter the velocities are determined at a number of different points and the total discharge of the entire stream computed in the same manner as in Art. 69. 71. Comments on Measurement of Water. The accurate measurement of rate of discharge is one of the most difficult problems in practical hydraulics. The only positive way of measuring rate of discharge is to weigh the amount of water dis- charged in a given time or to determine its volume in suitably calibrated tanks or reservoirs. The former method is applicable only for relatively small rates of discharge, and facilities for the latter are seldom to be had. Also in the latter method the effect of leakage, evaporation, and other factors may sometimes prove troublesome. The methods that are usually employed are the ones that have been given in this chapter. They are all indirect in that we as- sume the velocity or the rate of discharge to be a function of some other quantity which can be measured. The discharge of water from any tank can be measured by an orifice, tube, or nozzle. When a stream of water flows in an open channel it may be caused to flow over a weir or its velocity through- out any cross-section may be found by a current meter, by floats, or other means. For a stream of water confined within a closed pipe we may use a Pitot tube to determine the velocity across a cross-section or cause the water to flow through a Venturi meter. At the end of a pipe line we might place a nozzle which would also permit the rate of discharge to be obtained. ^ The discharge from a nozzle may be computed or it may be measured directly by determining the velocity of the jet with a Pitot tube. The means of measurement that is to be used depends upon the circumstances. In addition to the methods of measurement that have been described in this chapter, there are other methods, especially chemical methods. One of these is simply a matter of discharg- ing a small quantity of highly colored liquid into the intake of a pipe line and noting the time that it takes for the discoloration to APPLICATIONS OF HYDROKINETICS 94a be noted at the other end. Knowing the length o*f pipe it is easy to compute the velocity of the water. Another valuable method consists of adding a strong salt solution at a known definite rate. Samples of water are taken at a down- stream section and analyzed. Knowing the strength of the solution used, its rate of discharge, and the amount of dilution in the main stream the rate of discharge in the latter may be determined This method has been used in some cases with a high degree of accuracy and it may offer an easy, cheap, and convenient way of measuring rate of discharge of large quantities of water. 1 Where water flows over a spillway dam the latter may be used as a special type of weir. The same weir formula as given in equation (46) may be applied, if the proper value of the coefficient is known. Since the spillway crest may be of various shapes and dimensions, it is not a standard piece of apparatus like the sharp crested weir. Hence the value of the coefficient has to be deter- mined for each case either by calibrating the spillway in question or using the results of observations upon another spillway of similar form. 72. Discharge under Varying Head. If the head varies, the rate of discharge will likewise vary and the total discharge in a given time, or the time required for a given total discharge, must be determined as follows. Let Q = the total volume in cubic feet of any given body of water, while q = cu. ft. per second as usual. Then q = dQ/dt or dQ = qdt. Suppose that into this body of. water in question there is an inflow at the rate of qi cu. ft. per second, while water flows out at the rate of ^=^ D> an d hence we should determine the average velocity in each vertical line. This might be done by taking a num- ber of observations so that curves similar to that in Fig. 132 could be plotted. But a study of a number of such curves has shown that in general the average velocity in a vertical line is found at about 0.6 the depth. Hence if the current meter be set at that depth, the velocity determined by it may be assumed to be the mean velocity. Of course this is only an approximation. To insure a higher degree of accuracy than a single observation could give, measurements are often taken at 0.2 the depth and 0.8 the depth. The mean of these two values will be approximately the average velocity. Thus, in an actual stream gaging, observa- tions would be made at the points indicated by the circles in ^Tj^^r] Average Velocity ^TuT^IL FIG. 132. 138 HYDRAULICS Fig. 131. Further details of this topic are not within the scope of this text. 1 Sometimes floats are used but such procedure is less accurate. However they are often applicable when other methods are not feasible, such as during floods. If surface floats are used, the average velocity may ordinarily be taken as about 0.9 that of the surface velocity. But the velocity at the surface is greatly affected by the wind. 102. Rating Curve. If a natural stream is to be used for water supply or power purposes, it is necessary to determine the amount of water it can be depended upon to furnish. Since the flow will usually be subject to wide fluctuations during a long period of time it is necessary to make an extended series of observations upon it. Rate of Discharge FIG. 133. Rating curve. The level of the surface of the water in a stream is called the gage-height, and may be measured above any arbitrary point. Thus the gage-height does not necessarily coincide with the depth of the stream. It is apparent that for a given stream, the rate of discharge will be a function of the gage-height. If the rate of discharge of the stream be determined for several gage-heights a curve, such as in Fig. 133, may be constructed. This curve is called the rating curve, and from it the value of q for any height of water can be obtained. Thus in making a study of the stream it is necessary to make only a record of the gage-heights. From the rating curve the quantity of flow can then be determined. This gage-height might simply be read and recorded once a day by an observer, or by means of a float and clockwork a continuous record could be 1 See Hoyt and Grover, "River Discharge." UNIFORM FLOW IN OPEN CHANNELS 139 obtained which would show all the variations in the- flow. Fig. 87 shows such a gaging station. 103. PROBLEMS 1. A circular conduit of smooth cement is exactly half full of water. The diameter is 4 ft. and the slope is 1 ft. per 10,000 ft. Compute the rate of discharge by the formulas of KutJ;er, Manning, and Bazin. Ans. q = 8.38, 8.51, and 8.92 sec. ft. respectively. 2. A rectangular flume of timber slopes 1 ft. per 1,000 ft. Compute the rate of discharge if the width is 6 ft. and the depth of water 3 ft. Ans. 114'sec. ft. 3. What would be the rate of discharge in problem (2) if the width were 3 ft. and the depth of water 6 ft. ? Which of the two forms would require less lumber? 4. A rectangular channel of rubble ma- sonry is 6 ft. wide, the depth of water is 3 ft., and the slope of 1 ft. per l r OOO ft. Com- pute the rate of discharge and compare with that in problem (2). Ans. 65 sec. ft. 6. A semicircular channel of rubble ma- sonry with a slope of 1 ft. per 1,000 ft. will give what discharge when flowing full if its diameter is 6.55 ft.? Compare the cross- section areas and amounts of lining required with that in problem (4). Ans. 65 sec. ft. 6. A circular conduit of concrete (n = 0.012) is 10 ft. in diameter and slopes 1.6 ft. per 1,000 ft. (See Fig. 134). The following table gives values of wetted perimeter and area of water cross-section for various depths of water in the conduit. Find values of V and q for the various depths in the table. What value of y gives the highest velocity? What value of y gives the highest rate of discharge? FIG. 134. Depth, y Wetted perimeter Area, F m Vm c V Q. 1.0 6.44 4.09 3.0 11.59 19.82 5.0 15.71 39.27 8.0 22.14 67.36 9.0 24.98 74.45 9.5 26.91 77.07 10.0 31.42 78.54 7. The amount of water to be carried by a canal excavated in firm gravel is 370 sec. ft. It has side slopes of 2:1 (horizontal component is two times vertical component) and the depth of water is to be 5 ft. or less (Fig. 135). 140 HYDRAULICS If the slope is 2.5 ft. per mile, what must be the width at the bottom? (This problem can best be solved by trial.) 10 FIG. 135. CHAPTER IX HYDRODYNAMICS 104. Dynamic Force Exerted by a Stream. Whenever the velocity of a stream of water is changed either in direction or in magnitude, a force is required. By the law of action and reaction an equal and opposite force is exerted by the water upon the body which produces this change. This is called a dynamic force in order to distinguish it from static forces due to the pressure of the water. AV Let the resultant force exerted by any body upon the water be denoted by R and its components by R x and. R y . Let dR be the force exerted upon the elementary mass shown in Fig. 136. Then, since resultant force equals mass times acceleration, dV dR = dm dt The summation of the forces acting upon all such elementary masses along the path will give the total force exerted upon the water by the entire body at any instant. But it is necessary to express dm as a function of V or vice versa before this can be integrated. Let the time rate of flow be dm/dt. Then in an interval of time dt there will flow past any section the mass (dm/dt)dt, which will be the amount considered. Hence we may write But (dV/dt)dt = dV. Our discussion here shall be restricted to 141 142 HYDRAULICS the case where the flow is steady in which case dm/dt is constant and equal to W/g. Therefore dR = dV. In general these various elementary forces will not be parallel and, since integration is an algebraic and not a vector summa- tion, it is necessary to take components along any axes in order to integrate the above. Thus Q Now at point (1) the value of V x is FI cos a\ and at (2) it is F 2 cos 2 . Inserting these limits and noting from Fig. 136 that F 2 cos 2 Vi cos ai = AF X we have W W R x = (Vz cos 2 Vi cos ai) = AF* 9 9 If P indicates the value of the force exerted by the water, which is equal and opposite to R, we shall have W W P x = (Vi cos ! - F 2 cos ,) = - - AF X (81) y y In similar manner the y component of P will be W W P v = (Vi sin ai - F 2 sin 2 ) = - AF, (82) / J/ Since P = \/P* 2 + P y * and AF = \/AF x 2 + AF V 2 , the value of the resultant force is P= AF (83) The direction of 72 will be the same as that of AF and the direc- tion of P will be opposite to it. It is because P and AF are in opposite directions that the minus sign appears in the last terms of equations (81) and (82). Note that AF is the vector difference between V\ and F 2 . 104a. Dynamic Force (Second Method). The preceding derivation pictures the total dynamic force exerted by a flowing stream to be the vector sum of all the elementary forces exerted along its path at any instant. The following derivation makes it clear that the total force depends solely upon the initial and terminal conditions and is independent of the path. (Of course the numerical value of the terminal velocity would be affected by friction losses which might be different for different paths.) The former method is based on the principle that resultant force equals mass times acceleration. The second method is based on H YD ROD YNA MICS 1 42a the principle of force and momentum, which may be stated as follows: The time rate of change of the momentum of any sys- tem of particles is equal to the resultant of all external forces acting on the system. Thus instead of R mdV/dt we write R = d(mV}/dt. Consider the portion of a filament of a stream in Fig. 136a which is between two cross-sections A and B at the beginning of a time interval dt, and between the cross-sections A' and B' at the end of the interval. Denote by dsi and ds z the distances moved during the interval by particles at A and B at the begin- t . / FIG. 136a. ning. Let FI be the cross-section area at A, Vi the velocity of the particles, and i the angle between the direction of Vi and any convenient x axis. Let the same letters with subscript (2) apply to B. At the beginning of the interval the momentum of the portion of the filament under consideration is the sum of the momentum of the part between A and A' and that of the part between A' and B. At the end of the interval its momentum is the sum of the momentum of the part between A' and B and that of the part between B arid B r . In the case of steady flow the momen- tum of the part between A' and B is constant. Hence the change of momentum is the difference between the momentum of the part between B and B' and that of the part between A and A'. Noting that wF^Si = wF 2 ds 2 , since the flow is steady, the 1426 HYDRAULICS change in the x component of the momentum during dt is then t cos 0:2 K i cos If the rate of flow be denoted by W, then wFidsi == Wdt and the time rate of change of the x component of the momen- tum is W - (F 2 cos 2 Fi cos i). Denoting by R 2 the x component of the resultant force which changes the momentum, W W R x = (Vz cos 2 Vi cos i) = AF X . Q 9 From this point the treatment is the same as the last paragraph of the preceding article. This method has the advantage that it may readily be extended to the case where the flow is unsteady, if desired. 104b. Dynamic Action upon Stationary Body. In order to v find the dynamic force exerted by a stream upon a stationary object, we have merely to find the value of AF, assuming the rate of dis- charge to be known. The follow- FIQ. 137. m g special forms of equations (81) and (82) will often be more convenient. If the x axis be taken parallel to FI and the angle between V\ and F 2 be denoted by 6, Fig. 137 (0 = a 2 ai and cos i = 1.0), W W P x = - - AF X = (Vi - Vz cos 0) (84) y y W W P y = - AF, = - F 2 sin 6 (85) y y In certain special cases a stream will be equally divided so that the P y for one half will be equal and opposite to the P y for the other half. Hence in this special case P = P x . It may be noted also that friction in flow over the body tends to decrease the numerical value of F 2 . This increases the value of P x if is less than 90 but decreases it if is greater than 90. EXAMPLES 1. In Fig. 137 assume that 6 = 60, and that the stream striking the body is a jet 2 in. in diameter with a velocity of 100 ft. per sec. If the f rictional loss is such as to reduce the velocity of the stream leaving the HYDRODYNAMICS 143 body to 80 ft. per sec., find: (a) the component of the force ^n same direc- tion as the jet, (6) the component of the force normal to the jet, (c) the magni- tude and direction of the resultant force exerted by the water. Ans. (a) 254 Ib. (6) 293 Ib. (c) 388 Ib. at 49 08' with direction of jet. 2. Suppose the jet in problem (1) struck a flat plate normally, what would be the value of the force exerted upon the plate? Ans. 423 Ib. 3. Suppose the jet in problem (1) were completely reversed in direction, or that 6 = 180. If ,F 2 were lOOfft. per sec., what would be the component of the force in the same direction as the jet? (Compare with problem (2).) What would be the component normal to the direction of the jet? Ans. 846 Ib 4. Suppose that in problem (3) the value of V% were reduced to 80 ft. per sec. as in problem (1). What would be the value of the force exerted? (Compare with problem (3).) Ans. 761 Ib. 105. Force Exerted upon Pipe. When a flowing stream is confined there may be static forces due to pressure as well as dynamic forces due to changes in velocity. Con- sider the water to be flowing to the right in Fig. 138. -^> Since the velocity is increased from Vi to V*, the dynamic p IG 133 force exerted upon the water, according to equation (83), will be This force, producing the acceleration of the water, must be the resultant of all the forces acting. The real forces acting upon the volume of water shown are the pressures upon the two ends p'iFi and p' 2 Fz exerted by the rest of the water, and the force N exerted by the pipe walls. 1 If there were no friction this force would be normal to the walls, but actually it will be inclined somewhat from the normal because it must have a frictional component. Let the component of N parallel to the axis of the pipe be denoted by N x . It may be seen that R must be in the same direction as Vi and 7 2 in Fig. 138. Hence the sum 1 The N shown in Fig. 138 represents the force for an element only. For a pipe of circular cross-section the resultant force exerted by the walls must be axial 144 HYDRAULICS of all the forces parallel to the axis of the pipe must equal R. Therefore R = pVi - P'*F 2 - N x . (86) Inserting the value of R given above, it follows that N x = p' 1 F 1 - p' 2 F 2 - ~ (V 2 - Fi) It must be remembered that N x is assumed to be the axial com- ponent of the force exerted upon the water by the conical por- tion of pipe. The force exerted by the water upon the pipe is equal and opposite to this. That is, its magnitude is given by equation (86) but it acts toward the right. If the velocity of the water in a closed passage undergoes a change in its direction, as in the pipe bend shown in Fig. 139, the procedure would be similar to that in the preceding case. The forces acting on the water in the bend are the pressures p'iFi and p' 2 F 2 and the pressure exerted by the walls of the pipe, designated by N. By equation (83) the resultant force acting upon this volume of W water will be R = AF, but R is the resultant of the three forces just mentioned. Since these are vector quantities not in the same straight line, it will be better to take x and y com- ponents. Thus we should write FIG. 139. and R x = -- (F 2 cos e - F x ) = p y W R y = F 2 sin B = - p f 2 F 2 sin 6 + N v . - p f 2 F 2 cos e - N f Solving these equation we find that W N x = p'iF, - p f 2 F 2 cos + (Vi - V 2 cos 0) and W N v = p'zF 2 sin B -\ -- V 2 sin 6. y (87) (88) But again N represents the force exerted by the pipe bend upon H YD ROD YNA MICS 145 the water. The force exerted by the water upon the bend will be equal and opposite to this. It may be seen that these forces tend to move the portion of pipe considered. Hence a pipe should be " anchored" where such changes in velocity occur. EXAMPLES 1. On the end of a 6-in. pipe is a nozzle which discharges a jet 2 in. in diameter. The pressure in the pipe is 55 Ib. per sq. in. and the pipe velocity is 10 ft. per sec. The jet is discharged into the air. (a) What is the re- sultant force acting on the water within the nozzle? (6) What is the axial component of the force exerted on the nozzle? Ans. (a) 304 Ib. (b) 1,250 Ib. 2. Water under a pressure of 40 Ib. per sq. in. flows with a velocity of 8 ft. per sec. through a right-angle bend having a imiform diameter of 12 in. (a) What is the resultant force acting on the water? (6) What is the total force exerted on the bend? Ans. (a) 137.8 Ib. (6) 6,530 Ib. 106. Theory of Pitot Tube. The Pitot tube has been briefly described in Art. 69 and illustrated in Fig. 94. We shall now consider the dynamic action of the water upon it. In Fig. 140 a Pitot tube is placed with its opening facing upstream, the velocity of the water being denoted by V. The dotted lines in the figure are intended to represent an imaginary cylinder of cross-section area F equal to that of the mouth of the tube and extending to point (1) as far up stream as the influence of the tube is felt. 1 1 The method of derivation of the Pitot tube formula given here, as well as some interesting experimental results, will be found in a paper by L. F. Moody, "Measurement of the Velocity of Flowing Water," Proc. of the Engineers' Soc, of West. Penn., vol. 30, page 279 (1914). 146 HYDRAULICS At point (1) the water within this cylinder has a velocity V and, as it approaches the Pitot tube, its velocity continually decreases until it becomes zero at point (2). But if water flows into this cylinder, bounded by the dotted lines, it must also flow out. It does this along the sides, for since the area F is constant while the velocity is a decreasing quantity, it follows that q (within the cylinder) must become less as the Pitot tube is approached. The conditions for a certain mass of water are therefore as shown in Fig. 140 (a). As the velocity of the water decreases the cross-section area must increase for the same value of q. Referring to Fig. 140 (a), consider that water flows into this portion of the stream with a velocity V and leaves it with a velocity V AV'. If the cross-section area of the stream entering the section is F, we have W = wq = wFV. If the two faces of this volume be taken at an infinitesimal distance apart the velocity will decrease by an amount dV, hence the dynamic force exerted upon this small mass of flowing water will be W wF dR = - dV = - - - V'dV. Q 9 The value of V varies from V at point (1) to at point (2). Since F is constant we may write 9 The dynamic force exerted by the flowing water upon the body of still water within the Pitot tube is equal and opposite to R. If the force be represented by P, we have P = wF ^ (89) This is the value of the total force distributed over the area F. The intensity of pressure is p' = wV 2 /2g, or dividing by w we have intensity of pressure in feet of water so that h = ^'- (90) That this is true has been amply demonstrated by experimental evidence. If the water is under pressure, the Pitot tube will read the sum of the pressure head and the dynamic head, given by equation (90). It is therefore necessary to determine the HYDRODYNAMICS 147 pressure head separately or else use a differential fnanometer, one side of which shall be connected to the Pitot tube and the other to a piezometer tube. The chief source of error in the use of the Pitot tube lies in the measurement of the pressure. If the Pitot tube is used in a pipe, the pressure reading should be taken by a piezometer tube which does not project within the walls of the pipe and which is at right angles to it, as in the second tube of Fig. 94. If it is necessary, for some reason, to have the tube project into the stream, a correct reading may be obtained if the piezometer orifice is made in a flat plate, the plane of which may be parallel to the stream lines, as is shown by the third tube in Fig. 94. Or the orifice may be made in the side of a smooth tube, whose axis is parallel to the stream lines and whose closed upstream end is pointed so as to diminish eddy disturbances. 1 107. Water Hammer and Surges in Unsteady Flow. In all the rest of this book the treatment is restricted to cases of steady flow, but in the present article a brief description will be 1 The Pitot tube formula has often been derived by an incorrect ap- plication of the principles of Art. 104. If a jet of water with cross-section area F impinges normally upon a flat plate, the dynamic force will be p _ ,. 99 9 This is twice the value given by equation (89), and dividing this by the area of the Pitot tube orifice, which is also assumed equal to F, the intensity of V 2 pressure in feet of water is apparently h = . But this reasoning is in- correct; for a flat plate of an area the same as that of a jet would not be able to deflect all the water through an angle of 90. Experiment shows that the dynamic pressure exerted by a circular jet is distrib- uted over a circular area whose diameter is at least twice that of the -jet. Therefore if the entire stream of water is to be deflected through an angle of 90 the area of the plate must be at least four times that of the jet. Dividing P by 4F we should have Pressure Curve the average intensity of pressure to be p f = wV 2 /4:g. It is found experimentally that the maximum intensity of pressure at the FIG. 140(6). center of the plate in feet of water is V 2 /2g, and that this pressure diminishes in intensity as the outer margins of the area in question are approached, as shown in Fig. 140(6). See "Pitot Tube Formulas Facts and Fallacies" by B. F. Groat, Proc, of Engineers' Soc. of West. Penn., vol. 30, page 324 (1914), 148 HYDRAULICS given of the problems of unsteady flow that are of the most practical importance. An adequate mathematical treatment of unsteady flow would occupy too much space to warrant its inclusion here and no attempt will be made to do more than record some accepted results. In the event of a valve at C in Fig. 141 (a) being rapidly closed in a short interval of time At, the velocity of the water in the pipe will be abruptly reduced to zero. But in so doing there will be a considerable rise in pressure within the pipe, which may be much greater than any static pressure that could possibly exist in the given pipe. This high pressure lasts for an instant only, and then follows a periodic fluctuation of pressure which finally dies out, if the pipe does not burst in the meantime. This is known as water hammer. What happens is that the lamina of water next to the valve at C is brought to rest and is then compressed by the rest of the column of water flowing up against it. At the same time Normal Gradient d Load the walls of the pipe surrounding this lamina will be stretched by the excess pressure. The next lamina of water will be brought to rest by the first and so on. It is seen that the volume of water in the pipe does not behave as a rigid body but that the phenomena is affected by the elasticity of the water and the pipe. Thus the cessation of flow and the increase of pressure progresses along the pipe as a wave action. After a short interval of time the volume of water BC will have been brought to rest, while the water in the length AB will still be flowing with its initial velocity, and with its initial pressure. But the volume of water in BC will be under a much higher pressure due to the compression it is under and the pipe walls will be stretched. The excess pressure DE is the same for all portions of the pipe and is independent of the length of the pipe. Finally the pressure wave will have reached the reservoir and the entire volume of water will be at rest. But, owing to its HYDRODYNAMICS 149 compression as well as the tension of the pipe, the flow will tend to start toward the reservoir. Thus a wave of rarefaction proceeds from A to C, and so on until the waves die out. If the valve is alternately opened and closed at just the proper intervals of time it is possible to add one pressure wave on top of another, so that there is no limit to the maximum pressure that might be attained. The velocity with which this pressure wave progresses along the pipe will be given by the following formula: 1 F w =4700 /- - d (91) ^# + 300,000^- where V u = velocity of pressure wave in feet per second, E modulus of elasticity in tension of the material composing the pipe in pounds per square inch, and d/t is the ratio of the diameter of the pipe to the thickness of the walls, which means that both d and t must be in the same units. The values of E for steel, cast iron, and wood are about 30,000,000 Ib. per sq. in., 15,000,- 000 Ib. per sq. in., and 1,500,000 Ib. per sq. in. respectively. For pipes of ordinary dimensions the velocity of this pressure wave will be about 3,300 ft. per sec. In any event it will be less than 4,700 ft. per sec., which is the velocity of sound in water or the velocity with which a pressure wave would be propagated in water in a rigid pipe. See equation (91). The time required for a pressure wave to travel the length of the pipe, or the time that it takes for the entire mass of water to be brought to rest will be T = l/V u (92) The total force exerted may be determined by applying the principle that force equals mass times acceleration. Since the volume of water is a non-rigid body we must deal with the acceleration of the mass center. The pressure wave travels at a uniform rate; hence the velocity of the mass center is uniformly retarded. Therefore the acceleration may be deter- mined by dividing the change in velocity by the time required for the change to occur. The velocity of all the water, and 1 The complete expression involves the volume modulus of elasticity of the water, the density of the water, and the value of g. Using average values of these quantities the above is obtained. 150 HYDRAULICS hence that of the mass center, decreases from V to in the time T. Thus we may write p _wFlV wFl VV U = VVy g T~ g I g Dividing by the area F and also by w we obtain intensity of pressure in feet of water. If this excess pressure, due to water hammer, be denoted by p m , we have VV (93) It will be noted that this pressure increase is independent of the length of the pipe line. 1 The length of the pipe line enters into the problem in this way. The time required for a pressure wave to make the round trip from the gate to the reservoir and back is twice the value given by equation (92). It has been found that the pressure created is independent of the time of closure of the gate provided that it is closed in less time than it takes for a pressure wave to make the round trip. That is the gate must be closed in less time that 21 /V u . If the time is greater than this the pressure is reduced in proportion as follows : P = Pm ^7 (94) where T r is the time for a round trip of the pressure wave, T' is any time greater than this and p is the pressure that will be attained in such a case. It is seen that in a short pipe line the value of T r is so small that it is nearly impossible to close the gate quickly enough to pro- duce water hammer of maximum intensity. In a long pipe line it is necessary to close the gate slowly in order to prevent this and the longer the pipe line the slower the gate must be closed. For the sake of clearness in explanation it has been assumed in the preceding discussion that the velocity of the water has 1 The subject of water hammer has been experimentally investigated by Joukovsky of Moscow on pipes of 2-, 4-, 6-, and 24-in. diameter and with lengths ranging from 1,050 ft. to 7,007 ft. He found the results to agree with the formulas given. For a resume of his work see "Water Hammer" by Simin, Trans. Amer. W. W. Ass'n, 1904. HYDRODYNAMICS 151 been reduced to zero. But the results are true for any reduction in velocity, it being simply necessary to substitute AF for V. Water hammer may be prevented by the use of slow closing valves, or its effects diminished by the use of automatic relief valves which permit water to escape when the pressure exceeds a certain value. Also air chambers of suitable size provide cushions which absorb a great portion of the shock. But for water power plants a standpipe or surge chamber such as is shown in Fig. 141(6) has certain marked advantages. In the event of a sudden decrease in load on a water power plant it would be necessary for the governors to rapidly reduce the amount of water supplied to the wheels, if the speed of the latter is to be maintained constant. A surge chamber provides a place into which this excess water may flow and thus avoids water hammer in the supply pipe. The inertia of the mass of water flowing down this supply pipe may be such as to carry the water level above the static level and . produce an ascending hydraulic gradient. But this excess pressure acts as a retarding force on the mass of water in the. pipe line and thus reduces its velocity. In any event the temporary water level in this surge chamber will be higher than the normal value and hence it will reduce the velocity of flow too much. The result will be that there will be fluctuations of velocity in the pipe line accompanied by "surges" of the water level in the chamber until a condition of equilibrium is finally reached. The phe- nomenon is very similar to that of water hammer as there are periodic alternations in pressure and velocity, but the pressure variations are much less severe. The surge chamber fulfills another valuable function in that it not only takes care of excess water in case of a sudden re- duction of flow but it also provides a source of water supply in the event of a sudden demand. When the load on the plant increases it is necessary to supply more water to the wheels at once. If the pipe line is long it may take some time to accelerate the entire mass of water and in the meantime the head at the plant has dropped considerably in order to provide an ac- celerating force. But the surge chamber permits a certain amount of water to flow out during that period. To be sure enough flows out so that the hydraulic gradient drops below its normal level for the new load, but the effect is not as serious as if the surge chamber were absent. 152 HYDRAULICS In Fig. 195 is shown a surge chamber of large size. It is at the end of a pressure tunnel which is approximately 7.76 miles in length, with an average cross-section area of 100 sq. ft. and in which is a maximum velocity of flow of 10 ft. per sec. 1 EXAMPLES 1. A cast-iron pipe line is 24 in. in diameter and the metal is 0.75 in. thick. If the velocity of water in it is 6 ft. per sec., find the pressure that would be created by the instantaneous closure of a valve. Ans. 296.5 Ib. per sq. in. 2. If the pipe line in problem (1) were 500 ft. long, within what length of time must the valve be closed to produce the same pressure as an in- stantaneous closure? What would the length of time be if it were 5,000 ft. long? Ans. T = 0.27 sec., 2.7 sec. 3. If the pipe line in problem (1) were 7,000 ft. long what would be the time of closing the valve so that the pressure produced were only one- third of that in case of instantaneous closure? Ans. 11.43 sec. 108. Relation between Absolute and Relative Velocities. In much of the work that follows it will be necessary to deal with / M 8 ^ ' r s >j r*-~s-- i I FIG. 142. Relation between absolute and relative velocities. both absolute and relative velocities of the water. The absolute velocity of a body is its velocity relative to the earth. The rela- tive velocity of a body is its velocity relative to some second body which may in turn be in motion relative to the earth. The absolute velocity of the first body is the vector sum of its velocity relative to the second body and the absolute velocity of the lat- ter. The relation between the three is shown in Fig. 142. 2 *W. F. Durand, "Control of Surges in Water Conduits," Journal, A. S. M. E., June, 1911. See also, "The Differential Surge Tank" \y by R. D. Johnson, Trans. A. S. C. E., vol. 78, 760 (1915). FIG. 143. 2 A clearer idea of this relationship may be ob- tained from the illustration in Fig. 143. Suppose raft is moving downstream with a uniform velocity u. A man on the HYDRODYNAMICS 153 109. Dynamic Action upon Moving Body. The dynamic force exerted by a stream upon a moving object can be determined by a direct application of equation (83). The principal difference between action upon a stationary and upon a moving object is that in the latter case we need to deal with both absolute and relative velocities, and the determination of AF may be more difficult. Let us assume a stream of cross-section area FI and absolute velocity FI to flow upon a moving object. The rate of discharge will be FiVi so that W = wFiVi. But this may not be the amount of water that strikes the object per second. For instance, if the body is moving as rapidly as the stream and in the same direction it is clear that none of the water will strike it. The amount of water which will flow over any object is proportional to the velocity of the water relative to the object itself. If we denote by W the pounds of water striking the moving body per second, then W = wfiVi, where /i is the cross-section area normal to VL As a special case to illustrate the above let us consider a jet from a nozzle acting upon a body moving in the same direction as the jet with a velocity u. In this case, since u and V\ are in the same direction, FI = fi and vi = Vi u. But in general we should have a vector relationship as shown in Fig. 142. How- ever, for this particular case W = wF\.(V\ u). The reason that less water strikes the body per second than issues from the nozzle per second is that the body is moving away from the latter and there is an increasing volume of water between the two. If we consider an impulse wheel with a number of vanes around its circumference, the above is true for one vane only. But the wheel as a whole does not move away from the nozzle and hence the amount of water striking the wheel may equal that issuing from the nozzle. The explanation is that two or more vanes are acted upon by the water at the same time as is shown in Fig. 202. In order to find A V it is necessary to determine the magnitude raft at A walks over to the diagonally opposite corner at a uniform rate. But by the time he reaches B the latter point on the raft will have moved downstream to point C. Thus the path of the man relative to the raft is A B but relative to the earth it is AC. Since the velocities are all uniform they are all proportional to the distances traversed in this in- terval of time. 154 HYDRAULICS and direction of the absolute velocity at outflow, assuming the FI to be known. The direction of the relative velocity at out- flow is tangent to the surface of the body at that point. The angle between v 2 and the positive direction of u is denoted by a z . Assuming u and a 2 to be known we may proceed as follows. Solve the vector triangle (Fig. 142) for vi if FI, u, and AI are known. In flow over the object there may be a loss of energy due to friction such that v 2 is less than vi and hence we may write v 2 = nvi, where n is less than unity. Having now the values of v 2} u, and a 2 we may solve the vector triangle (Fig. 142) for F 2 and A 2 . This enables us to find the value of AF. As a special case to illustrate the procedure let us consider Fig. 144 where the jet strikes an object moving with a uniform FIQ 144 velocity in the same direction as the jet. Hence A l = 0. Let us assume the x axis parallel to the jet and use equations (84) and (85). W W P x = - AF X = (Fi cos A l - F 2 cos A 2 ) y y Fi COS AI = Fi Vz cos A 2 = u + v 2 cos a 2 from Fig. 142. = u + nvi cos a 2 Since AI = 0, v\ = Vi u and hence for this special case F 2 cos A 2 = u + n(Vi u) cos a 2 . Substituting this in the above and reducing we have W P x = (1 - n cos a 2 )(Fi - u) (95) = l (1 - n cos a 2 )(7i - u) 2 (96) The value of P y may be determined in a similar manner, noting that in this case F 2 sin A 2 = v 2 sin a 2 . It must be borne in mind that these equations" are true only for the special case considered and that they apply to a single moving object. The general case would differ from the above only in the fact that Vi is then the vector and not the algebraic difference between Fi and u. HYDRODYNAMICS 155 It may be seen that the magnitude of the force exerted by a jet depends both upon the shape and the velocity of the object struck. In fact the same value of AF might be had with either a stationary or a moving object or with moving objects having different velocities provided only that their shapes, which in this case means their values of a 2 , were suitable. As another illustration of the foregoing let us consider the dynamic force exerted by a*jet of water upon the moving body from which it issues. When a stream of water issues from any device, such as the vessel shown in Fig. 145, a force is required to accelerate the water and impart to it the velocity it has upon leaving. This force is exerted upon the particles of water flow- ing out the orifice by adjacent particles of water and ultimately by the walls of the vessel. By the law of action and reaction an equal and opposite force will be exerted upon the vessel. It is impossible to analyze this reaction in detail but we know that its total value will be given by an application of equation (83). Let us assume that the vessel in Fig. 145 moves to the left with a uniform velocity u, and that the orifice is so small com- pared to the size of the vessel that the relative velocity of the water in the latter may be neglected as may also the change in h. Then Vi = u. If the jet issues from the orifice with a velocity V 2 the absolute velocity of the jet will be F 2 = u v%. Hence AF = Vi - V z = u - (u - t; 2 ) = v 2 . Therefore W wF (97) This might have been determined more directly if another proposition had been previously established. That is that for any case whatever AF X = &v x + Au x , where the subscript x merely denotes a component along any axis. In this case, since u is constant, it may be seen that AF = Av and is independent of the velocity of the vessel. Since v% = c v \/2gh, we may write equation (97) as P = 2c v 2 wFh (98) FIG. 145. If losses of energy be neglected in both cases, it may be seen that the re- action of the jet in Fig. 145 is equal to the force of impact upon 156 HYDRAULICS a flat plate, normal to the jet, providing the area of the plate be large enough to deflect the water through 90. l EXAMPLES 1. A jet of water 3 in. in diameter has a velocity of 120 ft. per sec. It strikes a vane with an angle a 2 = 90 which moves in the same direction as the jet with a velocity u. Assume that the loss in flow over this vane is such that n = 0.9. When u has values of 0, 40, 60, 80, 100, and 120 ft. per sec., find values of: (a) W, (6) F 2 cos A 2j (c) P x . 2. If the jet in the preceding problem strikes a vane for which a 2 = 180, all other data remaining the same, find values of: (a) W, (6) t> 2 , (c) 7 2| 110. Impulse and Reaction of a Jet. When a stream of water strikes any object, the dynamic force exerted, due to the impact, is often termed the impulse of the jet. The dynamic force ex- erted by the jet upon the vessel from which it issues is often called the reaction of the jet. But in both cases the force is due to the change that is produced in the velocity of the water. 111. Distinction between Impulse and Reaction Turbine. The distinction between these two fundamental types of turbines according to the action of the water as defined in the preceding article was proper in primitive wheels. But in modern turbines the so-called impulse at entrance and reaction at exit may both be effective in either type. A better classification is as to "pres- sureless" and " pressure" turbines. Thus the water within the impulse wheel is not confined but is open to the air, while in the reaction turbine the wheel passages must be completely filled with water. In the former the pressure remains unchanged in flowing over the buckets, while with the latter the pressure decreases during flow through the runner. The energy delivered to the impulse turbine is all kinetic, while that delivered to the reaction turbine is partly kinetic and partly "pressure energy." 1 The hydrostatic pressure on an area equal to that of the jet F at a depth h is given by wFh. The fact that this is only half the dynamic pressure considered is of no significance. As has already been pointed out, the dynamic pressure on a plate is distributed over an area much larger than that of the jet and we have not increased the intensity of pressure in either case. HYDRODYNAMICS 157 But it is well to bear in mind that in both types tke essential thing is that the velocity of the water must be altered in order that a dynamic force may be exerted upon the wheel. And in both types it is necessary if high efficiency is attained that the absolute velocity of the water as it leaves the wheel be low, since this velocity represents so much kinetic energy that is not utilized. 112. Theorem of Angular Momentum. In Fig. 146 we will suppose a particle of mass dm to be located at a point whose co- ordinates are x and y and to be moving with a velocity V. The momentum of this particle will be dm.V. The moment of mo- mentum is called angular momentum. For this particle the angular momentum is dm.V X rcosA. Since the moment of \ X O FIG. 146. any quantity is the algebraic sum of the moments of its compo- nents, we may write dm.rV cos A dm.VyX dm.V x y dm(-rrx -37- y } . Differentiating the above with respect to time we obtain d(rVcosA) , fdy dx , d*y dx dy d*x - ' = dm(a y x a x y) where a denotes acceleration, with a x and a y as its axial com- ponents. (V x = dx/dt } a x = dV x /dt = d*x/dl 2 , etc.). If the re- sultant force acting on the particle be denoted by dR, dma v = dR v , and dma x = dR x . Thus dmd(rV cos A)/dt = dR v x dR x y. 158 HYDRAULICS The torque exerted upon the particle with respect to point is seen to be dR X I. By the principle of moments dR X I = dR y x - dR x y Thus, if T denotes torque so that dR X I = dT, (99) That is, the time rate of change of the angular momentum of any particle with respect to an axis is equal to the torque of the re- sultant force on the particle with respect to the same axis. 113. Torque Exerted upon Turbine by Water. When a stream flows through a turbine runner in such a way that its distance from the axis of rotation remains unchanged, the dynamic force can be computed by the principles of Art. 104. But when the KeiativePath of Water Absolute Path of Water FIG. 147. Hydraulic turbine. radius to the stream varies it is not feasible to compute a single resultant force. It is necessary to find the total torque exerted by summing up the elementary torques produced by all the ele- mentary forces. In Fig. 147 let MN represent a vane of a wheel which may rotate about an axis perpendicular to the plane of the paper. Water enters the wheel at M and since the wheel is in motion, by the time the water arrives at N on the vane that point of the vane will have reached position N r . Thus the absolute path of the water is really MN'. Let us consider an elementary volume of water forming a hol- low cylinder, or a portion thereof, concentric with 0. Let the HYDRODYNAMICS 159 time rate of mass flow be dm/dt. Then in an internal of time dt, there will flow across any section the mass (dm/dt)dt. Let' this be the mass of the elementary volume of water we are to consider. Substituting this value in equation (99) we have (dm \ d(rVcosA) (dm\ ,, T _ dT = -' dt - ~~ ' = ' d(rV cos A ) The above procedure is simitar to that in Art. 104, the only dif- ference being that here we are dealing with the moment of a force. In the case of steady flow (dm/dt) = (W/g) and thus d(rV cos A) wn ' VJi Integrating between limits we have the value of the torque ex- erted by the wheel upon the water, or by changing signs, the value of the torque exerted by the water upon the wheel. There- fore the torque exerted upon the wheel by the water is W T = (nVi cos Ai - r 2 V 2 cos A 2 ) (100) a It may be seen that V cos A is the tangential component of velocity. It is convenient to represent this by a single letter s and so T = (ni - rt8J (101) It is immaterial in the application of this formula whether the water flows radially inward, as in Fig. 147, radially outward, or remains at a constant distance from O. In any case r\ is the radius at entrance and r 2 is that at exit. 114. Torque Exerted upon Water by Centrifugal Pump. The derivation of an expression for the torque exerted by a pump impeller is exactly the same as in the preceding article except for the substitution of limits. As turbines are universally con- structed there are certain guide vanes surrounding the runner which give the water its initial direction A\. Thus any angular momentum which the water has as it flows into the runner is imparted by the guides. In some centrifugal pumps there are guide vanes within the "eye" of the impeller, which give the water a definite direction as it flows into the latter. For such pumps we should merely reverse equation (101), since we desire the torque exerted upon the water and not by it. But for the usual type of centrifugal pump there is nothing at 160 HYDRAULICS entrance to the impeller to give the water any definite direction. In fact the water enters a given impeller at different angles de- pending upon conditions of operation, and, while the ordinary pump is designed for a "radial entrance" this can be had only for the normal rate of discharge. But any angular velocity with which the water enters the impeller has been really derived from the impeller and so should be credited to the latter. Hence we should take as our lower limit of integration not (1) where the water enters the impeller but some point back in the suction pipe, where the angular momentum is zero. 1 Thus we should have Absolute Path. FIG. 148. Centrifugal pump. for the ordinary centrifugal pump under all conditions of opera- tion, W T = r 2 s 2 (102) 115. Power. If torque be multiplied by angular velocity the product represents power. Angular velocity must be expressed in radians per second, hence, if N = r. p. m. ; the horse- power will be If T has the value given by equation (101), the power will be less than that supplied to the turbine by the water, the difference being the power lost in hydraulic friction within the turbine case, runner, and draft tube. It is greater than the power delivered by the turbine by an amount equal to the losses in mechanical friction. It is the power that is actually delivered to the shaft 1 This point is fully discussed by the author in "Centrifugal Pumps," page 61. HYDRODYNAMICS 161 from the water, and is analogous to the indicated power of a steam engine. If T has the value given by equation (102) the power given by equation (103) will be less than that required to run the pump by an amount equal to the mechanical losses and it will be greater than the power delivered in the water by the amount of the hy- draulic losses. It represents 4he power actually expended by the impeller on the water and is analogous to the indicated power of a reciprocating pump. While equations (101) and (102) are true, they are of little real service because the proper values to use in them are often not known with exactness. The precise values of velocities and direc- tions of stream lines are difficult matters to determine. Since water does not fulfill the ideal conditions assumed, it will be found that these equations often yield numerical results that are considerably in error. 1 It should be noted that power can be expressed in the following forms; as well as by equation (103). WH wqH qH '"550 = ~ = In the last expression H may represent any head for which the corresponding power is desired. 116. Definitions of Heads. In turbine and pump practice we find the word "head" used to express several different physical quantities. The head h under which a turbine or pump actually operates is explained in Arts. 88 and 89. But there is energy lost in hydraulic friction within the runner or impeller and thus there is head lost. We shall designate this by h'. And in the turbine a portion of the energy of the water is delivered in the form of mechanical work and the head thus utilized by the runner we shall denote by h". In the centrifugal pump h" will represent the head actually imparted to the water by the impeller. Thus for the turbine we shall have h" = h h' and for the pump h = h" h'. 117. Definitions of Turbine Efficiencies. The word "effi- ciency" without any qualifying adjective is always understood x See "Hydraulic Turbines," page 83 and "Centrifugal Pumps," pages 76, 81, 82, and 84. 162 HYDRAULICS to mean gross or total efficiency. It is the ratio of the developed or brake horsepower to the power delivered in the water to the turbine. That is e = b.hp./w.hp. (106) Mechanical efficiency is the ratio between the power delivered by the machine and the power delivered to its shaft by the water. If q represents the total turbine discharge while q f equals the amount of leakage through the clearance spaces, the actual amount of water doing work is q q'. Hence (107) Hydraulic efficiency is the ratio of the power actually delivered to the shaft to that supplied in the useful water. That is e h = W (q - q')h"/w(q - q')h = h"/h (108) Volumetric efficiency is the ratio of the water actually used by the runner to total amount discharged. Thus, e v = (q~ q')/q (109) The total efficiency is the product of these three separate fac- tors. That is, e = e m , X e h X e v (110) 118. Definitions of Pump Efficiencies. The various pump efficiencies are similar to those for the turbine. 'The total effi- ciency is e = w.hp /b.hp. (Ill) The mechanical efficiency is (112) The hydraulic efficiency is e h = w(q+q')h/w(q + q')h" = h/h" (113) The volumetric efficiency is e, = q/(q + q') (114) As in equation (110) the total efficiency is the product of these three. j 119. Centrifugal Action or Forced Vortex. If a vessel con- taining a liquid is rotated about its axis, the liquid will tend to HYDRODYNAMICS 163 rotate at the same speed. If the vessel is open the free surface of the water will assume the curve shown in Fig. 149a. If the water is confined within a closed vessel, which it completely fills so that it cannot change its position, the pressure along a horizon- tal line will vary in the same way as in the preceding case. In fact if piezometer tubes were connected to the vessel the water would rise in them as shown in Fig. 1496. Since the hydraulic gradient represents the free surface that corresponds to the actual pressure conditions, it is seen that the two cases are equivalent. Such a rotation is sometimes described as a forced vortex because the water is forced to rotate by external forces. The variation in pressure in such a body of water may be (a) OPEN VESSEL (6) CLOSED VESSEL FIG. 149. Forced vortex. found in the following manner. If we take an elementary volume in Fig. 149 whose length along the radius is dr and whose area normal to the radius is dF, we have an elementary mass wdFdr/g moving in a circular path. This mass has an acceleration u z /r or w 2 r, directed toward the axis of rotation. Consequently the accelerating or resultant force is (wdFdr/g)u 2 r directed toward the axis. The intensity of pressure on the two faces of the ele- mentary volume differs by dp' = wdp r . The value of the resultant force is therefore wdp r dF. Consequently, wdp r dF = (wdFdr/g)u 2 r dp r = (< But this expression shows only the difference of pressure along 164 HYDRAULICS the radius and in the same horizontal plane. If we move along a path parallel to the vertical axis of rotation so that the radius is constant, the pressure decreases directly as the elevation in- creases. Thus, dp z = dz. The variation of the intensity of pressure in any direction may be found by combining the two preceding equations. Thus, in general, when both r and z vary, dp = - dz + (u*/g)rdr (115) To find the equation of the free surface or any surface of equal pressure we need only place dp equal to zero. We then have fdz = (o*/g)frdr z = r 2 co 2 /2<7 + constant. To determine the constant we may assume z = when r = 0. Thus the constant = so that z = rW/2g (116) From this it may be seen that the free surface or any surface of equal pressure is a paraboloid. To find the variation of pressure in the same horizontal plane we need only assume dz = 0, and integrating between limits we obtain ; P2.- .2?r = fa 2 - ri) 2 /20 = ( W2 2 _ u ^/2g (117) For the difference in pressure between any two points we must integrate equation (115) which gives (118) ^ ( -t p, T 120. Free Vortex. Where external forces are applied to the water, as in the preceding case, we have a forced vortex. Where no external forces are applied but the water rotates by virtue of its own angular momentum, previously derived from some source, we have a free vortex. A free circular vortex consists of a body of water in rotation without any appreciable flow so that the stream lines are con- concentric circles. Since no torque is exerted on the water, neglecting friction, it follows that there can be no change in angular momentum. Since angular momentum is proportional to HYDRODYNAMICS. rV cosA, it is apparent tljat V cos A varies as 1/r, as the angular momentum is constant. Since the stream lines are circles V cos A is the value of V itself. Since no energy is im- parted to the water, we will have, if friction is neglected, H = p 72 + TT = constant. (a,) FIG. 150. Free vortex. The free surface of such a vortex is shown in Fig. 150a. A familiar example of such a surface is when water entering a vertical pipe sets up a rotation and sucks air down the center, though of course that velocity then has a radial component. Since V X r = constant, it is seen that when r = 0, the value of V is infinity and p + z = infinity. Since this is impossible, we never have the free vortex exist with extremely small values of r. (If p is constant or equal to zero as in the case of the free surface, values of z will give the elevation of the surface. If z is constant values of p will give the hydraulic gradient, which is the same curve.) Considering a pure radial flow between two circular plates, either inward or outward, "as in Fig. 1506, and letting b equal the distance between the two plates we have q = 2irrbV. For steady flow q is constant and hence rbV is constant. And if the plates are parallel rV is constant. Thus V varies as 1/r, as in the preceding case. A free spiral vortex is a combination of radial flow and circular flow. The velocities in the two cases above are then merely components of the .velocity in the latter case. Since each com- ponent varies as 1/r it follows that the velocity in a spiral flow 1646 HYDRAULICS also varies as 1/r. Also since both components vary at the same rate, the angle A, which the velocity makes with the tangent to the circle, remains constant. Thus the free stream line is the equi-angular or logarithmic spiral. Since the total head is also constant here, neglecting friction, the free surface or the hydraulic gradient, as the case may be, is the same as shown in Fig. 150. Since H is constant we may write, considering the two points to be at the same elevation H = Pl + TV/20 = p 2 + TV/20 Pi - Pi = [1 " (ri/r,)']TV/20 (119) Of course the effect of friction is always to make p 2 smaller than would be given by the above, since H is not constant. (Note that the flow may be either inward or outward.) It may be seen that as r increases V decreases and p approaches H as a limit. The principal application of Arts. 119 and 120 is in the case of the centrifugal pump. In Fig. 221 may be seen a forced vortex in the impeller from (1) to (2) and a free vortex in the casing between (2) and (3). It may be added that the foregoing treat- ment may readily be extended to the case where the width b is variable. 120a. Flow Through Rotating Channel. We shall now extend the treatment of the forced vortex (Art. 119) to the more general case where the water flows through the rotating vessel. It has been seen that with the free vortex the hydraulic gradient or the resulting equation (119) is the same whether the water merely rotates in concentric circles or flows in spiral paths, but with the forced vortex the equation will be found to be somewhat different when flow occurs. The reason is that in the free vortex no energy, save that lost by friction, is imparted to or taken from the water; but in case water flows through a rotating vessel, energy is delivered either to it or by it. The torque exerted by the water on a moving object is given by equation (100). 'When multiplied by the angular velocity, such that (120) (121) It may be seen that c v is the velocity coefficient of the nozzle, the value of which is constant for any setting of the needle. Thus for any given value of the relation between Vi and u\ is known at once regardless of the value of h. THEORY OF THE IMPULSE WHEEL 211 FIG. 203. <^ U\ V 1 ll Vi v W, ~~ Vi U MS 0=0.0 0=0.20 0=0.45 = 0.55 FIG. 204. Velocity diagrams for different speeds. From a photograph by the author. FIG. 205. A 42-in. Pelton-Doble impulse wheel. 212 HYDRAULICS From a photograph by the author. FIG. 206. Showing discharge from buckets when wheel is at rest, or = 0.0. From a photograph by the author. FIG. 207. Wheel running at slow speed. = 0.20. THEORY OF THE IMPULSE WHEEL 213 From a photograph by the author. FIG. 208. Wheel running at normal speed. = 0.45. From a photograph by the author. FIG. 209. Wheel running at higher speed. = 0.55. 214 HYDRAULICS The absolute path of the water and the velocity vectors at discharge from the buckets may be seen in Fig. 203. For dif- ferent wheel speeds under the same head, which means different values of 0, we should have such diagrams as are shown in Fig. 204. As the speed of the wheel increases from zero, the angle of deflection of the jet continually decreases. It may also be seen from the diagrams that the value of 7 2 is relatively high when the wheel is at rest, that it becomes a minimum at such a From a photograph by the author. FIG. 210. Wheel at run-away speed.

, it may be seen that it is zero when the wheel is at rest, though the torque is then a maximum, and it is also zero when the speed is a maximum for the torque is then zero. B.P.M. FIG. 211. Relation between torque and speed. 1 The maximum power will be obtained for some speed between these two extremes, as shown by Fig. 212. Since for a given head and nozzle opening the power input is constant regardless of the speed of the wheel, it follows that the efficiency is directly proportional to the power developed. But it should be noted that the power delivered in the water increases with the nozzle opening so that the needle setting that gives the largest power is not necessarily the most efficient. 138. Speed. From equation (125) we should conclude that P would become zero when Ui = Vi or when = c v , the value of which would be about 0.98. Also if we should multiply this 1 From the test of a 24-in. wheel by F. G. Switzer and the author. 218 HYDRAULICS equation by HI the power would be seen to be a maximum when u\ = 0.5Fi. But equation (125) is only an approximate representation of the actual facts. Because of the large amount of water that is not utilized at high values of 0, and also because 148 Turns B.P.M. FIG. 212. Relation between power and speed. the bearing friction and windage prevent the torque from ever being reduced to zero, the actual maximum speed attained by the tangential waterwheel is such that < is approximately equal to 0.80. In like manner the maximum power, and hence the maximum THEORY OF THE IMPULSE WHEEL 219 efficiency for a given nozzle opening, is also attained when the wheel speed is something less than 0.5W Thus in actual practice we have for the best efficiency (j) e = 0.43 to 0.48. In practical applications we are usually interested in the performance . of a wheel at a constant speed under a constant head. Values for this may be obtained from Figs. 211 and 212 by following along any vertical line. Generally the vertical line should be the one for the speed at which the maximum efficiency is found. The resulting curves for the impulse wheel would be very similar to those for the reaction turbine shown in Fig. 216. FIG. 213. 139. Head on Impulse Wheel. The nozzle is considered an integral part of the impulse wheel and hence the head under which the wheel is said to operate must include it. If C in Fig. 213 indicates a point at the base of the nozzle, h = H c = Pc + 20 (126) It is this value of h that should be used in determining the efficiency of the wheel. This value of h is the total fall from headwater to nozzle minus the head lost in the pipe line. The energy supplied at this point is expended in four ways. A small amount is lost in flow through the nozzle, a portion is expended in hydraulic friction and eddy losses within the buckets, kinetic energy is carried away in the water discharged from the buckets, while the rest is delivered to the wheel to do useful work and overcome mechanical friction and windage losses. Calling h" the head delivered to the buckets we may write /"I .xFi 2 . 7 V2 2 , F 2 2 h = &r. l )2T + k 9J + '*** r ' 220 HYDRAULICS It may also be noted that Wh" = Pui, hence h" can be obtained from equation (122) by merely substituting Ui for W. See also Art. 120a. 140. PROBLEMS 1. A nozzle having a velocity coefficient of 0.98 discharges a jet 6 in. in diameter under a head of 800 ft. This jet acts upon a wheel with the follow- ing dimensions: diameter 6 ft., A\ = 10, a 2 = 165, and it is assumed k = 0.70. Find the force exerted upon the buckets when = 0.45. 2. Solve problem (1), assuming that AI =0. 3. Find the power developed upon the buckets of the wheel in problem (1), assuming AI = 0. What is the hydraulic efficiency of the wheel? 4. If the mechanical efficiency of the wheel is 0.97, what is the gross efficiency in problem (3)? 5. Assuming AI =0 in problem (1), find the power lost in hydraulic friction within the buckets. Find the value of V z and determine the power carried away in the water discharged from the turbine. 6. Is the hydraulic efficiency of an impulse wheel dependent upon the head under which the wheel is run? What equation would express the value of the hydraulic efficiency for any tangential waterwheel? 7. What would be the proper r.p.m. of the wheel in problem (1) ? 8. A good proportion between jet and wheel is that the diameter of the wheel in feet should equal the diameter of the jet in inches. Using this ratio, what size wheel would be required to deliver 5,000 hp. under a head of 1,400 ft., assuming an efficiency of 82 per cent.? What would be the speed of the wheel? 9. The maximum speed attained by the wheel of Fig. 211 was 475 r.p.m. under a head of 65.5 ft. What was the value of ,t CHAPTER XIV THEORY OF THE REACTION TURBINE 141. Introductory Illustration. The reaction turbine is so called because an important factor in its operation is the reac- tion of the streams of water discharged from the runner. It is well to bear in mind, however, that the total dynamic effort is due to the entire change in the momentum of the water just as in the impulse turbine. As an illustration, consider the vessel ABC of Fig. 214 into which water enters across A B with a velocity Vi and is dis- charged at C with a velocity F 2 . Now the reaction of the jet alone could be determined . & by an application of Art. 110. But the total force is due not only to this reaction but also to the impulse of the water entering at AB. It is not FlG 2 i4. feasible to separate the effect of impulse from that of reaction, neither is it necessary to do so. The horizontal component of the total dynamic force is obtained directly by p = ^(y lCOsAl +y 2 ). t? Suppose now that this vessel moves to the left with a uniform translation u. Assume that in some way the water is still supplied to it with a velocity Vi. This might be the case if the vessel passed under a series of stationary passages each of which in turn was permitted to discharge water into it. The value of the absolute velocity of discharge is now V 2 = v 2 u. Inserting this value above we have W P = (Vi cos Ai + v 2 u) 221 222 HYDRAULICS This equation indicates that ' P decreases as the speed increases, just as in the case of the impulse turbine. Also the water enter- ing the vessel at A B is under pressure and is not a free jet. Therefore V\ must be less than y/2gh. Since all the passages are completely filled with water the equation of continuity can be applied, and it will show that V\ and v z are inversely pro- portional to the areas of their respective streams. But the value of v 2 depends upon the losses of head, and in a real turbine these Values of 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 100 200 300 400 500 600 700 800 900 1000 1100 Revolutions per Minute FIG. 215. Test of 27 in. I. P. Morris turbine. Head and gate opening constant. Speed variable. 1 hydraulic losses vary with the speed. Since Vi is proportional to 02, it follows that V\ varies with the speed of the wheel. Thus some fundamental differences between impulse and reaction turbines are that in the former V\ = c v \/2gh, where c v is a velocity coefficient nearly equal to unity. This velocity, and hence the amount of water discharged by the nozzle, is 1 Figs. 215 and 216 are from the test of a reaction turbine in the Cornell University power plant. See "Investigation of the Performance of a Reaction Turbine" by R. L. Daugherty, Trans. A. S. C. E., vol. 78, p. 1270 (1915). THEORY OF THE REACTION TURBINE 223 entirely independent of the design of the wheel and its operation. But for the reaction turbine Vi = cVlgh (127) where c is not a velocity coefficient but a factor whose value varies from about 0.6 to 0.8 for ordinary designs. The value of c depends upon the design of the wheel and the speed at which it is run under a given head. This means that c is also a function of <, where has the meaning given by equation (121). With the radial-flow type of turbine centrifugal force also causes the value of c to vary with the speed of the wheel, the head remaining constant. The centrifugal force opposes the flow of water in the case of the inward-flow turbine so that, as the speed increases under a constant head, the discharge tends to decrease as shown in Fig. 215. But there are other influences at work also, so that for some inward-flow turbines the value of q actually increases somewhat as the speed increases above zero, but after a certain speed is exceeded the rate of discharge falls off again. 142. Torque Exerted. The preceding article merely illustrates a few fundamental points regarding the reaction turbine. Since with the real machine the radii of the water at inflow and outflow differ materially, it is not feasible to compute the force exerted by the water and we must get the torque instead. Before pro- ceeding any further with the theory it should be noted that, while our equations are rational, they must assume that all particles of water move in similar paths with equal velocities. Actually we have to deal with average values. But we do not know these average values with any precision. For example, we have no assurance that the angles A\ and a^ are the same as the angles of the guide vanes and the runner vanes respectively. In fact we have some evidence to indicate that they differ by as much as 5 or 10. The same condition exists with regard to the areas of the streams and all other dimensions used. Thus the numerical results of such computations cannot be expected to agree precisely with actual facts. Despite this the theory has its value. It serves to explain the principles of operation of such machines, to indicate the nature of their actual characteristics, and to account for numerous observed facts. In design the theory shows what proportions are desirable and what the effect of certain changes of dimensions 224 HYDRAULICS would be. Thus if we have some actual test data to work from, the theory would enable us to alter existing designs with some degree of assurance. In order to compute the torque exerted upon the runner by the water we should take the fundamental formula of Art. 113, W T = (nFi cos Ai r 2 F 2 cos A 2 ). Just as in the case of the impulse turbine in Art. 136, the values of Vz and A 2 are variable and unknown, and it is necessary to replace them in terms of known quantities. It is assumed that all the dimensions of the wheel and the values of V\ and Ui are known. From the vector diagram it may be seen that Vz cos A 2 = Uz + v z cos a 2 . But Uz = (r2/ri)wi, and, since the passages are completely filled with water in the reaction turbine, the equation of con- tinuity gives q = FiVi = f 2 v 2 , or v 2 = (F l /f 2 Wi (128) (Contrast this procedure with that for the impulse turbine in Art. 136 and note that equation (122) does not apply here.) Making the proper substitutions we easily derive cos Al ~ ' cos tt2 Vl " * 1 (129) = 7 ri [ ( In the use of equation (129) we should have to determine the value of W for any value of u\, either by experiment or by computing the rate of discharge by theory. 143. Power. The power developed by the water is deter- mined by multiplying T by the angular velocity. The torque actually exerted by the shaft and the power delivered are ob- tained by multiplying these values by the mechanical efficiency. The hydraulic efficiency of the turbine is obtained by Tu _ Wh" _ h^ eh ~ Wh~ Wh '"' h' It is difficult to obtain the hydraulic efficiency by test as it is necessary to determine the bearing friction and also the disk friction due to the drag of the runner through the water THEORY OF THE REACTION TUR&tNE 225 in the clearance spaces. But these losses may be allowed for and the hydraulic efficiency then secured approximately. Because of the necessary defects of the theory, the hydraulic efficiency may be assumed with less error than is usually in- volved in computing T. For turbines of rational design and running at their proper speeds the value of the hydraulic ef- ficiency may range from Or80 to 0.95. The higher values are found only in large turbines and with favorable proportions. Only experience can enable one to select the proper value be- 100 200 300 400 B.HP. =HP. Output 500 FIG. 216. Test of 27 in. I. P. Morris turbine. Head and

2 may easily be as large as in the case of the impulse wheel. And if u 2 = v z , it may be seen that u 2 will be about the same in either type. But with the inward flow reaction turbine HI is greater than HZ, and therefore the peripheral velocity of the reaction turbine is greater than that of the impulse wheel. Not only is the peripheral speed higher for maximum efficiency but also the runaway speed is higher. The maximum value of $ for the reaction turbine is about 1.30, though with some it may easily exceed this value. And for the normal speeds at which the maximum efficiency is obtained we have e = 0.60 to 0.90, the exact value for a given wheel depending upon its design. 145. Values of c e and e for Maximum Efficiency. The turbine should run normally at such a speed under any head that the maximum efficiency will be attained. It will be assumed that this speed is such that the discharge is "radial" or A 2 = 90. The angle which the runner vane at entrance makes with HI will be denoted by a'i. As the turbine is ordinarily designed, the value of the vane angle would be such that it would agree with ai as determined by the vector diagram for this same speed. But at any other speed the value of a\ would be different THEORY OF THE REACTION TURBINE 227 from a'i, hence there would be an abrupt change in the direction of the water entering the runner giving rise to what is known as " shock loss." The following expressions therefore apply only to the special case where A 2 = 90 and a 'i = a\. From the vector diagram of velocities we have Vi sin A] = Vi sin ai = Vi sin a\ Vi cos Ai = U} + Vi cos ai = u\ + v\ cos a'\ Eliminating Vi between these two equations we have as the relation between u\ and Vi when there is no loss at entrance to the runner. The power delivered by the water to the runner may be expressed as Tco = W h" = e h Wh, where T has the value given by equation (100). Thus W u Wh" = (riVi cos AI rzVz cos A 2)-* If the discharge is "radial," A 2 = 90 and hence F 2 cos A 2 = 0. Therefore Solving equations (130) and (131) simultaneously, we have e h 2gh sin 2 From this it follows that 2 sin (a'i A\) cos 2gh sin (a'i / _ e h sin a'i _ '\2sm(a f l -A 1 )cosAi le h sm(a'i - AI) ** = \Trin a'i cos Ax' It must be borne in mind that equations (132) and (133) can be 228 HYDRAULICS applied only for the special case stated. For any other speed a different procedure would be necessary but it will not be given here. 1 The speed desired is the one for which the gross efficiency is a maximum and this may not be quite the same as the one for which the hydraulic efficiency is the highest. Hence the true value of fa may differ slightly from the value given by equation (133). These equations appear to be independent of conditions at outflow from the runner. But it must be noted that they are to be used only upon the assumption that the dimensions used at exit will be such as to make A 2 = 90, when c e and e have the values given. An interesting result may be obtained by multiplying equations (132) and (133). This gives (134) This would indicate that all other things being equal, the higher the value of fa the smaller the value of c e . With the impulse turbine, to which these equations apply also, fa is small but c = c v and is near unity. With the reaction turbine fa is larger than for the impulse wheel but c is smaller. 146. Theory of th,e Draft Tube. If the draft tube is properly designed its area next to the runner should be such that the velocity in it is equal to Vz, the absolute velocity of discharge, otherwise there will be an abrupt change of velocity involving losses. For Fig. 217 we may write #2 = Pz + z 2 + V z */2g, and H 4 = 0. The losses between points (2) and (4) are made up of the friction losses within the tube, H' f , and the discharge loss at (3). Apply- ing the general equation bet ween "points (2) and (4) we have 7 2 2 7, 2 ""r*~j& +jr/ ** (135) The larger the diameter of the tube at its mouth the less will be the value of F 3 and hence the less the pressure at the point of discharge from the runner. But if too great a rate of diffusion is provided for in the tube the flow in it will be unstable and *A general relation between c and for all conditions will be found in the author's "Hydraulic Turbines," Chap. VIII. THEORY OF THE REACTION TURBINE 229 the friction loss H'f will be increased. The pressure at the top of the draft tube should not be made less than about 5 ft. absolute, and the value of z z determined accordingly. A " high-speed " turbine runner with a large value of F 2 cannot be set as far above the water level as a " low-speed" turbine with a smaller value of F* If H'f be assumed equal for both straight and flared tubes, it may be seen that the diverg- ing draft tube increases the head utilized by the turbine by 147. Head on Reaction Turbine. For a reaction turbine the draft tube is an integral part of the machine, hence (Fig. 218) the head under which it operates is given by h = H c - H f = 20 (136) This is the value of h upon which computations are based, and it is the one to be used in determining the efficiency of the turbine. FIG. 218. However, though the turbine maker usually constructs or designs the draft tube also, he is often limited by the conditions of the setting and may not be able to use the proper proportions. In order to eliminate this defect in the setting, over which he has no control, the velocity head at (E) is sometimes deducted from the value given by equation (136). If it were feasible to 230 HYDRAULICS eliminate the friction in the draft tube as well we should then have the efficiency of the runner alone, which is independent of the draft tube. But what we usually desire is the efficiency of the entire unit from the intake of the casing to the tailrace. 148. PROBLEMS 1. A certain reaction turbine was found by actual test to have a hydraulic efficiency of 0.83 when < = 0.670 and c = 0.655. The angles were: Ai = 13 and a'i = 115. Compute the values of e and c e and compare with the actual values. (The slight discrepancy between the two is largely due to the fact that shockless entrance and radial discharge were not obtained at exactly the same speed.) Ans. e = 0.678, c e = 0.628. 2. For a reaction turbine the dimensions are: Ai = 35, o'i = 136, h = 0.845. Compute the values of e and c e . Ans. e = 0.85, c e = 0.60. 3. In the test of the Cornell University turbine the pressure was read by a mercury manometer attached near the intake flange where the diameter was 30 in. At full load when the discharge was 44.5 cu. ft. per sec., the manometer read 9.541 ft. of mercury, the top of the shorter mercury column being 0.500 ft. above the intake. If the elevation of the intake above the water level in the tailrace is 9.230 ft. find the head on the turbine. Ans. 140.5ft. 4. In the turbine of problem (3) the diameter of the draft tube at the upper end is 24.5 in. and at the bottom it is 42 in. Find the gain in head due to its use when the discharge is 44.5 cu. ft. per sec. 5. The top of the draft tube in problem (4) is 10.0 ft. above the level of the water in the tailrace. Neglecting the friction in it, but considering the discharge loss at the bottom, find the pressure at its top. 6. A reaction turbine by test was found to discharge 31.8 cu. ft. of water per sec. when running at 600 r.p.m. under a head of 143.1 ft. If FI = 0.535 sq. ft. and D = 27 in., find values of c and <. CHAPTER XV TURBINE LAWS AND FACTORS 149. Operation under Different Heads. In the entire dis- cussion in the two preceding chapters we have assumed that the head remained constant though the other quantities might vary. But a turbine may be installed in a plant where the head changes from time to time, and also a given design of turbine might be used in different plants under a wide range of heads. Thus we desire to investigate this phase. Let us recall the expression, Ui = 4>\/2gh. Suppose now that a turbine is compelled to run at a constant speed while the head varies. It is clear that < also varies then just as it would in the preceding case. But it would be possible under some circumstances to change the speed as well in such a way as to keep a constant. Hence we need to consider two distinct cases when the head changes ; one is where is constant, and the other is where also changes. If remains constant, the wheel speed must vary as \fh. But a definite value of is accompanied by a definite value of c. Hence the rate of discharge must also vary as \/Ti, since Vi c\^2gh. Now the power of the water is proportional to the product of q and h. Since q varies as \ / Ji, it follows that the power varies as h?*. And in similar fashion it may be shown that the torque varies as h. The hydraulic efficiency is a function of c, 0, and the turbine dimensions. As long as remains constant the hydraulic efficiency remains the same regardless of the head. This must be true because the hydraulic losses may all be shown to vary as h , just as the power input. But the friction of the bearings and the windage or the disk friction of the runner do not vary in the same way. It is not possible to formulate an exact law for this but they may be said to vary between N and N 2 . Since N varies as yT, they must vary between h H and h. Hence the mechanical losses become a smaller percentage of the total as 231 232 HYDRAULICS the head increases. 1 But, except for very low heads, the dif- ference in the efficiency is usually a matter of not more than 2 or 3 per cent, at most. See Fig. 219. Now if the speed remains constant while the head changes, or if it does not vary as \/h, the value of will change. Re- ferring to Fig. 215, it may be seen that this means a change of c also. Hence the efficiency will change. Thus none of the simple proportions that have just been stated will be true in such a case. It is impossible to calculate the new results unless Hydraulic Efficiency Gross Efficiency = Constant Head FIG. 219. Effect of head upon efficiency of a given turbine. curves, such as those of Fig. 215, are available, or unless we have some complex equations which will give values of all these quantities for any value of . 150. Different Sizes of Runner. If a series of runners are all built of the same design with the same angles and proportions so that one is simply an enlargement or reduction of another, they should all have the same values of e and c e . Since their peri- pheral speeds would all be the same under a given head it follows that their rotative speeds would be inversely proportional to their diameters. And the area FI would be proportional to D 2 . Hence their capacity and power would vary as D 2 . Thus if the performance of one runner is known, that of the rest of the series 1 An impulse wheel should be set with sufficient space on either side of the buckets at discharge so that the water rebounding from the walls will not strike them. The velocity with which the water rebounds is pro- portional to Vz and hence to \/Ji. Therefore if this space is not ample for all values of h, a point may be reached where this action would decrease the efficiency. TURBINE LAWS AND FACTORS 233 may be predicted with some assurance, due allowance being made for slight increases in efficiency as the size increased. We may express these statements algebraically as follows: D where 1,840 is a constant which we obtain when we solve for N in terms of peripheral speed, the latter being given by equation (121). For the two types of turbines in common use we have: Impulse wheel <$> e = 0.43 to 0.48 Reaction turbine e = 0.60 to 0.90 according to design. And as to capacity q = KiD*\fh (138) where K\ has the following range of values: Impulse wheel KI = 0.0002 to 0.0005 Reaction turbine KI = 0.0014 to 0.0360 It must be understood that these constants are based upon values corresponding to e and that the speed of the wheel must be such that e is obtained if they are to apply. Making a suitable allowance for the efficiency the power delivered by the turbines can be determined when the discharge is known. It may be seen that the peripheral speed of the reaction turbine is higher than that of an impulse wheel and that it may be varied through a wider range by changes in the design. Also the values of KI show that for a given diameter a reaction turbine can dis- charge more water, and hence develop more power, than an im- pulse turbine. That means that if they are to deliver the same power the diameter of a reaction turbine will be much less than that of a corresponding impulse wheel. Thus for a given head and power the rotative speed of the reaction turbine will be much , higher than that of the impulse wheel due, both to its higher peripheral speed and to its much smaller size. 161. Specific Speed. A useful factor in turbine work is one that will now be derived. It involves the head, speed, power, and efficiency. 234 HYDRAULICS Since power is proportional to D 2 and to h % we may write, b.hp. = KzD 2 h 3/i . This may be rewritten as Inserting the value of D as given by equation (137) we have Vb.hp. Rearranging this and letting \/^2 1,8400 = N,, we have (139) While any value of N might be used, the expression has but little meaning unless a particular value is employed. That is generally understood to be N e , the value of N at which the maximum efficiency under a given head is attained. As to the value of b.hp. it should logically be the one for which the maxi- mum efficiency is obtained under the given head. But in some cases the value of the maximum power at this same speed is used. The quantity N a is generally known as specific speed. Other names applied to it are unit speed, type characteristic, and characteristic speed. Its value indicates the class to which a turbine belongs. Thus we have seen that for a given head and power the impulse wheel runs at a relatively low r.p.m. There- fore it would have a low value of N 8 . For an impulse wheel under a given head at a given speed the power would increase with the size of the nozzle used. Thus there need not be any lower limit to the value of N a but the upper limit would be the one for which we had the maximum size jet that could be employed. We find that the efficiency is not appreciably reduced until after we pass a value of about 4.5 for N a and after a value of 6 the jet is so large for the size of the wheel that the efficiency drops off materially. But any value above 4.5 involves some sacrifice of efficiency. For the reaction turbine we have limits in both directions as indicated below, though these may be extended in future designs. The values of the specific speed are *h* = h Xh^ TURBINE LAWS AND FACTORS 235 Impulse wheel N 8 = to 4.5 (6 max.) Reaction turbine N 8 = 10 to 100. For a given turbine the value of N 8 is naturally a constant' but it is also practically constant for a whole series of runners of the same design regardless of size. The larger the diameter of a runner the greater its power but the less the value of N for a given head. Hence the product remains constant. Values of N 8 given for the impulse wheel are for a single jet upon a single wheel. When two or more jets are used the power is naturally increased without changing the speed. This enables values between 6 and 10 to be obtained, if necessary. For values above 100 the conditions are impossible. Either the power or the speed of the unit must be decreased. The specific speed factor shows that the impulse wheel is a low-speed, low-capacity turbine and the reaction turbine is a high-speed, high-capacity turbine. The use of these words is relative rather than absolute. Thus the turbine in Fig. 170 runs at only 55.6 r.p.m. but its specific speed is 82.3, thus in- dicating that it is a high-speed wheel. For the speed is high as compared with that of other turbines of the same power under that head. For instance the speed of an impulse wheel for similar conditions would be only 4 r.p.m. And the specific speed of the highest head impulse wheels in the world (Art. 125) is only 0.592 though they run at 500 r.p.m. But a slow- speed reaction turbine under the same conditions would run at 8,450 r.p.m. at least, and a high-speed reaction turbine such as those at Cedars Rapids would run at 69,300 r.p.m. Of course these values are absurd and simply demonstrate the fitness of each type for its own field. 152. Uses of Specific Speed. The values of N a , as of all other factors in this chapter, are supposed to be obtained from test data, not computed by theory. They serve to classify a turbine and indicate to what type it belongs. They are useful in selecting units for a prospective plant. For such a case the head is known but the size and speed of the units is not. If it is desired to use wheels of a certain type, that fixes the value of N a between narrow limits, and it is easy to compute the com- binations of speed and power that can be produced. Or, if the speed and power be fixed, it may at once be found what type of turbine would be required. J 236 HYDRAULICS 153. Factors Affecting Efficiency. The efficiency of the im- pulse wheel is practically independent of the size of the wheel. The author makes this statement after testing sizes from 12 in. to 84 in. in diameter and comparing all the other test data which is accessible. It would seem reasonable that this should be so, for there is no loss in connection with the impulse wheel which would not vary in proportion to the power of the wheel. Aside from questions of design and workmanship the efficiency would appear to be a function of the specific speed. Too low a value of the specific speed would mean a large diameter of wheel for a given power output with a consequently large friction and windage loss. Too high a specific speed would mean that the jet was too large for the wheel and buckets with a consequent lowering of the hydraulic efficiency. The most favorable value Impulse Wheel Size of Unit FIG. 220. Effect of size of turbine upon its efficiency. of N a is about 4.0 and the best efficiency that is obtained is about 82.0 per cent. This is slightly exceeded at times and values below it are often obtained. With the reaction turbine the efficiency is a function of its size. This is partly due to the fact that the hydraulic ef- ficiency increases with the size, but more to the fact that the volumetric efficiency increases. With a reaction turbine there is always a certain amount of leakage between the guides and the runner so that a portion of the water escapes through the clearance spaces and does not pass through the wheel. The area of these clearance rings would naturally be less in proportion to the area of the wheel passages as the size of the wheel in- creases. Hence a much larger per cent, of the water is made to deliver its power to the runner. Such a condition does not TURBINE LAWS AND FACTORS 237 exist with the impulse wheel. This leads to comparative values for the two types as shown in Fig. 220. Another distinction between the two types of turbines is that the reaction turbine suffers certain hydraulic losses on part gate that are lacking in the other. Hence, although in some cases the maximum efficiency of a reaction turbine is greater than that of the impulse wheel, the efficiency on a light load might not be as good. Like the impulse turbine the efficiency of the reaction turbine also depends upon the specific speed, being less at either extreme. The best efficiencies are obtained with values of N 8 ranging from 30 to 60. The efficiency of a turbine of good design and work- manship depends upon size, specific speed, and other factors to such an extent that definite values cannot be given, but for fair size units it should range from 80 to 90 per cent, and occa- sionally more. For small wheels, especially with unfavorable specific speeds, a value of from 60 to 80 per cent, is all that should be expected. 154. PROBLEMS 1. The turbine, whose performance is shown in Fig. 216, developed its maximum efficiency of 88.0 per cent, when delivering 550 hp. at 600 r.p.m. under a head of 141.8 ft. The water consumed was 38.8 cu. ft. per sec. What would be its proper speed under a head of 283.6 ft.? What would then be the rate of discharge and the horsepower? 2. In Fig. 215 the turbine delivered 617 hp. when running at 600 r.p.m. under a head of 140.5 ft., the rate of discharge being 44.5 cu. ft. per sec. and the efficiency 87.0 per cent. If the speed is maintained at 600 r.p.m. when the head is 70.2 ft., fincl values of discharge, power delivered, and efficiency. (Note: This can be determined only by making use of the curves for this particular turbine. The procedure would be to find the value of for the new conditions and then take values of q, hp., and e from the curves. These quantities would then have to be reduced to the proper values for the new head.) 3. Compute the factors < and N s for the runners shown in Figs. 152, 157, 167, 170, and 172. Compare these values with each other. 4. It is desired to develop 6,000 hp. at 514 r.p.m. under a head of 625 ft. Will an impulse or reaction turbine be required? (This can be deter- mined by computing the specific speed.) 6. If only 900 hp. is to be developed for the conditions given in problem (4), what type of turbine will be required? 6. It is desired to use a type of turbine whose specific speed is 30 to deliver 100 hp. under a head of 100 ft. What will be the proper r.p.m. for the unit? 7. An impulse wheel is to be used for 625 hp. under a head of 144 ft. 238 HYDRAULICS What will be the maximum rotative speed at which it can be run without material sacrifice of efficiency? What will be the approximate diameter of the wheel? 8. What would be the minimum speed for a reaction turbine for the conditions of problem (7)? If = 0.60, what would be the diameter of the runner? 9. What would be the maximum speed for a reaction turbine in problem (7)? Assuming < = 0.85, what would be the diameter of the runner? 10. The runner in the Cornell University turbine is 27 in. in diameter. The wheel develops 550 hp. when running at 600 r.p.m. under a head of 141.8 ft. What would be the speed and power of a 54-in. runner of the same type under the same head? Would the specific speed of these two be the same? CHAPTER XVI THE CENTRIFUGAL PUMP 155. Definition. Centrifugal pumps are so called because of the fact that centrifugal force or the variation of pressure due to rotation is an important factor in their operation. 1 In brief, the centrifugal pump consists of an impeller rotating within a case as shown in Fig. 221. Water enters the impeller at the center, flows radially outward, and is discharged around the circumference into the case. During flow through the impeller the water has received energy from the vanes resulting in an increase both in pressure and velocity. Since a large part of the energy of the water at discharge is kinetic, it is necessary to conserve this kinetic energy and transform it into pressure, if the pump is to be efficient. As a matter of convenience in illustration, the water is repre- sented as entering the impeller in Fig. 221 with a positive pres- sure. However, the pump is usually set above the level of the water from which it draws its supply, in which case the pressure at this point would be negative. Likewise, the axis of rotation need not be vertical as shown. 156. Classification. Centrifugal pumps are broadly divided into two classes: 1. Turbine pumps. 2. Volute pumps. While there are still other types these two are the most funda- mental. Also, as we shall see, these may in turn be subdivided in other ways. The turbine pump is one in which the impeller is surrounded by a diffuser containing stationary guide vanes as shown in Fig. 222. These provide gradually enlarging passages whose function it is to reduce the velocity of the water leaving the impeller and thus efficiently transform velocity head into pressure head. The casing surrounding the diffuser may be either circular and con- 1 For a more complete treatment of this entire subject, either descriptive, theoretical, or practical, see "Centrifugal Pumps," by R. L. Daugherty. 239 240 HYDRAULICS centric with the impeller or it may be spiral like the cases of some reaction turbines. The volute pump, shown in Fig. 223, is one which has no diffusion vanes but, instead, the casing is of a spiral type so FIG. 221. proportioned as to produce an equal velocity of flow all around the circumference and also to gradually reduce the velocity of the water as it flows from the impeller to the discharge pipe. Thus the energy transformation is accomplished in a slightly THE CENTRIFUGAL PUMP 241 different way. This spiral is often called a volute, whence the name of the pump. Occasionally pumps have been built with a whirlpool chamber as shown in Fig. 221. This produces a free spiral vortex, the nature of which has been shown in Art. 120. FIG. 222. Turbine pump. 157. Description of the Centrifugal Pump. The centrifugal pump is similar to the reaction turbine both in its construction and in its theory. However, one is not the reverse of the other, and their differences are as striking as their similarities. FIG. 223. Volute pump. The rotating part of the pump which is instrumental in delivering the water is called the impeller. Impellers may receive water on one side only or, as in Fig. 224, from both sides, in which case they are known as double-suction im- pellers. Fig. 225 gives a view of the pump whose section is 16 242 HYDRAULICS seen in Fig. 224. It may be seen that this impeller is relatively narrow as compared with its diameter, while the opposite type is shown in Fig. 226. For the same rotative speed the latter will discharge more water than the former but at a lower head. For high heads it becomes desirable to place impellers in series, in which case we have the multi-stage pump, such as is shown in Fig. 227. Multi-stage pumps may be either of the turbine or the volute type. The former may be seen in Fig. 227 and the latter in Fig. 228. The addition of guide vanes so as to produce a turbine pump results in a much more complex con- Courtesy of the Allis-C halmers Mfg. Co. FIG. 224. Double-suction volute pump. struction, as Fig. 229 will show. The water in a multi-stage turbine pump usually passes from one impeller to the next through passages which are like those shown in Fig. 230. There are other arrangements besides this, but they will not be de- scribed here. 158. Conditions of Service. Centrifugal pumps are used for lifting water a few feet only or as much as several thousand feet, if necessary. Several such pumps have been built for heads of 2,000 ft. The capacities of centrifugal pumps range from very small quantities up to as high as 300 cu. ft. per sec. (134,500 G.P.M. or 194,000,000 gal. per 24 hr.). The 1. P. Morris Co. has built several of the latter for a head of 16 ft. at 77.5 r.p.m. THE CENTRIFUGAL PUMP 243 Courtesy of the Allis-Chalmers Mfg. Co. FIG. 225. Double-suction volute pump. Courtesy of Platt Iron Wks. FIG. 226. Double-suction volute pump. 244 HYDRAULICS Courtesy of Chicago Pump Co. FIG, 227. Two-stage turbine pump. Courtesy of Platt Iron Wks. FIG. 228. Three-stage centrifugal pump without diffusion vanes, THE CENTRIFUGAL PUMP 245 Courtesy of Platt Iron Wks. FIG. 229. Two-stage centrifugal pump with diffusion vanes. Discharge Suction FIG. 230. Worthington two-stage turbine pump. 246 HYDRAULICS The greatest power of any centrifugal pump is that of a pump installed by Sulzer Bros, in Italy. A single-stage pump running at 1,002 r.p.m. delivers 32,530 G.P.M. at a head of 498.6 ft. with an efficiency of 81.0 per cent. The water horse- power is 3,590 and the power required to run it is 4,430 hp. For most pumps the power required is less than 500 hp. Rotative speeds may vary all the way from 30 to 3,000 r.p.m. in ordinary practice according to circumstances. The highest speed ever employed was 20,000 r.p.m. for a single-stage volute Courtesy of Allis-Chalmers Mfg. Co. FIG. 231. 72-in. centrifugal pump for drainage at Memphis, h = 15'; N = 100. Capacity, 194,000,000 gal. per day. pump with an impeller 2.84 in. in diameter. The pump delivered 250 G.P.M. against a head of 700 ft. with an efficiency of 60.0 per cent. The highest peripheral speed used was with a single- stage pump with an impeller 3.15 in. in diameter. At 18,000 r.p.m. it delivered 189 G.P.M. against a head of 863 ft. and for a smaller discharge it developed a head of 995 ft. Centrifugal pumps have been built with as many as 12 stages. It is customary to limit the head per stage to a value of not more than 100 to 200 ft., but this has been greatly exceeded in several cases mentioned above. Water turbines are rated according to the diameters of their runners, but the size of a centrifugal pump is usually designated THE CENTRIFUGAL PUMP 247 by giving the diameter of the discharge pipe. The rated head and discharge for a centrifugal pump are the values for which the efficiency is a maximum under a given speed. This value of the rate of discharge is often designated as the normal dis- charge. These values will be different for different speeds. 159. Head Developed. The head developed by a centrifugal pump when no flow occurs is called the " shut-off head" or the Pressure Chamber ^==^ FIG. 232. Crude centrifugal pump. "head of impending delivery." Its value may be found by applying the principles of Art. 119/ If water in a closed chamber be set in motion by a paddle- wheel as in Fig. 232, there will be an increase in pressure from the center to the circumference. If the water is assumed to rotate at the same speed as the impeller, the peripheral velocity of which is u 2 , it may be seen from equation (117) that p% p\ = u 2 */2g, where pi denotes the pressure at the center. If this water ?.s in communication with a pressure chamber to which a pie- 248 HYDRAULICS zometer tube is attached, as in Fig. 232, water will rise in the latter to such a height that h = u 2 z /2g (140) If the height of the tube were less than this, water would flow out and we should have a crude centrifugal pump. Actually there are certain influences at work in the real pump which affect this relation slightly. Some of these factors tend to increase the head and others to decrease it. The net effect is that for the usual type of centrifugal pump the head of impending delivery is h = 0.85 to 1.10 uj/2g. (141) 130 120 110 100 < 90 80 70 60 50 40 30 20 10 / Risln istl-c ^"^ ^N s. / / \ \ X / Fh t Chi ractei .istic \ \ ***** -ssj ~r^r- o""" 1 - < \ \ r 0j ^ 5^ ^ N \ \ S! ^ x s \ \ v \ \ 1 "\ ^ \l o '3 s W | 1 \ Spee d Con rtant M a S \ \ \ 102030406060708090100110120130140 Percent of Normal Discharge FIG. 233. Head-discharge characteristics of different pumps. But as soon as flow occurs the above relation is no longer true, as may be seen in Figs. 233 and 234. When water is being delivered, the head may be either greater or less than the shut-off head, according to the design of the pump. We shall now derive a general relation between head, impeller speed, and rate of discharge for all conditions of operation. THE CENTRIFUGAL PUMP 249 As the water in the suction pipe approaches the impeller it may have a rotary motion imparted to it before it ever reaches the latter, due to the viscosity of intervening particles of water. Hence, we shall write equations between points (2) and (s) in Fig. 221, the latter point being removed far enough from the impeller so that the water has no rotational flow imparted to it. In Fig. 234a is shown the hydraulic gradient in the case of Uz 2 zero flow and, as in equation (140), h = ^~- ^9 When flow occurs there will be a drop in pressure at (2), which is just within the impeller, due to the velocity head at that point 60 30 W 2 Efficien De jLav Q W.H.P. 10 0.2 0.4 0.6 P.8 1.0 1.2 1.4 L6 1.8 2.0 2.2 2A 2.6 Discharge-Ou. Ft. p.er Sec. FIG. 234. Characteristics of a 6-in. pump at a constant speed. and also due to the loss of head within the impeller passages or from (s) to (2). If k" is the coefficient of loss the drop in pressure at the outlet of the impeller will be (1 + k"}v^/2g. 1 But the pressure at (s) likewise decreases by an amount equal to V s 2 /2g. Also as the water flows from (2) to (d) in Fig. 234a, there is a reduction in velocity head from V 2 2 /2g to V d */2g. This means. 1 As an illustration consider a hose with a nozzle on the end. When the nozzle is opened so that water may flow, the pressure at the base of the nozzle is decreased below the value obtained when it is closed, by an amount equal to the velocity head at that point and to the friction losses up to that point. If next the hose should be moved around, this pressure drop would not be affected in the least, for it is a function of the velocity of flow within the hose, which is the relative velocity, and does not depend upon the velocity of the water with respect to the earth. 250 HYDRAULICS a corresponding gain in pressure, though not without some loss. Thus ra TV/20 is converted into pressure plus V d 2 /2g, the loss being (1 - m)V 2 2 /2g. From equation (145) the total head developed by the pump, including impeller and case, is (see Fig. 234a) h = H d - H s = ( Pd + z d + TV/20) - (p. = h Q - (1 + k")v 2 2 /2g + ra TV/20 _ u 2 2 v 2 2 TV 20 2g 20 V 2 /2g] (143) FT i FIG. 234a. In reality there will be a further drop in pressure at (s) when flow takes plaice, due to the loss of head in friction in the suction pipe. However, this would also have the effect of decreasing the pressures at (2) and (d) by the same amount. Hence the difference in pressure, with which we are here concerned, would be exactly the same. It must be noted that the quantity m is a variable. When the discharge from the impeller is such that the angle A 2 (see Fig. 235) agrees with the angle of the diffusion vanes of a turbine pump, or the velocity Vz is the proper value for a volute pump, the maximum proportion of the velocity head will be saved. For larger or smaller discharges than this there will be additional losses attending this conversion. For a turbine pump the maximum value of m is about 0.75 and for a volute pump it is somewhat less. With a centrifugal pump the impeller areas are fixed and THE CENTRIFUGAL PUMP 251 constant in value and hence it is convenient to express the rate of discharge as 0. = /a2. (144) FIG. 235. Velocity diagrams. FIG. 236. Stream lines for three different rates of discharge. Now referring to the vector diagrams shown in Figs. 235 and 236 it may be noted that as the rate of discharge varies the values of Vz and A 2 change. It may be seen that as q ap- proaches zero, v 2 and Az approach zero, while Vz approaches uz. Hence for an infinitesimal discharge the value of Vz may be regarded as equal to u 2) while the velocity of the water in the case surrounding the impel ] er is practically zero. Therefore, a particle of water leaving the impeller with a high velocity enters 252 HYDRAULICS a body of water at rest and loses all of its kinetic energy. Thus, as the rate of discharge approaches zero, the factor m approaches zero. Hence it may be seen that when we have zero discharge the value of h in equation (143) reduces to that given by equa- tion (140). An inspection of equation (143) serves to explain the rising or falling characteristics of Fig. 233. If the increase of pressure due to the conversion of the velocity head of discharge is more than enough to offset the decrease due to the velocity and the losses within the impeller, we have a rising characteristic. If they are about equal we have a flat characteristic, and if the quantity mVz 2 is less than (1 + k")v 2 , and V% vary as Hence we shall find it convenient to write 1 u z = V2h (147) (148) We find that a certain value of is required to obtain the maximum efficiency just as with the turbine. And a definite value of c is associated with every value of as may 1 For the pump has the same meaning as in the inward flow reaction turbine since it gives the peripheral speed in both cases. But c has a different meaning, since we find it more convenient to deal with v 2 rather than with F 2 - THE CENTRIFUGAL PUMP 257 i be seen from equation (143). For ordinary types of pumps we find the following values of these factors: For shut-off = 0.95 to 1.09 For normal discharge $ e = 0.90 to 1.30 For normal discharge c e = 0.10 to 0.30 The value of 6 will depend upon the design of the pump. Thus the smaller the angle 0,2 and the fewer the number of impeller vanes, the larger the value of e . Just as in Art. 150, it may be shown that - : . (137) 163. Specific Speed. The specific speed factor for turbines involves the developed horsepower, since that is the quantity with which we are concerned. But with centrifugal pumps we are primarily interested in their capacity and it will be more useful if we derive a similar expression giving N 8 in terms of discharge. Since power and discharge are really proportional to each other it may be seen that we are merely expressing the specific speed for the pump in terms of different units. Proceeding just as in Art. 151, except that we use equation (138) direct, we obtain for the centrifugal pump (149) The capacity of a centrifugal pump is generally expressed in gallons per minute rather than in cubic feet per second. Thus the expression will probably be the handiest in the above form. (1 cu. ft. = 7.48 U. S. gal.). For an impeller, either single-suction or double-suction, values of specific speed may be found between the following limits : N 8 = 500 to 8,000. For special constructions even higher values may be attained. It must be noted that these apply only to single stages. For a multi-stage pump it is necessary to divide the total head by the number of stages to obtain the proper value of h for use in the equations in this chapter. *Note that h& = h + h y * = h -i- \/V^. Some values of h*. will be found on page 263. 17 258 HYDRAULICS Values of specific speeds are obtained from tests of actual pumps and they may then be applied to other pumps of the same type. For N 8 is an index of the type of pump just as it is in the case of the turbine. Its great value is that it enables us to determine the combinations of speed, capacity, and head per stage that are possible or desirable. And if we desire to employ a certain type of pump with a definite value of N 9t we may then find the combinations of these factors that are required. 164. Operation at Different Speeds. In this chapter we have shown characteristics of centrifugal pumps operating under variable heads at constant speeds. We may now desire to know how the pump is affected by a change in speed. This is shown by equations (147) and (148). To obtain similar conditions of operation it is necessary that the values of and c be maintained constant. If they are, it may be seen that both the speed and discharge of the pump will vary as the square root of the head. But if $ and c are not constant then we have no simple index to the variation of the equantities. We can then only resort to some second degree equation of the form shown in Art. 159. Hence if the head is varied due to a change in speed it must be understood that the rate of discharge varies also if the following simple ratios are to apply. From equation (147) it may be seen that 1 U 2 2 . . h = ^w Which shows that if remains constant, the head developed varies as the square of the pump speed. From equation (148) we may obtain, after substituting the value of h given by equa- tion (150), v z = -Uz 9 Hence it follows that if < remains constant, c will also remain constant and the rate of discharge must vary directly as the speed. Since power is a function of the product of h and q it may be seen that it will vary as the cube of the speed. Just a& in the case of the turbine, the hydraulic efficiency of a centrifugal pump is independent of the speed, within reasonable limits, as long as is constant. But the maximum gross efficiency of a given pump will increase slightly as higher speeds are attained. THE CENTRIFUGAL PUMP 259 165. Factors Affecting Efficiency. The considerations of Art. 153 apply here also. The most important factor in deter- mining the efficiency of a centrifugal pump is its capacity, as may be seen in Figs. 240 and 241. A pump of small capacity will have a low volumetric efficiency because of the relatively large per cent, of the water which will leak back into the 1,000 2,000 3,000 4000 5,0j ~ m 6,000 ,1,000 8,000 9,000 10,000 Turbine Pumps, Discharge in GalU'peE'Min. FIG. 240. Efficiency as a function of capacity. suction side through the clearance rings. Also the disk fric- tion of such a pump is a greater percentage of the total power expended. It may be shown that the head per stage has only a slight effect upon the efficiency of the pump, providing the design is carefully made. oiwi 90 "- . ' Vf " ' ' H ( W 3 I 30 W p 2,000 4,000 6,000 8,000 .10,000 12,000 14,000 16,000 18,000 20,0 Volute Pumps, Discharge in Gal. per Min. FIG. 241. Efficiency as a function of capacity. For a given capacity, however, the efficiency will be found to differ with different pumps, due not only to variations in work- manship and construction but also to the other factors such as speed and head. Since these are all involved in the specific speed, it would seem reasonable that efficiency may be expressed 260 HYDRAULICS as a function of the latter. Figs. 242 and 243 show the relation between efficiency and specific speed for a large number of turbine and volute pumps. But it should be borne in mind that for any given specific speed the larger the capacity the higher the efficiency. Hence we can have no single curve that will enable us to select definite values for any case. 90 I- 200 400 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 360Q Tutblne Pumps Specific Speed y 3= N Y G - P -^L- FIG. 242. Efficiency as a function of specific speed. 40U 800 1200 1600 2000 2400 2800 3200 3600 4000 4400 4800 6200 660060006400 6800 7200 7600 8000 Volute Pumps. Specific Speed, N ' = W ^ FIG. 243. Efficiency as a function of specific speed. 166. PROBLEMS 1. The curves of Fig. 239 are for a single-stage pump in which D = 9.12 in., / 2 = 0.0706 sq. ft., a 2 = 27. At 1,700 r.p.m. when q = 1.315 cu. ft. per second, h = 55.7 ft. If it be assumed that m = 0.50, find the value of k". 2. A two-stage turbine pump running at 1,700 r.p.m. delivered 0.429 cu. ft. per sec. at a head of 225 ft. The essential dimensions were: D = 12 in., / 2 = 0.0244 sq. ft., a 2 = 26. K it be assumed that m = 0.70, compute the value of k". 3. The curves of Fig. 234 were obtained from the test of a single-stage pump having the following dimensions: D =9.12 in., / 2 = 0.0706 sq. ft. Find the values of t and c e . 4. If it be assumed in Fig. 233 that the value of for shut-off is 1.0, what is the value of < for the maximum lift of the pump with the rising char- acteristic? What is the value of for maximum efficiency in each case? THE CENTRIFUGAL PUMP 261 6. The diameter of a pump impeller is 10 in. The speed is to be 1,200 r.p.m. If = 1.20, what is the value of hi 6. Compute the value of the specific speed for the pump shown in Fig. 231. 7. Compute the value of the specific speed for the pump whose dimensions are given in problem (2). 8. What would be the capacity, head, and power of the pump whose performance is shown in Fig. 234, if it were run at a speed of 1,000 r.p.m.? 9. What speed would be necessary to double the capacity of the pump whose curves are shown in Fig. 234? What speed would be required to double its lift? 10. If the speed of the pump of Fig. 234 were doubled, what would be the head for a discharge of 2.4 cu. ft. per sec.? What would be the ef- ficiency for this rate of discharge at the higher speed? 11. It is desired to deliver 1,600 G.P.M. at a head of 900 ft. with a single- stage pump. What would be the minimum rotative speed that could be used? 12. If a speed of 600 r.p.m. is desired in problem (11), how many stages must the pump have at least? 13. It is desired to use a type of pump whose specific speed is 2,000 under a head of 16 ft. If the speed is to be 1,800 r.p.m., what will be the capacity? 14. Compute the specific speeds of the pumps for which data are given in Art. 158. 262 HYDRAULICS APPENDIX TABLES TABLE 9. AREAS OF CIRCLES Diameter Area Diameter Area Inches Feet Square inches Square feet Inches Feet Square inches Square feet H 0.0021 0.0491 0.00034 30 2.500 706.9 4.90 K 0.0042 0.1963 0.00136 32 2.667 804.3 5.58 X 0.0062 0.4417 0.00306 34 2.830 907.9 6.30 i 0.083 0.7854 . 00545 36 3.000 1,018.0 7.07 IK 0.104 1.227 0.00853 38 3.17 1,134.0 7.88 IK 0.125 1.767 0.0123 40 3.44 1,257.0 8.72 1% 0.146 2.405 0.0167 42 3.50 1,385.0 9.62 2 0.167 3.142 0.0218 44 3.67 1,521.0 10.57 2y 2 0.208 4.909 0.0341 46 3.83 1,662.0 11.53 3 0.250 7.069 0.0492 48 4.00 1,810.0 12.56 3K 0.292 9.621 0.0668 50 4.17 1,964.0 13.63 4 0.333 12.566 0.0872 52 4.33 2,124.0 14.75 4^ 0.375 15.909 0.1105 54 4.50 2,290.0 15.90 5 0.417 19.635 0.1362 56 4.67 2,463.0 17.10 6 0.500 28.27 0.196 58 4.83 2,642.0 18.35 7 0.583 38.48 0.267 60 5.00 2,827 . 19.62 8 0.667 50.26 0.349 62 5.17 3,019.0 20.93 9 0.750 63.62 0.442 64 5.33 3,217.0 22.3 10 0.833 78.54 0.545 66 5.50 3,421.0 23.8 12 1.000 113.1 0.785 68 5.67 3,632.0 25.2 14 1.167 153.9 1.068 70 5.83 3,848.0 26.7 16 1.333 201.1 1.395 72 6.00 4,072.0 28.3 18 1.500 254.5 1.765 76 6.33 4,536.0 31.4 20 1.667 314.2 2.18 80 6.67 5,027.0 34.9 22 1.833 380.1 2.64 90 7.50 6,362.0 44.2 24 2.000 452.4 3.14 100 8.33 7,854.0 54.5 26 2.164 530.9 3.68 110 9.17 9,503.0 66.0 28 2.332 615.8 4.27 120 10.0 11,310.0 78.5 THE CENTRIFUGAL PUMP 263 TABLE 10. STANDARD WROUGHT-IRON PIPE SIZ'ES Diameter Internal area Diameter Internal area Nominal, inches Actual internal, inches Square inches Square feet Nominal, inches Actual internal, inches Square inches Square feet H 0.27 0.0573 0.0004 3H 3.548 9.887 0.0687 H 0.364 0.1041 6.0007 4 4.026 12.73 0.0884 H 0.494 0.1917 0.0013 4>^ 4.508 15.96 0.1108 H 0.623 0.3048 0.0021 5 5.045 19.99 0.1388 H 0.824 0.5333 0.0037 6 6.065 28.89 . 2006 i 1.048 0.8626 0.0060 7 7.023 38.74 0.2690 iK 1.380 1.496 0.0104 8 7.982 50.04 0.3474 i 1.611 2.038 0.0141 9 8.937 62.73 0.4356 2 2.067 3.356 . 0233 10 10.019 78.84 0.5474 23^ 2.468 4.784 0.0332 11 11.000 95.03 0.6600 3 3.067 7.388 0.0513 12 12.000 113.1 0.7854 TABLE 11. VALUES OF m m*> m * m m ? * m H 0.2 .342 .2 1.69 4.0 2.52 8.0 4.00 0.4 .543 A 1.79 .2 2.60 .5 4.17 0.6 .712 .6 1.89 .5 2.73 9.0 4.33 0.8 .863 .8 1.98 5.0 2.92 10.0 4.63 1.0 1.000 3.0 2.08 .5 3.12 11.0 4.93 .2 1.13 .2 2.17 6.0 3.29 12.0 5.22 .4 1.25 .4 2.26 .5 3.48 13.0 5.52 .6 1.37 .6 2.35 7.0 3.66 14.0 5.80 .8 1.48 .8 2.44 .5 3.83 15.0 6.10 2.0 1.58 TABLE 12. VALUES OF h 1* h 1* i ** h ** 10 5.62 25 11.18 70 24.20 140 40.6 11 6.03 30 12.82 80 26.77 150 42.8 12 6.45 35 14.38 90 29.33 170 47.1 13 6.85 40 15.90 100 31.6 200 53.2 14 7.24 45 17.38 110 33.9 230 59.0 16 8.00 50 18.80 120 36.2 260 64.8 18 8.73 ' 60 21.25 130 38.5 300 72.0 20 9.45 ! 264 HYDRAULICS FUNDAMENTAL TRIGONOMETRY In a right angle triangle, such as Fig. 244 : sin A = a/c sec A = c/b cos A = b/c tan A a/6 esc A cot A c/a b/a Any function of A is the same numerically as the co-function of any combination of A with an odd multiple of 90, Thus: sin A = cos (90 A) = cos (270 A). Any function of A is the same numerically as the function of any combination of A with an even multiple of 90. Thus: sin A = sin (180 A). The sign of the function depends in any case upon the quadrant in which the angle itself lies. b FIG. 244. For the solution of an oblique triangle, such as that shown in Fig. 245, we have sin A _ sin B _ sin C a b c a 2 = b 2 + c 2 - 26c cos A. a* = (b - c) 2 + 46csin 2 ^ a 2 = (b + c) 2 - 46c cos 2 ^ a 2 = (b sin A) 2 + (c cos A - b) 2 This is as much as is required for the solution of the vector triangles that will be encountered with turbines and centrifugal pumps. INDEX Absolute path, 153, 158, 240 pressure, 12 velocity, 152 Air chamber, 151 Air in pipes, 105 Angular momentum, 157 Atmospheric pressure, 11 Automatic crest, 39 B Barometer, 11 Bazin, 87, 132 Bends, pressure at, 144 Bernoulli's theorem, 50 Branching pipes, 117e Buckets, 166 Buoyancy, center of, 31 force of, 30 Case, 189 Center of pressure, 20 Centrifugal action, 161 Characteristics, pump, 218 Chezy's formula, 107 Cippoletti weir, 89 Coefficient of contraction, 63 of discharge, 64 of velocity, 62 Compound pipes, H7d Compressibility of water, 2 Continuity, equation of, 48 Contracted weir, 84 Critical velocity, 46 Current meter, 93 Dams, 32 Density of water, 3 Differential manometer, 16 Diffuser, 239 Discharge loss, 100 Discharge, measurement of, 92 rate of, 48 Draft tube, 185 Dynamic force, 141, 153 E Effective head, 52 Efficiency, factors affecting, 236, 259 as functions of size, 236, 259 of pipe line, 113 pump, 162 turbine, 162 Energy, 54 Entrance losses, 97 Flashboards, 39 Flow, line, 199 non-uniform, 135 steady, 47 uniform, 127 unsteady, 148 Fluid, 2 Force exerted, 104, 143, 215 Forced vortex, 162 Forebay, 201 Francis turbine, 177 Francis weir formula, 86 Free surface, 2 Free vortex, 164 Friction factors, 110 G Gage height, 138 Gage pressure, 12 Gates, 182 Governing, 172, 185 Gradient, hydraulic, 104 265 266 INDEX Graphical integration, 23a Guide vanes, 178, 183 Hazen- Williams, 108 Head, developed by pump, 114, 251 for impulse wheel, 219 for reaction turbine, 229 losses of, 96 meaning of, 52 measurement of pump, 252 shut-off, 247 Hook gage, 80 Hydraulic gradient, 104, 128 mean depth, 97 radius, 97 slope, 107 Manometric coefficient, 256 Measurement of head, 14, 252 Metacenter, 31 N Needle nozzle, 170 Non-uniform flow, 135 Nozzles, 71, 170 efficiency of, 73 loss in, 100 Open channel, 125. 132 Orifice, 65 coefficients, 69 Ideal velocity, 65 Impeller, 241 Impending delivery, 247 Impulse of jet, 156 Impulse turbine, 155, 165, 209 Intensity of pressure, 7 Jet, coefficients of, 63 definition of, 62 force of, 153 power of, 55 - Kinetic energy, true, 57 Kutter's formula, 129 Laterals, pipe with, 1170 Liquid, definition, 2 Losses of head, 96 M Manning's formula, 131 Path of water, 153, 158 Pelton wheel, 165 Penstock, 201 Piezometer, 14 Pipe efficiency, 113 friction, 97, 108 Pitot tube, 91, 145 Power, 160, 217 delivered by pipe, 113 meaning of head, 55 plants, 198 Press, hydraulic, 17 Pressure, absolute, 17 on area, 19, 23 gage, 17 intensity, 7 negative, 13 wave, 148 Pump, characteristics, 218 service, 242 turbine, 239 volute, 240 R Radius, hydraulic, 97 Rate of discharge, 48 Rating curve, 138 Reaction of jet, 156 INDEX 267 Reaction turbine, 155, 177, 221 Relative velocity, 152 Runner, 179 S Settings, turbine, 188 Shut-off head, 247 Siphon, 123 Size of pipe, 110, 117a Slope, hydraulic, 107 Specific speed, 234, 257 uses of, 235 Speed, 218, 226, 258 Standard orifice, 67 Steady flow, 47 Stream gaging, 136 Suppressed weir, 81 Surge, 148 Surge chamber, 151 Tangential wheel, 165 Torque, 158, 159, 224 Trapezoidal weir, 84 Triangular weir, 81 Tubes, 70 Turbine case, 189 Turbine pump, 239 Turbine setting, 188 U Uniform flow, 127 Units, 5 Unsteady flow, 148 Vacuum, 12 Varying head, 94a Velocity, absolute, 152 diagrams, 211 critical, 46 head, 52 measurement of, 93 relative, 152 Venturi meter, 74 Volute pump, 240 Vortex, 161, 164 W Water barometer, 1 1 Water hammer, 107, 148 Water, properties of, 3 Weirs, 79 Wetted perimeter, 128 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY w"Ll INCREASE TO so CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. MAY 6 i efee-i SEP 1H 943 NOV 22 DEC 24 ^WT'W ..