A TEXT-BOOK ON HYDRAULICS INCLUDING AN OUTLINE OF THE THEORY OF TURBINES BY L. M. HOSKINS ' Professor of Applied Mathematics in the Leland Stanford Junior University NEW YORK HENRY HOLT AND COMPANY 1907 Copyright, 1906, BY HENRY HOLT AND COMPANY ROBERT DRCMMOND, PRINTER. NEW YORK PEEFACE, THIS book is designed primarily for the use of students of engineering in universities and technical colleges. In its prep- aration the aim has been to present fundamental principles in a manner both sound and as simple as possible. The treatment presupposes a good elementary knowledge of the principles of mechanics, and a working knowledge of the elements of cal- culus; but to the student thus equipped, who is also well trained in arithmetic, algebra and trigonometry, it presents little mathe- matical difficulty. Many numerical examples Are introduced, the complete solution of which should form an important part of the work of the student. It is perhaps not too much to say that the key to a correct understanding of all problems in the steady flow of liquids is supplied by Bernoulli's theorem, or, as it is usually called in the text, the general equation of energy. Familiarity with this principle is therefore much more important than a memory- knowledge of special rules, and for this reason the explanations of particular cases of flow have in most cases been based directly upon the fundamental equation. The meaning and importance of the term representing lost energy in this equation have also been emphasized. The corresponding theory applied to gases is given in Appendix A. In the presentation of working rules for estimating flow in the various practical cases met by the engineer it has been aimed in every case to give a clear statement of the rational basis of the formula adopted, and also to make clear to what extent the theory is defective and the formula therefore empiri- cal. It has also been attempted to avoid the appearance of in 161891 iv PREFACE. precise knowledge where the reality is absent. For example, no elaborate tables have been given purporting to show accu- rately how frictional loss of head in pipes depends upon velocity and diameter, or giving precise values of the friction factor for pipes of different kinds. A somewhat careful study of experi- mental data has failed to convince the author of the reliability of any such tables. The treatment of turbines and water wheels has been re- stricted to an outline of the theory, but several illustrations showing typical American practice have been included. For these the author is indebted to the courtesy of manufacturers, to whom credit is in every case given in the text. The aim has been to unify the theory, the treatment of all specific cases being based upon the same general principles and equations, and a general notation for velocities and their direction-angles being adopted which it is hoped will be found simple and helpful. This unification includes the theory of turbine pumps. In various discussions throughout the book reference is made to the author's text-book on Theoretical Mechanics for a fuller explanation of basal principles. L. M. H. PALO ALTO, CAL., June, 1906. CONTENTS. CHAPTER PAGE I. PRELIMINARY DEFINITIONS AND PRINCIPLES 1 II. HYDROSTATICS . 7 III. FLOW OF WATER THROUGH ORIFICES. TORRICELLI'S THEO- REM 30 IV. THEORY OF ENERGY APPLIED TO STEADY STREAM MOTION. ... 43 V. APPLICATION OF GENERAL EQUATION OF ENERGY, NEGLECTING LOSSES BY DISSIPATION 52 VI. APPLICATION OF GENERAL EQUATION OF ENERGY, TAKING AC- COUNT OF LOSSES 64 VII. GENERAL EQUATION OF ENERGY WHEN PUMP OR MOTOR is USED 85 VIII. FLOW IN PIPES: SPECIAL- CASES 90 IX. FRICTIONAL Loss OF HEAD IN PIPES 102 X. EQUATION OF ENERGY FOR STREAM OF LARGE CROSS-SECTION. 116 XI. UNIFORM FLOW IN OPEN CHANNELS 122 XII. OPEN CHANNELS: NON-UNIFORM FLOW. 135 XIII. THE MEASUREMENT OF RATE OF DISCHARGE 142 XIV. DYNAMIC ACTION OF STREAMS 160 XV. THEORY OF STEADY FLOW THROUGH ROTATING WHEEL 176 XVI. TYPES OF TURBINES AND WATER WHEELS 185 XVII. THEORY OF IMPULSE TURBINE 190 XVIII. THE TANGENTIAL WATER WHEEL 199 XIX. THEORY OF THE REACTION TURBINE 208 XX. TURBINE PUMPS 224 APPENDIX A. STEADY FLOW OF A GAS 249 APPENDIX B. RELATIVE MOTION 261 APPENDIX C. CONVERSION FACTORS 264 INDEX 267 v XA ; I UNIVERSITY 1 >K r\c ^^ HYDRAULICS CHAPTER I. PRELIMINARY DEFINITIONS AND PRINCIPLES. 1. Definition of Subject. The mechanics of fluid bodies is called Hydromechanics. It embraces Hydrostatics, dealing with the principles of fluid equilibrium, and Hydrokinetics, dealing with the laws of fluid motion. Hydraulics, the subject of this book, may be defined briefly as practical Hydromechanics. It deals especially with the flow of water in streams of various kinds, but may be taken to in- clude all the principles and applications of Hydromechanics that bear directly upon problems of practical utility. Many of the laws of Hydraulics are largely empirical, but certain fundamental dynamical principles, especially the law of energy, serve to unify the subject and to put it upon a scientific basis. The bodies dealt with in Hydromechanics may be either liquids or gases, Hydraulics deals mainly, but not exclusively, with liquids, and especially with water. 2. Distinction between Solid and Fluid Bodies. A solid body can permanently resist change of shape; a fluid body cannot. A fluid is either liquid or gaseous. A gas tends to expand indefinitely, so as to fill any continuous closed volume in any portion of which it may be placed. A liquid changes its vol- ume only slightly under changes of pressure; a given portion may be wholly freed from external pressure without expanding beyond a certain volume. PRELIMINARY DEFINITIONS AND PRINCIPLES. The distinction between solids and fluids may be made more definite by a consideration of internal stresses. 3. Internal Stresses in a Body. The two equal and oppo- site forces exerted by two portions of matter upon each other constitute a stress. The forces making up a stress are thus the "action and reaction " of Newton's third law of motion. If the two portions of matter are parts of the same body, the stress is internal with reference to that body. Internal stresses, acting between adjacent portions of a body, are called into action whenever external forces tend to change the shape or size of the body. If a plane surface be conceived to divide a body into two contiguous parts, the forces which these parts exert upon each other will for convenience be regarded as resolved into com- ponents normal and tangential to the plane. The stress com- posed of these forces is thus resolved into (a) A normal stress, which resists whatever tendency there may be for the two parts of the body to approach or recede from each other in the direction of the normal to the plane of separation, and (6) A tangential stress, which resists any tendency to slid- ing, or relative motion parallel to the plane. 4. Mathematical Definitions of Solid and Fluid. A fluid body is one in which tangential stress cannot act, except while the shape of the body is changing. If, by reason of external forces, any two adjacent portions of a fluid tend to slide over each other, tangential stresses come into action to resist such sliding. The sliding is not pre- vented, however, but continues until a condition of equilibrium is attained; in this condition the tangential stress on every plane vanishes. A solid body is one in which tangential stresses can act permanently to resist change of shape. A " perfect" fluid may be defined as one which offers no resistance to change of shape. In other words, no tangential PRESSURE; INTENSITY OF PRESSURE. 3 stress acts in a perfect fluid even while the particles are sliding over one another. No known fluid is perfect in this sense. The laws of equilibrium are the same for an actual fluid as for a perfect fluid, since it is only when the parts of a body of fluid move relatively to one another that tangential stresses act. In other words, the statics of actual fluids is the same as the statics of perfect fluids. 5. Pressure ; Intensity of Pressure. The normal stress be- tween two adjacent parts of a body may be either tensile or compressive. Tensile stress (or tension) resists a tendency of the two por- tions of the body to separate. Compressive stress (or pressure) resists a tendency of the two portions of the body to approach each other. In Hydromechanics we are concerned mainly with pressure, since a fluid body can sustain only a slight tensile stress. We shall have to consider not only internal pressure, but also pres- sure acting between a body of water and other bodies in con- tact with it. Intensity of pressure means pressure per unit area. If, on any surface subject to pressure, the pressures upon any two elementary areas, however small, are proportional to the areas, the intensity of pressure is uniform over the surface. In this case its value is at every point equal to the total pressure divided by the total area. Algebraically, let P = total pressure on area F\ p= intensity of pressure at any point of the area; then P If the intensity of pressure has not the same value at all points of the area, we may regard P/F as its average value for the area F. The true value of p at any given point may be expressed approximately as its average value over a small area containing the point. If 4F is the area of this element PRELIMINARY DEFINITIONS AND PRINCIPLES. and AP the total pressure on the element, we have approxi- mately * p ' the approximation being closer the smaller the element. Taking AF smaller and smaller with limit 0, but always containing the point at which the value of p is to be expressed, we have as the exact value of the intensity of pressure at that point AP dP For brevity the word " pressure " is often used instead of " intensity of pressure." This abbreviation will sometimes be employed in the following discussions, but should be avoided when it is liable to cause ambiguity. Although the foregoing discussion of intensity of stress has referred to normal stresses only, similar considerations hold for tangential stresses. 6. Elasticity. Elasticity is the property by virtue of which a body regains its original size and shape (in whole or in part) after these have been changed by the action of external forces. Both elasticity of volume and elasticity of shape are pos- sessed in very different degrees by different bodies. Fluid bodies possess practically perfect elasticity of volume, but no elasticity of shape. Since change of shape always involves sliding, or tangential motion of the parts of a body relative to one another, the body cannot of itself regain its* original- shape except by the con- tinued action of the tangential stresses which resist such motion. A fluid body, having been deformed from one form of equilib- rium to another, cannot of itself return to the original shape, because after equilibrium is attained no tangential stresses are in action. DENSITY AND COMPRESSIBILITY OF WATER. 7. External Forces. A force acting upon any portion of a body is called external if the portion of matter exerting the force is not a part of the body. The external forces acting upon any body of fluid are of two classes: (a) surface forces and (6) bodily forces. (a) A surface force is one whose place of application is some portion of the bounding surface of the body. Such forces are exerted by other bodies in contact with the given body. (6) A bodily force is one which is applied throughout some definite volume of the body. Such a force does not depend upon contact between the two bodies concerned; it is of the kind called "action at a distance." .An example of a bodily force is gravity, acting upon every particle of a body of fluid near the earth's surface. In practical Hydraulics this is the only bodily force to be considered. 8. Density and Compressibility of Water. Practical Hy- draulics deals mainly with water. The properties of water with which we shall chiefly be concerned are its density and com- pressibility. Density. The density (or mass per unit volume) of water varies but little with pressure and temperature within ordinary ranges. The density of pure water at several different tem- peratures and under ordinary atmospheric pressure is given in the following table: TABLE I. Temp, Fahr. Density in Ibs. per cu. ft. Temp. Cent. Density in kgr. per cu. met. 32 62.42 999.9 39.3 62.424 4 1000.0 50 62.41 10 999.8 60 62.37 15 999.2 70 62.30 20 998.3 80 62.22 25 997.1 90 62.12 30 995.8 100 62.00 35 994.6 110 61.86 40 992.4 The density of ordinary terrestrial water is increased very slightly by the presence of various substances in solution. In 6 PRELIMINARY DEFINITIONS AND PRINCIPLES. most ordinary computations the convenient numbers 62.5 Ibs. per cubic foot and 1000 kilograms per cubic meter may be em- ployed; 62.4 Ibs. per cubic foot is, however, more accurate than the former value. The density of sea- water is about 2.6 per cent greater than that of pure water. Compressibility. The compressibility of water is so small that for ordinary purposes it may be neglected. The ratio to the original volume of the decrease in volume caused by a given pressure (of uniform intensity over the whole bounding surface) may be taken as a measure of the compressibility. The value of this ratio for a pressure of 1 atmosphere * varies somewhat with the temperature. Near the freezing-point it is about 0.000050; at 77 Fahr. (25 Cent.) it is about 0.000045. * See Art. 14. CHAPTER II. HYDROSTATICS. 9. Pressure on Different Planes Passing Through a Point. From the fact that no tangential* stress exists on any plane in a body of fluid in equilibrium, it follows that the intensity of normal pressure has, at a given point, the same value for all planes passing through that point. Consider any two planes passing through the point (Fig. 1). Let OM, ON be their traces on a plane perpendicular to both which is taken as the plane of the figure. Take OA = OB, and \ / let AB be the trace of a third \ _ / plane which is perpendicular to \ /w, x*z sina cosa the plane of the figure. Consider the body of fluid bounded by five planes as follows: The plane of the figure ; a plane parallel to it at a distance 2; planes perpendicular to the plane of the figure having traces OA, OB, AB. The forces acting upon this body are the normal pressures exerted upon every part of its bounding surface by the surrounding fluid, and whatever bodily forces may be acting upon it. Let the average value of the bodily force per unit volume, for the body just described, be w', and let Wi =< component of w' in direction AB', pi = average intensity of pressure on the face OA; p 2 = average intensity of pressure on the face OB; 2a= angle AOB; x = OA=OB. 7 8 HYDROSTATICS. Then pixz = total pressure acting on face OA ; p 2 xz= " " " " " OB; WiX 2 z sin a cos a = total bodily force in direction AB. These are the only forces acting on the body which are not perpendicular to AB; hence for equilibrium, resolving in direction AB, PIXZ cos a p 2 xz cos a +WiX 2 z sin a cos a =0, or pi p 2 +WiX sin a = 0. This equation is true for any values of x and z. If both be made to approach 0, the third term of the equation approaches 0, so that limit pi = limit p 2 . But the limiting values of pi and p 2 are the true values of the intensity of pressure at the point on the planes OA and OB respectively. And since these may be any two planes passing through without changing the reasoning, the proposition is established. 10. Normal Pressure in a Fluid Free from Bodily Forces. Although in all the practical problems of Hydraulics and Hydro- statics the fluids considered are acted upon by the force of gravity, it is instructive to consider the ideal case of a fluid not acted upon by any bodily force. It is easily shown that for such a body in equilibrium the intensity of pressure has the same value at all points. Let A and B (Fig. 2) be any two points such that the straight line AB lies wholly within the fluid. Consider an elementary prism of small cross-section F, so taken that A and B lie in its two bases. Let pi and p 2 denote the values of the intensity of pressure at A and B respectively, and let all forces acting upon the prismatic element be resolved parallel to FIG. 2. its axis. The only forces not normal to the direc- tion of resolution are the pressures on the end elements at A and B; hence for equilibrium we have p 2 F = 0, or pi p 2 . PRESSURE IN FLUID ACTED UPON BY GRAVITY. 9 That is, the intensity of pressure has the same value at A and at B. This result may obviously be extended to any two points in a continuous body of fluid free from the action of bodily forces. 11. Variation of Pressure in Fluid Acted upon by Gravity. In case of fluids at the earth's surface the only bodily force to be considered is the attraction of the earth upon every particle. For a fluid of uniform density this attraction (per unit volume) has practically the same magnitude and direction at all points. If the weight of a pound mass (called a pound force) is taken as the unit force, the force per unit volume has the same value as the mass (in pounds) per unit volume. The law of variation of pressure in a body of fluid acted upon by gravity may be determined as follows: Let A and B (Fig. 3) be any two points such that the straight line AB lies wholly in the fluid. Let BC be horizontal and AC vertical, and let AB = l, AC=z. Consider an elementary right prism with axis AB and cross-section F, the bases con- taining the points A and B. Let all forces acting upon this prism be resolved parallel to AB. Let pi = intensity of pressure at A ; p 2 = in tensity of pressure at B, w = weight of fluid per unit volume; a = angle BAG. For equilibrium, p l F-p 2 F+wFlcosa=Q. Or, since I cos a = z, 9 p 2 -pi=wz. This result may be extended to any two points of a con- nected fluid, and may be stated in general terms as follows: In any body of homogeneous fluid in equilibrium the pres- 10 HYDROSTATICS. sure varies in direct ratio with the depth. The difference between the values of the pressure (per unit area) at any two points is equal to the weight of a prism of the fluid of unit cross-section and of length equal to the vertical distance be- tween the two points. The following discussions will refer usually to the case of a liquid acted upon by gravity. Units. It is to be remembered that a consistent system of units must be used in applying the above formula. Through- out this book the foot will nearly always be used as the unit length, and the pound as the unit force. The value of w is therefore 62.5 (more accurately 62.4) pounds per cubic foot, z is to be expressed in feet, and p in pounds per square foot. With French units, if the meter is taken as the unit length and the kilogram as the unit force, the value of w is the weight in kilograms of a cubic meter of water, or 1,000. 12. Surface of Equal Pressure. As a special case of the above proposition it is seen that if the intensity of pressure has the same value at any two points, these must lie in the same horizontal plane. Any horizontal plane is, in fact, a surface of equal pressure. A surface of equal pressure is called a level surface. Free surface. The bounding surface of a liquid is said to be free if not sustaining pressure. Ordinarily the upper sur- face of a body of water is called free although under atmos- pheric pressure. Whether under zero pressure or any uniform pressure, the free surface of a liquid in equilibrium is obviously a horizontal plane * if gravity is the only bodily force. 13. Pressure Expressed in Terms of Height of Liquid Column. Fluid pressure is often estimated in terms of the " equivalent height" of some specified liquid. Thus, a column of water 1 foot high is said to be " equivalent to " a pressure of about * The accurate statement is that the free surface (or any level surface) is everywhere normal to the direction of gravity, and is approximately a spherical surface concentric with the earth. ATMOSPHERIC PRESSURE. 11 62.5 pounds per square foot, since the intensity of pressure at the base of such a column exceeds that at the top by the amount stated. In discussions in Hydraulics it is quite com- mon to express pressures in this way. In scientific investigations mercury is often taken as the standard fluid, pressures being expressed as so many inches, or centimeters, of mercury. Since the density of mercury is about 13.6 times that of water, the height of the water column equivalent to a given pressure is about 13.6 times as great as that of the correspond- ing mercury column. 14. Atmospheric Pressure. The air exerts upon all terres- trial bodies a pressure whose intensity is equal to the weight of a column of air of unit cross-section extending upward com- pletely through the atmosphere. In many hydraulic problems this pressure may be disregarded because its effects at different points counterbalance. It is quite common to reckon pressures from atmospheric pressure as zero, pressures of less intensity being regarded as negative. An actual negative pressure (i.e., a tensile stress) of any considerable intensity cannot exist in a liquid, any tendency to such a stress resulting in a separation of the parts of the liquid. The intensity of atmospheric pressure at any locality varies somewhat with weather conditions, and at different places it varies with the elevation. At sea-level under ordinary condi- tions the value is about 14.72 pounds per square inch, or 2120 pounds per square foot. The height of the equivalent water column is very nearly 34 feet (10.34 meters), and that of the equivalent mercury column 30 inches (76 centimeters). At any height above sea-level the corresponding values may be found by multiplying the above numbers by the proper factor taken from the following table, which gives the ratio of atmospheric pressure at any elevation to its value at sea-level. It must be understood that the results will be only approximate, the table being computed for the 12 HYDROSTATICS. ideal case of a perfect gas at the uniform temperature of Cent. TABLE IT. Elevation above sea-level in feet. Ratio of atmos- pheric pressure to its value at sea- level. Elevation above sea-level in feet. Ratio of atmos- pheric pressure to its value at sea- level. 500 .9811 8,000 .7370 1,000 .9626 8,500 .7231 1,500 .9444 9,000 .7094 2,000 .9265 9,500 .6960 2,500 .9090 10,000 .6829 3,000 .8918 10,500 .6699 3,500 .8750 11,000 .6573 4,000 .8584 11,500 .6449 4,500 .8422 12,000 .6327 5,000 .8263 12,500 .6207 5,500 .8107 13,000 .6090 6,000 .7954 13,500 .5975 6,500 .7804 14,000 .5862 7,000 .7656 14,500 .5751 7,500 .7512 15,000 .5642 15. Resultant Pressure. The resultant pressure on any sur- face, plane or curved, is found by combining the pressures on the elementary portions of the surface, having regard for direc- tion as well as magnitude. The point of application of the resultant pressure, for a given surface, is called the center of pressure. 16. Resultant Pressure on Horizontal Plane Area. Let F = area of a horizontal plane surface ; z =its depth below free surface of liquid: p = intensity of pressure at any point; P= resultant pressure on area F. Then P=Fp=wzF. The center of pressure (or point of application of P) is evi- dently coincident with the centroid of the area; for the forces RESULTANT PRESSURE ON PLANE AREA. 13 of which P is the resultant are parallel, and are proportional to the elementary areas on which they act. 17. Magnitude of Resultant Pressure on any Plane Area. If a submerged plane area is not horizontal, the magnitude of the resultant pressure may be computed as follows: Let AB and A'B' (Fig. 4) represent two vertical projections of the surface, its inclination to the horizontal being 0. Let 2= depth of an elementary area dF below free surface of liquid; P= resultant pressure on whole area F. B FIG. 4. Then the total pressure on the element dF is hence P = fpdF = wfz dF, the integration being extended over the whole area F. If z is the value of z at the centroid of the area F, we have f z dF=zF; therefore P = wzF. 18. Center of Pressure of Plane Area. To determine the point of application of the resultant pressure P, the principle of moments must be employed. Referring to Fig. 4, let the line of intersection of the plane AB with the free surface be taken as axis of moments, and let the moment of the resultant pressure be equated to 14 HYDROSTATICS. the sum of the moments of the pressures on all elementary areas. Let the distance of an elementary area dF from the axis of moments be denoted by y, the corresponding vertical distance being z, so that z = y sin 6. Let z, y refer to the centroid of the area F, and z' ', y' to the center of pressure. The equation of moments is the integration covering the whole area F. Since dP = wz dF =wy sin 0-dF, and P = wzF=wj sin 0-.F, the equation may be written wF yy f sm6 = w sin f y 2 dF\ from which The numerator of this value is equal to the moment of iner- tia of the area with respect to the line in which its plane inter- sects the free surface of the water; the denominator is the statical moment of the area with respect to the same axis. Denoting these quantities by / and G respectively, we have If the area has an axis of symmetry which is perpendicular to the line in which its plane intersects the water surface, the center of pressure lies in this axis and is completely determined by equation (1). If this is not the case, a second moment- equation is required for the complete location of the center of pressure. Taking as axis of moments a line lying in the area and perpendicular to the axis from which y is measured, let x denote CENTER OF PRESSURE OF PLANE AREA. 15 the distance from this axis of an element dF; then the equation of moments is Px'= xP, which may be written wF yx r sin = w sin j xy dF\ from which jJ^J- (2) yF ~G ........ (2) Here J denotes the product of inertia of the area F with respect to the axes from which x and y are measured. EXAMPLES. Find the .magnitude and point of application of the resultant pressure on the area described in each of the following examples, the liquid being water. 1. A rectangle of sides 4 ft. and 6 ft., placed vertically, (a) with shorter side in water surface, (b) with longer side in water surface. Ans. (a) P = 4500 Ibs.; t/' = 4 ft. 2. A rectangle with sides b, d, placed vertically^ with the side b in the water surface. Ans. P = wbd*/2; y' = %d. 3. A circle of diameter d } its plane being inclined at angle to the vertical, and the center being distant a vertically below the surface. Ans. P = wnd*a/4i', y' = asec6+d 2 /16asecd. 4. A circle of 2 ft. diameter, the highest point being 6 inches below the surface, and the plane inclined 60 to the vertical. Ans. P = 196 Ibs.; y' = 1\ ft. 5. A semicircle with plane vertical and diameter in the surface. 6. A triangle of base 2 ft. and altitude 3 ft., the plane being vertical, (a) with vertex in water surface and base horizontal, (6) with base in surface. Ans. (a) y' = ld. (b) y' = \d. 7. An area F, whose radius of gyration about a horizontal central axis is k, placed vertically, with centroid at depth a below water surface. Ans. P=wFa-, y' = 16 HYDROSTATICS. 8. The vertical plane area shown in Fig. 5. 5' 5' 15' FIG. 5. FIG. 6. 9. The vertical plane area shown in Fig. 6. Ans. P = 14,000 Ibs.; x' = 3.7l ft., 5.71 ft. 19. Pressure Resolved in Given Direction. The component, in any direction, of the resultant pressure on a plane area may be found by the following rule : Pass a plane through the centroid of the area perpendicular to the given direction, and project the area orthographically upon it; the pressure on this projected area is equal to the required component of the resultant pressure on the given area. Thus, let it be required to find the component, in the direc- tion MN, of the resultant pressure on the area AB (Fig. 7). FIG. 7. The value of the resultant pressure is P = wFz, z being the depth below the free surface of C, the centroid of the area F. Through C pass a plane perpendicular to MN, and let A'E f be the projection of AB on this plane. The projected area is F'=Fcos6, where 6 is the angle between P and MN] the HORIZONTAL PRESSURE ON CURVED SURFACE. 17 centroid of the projected area coincides with that of the given area AB. Resolving P in the direction MN, we have P cos 6 = wF z cos 0, while the resultant pressure on A'E' is wF'z=wFz cosO. These values being equal, the proposition is proved. EXAMPLES. 1. Compute the horizontal and vertical components of the resultant pressure on a rectangular area 6 ft. by 8 ft., inclined 30 to the vertical, one of the longer edges being in the water surface. Ans. Hor. comp.=6750 Ibs.; vert. comp.=3900 Ibs. 2. Compute the horizontal and vertical components of the resultant pressure on a circular area 4 ft. in diameter, inclined 20 to the vertical, the center being 5 ft. below the water surface. Ans. Hor. comp.=3690 Ibs.; vert. comp.=1345 Ibs. 20. Horizontal Pressure on Curved Surface. In case the surface is not plane, the resolved pressure in a given direction cannot in general be computed by the rule above given for a plane surface (Art. 19). The rule does hold, however, if the direction of resolution is horizontal. Thus, consider the pressure upon the surface AB of the submerged body X (Fig. 8). This pressure is identical with FIG. 8. that upon the surface AB of the liquid which would replace X if removed. HYDROSTATICS. Let A'B' be the orthographic projection of the surface AB upon any vertical plane, and consider the body of fluid ABB' 'A' '. The horizontal component of the pressure on A B is counter- balanced by the pressure on A'B', and must therefore have the same magnitude and line of action. EXAMPLES. 1. A circular cylinder 4 ft. long and 2 ft. in diameter is placed with axis horizontal, and is filled with water to a depth of 18 inches. Com- pute the magnitude and line of action of the horizontal thrust of the water in a direction perpendicular to the axis of the cylinder. Ans. 281 Ibs. acting in line 1 ft. below water surface. 2. A hemispherical bowl 2 ft. in diameter is filled with water. Deter- mine the magnitude and line of action of the resultant thrust of the water on the bowl in a given horizontal direction. Ans. 41.7 Ibs.; .59 ft. below surface. 3. A cylindrical barrel 2 ft. in diameter is filled with water to a depth of 30 inches. Determine the magnitude and line of action of the result- ant horizontal thrust on half the interior surface Ans. 391 Ibs.; 20 inches below water surface. 21. Vertical Pressure on Curved Surface. Let AB (Fig. 9) be a portion of the surface of a submerged body X, and let it be required to compute the resultant of the vertical components A' of the pressures exerted by the water upon the elements of AB. Let A'B' be the orthographic projection of AB upon the plane of the water surface. The volume of water ABB' A' is in equilibrium under the action of its weight and the pres- PRESSURE ON MASONRY DAM. 19 sures acting on its bounding surface. The surface A 'B r being assumed free from pressure,* the resolution of these forces vertically shows that the vertical component of the pressure on AB is equal to the weight of the body of water ABB' A', and that its line of action passes through the center of gravity of this body. In the case shown in Fig. 9, the resultant vertical pressure of the water on the surface AB of the body X acts down- ward; while in such a case as that shown in Fig. 10 it acts upward. But in the latter case, as in the former, the vertical pressure is equal in magnitude to the weight of a body of water ABB' A' . 22. Resultant Pressure on Curved Surface. The resultant pressure on a curved surface is not, in the most general case, a single force, since a system of forces in three dimensions is not generally equivalent to any single force. In practical prob- lems it will usually suffice to determine the effective pressures in horizontal and vertical directions, by the methods just explained. If it is required to carry the reduction of the system farther, this may be done by methods explained in works on Statics. f It is sometimes evident from symmetry that the total pressure is equivalent to a single force. This is true in the following case. 23. Pressure on Masonry Dam. In discussing the stability of a dam it is necessary to compute the resultant pressure exerted upon it for a given length, say one foot. The horizontal and vertical components of this pressure may be computed by the rules above given, and since these components act in the same vertical plane, they have a single resultant which may readily be determined. * Atmospheric pressure is usually neglected, because in practical cases it is commonly the excess of pressure above that due to the atmosphere that is of importance. It may, however, be taken account of in the present case by conceiving the column of water ABB' A' extended to a height p /w above the actual water surface, p being atmospheric pressure. f Theoretical Mechanics, Chapter X. 20 HYDROSTATICS. FIG. 11. In Fig. 11, the total horizontal pressure H on the face AB is equivalent to the resultant pressure upon A'B, the projection of AB on a vertical plane; while the vertical pressure V is equal to the weight of the body of water A A'B. The line of action of H passes through the center of pres- sure of A'B, while that of V passes through the center of gravity of A A'B', their intersection gives a point in the line of action of P, the resultant of H and V. In masonry dams as actually constructed, the profile AB often differs but little from a vertical straight line, and the vertical component of the pressure is neglected in discussing the stability of the dam. This is an error on the side of safety in the design. Upward pressure on base of dam. If a masonry dam rests wholly or partly upon a bed of porous material such as sand or gravel, which is continuous with the bed of the reservoir so that water freely enters and saturates it, an upward pressure results which may have an important effect upon the stability of the dam. If the water is at rest throughout the porous bed, the upward pressure must be computed as due to the column of water A'B (Fig. -11). If the water flows through the gravel, as it will unless intercepted by an impervious barrier, the pres- sure will decrease in the direction of flow (Arts. 90-93). Even if the dam rests upon impervious rock, a hori- zontal fissure in the masonry may permit the entrance of water, thus causing an upward pressure upon the masonry above. The following examples refer to a dam of the cross-section shown in Fig. 11, the dimensions being as given below. In all cases the required pressures are to be computed for one linear foot of the dam. Referring to the figure, let y denote depth below water sur- face, x the corresponding horizontal distance from A'B to AB, CURVED SURFACE UNDER UNIFORM PRESSURE. 21 and x r the horizontal thickness of the masonry. The values given in the table are in feet. y X *' y X x' -10 34.8 17.5 80 29.8 57.5 34.3 19.2 90 28.5 66.0 10 33.8 21.7 100 26.6 75.7 20 33.4 24.6 110 24.2 86.5 30 32.9 28.5 120 21.2 98.2 40 32.4 33.0 130 17.6 111.0 50 31.9 38.2 140 12.9 125.3 60 31.5 44.0 150 7.2 140.7 70 30.6 50.4 160 0.0 159.5 EXAMPLES. 1. Determine the magnitude, direction, and line of action of the resultant pressure on the surface AB. 2. Assuming the dam to rest upon saturated gravel for its entire thickness BD, compute the total upward force if the pressure is every- where due to the head A'B. 3. Compute the total upward force if the pressure head varies uni- formly from at D to A'B at B. 4. If the material of the dam weighs 150 Ibs. per cu. ft., compute the magnitude and line of action of the weight per linear foot. Deter- mine the magnitude, direction, and line of action of the resultant of the water pressure and the weight of the dam; also the moment of this resultant about D. (Solve on each of the above assumptions as to the pressure on the base.) 24. Curved Surface under Pressure of Uniform Intensity. If a body is submerged to a depth which is great in comparison with the vertical dimension of the body, the variation of the intensity of pressure on its bounding surface will be small in comparison with its actual value, and for practical purposes this variation may usually be disregarded. If a surface is under pressure of uniform intensity, the total resolved pressure in any direction is equal to the resultant pressure on the orthographic projection of the surface upon a plane perpendicular to that direction. 22 HYDROSTATICS. Thus, consider the pressure on the part AB of the surface of the body X (Fig. 12). If it is required to compute the resolved part of this pressure in any given direction, let A'E' be the ortho- graphic projection of AB upon a plane perpendicular to that direction. If X were removed, the body of water ABB'A' would be in equilibrium under the action of the pressures exerted by the surround- Fra 12 * n ^ ^ U ^ ^^ e we *gkt f this kdy Dem g by supposition negligible in comparison with these pressures). Resolving perpendicularly to A'B 1 ', the resolved pressure on AB exactly balances the pressure on A'&. EXAMPLES. 1. A 6-inch pipe carries water under a head of 200 ft. Compute the total pressure per foot of length on one half the interior surface. Ans. 6250 Ibs. 2. Compute the resultant pressure on a hemispherical surface 2 ft. in diameter subjected to a pressure of 5 atmospheres. Ans. 33,300 Ibs. 25. Resultant Pressure on Submerged Body. The resultant pressure exerted by a fluid upon a body which is wholly or partly submerged in it is a force equal and opposite to the weight of the displaced fluid, and its line of action passes through the center of gravity of the displaced fluid. It is obvious that, if the body were removed, the body of fluid which would replace it would be subjected to exactly the same pressure as that actually exerted upon the submerged body. But the body of fluid would be in equilibrium under the action of two sets of forces: (a) the weight of every particle, with resultant acting through the center of gravity; (6) the pressure of the surrounding fluid upon every element of the bounding surface. The resultants of these two sets of forces must be equal in magnitude, opposite in direction, and col- linear. The resultant pressure of a liquid upon a body wholly or CONDITIONS OF EQUILIBRIUM OF A FLOATING BODY. 23 partly submerged in it is called the " buoyant force." The cen- troid of the submerged volume (the point of application of the buoyant force) is the " center of buoyancy." If a body is submerged partly in one fluid and partly in another, the resultant pressure exerted upon it by both is equal to the total weight of both fluids displaced. Thus, a body float- ing at the surface of water displaces a certain body of water and a certain portion of air. The weight of the displaced air is so small in most ordinary problems that it is often disre- garded. In the following discussions relating to floating bodies the pressure of the air will not be considered unless specifically mentioned. 26. General Conditions of Equilibrium of a Floating Body. If the only forces acting upon a floating body are its weight and the pressure of the liquid, these two forces (or strictly sets of forces) must balance each other. This requires (1) That the weight of the displaced water shall equal the weight of the body, and (2) That the center of gravity of the body and that of the displaced water shall lie in the same vertical line. From these two conditions it is possible to determine what volume of water will be displaced by a body of known weight; and in the case of bodies of regular shape to determine by inspection some or all of the possible positions of equilibrium. EXAMPLES. 1. A plank 2" by 12" by 16', weighing 40 Ibs. per cu. ft., can carry what weight without sinking? Ans. 60 Ibs. 2. A box closed on all sides is made of lumber 1" thick weighing 45 Ibs. per cu. ft. Its outside dimensions are 2' by 3' by 6'. If half filled with water, at what depth will it float? ^ 3. A cone of specific gravity 0.5 floats with axis vertical and apex downward. The altitude being h and radius of base a, compute the depth of flotation. 4. Compute the depth of flotation of the cone described in Ex. 3 if floating with vertex upward. 24 HYDROSTATICS. 27. Floating Body Acted Upon by Any Forces. If a float- ing body is in equilibrium under the action of forces additional to the pressure of the liquid and the weig^f ^ie body, such additional forces must be included in of equilibrium. Since the weight of the body and the force are both vertical, it is evident that, for equilibrium, the additional forces must be equivalent either to a vertical fS^^pr to a couple. EXAMPLES. 2' b^> 1. A homogearneous pallelopiped 1' by 2' by 3^p>f specific gravity 0.25. is half submerged in water, one face being horizontal. What force, besides its weight and the pressure of the water, must be^J|y^to hold it in equilibrium? 2. If the same body is in equilibrium with a diagonal pi plane of the water surface, determine the magnitude, directiol line of action of the resultant force acting upon the body in addition to* its weight and the buoyant force. 3. A right prism whose bases are equilateral triangles floats in such a position that a median plane coincides with the plane of the water surface. If the specific gravity is 0.8, what force, besides gravity and the buoyant force, must be acting upon the body? 28. Stability of Equilibrium. The equilibrium of a floating body is stable, unstable, or neutral, according as the body tends, after being slightly displaced, to return to the original position, to depart farther from it, or to remain in the new position. The brief discussion of stability which follows is limited to the case in which the only forces acting on the body are its weight and the pressure of the liquid. A complete discussion of stability would require a consid- eration of all possible displacements. These displacements may be resolved into translations and rotations. For translations the nature of the equilibrium is obvious, being stable for vertical displacements and neutral for horizon- tal displacements. For rotations the nature of the equilibrium cannot be determined so simply. METACENTER. 25 29. Metacenter. Let a floating body be displaced in such a way that the submerged volume remains constant. Let the body in the new position be represented by KMN (Fig. 13), K LN being the water surface; and let UN' be that plane in the body which coincided with the water surface in the original position of equilibrium. Let A f be the centroid of the volume L'M'N' (being there- fore the position, in the body, of the center of buoyancy in the position of equilibrium), and A the centroid of the volume LMN (being therefore the center of buoyancy in the displaced position) . Let E' be the center of gravity of the floating body. In the original position the line A'B f is vertical. In the new position let a vertical line through A intersect A'E' (pro- duced if necessary) in C. If the angle of displacement be taken smaller and smaller with zero as limit, C approaches a definite limiting position. This limiting position is called the "meta- center." The stability of the equilibrium can be tested by determin- ing the position of the metacenter. It is seen that in the displaced position the forces acting upon the body are equivalent to the following couple: the weight of the body acting downward through B', and the buoy- ant force acting upward through A. If the metacenter C falls above B', the couple tends to bring the body back to the original position of equilibrium; if C falls below B' , the couple tends to displace the body still farther. Hence in the former case the equilibrium is stable and in the latter unstable; while if the metacenter coincides with the center of gravity of the body the equilibrium is neutral. 26 HYDROSTATICS. < pi *~- 1 -1*rJ f, ^B j T fi j j A simple rule may be deduced for determining the position of the metacenter, and thus testing the stability of the equilib- rium. In studying the stability of ships, however, it is not enough to test whether the equilibrium is stable for small dis- placements, but the degree of stability for both small and large displacements must be determined. The moment of the couple consisting of the weight of the body and the buoyant force, for any displacement, is a measure of the degree of stability. For a vessel of known shape and weight, the value of this moment may be computed for any angular displacement. 30. Variation of Pressure in Rotating Liquid. If a body of water is forced to maintain a con- dition of uniform rotation about a vertical axis, the pressure in- creases with the distance from the axis of rotation in a manner which can be estimated as follows : Let Fig. 14 represent horizontal and vertical sections of a cylin- drical vessel filled with water. If the vessel is caused to rotate uni- formly about its axis of figure, the water will soon take up a rota- tional motion, and the vessel and the water will rotate together as if forming one rigid body. In order to determine the variation of pressure with the dis- tance from the axis of rotation, consider a prismatic element of water of uniform cross-section F whose axis is horizontal and co- planar with the axis of rotation. Such an element is represented in vertical projection at AB and in horizontal projection at A'B' '. Let 7*1 = distance of (A, A') from axis of rotation; r 2 = " " (,') " " " pi = in tensity of pressure at point (A, A'); FIG. 14. co = angular velocity of rotation. VARIATION OF PRESSURE IN ROTATING LIQUID. 27 The fundamental equation of dynamics, force = mass X acceleration, applies to the motion of any body whatever,* force meaning the resultant of all external forces acting on the body, mass the total mass of the body, acceleration the acceleration of its mass- center. Applying this to the elementary prism AB, it is seen that the acceleration of the mass-center and the resultant force have the direction BA, since the mass-center describes a circle whose center lies in the axis of rotation. The radius of this circle being J(ri +r%) t the acceleration has the value The only forces not perpendicular to AB are the pressures on the ends of the element, and the resultant of these is The mass of the element is The dynamical equation therefore reduces to the form w w 2g The variation of the pressure in the vertical direction follows the same law as if the body of water were at rest. Thus, con- sider a prismatic element CD (Fig. 14), whose axis is vertical. If force and acceleration be resolved vertically, it is seen that, since the acceleration in the vertical direction is 0, the sum of the vertical components of all forces acting upon the element must be 0. The forces having vertical components are the nor- mal pressures on the ends of the prism and the weight of every particle of the water. If z\ and z 2 are the heights of D and C respectively above any horizontal plane, pi and p 2 the corre- sponding values of the pressure-intensity, and F the cross-sec- tion of the element, we have for the total downward force * Theoretical Mechanics, Art. 380. 28 HYDROSTATICS. Equating this to 0, P2 Pi =Z\ W W (2) The results expressed by equations (1) and (2) may be com- bined into a single equation as follows : Let A and B be any two points in the rotating liquid; zi, Z2 their ordinates from some horizontal plane; r\, r 2 their dis- tances from the axis of rotation; p i} p 2 the pressures at the two points respectively. Then p 2 - or, in symmetrical form, pi w w (3) . .. . (4) It is not difficult to show that this equation holds even if the axis of rotation is not vertical. In such a case the value of z for a given particle will continually change, since it must be measured from a horizontal plane, irrespective of the direc- tion of the axis of rotation. The form of the containing vessel is obviously of no consequence. 31. Form of Free Surface of Rotating Liquid. It may be shown that, if a body of liquid rotates uniformly about a ver- tical axis, the upper surface, if free, will assume the form of a paraboloid of revolution. Let A (Fig. 15) be the point in which the axis of rotation pierces - the free surface, and B another point of the free surface. Let z denote the height of B above a FlG> 15 - horizontal plane through A, and r the distance of B from the axis of rotation. Then in equation A r ; FORM OF FREE SURFACE OF ROTATING LIQUID. 29 (4) we may put z\ = 0, 22 = z, p\ =p2 = 0, r*i =0, r^ =r; and the equation becomes (5) Evidently r, 2 are the coordinates of the curve cut from the free surface by a plane containing the axis of rotation. Equa- tion (5) represents a parabola with vertex at A and principal diameter vertical. Obviously the same equation results if the pressures at A and B have any equal values; all surfaces of equal pressure are therefore alike. EXAMPLES. 1. A body of water rotates uniformly about a vertical axis, making 80 revolutions per minute. If the upper surface is free, what is the difference in level between its lowest point and a point 16 inches from the axis? Ans. 1.95 ft. 2. Does the variation of pressure due to rotation depend upon the density of the liquid? What is the result of Ex. 1 if the liquid is mer- cury (specific gravity 13.6)? CHAPTER III. FLOW OF WATER THROUGH ORIFICES. TORRICELLI'S THEOREM. 32. Stream of Water with Steady Flow. The streams of water with which hydraulic discussions and experiments are concerned may be either confined in pipes, partly confined in open channels, or wholly unconfined. If, at every cross-section of a stream, the velocity of flow and the form and size of the cross-section remain constant (the conditions at different sections, however, not necessarily being alike), the flow is said to be steady. The most important practical cases are of the kind thus described, and to such the discussion will for the most part be restricted. 33. Rate of Discharge. By rate of discharge of a stream is meant the quantity of water passing a given cross-section per unit time. It will usually be expressed in cubic feet per second. It is evident that, so long as the condition of flow remains steady, the rate of discharge has the same value at all cross- sections, as well as a constant value at any given section. 34. Velocity in a Cross-section. It is impossible to deter- mine, either in magnitude or in direction, the velocities of all the various particles which, at any instant, are passing a given cross-section of a stream. Even at a section where the stream is neither converging nor diverging (as at A, Fig. 16), the direc- tions of motion of the different particles doubtless differ, at least slightly, from the direction of the axis of the stream, 30 MEAN VELOCITY IN A CROSS-SECTION. 31 although the predominating motion has that direction. Still more important, probably, are the irregularities in the mag- nitudes of the velocities; especially in case of a confined stream, in which the particles adjacent to the confining surface are retarded by friction. In a section where the stream converges or diverges (as at B or D) the variation in direction of the velocities of particles in different parts of the cross-section is doubtless still more important. The component of velocity in the direction of the axis of the stream is the only component usually considered. In the following discussions, therefore, it is commonly to be under- stood that "velocity" means "axial component of velocity." o FIG. 16. 35. Mean Velocity in a Cross-section. Mean velocity in a cross-section may be defined as the quotient of the rate of dis- charge by the area of the cross-section. Let q =rate of discharge (cu. ft. per sec.); F = area of cross-section (sq. ft.); v=mean velocity in cross-section (ft. per sec.); then v = p> or q = Fv> 36. Values of Mean Velocity in Different Cross-sections. In case of steady flow, the mean velocity remains constant at any given cross-section, but has different values at different sections if the cross-sectional areas are unequal. If the values of the area at different sections are denoted by F\, F'%, . . . , and the corresponding values of the mean velocity by vi, v-2, . . . , we have FiVi =^2^2= =q= constant. The mean velocity is thus inversely proportional to the area of cross-section. The above equation is called the equation of continuity. 32 FLOW OF WATER THROUGH ORIFICES. 37. Torricelli's Theorem. If a small orifice be opened in the side of a vessel containing water, the velocity of the escaping jet will be nearly equal to the velocity acquired by a body falling freely from rest through a vertical distance equal to the depth of the orifice below the free surface of the water. Let h denote the depth of the orifice below the free surface, and v the velocity of the jet; then the proposition states that the following equation is nearly satisfied : the actual value of v being a little less than that given by the equation. The truth of the proposition is known from experiment, and it is inferred that, if frictional resistances could be eliminated, the equation v 2 = 2gh would be exactly satisfied. It will be shown later that this conclusion follows from the principle of energy. The depth of the center of a small orifice below the free sur- face of the water is called the head on the orifice. 38. Actual Velocity of Jet. The actual mean velocity of the jet is always less than that computed from the formula v 2 = 2gh' t how much less depends upon the nature- of the orifice. In Fig. 17 are represented four cases. The orifices are all supposed to be circular, the smallest diameters being equal, and all are supposed to be under the same head. In case of the sharp-edged orifice shown at A, the stream converges as it passes the plane of the orifice. The smallest cross-section of the jet is found at S, and at this section the mean axial velocity has its FlG - 17. greatest value. For such an orifice the value of the mean velocity is but slightly less than the value computed from the above formula. This is because ACTUAL VELOCITY OF JET. 33 the frictional resistances (which depend upon the velocities with which adjacent portions of fluid slide over each other or over other bodies) are in this case comparatively small. Within the vessel the velocities of the particles are small, and the only place where the friction becomes important is near the orifice. The surface of contact between the jet and the vessel is so small that the retarding effect is slight. A cylindrical orifice in a plate, such as is shown at B, Fig. 17, gives practically the same result as the sharp-edged orifice at A, if the thickness of the plate is small. If, however, the thickness is increased, the condition of the jet changes mate- rially. Thus, consider a cylindrical orifice in a thick plate, as shown at (7, Fig. 17. As the jet enters the orifice it tends to converge (as in preceding cases) . The air surrounding the jet at the con- tracted portion within the orifice is, however, quickly carried out by friction, thus reducing the pressure below that of the atmosphere; this causes the stream to expand and fill the ori- fice. Whether this result is complete or partial will depend upon the thickness of the plate. The frictional resistances between the particles of water (due to the irregular motions set up within the orifice), and also the friction against the cylindrical surface of the orifice, are considerably greater in this case than in the preceding, and the velocity of the jet is correspondingly less. In case of an orifice with rounded inner edge (D, Fig. 17), the frictional resistance due to the surface of the orifice is greater than in the case of the sharp-edged orifice, because the area of contact is greater. The internal friction due to irregularities of motion of the particles is, however, less than in case C. The velocity of the jet in the case shown at D is greater than in case C, but less than in cases A and B. In all cases, the " velocity of the jet" is taken to mean the average velocity in a section outside the orifice at a point where the jet has become cylindrical (sections such as are marked S in the four cases shown in Fig. 17). 34 FLOW OF WATER THROUGH ORIFICES. 39. Rate of Discharge from Small Orifice. The rate of dis- charge from an orifice is equal to the product of any cross- sectional area of the stream into the mean axial velocity in that cross-section. Although this is true for any section, it is com- mon to consider the section at which the stream has become cylindrical.' If the four cases above discussed be compared with refer- ence to the rate of discharge (the smallest cross-section of the orifice having the same value in all cases), it will be seen that the greatest discharge does not necessarily accompany the greatest velocity. Thus, experiment shows that case D gives the greatest discharge and case C the next greatest. The reason is that the greater size of the jet at S in these cases more than counterbalances the greater velocity found in cases A and B. 40. Coefficient^? Velocity. The factor which, applied to the ideal velocity \/2gh, gives the true mean velocity of the jet is called the coefficient of velocity. This coefficient is an abstract number less than unity. 41. Coefficient of Contraction. The ratio of the area of the cross-section of the jet to that of the orifice is called the co- efficient of contraction. This is also an abstract number, and in most cases of practical importance its value is less than unity. 42. Ideal Velocity and Discharge. Practical formulas for velocity and discharge from small orifices are obtained by applying experimental coefficients to formulas which would hold in a certain ideal case. This ideal case is one in which there are supposed to be no frictional resistances to affect the velocity, and in which the cross-section of the jet is supposed to be equal to that of the orifice. If F denotes the area of the orifice and h the head on its center, the ideal velocity is given by the formula ACTUAL VELOCITY AND DISCHARGE. 35 and the ideal discharge (per unit time) by the formula 43. Actual Velocity and Discharge. If c' denotes the coeffi- cient of velocity and c" that of contraction, the true values of the mean velocity of the jet and the rate of discharge may be written v=c f \/2gh; q=c"Fv=c'c"F\/2gh. 44. Coefficient of Discharge. The ratio of the actual value of the rate of discharge to its ideal value is called the coefficient of discharge. If the value of this coefficient is c, we have from the defini- tion and therefore c=c'c". 45. Circular Standard Orifice. An orifice with sharp edge (as A, Fig. 17), or an orifice in a thin plate (as B, Fig. 17), is called a standard orifice. The values of the coefficients of veloc- ity and contraction for small circular orifices of this kind have been fairly well established by experiment. The following may be taken as sufficiently near the true values : c'=0.98; c"=0.62; c=cV'=0.61. 46. Discharge from Large Orifice. If an orifice is not small in comparison with the head on its center, it may be necessary, in estimating the discharge, to take account of the different values of the head for different parts of the cross-section. For the ideal case described in Art. 42 the value of the discharge per unit time may be found as follows: Let dF = area of a differential element of the cross-section; z =head on element dF; then the discharge per unit time through this element is V2gz dF, 36 FLOW OF WATER THROUGH ORIFICES. and the discharge per unit time for the whole orifice is the integral of this expression for the entire area F. For the actual case, the rate of discharge is found by apply- ing a coefficient to this ideal value. The result may be written q=cV2gfz*dF. 47. Large Horizontal Orifice. If the plane of the orifice is horizontal, z is constant and equal to h. The formula therefore reduces to q = cF\/2gh, identical in form with the formula for the case of a small orifice. The value of c is found to vary with the form of the orifice; and for circular orifices it doubtless differs somewhat from the value applying to very small orifices. 48. Large Vertical Orifice. In order to apply the above general formula to the case in which the plane of the orifice is not horizontal, the form and dimensions of the orifice must be given. The most important case practically is that of a rect- angular orifice with one pair of edges horizontal. 49. Rectangular Orifice. Consider a rectangular orifice of horizontal width b and depth d (Fig. 18). Let hi = head on upper edge ; h 2 =head on lower edge; h = head on center. To apply the general formula of Art. 46, let the differential area be an elementary strip of length b (horizontal) and width dz (ver- tical). Then q=cV2gb * If the head on the center is great in comparison with the RECTANGULAR NOTCH OR WEIR. 37 vertical dimension of the orifice, a simpler approximate expres- sion may be used. Thus we have , j d -. , d n 2 =fi-l--; hi=h~-~ Expanding (h +d/2) f and (h-d/2)* by the binomial theorem and substituting in the above formula, the result becomes in which the series converges rapidly except for relatively large values of d/h. If all terms except the first be neglected, we have q = d)d\' 2gh, which is identical with the formula for discharge through a small orifice. Whatever the form of the orifice, the formula obtained by taking account of the actual head on every part of the orifice differs little from that obtained by using the head on the cen- troid as applying to all parts of the area, if h exceeds a small multiple of the vertical dimension of the orifice. 50. Rectangular Notch or Weir. In practical Hydraulics the most important case of rectangular orifice is that in which the upper side is open. The formula for this case is obtained by putting hi = 0. Thus, using a coefficient of discharge, and re- placing h 2 by H, the actual value of q may be written 51. Triangular Notch. Another case of some importance is that of a triangular notch (Fig. 19). Let b = width of notch at water surf ace; H = head on vertex. Take as elementary area a horizon- tal strip at depth z, of vertical width FIG. 19. dz. The length of this strip is b(H-z)/H, and 38 FLOW OF WATER THROUGH ORIFICES. The value of the rate of discharge for the whole area is or c being the coefficient of discharge. The values of the coefficient of discharge for this case and the preceding cases will be considered in Chapter XIII, in which the use of orifices and weirs for the measurement of rate of dis- charge is discussed. 52. Unequal Pressures on Water-surface and Jet. In the ordinary experiments upon which Torricelli's theorem (Art. 37) is based, the surface of the water is under atmospheric pressure, and the jet discharges into the atmosphere. The theorem still holds if the pressures at these two points differ from that of the atmosphere, so long as the two are equal. If, however, unequal pressures exist at the water surface and the point of discharge, the theorem must be modified. Consider first the case in which the pressure at the water surface exceeds that of the atmosphere by p 1} while the jet discharges into the atmosphere. Thus the space above the water (at A, Fig. 20) may contain compressed air. Evidently the effect of the pressure pi on the conditions existing within the body of water is the same as that of an additional depth of FIG. 20. FIG. 21. water. The height of water column necessary to produce a pressure of intensity pi is pi/w; hence the discharge takes place as if under a head h+pi/w with equal pressures at the water surface and jet. SUBMERGED ORIFICE. 39 Next suppose the discharge to take place into a closed chamber (Fig. 21) in which the pressure exceeds that of the atmosphere by p 2 . The effect of this pressure p 2 is the same as that of a diminution of the head on the orifice by the amount p 2 /w. It appears, therefore, that Torricelli's theorem may be applied to the case of unequal pressures if the head on the orifice be corrected for the difference of the pressures. Thus, if the pressures at A and B (Fig. 21) are pi and p 2 respectively, the value of the ideal velocity is w EXAMPLES. 1. If the depth of the orifice below the water surface (Fig. 21) is 4 ft., the pressure at A atmospheric, and that at B absolute zero, com- pute the velocity of the jet. What head would produce the same velocity if the pressures at A and B were equal? Ans. v = 49A ft. per sec. 2. If in Fig. 21 h = 40 ft., pi = absolute zero, p 2 = atmospheric pressure, compute the velocity of the jet. 53. Submerged Orifice. If the discharge takes place as in Fig. 22, the orifice being below the surface of the water in the receiving chamber, the jet at B is under a pressure wh 2 in addition to the pressure existing at the surface D. The head on the orifice must therefore be corrected by the subtraction of h 2 , in applying Torricelli's theorem. That is, FIG. 22. the ideal velocity is D v=\ / 2g(h 1 -h 2 ). In applying this formula to actual cases, the coefficients of velocity and discharge must be given different values from those applying to similar orifices in case of discharge into air. The actual values of coefficients of discharge for some of 40 FLOW OF WATER THROUGH ORIFICES. the cases of most practical importance will be given in Chapter XIII. The foregoing principles will receive theoretical justification in the following chapters, in which the theory of energy is applied to all cases of steady flow. 54. Discharge under Varying Head. If the head on an orifice varies, the total discharge in a given time, or the time required for a given total discharge, can be computed only by integra- tion. Let the head on the orifice at any instant be y, the rate of discharge at that instant being q, and let c be the coefficient of discharge; then q=cFV2gy. If dQ denotes the volume discharged in the time dt, dQ=qdt=cF\/2gy-dt. In a particular case dQ can be expressed in terms of y and dy and the equation can be integrated. 55. Time of Emptying a Reservoir. Consider a vessel or reservoir of any form, filled with water to a certain level, and let it be required to determine the total time required for the surface to fall any given amount. It will be assumed that the coefficient of discharge does not vary with the head. (See Fig. 23.) 23. Let y= head on center of orifice at time t; 2/i = initial value of y; 2/2 = final value of y; A = horizontal cross-section of reservoir at water surface (variable) ; TIME OF FILLING A RESERVOIR. 41 F = area of orifice ; Q = total volume discharged from some assumed instant up to the instant t\ q= rate of discharge at time t; c = coefficient of discharge, assumed constant. Then, as in the preceding article, But also dQ=qdt=cFV2gy-dt. dQ=-Ady (the minus sign being used because Q increases as y decreases) hence cFV2gy-dt = -Ady. If A is variable, it must be known as a function of y in order that the solution may be completed. Reservoir of uniform horizontal cross-section. Let A be con- stant, and let T denote the time required for y to change from ?/i to y 2 . Integrating the last equation between the stated limits, 56. Time of Filling a Reservoir. If water flows from one reservoir or chamber into another through a submerged orifice, the pressure against which the flow takes place increases as the water surface rises in the receiving chamber. If the horizontal area of the supply reservoir is great in comparison with that of the receiv- ing chamber, the drop in the sur- face of the former will be inap- p IG 2 4 preciable. In this case let y denote the difference in level between the water surfaces in the two reservoirs at any instant, the remaining notation being as 42 FLOW OF WATER THROUGH ORIFICES. in the preceding case. Then the reasoning of Art. 55 applies without change, leading to the same formula. 57. Canal Lock. A canal lock consists of a chamber or com- partment which is placed in communication alternately with two bodies of water whose surfaces are at different levels. The time of emptying or filling the lock may be estimated by the above formula, with proper value of c. If the discharge into or from the lock takes place through a tunnel, the coefficient of discharge will be much less than for a simple orifice. Its value will depend upon the length of the tunnel, the form and size of its cross-section, and the roughness of its surface. The problem will be analogous to that of esti- mating the effect of friction on the flow in a pipe, to be discussed in Chapter IX. EXAMPLES. 1. A reservoir of 1000 sq. ft. horizontal cross-section is emptied through an orifice of area 2 sq. ft. Taking the head on the center of the orifice as initially 19 ft., compute the times required for the surface to drop 1 ft., 3.5 ft., and 7 ft. respectively. Take c = 0.65. Ans. 22.2 sec.; 81.0 sec.; 172 sec. 2. With initial conditions as in Ex. 1, compute the drop of the water surface in 3.5 min and in 7 min. Ans. 8.3 ft.; 14.3 ft. 3. Two reservoirs of horizontal cross-sections A', A", are connected by a submerged orifice. If the difference in level of the two water sur- faces at any instant is denoted by y, show that the time required for y to change from y\ to y. 2 is given by the formula in which A = A'A"/(A'+A"). This includes the two cases above treated, reducing to one of them if A' or A" is infinite. 4. As a particular case of Ex. 3, let A = 24 sq. ft., 4" = 10 sq. ft., F = 0.5 sq. ft., y l = 15 sq. ft., y z = 0, c = 0.6. Determine T. CHAPTER IV. THEORY OF ENERGY APPLIED TO STEADY STREAM MOTION.* 58. Transformation and Transference of Energy in Steadily Flowing Stream. Let AB (Fig. 25) represent a portion of a steady stream, the direction of flow being from A toward B. Consider the transformations and transferences of energy in which any given particle of water is concerned. The energy possessed by a particle at any instant is in gen- eral part potential and part kinetic. A particle of mass m having velocity v possesses mv 2 /2 units of kinetic energy. If v is in feet per second and m in engineers' kinetic units f (the unit mass being equal to 32.2 pounds-mass), mv 2 /2 is in foot-pounds. In estimating potential energy due to gravity a horizontal reference plane must be chosen. If a particle weighs W pounds and is z feet above the reference plane, it possesses Wz foot- pounds of potential energy. J Thus, the potential energy of any particle of water moving with the stream increases or decreases with the height of the particle above datum, and its kinetic energy increases or de- creases with its velocity. It is instructive to consider some- what definitely how these changes occur in the case of steady flow. (1) Any small portion of the fluid is continually receiving energy from certain adjacent particles and giving up energy to * The theory of steady flow of gases is treated in Appendix A. f Theoretical Mechanics, Art. 218. J Strictly it should be said that this potential energy is possessed by the system consisting of the body and the earth. See Theoretical Mechanics, Art, 362. 43 44 THEORY OF ENERGY APPLIED TO STEADY FLOW. others. Thus, consider a small body of the water between two transverse planes (as X, Fig. 25). This body X is acted upon by pressures exerted by the adjacent B bodies of water P and Q. Let F f be the area of the cross-section between P and X, and p f the normal pressure per unit area exerted across this sec- tion; let F" be the area of the cross- section between X and Q, and p" the intensity of pressure in this section; and let v', v" be the velocities in the cross- sections F', F" respectively.* In a short interval of time Jt the body X receives an amount of energy p'F'v'M by reason of the positive work done upon it by the total pres- sure F r p'\ the body P loses an equal quantity of energy because an equal amount of negative work is done upon it. During the same time the body X loses an amount of energy p"F"i/'dt because of the negative work done upon it by the pressure p"F"', the body Q gaining an equal quantity of energy. In other words, during the time At the body X receives from P a quantity of energy p'F'tfM, and gives up to Q a quantity p"F"i/'M. Since F'tf =F"v" J these two amounts of energy will be equal if p f =p"', hence in that case the body X neither gains nor loses energy by reason of the pressures, but acts as a transmitter of energy f from P to Q. If, how- ever, p' and p" are unequal, the body X receives from P either more or less energy than it gives to Q. But in any case, so far as the process here considered is concerned, one portion of water gains exactly as much energy as other parts lose; the energy possessed by the whole stream remains constant, while that of individual particles continually varies. (2) If two adjacent portions of the water slide over one another, the tangential forces which resist such sliding result in a twofold energy changs. So far as the two portions have * The velocity is assumed uniform throughout each cross-section. If this is not true, the reasoning still holds for an elementary portion of the stream. f This is analogous to the transmission of energy from one pulley to another through a belt. ENERGY PASSING ANY GIVEN SECTION. 45 a common component of motion parallel to the sliding, there is a transfer of energy from one portion to the other exactly similar to that just explained as due to normal pressure. But .since the two portions move unequally in the stated direction, the energy lost by one is not equal to that gained by the other, The negative work done on one portion is always greater than the positive work done on the other, so that the net result is a loss of energy. The mechanical energy thus lost is trans- formed into molecular energy or heat. This transformation of mechanical energy into heat is called dissipation of energy. Energy thus dissipated is practically lost, so far as the possi- bility of its utilization is concerned. (3) Particles adjacent to the surface of the pipe (or what- ever body encloses the stream) lose energy by reason of the negative work done upon them by the frictional forces exerted by the pipe surface. The energy thus lost is dissipated into heat. It is thus seen that in any steady stream there is in general a continual transference of energy in the direction of the flow, accompanied by a dissipation of mechanical energy into heat. Because of this dissipation, the energy transferred across a sec- tion A (Fig. 25) is always greater than that transferred across a section B down-stream from A. This will be made definite in the following article. 59. Energy Passing Any Given Section. A useful expression may be deduced for the total quantity of energy passing any cross-section of a steady stream in a given S time. ^r^JZ ["~f~^~- Consider any section of the stream, as Let p = intensity of pressure at every point in the section (Ibs. per sq. ft.); J^=area of cross-section (sq. ft.); 2=ordinate of centroid of section (ft.) above a horizon- tal datum plane, taken as reference plane in esti- mating potential energy; v = velocity of flow across the section (ft. per sec.); 46 THEORY OF ENERGY APPLIED TO STEADY FLOW. W= weight of water passing the section per unit time (Ibs. per sec.) ; w = weight of unit volume of water (Ibs. per cu. ft.) ; f) W X ~= volume of water passing the section per unit time (cu. ft. per sec.). Energy passes the section S in two ways : (a) Each particle of water passing the section possesses a certain quantity of energy, part potential and part kinetic; (6) the particles on one side of the section are at every instant giving up energy to the particles on the other side by reason of the pressure and motion. (a) During a short interval of time At, WAt pounds of water pass the section. This water possesses potential energy of amount WAt-z, and kinetic energy of amount ~ > i- e -> a total quantity of energy (6) Designating by X and Y the bodies of water adjacent to the section and separated by it (Fig. 26), it is seen that, by reason of the pressure exerted by X and Y upon each other, the body X loses and the body Y gains a quantity of energy which may be computed as follows: The total pressure resist- ing the motion of X is Fp' } an equal and opposite force acts upon Y in the direction of the motion. In a time At, the bodies upon which these forces act move a distance vAt. The force Fp acting upon X does an amount of work FpvAt, and the force Fp acting upon Y does an amount of work + FpvAt. That is, X loses a certain amount of energy and Y gains an equal amount; or there is a transfer from X to Y of an amount of energy FpvAt. Since W = wFv, the quantity of energy passing the section in time At by reason of the transfer of energy from X to Y may GENERAL EQUATION OF ENERGY IN STEADY FLOW. 47 be written in either of the equivalent forms w Combining the values found in paragraphs (a) and (6), there results for the total energy passing the section S in the time At the value w The quantity of energy passing the section per unit time is evidently w Stated in still another way, it is seen that for every unit weight oj water passing any section a quantity of energy y w passes the same section. The quantities of water passing dif- ferent sections in any time are equal, but the quantities of energy are unequal, because p and v in general change from section to section. 60. General Equation of Energy in Steady Flow. An im- portant equation, often called Bernoulli's theorem, is obtained by applying the theory of energy to a portion of a steadily flowing stream. Referring to Fig. 25, consider the volume included between two fixed cross-sections A and B. Let pressure, velocity, area of cross-section, height above datum, etc., be represented by the same symbols as above, with suffix d) for the up-stream section A and ( 2 ) for the down-stream section B. If the flow is steady, it is evident that the total quantity of mechanical energy contained in the volume AB remains 48 THEORY OF ENERGY APPLIED TO STEADY FLOW. constant. For, considering any elementary volume in a fixed position in the stream, the water occupying it at any instant has the same mass and velocity as that which has replaced it at any succeeding instant, and the two elements therefore pos- sess equal quantities of energy, both kinetic and potential. The total mechanical energy gained by the volume AB during any time must therefore equal the total mechanical energy lost by it during the same time. Consider the energy gained and lost by the volume AB while W pounds of water pass any section of the stream. The only gain of mechanical energy is that passing the sec- tion A\ its value is The mechanical energy lost is made up of two parts, that passing the section B } and that dissipated into heat. The value of the former part is w The value of the energy dissipated will be represented by K. Equating the total energy gained by the volume AB to the total energy lost by it, we have -a- ..... 0) TT in which H 7 = ^. Equation (1) expresses Bernoulli's theorem, or the general equation of energy for steady flow. 61. Meaning of " Head." The word head is used by writers on Hydraulics in a somewhat indefinite way. In all cases it EFFECTIVE HEAD. 49 means the height of a column of water, either actual or ideal. Thus, "head on an orifice/' or on any point of an orifice, has in preceding discussions been used to designate the vertical height of the free surface of water above the point under consideration. Most writers at the present time use the word head in the following senses : The pressure head at any point in a body of water is the height of a column of water which, in equilibrium, would by its weight produce the pressure existing at the point. The pressure head corresponding to a pressure of intensity p is therefore equal to p/w. The velocity head at any point of a stream is the height through which a body must fall from rest under gravity to acquire a velocity equal to that existing at the given point. The velocity head corresponding to a velocity v is therefore v 2 /2g. The sum of the pressure head and the velocity head at any point of a stream is often called the effective head at that point. It seems desirable, however, to modify the meaning of this term in the following manner. 62. Effective Head. It has been shown (Art. 59) that the total energy passing a given section of a steady stream, per unit weight of water discharged, is equal to g w Each of the three terms in this expression represents a linear magnitude, and each may be called a "head." The second and third terms may be called velocity head and pressure head re- spectively, in accordance with usage as explained above. The term z has been called elevation head, and also potential head. The former term will be here adopted. Effective head will here be defined as the sum of the pressure head, velocity head, and elevation head. It seems desirable to include the term z in the definition of effective head, for the reason that this term has equal signifi- 50 THEORY OF ENERGY APPLIED TO STEADY FLOW. cance with the other two in estimating the total energy deliv- ered at a given point in the stream. The symbol H will hereafter be employed to designate the value of the effective heacl at any point of a stream. That is, 2g w The values of H at different cross-sections, like those of z, p, and v, will be distinguished by suffixes. 63. Lost Head. The general equation of energy (Art. 60) shows that the effective head decreases along the stream in the direction of the flow. Thus, with the above notation, the equa- tion may be written The quantity H' is a linear magnitude, and expresses the amount by which the effective head decreases from A to B (Fig. 25) ; H' may therefore be called the loss of head between A and B. That is, The loss of head between any two sections of a stream is defined as the amount by which the effective head at the up-stream section exceeds that at the down-stream section. The reasoning by which the general equation of energy was deduced shows that H' has an important meaning as energ}^. Since H' = K/W, in which K denotes the energy lost by dissi- pation per unit time between A and B, and W the weight of water discharged (across any section) per unit time, it is seen that The loss of effective head between any two sections of a steady stream is equal to the energy lost by dissipation between those sections per unit weight of water discharged. In practical applications, the value of the lost head between two sections is sometimes found by determining HI and H 2 and taking their difference; and sometimes its value is estimated from a consideration of its energy meaning as just explained. LOST HEAD. 51 It is instructive to devote some time to the application of the equation of energy on the assumption of no loss of head. In some cases the results have practical value, the actual losses being small. In other cases such a treatment is useful only in illustrating general principles. EXAMPLES. 1. In a horizontal pipe 6 inches in diameter water flows with a velocity of 12 ft. per sec. At a section A the mean pressure is 50 Ibs. per sq. in., and at a section B it is 40 Ibs. per sq in. Compute the quantity of energy passing each of these sections in one second, estimating potential energy with reference to a datum plane 18 ft. below the axis of the pipe. Ans. 18,950 foot-pounds and 16,540 foot-pounds. 2. In Ex. 1 what is the value of the lost head between A and J5? In which direction is the flow? 3. The diameter of a pipe is 12 inches at a section A, 16 ft. above datum, and 8 inches at a section B, 8 ft. above datum. At A the velocity is 12 ft. per sec. and the pressure head 12.6 ft. Assuming no dissipation of energy between A and B, compute the velocity and the pressure head at B. Ans. Velocity = 27 ft. per sec. ; pressure head = 10.9 ft. CHAPTER V. APPLICATION OF GENERAL EQUATION OF ENERGY, NEGLECTING LOSSES BY DISSIPATION. 64. General Method. Assuming no loss of energy by dissi- pation in any part of the stream, the general equation of energy may be written v 2 p 2 + 77- H =H = constant. 2g w The general method of applying this equation is as follows : (1) Choose a datum plane. This fixes the value of z at every point of the stream. (2) Notice at what points the velocity is known. If at any point the cross-section is very great in comparison with its value at other points, the velocity will be so small that the velocity head at that point may be neglected. (3) Notice at what points of the stream, if any, the pressure is known. (4) Notice whether there is any point of the stream at which the three parts of the effective head elevation head, pressure head, and velocity head are all known. If there is, the value of H becomes known at that point, and therefore at all sections of the stream. (5) If there is no point at which elevation, pressure, and velocity are all known, note the points at which the value of H can be expressed with the use of the fewest unknown quantities. By writing expressions for H for two or more sections, it may be possible to solve the resulting equations and determine the unknown quantities. 52 OF iv* 'F- FLOW FROM A RESERVOIR THROUGH AN ORIFICE. 53 65. Flow From a Reservoir Through an Orifice. Consider a small orifice in the side of a vessel (Fig. 27), from which a jet is discharged into the atmosphere. Let the center of the smallest cross-section of the jet (S) be at a distance h below the level of the free surface of the water in the vessel. The particles of water approaching the orifice doubtless come from various parts of the reservoir; but it will appear presently that the same solution of the problem will result whatever point in the reservoir be taken as a point of the stream. (1) Choose as datum the horizontal plane through the cen- ter of the stream at S. (Any other horizontal plane might be chosen.) (2) The velocities of particles within the reservoir are very small, except in the neighborhood of the orifice. For any point at some distance from the orifice the velocity may therefore be taken as zero without sensible error. (3) The jet being surrounded by the atmosphere, the pres- sure within the stream at S is atmospheric. Atmospheric pres- sure also exists at the water surface A. At points within the vessel (except near the orifice where the particles have sensible velocities) the pressure follows the hydrostatic law; that is, at a depth y below the free surface the pressure exceeds atmos- pheric by wy. (4) At any point within the reservoir at which the velocity is sensibly zero the value of the effective head is known imme- diately. Consider a point B (Fig. 27). Whatever the value of z for this point, we have for the value of the pressure head (call- ing atmospheric pressure p ) w w The velocity head being sensibly zero, w w w 54 EQUATION OF ENERGY WITHOUT LOSSES. That is, w so that the value of the effective head at all points becomes known. To complete the solution, let v denote the velocity of the jet at S. The values of elevation head, pressure head, and velocity head for this section are 0, , and ^-; that is, rr ti = rrr . w 2g Equating the values of H, V 2 It is thus seen that, neglecting frictional losses of energy, Torricelli's theorem (Art. 37) is a direct consequence of the theory of energy. 66. Case of Unequal Pressures. If the pressures at the sur- face of the reservoir and at the section S of the jet are unequal, let their values (per unit area) be p\ and p 2 respectively. Then for the point B (Fig. 28) we have FIG. 28. and for the point S Equating these values, _>+(*-*); H^ w w w rr i, ti = r;r . w 2g w w which agrees with Art. 52. VELOCITY OF APPROACH. 55 67. Velocity of Approach. If the cross-section of the reser- voir is not so great that the downward velocity of the water may be neglected, another term enters the formula. Let v r = velocity of water at point B (Fig. 27); v = velocity of jet at section S. Then, assuming atmospheric pressure at A and at S, we have for the point B ~w 2g' and for the point S w 2g" Equating these values of H, v' 2 v 2 Let F = cross-section of jet at 8', F f = cross-section of vessel at B. F'v'==Fv, or v' = (j^v. Substituting this value of t/ in the foregoing equation and solv- ing for v, we have If the pressures at A and S are unequal, the above reasoning ieads to an equation similar to the above with h4-p\/wpz/w substituted for h. EXAMPLES. 1. If the diameter of the jet is 0.5" and that of the vessel 2", both being circular in cross-section, what percentage of error is introduced by neglecting velocity of approach? Ans. About 0.2 per cent. 2. For what value of the ratio F/F' will an error of one per cent be introduced by neglecting velocity of approach? 56 EQUATION OF ENERGY WITHOUT LOSSES. 68. Effect of Velocity of Approach in Case of Rectangular Weir. The foregoing discussion has referred to an orifice so small that no important error results from regarding the head on the center as applying to all parts of the orifice. The effect of velocity of approach in case of large vertical orifices needs special consideration. It will be sufficient to consider a rect- angular orifice, since it is this form that possesses most prac- tical importance. Consider a stream flowing horizontally in an open channel (Fig. 29) and discharging at the end of the channel through a FIG. 29. rectangular orifice of width 6, the upper and lower edges being at depths hi and h% respectively below the surface of the water in the channel. Consider the ideal case in which there is no contraction of the stream and no loss of energy by dissipation, and assume that the velocity of the approaching stream has the same value throughout all parts of any given cross-section. Let ?/ be the value of this velocity. At the cross-section S of the issuing stream, let v= velocity at a point C whose depth below the water surface in the chan- nel is x. The value of v is found by applying the equation of energy to the point C and a point B of the approaching stream. Let z\ and 22 denote the ordinates of B and C respectively above any assumed datum; then the depth of B below the water surface is x+z^zi, and the pressure head at B is x +2 2 -2i+ , w' p Q being atmospheric pressure. Hence DISCHARGE FROM RESERVOIR THROUGH PIPE. 57 For the point C, the pressure being that of the atmosphere, we have w Equating these values of H, v 2 v' 2 , Taking an elementary area of length b and width dx, we have for the discharge per unit time through that element dq=bvdx=b (v' 2 + 2gx) *dx. Integrating, Inspection of this result shows that the velocity of approach, so far as its effect on the rate of discharge is concerned, is equivalent to an increase of v' 2 /2g in the head on every part of the orifice. In case of a weir, the value of hi is zero. If H denotes the "head on the crest," i.e., the depth of the bottom edge (or crest) of the orifice below the surface of the approaching stream, and h f is written for v' 2 /2g (the head " equivalent to" the veloc- ity of approach v f ), the formula becomes q = %V2g-b[(H+h'f-h'*]. 69. Discharge from Reservoir Through Tube or Pipe. If a pipe or tube, or any system of pipes, leads from a reservoir and discharges a stream at any point, it is evident that the velocity of discharge (on the assumption that no energy is dissipated) is given by the same formula as that applying to discharge from an orifice. B 58 EQUATION OF ENERGY WITHOUT LOSSES. Thus, let any system of pipes lead from a large reservoir (Fig. 30) and discharge into the atmosphere at a point whose depth below the water surface is h. The reasoning employed in the discussion of discharge through an orifice applies un- r IG. 30. 1111 changed, and the velocity of the stream at S is given by the formula v=V2gh. If the cross-section of the issuing stream is known, the rate of discharge is known from the relation q=Fv. 70. Pressure at Any Section. After the value of q has been determined the velocity and pressure at any point in the stream may be computed, if the cross-section at that point is known. Thus, in Fig. 30, let h =40 ft., and let the diameter of the jet at S be 1 inch. Let it be required to determine the velocity and pressure at a point C, 10 ft. higher than S, the pipe at C being 1.25 inches in diameter. Let v 2 = velocity at S; Vi = velocity at C; pi = pressure at C. As above, we have !-*-. Taking datum plane through the center of the jet at S, we have w 2g w For the point C, w 2g Equating values of H, w PRESSURE CAN NEVER BE NEGATIVE. 59 Pi Po , qn ^ or = -- r oU pr~. w w 2g /5\ 2 v-i 2 /4\ 4 v% 2 But vi=v 2 , or = ' = 16A ' hence That is, the pressure at C exceeds that of the atmosphere by the equivalent of 13.6 ft. head. It will be noticed that, if different points of the stream at the same level be compared, the pressure is greatest where the velocity is least. Since a contraction of the stream increases the velocity, it causes a decrease of pressure. More generally, consider what is implied by the equation J) V^ z + -\-JT- =H = constant. w 2j (1) For a stream of uniform cross-section and consequently having equal velocities at all sections, the pressure head in- creases as the elevation decreases, and vice versa; the variation of pressure follows the hydrostatic law. (2) For a stream all parts of which are at the same level the pressure head increases as the velocity head decreases, and vice versa. When losses of energy by dissipation are considered, these principles are greatly modified. But it is instructive to con- sider carefully the conditions in the ideal case of no loss of energy. 71. Pressure Can Never be Negative. In such a case as that shown in Fig. 30, the velocity of the jet at is independent of the dimension's or elevation of any part of the pipe leading from the reservoir to the point of discharge. The only condi- tions affecting the velocity of outflow are the pressure and ele- vation at A and S, and the relation between the cross-sectional areas at these points. Between these sections the pipe may 60 EQUATION OF ENERGY WITHOUT LOSSES. vary in diameter in any way, and its elevation may vary in any way without affecting the velocity of discharge. There is, however, an important limitation on the application of this theory, even without the consideration of losses of energy. From the equation n v 2 z+ \-JT =H= constant w 2g it is evident that, by increasing v or 2, the pressure p may be decreased to any assigned value. The velocity of outflow, and therefore the rate of discharge, being fixed, if at any point the cross-section is made very small the velocity becomes very great. Since v=q/F, there is no limit to the value which may be given to v by decreasing F. For any value of 2, therefore, p/w may be made zero, or even negative, by decreasing the cross-section of the pipe. A negative pressure would mean a tension, and the nature of a liquid is such that a tendency to tension causes separation of the particles.* Therefore if the pressure at any point of the stream, as computed by the foregoing theory, is found to be negative, the conclusion to be drawn is that the stream will break at that point, and the solution must be modified. If, at the point where the pressure as computed from the formula has the least (algebraic) value, this value is negative, the actual pressure may be put equal to zero; and the velocity computed on this assumption may be regarded as the greatest possible value of the velocity at that point on the supposition of no dissipation of energy. EXAMPLES. 1, In Fig. 31, let the diameter of the pipe have the following values at different points: at B, 3"; at C, 2.5"; at D, 2.5"; at E, 2". Let the diameter of the jet at F be 1.75". Determine the velocity and pressure at each of the four points B, C, D, E. 2. Keeping the size of the pipe unchanged, how high can the pipe be carried at D without reducing the pressure to absolute zero? * Strictly, it should be said that under certain conditions a slight tension can exist in liquids; but the statement above made is practically true. EXAMPLES. 61 3. With the pipe D at the elevation shown in Fig. 31, how small may the diameter be without reducing the pressure to absolute zero? What will happen if the pipe is made smaller than this limiting size? Ans. Diameter at D = 1.80". FIG. 31. 4o What will be the velocity of discharge from the siphon shown in Fig. 32? Ans. 11.3 ft. per sec. 5. Discuss the pressure throughout the tube in Fig. 32, assuming the diameter to be uniform. How small may the pipe be at A (as com- pared with its size at B) without making the pressure at A absolutely zero? Diameter at A Ans - Diameter at B = - 51 (llmltm S value) ' FIG. 32. _l l_ FIG. 33. 6. In Fig. 33, let the diameter of the pipe be 1" and that of the jet at B 1.75". Compute the pressure at the point A. At what points in the pipe has the pressure the least and greatest values? Ans. At A, p/w = p /w-4.3. 7. In Fig. 34, let the diameter of the diverging tube be 1" at B and 1.5" at Cj the jet at S having the same size as the tube at C. Compute velocity and pressure at B and at C. Ans. At B, v= 51.2; p/w=p /w-32.5. 8. In Fig. 34, if the tube were cut off at B, how would the discharge be affected? 62 EQUATION OF ENERGY WITHOUT LOSSES. 9. In Fig. 34, what is the effect of increasing the diameter of the tube at C, that at B remaining unchanged? What is the greatest pos- sible value of the velocity at B? Ans. 51.9 ft. per sec. 10. In Fig. 35, let the diameter of the pipe be 2" at A and 1" at B, its axis being horizontal. The pressures at the two sections are meas- ured by the heights of water standing in the vertical tubes A A' and BB'. If the water surface is 2' higher at A f than at B', what is the rate of discharge? Ans. 9 = .064 cu. ft. per sec. 11. Fig. 36 represents a pipe in which water flows in the direction AB. In order to measure the difference between the pressures at the FIG. 35. sections A and B, tubes A A' and BB' are inserted at the two sections and connected with the two branches of a U tube M , containing mercury. If mercury fills the portion of the tube A'B' and water the portions AA' and BB', the difference between the heights of A' and B' indicates the difference between the pressures in the sections A and B. The specific gravity of mercury may be taken as 13.6. (a) If the vertical height of B' above A' is h, what is the difference "between the pressures in the tube at A' and B' } expressed in "head' or height of water column? Ans. 13. 6A. (6) What is the difference between the pressure at B' and that at A" } on a level with B"! Ans. 12 M. (c) What is the difference between the values of the pressure head at C and D, these points being at the same level*? 12. Wit^h the arrangement shown in Fig. 36, let the diameter at A be 3" and that at B 2". If h = 6", what is the velocity e>f flow at 5? Ans. 22.5 ft. per sec. 13. In Fig. 36, let the cross-sectional a'reasfat A and B be Ft and F 2 , and the. corresponding velocities Vi and v , and let h be the difference of level of the mercury -surfaces at A' and B'. Deduce the following formulas (s denotes specific gravity of mercury) : A" -r B' < __j_ A' M C - A \ R ->- FIG. 36. EXAMPLES. 63 \2(s-l)gh 14. In Fig. 35, let the diameters at A and B be 4" and 3" respectively and suppose pistons to be fitted into the pipe at the two sections, the intervening space being filled with water. If the piston at A advances at the rate of 6 ft. per sec., and the height of A' above the center of the pipe is 7', compute the difference between the total pressures upon the two pistons. Neglect friction. Ans. 75 Ibs., neglecting atmospheric pressure. CHAPTER VI. APPLICATION OF GENERAL EQUATION OF ENERGY, TAKING ACCOUNT OF LOSSES. 72. Methods of Estimating Loss of Head. When losses of energy by dissipation are taken into account, it is necessary to estimate the value of H' in the general equation of energy, H i HZ =H . As explained in Art. 63, this term is called the loss of head between the two sections at which the effective head has the values Hi, H 2} respectively, the former referring to the up- stream section. The energy meaning of H', as stated in Art. 63, should be kept clearly in mind. It is the energy lost by dissipation between the two sections considered, per pound of water discharged. The value of H' in particular cases may be estimated either (a) theoretically or (b) experimentally. (a) If the loss of energy occurring between the two sections can be estimated theoretically, the value of H' can be com- puted from its energy meaning as just stated. (b) If the values of HI and H 2 can be determined by actual measurement, the value of H f becomes known from the equa- tion H' = H l -H 2 . In most cases in which the first method is employed the theoretical value of H' involves coefficients whose values can be known only by experiment. 73. Experimental Determination of Lost Head. The deter- mination of the effective head at any section requires the 64 WATER PIEZOMETER. 65 measurement of z, p, and v. The elevation above datum may be found by direct measurement or leveling. It remains to consider how pressure and velocity may be determined. The mean velocity in any section is known from the relation v = q/F, as soon as the rate of discharge is known. Methods of measuring q will be discussed in Chapter XIII. If the tvvo cross-sections to be compared are equal, the values of v are equal, and need not be known in order to determine the loss of head between the sections. The pressure at any section of a pipe may be determined by tapping the pipe and attaching a pressure-gauge. Several forms of pressure-gauges are used. 74. Water Piezometer. The simplest form of pressure- gauge is a tube inserted in the pipe at right angles to its axis and carried up to such height as may be necessary to prevent overflow, the top being open. Thus, in Fig. 37, the pressure at any point A in the cross-section S is equal to the pressure due to the height of the column of water AA' plus atmospheric pressure; that is, the pressure at any depth h below the top of the column is po + wh, this law holding from the top of the column to the bottom of the cross-section S. That it holds throughout the cross-section follows from the fact that the direction of motion of the water is at right angles to the plane of the cross-section. The reasoning of Art. 11 applies if the S FIG. 37. points A and B (Fig. 3) are in the plane of the cross-section since the elementary prism AB has no acceleration in the direc- tion of its length.* * This reasoning is rigorously true on the assumption that all particles 66 EQUATION OF ENERGY WITH LOSSES. If the pipe is not horizontal (Fig. 38), the law p = still holds throughout the tube and the cross-section S, h being measured vertically. It obviously makes no difference whether the tube com- municates with the cross-section S at the highest point or at some other part of the perimeter. Thus, if Fig. 39 is a cross- sectional view, and if the tubes A A' and BE' are inserted at different points, the tops of the columns (A f and B f ) will be at the same level. In order that the piezometer may give a reliable indication of the pressure in the pipe, the axis of the tube at the point A .. B C' FIG. of attachment must be at right angles to the axis of the pipe. Thus, if three tubes are inserted, as at A, B, and C (Fig. 40), the friction of the water passing the orifice will increase the pressure in the tube A and decrease it in the tube C, while the column BB f will indicate the pressure correctly. The tops of the columns will stand at different levels, as shown. The foregoing discussion applies only to the case in which the pressure throughout the cross-section of the pipe is greater than that of the atmosphere. If, at the point A (Fig. 39), the pressure within the pipe is less than that of the atmosphere, no column of water will of water move in straight lines parallel to the axis of the pipe, and in straight pipes the conclusion undoubtedly holds within the limits of accuracy ordinarily attainable in hydraulic measurements. For the most accurate work it is best to connect the piezometer tube with a chamber communicating with the cross-section of the pipe by small orifices at several points of the circum- ference. MERCURY GAUGE. 67 FIG. 41. be sustained in the tube, but the external pressure will con- tinually force air into the pipe through the tube. The pressure may in such case be measured by what is called a "vacuum piezometer " (Fig. 41). The tube inserted at A is carried upward, then bent down, the end dipping into an open vessel of water at C. The pressure within the bent tube, after a condition of equilibrium has been established, will be less than atmospheric, and water will rise from the vessel to some height CC". In the othei branch of the tube, water will stand at some height A A', which may or may not be zero, depending upon the initial conditions. Neglecting the weight of the air in the tube, the pressure within the air-space in the tube is uniform. Therefore, at a vertical distance below A' equal to CC' (in the column AA f or the section S), the pressure is atmospheric. The pressure at any point in the section S can therefore be determined if the tops of the two columns A A! and CC' can be observed. 75. Mercury Gauge. For high values of the pressure the piezometer tube would need to be so long as to render its use impracticable. In such cases it may be possible *to use a mercury gauge, such as is represented in Fig. 42. The tube P communicates at A with the cross-section of the pipe at which it is desired to measure the pressure, and also communicates with the reser- voir R at the top. This reservoir is partly filled with mercury, the tube P and the space in the reservoir above the mercury being filled with water. The vertical tube T, open at both ends, is fitted tightly into the reservoir at the top, and projects far enough down so that the lower end is always immersed in mercury. FIG. 42. 68 EQUATION OF ENERGY WITH LOSSES. Throughout the space above the mercury in the reservoir, the tube P, and the cross-section S of the pipe, the pressure varies according to the hydrostatic law; the pressure at the mercury surface B being equal to that at points at the same level in the cross-section S or in the tube. If the pressure at B is greater than atmospheric, mercury rises in the tube T until the pressure at B due to the mercury column BC, plus atmospheric pressure, is equal to the pressure at the same point communicated through the tube P. If, by means of a fixed vertical scale, the height of the column BC can be meas- ured, the pressure at B, and therefore at any point in the sec- tion S, can be determined. Thus, let h = height of column BC] s = specific gravity of mercury; p = atmospheric pressure; p' = pressure at surface of mercury in reservoir. Then w w If p denotes the intensity of pressure at a point in the cross- section S whose vertical distance below the horizontal plane through B is x, WWW The tube P must be kept free from air, for which purpose an air-cock should be provided at its highest point. 76. Bourdon Gauge. Where precision and accuracy are not required the form of gauge ordinarily used with steam boilers is often employed for the measurement of water pressure. EXAMPLES. 1. In order to determine the loss of head between two sections A and B of a straight pipe of uniform cross-section, piezometers were attached at the two sections. The top of the column at A stood 150' above a horizontal datum plane, and that of the column at B 140' above the same plane. The center of the pipe at A was 80', and that at B 60', above datum. LOSS OF HEAD IN STANDARD ORIFICE. 69 (a) What was the total loss of head between A and B? (6) What was the pressure at the center of each of the cross-sections A and ? (c) In which direction was the flow? (How can the direction of flow always be determined from an experiment of this kind?) 2. With the pipe as in Ex. 1, suppose the piezometers to be replaced by mercury gauges. Let the surfaces of mercury in the reservoirs of the two gauges be 82.63' and 59.80' respectively above datum, and let the vertical heights of the mercury columns be 12.86' and 13.46' respect- ively. (a) Compute the loss of head between A and B. (6) Compute the pressure at the center of each of the two sections. Ans. (a) 14.67ft. 3. Take data as in Ex. 2, except that the size of the pipe is not uniform, (a) What additional data must be given in order that the loss of head may be computed? (6) If the diameter is 12" at A and 10" at B, and the rate of dis- charge is 6 cu. ft. per sec., what is the loss of head between A and B when the gauges read as stated in Ex. 2? Ans. (6) 13.70 ft. 77. Loss of Head in Standard Orifice. The velocity of efflux from a standard sharp-edged orifice (Art. 45) is v = c f V2gh } in which the coefficient of velocity c f has a value fairly well established by experiment. Comparing the values of the effective head at two points, one of which is within the reservoir where the velocity is practically zero, and the other in the contracted section of the iet (as the points B and S, Fig. 43), we have FIG - 43 ' - -- "T p being atmospheric pressure and v the velocity of flow at section S, and the datum plane being taken at the level of the 70 EQUATION OF ENERGY WITH LOSSES, center of the orifice. Hence ,,2 i_ c /2 V 2 The value of H f is thus expressed in terms either of the head on the orifice or of the velocity of the jet. If c' = 0.98 (Art. 45), the values become , H' = 0.0407t = 0.041 -V 78. Loss of Head in Short Tube. Consider next the loss of head in case of discharge from a short cylindrical tube. The length is assumed to be just sufficient so that the stream, after converging as it enters the tube, expands and fills the tube at the outer end (Art. 38). The inner edge of the tube is sup- posed to be square (Fig. 44). * FIG. 44. FIG. 45. The expression for H r in terms of the coefficient of velocity is the same as in the case of the orifice; but the value of this coefficient is different. As determined experimentally it is which gives c' = 0.82 (nearly), ' = 0.33/1 = 0.49. If the tube projects within the vessel, as in Fig. 45, the velocity of the jet is less than in the preceding case. Experi- ment gives ^ = 0.72 (about) LOSS OF HEAD IN UNIFORM STRAIGHT PIPE. 71 and i v 2 #'=0.48/i = 0.932-. 79. Loss of Head in Straight Pipe of Uniform Cross-section The loss of head in a given length of a uniform straight pipe depends upon the nature of the inner surface, the diameter, and the velocity of flow.* Experiment indicates that it is independent of the pressure existing in the pipe. Since loss of head is proportional to energy dissipated (Art. 63), it obvi- ously depends upon the friction between the water and the pipe, as well as upon the friction and impact of the particles of water among themselves. No theoretical formula has been proposed which can be depended upon to give more than a rough approximation to the actual losses of head in pipes. The following formula, applicable to a uniform pipe of any form of cross-section, is often employed, and is the basis of most of the formulas that will be considered in the following pages : Let F = cross-sectional area; C = length of perimeter of cross-section; F r =C> #'=loss of head in length I of pipe; i>=mean velocity of flow. Then H '=f7Yg- In this formula f is a coefficient whose value depends upon the roughness of the surface of the pipe, and also (although to a less degree) upon the size of the pipe and the velocity of flow. The quantity r is a length, and is called the hydraulic ridius of the cross-section; its value depending only upon the form and size of the section. * Doubtless, also, upon the temperature of the water; but regarding this little experimental knowledge is available. 72 EQUATION OF ENERGY WITH LOSSES. The theoretical basis of this formula will be considered in Chapter IX. 80. Cylindrical Pipe. If the cross-section of the pipe is a circle of diameter d, we have nd 2 F d Hence (writing / for 4/') the formula becomes Experiment indicates that the coefficient or "friction factor " / in this formula is not constant for a given kind of pipe but varies with both d and v, decreasing as each of these quantities increases. The variation with v appears, however, to be less important than that with d. For new or clean cast-iron pipe, and for velocities within the range ordinarily occurring in practice, Darcy recommended a formula equivalent to the fol- lowing (d being in feet) : For old pipe these values should be doubled.* This formula may be used in solving the examples which follow. The values of / for pipes of different kinds will be considered further in Chapter IX. 81. Loss of Head at Entrance to Pipe. If water is con- ducted from a reservoir by a straight pipe, the conditions of flow are less uniform near the entrance to the pipe than they become a little farther on. The contraction and expansion of the enterirg stream result in a loss of head near the entrance * Recherches expe>imentales relatives au mouvement de 1'eau dans bs tuyeaux. Chapter IV, also p. 228. LOSS DUE TO SUDDEN ENLARGEMENT OF PIPE. 73 that is much greater than occurs in an equal length where the flow has become uniform. The entrance loss is, however, small in comparison with the total loss unless the pipe is short, and for very long pipes it may be neglected. For those cases in wilich entrance loss is important enough to be considered, a formula for it is obtained by assuming that the conditions in the entrance portion of the pipe are similar to those existing in a short tube discharging into the atmosphere. It is assumed that, for any given pipe, the loss of head at entrance varies mainly with the velocity of flow, and that it has the same value as for a short tube of the same diameter, provided the conditions are such that the velocities are equal in the tube and the pipe. The entrance loss may therefore be expressed in terms of the velocity by the formulas already given (Art. 78) for the case of a short tube. Practi- cally, the following may be taken as sufficiently correct : v 2 H' = .5 -y- for pipe not projecting into reservoir; v 2 * H =2~ for projecting pipe. 82. Loss of Head Due to Sudden Enlargement of Pipe. If the cross-section of a pipe enlarges suddenly from an area FI to an 'area F 2 (Fig. 46), the irreg- ular motions of the particles of water due to the sudden expansion of the stream cause a dissipation of energy and. consequent loss of head. If vi and v 2 denote the values of the FlG 46 velocity at the two sections, the loss of head is found by experiment to be given approximately by the formula H ,_(vi-v 2 ) 2 Since FIV\ =F 2 v 2 , this formula may be written 74 EQUATION OF ENERGY WITH LOSSES. 2g- Special Case. If the ratio F\/Fz is very small, the above formula reduces to 77' - V]2 ~ 9 ' This applies to the case in which a pipe discharges into a reser- voir at a point below the water surface. 83. Loss of Head Due to Sudden Contraction of Stream. If the cross-section of the pipe decreases suddenly in the direc- tion of the flow (Fig. 47), there is a loss of head, but less than in the case of sudden enlargement. The value of the loss depends upon the ratio of the two cross-sections in a way that can be known only by experiment. Let F\ and F 2 denote the larger and smaller cross-sections respectively, Vi and v 2 being the corresponding velocities; then the loss of head may be expressed by the formula FIG. 47. _ k being an experimental coefficient depending upon F 2 /Fi. The following values of k are based upon data given by Weisbach : F 2 FI .10 .20 .30 .40 .50 .60 .70 .80 .90 1.00 k .362 .338 .308 .267 .221 .164 .105 .053 .015 84. Loss of Head Due to Obstruction in Pipe. If the cross- section of a pipe at any point is partly closed by an obstruction, it may be assumed that the loss of head is due mainly to the LOSS OF HEAD CAUSED BY BEND. 75 sudden expansion of the stream just beyond the obstruction, and the loss may be estimated as in the case of a sudden enlarge- ment in a pipe (Art. 82). Thus, let F = cross-section of pipe; -F' = area at obstructed section; v = velocity where cross-section is F] The formula becomes 2g F' 85. Loss of Head Caused by Bend. The loss of head in a curved pipe is greater than that in an equal length of straight pipe. No rational formula can be given for estimating its value, and the empirical rules commonly given are probably not very reliable. It is reasonable to suppose that for pipe of a given size the loss per unit length is greater as the radius of the curve is less, and that the total loss due to a curve of given radius increases with the total angle of deflection, i.e., with the length of the curve. For curves of short radius, having the same total angle of deflection but different radii, the usual assumption is that the loss of head increases as the ratio of the radius of the curve to that of the pipe decreases. For larger values of the radius, however, the total loss for a curve of given total deflection increases with the radius because of the increased length of the curve. For 90 curves of short radius, the formula usually given is that of Weisbach, who assumed and gave for k the empirical formula in which r is the radius of the pipe and R that of the bend 76 EQUATION OF ENERGY WITH LOSSES. That for a given value of R the loss of head should increase with the diameter of the pipe seems hardly reasonable, even when R is small. At all events, Weisbach's formula should not be used when the radius of the curve is greater than four or five times the diameter of the pipe. For curves of longer radius experimental data are too lim- ited to form the basis of any general formula.* 86. Hydraulic Gradient. If piezometers be inserted into a pipe at various points of its length, the line joining the highest points of all the piezometric columns is called the hydraulic gradient. In Fig. 48, A'E r represents the hydraulic gradient for the pipe AB. Datum FIG. 48. The general nature of the curve of the hydraulic gradient may be inferred from the equation of energy. Thue, in Fig. 48, let zi, Vi, pi refer to some given section A, and 2, v, p to any section B, down-stream from A. Let y denote the eleva- tion above datum of the top of the piezometric column at any section, y\ being the value of y at A. The equation of energy * Experiments on the loss of head caused by 90 bends in pipes of 12", 16", and 30" diameter, made by Gardner S. Williams, Clarence W. Hubbell and George H. Fenkell, are described in Vol. XLVII of the Transactions of the American Society of Civil Engineers. The loss of head was in each case measured for a total length of pipe which included two tangents connected by the bend, and the effect of the bend was estimated by comparison with the loss in an equal total length of straight pipe. Those of the results show- ing the greatest regularity indicated that the total loss of head due to each bend in the 30" pipe was approximately three times the loss caused by an equal length of straight pipe. HYDRAULIC SLOPE. 77 gives 9 ,PI , y i 2 v2 w Z-\ -- =Z\-\ --- HO -- 7;; -- H- i w w 2g 2g if H r is the head lost between A and B. Since y = z + p/w, the equation may be written Consider the following special cases : (1) Suppose the cross-section of the pipe is constant. Then v = vi and y-yi-H'. That is, the hydraulic gradient falls, between any two sections, by just the amount of the head lost between those sections. On the assumption of no loss of head, the hydraulic gradient, in case of uniform cross-section, would be a horizontal line. In a straight pipe of uniform interior roughness the loss of head is proportional to the length, and hence the hydraulic gradient has a uniform downward slope in the direction of flow. A bend or obstruction would cause a sudden drop in the hydraulic gradient. (2) Suppose the cross-section of the pipe variable. Then, in addition to the effect of loss of head, the hydraulic gradient will rise or fall with varying velocity; rising as the velocity decreases and falling as it increases. 87. Hydraulic Slope. The fall of the hydraulic gradient per unit length of the pipe is called the hydraulic slope. If I denotes the length of the pipe measured from a fixed section (as A, Fig. 48), and s the hydraulic slope at the point considered, y being the ordinate of the hydraulic gradient at that point (as B, Fig. 48) , we have dy 78 EQUATION OF ENERGY WITH LOSSES. If the hydraulic slope is constant from A to J3, - If the cross-section is constant, so that the fall of the hydrau- lic gradient is wholly due to loss of head (as in case (1), Art. 86), i/i -?/=#' and H' S= T' 88. Applications of Theory. In the preceding chapter was treated the method of applying the general equation of energy to the solution of problems in steady flow, neglecting losses of energy by dissipation. The general method there outlined may readily be extended so as to take account of losses of head, the value of the term H f in the equation of energy being expressed in accordance with the foregoing principles.* This will be illustrated by the solution of an example. EXAMPLES. 1. The pipe AB (Fig. 49) is 1000' long and 6" in diameter, and the length AC is 550'. Compute (a) the rafe of discharge; (b) the pressure at the center of each of the cross-sections A, B, C. M I Datum FIG. 49. Solution. (a) First apply the equation of energy, comparing the points M and N, at the surfaces of the two reservoirs; //i denoting the * In solving the following examples, the coefficient /, for estimating the frictional loss of head in pipes, may be determined by Darcy's formula (Art. 80). UNIVERSITY I EXAMPLES. 79 effective head at M, and H 2 that at N t while R f denotes the total head lost from M to N. Taking datum plane as shown in the figure, #,=100, tf 2 =60. The value of H' is made up of three parts, each of which may be ex- pressed in terms of the velocity of flow in the pipe. Let v denote this velocity; then we have: (1) The loss at entrance to the pipe (Art. 81) may be taken as (2) The loss between A and B due to friction (Art. 80) is I v 2 Darcy's formula gives / = .0232, which reduces the expression for the f fictional loss to ' (3) The loss of head due to the sudden destruction of the velocity at the point of discharge into the reservoir N is *(by Art. 82, putting F,/F-Q) Combining the three losses, #'=47.9-, V and the equation #1 -H 2 = H' becomes from which 100-60=47.9^, v 2 =0.833; v = 7.32 ft. per sec. The rate of discharge is g = Fv =0.196X7.32 = 1.43 cu. ft. per sec. (6) Applying the energy equation to sections M and A (the latter 80 EQUATION OF ENERGY WITH LOSSES. being a short distance within the pipe), the value of H' is Q.o(v*/2g) =0.42. Also, with datum as before, arid the equation HiH 2 =H' gives = 78.7 ft. = pressure head at A. The absolute pressure at A is p 2 + p , since in this solution atmospheric pressure has been called zero. By a similar method the pressure head at C is found to be 46.3 ft. The hydraulic gradient is in this case a straight line except near the entrance to the pipe. Starting at the surface of the reservoir M, it falls by an amount l.5(v z /2g) = 1.25 ft. by reason of the entrance loss and the velocity head, then slopes uniformly by reason of the frictional loss, falling .03875 ft. per foot of length of the pipe until the surface of the reservoir N is reached. 2. In Fig. 49, how large must the pipe be in order that the rate of discharge may be 5 cu. ft. per sec.? In that case, what pressure exists in the pipe at C? A ns. d = .82 ft. Pressure head at C =46.4 ft. above atmosphere. 3. A pipe AB, 5000' long, is to discharge 5 cu. ft. per sec. in the direction AB. The point B is higher than A, and it is required that the pressure head at A is to be 200', and that at B 100'. What must be the diameter of the pipe? Ans. Nearly 1 foot. 4. A pipe AB is 6000' long and 8" in diameter, the point B being 80' higher than A. If the pressure head is 100' at B and 190' at A, in which direction is the flow, and what quantity is discharged per sec.? Ans. q = .623 cu. ft. per sec. in direction AB. 5. In Ex. 4, if the discharge is to be 3 cu. ft. per sec. in the direction AB, what must be the excess of pressure at A over that at 5? Ans. 84.3 ft. head. 89. Loss of Head in Capillary Tubes. Experiment indicates that the loss of head in tubes of very small diameter varies according to quite different laws from those applying to ordinary pipes. Poiseuille gave a formula equivalent to the following, deduced from experiments on glass tubes from .014 mm. to .65 mm. in diameter: H'-k- H -A. FLOW OF WELLS. 81 The coefficient k depends upon the viscosity of the water, and decreases as the temperature increases. 90. Flow Through Gravel. When water flows through porous earth or gravel, the velocity is in general small, and the loss of head is found to vary approximately as the first power of the velocity. The loss per unit distance along the stream depends upon the character of the material through which the water is flowing, being greater as this is finer and more compact. 91. Flow of Wells. A bed of gravel or other porous material lying between two layers of clay or impervious rock may carry water under pressure. Such a case is represented in Fig. 50, A' B' FIG. 50. which shows two tubular wells A and B which have been driven from the surface of the ground down to and through the water-bearing stratum WW. The portion of each tube within the porous bed is perforated so as to permit free entrance of water. If the same stratum is tapped by several such tubes, the water will rise in all to the same level if the water in the stratum is at rest. If, however, as is generally the case, the water is flowing through the porous layer, different pressure columns will stand at different levels. 82 EQUATION OF ENERGY WITH LOSSES. Considering a case in which the water is at rest, let the two tubes A, B (Fig. 50) be carried high enough to prevent over- flow, and let A'E' be the horizontal plane to which the water rises. Now suppose water to be drawn from the tube A at a uniform rate, either by pumping or by cutting the tube off at some distance below A' so as to permit overflow. Flow at once begins into the tube and toward it from all directions through the gravel, and a steady condition ensues in which the quantity entering the tube in a given time is equal to the quan- tity flowing out. In this steady condition there is a definite relation between the rate of discharge and the drop of the water surface below the plane A'E'. The nature of this rela- tion depends upon the way in which loss of head varies with velocity of flow both in the porous stratum and in the tube. 92. Relation between Rate of Discharge and Fall of Water Surface. Assuming the horizontal plane A'E' as datum, let the equation HiHz^H' be applied, taking the up-stream point at a great distance from the well and the down-stream point at the water surface in the tube. Let v = velocity of flow in the tube; q = rate of discharge ; u = drop of water surface = A A" . Then #!=0, H 2 =-u + ^-. To express H f in terms of v we assume the law stated in Art. 90, that the loss of head due to flow through a porous stratum varies directly as the velocity, other conditions remain- ing constant. Although the velocity has different values at different points, the values at all points may be assumed to vary directly as v, so that the total loss of head in the gravel may be expressed by a term Civ, c\ being a constant. The loss in the tube is doubtless ordinarily small in comparison with that in the gravel, but to take account of this and the loss in entering the tube a term cv 2 may be introduced. The total loss of HYDRAULIC GRADIENT IN CASE OF FLOWING WELL. 83 head thus takes the form and the equation HiH 2 = H' becomes v 2 U- y from which d and c 2 being constants. Since v varies directly as q, ki and k 2 being constants whose values for any given well must be determined by experiment. The simpler equation is often sufficiently correct. These results are verified by experiments on the actual discharge of wells for different heights of overflow. 93. Hydraulic Gradient in Case of Flowing Well. Since the hydraulic gradient falls in the direction of flow by the amount of the loss of head (the change in the velocity head being in this case inappreciable) , the gradient is a surface which rises in all directions in going from the well, approaching tan- gency with the horizontal surface A'B 1 '. The position of the water surface in the tube B indicates the height of the hydraulic gradient at that point. The difference of level of the water surfaces in the two tubes (corrected slightly for the velocity head at A" and the loss of head due to the tube) is the loss in the gravel from B to A. The less the porosity of the gravel the greater is the value of this loss, i.e., the rate at which the hydraulic gradient rises in passing away from the well is greater as the stratum is less pervious. 84 EQUATION OF ENERGY WITH LOSSES. If water be drawn from both wells at the same time, the drop in the surface of each will be due in part to its own dis- charge and in part to that of the other well. The quantity of water that can be drawn from a well with a given depression of its surface below the static level A'B' (Fig. 50) may thus be influenced in an important manner by the flow of other wells in the vicinity. CHAPTER VII. GENERAL EQUATION OF ENERGY WHEN PUMP OR MOTOR IS USED. 94. Equation of Energy for Stream Flowing Through Motor. In deducing the general equation of energy (Art. 60), it was assumed that no mechanical energy is imparted to the water between the two sections A and B (Fig. 25), and that no energy is taken from it except such as is dissipated by reason of fric- tion between the particles of water and the pipe, and friction and impact among the particles of ^ water themselves. T - If a motor is driven by the stream, mechanical energy is taken by the j, motor from the water, and this must j be taken into account in forming the equation of energy. This case may be J _ represented by Fig. 51, in which M is FlG - 51 - a motor through which water flows in the direction AB. Let H f now mean the total mechanical energy lost by the water between A and B, per pound of water discharged, and let (1) in which h' is the part of H f corresponding to energy dissipated and h" the part corresponding to energy received by the motor. Then the energy gained by the volume AB per unit time is W 2 v 85 86 EQUATION OF ENERGY WITH PUMP OR MOTOR. and the energy lost is Equating, or " = H l -H 2 -h'. (2) 95. Formula for Energy Transferred to Motor. The amount of energy received by the motor from the stream per unit time is and the horse-power imparted to the motor is (4) W being in pounds per second and HI, H 2 , h' in feet. Suppose the stream passing through the motor- flows from one reservoir to another, the sur- face of the discharge reservoir being h ft. lower than that of the supply reservoir (Fig. 52) . Tak- ing sections A and B at the reser- voir surfaces, we have - FIG. 52. Hence in this case we may write (5) Wh' being the energy dissipated per second in the entire stream from A to B. The effect of the frictional losses of energy is thus to decrease by Wh' the energy imparted to the motor per second, which amounts practically to decreasing the available fall of the water by h'. STREAM FLOWING THROUGH PUMP. 87 If the motor discharges directly into the air, equations (3) and (4) still apply, but instead of reducing this to the form (5) , it is more convenient to change the notation. If h now denotes the fall from surface of supply reservoir to point of outflow from motor, and v- 2 the velocity of outflow, and (3) becomes V . :.- wv.w^-V-*?) (6) 96. Equation of Energy for Stream Flowing Through Pump. If the flow of a stream is maintained or aided by a pump, the mechanical energy given up by the pump to the water must be taken ^ p r into account in forming the equation /f of energy. Let this case be repre- sented by Fig. 53, P representing the pump, and AB being the direction of flow. ' ' L \ Let ft'" denote the energy received by the water from the pump per FIG. 53. pound of water discharged, and let h f denote as above the energy lost by dissipation between A and B per pound of the discharge. Then H' being the net loss of head from A to B, H' = h'-h m ........ (7) Reasoning as in Art. 60, we may write at once from which ft'"=# 2 -#i+ft'. ...... (8) It is seen that the action of the pump causes the effective head to increase in the direction of flow by the amount h'" '. If h'" is greater than h' ', H 2 is greater than Hi, i.e., the total loss of head H r is negative. 88 EQUATION OF ENERGY WITH PUMP OR MOTOR. 97. Formula for Energy Used in Pumping. The total me- chanical energy supplied by the pump per second is = W(H 2 -Hi+k'). (9) Here h' must be understood to include energy lost by hydraulic friction within the pump as well as in other parts of the stream. The rate of working of the pump, in horse-power, is H.P. Wk"> "550" W I. ... . (10) Suppose water is pumped from one reservoir to another (Fig. 54), the vertical lift between reservoir surfaces being h. B Taking the sections A and B at the two surfaces, we have and (9) may be written in the form The work done in pumping is thus FIG. 54. . Till "'' the same as would be done in lifting the water a height k + k' against gravity if all waste of energy could be avoided. EXAMPLES. 1. Water is supplied to the motor M (Fig. 55) from the head-race A and conducted from M to the tail-race D. The supply-pipe and waste- pipe are each 6" in diameter. Suppose that no energy is lost by dissipation in any part of the apparatus, and that the discharge is 3 cu. ft. per sec. Compute the pressure at B and at C, and the H.P. transmitted to the motor. Ans. Pressure head at B= p /w + 6.4'; pressure head at C=p,/w- 17.6'; H.P. = 15. 14' p[ 2. Water is pumped from one reservoif into another through 1000' of 6" pipe. The surface of the second reservoir is 20' higher than that of the first. If the quantity delivered is 3 cu. ft. per sec., (a) at what rate is energy imparted to the water by the pump? (6) What fraction of this energy is utilized in lifting the water? (c) The losses Cp C 10' T FIG. 55. EXAMPLES. 89 of energy in the flow are equivalent to what added lift? (d) Draw hydraulic gradient. Arcs, (a) 66.8 H.P. (6) About 10%. (c) 176 ft. 3. Water flows from one reservoir into another, the surface of the second being 10' lower than that of the first. The flow takes place through a pipe 8" in diameter and 1600' long, the intake end projecting into the reservoir, while the other end discharges below the water surface, (a) If the flow is due to gravity alone, what quantity is discharged per second? (6) In order to double the rate of discharge, at what rate must energy be given to the water by the pump? Ans. (a) 1.19 cu. ft. per sec. (6) 8.12 H.P. 4. In Fig. 56, C is the cylinder of a piston-pump which is lifting water from the reservoir B and discharging it at D. The diameter of the pipe is everywhere 2", and the radius of the bend is 3". (a) At what maximum rate can water be pumped without causing a pressure head of less than 4' (absolute) at any point in the pipe? Where will the mini- mum pressure occur? (6) Compare the pressures just below and just above the pump cylinder, (c) Compute the H.P. of the pump. [Estimate the loss due to a bend by formula of Weisbach. Solve, also, disregarding this loss and see whether it is important.] Ans. (a) 0.257 cu. ft. per sec. (6) Below, p/w=Q (absolute); above, p/w =35.8' above atmosphere, (c) 2.06 H.P. The pressure head above the cylinder is increased 0.4' by the bend, and the H.P. is increased about 0.5%. , 50- Di FIG. 56. CHAPTER VIII. FLOW IN PIPES: SPECIAL CASES. 98. Flow From Reservoir Through Pipe. If a pipe leads from a reservoir and discharges into the atmosphere, formulas for the velocity of flow and the rate of discharge may be obtained by applying the general equation of energy Hi-H 2 = H'. Referring to Fig. 57, let FIG. 57. h = total fall from reservoir to center of outflowing stream; v = mean velocity in pipe ; i/= " of outflowing stream; / = length of pipe; d = diameter of pipe. Writing the equation of energy for the stream AB, taking datum plane through the center of the stream at B, Hi=h = effective head at A; v' 2 #2 = o~ = effective head at B' t 90 FLOW FROM RESERVOIR THROUGH PIPE. 91 If the pipe is of uniform section, straight and unobstructed, H f is the sum of two parts, the loss at entrance, and the loss by friction in the pipe. If there are bends or obstructions of any kind, or expansions or contractions of the cross-section, these cause additional losses which may be taken as approxi- mately proportional to the square of the velocity. CASE I. Let the cross-section of the outflowing stream or jet be equal to that of the pipe, so that i/ = v; then The losses of head may be expressed as follows : v 2 The entrance loss is ra^-, where w = l for projecting and 0.5 for non-projecting pipe (Art. 81). I v 2 The frictional loss in the pipe is / -y- 5-, in which / has the meaning explained in Art. 80. The loss due to bends or obstructions is expressed by a v 2 term n~-, the value of n depending upon the circumstances of each particular case. Combining all losses, The equation Hi~H 2 = : H / now becomes The method of solving this equation in particular cases will be illustrated below. CASE II. Let the cross-section of the jet be less than that of the pipe. If F = cross-section of pipe and F' = that of jet, let F=cF'. Then v' = cv and (cv) 2 Hi-Hz-h-^-. 92 FLOW IN PIPES: SPECIAL CASES. To the losses of head occurring in Case I must be added a loss due to the contraction of the stream at the point of out- v' 2 flow. This loss may be expressed by a term k ~-, the value of k depending upon the ratio of contraction and the construction of the pipe and orifice. Combining all losses, and the equation of energy becomes (cv) 2 / f l\v 2 . .. 2g ' d/2g 2g Equation (1) is a special case of (2), in which c = l, Equation (2) may be written in the torm (2) - ..... (3) From it may be computed any one of the quantities v, h, d, if the others are known. For certain applications it is conve- nient to introduce the rate of discharge q instead of v. Thus, suppose it is required to estimate the size of the pipe which will give a certain rate of discharge, the total fall and length of pipe being known. Since v = q/F -=4q/nd 2 , equation (3) may be written from which d* = ^- h ([(l+k)c 2 +m+n]d+fi), .... (4) or d 5 = Ad + B, ....... (4 r ) A and B being constants. FLOW FROM RESERVOIR THROUGH PIPE. 93 In many cases, especially for long pipes discharging as in Case I, the term Ad is small in comparison with B, and the solution may be made by successive approximations. As an example, let h = 5Q', = 2500', g = 10 cu. ft. per sec., c = l ; & = 0, m = l, n = Q. These values reduce equation (4) to For a first approximate solution assume /=.02 and neglect the term 2d, which is small in comparison with 2500/. This gives which water is to be delivered, the pressure head there being required to have the value y. Let / = length of pipe to B, d = diameter, /i*= depth of B below reservoir surface. Taking datum plane through the center of the pipe at B, the values of the effective head at A and B are v 2 Hi=h, H. 2 = y+7r; while the loss of head between A and B is l\ v 2 The equation of energy therefore becomes v 2 I l\ v 2 This is identical with equation (1) with h y substituted for h. BRANCHING PIPE. 95 EXAMPLE. Water is to be delivered through a pipe 1800' long to a point 70' below the level of the reservoir surface. The rate of discharge is to be 15 cu. ft. per sec., and the pressure head at the point of delivery 40'. What should be the size of the pipe? Ans. d = 18". 100. Branching Pipe. If water is delivered through a main pipe with branches, the equation of energy must be applied to each portion of the pipe separately, sufficient data being known or assumed to make the problem of flow determinate. Thus, as a simple case, consider the main pipe AB with two branches BC, BD (Fig. 59), and let the given data be the FIG. 59. rate of discharge at C and at D, and the pressure at each of the three points B, C, D. The method of treatment may be illustrated by a numerical case. Let 8 cu. ft. per sec. be delivered at C and 10 cu. ft. per sec. at D;. let the depth of B below the reservoir surface be 100', that of C 80', and that of D 110'; let the length of the main pipe to B be 2000', that of BC 1200', and that of BD 1600'; and let the pressure head have the value 60' at B, 30' at C, and 50' at D. Let d, d', d" be the diameters of the three pipes AB, BC, CD; and v, v f , v" the corresponding velocities of flow. The losses of head at entrance to the branch pipes cannot be accurately estimated, but will be small in comparison with the frictional losses in the pipes and will be neglected. For the pipe AB the equation is 40 _ ~ 2000X1* d >> 96 FLOW IN PIPES: SPECIAL CASES. 1 8 from which, since v= 2 > the va ^ ue f ^ ma y be found as in Art. 98. For the pipe BC we have ' 2 12001/2 v 2 v' 2 1200/2 hence 77- + 10 ^ = / -rr- . 2g 2g d' 2g The value of d having been previously determined, v is known. o And since i/ = 2/ :, d' may be determined by solving the last equation in the usual way. The branch BD may be treated in the same manner. EXAMPLE. Complete the solution of the above numerical case. 101. Relation of Pipe to Hydraulic Gradient. For a uni- form pipe, straight and unobstructed, the hydraulic gradient (Art. 86) is uniform for the entire length. If the stream dis- charges into the air without contraction, the pressure at the outlet end is atmospheric, while at a short distance within the intake end it is less than atmospheric by a small amount, equiv- alent to the velocity head plus the entrance loss. Between these points the hydraulic gradient is a straight line, as shown in Fig. 57. Gradual curves in the pipe line will not materially affect the uniformity of the hydraulic slope, but the velocity which corresponds to a given total fall will be less when there are bends than in a straight pipe of the same length. In laying a long pipe line, both horizontal and vertical curves must often be introduced because of the contour of the ground. In planning the line it is of especial importance to see that EFFECT OF AIR IN CHECKING FLOW. 97 the pipe shall everywhere lie below the hydraulic gradient as it will exist when the rate of discharge has its greatest value. Fig. 60 shows a case in which a large part of the pipe is above the hydraulic gradient AB. Assuming full flow to exist, the pressure in the pipe everywhere between C and B is less than atmospheric. The resultant pressure upon the pipe is thus from without, but (aside from the danger of collapsing the pipe) the decrease of pressure will not of itself interfere with the flow unless the pipe goes high enough to reduce the pressure to absolute zero. Practically, however, it would be difficult to start full flow under such conditions, and even if started it would not continue unless the pipe were absolutely air-tight FIG. 60. and the water free from air. Since air is always carried by natural waters, there would soon be an accumulation of air in the higher parts of the pipe which would check the flow even if there were no leakage from without. 102. Effect of Air in Checking Flow. The collection of air at a summit may seriously interfere with the discharge of a pipe even when everywhere below hydraulic gradient, but so long as the pressure is above that of the atmosphere the air may be removed by means of a valve provided for the purpose. To understand the action of air in checking the flow, con- sider the case represented in Fig. 61. Here the hydraulic gra- dient for full flow is everywhere above the pipe. But if air collects in the bend C, the stream will finally be divided into two parts, the surface beyond C being forced down toward 7), 98 FLOW IN PIPES: SPECIAL CASES. as shown at Y. The flow will be retarded, but will not be stopped unless the pressure of the confined air becomes great enough to balance the static pressure due to the water column AX plus atmospheric pressure. If D is far enough below B, the condition represented in Fig. 62 will finally ensue and flow will cease. The pressure of the confined air will then just bal- ance each of the two equal * columns AX, YB. A FIG. 61. The effect of air upon the hydraulic gradient is shown in Figs. 61 and 62. With full flow the gradient would be a line of nearly uniform slope from A to B. As air accumulates, the pressure increases in the up-stream part of the pipe and de- creases in the down-stream part, piezometer columns standing FIG. 62. at equal heights above the surfaces X and Y as shown in Fig. 61. When the flow is stopped, as in Fig. 62, the gradient con- sists of two horizontal lines, one lying in the plane of the reservoir surface, the other in a horizontal plane as far above Y as X is below the reservoir surface. * This neglects the weight of the confined air, by reason of which the pressure at Y would be slightly greater than that at X. CHEZY'S FORMULA. 99 103. Long Pipe with Full Discharge. In case of a long pipe I v 2 the friction loss, expressed by the term / -7 75-, is very great in comparison with the other losses of head, and also in compari- son with the velocity head. Equation (1) therefore reduces practically to the form f =f d2g' and equation (5) to the form In the majority of practical problems relating to the design of water supply systems these simplified equations are used, the value of the neglected terms being within the limits of relia- bility of the data. 104. Che"zy's Formula. The formula for loss of head in a uniform straight pipe, given in Arts. 79 and 80, is often written in another form. Introducing the hydraulic radius (the ratio of the area of the cross-section to its circumference) instead of the diameter, the formula I V 2 becomes F/== Solving for v, writing s for -r, and introducing a new coefficient c such that we have v = c\/rs, which is known as Chezy's formula.* * This is the formula commonly used in the discussion of flow in open channels. See Chapter XI. 9 100 FLOW IN PIPES: SPECIAL CASES. If there are bends in the pipe, or if the roughness of the surface varies in different parts, the losses of head in different equal lengths are unequal, and the hydraulic slope s varies along the pipe. This is usually the ease in practice. In such cases the formula v = cVrs is still often employed, s being regarded as the average hydraulic slope for the entire length, i.e., s = H f /I, where H f is the total loss of head in the length /. The value of c must be taken less than in the case of straight pipe. The formula for / given in Art. 80, based upon Darcy's results, is equivalent to the following formula for c: Further discussion of the values of / and c is given in Chap- ter IX. The two formulas are, as above shown, identical. Although the latter has been much employed, the former is preferable for the reason that / is an abstract number, while c depends upon the system of units employed. In solving the following examples, let Chezy's formula be employed, the value of c being determined from the formula above given. EXAMPLES. 101 EXAMPLES. 1. A pipe 18" in diameter will discharge what quantity of water per second if the hydraulic gradient falls 10' in a length of SOOO'? Ans. = wF(zi - z 2 ) . Equating to the sum of the resolved forces in the direction of the flow, piF-p 2 F+wF(zi-z 2 )-P=Q. Since pi/w + Zi = yi and p 2 /w + z 2 = y 2 , and since y\-y 2 = H' = loss of head between A and B, the equation may be written Thus the loss of head is expressed in terms of the total frictional force exerted by the pipe surface upon the water. If the frictional force per unit area is P', so that P = CIP', we may write OF = j_ P_ wF r w' r being the hydraulic radius as already defined (Art. 79). Experiment indicates that the value of P' is independent of the pressure, but varies with the velocity of the relative motion. If the law of variation were known, the above formula would MEAN VELOCITY IN PLACE OF SURFACE VELOCITY. 107 give the loss of head in terms of the surface velocity. Thus, if vi is the surface velocity, and if the formula becomes 109. Introduction of Mean Velocity in Place of Surface Velocity. Equation (1) would be of little use even if the form of the function <(i?i) were known, since the quantity of impor- tance is not the surface velocity v\, but the mean .velocity v. If it be assumed that there is a fixed relation between v\ and v which is independent of the size of the pipe, there results an equation of the form . ,: .--' < . . (2) If this assumption were correct, the dependence of H f upon the size of the pipe would be expressed by a very simple rela- tion. Thus, since for circular pipes r = d/4, loss of head would vary inversely as diameter. The fact that this simple law is not verified by experiment indicates that the relation between mean velocity and surface velocity is not independent of the size of the pipe. Some of the earlier writers on Hydraulics, however, assumed that mean velocity varies directly as surface velocity, and thus deduced the form of the function F(v) in equation (2) from the experimental law governing the friction between a solid and a fluid. 110. Application of Experimental Law of Fluid Friction. The frictional force per unit area exerted by a solid body upon a fluid flowing uniformly over its surface is found (a) to be independent of the pressure and (b) to vary approximately as the square of the velocity of flow, except for quite small velocities, 108 FRICTIONAL LOSS OF HEAD IN PIPES. and (c) to vary nearly in direct ratio with the velocity of flow if this is very small. Applying this to the pipe problem, the value of P f may be expressed approximately in the following form: (3) Assuming that v\ varies directly as v t equation (2) may be written H' = (av+bv 2 ) ....... (4) t 111. Formula of Proqy. Equation (4) is the formula given by Prony, who, from such experimental data as were available, deduced numerical values of a and b which were supposed to hold for all sizes of pipe. It was supposed, also, that the char- acter of the surface tif the pipe had little influence upon the values of these coefficients. Formula of Darcy. Equation (4) was also employed by Darcy, who found, however, that the coefficients a and b vary greatly with the character of the pipe surface,* and also that they are not independent of the diameter. From his experiments he deduced empirical formulas of the forms B (5) giving numerical values for a, /?, a:', /?', applicable to pipes of about the degree of smoothness of new cast iron. Darcy's experiments, as well as many made subsequently, indicate not only that the term av in equation (4) is unimpor- tant in comparison with bv 2 except for quite small velocities, but also that its importance is less the rougher the surface of * Being, for example, about twice as great for cast-iron pipes fouled by some years of use as for the same pipes new or thoroughly cleaned. CHEZY'S FORMULA. KUTTER'S FORMULA. 109 the pipe. For pipes in practical use, and for the range of velocities ordinarily existing, Darcy recommended as sufficiently correct a formula of the form fl"-7*i* ....... (6) And for values of d within the range of his experiments (from 0.5 inch to 20 inches approximately) he gave the following for- mula for the coefficient 61 : The equivalent of this formula, with numerical values of the coefficients, is given in Art. 80. / 113. ChSzy's Formula. It will be seen that equation (6) is really the same as the formula already given in two forms (Arts. 80, 104) : v = cVrs. .;,'... . . (9) In the latter form it is often known as Chezy's formula, and is identical with the formula commonly employed in estimat- ing the discharge of streams flowing in open channels.* 114. Kutter's Formula. Kutter's formula is an empirical expression for the value of c in terms of r, s, and a certain quan- tity n called the coefficient of roughness, whose value depends upon the character of the surface over which the water is flow- ing. The relation is as follows: I +41.654^ r being in feet and v in feet per second. * See Art. 128. 110 FfUCTlOiNAL LOSS OF HEAD IN PIPES. Although this complex formula was based upon experi- mental data for flow in open channels, it has often been applied to pipes and closed conduits, especially those of large size. Its use cannot, however, be recommended, except in the absence of experimental knowledge directly applicable to the case in hand. Some indication of the values of n to be used for pipes of different kinds will be given below. 115. Exponential Formula. The foregoing discussion has brought out the fact that the loss of head in a given length of pipe varies approximately, but not exactly, as the square of the velocity of flow in pipes of the same size, and inversely as the diameter in pipes of different sizes; all being assumed alike in the character of the interior surface. It was found by Osborne Reynolds,* in experiments with lead tubes of one- fourth inch and one-half inch diameter, that the loss of head varied as a power of the velocity whose index was less than 2. A similar relation was observed by Lampe in experiments with a pipe about 16.5 inches in diameter. From a study of the experimental results of Darcy and Poiseuille, Reynolds found a similar exponential relation, but with the index varying with the character of the pipe. A study of other experimental data obtained with both small and large pipes shows that the variation of loss of head with velocity in a given pipe may often be closely represented by the formula with constant values of A and m. The value of m is usually less than 2, but varies considerably for different pipes, and in some cases has been found to be greater than 2. There is some evidence tending to show that for a given character of pipe surface m is independent of the diameter, and that it increases with the roughness of the surface. Thus a value of 1.72 was found by Reynolds for the small lead tubes, and about the same value has been found for other quite smooth * Philosophical Transactions of the Royal Society, 1883, Part III, p. 975. EXPONENTIAL FORMULA. Ill pipes,* while for pipes quite rough by incrustation or other- wise a value of about 2 has been found.! Whether the character of the pipe surface is the only im- portant factor affecting the value of the exponent cannot be determined from known experimental data, but it seems likely that for pipes laid under the conditions of practice it may vary with the curvature of the pipe line, the temperature of the water, and perhaps other factors. The value of A probably depends upon the character of the pipe surface, as well as upon the diameter. A series of experi- ments upon pipes alike in all respects except in size would doubtless show a regular variation of A with the diameter. It has been supposed that this also may prove to be an exponen- tial relation, but for large pipes experimental data for the establishment of such a law are almost wholly lacking. Such evidence as there is seems to show that if A varies with a power of d, the exponent will be negative and numerically greater than unity, t It was pointed out by Reynolds that the above formula is a convenient one for representing experimental data because of the ease with which the constants can be determined graphic- ally. Taking logarithms, the formula becomes log H f = log A+m log v. If a series of experimental values of H' and v satisfies this for- mula, the locus whose coordinates are log H f and log v will be a straight line. The slope of this line determines the value of m, and its intercept on the axis along which log H f is meas- ured gives the value of log;!. The plotting is facilitated by the use of logarithmic cross-section paper. The plotting of the results in this manner shows at once whether the assumed form of formula agrees well with the experimental data, and also gives a simple determination of the. constants in case it does. * See, for example, Trans. Am. Soc. C. E., Vol. LI, p. 52. t Phil. Trans. Roy, Soc., 1883, Part 111, p. 981. Trans. Am. Soc. C. E. Vol. XXXV, p. 258. J Tables for practical use, based upon an exponential formula, have recently been published by Gardner S. Williams and Allen Hazen. 112 FRICTIONAL LOSS OF HEAD IN PIPES. 116. Critical Velocity. The experiments of Reynolds showed that there was a certain velocity below which the loss of head varied nearly as the first power of the velocity, and above which it varied as a higher power. The value at which the law changes he called the critical velocity. By observations of the flow in glass tubes he found that for very low velocities the flow took place quite accurately in parallel filaments, while above the critical velocity the stream became turbid. In practical Hydraulics velocities below the critical value rarely or never need be considered. It is to the range of veloc- ities above the critical point that the above discussion of the exponential formula refers. 117. Formula Adopted. For the purpose of expressing working rules for estimating loss of head, the equivalent for- mulas v = cv rs, will here be employed. For convenience values both of the friction factor / and of the coefficient c will in most cases be given. When one of these quantities is known the other can be computed from the relation As already pointed out, / is an abstract number and therefore independent of the system of units employed, while c depends upon the units of length and time, c 2 being of the same dimen- sions as g. Owing to the imperfection of the theory upon which the above formulas are based, / and c cannot be regarded as con- stants for a given kind of pipe, but vary with both diameter and velocity. 118. Values of Friction Coefficients. Many attempts have been made, by study of known experimental data, to establish OF THfe UNIVERSITY OF VALUES OF FRICTION definite values of the friction factors for the kinds of pipe used in practical hydraulic works. Formulas or tables professing to give such values with great accuracy cannot, however, be accepted with confidence, because of the great uncertainties in the reliability of the data upon which they are necessarily based. The experiments of Darcy * are probably the most trust- worthy series yet made covering any considerable range of diameters and velocities. Upon the results he based two for- mulas, one expressing as accurately as possible the variation of the friction factor with both diameter and velocity, the other neglecting the variation with velocity. The latter, which was recommended by its author for practical use, was equivalent to the following, in which d is the diameter in feet : This is for pipes of new or clean cast iron, or other pipes of equal smoothness. For pipes tuberculated or otherwise rough- ened by some years of use Darcy recommended that the values of / given by this formula be doubled. The pipes used in Darcy 's experiments ranged in diameter from 0.0122 meter to 0.5 meter. The most reliable experiments made since by others indicate that his formula gives safe values, not only within this range, but for larger sizes, and the formula has often been used for diameters as great as 4 feet. Table III (page 114), based upon the formula, is carried to about double the largest size experimented upon by Darcy. There is some evidence that the coefficients in column 3 of the table are somewhat too great for pipes as smooth as ordinary clean cast-iron water pipes. It is, however, impor- tant in engineering design to allow a margin of safety to cover both the large element of uncertainty in existing knowledge regarding friction losses in pipes of any given character, and the uncertainty in any given case regarding the actual present * Recherches experimentales relatives au mouvement de 1'eau dans les tuyeaux. Henry Darcy. Paris, 1857. 114 FRICTIONAL LOSS OF HEAD IN PIPES. and future character of the particular pipe under consideration. The tabulated coefficients, both for smooth and for rough pipes, probably err on the side of safety, unless it be for the case of pipes which have suffered exceptional deterioration with long use. TABLE III. VALUES OF THE FRICTION FACTOR / AND OF THE COEFFICIENT c, BASED UPON THE FORMULA OF DAR Y. 1 2 3 4 5 6 Diameter. Smooth Pipe Rough Pipe Inches. Feet. / c / c 1 .0833 .0398 80 .0796 57 2 .167 .0298 93 .0596 66 3 .250 .0265 99 .0530 70 4 .333 .0248 102 .0496 72 6 .500 .0232 105 .0464 74 8 .667 .0224 107 .0448 76 10 .833 .0219 108 .0438 77 12 1.000 .0216 109 .0432 77 14 1.167 .0213 110 .0426 78 16 1.333 .0211 110 .0422 78 18 1.500 .0210 111 .0420 78 24 2.000 .0207 111 .0417 79 30 2.500 .0206 112 .0412 79 36 3.000 .0205 112 .0410 79 The degree of roughness or smoothness of any given pipe must be a matter for the judgment of the engineer. But, except in the rare cases in which pipes are laid for temporary use only, the design must be governed by probable future conditions rather than by the character of the pipe when new. 119. Large Pipes. Although there is little experimental evidence regarding the variation of / and c with the diameter in the case of large pipes, this variation appears to be unim- portant. As regards variation with the velocity, there is evi- dence that, except for quite small velocities, this is of little practical importance even with smooth pipes, and is quite inappreciable in the case of rough pipes, such as those of riveted steel or almost any pipe after long use. For pipes from 3 ft. FRICTION FACTORS FOR LARGE PIPES. 115 to 6 ft. in diameter it is therefore sufficient to use values of / and c which are independent of velocity and diameter, varying only with the character of the pipe. The general range of these values is indicated in the follow- ing table. It is of course to be understood that actual pipes present all gradations of roughness, and the four cases specified in the table can only serve as a very general and approximate guide. The values tabulated (except for Case I) are designed to allow for the resistance due to such curves as are likely to exist in practical cases. TABLE IV. FRICTION FACTORS AND COEFFICIENTS FOR LARGE PIPES. (Diameter from 3 ft. to 6 ft.) Case Typical Pipe. / c n I. (Very smooth.) Exceptionally smooth cast-iron pipe, new or thoroughly cleaned, without bends 013 140 Oil II. (Smooth ) Ordinary new cast-iron or wooden pipe with some bends 018 120 0125 III. (Rough.) New riveted pipe, with some bends, or any pipe after some years of use. . . .023 106 .014 IV. Any pipe after long use 046 75 019 (Very rough.) 120. Kutter's Coefficient of Roughness. Kutter's formula (Art. 114) has come into quite general use as a practical guide in estimating the discharging capacity of large pipes. This formula assumes to take account of the roughness of the sur- face by the value assigned to the coefficient n. It also makes c vary with diameter and velocity. The values of n given in the last column of the above table are about the average values implied by the corresponding values given for c. CHAPTER X. EQUATION OF ENERGY FOR STREAM OF LARGE CROSS- SECTION. Streams of Large Cross-section. The theoretical dis- cussion given in Chapter IV, leading to the general equation of energy for a steady stream, involved the assumption that the intensity of pressure may be regarded as uniform throughout any cross-section, and that all particles passing the section have equal velocities. These assumptions cannot be supposed to be even approximately true when the cross-section is large, as in the case of rivers and canals. Let us consider whether the theory requires important modification when variations of pressure and of velocity in a cross-section are taken into account. Variation of Pressure Throughout a Cross-section. If all particles move in straight lines parallel to the axis of the stream, the pressure in any normal cross-section varies with the depth according to the hydrostatic law. For let A and B be two points such that AB is perpendicular to the direction of flow, and consider a prism of water of small cross-section whose axis is AB. Resolving all forces acting upon this prism in the direction AB, the sum of such resolved forces must be zero, since there is no acceleration in this direction. Hence the reasoning of Art. 11 may be applied, leading to the same conclusion. 123. Variation of Velocity in a Cross-section. The way in which the velocity varies throughout the cross-section has been the subject of considerable investigation, both experimental and theoretical. The following discussion is, however, inde- pendent of any assumption as to the actual distribution of 116 ENERGY PASSING A CROSS-SECTION. 117 velocities ; though, as will be seen, a certain term in the equation of energy cannot be evaluated unless the law of distribution is known. 124. Energy Passing a Cross-section. To determine the energy passing any cross-section per unit time, we may reason substantially as in Art. 59, introducing such changes as are necessitated by the variation of pressure and of velocity through- out the section. Referring to the cross-section A (Fig. 65), let v = velocity at any point in the section; F = whole area of the section; g= rate of discharge across area F; v f = -p = mean velocity ; p = pressure at point whose velocity is #; z = height of that point above datum. Energy transferred across the section by pressure. The energy transferred across an element dF of the cross-section by reason of the pressure and velocity may be computed as in Art. 59. Its value per second is vpdF, and therefore the energy thus transferred across the entire section per second is fpvdF, (1) the integration covering the whole section F. Potential energy carried across the section. The potential energy carried across an elementary area dF per second is wzv dF, 118 EQUATION OF ENERGY: LARGE STREAM. and the potential energy passing the whole area F per second is yfzvdF. . . . . . . * (2) w Sum of potential energy and energy transferred by pressure. The sum of the values (1) and (2) may be written w Now since the pressure varies throughout the section according to the hydrostatic law, in passing to different points in the section p/w increases (or decreases) by just the amount of decrease (or increase) of 2; that is, z + p/w is constant through- out the section. Denoting this sum by y, it is seen that y is equal to the height above datum of the top of a piezometer column communicating with the pipe at some point of the given cross-section. Since y is constant in the integration, the value of the in- tegral becomes (3) Kinetic energy passing the section. The amount of kinetic energy passing the elementary area dF in one second is hence the amount passing the whole section F is w This integral cannot be evaluated unless the variation of v throughout the section is known. It may, however, be shown that it is greater than the value which would be obtained if ENERGY PASSING A CROSS-SECTION. 119 the mean velocity ?/ were assumed to apply throughout the section. That is, Thus, let v = v'+u; then fu dF + But since it denotes the excess of actual velocity at any point over mean velocity. We may therefore write The last integral is positive; for and since v f -\-u or v is always positive even when u is negative, it follows that v'H--^ must be positive for all values of u. The o inequality (5) is thus proved. The total kinetic energy carried across the section per unit time may therefore be written K being a positive quantity equal to the last term of (6). Total energy parsing the section. The total energy passing 120 EQUATION OF ENERGY: LARGE STREAM. the section in one second is the sum of the values (3) and (7). It may be written in which k=K/wq. The value of the energy passing any cross-section per unit weight of water discharged may therefore be expressed in either of the following forms: V 2 V V 2 in which v is now written for the mean velocity in the cross- section. As shown above, z and p may refer to any point in the section, their sum being constant. Comparing this result with that reached in Art. 59 it is seen that the effect of the variation of velocity is represented by the term k. 125. Equation of Energy in Case of Large Stream. The reasoning of Art. 60 may be applied to the case of a large stream, using the expression above deduced for energy passing a section. With suffixes (i) and ( 2 ) to refer to up-stream and down-stream sections respectively, we have in which the values of z and p refer to any point of the section. Instead of z-\ p/w we may write y; y being constant for each section, and meaning the height above datum of the top of a piezometer column communicating with the pipe at any point of the perimeter of the section. As shown above, k\ and k% are positive quantities, the value of each depending upon the distribution of velocities in the corresponding section. The quantity k cannot be zero unless the velocity has the same value throughout the section, and will be great in proportion as the variation of velocity is great GENERAL EQUATION OF ENERGY. 121 If the two sections compared have the same size and shape, it may reasonably be assumed that ki=k-2\ in this case the equa- tion of energy takes the same form as if the velocity were uniform throughout every cross-section. For sections of un- equal size and consequent unequal mean velocities no such assumption can rationally be made. Ordinarily, for circular pipes, no great error results from using the equation of energy in the ordinary form, as above in the discussion of small pipes. This statement is justified by experiment; and it may be concluded either that the value of k is usually small, or that its values for sections of different size are nearly equal. CHAPTER XI. UNIFORM FLOW IN OPEN CHANNELS. 126. General Principles. The principles employed in the discussion of flow in pipes apply in the main to streams in open channels. The chief difference arises from the fact that in the case of a stream in an open channel the upper surface is in contact with the atmosphere. From this it follows (a) that the pressure at this surface has a known constant value, and (&) that the frictional resistance to flow at this surface is much less than at the surface of the channel itself. The way in which the velocity of flow varies throughout any cross-section is in general unknown, and can be determined only by experiment. The effect of this variation upon the general equation of energy for steady flow has been considered in Chapter X, the discussion there given applying to open as well as to confined streams. In the following discussion it is usually the mean velocity in a cross-section, rather than the actual velocities in different parts of the section, that will be considered. If this mean velocity is denoted by v, the relation = constant for all cross-sections holds for an open channel, if the flow is steady. The following discussion will be restricted to the case of steady flow in a channel of uniform cross-section and slope. In the cases of most importance it will further be true that the depth of the water is the same at all sections, so that the water cross-sections are all alike in size and shape. The flow 122 FORMULA FOR MEAN VELOCITY. 123 is then said to be uniform. The problem of non-uniform flow will be considered in the next chapter. 127. Uniform Flow. Consider the case of a channel of uni- form cross-section and slope and of considerable length, into the upper end of which water is admitted at a constant rate. In a short time the flow will become steady, and it will also become practically uniform throughout the greater part of the straight and uniform channel under consideration. When this condition is reached, the velocity of flow will depend upon (a) the slope of the channel, (b) the character of the surface over which the water flows, and (c) the dimensions of the cross- section. (a) The velocity is greater as the slope is greater. (b) The velocity is greater as the channel surface is smoother. (c) The greater the length of the wetted perimeter of the cross-section in comparison with its area, the greater the re- tarding effect of friction, and therefore the less the velocity. 128. Formula for Mean Velocity in Case of Uniform Flow. No theoretical discussion of flow in open channels, even in the simplest case of uniform flow, can serve as more than a rough guide in the solution of practical problems. The formula most commonly employed in practice is that of Chezy, already given as applying to pipes. The reasoning of Arts. 108-110 may be applied with slight change to the case of an open channel. Let Fig. 66 represent a longitudinal section of the stream, and consider the body of water between two cross-sections at A FIG. 66. A and B. The flow being steady, this body has no acceleration, and the forces acting upon it form a system in equilibrium. 124 UNIFORM FLOW IN OPEN CHANNELS. Let these forces be resolved in the direction of the flow. The forces are the following: (a) The weight of the body, equal to wFl, F being the cross-sectional area and / the length AB. (6) The pressures of the adjacent water upon the cross- sections at A and B. Since the two sections are alike in all respects, and since the upper surface is under uniform pressure, the total pressure upon the cross-section A is equal and oppo- site to that upon B, each having the value pF, the product of the cross-sectional area into the intensity of pressure at the centroid of the area. (c) The frictional forces exerted in the direction BA upon the water flowing over the channel surface. Computing it as if all particles of the water had equal velocities, and assuming the friction per unit area P f to be independent of the pressure and proportional to the square of the velocity, the total fric- tional force would be C being the length of the wetted perimeter, and k a constant depending upon the roughness of the surface. (d) The frictional retardation due to air at the upper sur- face. This is assumed to be negligible. Equating to zero the sum of the resolved parts of these forces in the direction AB, and representing by s the sine of the angle between the water surface and the horizontal, we have wFls-kClv 2 = Q. Introducing r= F/C = hydraulic radius of the cross-section, the equation may be written in which c is a coefficient whose value depends upon the rough- ness of the channel, and would be constant for any given kind of channel if the assumptions above made were rigorously true. In comparing this result with that previously obtained for pipes (Art. 113), it will be noticed that s here means the slope VARIATION OF COEFFICIENT IN CHEZY FORMULA. 125 of the water surface, while in the previous case it means the hydraulic slope. It is obvious, however, that the hydraulic slope of a stream having a free upper surface is identical with the slope of that surface. If a piezometer be introduced at any cross-section, as in Fig. 67, the water will rise in it to the level of r - the surface of the stream. Fig. 66 Tf shows at once that * " i -= i i' so that the above equation might be written in either of the two forms given in Art. 113. 129. Variation of Coefficient in Chezy Formula. The above theory is obviously seriously defective, but no general formula having a better basis in theory has been proposed. In accepting this formula as a practical guide, the coefficient c must be regarded as dependent not merely upon the character of the channel surface, but upon various other conditions. The practical rules which have been given for estimating the value of c, based upon experiment, usually express it as a function of one or more of the three quantities v, r, and s, in addition to the rough- ness of the channel. It is altogether improbable that any such assumption can properly be made (except as a rough approxi- mation) in comparing streams having very different forms of cross-section, since the form of the section probably affects the flow in a manner which is not dependent merely upon the value of r, and which in fact cannot be expressed accurately in any simple way. In solving practical problems in flow, however, a rough approximation to the true result is often all that it is possible to attain. In spite of the defects in the theory above given, it is still true that the most important factor affecting the value of c is the character of the channel as regards roughness. The range of values of the coefficient for different cases may be stated roughly as follows : 126 UNIFORM FLOW IN OPEN CHANNELS. c = 100 to 140 for timber or cement-lined conduits. c= 80 " 110 " conduits of smooth masonry. c= 60 " 90 " conduits of rough masonry (rubble). c= 50 " 80 " ditches in clean earth or gravel. c= 20 " 40 " ditches or canals in bad order. The experimental determination of c is much more difficult for open channels than for pipes. An accurate determination requires that the channel under experiment shall be quite truly uniform in slope; otherwise the cross-section of the stream will vary even if that of the channel does not. Unless the condition of uniform flow is very accurately maintained, no reliable value of c can be deduced from the experiment. 130. Experiments of Bazin. An extensive series of experi- ments was performed by Bazin for the purpose of determining how the flow is influenced by the character of the surface, the slope of the channel, and the size and shape of the cross-section. From these experiments the author drew the conclusion that the value of c varies but little with s, and that, so far as the dimensions of the cross-section affect it, c is mainly a function of r. The conclusions reached were embodied in a formula equivalent to the following: The empirical constants c f and a depend upon the roughness of the channel, and Bazin gave for four typical cases the values shown in the accompanying table. TABLE V. VALUES OF COEFFICIENTS IN FORMULA OF BAZIN FOR OPEN CHANNELS. Kind of Channel. c' a Very smooth . . . . ' 143 0.10 (Cement or planed timber.) Smooth 127 0.23 (Ashlar, brickwork, rough timber.) Rough 113 0.82 (Rubble masonry.) Very rough 104 4.10 (Ditches or canals in bad order.) KUTTER'S FORMULA. 127 It will be noticed that c' is the limiting value approached by c as r increases. The range of the experiments does not, how- ever, justify the use of the formula for values of r greater than 2 or 3 feet. , 131. Kutter's Formula. The only wholly general rule for choosing the value of c which has been widely used is the formula proposed by Ganguillet and Kutter. This formula is the result of a careful study of all the experimental data regard- ing flow in open. channels which was known to its authors. It is designed to apply to the widest range of cases, from the smallest of artificial channels up to large rivers, and to take account of all factors affecting the value of the coefficient. In Kutter's formula c is expressed in terms of the slope s, the hydraulic radius r, and a third quantity n called the coeffi- cient of roughness. For English units the expression is n The authors of the formula gave values of n for six typical kinds of channel surface, ranging from .010 for flumes lined with well-planed timber or neat cement to .030 for streams impeded by detritus or aquatic plants. The following more detailed series of values is often given : Nature of Channel. n Well-planed timber 009 Neat cement 010 Cement one-third sand. Oil Ashlar and brickwork 013 Canvas on frames 015 Rubble masonry 017 Canals in very firm gravel 020 Rivers and canals in perfect order, free from stones or weeds .025 Rivers and canals in moderately good order 030 Rivers and canals in bad order, with weeds and detritus 035 Torrential streams encumbered with detritus ' 050 While these values may be fairly reliable for channels of uniform slope and alignment, greater values should probably 128 UNIFORM FLOW IN OPEN CHANNELS. be used in designing flumes and ditches under ordinary prac- tical conditions. The range of values given in Fig. 69 may be suggested. 132. Reliability of Kutter's Formula. This formula has gained wide acceptance as a guide in choosing values of c for all cases of open channels, and also for large pipes conveying water under pressure. It is undoubtedly a formula of great value, since it provides a rule applicable to cases for which otherwise there would be no guide whatever because of the lack of experimental data. The student should, however, be warned against the too implicit acceptance of the results of this or any other empirical formula in hydraulics, and espe- cially against the supposition that the estimated values of c (and the quantities computed from it) can be regarded as more than approximations subject to a considerable percentage of uncertainty. It is quite common for writers to give values of c to four significant figures, when in fact they can hardly be supposed to be certainly correct to within 10 per cent. Here as elsewhere in engineering computations it is important for the student to acquire the habit of putting a reasonable esti- mate on the degree of reliability of his data and formulas. 133. Tables Computed from Kutter's Formula. To facilitate the use of Kutter's formula, values of c computed from it are given in Table VI. Values not given directly in the table may be found with sufficient accuracy by interpolation. 134. Graphical Representation of Kutter's Formula. The formula of Ganguillet and Kutter may be written in the form y _ yVr in which TABLE COMPUTED FROM KUTTER'S FORMULA. 129 TABLE VI. VALUES OF c IN THE FORMULA v = c\/rs, COMPUTED FROM FORMULA OF GANGUILLET AND KUTTER. 8 n r (feet). .2 .3 .4 .6 .8 1.0 1.5 2.0 3 4 6 8 10 15 .00005 .010 87 99 109 122 133 140 154 164 178 187 199 206 212 222 .012 68 79 88 98 107 114 126 135 148 156 168 175 181 190 .015 51 59 66 76 83 89 99 107 118 126 137 144 149 158 .020 35 41 46 53 59 64 72 79 88 95 105 111 116 125 .025 26 31 35 41 46 49 57 62 71 77 85 91 96 104 .030 21 25 28 33 37 40 47 51 59 64 72 78 82 90 .035 18 21 24 28 31 34 40 44 50 56 63 68 72 80 .040 15 18 20 24 27 29 34 38 44 49 56 61 64 72 .0001 .010 98 109 119 131 140 147 159 168 178 186 195 201 205 212 .012 76 87 95 105 114 120 130 138 149 155 164 170 174 181 .015 57 65 72 81 88 93 103 109 119 125 134 139 143 150 .020 39 45 50 57 63 67 75 81 89 94 102 107 111 118 .025 29 34 38 44 48 52 59 64 71 76 84 88 92 98 .030 23 27 31 35 39 42 48 53 59 64 71 75 78 85 .035 19 22 25 30 33 35 41 45 51 55 61 66 69 75 .040 16 19 22 25 28 31 35 39 45 49 54 59 62 68 .0002 .010 105 116 125 138 145 151 162 170 179 185 193 198 201 207 .012 83 92 100 111 118 123 133 140 149 155 162 167 170 176 .015 61 69 76 85 91 96 105 111 119 125 132 137 140 145 .020 42 48 53 60 65 69 77 82 89 94 100 105 108 113 .025 31 36 40 46 50 54 60 64 72 76 82 87 89 95 .030 25 29 32 37 41 44 49 54 59 63 69 73 76 82 .035 21 24 27 31 34 37 42 45 51 55 60 64 67 72 .040 17 20 23 26 29 32 36 40 45 48 53 57 60 65 .0004 .010 110 120 129 140 148 154 164 170 179 184 191 196 199 204 .012 87 96 104 113 121 125 135 141 149 154 161 165 168 174 .015 65 73 79 87 93 98 106 112 119 124 130 135 138 143 .020 44 50 55 62 67 70 78 83 89 94 99 104 107 112 .025 32 37 42 47 51 55 60 65 71 76 81 85 88 94 .030 25 30 33 38 42 45 50 54 59 63 69 73 75 81 .035 21 24 27 31 35 37 42 45 51 55 60 64 66 72 .040 18 21 23 27 30 32 37 40 45 48 53 57 59 65 .0010 .010 113 124 131 142 150 155 165 171 179 184 190 194 197 202 .012 89 98 105 115 122 127 136 142 149 154 160 164 167 172 .015 66 74 80 88 94 99 108 112 119 124 130 134 136 141 .020 45 51 56 63 68 71 78 83 89 93 99 103 105 110 .025 34 39 43 48 52 56 62 66 71 75 81 85 87 92 .030 27 30 34 39 42 45 50 54 59 63 68 72 74 79 .035 22 25 28 32 35 3 43 46 51 54 59 63 65 70 .040 18 21 24 27 30 33 37 40 45 48 52 56 58 63 130 UNIFORM FLOW IN OPEN CHANNELS. If s be eliminated between these equations, there results an equation between x, y, and n. For a given constant value of n this equation may be represented by a curve with x and y as rectangular coordinates. A series of such curves may thus be drawn, each corresponding to a definite value of n. In like manner a second series of curves may be determined, each corresponding to a definite value of s. Thus, eliminating s, x=ny -1.811. Eliminating n, .00281 (4) (5) Taking rectangular axes OX, OY (Fig. 68), let equation (4) be plotted for some definite value of n. The resulting locus is 1 811 a straight line AB, such that OA = 1.811, OB = - . For a Tl second value of n another straight line is obtained passing through the same point A. Next consider the curve represented by (5) when s has any constant value. This is seen to be a rectangular hyperbola Y Q FIG. 68. M whose asymptotes are OY and a line parallel to OX at a dis- 00981 tance from it 0^=41.65 + ^=^. s Let lines such as AB be drawn for a series of values of n f and curves such as CD for a series of values of s. The point CHANNEL OF SMALLEST CROSS-SECTION. 131 in the plane OXY which corresponds to any definite values of n and s can then* be located by inspection. The values of x and y are thus known, and c may be found by the following graphical construction. For any given value of r, let OM (Fig. 68) be taken equal to vV. Let N be the point whose co-ordinates are x, y, deter- mined from any given values of n and s. Draw MN, and let P be its point of intersection with OY. From similar triangles, OP OM Qp _OMxQN_jV7 QM Vr+x Comparing with (1), From such a diagram it is possible not only to determine c when r, s, and n are given, but to determine any one of the four quantities when the other three are given. For accurate computations the diagram should be drawn carefully to a large scale. But even the diagram shown in Fig. 69 suffices for mak- ing computations with less uncertainty than that of the data usually involved in practical hydraulic problems.* 135. Smallest Cross-section for Given Rate of Discharge. If, with a given slope, a channel is to discharge water at a given rate, the cross-sectional area will depend upon the form of section adopted. Let it be required to make this area as small as possible. In the equation _ _ ET| q = Fv = Fc\/rs = c\/ the conditions of the problem make s constant, and it will be as- sumed that c is constant. Then q being expressed as a function of two variables F and C, it is required to satisfy the conditions q = constant, F = a minimum. * The computation of velocity for any values of r s, and n is greatly facil- itated by the use of diagrams published by Prof. I. P. Church. 132 UNIFORM FLOW IN OPEN CHANNELS. c . T-H T-H s r r "3 T-H o o T-H o J 1 ]3 Q | g bfi 8 8 a I a j o. 1 2 - o i h O VH i b s 1 i 1 .0 S s mason J .g o .3 | i 1 3 3 i -i g a ? JD a S e 5 a I i I I I jo S9n[t?^ w I i I i i I l I I l I I CHANNEL OF SMALLEST CROSS-SECTION. 133 From the form of the above value of q it follows that, q being constant, C is a minimum when F is. The values of F and C admit of infinite variation, depend- ing upon the form chosen for the cross-section. The discussion will be restricted to the case of a trapezoidal section, represented in Fig. 70. Trapezoidal section. Let x= FIG. 70. bottom width, y = depth of water, 6 = angle between side slope and horizontal; then F and C can be expressed in terms of x and y. For any fixed value of let it be required to determine what relation between x and y will give minimum values of F and p. We have F = y(x + y cotan 6) ; C = x + 2y cosec 6. The conditions to be satisfied are Differentiating, dF = ydx + (x + 2y cotan dC = dx + 2 cosec 6-dy = . Eliminating dy/dx between these equations, 2(1-0080) ff = _ _ _ L f>/ sin ^ which is the relation between x and y for minimum cross-sec- tion and wetted perimeter. By means of this relation the values of F, C, and r may be expressed in the forms cotan 6) = (2 cosec 6 cotan d)y 2 , C=x + 2y cosec 6 = 2(2 cosec 6 - cotan 6) y, 134 UNIFORM FLOW IN OPEN CHANNELS. Rectangular section. If = 90, the trapezoid reduces to a rectangle, and the solution for minimum cross-section and wetted perimeter becomes EXAMPLES. In the following examples let c be determined by Kutter's formula. 1. A rectangular flume of rough plank is to be laid on a slope of 1/1000, and is to discharge 10 cu. ft. per sec. Estimate the width and depth for minimum cross-section and wetted perimeter. Ans. y = 1.3',x=2.G'. 2. A ditch with trapezoidal section, side slopes 45, slope of bed 1/1000, is to discharge 40 cu. ft. per sec. Required the best values of the width and depth. Assume n = .025. Ans. y=3.6'. 3. Compute the rate of discharge of a rectangular flume of brick masonry, 3.5 ft. wide, sloping 1 ft. in 3000 ft., the depth of the water being 2 ft. Ans. About 15.8 cu. ft. per sec. 4. Prove that, of all trapezoidal sections giving a certain rate of dis- charge, the section of minimum area has side slopes of 60. CHAPTER XII. OPEN CHANNELS: NON-UNIFORM FLOW. 136. General Equation of Energy for Stream of Variable Cross-section. The reasoning by which the general equation ~ w 2g - w 2g was established (Arts. 59, 60) is valid in the case of a stream with a free upper surface. For such a stream the hydraulic gradient coincides with the surface of the stream, and z -f p/w or y denotes the height of the water surface above a horizontal datum plane. (See Fig. 67.) In Fig. 71 let A and B be any two cross-sections of a stream, the flow being from A toward A B] let 2/1, vi be the values of y and v at A } and y 2 , v 2 the values at B] and let H f denote the loss of head between A and B (i.e., the energy lost between A and B for every pound of water passing any cross-section of the FIG. 71. stream) . Then / 11, 2\ / ,,^2\ -H' (1) 137. Value of Lost Head. In the case of uniform flow the loss of head in any given length I of the stream may be ex- pressed by means of the formula v = cVrs. 135 136 OPEN CHANNELS: NON-UNIFORM FLOW. For in this case, Vi and v 2 being equal, and so that c 2 r (2) Let this same expression for H f be assumed for the case of variable flow, it being understood that v and r have values intermediate between those applying to the sections A and B. Equation (1) then becomes 138. Differential Equation for Surface Curve. Consider a longitudinal section of the stream by a vertical plane, and let x, y be the coordinates of the surface curve referred to axes Xj OY, the former being horizontal and the latter vertical Y FIG. 72. (Fig. 72). If the above equation be applied to the portion of the stream between two sections A and B infinitely near together, we must put l=dx, y 2 -yi=dy, CHANNEL OF UNIFORM SHAPE AND SLOPE, 137 so that the equation becomes v v 2 -0 (4} \Jb V I *-> \AS**S V/ \^7 g c z r 139. Channel of Uniform Shape and Slope. Let the channel have a uniform slope and uniform cross-section, the cross-sec- tion of the stream, however, varying with the depth of the water. Let i = slope of bed ; u = depth of stream at point (x, y) ; b = width of stream at water surface. Let y be replaced by u, by means of the evident relation dy = du-i dx. Also, we have v = -^ (q being constant) ; q dF _ qbdu ~ = ~ Equation (4) may therefore be written Now let q be replaced by its value in terms of the dimen- sions of a uniform stream in the same channel. Thus, if the channel were unobstructed for a great distance, the surface would assume some position as MN (Fig. 72) parallel to the bed. Let u', F', r' be the values which u, F, r would have in such a case; then and equation (5) takes the form (^ r ' F/2 V,7 /i c*i, (I- )t dx ={i-.- du ..... (6) 138 OPEN CHANNELS: NON-UNIFORM FLOW. When the form of the channel is given, F, b and r can be ex- pressed in terms of u, and the relation between u and x is then to be found by integrating equation (6) . In certain ideal cases the integration is fairly simple. The one most commonly treated is that in which the cross-section is assumed to be of uniform depth and of great width. 140. Cross-section of Great Width and Uniform Depth. For a rectangular cross-section b is constant and u' being the depth corresponding to cross-section F'. If b is very great in comparison with u, we have approxi- mately r = u, r' = u', and equation (6) becomes Let u/u'=z; then du = u'dz, and - , i\ dz or -,dx = dz+(l- ) - (8) For brevity let 2 3 -l then the integration of (8) gives 9 jT / (* "9 \ = z- (1 - } (f)(z) + constant (9) BACKWATER. 139 Let the integration be taken between limits x*=x\ and x = X2, where x% o;i = / = distance between any two definite cross-sections of the stream, z lt z 2 being the corresponding values of z. Then . . . . (10) In order to apply this equation it is necessary to compute values of (z) for a series of values of z. The general value of the function is found by integration * to be r dz 1 > -J ^r6 From this are computed the values entered in Table VII. For very exact computations a fuller table is required. Intermediate values may, however, be found by interpolation with less error than that due to the defects in the above theory and the uncertainty always existing in the data of hydraulic problems. TABLE Vll. = - / -; 7. J z i VALUES OF z .) z *.) z to) z *(.) .00 oo 1.10 .680 1.30 .373 1.65 .203 .01 1.419 1.12 .626 1.32 .357 1.70 .189 .02 1.191 1.14 .581 1.34 .342 1.80 .166 .03 1.060 1.16 .542 1.36 .328 1.90 .147 .04 .967 1.18 .509 1.38 .316 2.00 .1318 .05 .896 1.20 .480 1.40 .304 2.10 .1188 .06 .838 1.22 .454 1.45 .278 2.20 .1074 .07 .790 1.24 .431 1.50 .255 2.30 .0978 1 08 .749 1.26 .410 1.55 .235 2.40 .0894 1 09 713 1 28 .390 1.60 .218 2 50 .0822 141. Backwater. If the surface of a stream at a certain point be raised by a dam or other obstruction, the effect will extend up-stream for some distance, decreasing as the distance * See Williamson's Integral Calculus, p. 59. To the value there given is added such a constant that (zO = .680, (z 2 ) = .255, hence Z = 100000[(1.5-l.l) + 960(.680-. 255)] = 80800 ft. If, instead of z\, I is given and z\ is required, the solution is not so direct. Thus, suppose in the above case it is required to determine the increase in depth at a section 25000 ft. above the point where the depth is 30 ft.; that is, Z= 25000 and z\ is required. Substituting in equation (10), i-.960^(2i) = 1.005. By taking trial values from the table it is found that the equa- tion is nearly satisfied by z\ = 1.34. The depth at the specified section is therefore 1.34x20 ft. =26.8 ft. In applying the above theory to actual streams for which the assumption of uniform depth throughout the cross-section is not even roughly true, u may be taken as the average depth for the cross-section, computed from the formula u = This method may be applied to Ex. 4 of the following list. BACKWATER. 141 EXAMPLES. 1. A wide stream 12 ft. deep is obstructed so that the depth at a certain section becomes 16 ft. The slope of the bed is 2 ft. per mile. Compute the depth (a) 2 miles up-stream and (b) 2 miles down-stream. [Takec=65.J Ans. (a) 14.0ft. (6) 18.7ft. 2. In Ex. 1, determine the positions of sections at which the depth has values 13 ft., 15 ft., and 16 ft. Ans. The depth would be 13 ft. at about 3.73 miles up-stream. 3. A wide stream 20 ft. deep is obstructed so that the depth at a certain section becomes 25 ft. If i = .0002 and c = 70, determine (a) the depth 10000 ft. up-stream from the given section, and (6) where the depth will be 21 ft. Ans. (a) 24.1ft. (6) About 66000ft. up-stream. 4. A stream whose cross-sectional area is 5400 sq. ft. and width 350 ft. when unobstructed has its surface raised 10 ft. at a certain section by a dam. The slope is 1.6 ft. per mile, and the value of c is 75. (a) Compute the rise of the surface at distances of 2 miles and 5 miles above the given section, (b) Where will the increase of depth be 5 ft.? Ans. (a) 7.6 ft. and 4.6 ft. (b) At 4.53 miles up-stream. CHAPTER XIII. THE MEASUREMENT OF RATE OF DISCHARGE. 142. General Methods. One of the most important of the problems of practical Hydraulics is the determination of the rate of discharge of streams. The conditions of the problem vary greatly in different cases, and the methods employed must vary correspondingly. The quantities to be measured vary from the discharge of a small pipe to that of a large river. The methods employed may be classed as (1) direct and (2) indirect. (1) An actual measurement may be made of either (a) the total quantity discharged in a given time, or (6) the velocity of 'flow. (2) Measurement may be made of certain quantities upon which the rate of discharge is known to depend, its value being computed from the data thus determined. 143. Direct Measurement of Total Discharge. The total quantity discharged in a given time may be determined either by weighing the water or by measuring its volume in a prop- erly calibrated vessel or reservoir. Accurate measurement by weighing will in general be pos- sible only when the rate of discharge is small, since for accurate results the experiment must extend over a considerable time. A discharge of 1 cu. ft. per second would give in 1 minute a total discharge of 3750 Ibs. If a reservoir is available whose volume for given depths is accurately known, the total discharge in a given time may be measured by collecting the water in this reservoir. Much larger quantities can thus be measured than it is practicable to weigh. 142 METHODS OF MEASURING VELOCITY. 143 This method is sometimes employed in estimating the rate of discharge of the conduit of a water supply system, the discharge for a period of several hours or even a day or more being col- lected in a large storage reservoir whose horizontal area at dif- ferent levels is accurately known. Such measurements are liable to uncertainty because of evaporation and leakage, and also because a small error in measuring the depth of the water results in a relatively large error in the total quantity. 144. Discharge Computed from Direct Measurement of Velocity. If the velocity of flow be measured at many differ- ent points in a given cross-section of a stream, the rate of dis- charge for the whole section may be closely estimated. Let the total cross-section be F } and let it be divided into parts whose areas are FI, F 2 , . . . , the values of the mean velocity in these parts being i? 1; v 2 , . . . , and the mean velocity for the entire section v. Then the total rate of discharge is This method is especially applicable to open streams of consid- erable size. 145. Methods of Measuring Velocity. Of the various devices that have been employed for measuring velocity of flow there may be mentioned floats, the. tachometer or current-meter, and the Pitot tube. Floats are especially applicable to streams of considerable size. Under favorable conditions they may be used for measur- ing the velocity not only at the surface but at any desired depth, being suspended by wires from surface floats of such shape and size as to be little influenced by the surface velocity. Floats are also sometimes made in the form of tubes so weighted as to float in an upright position and extend from the surface nearly to the bottom, the motion of a float thus indicating a mean of the velocities at different depths. The field operations required in this and other methods of gauging the flow of large streams are quite elaborate and will not be here described.* * Con suit Physics and Hydraulics of the Mississippi River, by Humphreys and Abbot. 144 THE MEASUREMENT OF RATE OF DISCHARGE. The current-meter consists of a vaned wheel which is rotat- ed by the current at a rate depending upon the velocity. The total number of revolutions in a given time is counted, usually by means of an automatic register. The relation between the velocity of flow and the number of revolutions per unit time must be determined for any given instrument by experiment. This is called "rating" the meter. Pitot's tube, in its simplest form, is an open tube one end of which is bent at right angles to the length of the main por- tion. The bent end is submerged and placed with the opening facing the cur- - rent, while the other end projects above \ the free surface (Fig. 73). The water = in the tube rises above the surface of -y_-f:j |^~: : the stream to a height depending upon ^ the velocity of flow at the lower end. IG. 73. No exact relation between the velocity v and the height h of the column above the water surface can be determined from theory alone. It would seem that the increase of pressure at the lower end of the tube will be proportional approximately to the square of the velocity, so that k being a coefficient whose value for any given instrument must be determined experimentally. (See Art. 171.) The Pitot tube has also been employed for measuring veloci- ties in pipes carrying water under pressure. For such use the instrument consists of two tubes whose lower ends are open, but so placed that one opening faces the current, while the other is at right angles to it. If the upper ends are open, water will rise in the two tubes to different heights, the difference being a measure of the velocity of the current. If the pressure is too great for the use of open piezometers, a difference-gauge, such as is shown in Fig. 36, may be connected with the two tubes of the instrument. Instead of mercury a fluid of less density may be used if increased sensitiveness is required. MEASUREMENT BY ORIFICES. . 145 146. Indirect Methods of Measuring Rate of Discharge. Of the methods which above were called indirect, the most important are three : by orifices, by weirs, and by the Venturi meter. These will be considered in order. 147. Measurement by Orifices. If a stream be conducted into a tank or reservoir in the side of which is an orifice, the surface of the water will assume such a position that the rate of discharge through the orifice becomes equal to the rate at which water flows into the reservoir. The rate of discharge of an orifice of cross-section F, under a head h on its center, is q = cFV2gh, ....... (1) c being the coefficient of discharge . (Art. 44). This formula may be used unless the dimensions of the orifice are relatively large in comparison with the head. For a large orifice whose plane is not horizontal the form of the formula depends upon the shape and cross-section of the orifice, as shown in Art. 46. It is rarely needful, however, to employ the theoretically more accurate formula, since the difference between the results given by it and by the approximate formula is usually much less than the uncertainty in the value of the coefficient of discharge. Thus for a circular orifice in a vertical plane the accurate formula may be developed in a series of which the first two terms are as follows: d being the diameter. If d/h = l, the second term is less than one per cent of the first, and the series converges rapidly. For a rectangular orifice of width b the exact formula (Art. 49) is W), ...... (3) in which hi and h 2 are the depths of the upper and lower edges below the water surface. Putting h 2 = h+d/2 and hi=h d/2, this can be expressed as a series of which two terms are as follows : 146 THE MEASUREMENT OF RATE OF DISCHARGE. . . - (4) If d/h = l, the second term is a little greater than one per cent of the first, and the series converges quite rapidly. The coefficients of discharge for orifices made in a standard way are fairly well known by experiment. The accompanying tables * give values of c for circular standard orifices, square orifices, and rectangular orifices one foot wide. In every case the orifice is supposed to be formed with a sharp inner edge (as at A, Fig. 17), so that full contraction of the jet occurs. It is also needful that the orifice shall be at a considerable distance from the sides and bottom of the reservoir; otherwise the approaching particles of water will be so guided that full con- traction will be prevented and the value of c will be uncertain. The horizontal lines in the tables indicate the limiting values of h below which the approximate formula (1) should be re- placed by (2) or (3). TABLE VIII. VALUES OF COEFFICIENT OF DISCHARGE FOR CIRCULAR VERTICAL ORIFICES h (feet). d (feet). 0.02 0.04 0.07 0.1 0.2 0.6 1 0.4 0.637 0.624 0.618 0.6 0.655 .630 .618 .613 0.601 0.593 0.8 .648 .626 .615 .610 .601 .594 0.590 1. .644 .622 .612 .608 .600 \595 .591 1.5 .637 .618 .608 .605 .600 .596 .593 2. .632 .614 .607 .604 .599 .597 .595 2. .629 .612 .605 .603 .599 .598 .596 3.5 .627 .611 .604 .603 .599 .598 .597 4. .623 .607 .603 .602 .599 .597 .596 6. .618 .607 .602 .600 .598 .597 .596 8. .614 .605 .601 .600 .598 .596 .596 10. .611 .603 .599 .598 .597 .596 .595 20. .601 .599 .597 .596 .596 .596 .594 50. .596 .595 .594 .594 .594 .594 .593 100. .593 .592 .592 .592 .592 .592 .592 * These tables are taken from Hamilton Smith's Hydraulics. MEASUREMENT BY ORIFICES. 147 TABLE IX. VALUES OF COEFFICIENT OF DISCHARGE FOR SQUARE VERTICAL ORIFICES h (feet). d(feet). 0.02 0.04 07 1 02 0.6 1.0 0.4 0.643 0.628 0.621 0.6 0.660 .636 .623 .617 0.605 0.598 0.8 ' .652 .631 .620 .615 .605 .600 0.597 1. .648 .628 .618 .613 .605 .601 .599 1.5 .641 .622 .614 .610 .605 .602 .601 2. .637 .619 .613 .608 .605 .604 .602 2.5 .634 .617 .610 .607 .605 .604 .602 3. .632 .616 .609 .607 .605 .604 .603 5. .628 .614 .608 .606 .605 .603 .60^ 6. .623 .612 .607 .605 .604 .603 .602 8. .619 .610 .606 .605 .604 .603 .602 10. .616 .608 .605 .604 .603 .602 .601 20. .606 .604 .602 .602 .602 .601 .600 50. .602 .601 .601 .600 .599 .600 .599 100. .599 .598 .598 .598 .598 .598 .598 TABLE X. VALUES OF COEFFICIENT OF DISCHARGE FOR RECTANGULAR ORIFICES 1 FT. WIDE. ft (feet). d (feet). 0.125 0.25 0.50 0.75 1.0 1.5 2.0 0.4 0.6 0.8 1. 1.5 2. 2.5 3. 4. 6. 8. 10. 20. 0.634 0.633 .633 .633 0.622 .619. .618 .618 .618 0.614 .612 .612 .611 .611 .611 0.608 .606 .605 .605 .605 .605 0.626 .626 .624 .616 .614 .612 0.628 .630 .627 .619 .616 .610 .633 .633 .632 .630 .629 .628 .627 .624 .615 .609 .606 .632 .631 .620 .628 .627 .624 .615 .607 .603 .617 .616 .615 .614 .609 .603 .601 .610 .609 .604 .602 .601 .601 .605 .602 .601 .601 .601 .606 .6Q2 .601 .601 .604 .602 .602 148 THE MEASUREMENT OF RATE OF DISCHARGE. 148. Miner's Inch. Where water is sold for hydraulic min- ing or for irrigation the unit of measurement commonly em- ployed is the miner's inch. This is usually understood to mean the discharge through an orifice one inch square under some specified head; multiples of the " inch " being obtained by increasing the horizontal dimension of the orifice while leaving the head unchanged. Various definitions have, however, been accepted at different times and in different localities. In California a common definition of the miner's inch has been the discharge from an orifice one inch square when the head on the upper edge is four inches, and hydraulic engineers have quite generally accepted 1.2 cubic feet per minute as its equivalent, corresponding to a value of about 0.59 for c. The legal equivalent* is, however, 1.5 cubic feet per minute. 149. Measurement by Weirs. A weir is a notch in the side of a vessel or reservoir. Such notches are used for the meas- urement of rate of discharge, and it is essential that their con- struction shall conform to accepted standards in order that reliable results may be obtained. The most common form of weir is rectangular. The standard rectangular weir is constructed with the lower edge or " crest" sharp (like the edge of a standard orifice), so as to permit full "crest contraction" of the stream. The vertical edges should either be sharp or else should lie in the vertical sides of the channel of approach. In the former case there is full "end contraction" of the stream; in the latter there will be no lateral contraction at the ends, which is commonly expressed by saying that the end contractions are "suppressed." It is essential also that the depth of the channel of approach shall be sufficient so that full crest contraction is not interfered with. Also, unless the end contractions are completely sup- pressed, the channel of approach should be considerably wider than the length of the weir, in order that full contraction may occur at the ends. * * Since 1901. THE FRANCIS WEIR FORMULA. 149 The formula for the rate of discharge over a rectangular weir is given in Art. 50. It is c being a coefficient whose value must be determined by experiment. The "head on the crest" (H) must be measured to the level of the surface of still FIG. 74. water back of the weir (Fig. 74). 150. Thfc Francis Formula. Elaborate experiments were made by J. B. Francis * for the purpose of determining the value of the coefficient c in the above formula, and also the effect of end contractions. The weirs used in these investiga- tions ranged in length from 4 ft. to 10 ft., and H ranged from 0.6 ft. to 1.6 ft. As a result of these experiments their author adopted an average value 0.622 for the coefficient c. His formula, for the case of suppressed end contractions, is (6) b and H being in feet. As regards end contractions, Francis concluded that each contraction decreases the effective length of the weir by an amount proportional to H, and gave the formula (7) in which n is the number of end contractions, usually two. Velocity of approach. If the channel of approach is so small that the velocity of the approaching water is appreciable, the formula of Art. 68 is taken as the basis, giving the following: q = 3.33(b-Q.lnH)[(H + h') 3 *-h'i], ... (8) i/ 2 in which h' = -^-, v' being the velocity of approach. In prac- Z 9 tice !/ is taken as the average velocity in the channel, computed from the cross-section and the rate of discharge. * Lowell Hydraulic Experiments. 150 THE MEASUREMENT OF RATE OF DISCHARGE. It is probable that for small values of H the constant co- efficient adopted by Francis gives results a little too great. In practice, however, cases are rare in which measurements can be made with a degree of reliability warranting the use of formulas or coefficients claiming greater accuracy than those of Francis. Fanning * recommends, for use in the formula of Francis, values of the coefficient varying with H as in the following table : TABLE XI. H (feet). c cv/ 2 // (feet). c ciVjfc .083 .6100 3.263 .833 .6240 3.338 .125 .6120 3.274 1.000 .6241 3.339 .167 .6140 3.285 1.167 .6242 3.339 .250 .6170 3.301 1.333 .6243 3.340 .333 .6195 3.314 1.500 .6242 3.339 .500 .6223 3.329 1.667 .6241 3.339 .667 .6235 3.336 2.000 .6240 3.338 151. Method of Hamilton Smith. The formula adopted by Smith differs from that of Francis in the method of allowing for end contractions and also in the correction for velocity of approach. Smith also adopts values of c varying with the head, even when the end contractions are suppressed. The formula (9) is used for both contracted weirs, and weirs with end contrac- tions suppressed, but the values of c are different for the two cases, as shown in Tables XII and XIII. Velocity of approach. If the velocity of approach is not negligible, let h' denote the head equivalent to the mean veloc- ity in the channel above the weir, and substitute for H in equation (9) for weir with end contractions suppressed, Ah' " " " full end contractions. * A Practical Treatise on Hydraulic and Water-Supply Engineering. WASTE WEIRS. 151 TABLE XII. VALUES OF THE COEFFICIENT OF DISCHARGE (c) FOR WEIRS WITHOUT END CONTRACTIONS. H (feet). Length (6) in Feet. 0.66 2 3 4 5 7 10 19 0.1 0.675 0.659 0.658 0.658 0.675 0.15 .662 0.652 0.649 0.647 .645 .645 .644 .643 0.2 .656 .645 .642 .641 .638 .637 .637 .635 0.2 5 .653 .641 .638 .636 .634 .633 .632 .630 0.3 .651 .639 .636 .633 .631 .629 .628 .626 0.4 .650 .636 .633 .630 .628 .625 .623 .621 0.5 .650 .637 .633 .630 .627 .624 .621 .619 0.6 .651 .638 .634 .630 .627 .623 .620 .618 0.7 .658 .640 .635 .631 .628 .624 .620 .618 0.8 .656 .643 .637 " .633 .629 .625 .621 .618 0.9 .645 .639 .635 .631 .627 .622 .619 1.0 .648 .641 .637 .633 .628 .624 .619 1.2 .646 .641 .636 .632 .626 .620 1.4 .644 .640 .634 .629 .622 1.6 .647 .642 .637 .631 .623 TABLE XIII. VALUES OF COEFFICIENT OF DISCHARGE (c) FOR WEIRS WITH END CONTRACTIONS. H (feet). Length (6) in Feet. 0.66 l 2 3 5 10 19 0.1 0.632 0.639 0.643 0.652 0.653 0.655 0.656 0.15 .619 .625 .634 .638 .640 .641 .642 0.2 .611 .618 .626 .630 .631 .633 .634 0.25 .605 .612 .621 .624 .626 .628 .629 0.3 .601 .608 .616 .619 .621 .624 .625 0.4 .595 .601 .609 .613 .615 .618 .620 0.5 .590 .596 .605 .608 .611 .615 .617 0.6 .587 .593 .601 .605 .608 .613 .615 0.7 .590 .598 .603 .606 .612 .614 0.8 .595 .600 .604 .611 .613 0.9 .592 .598 .603 .609 .612 1.0 .590 .595 .601 .608 .611 1.2 .585 .591 .597 .605 .610 1.4 .580 .587 .594 .602 .609 1.6 .582 .591 .600 .607 152. Waste Weirs. A waste weir is a notch left at the crest of a dam for the purpose of discharging flood water with- out injury to the dam. To choose the proper length and depth of the waste weir, 152 THE MEASUREMENT OF RATE OF DISCHARGE. knowing the greatest rate of accumulation of flood water, the general weir formula may be used. Since a waste weir will not in general have a sharp crest, and since the other conditions are not the same as in the case of weirs constructed for the purpose of accurate measurement of rate of discharge, no pre- cise estimate can be made of the rate of discharge of a waste- weir under a given head. Francis gave the empirical formula (10) in which b and H are in feet, and q is in cubic feet per second. The ordinary weir formula, with the coefficient determined by Francis, may be used with results differing little from those given by equation (10). That is, (11) 153. Triangular Weir. Although the rectangular form of weir is generally the most convenient, a triangular notch may be used when the quantity to be measured is not too great. Theoretically the triangular form has the advantage that the shape of the water section does not change with the head, so that the coefficient of discharge would be expected to be more nearly constant with varying head than in the case of the rect- angular form. This is confirmed by experiment. The formula for the rate of discharge over a triangular notch, as deduced in Art. 51, is . , ; . . , (12) If the angle at the vertex is 90, 6 = 2H, and the formula becomes (13) If the edges are made sharp, as in the case of the standard rect- angular weir, the value of c for heads under 0.8 ft. may be SUBMERGED WEIR. 153 taken as .592. With the foot as unit length the formula may be written (14) EXAMPLES. 1. Compute the rate of discharge of an orifice 6" square when the head on the center is 18 ft. Ans. 5.12 cu. ft. per sec. 2. Estimate the diameter of a circular orifice to discharge 1 cu. ft. per sec., the head on the center being 5 ft. Ans. 0.345 ft. 3. A weir without end contractions is 4 ft. long, the bottom of the channel of approach being 3 ft. below the crest. Compute the rate of discharge when H =0.75 ft. (a) by formula of Francis and (6) by formula of Smith. What percentage of error is caused by neglecting velocity of approach? Ans. (a) 8.73 cu. ft. per sec. (6) 8.90 cu. ft. per sec. 4. A contracted weir 9 ft. long is to discharge 60 cu. ft. per sec. Esti- mate the head on the crest (a) by Francis' formula and (6) by Smith's formula. Ans. (a) H =1.630 ft. (6) H = 1.572 ft. 5. Water enters a reservoir at the rate of 20 cu. ft. per sec. In the side is a rectangular notch 3.5 ft. long, with ends and crest so formed as to allow complete contraction. Estimate the head on the crest when a steady condition is attained. Ans. H = 1 .52 ft. 6. A rectangular orifice 1 ft. wide and 4 in. deep is formed in the vertical side of a reservoir into which water flows at the rate of 5 cu. ft. per sec. How high above the center of the orifice will the water surface rise? Ans. About 9.6 ft. 7. Compute the rate of discharge of a triangular notch having a vertex angle of 90 when the head is 0.46 ft. Ans. 0.365 cu. ft. per sec. 154. Submerged Weir. A weir is said to be submerged if the surface of the stream below is higher than the crest of the weir. To deduce a formula for the rate of discharge in such a case, let H, h denote the elevations of the up-stream and down- stream surfaces respectively above the crest, and b the hori- zontal length of the weir (Fig. 75). The total discharge orifice ABFE consists of two parts, ABDC, CDFE, of which the former may be treated as an ordinary weir and the latter as a sub- merged orifice. Omitting coefficients of discharge, the formula for the ordinary weir ABDC is 154 THE MEASUREMENT OF RATE OF DISCHARGE, and that for the submerged orifice CDFE is q = bhV2g(H-h). Combining these, and introducing the coefficient of discharge c, the formula for the rate of discharge of the submerged weir becomes (15) , _A| B i \_ ,_ ._CL_ _D x_-x*^ ^ " * B 6 _Y.v L I FIG. 75. Equivalent .ordinary weir. If H r is the head on an unsub- merged weir whose rate of discharge is q, we have (16) Assuming the coefficients of discharge for the two cases to be equal, the values of q given by (15) and (16) will be equal if or Writing n for H' /H, and assuming c to have the mean value found by Francis for ordinary weirs, trie formula for the rate of discharge over a submerged weir without end contractions may be written (18) in which n is to be computed from (17). SUBMERGED DAM. 155 The relation between h/H and n as given by equation (17) is shown by the curve (B) in Fig. 76. Curve (A) shows the .2 .4 __.G .8 1.0 ,(Bj \ \ .4 .2 .8 1.0 FIG. 76. relation deduced by Herschel * from experiments made by Francis and by Ftely and Stearns. These results are also given in Table XIV. 155. Submerged Dam. If the channel of a stream is ob- structed by a dam, the crest of which is lower than the original surface of the stream, the surface up-stream from the obstruc- tion will be raised an amount depending upon the position of the crest, the length of the overfall, and the rate of discharge. An approximate estimate of the effect may be made by the formula above given for submerged weirs. * Transactions American Society of Civil Engineers, Vol. XIV, p. 189. 156 THE MEASUREMENT OF RATE OF DISCHARGE. TABLE XIV. n n n h h h H Theoreti- Experi- H Theoreti- Experi- H Theoreti- Experi- cal. mental. cal. mental. cal. mental. 1.000 1.000 .45 .959 .912 .84 .724 .656 .05 1.000 1.007 .50 .945 .892 .86 .697 .631 .10 .999 1.005 .55 .928 .871 .88 .666 .604 .15 .997 .996 .60 .908 .846 .90 .631 .574 .20 .995 .985 .65 .883 .818 .92 .589 .539 .25 .991 .972 .70 .853 .787 .94 .538 .498 .30 .986 .959 .75 .816 .750 .96 .473 .441 .35 .979 .944 .80 .770 .703 .98 .378 .352 .40 .970 .929 .82 .748 .681 1.00 .000 .000 As an example, suppose a stream discharging 600 cubic feet per second is obstructed by a dam with an overfall 40 feet long, the crest being 2 feet below the original water surface. The increase in depth above the dam may be estimated as follows: With the notation of Art. 154, the value of H' ' , the head on the equivalent weir, is found from the equation 600 = 3.33 X40#' s , or Also, since h = 2 ft., nH 2.728 1.364, or 1.364^. By trial substitutions in the table, using the experimental values of n, it is found that h/H = .6l5 very nearly, from which H = 3.25 ft. If the curves in Fig. 76 be carefully drawn on cross-section paper, the above solution may be shortened. The equation n- 1.36477 represents a straight line whose intersection with curve (A) determines n and h/H at once. VENTURI METER. 157 As a second example, let it be required to determine where the crest of the dam should be placed in order to raise the sur- face 1 foot, the data being otherwise as above given. From the condition stated, and as before H' = nH = 2.728 ft., 2.728 or H = n From these equations, H 2.728' By trial substitution in the table it is found that n = .775, from which # = 3.52 ft. and h = 2.52 ft. This case may also be solved graphically, by finding the in- h Ti tersection of the straight line 1 77 = 70Q with curve (A) in n Z./Zo Fig. 76. EXAMPLES. 1. A stream discharging 550 ou. ft. per sec. is obstructed by a dam of which the crest is 2.5 ft. below the original water surface, the length of the overfall being 50 ft. What will be the effect on the water surface above the dam? Ans. H-h=O.Q3 ft. 2. Where must the crest of a dam be placed in order that the water may be raised 2 ft., the length of the overfall being 30 ft. and the rate of discharge 500 cu. ft. per sec.? Ans. h = 1.10 ft. 156. Venturi Meter. The essential features of the Venturi meter have already been shown in Figs. 35 and 36, illustrating examples given at the end of Chapter V. The instrument con- sists merely of a converging tube with piezometers connected substantially as shown in Fig. 77, or a difference-gauge, as in Fig. 78. 158 THE MEASUREMENT OF RATE OF DISCHARGE. Referring first to the arrangement in Fig. 77, let the usual notation be used for cross-sectional areas, velocities, and heights of piezometer columns, and let the equation of energy be written, A being the up-stream and B the down-stream section. Then, A' T~ b B' ^ A B FIG. 77. A" i B' li L A' M C B FIG. 78. whether the axis of the pipe be horizontal or not, we have, as in Art. 125, H' being the head lost between A and B, and ki, k 2 being posi- tive quantities whose values depend upon the distribution of velocities throughout the cross-sections. Neglecting ki, k 2 and H', and introducing the relation FiVi=F 2 v 2 , there results the equation in which c is a coefficient whose value would be 1 if the theory were perfect, and whose value as found by experiment usually ranges from 1 to 0.97. The value of c decreases with increasing velocity. The factor y\y^ in the above formula is the difference in level between the tops of the two piezometer columns A f and B' (Fig. 77). If the pressure is too great for the use of open piezometers, a difference-gauge may be connected as in Fig. 78. If the specific gravity of the mercury or other fluid in the U tube is s, and if h is the difference in level of the two columns, VENTURI METER. 159 as indicated in Fig. 78, the value of yi -y 2 in the formula for v 2 is (s-l)h. For mercury the value of s is very nearly 13.6. EXAMPLES. 1. A Venturi meter whose two diameters were 48 inches and 16 nches was furnished with piezometers as in Fig. 77. The difference in level between the tops of the two columns was found to be 1.27 ft. in a certain case. Compute the rate of discharge, assuming c =0.985. Ans. 12.5 cu. ft. per sec. 2. A Venturi meter of the dimensions given in Ex. 1 was furnished with a mercury difference-gauge, as in Fig. 78. In a certain experiment the difference in elevation of the two mercury columns in the U tube was 0.625 ft. Estimate the rate of discharge, taking c =0.98. Ans. 31.0 cu. ft. per sec. CHAPTER XIV. DYNAMIC ACTION OF STREAMS. 157. Meaning of Dynamic Action. The object of the pres- ent chapter is to investigate the forces which are exerted by a stream upon bodies which constrain its motion, and the reac- tions exerted by these bodies upon the stream. Thus, if a free jet of water is deflected, forces are called into action between the particles of the jet and the body causing the deflection. The same is true in the case of a confined stream whose velocity varies in magnitude from section to section because the cross-section of the pipe changes, or in direction because the pipe bends. These are illustrations of dynamic action in cases of steady flow. Another important case is that in which the rate of dis- charge of a confined stream, and therefore the velocity at every section, are suddenly changed, as by the closing of a valve. In all cases, dynamic effects are to be determined by apply- ing the fundamental laws of motion or the equations based upon them. 158. Principle of Momentum. If a constant force P acts upon a body of mass m for a time M, thereby giving its veloc- ity * the increment 4V, we have * Throughout the remainder of this book it is necessary to distinguish between absolute and relative velocities of a stream. We therefore use V for the former and v for the latter. See Appendix B for a discussion of absolute and relative motion. 160 FORCE PRODUCING VELOCITY OF JET FROM ORIFICE. 161 If the force is not constant, the equation gives its average value during the time At. Thus if a body of 12 Ibs. mass (m = 12/0 = .373 if force is to be expressed in pounds) has its velocity changed in .1 sec. from 18 ft. per second in a given direction to 23 ft. per second in the same direction, the average value of the force during that time is _ If At = l sec., the equation is (2) That is, the change of momentum produced by a force in one second is equal to the value of the force if constant, or to its average value for that second if variable. 159. Force Producing Velocity of Jet from Orifice. Let W Ibs. of water per second flow from an orifice in the side of a reservoir, V being the velocity of the jet. The velocity of each particle is then changed by the amount V during its passage from the reservoir to the smallest cross-section of the jet. The force causing this change is exerted directly by contiguous particles, but indirectly by the walls of the reservoir. The values of the forces exerted upon individual particles cannot be determined, but the sum of the average forces for all particles can be computed from the principle of momentum above stated. In one second the mass of water flowing from the orifice is TF/0, and the total momentum produced in one second by the action of the reservoir upon the water is WV/g. This is there- fore the average value of the total force continually exerted by the reservoir upon the stream. This force is practically con- stant since the flow is steady, hence its average value is its actual value. If F is the cross-section of the jet, W = wFV f and the force is 162 DYNAMIC ACTION OF STREAMS. 160. Reaction of Jet upon Reservoir. By the law of action and reaction, the particles of the jet exert upon the reservoir forces whose resultant is equal and opposite to the force just computed. EXAMPLES. 1. Determine the reaction upon the reservoir due to a jet from a standard circular orifice I" in diameter under a head of 3'. Ans. 1.21 Ibs. 2. Show that if there were no loss of energy, the reactive force exerted by a jet from an orifice would be double the total static pressure upon an area equal to the cross-section of the jet, due to the head on the orifice. 161. Jet Striking a Fixed Surface Normally. If a jet is intercepted by a plane surface perpendicular to the axis of the jet, the resultant action and reaction between the stream and the body will be directed normally to the surface. If V is the velocity of the jet, the increment of velocity for each particle, resolved in the direction of the normal, is 7, so that the resultant force exerted by the body upon the jet is pJ^p^.^T. . . (4) in the direction opposite to the motion of the jet. The reaction of the jet upon the fixed body is equal and opposite to this. FIG. 79. Jet Striking a Moving Surface Normally. Let a jet whose velocity is V strike normally against the surface of a body which is itself moving with velocity u in the same direction as the jet. The increment of velocity for each particle, resolved normally to the surface, has the magnitude V u. If W denotes the number of pounds of water striking the surface per second, the force exerted upon the moving body has the value W JET WATER WHEEL WITH FLAT VANES. 163 Or, since W' = wF(V-u), P'^-(V-u? ....... (5) Work done upon moving body. The work done in one sec- ond by the force P' is P'u = ^-u(V-u) 2 ....... (6) t7 163. Jet Water Wheel with Flat Vanes. Fig. 81 repre- sents a wheel provided with flat vanes against which a jet of water is directed. The results of Art. 162 would be strictly applicable if the vanes always received the jet normally. Al- though this condition cannot be realized, the above results may be used as an ap- proximation, and equations (5) and (6) may be employed to compute the force exerted upon a vane, and the work done by this force per seqond. The total action of the jet upon the wheel is not, however, the same as its action upon one vane, since more than one vane will be receiving the jet at the same time. Replacing W by W , the weight of water discharged by the jet per second, we get for the force exerted upon the wheel (7) and for the work done upon the wheel per second (8) Maximum value of work done upon wheel. If u varies, the value of the work varies, having its maximum when u = V/2, which gives wFV 3 WV 2 Maximum = -- = = ...... (9) 164 DYNAMIC ACTION OF STREAMS. Since the kinetic energy of W Ibs. of water in the jet is WV 2 /2g, it is seen that not more than half of this can be utilized by a water wheel with flat vanes. 164. Force Causing Deflection of a Jet. If the velocity of a particle changes in direction, the average value of the force acting upon it during a time Jt is still given by the formula mAV At ' (10) B but AV is now the vector increment * of velocity. In Fig. 82 is represented a jet of water which is deflected by a curved vane MN, the velocity of every particle being changed from Vi (represented by the vector OA) to V% (repre- sented by the vector OB). The increment of velocity is then rep- resented by the vector AB. If the discharge of the jet is W Ibs. per second, the change of momentum produced in one sec- ond by the action of the deflect- ing surface is W JV, 9 ' which is therefore the value of the resultant force constantly exerted upon the jet by the deflecting body. If V lt V 2 and the angle of the deflection a are known, the magnitude and direction of AV can be computed by solving the triangle OAB. The line of action of the resultant force passes through the intersection of the two lines which coincide with the axis of the jet before and after the deflection, as indicated in Fig. 82. If the magnitude of the velocity is not changed, so that FIG * Theoretical Mechanics, Art. 253. VELOCITY OF STREAM REVERSED. 165 W ... 2WV a 2wFV 2 . a and p=-jy = sm2= sm^, . . (11) a being the angle of deflection, i.e., the angle between V l and V 2 . In any case, the reaction of the jet upon the body which deflects it is equal and opposite to the force exerted upon the jet. EXAMPLES. 1. A jet of 2 sq. in. cross-section, having a velocity of 40 ft. per sec., strikes a curved surface which deflects it 25 and changes its velocity to 35 ft. per sec. Determine the magnitude and direction of the force exerted by the jet upon the deflecting body. Ans. A force of 9.15 Ibs. making angle of 60 46' with jet. 2. If the jet in Ex. 1 is deflected 25 without diminution of velocity, determine the force exerted upon the deflecting body. Ans. 9.34 Ibs. making angle of 77 30' with jet. 165. Case in which the Velocity of the Stream is Reversed. If the deflection of the stream amounts to 180, as in Fig. 33, the increment of velocity is Vi + V 2 , and the force exerted by the deflecting body is W * P = JV 1+ y 2) IVM+VJ, . . . (12) if FI is the cross-section of the stream where its velocity is V FIG. 83. FIG. 84. If the jet could be reversed without diminution of its velocity (Fig. 84), so that V 2 = Vi = V, the force would have the value The line of action of P is found from the law of composition of parallel forces. The momentum WV 2 /g is the resultant of the 166 DYNAMIC ACTION OF STREAMS. momentum WVi/g and that due to the force P acting for cne second. The distance between the lines lying in the axis of the jet before and after deflection is therefore divided by the line of action of P in the inverse ratio of V\ and V 2 . 166. Reversal of Jet by Moving Vane. In the preceding case suppose the curved vane which deflects the jet to be moving with velocity u in the same direction as the impinging jet. If V\ and V 2 are the values of the absolute velocity of the stream just before striking the vane and just a ^ ter ^ eavm ^> anc ^ if ^ 7/ denotes the weight of water striking the vane per second, the force exerted by the jet upon the vane is The values of V 2 and of W will depend upon u. Let v denote the velocity of a particle relative to the vane at a point where its absolute * velocity is V ; v\ being the value of v at the point where the stream is about to strike the vane, and v 2 its value at the point where the stream leaves the vane, Then If there were no loss of energy in the flow over the vane, v 2 would equal vi. Assuming this to be true, Substituting in the above value of P', and noticing that we have P-(Fi-w)* ....... (14) y * See Appendix B. JET WATER WHEEL WITH CURVED VANES. 167 Work done on moving vane. The work done per second by the force P* is (15) Comparing with Art. 162, it is seen that these values of the force and the work are double those found in the case of a flat vane. 167. Jet Water Wheel with Curved Vanes. If the flat vanes in Fig. 81 were replaced by curved vanes so arranged as to receive and discharge the jet substantially as in Fig. 85, the above reasoning would hold approximately, equations (14) and (15) representing the force exerted upon a single vane and the work done by that force per second. Since, however, more than one jet would be receiving water at the same time, so that the wheel would be acted upon by all the water discharged by the jet, W must be replaced by W to get the total action upon the wheel, in which That is, the total force exerted upon the wheel would be P-y(F, + F 2 ) -F,(Fi + Fa) --Fi(F, -u) ; (16) y y y and the work done upon the wheel per second would be 1 (7 1 -u)w ..... (17) y These values are double those obtained in Art. 163 for a wheel with flat vanes. The maximum value of the work, given by u = Vi/2, is WV } 2 Maximum L = ^ , ..... (18) J/ which is equal to the whole kinetic energy of the jet. This form of water wheel is discussed in Chapter XVIII. 168 DYNAMIC ACTION OF STREAMS. , 168. Principle of Angular Momentum. From the principle stated in Art. 158 another of equal importance may be deduced. Since the total increment of momentum of a particle per second, is equal to the average value of the force acting upon it, it follows by taking moments about any axis that the total increment of the moment of momentum (or "angular momen- tum") per second is equal to the average value of the moment of the force. This principle is of use in the following discussion. 169. Action of Stream upon Rotating Vane in General Case. All the special cases above considered are covered by the following general discussion. Let MN (Fig. 86) represent a curved vane which rotates uniformly, with angu- lar velocity cu, about an axis at per- pendicular to the plane of the figure. Let the vane receive a stream at M and discharge it at N. The point M of the vane moves at every instant at right angles to OM with velocity HI, and / the point N moves perpendicularly to FIG. 86. ON with velocity u^. Let Vi, Vi= absolute and relative * velocities of a particle of water at M, just before striking the vane; Y 2j V2 = absolute and relative velocities of a particle at N, just leaving the vane; AI, a\ = angles between FI, v\ respectively and u\ , A <* ll T7ii (( (f ni Az, a 2 = v 2, V2 u%. In order to compute the action of the jet upon the vane, we may apply the principle of angular momentum, taking the axis of rotation as axis of moments. Let OM = ri, ON = r 2 . The angular momentum of a particle of water of mass m just before striking the vane is mV\r\ cos A\, *By " relative velocity" will always be meant " velocity relative to the rotating body," as described in Appendix B. FORCE EXERTED BY CONFINED STREAM. 169 and its angular momentum just as it leaves the vane is mV 2 r 2 cos A 2 , so that the action of the vane changes the angular momentum of each particle by the amount m(V 2 r 2 cos A 2 V\r\ cos A\) . The total change of angular momentum due to the action of the vane for one second is (V 2 r 2 cos A 2 V\r\ cos AI), if W f is the weight of water striking the vane per second. This is therefore also the average value of the total moment of the forces exerted by the vane upon the water. If there is a succession of similar vanes forming a water wheel, the total action upon the wheel is found by replacing W r by W, the total weight of water used by the wheel per second. If G denotes the total moment of the forces exerted by the water upon the wheel, we have W G= (V 2 r 2 cos A 2 ViTi cos AI), y W or G= (Fin cos AI V 2 r 2 cos A 2 ). . .. . (19) Applications of this result will be given in the following chapters. 170. Force Exerted upon Pipe by Confined Stream in Steady Flow. Let A B (Fig. 87) represent a portion of a pipe carrying a steady stream, FI, FI being the cross-section and velocity at A and F 2 , V 2 the like quantities at B. Let the rate of discharge be W pounds per second. In passing from A to B a particle of mass m receives an increment of momentum mJF, in which AV = A'B f (Fig. 87), FT and V-z being represented by the vectors OA f , OB'. Hence the average 170 DYNAMIC ACTION OF STREAMS. force exerted upon the particle is mAV ~7T' if At is the time required for this change of momentum. (Art. 158.) The resultant of such average forces for all particles will be constant, and may be computed by considering the body of water which at any instant occupies the volume AB } and following its motion for the time dt. During this time equal weights of water Wdt pass the sections A and B, the momentum W W of the former being Vidt, and that of the latter V> 2 dt. The flow being steady, the vector difference between these values, WAV or - dt, is the total change of momentum of the body under consideration during the time dt. The resultant force acting upon the body is therefore WAV W P=H-2L.= (vector A'ff) (20) .7 y The forces which make up this resultant are the weight of the water, the pressures exerted by the adjacent water upon the cross-sections A and B, and the forces exerted by the portion of the pipe between A and B. The resultant of these latter forces can therefore be determined if the pressures in the stream at A and B are known. The equal and opposite reaction to this force is the force exerted by the stream upon the portion AB-oi the pipe. EXAMPLES. 1. Let the axis of the pipe in Fig. 88 be horizontal, the diameter at A. A being 1 ft. and at B .5 ft. Assume the rate EJrEfE^Eg^ggg^^^JjjLs of discharge to be 2 cu. ft. per second in the ^^^^^^^= direction AB, and the pressure head at the center of the section A to be 12 ft. Compute FIG. 88. t ne resultant force exerted upon the pipe by this portion of the stream, on the assumption of no loss of head. From the given data V^ =2.55, Fa = 10.2, Ft 2 /20 = .101, VS/2g = 1.62. Assuming no loss of head, tL+!j*+ZL w 2g w 2g' from which p 2 /w 10.5 ft. THEORY OF PITOT'S TUBE. 171 The increment of momentum per second for the body AB is W (V V} 62.5 X2X (10.20 -2.55) g (V * Yl) ~~32.2 hence the resultant of all forces acting upon this body is 29.7 Ibs. in the direction AB. This is made up of the pressures on the two cross-sections A and B and the forces exerted by the pipe. Calling the resultant of the latter P (so that P is the force exerted by the stream upon the pipe), we have or P -piFi -p*F 2 - 29.7 =589 - 129 - 29.7 =430 Ibs. 2. Generalizing the preceding example, show that the resultant force exerted by the stream upon the pipe A B, neglecting loss of head, is -w(Fi -'>[$-(-)!'] 3. Solve with data as in Ex. 1, except that the flow is in the opposite direction. 4. A stream is discharged into the atmosphere through a nozzle which reduces the diameter from 2" to .5". If the pressure head at the nozzle entrance is 100' above atmospheric, compute the resultant force exerted by the stream upon the nozzle, assuming no loss of head. Ans. 1221'bs. 5. A straight horizontal pipe 1' in diameter is discharging 1 cu. ft. per sec. Assuming that the loss of head is given by Darcy's formula, determine the resultant force exerted by the stream upon the pipe in a length of 100'. Ans. 26.7 Ibs. 171. Theory of Pitot's Tube. The measurement of the velocity at any point in a stream by means of Pitot's tube has been referred to in Chapter XIII. Fig. 89 represents such a tube placed with the lower end facing the current, the main part of the tube being ver- tical and the upper end projecting above the water surface. Let V denote the velocity of the current passing the lower end of the tube, and h the height to which the water in the tube rises above FlG - 89 - the surface of the stream. Then h measures the excess of the 172 DYNAMIC ACTION OF STREAMS. pressure at any point within the tube above the pressure at the same level in the stream outside the tube. If F is the cross-sectional area of the tube at the end facing the stream, the dynamic action of the stream upon this area must amount to a force whF in order to maintain the column within the tube in equilibrium. The effect of the tube is to deflect a certain part of the current, and the total change of momentum thus produced in one second is equal in magnitude to the force exerted upon the tube and its contained water by the water which is deflected. (Art. 158.) If W is the weight of water deflected per second and AV its average increment of velocity, the force would be It seems reasonable to assume that AV is proportional to F, and that the part of W whose deflection is due to impingement against the water at rest in the tube is proportional to wFV, so that the force exerted by the current upon the area F is proportional to wFV* 9 Comparing with the value given above, wFV* whF varies as , F 2 or fc-fer, in which & is a coefficient which would be constant if the theory were exactly true. The values of k will depend upon the form of the tube, and RAM PRESSURE. 173 FIG. 90. must be determined by experiment for every instrument. For the three cases shown in Fig. 90 it would seem that (A) would give the least and (C) the greatest value of AV for a given value of V, and therefore that the coefficient k would be least in the former case and greatest in the latter. This is verified by experi- ment. Values of k ranging from 1 to 2 have been found by different experimenters for tubes of different designs. 172. Ram Pressure. The sudden stoppage of the flow in a pipe may cause pressures much greater than those due to the static head. Pressure due to this cause is called ram pressure, or " water-ram." In Fig. 91 let the flow in the pipe be steady, the velocity being F, and suppose a valve at C to be suddenly closed, thus checking and finally stopping the flow. At any point up- stream from the valve the pres- sure is increased by an amount which is less as the distance from the valve is greater. The amount of increase at any point depends upon the rate of decrease of the velocity. This effect is strictly analogous to that due to a moving body which is brought to rest by striking a fixed obstacle; it exerts upon the obstacle a force proportional directly to the mass of the moving body and to the rate at which its velocity is de- creased. The value of the ram pressure at any point, in the ideal case of an incompressible fluid and an unyielding pipe, may be expressed in the following manner. Let AB be any straight portion of the pipe of length Z, and let pi, p 2 denote the values of the pressure at A and B. If m is the total mass of the cylinder of water AB, the resultant of all forces acting upon it at any instant is (by the fundamental equation of motion) FIG. 91, (22) 174 DYNAMIC ACTION OF STREAMS. While the flow is steady this" resultant is zero, since dV/dt=Q, but while the velocity is changing P is not zero. In the present problem the forces called into action by the sudden closing of the valve are great in comparison with those acting when the flow is steady, and the latter may be neglected as in all cases in which impulsive forces are considered.* The only forces to consider are, therefore, the pressures caused by the sudden stoppage of the flow. The resultant of these pressures acting upon the body AB amounts to a force in the direction AB, F being the cross-section of the pipe. But also, as above, dV wFl dV wl dV therefore p 2 - p l = --- i/ .-L=-- ....... (24 ) w w g dt Since dV /dt is negative (because V decreases) , the second mem- ber of this equation is really positive, and p 2 is greater than p\ . This formula shows that the ram pressure varies along the pipe directly as the length, and is independent of the diameter for a given value of dV /dt. The result is seen to hold even if the pipe is curved, since the length may be subdivided into elements in applying the above reasoning. If the pipe leads from a reservoir, as in Fig. 91, the ram pressure will be zero at the intake end, so that for a point dis- tant I from the intake end we may write (pressure being ex- pressed in terms of equivalent water column) I fly Ram pressure = --- -77 ...... (25) * Theoretical Mechanics, Art. 321 ! RAM PRESSURE. 175 No practical use can be made of this formula unless dV/dt can be estimated. Experiment indicates that the closing of a valve produces no ram pressure of importance except during the last part of the closing; that is, the rate of change of the velocity does not become great until near the very end of the process. Dangerous ram pressures may therefore be avoided by using valves so constructed that the last part of the closing must be slow. The above results are doubtless modified in an important degree by the elastic yielding of the pipe and of the water. It is doubtful whether a theoretical discussion taking account of these factors can be given which will accord closely with prac- tical conditions. EXAMPLE. 1. If the velocity of flow in a pipe is changed in .25 sec. from 2 ft. per sec. to by the closing of a valve 1000 ft. from the reservoir, esti- mate the average value of the ram pressure close to the valve. Ans. p/w=248 ft. CHAPTER XV. THEORY OF STEADY FLOW THROUGH ROTATING WHEEL. 173. Statement of Problem. Wheels designed for the utilization of the energy of streams of water are of various forms, and may be divided into several classes possessing some- FIG. 92. what distinct characteristics. The same basal theory, how- ever, applies to practically all the important types. This theory it is the object of the present chapter to present; and for illustration reference will be made to the arrangement shown in Fig. 92. The figure shows an elevation and vertical section (A) and 176 NOTATION. 177 a plan and horizontal section (B). The " penstock " or supply- pipe P leads from the head race or supply reservoir and ter- minates in the guide passages g g. These are so formed as gradually to deflect the water outward and forward (i.e., in the direction of the rotation). A particle of water when leaving a guide passage is moving in a horizontal plane and in a direc- tion determined by that of a guide vane. Within the wheel the path of a particle relative to the wheel is determined by the form of the wheel vanes. Arriving at the outlet of the wheel, the relative velocity of a particle has a direction determined by that of a wheel vane, while its absolute velocity depends upon this relative velocity and also upon the rotation of the wheel. The problem whose solution is the basis of turbine theory is the following: Given all dimensions and the rotation speed of the wheel, required to determine the rate of flow through the wheel. If the wheel were at rest, this would be a problem in ordinary hydraulics, to be solved by the methods illustrated in Chapter VI. It is to be shown how the rotation of the wheel affects the solution. The explanation involves the following points : (1) The relation between the absolute and relative velocities of a particle; (2) the meaning of the general equation of energy in such a case as this; (3) the computation of the energy given up by the stream to the wheel; (4) the deduction from these results of a special form of the energy-equation involving rela- tive instead of absolute velocities. In the present chapter this problem will not be completely solved, since the details of the solution must be different for different types of wheel. In all cases, however, the solution involves certain general steps which are here outlined. 174. Notation. The following notation will be employed throughout the entire discussion of turbines and water wheels : r distance of a particle from the axis of rotation; 2= its height above a horizontal datum plane; v=its velocity relative to the wheel; F=its velocity relative to the earth; 178 STEADY FLOW THROUGH ROTATING WHEEL. [Briefly, v and V will usually be called simply relative velocity and absolute velocity.] aj = angular velocity of wheel (radians per second) ; u = raj = linear velocity of point of wheel momentarily coinciding with a particle of water under considera- tion; s = resolved component of V in direction of u (called briefly tangential component of V) ; a = angle between v and u\ A = angle between V and u\ q = volume of water discharged per second ; W = weight of water discharged per second; w= weight of unit volume of water; F = cross-section of stream at any point in the stationary part of the passages; / = cross-section of stream at any point within the wheel; L = energy imparted by water to rotating wheel per second. The values of any of the foregoing quantities for different sections of the stream will be distinguished by suffixes. Thus suffix d) will refer to the stream leaving the guide passages, just before its motion is influenced by the wheel, and suffix ( 2 ) to the stream leaving the wheel. In many cases the stream is divided into several parts in passing through the wheel, also as it approaches the wheel through the guide passages. By F and / will be meant the sum of the cross-sections of all the partial streams taken at corresponding points in the different passages. Thus FI = total cross-section of streams leaving guide passages, measured everywhere normally to the direction of FI ; / 2 = total cross-section of streams leaving wheel passages, measured everywhere normally to the direction of v 2 . The equation of continuity (Art. 36) holds throughout the entire series of passages; thus APPLICATION OF GENERAL EQUATION OF ENERGY. 179 In considering the relation between absolute and relative velocities it is often convenient to use the language of vector addition. For this purpose the vector value of a velocity will be represented by the use of brackets: [u], [v], [V] = vector values of velocities whose magnitudes are u, v, V. 175. Relation between Absolute and Relative Velocities.* The three velocities u, v, V are related in the manner shown by the vector triangle in Fig. 93. This relation is expressed by the statement that V is the vector sum of u and r, or ...... (1) Algebraically, any of the equations given by elementary trigonometry may be used for com- puting one of these vectors when the other two are known. The following are especially useful. Resolving along and per- pendicularly to u (Fig. 93), V cos A = u + v cos a, 1 V sin A = v sin a. j ' The tangential component of V is evidently s = V cos A = u + v cos a ....... (3) These relations being true for any particle, are true when suffix d) or suffix ( 2 ) is attached to each symbol. 176. Application of General Equation of Energy. The general equation of energy may be applied to any case of steady flow. In this equation H' means the quantity of energy lost by the water, between the sections d) and ( 2 ), per pound of water discharged. (Art. 63.) In most of the applications hitherto considered this loss has been wholly due to dissipation of energy into heat: If the water gives up energy in any other way, the equation still holds, * See Appendix B. 180 STEADY FLOW THROUGH ROTATING WHEEL. provided H' be made to include such loss as well as that due to dissipation into heat. Thus, in the case shown in Fig. 92, let the section d) be taken where the water is leaving the guide passages, just before its motion is influenced by the rotating wheel, and the section (2) at the point of outflow from the wheel. Then H' is made up of two parts, corresponding respectively to the energy lost by dissipation within the wheel and that given up to the wheel as mechanical energy. Calling these parts h r and h" respect- ively, we have 1 = 2g + Zl > and the equation may be written +* -+ -*+* > The equation applies also if the section d) be taken elsewhere, for example at the surface of the head race, proper values being given to FI, pi and z\, and it being understood that h' includes energy dissipated throughout the entire series of passages between d) and (2), while the meaning of h" is unchanged.* 177. Computation of Energy Given Up to Wheel. The value of the energy given up to the wheel by the water may be expressed by a simple formula which will now be deduced. The flow being steady and the rotation of the wheel uniform, the water exerts upon the wheel forces whose turning moment about the axis of rotation is constant. If G denotes this mo- ment, the work done by the forces upon the wheel f while the latter turns through an angle 6 (radians) is GO. * See Art. 94. f Theoretical Mechanics, Art. 508. NEW FORM OF GENERAL EQUATION OF ENERGY. 181 The value of G may be determined from the principle of angular momentum, as in Art. 169. The value there found may be written (noticing that FI cos AI =Si and 2 cos A 2 = s 2 ) W G =(r 1 s 1 -r 2 s 2 ) ....... (5) t/ The angle turned through in one second being to, the work done on the wheel per second is Ga>; or, denoting by L the energy imparted to the wheel in one second, and noticing that W W L= (riSir 2 s 2 )to = (uis\ u 2 s 2 ). ... (6) y y 178. New Form of General Equation of Energy. From the meaning of h" (Art. 176) it is evident that so that and equation (4) may be written This may be simplified by introducing relative instead of abso- lute velocities. Thus, from the vector relation between F, v and u (Fig. 93), V 2 2 = v 2 2 + u 2 2 -f 2u 2 v 2 cos a 2j SI=MI +vi cos ai, COS C1 2 . 182 STEADY FLOW THROUGH ROTATING WHEEL. The substitution of these values in (7) reduces it to the form (8) This may be called the equation of energy for the relative motion. It may be noticed that this equation includes as a special case the equation of energy in its ordinary form for flow through stationary passages, reducing to it when o; = 0; this makes HI and u 2 zero, and v\ and v 2 become absolute velocities. The meaning of h' should be kept clearly in mind. It is the energy lost by dissipation within the wheel, per pound of water discharged. 179. Summary of Principles. The principles to be employed in the theory of turbines are embodied in the general formulas already deduced, which may be summarized here for convenience of reference. The equation of continuity: q = FV = fv = constant; ..... (I) and in particular (I') The relation between absolute and relative velocities of a particle, expressed by the vector equation and by the algebraic equations V cos A = u + v cos a, 1 Ft ' ( sin A = v sin a, the two algebraic equations (II') being equivalent to the single vector equation (II) . EXAMPLES. 183 The equation of energy for steady flow in the ordinary form: The formula for energy imparted to the wheel: W L =(u l s l -u 2 s 2 ) ....... (IV) The equation of energy for the relative motion: 180. Application of General Theory. The application of the foregoing principles and formulas to particular cases involves special data pertaining to each case. The two cases of greatest practical importance are illustrated in the following examples, and the complete theory is treated in the succeeding chapters relating to the different types of turbines and water wheels. The following examples refer to the arrangement shown in Fig. 92. EXAMPLES. 1. A particle of water leaves the guide passage with a velocity of 22 ft. per sec. directed at angle 18 with the velocity u\. The distance from the axis of rotation is 3.5 ft., and the wheel rotates at the rate of 90 R.P.M. Determine the magnitude and direction of the velocity of the particle relative to the wheel. Am. v, =13.85 ft. per sec.; a, = 150 36'. 2. In order that the particle (Ex. 1) shall not be suddenly deflected as it enters the wheel, what should be the direction of the wheel vane at the point where the water enters? 3. Using the data of Ex. 1, and assuming that the particle moves in a horizontal plane, that no frictional loss of energy occurs within the wheel, and that the pressure is constant throughout the wheel and at guide outlets, what will be the magnitude of the relative velocity of a particle when 4 ft. from the axis? What will determine the direction of this relative velocity? Ans. v =22.91 ft. per sec. 184 STEADY FLOW THROUGH ROTATING WHEEL. 4. Suppose the radius of the wheel at outlet is 4.5 ft., and a 2 = 162. Making the same assumptions as in Ex. 3, determine the magnitude and direction of the relative velocity and of the absolute velocity of a par- ticle leaving the wheel. Ans. v 2 = 30.05 ft. per sec.; V. = 16.66 ft. per sec.; A 2 =33 52'. 5. With data as in Ex. 4, (a) how much energy does the wheel receive from the water for each pound of discharge? (6) If F\ =6 sq. ft., com- pute the H.P. imparted to the wheel, (c) The energy utilized is equiva- lent to what fall? Ans. (a) 3.21 foot-pounds. [In the following examples, instead of supposing the pressure to be uniform throughout the wheel, assume that the wheel passage is com- pletely filled at every cross-section, and that F,// 2 = 1.5, all other data being as in preceding examples. Answer the following questions.] 6. What further data are required for the determination of the rela- tive and absolute velocities of a particle 4 ft. from the axis? 7. Determine the magnitude and direction of the absolute velocity and of the relative velocity of a particle as it leaves the wheel. Ans. ^=33 ft. per sec.; F 2 = 15.03 ft. per sec.; A 2 =42 43'. 8. If the discharge from the wheel takes place into the atmosphere, what pressure exists at guide outlets? Ans. p\/w p^/w =2.9 ft. 9. If F> =6 sq. ft., compute H.P. imparted to wheel. What is the value of the "utilized head "? Ans. H.P. =103.5; utilized head =6.90 ft. CHAPTER XVI. TYPES OF TURBINES AND WATER WHEELS. 181. Definition of Turbine. There is no general agreement upon an exact definition of a turbine. By Bodmer * a turbine is defined as "a water wheel in which a motion of the water relatively to the buckets is essential to its action." Such a definition includes practically all modern wheels for the utiliza- tion of water power. The same author uses the term water wheel to designate the old-fashioned overshot, undershot, and breast wheels, which were usually of large diameter relatively to the fall utilized, and which were actuated either by the weight of the water directly or by the impact of a stream against flat vanes. These definitions are not in conformity with present-day usage in America. This usage can be best explained after the different types of wheels have been described. f The word tur- bine will, however, generally be used with the broad meaning above given. Classification of Turbines According to Direction of Flow. The direction of flow of the water through a turbine may be either radial, axial, or mixed. Radial flow means that the path of a particle within the wheel lies in a plane perpendicular to the axis of rotation. The direction of flow may be either outward (Fig. 92) or inward (Fig. 105). Axial flow means that the distance of a particle from the * Hydraulic Motors, p. 24. t See Art. 191. 185 186 TYPES OF TURBINES AND WATER WHEELS. axis of rotation remains constant during its passage through the wheel (Fig. 104). Mixed flow is a combination of radial and axial flow. It is usually inward and axial, as in Fig. 106. 183. Classification into Impulse Wheels and Reaction Wheels. If the total cross-section of the streams entering the wheel is so small that the wheel passages are not filled, and if air enters freely so that the entire stream within each wheel pas- sage is under atmospheric pressure, the turbine is called an impulse wheel. If the wheel passages are completely filled by the streams flowing through them, the turbine becomes a reaction wheel. This is the most fundamental distinction between different wheels of modern type. 184. Complete and Partial Admission. Water may be admitted to all the wheel passages at once, or to a limited num- ber of them. Reaction wheels necessarily have complete admission, while impulse wheels may have partial admission with little, if any, sacrifice of efficiency. 185. Conditions of Discharge. Impulse wheels always dis- charge into the air. A reaction wheel may discharge either (a) into the air, (6) into a body of free water, or (c) into a suction tube. When the discharge is into the atmosphere, the fall from the point of discharge to the tail race is lost. Such loss does not occur if the wheel is submerged, or if it discharges into a suction tube (provided the conditions are such that water fills the tube throughout its length) . The action of a suction tube depends upon atmospheric pressure. The pressure in the tube at the upper end (at Y, Fig. 104) is less than atmospheric by an amount depending upon the height of this point above the surface of the tail race. This decrease in the pressure at the point where the wheel dis- charges has the same effect upon the flow as an equal increase at X, the point of inflow. The fall z (Fig. 104) below the wheel, DIFFERENT FORMS OF TURBINES. 187 which would be lost if the discharge occurred at the same point but under atmospheric pressure, is thus exactly compensated. If the wheel be placed lower or higher, the pressures at X and Y will change equally, so that the operation of the wheel will be unchanged. Of course z must not be so great that the pressure at Y is reduced to absolute zero.* 186. Girard Impulse Turbine. Impulse turbines of the type known as the Girard have been extensively used in Europe, also to some extent in America. The flow may be either radial or axial, and admission may be either complete or partial. The case of radial flow with complete admission may be represented as regards general arrangement by Fig. 95. The arrangement of a Girard turbine with axial flow is shown in Fig. 96. Girard introduced the feature of ventilating the wheel passages by orifices for the free admission of air. 187. American Tangential Water Wheels. The type of motor shown in Figs. 98 and 99 is common in America, especially in mountainous regions where high falls are utilized. The wheel vanes or buckets receive the water in a cylindrical stream from a nozzle, the axis of the stream being tangent to the circle described by a certain point in each bucket. Buckets have been made of many forms, but in most cases the " split " bucket is used, the stream being divided by a sharp edge into two streams which follow the opposite but similar bucket surfaces. Fig. 97 represents a section of such a bucket by a plane parallel to the axis of rotation and to the stream from the nozzle. Although differing greatly in form from the turbines above mentioned, the tangential water wheel falls under the defini- tion of turbine above given, and is of the impulse type with approximately axial flow. 188. Fourneyron Turbine. This name is usually given to reaction turbines with radial outward flow. The discharge may be either into the air or into a body of water. In either | * A discussion of the suction tube is given in Art. 228. 188 TYPES OF TURBINES AND WATER WHEELS. case the general arrangement may be represented by Fig. 92. A suction tube cannot conveniently be used with this form of wheel. 189. Reaction Turbines with Inward Flow. Wheels similar in principle to the Fourneyron but with inward flow have been designed by Thomson, Francis, and others. An inward-flow wheel with suction tube is shown in Fig. 105. This type is now quite generally known as the Francis turbine. 190. Jonval Turbine. This is a reaction turbine with axial flow. The discharge may be either into the air, directly into the tail water, or into a suction tube (Fig. 104). 191. American Reaction Turbines. The most common forms of reaction turbine used in America are of the mixed- flow type, having inward admission and axial discharge (Fig. 106). In the United States the name turbine is usually confined to the wheels above called reaction turbines, while the " tan- gential " wheels described in Art. 187 are called water wheels. 192. Theory of Turbines. The principles developed in Chapter XV furnish a basis for the theory of turbines and water wheels of all the foregoing types. In the following chap- ters will be given the outline of the theory for the impulse turbine, the American tangential water wheel, and the reaction turbine. A discussion will also be given of turbine pumps, of which the theory will be seen to be closely similar to that of reaction turbines. The following definitions of available energy and efficiency apply to all types of water wheels and turbines acting as motors, but not to turbine pumps. 193. Available Energy. The available energy of a stream of water * depends upon the weight of the water and the avail- * Strictly this energy is possessed not by the water alone but by the system consisting of the water and the earth, being due to the gravitational attraction between the earth and the water. (Theoretical Mechanics, Art. 362.) The kinetic energy due to the velocity of flow in the stream is so small EFFICIENCY. 189 able fall; thus Wh foot-pounds of potential energy are given up by W pounds of water in descending h feet. The object of a motor is to receive this from the water as mechanical energy, with as little loss by dissipation as possible. The available power is the energy available per unit time. If the rate of discharge of the stream is q cu. ft. per sec. and the fall h ft., the available power is wqhft.-lbs. per sec.=T H.P. 194. Efficiency. In estimating efficiency we may be con- cerned with the whole apparatus, including the pipe or channel leading to the motor, the motor proper, and the pipe (if any) leading from the motor to the tail race, or we may be con- cerned only with the efficiency of the motor proper. Again, we may consider the energy imparted to the motor by the water to be the useful effect, or we may regard the energy ob- tained from the motor in doing external work as the useful effect. In any case, the efficiency is defined by the equation energy utilized e == energy given up by water' but the numerator and denominator may each have two differ- ent values, according to the point of view as just explained. Gross efficiency is computed by taking the external work done by the motor as "energy utilized." Hydraulic efficiency^ computed by taking energy imparted to the motor as "energy utilized." These two efficiencies differ because of the energy dissipated by mechanical friction of the moving parts of the motor. Hy- draulic efficiency corresponds to the "indicated" efficiency of a steam engine, since the indicated work is the work actually done on the piston by the steam pressure. in comparison with the potential energy due to the fall, that it is neglected in estimating the available energy. CHAPTER XVII. THEORY OF IMPULSE TURBINE. 195. Given Dimensions and Data.* In the theory of tur- bines certain dimensions and other data may be taken as known. Thus in an impulse turbine Ti,.the velocity of outflow from the guides, is fixed independently of the construction of the wheel and of its speed of rotation. Since the discharge from the guides occurs under atmospheric pressure, V\ depends only upon the fall to the point of outflow and the resistance occurring in the passages leading from the head race or reservoir. The angles AI and a 2 will also be taken as known. It may be seen that A i and 180 0,2 should both be small. That a 2 should approach 180 is seen from the fact that the absolute velocity of outflow from the wheel (2) should be as small as possible, since the kinetic energy represented by this velocity is wholly lost. Since V 2 is the vector sum of u 2 and v 2 , the more nearly opposite the directions of u 2 and v 2 are taken the smaller V 2 may become. This require- ment is, however, limited by the fact that the smaller the angle 180 -a 2 , the smaller the cross-section of the wheel passages. Thus, let Fig. 94 represent a section of several vanes of an outward-flow turbine, and let d 2 I \/ \ denote the vertical dimension of the wheel ' * \ passages at the point of outflow. If n is the total number of vanes, 2nr-2/n is the length of the portion of the wheel circumference corresponding to one passage, and the cross- section of a passage is approximately (2xr 2 d 2 sin a 2 ) /n, so that * For notation, see Art. 174. 190 GIRARD TURBINE WITH RADIAL FLOW. 191 approximately That AI should be small is seen from formula (IV) (Art. 179), which gives the energy imparted to the wheel per unit time. Neglecting the term u 2 s 2 , which is to be made small, and remembering that s L = Vi cos AI, it is seen that for a given power HI must vary inversely as V\ cos A i . Hence, to avoid an excessively high wheel speed, V\ cos^i should be as large as practicable. And since V\ is -fixed independently of the design of the wheel, cos AI should be as great as practicable. This requirement is, however, limited by the fact that F\ de- creases as A\ decreases, other dimensions remaining the same. In the following theory the quantities taken as known are Vi, AI, a 2 , r 2 /ri (which will be called c), z\ -z 2 . The discussion eads to a formula for efficiency, and to the determination of HI and a\ for highest efficiency. These results are found to be independent of the remaining dimensions of the wheel. (A) GIRARD TURBINE WITH RADIAL FLOW. 196. General Arrangement of Radial-flow Girard Turbine. The general arrangement of a radial-flow Girard turbine with axis vertical and with complete ad- mission is represented in Fig. 95. The wheel is placed as near the sur- face of the tail race as practicable. While the vertical widening of the wheel passages permits the particles of water to fall slightly during their passage through the wheel, the amount of this fall will be so small as to be = =/^^^^^7^^~7^ // . unimportant, and will be neglected / ^ ///////// / / ^ /// / / ^ /// in the following theory. The figure shows a cylindrical gate so placed as to regulate the size of the guide outlets. The value of Vi is not materially changed by varying the position of the gate, so that no serious loss of energy results from this method of regulating the supply of water. 192 THEORY OF IMPULSE TURBINE. Girard turbines are often arranged for partial admission. The various forms which have been made, and the devices for regulating the admission of water to the buckets, will not be considered here, since these modifications do not affect the theory as here presented. Governing, or the maintaining of a nearly uniform wheel speed in spite of fluctuations in the power consumed (or "load "), is accomplished by means of the regulating gate. This is con- nected by suitable mechanism with some form of centrifugal governor, so that an increase in the wheel speed causes a decrease in the gate opening. 197. Relation between Power and Wheel Speed. Since the quantity of water used par second is independent of the speed of rotation, the efficiency is proportional directly to L, the energy imparted to the wheel by the water per second. It is necessary, therefore, to express L as a function of tha velocity of rotation, and then consider what value of this velocity makes L a maximum. In the general formula (IV) (Art. 179) the variable velocities in the second member may all be expressed in terms of u\ in the following manner. We may write at once To determine s 2 it is necessary to solve the problem of flow through the wheel, which may be done as follows: From the vector relation between V \, u\ and Vi (Fig. 93), vi 2 = V? + Ui 2 - 2ViUi cos Ai ..... (1) The value of v\ thus determined is to be substituted in formula (V) * for the purpose of determining v 2 . Since in the present case z\ =Z2 and p\ p 2 , formula (V) takes the form It is necessary to estimate the value of h'. * Art. 179. GIRARD TURBINE WITH RADIAL FLOW. 193 The losses of energy by dissipation between the sections d) and ( 2 ) are due mainly to two causes: (a) the interference of the inner edges of the wheel vanes with the stream from the guides, and (b) hydraulic friction within the wheel. To reduce the former as much as possible the edges of the vanes against which the stream impinges should be sharp, and the direction of these wheel vanes should be such as to agree with that of the relative velocity vi. This direction of the wheel vanes can be adjusted to only one particular value of u\, which should be that giving best efficiency, as yet unknown. For the purpose of this discussion, however, the direction of the vane may be regarded as varying with u\ so as always to be parallel to v\, leaving the actual value of the vane angle to be assigned when the best value of HI becomes known. Loss (a) will thus depend upon the relative velocity of the stream entering the wheel, while loss (6) will depend upon the relative velocity of flow through the wheel passages. At best only an approximate estimate of these losses is possible, and it will suffice to express the entire loss la' by a single term: (3) in which & is a coefficient treated as constant, whose value cannot be accurately estimated apart from experiment. Equa- tion (2) thus becomes , .V . . (4) which with (1) gives . . . . (5) Using this value of v 2 , s 2 may be expressed in terms of u\ and constants as follows : \r -L i K (6) 194 THEORY OF IMPULSE TURBINE. We now have the values of Si, $2 and u 2 , all in terms of MI, for substitution in formula (IV). The result is W L = UiVi cos AI -c 2 ui 2 c cos a 2 Vl u 1 \/V 1 2 -2u l Vi cosAi+c 2 ^ 2 . (7) 198. Condition for Maximum Power and Efficiency. The mathematical condition for maximum L is dL/dui=0. This leads to an equation of the fourth degree for determining u\. The deduction of this equation and its solution in particular cases would involve a large amount of labor. In view of the imperfections in the foregoing theory, the following approxi- mate treatment of the problem is probably as satisfactory a guide to design as the exact solution would be. The greatest value of L results when the losses of energy are least. The loss which varies most with varying velocity is that due to the kinetic energy of the water leaving the wheel. Since V 2 is the vector sum of u 2 and v 2 , it appears that when V 2 is small u 2 and v 2 will be nearly equal in magnitude. An approximate solution of the problem of maximum efficiency will therefore result from the assumption V 2 U 2 = CUi. -..'..... (8) This value of v 2 substituted in (5) gives 7,2-27^! COS At -&C 2 Wl 2 = 0, .... (9) which determines the value of u\. 199. Value of Efficiency. The hydraulic efficiency of the wheel is the ratio of L to the available energy per unit time, WVi 2 /2g. The efficiency for any velocity may thus be com- puted from the general value of L given by (7). By substitut- GIRARD TURBINE WITH RADICAL FLOW. 195 ing the value of HI given by (9) , the highest efficiency (according to the foregoing approximate solution) is found. The following convenient forms for the values of maximum power and effi- ciency are easily obtained by combining (7) and (9) : W L m =[u l ViGosA l -c 2 (l+cosa 2 )ui 2 l . . (10) t7 _ O 1 e m =2j Y cos 4i -c 2 (l+ 00302)^2 J > . . (11) in which Ui/V } must have the value given by (9). 200. Best Vane Angle. The direction of a wheel vane at point of inflow should agree with that of the relative velocity i'i possessed by the water leaving the guides, in order to prevent sudden deflection of the water and consequent loss of energy and decrease of efficiency. Since the angle A\ is fixed, the vector triangle for u\, v\, V\ has all its angles determined when is known. In fact equations (II') * give u\ cotanai =cotan A\ y- cosec AI. . . . (12) Simple approximate rule. If k be assumed zero, equation (9) gives Vi =2ui cos AI, showing that the triangle whose sides are Ui,Vi, V\ is isosceles, Vi being equal to u\, and therefore (12a) This is often given as the rule for best value of a\. * Art. 179. 196 THEORY OF IMPULSE TURBINE. (B) GIRARD TURBINE WITH AXIAL FLOW. 201. General Arrangement. A Girard turbine with vertical axis and axial flow is represented in general arrangement in Fig. 96. The wheel discharges as near the surface of the tail- water as prac- ticable, and in order that it may act strictly as an impulse wheel the buckets should be provided with orifices for the free admission of air. The features which dis- tinguish this case from the pre- ceding, as regards the theory, are the equality of r\ and r 2 , and the inequality of z\ and z 2 . 202. Best Velocity. Following the same general method as in the case of radial flow, formula (V) now takes the form FIG. 96. ___ (13) which replaces equation (2). Making the same assumption as to the value of h f as in the preceding case, and also assuming that u 2 = v 2 for highest efficiency, we are led to the equation in which u is written for each of the equal velocities u\ and u 2 . This replaces equation (9) . In many cases 2122 is a small fraction of the total fall and may be neglected; but for very low falls it must be retained. 203. Efficiency. The available energy in this case includes that due to the fall within the wheel, 2i-2 2 ; therefore when this is an important fraction of the total fall, the general for- mula for efficiency is L & -IT n 1 GIRARD TURBINE WITH AXIAL FLOW. 197 and the maximum efficiency is found by using the value of L corresponding to the best speed as computed from (14). This may be written in the form W L m =[uVi cos AI - (1 +cos a 2 )u 2 ]. . . . (15) y Equations (14) and (15) correspond to (9) and (10) of the pre- ceding case. 204. Best Vane Angle. When z\z% is not negligible, the best value of the vane angle depends not merely upon the ratio of u and V\ (as in the preceding case), but upon their absolute values, since equation (14) does not give a value of u/Vi which is independent of the actual value of V\. The best value of the vane angle cannot, therefore, be determined except by assuming the ratio of z\Z2 to Vi 2 /2g. Strictly speaking, then, a given wheel cannot work with equal efficiency under different falls. This consideration is of little importance if the fall exceeds a very few feet. When u and V\ are known, a\ may be computed from equa- ion (12). EXAMPLES. 1. Take data as follows for an impulse turbine with radial outward flow: Ai =20, a? = 160, r,/r, =1.2. Assuming k =0, determine the best speed, best vane angle, and highest efficiency. Solution. The best speed is determined from equation (9), in which c = 1.2, k=Q, giving u\/Vi =.532. From equation (11) the maximum efficiency is .951. The best value of a\, determined from equation (12) with the above value of Ui/Vi, is 40. 2. Solve Ex. 1 assuming k = .2. Ans. WF,=.495; e = .890; ai=3735'. 3. With data as in Ex. 1, determine what value of k reduces the greatest efficiency to .80, and find the corresponding values of Ui/Vi and a\. Solution. This must be solved by trial. Trying k = .5, there result the values u\/V\ =.456, e = .820, ai =35 15'. Comparing with preceding results it may be estimated that the required value of k is about .6. Solving with this value, Ui/V, =.442, e = .80, a, =34 30'. 193 THEORY OF IMPULSE TURBINE. 4. Instead of the assumption w 2 = v 2 , upon which the above solution for maximum efficiency is based, it is sometimes assumed that the best velocity is that which makes A 2 =90. Show that this leads to an equa- tion like (9) with k' substituted for k, where k f = (! + &) sec 2 a 2 1. 5. Show that the assumption A? =90 leads to e=2(u\/V\) cos A\. 6. Take data as in Ex. 1 and solve on the assumption A 2 =90, for fc=0, .2, and .6. Estimate what value of k will give an efficiency of .80. Ans. For k= 0, W7,=.507, e = .962, a, =38 19'. " fc = .2, W7,=.471,e = .893, a,=36 5'. " /b = .6, wi/7, =.422, e = .798, a, =33 28'. 7. Assume dimensions as in Ex. 1, and k = .6. If the quantity of water to be utilized is 16 cu. ft. per sec., and the total fall is 600 ft., 10 per cent of which is lost by friction in the supply-pipe, determine the values of F t and 7i. Ans. 7, = 186.4. Fi = 16/186.4 = .0858 sq. ft. 8. In Ex. 7 determine highest efficiency and power. Ans. e = .80, L =432,000 ft.-lbs. per sec. H.P.=785. 9. An impulse turbine with vertical axis and axial flow is to utilize a fall of 10 ft. Take Zi-z z = l ft., Ai =26, a 2 = 156, and assume that the loss of head in the supply-pipe and guide-passages is negligible. Determine u/Vi, e, and a\, if k=Q. Ans. 7i=25.4, u = l5.5, w/7i-.61, e = .935, a, =56 40'. 10. With data of Ex. 9, find what value of k will make e = .80, and determine the corresponding values of u/V\ and a\. Ans. /c = .65, w = 13.1, w/7,=.515, e = .80, a, =48 50'. 11. In Ex. 10, if the wheel is to use 80 cu. ft. of water per sec., what must be the value of Fit What H.P. will be realized? CHAPTER XVIII. THE TANGENTIAL WATER WHEEL. 205. Introductory Illustration. Let Fig. 97 represent a sec- tion of a vane or bucket receiving a jet of water from a nozzle, and let the vane be moving in the same direction as the jet, the velocity of the water being V\ and that of the vane u. The vane consists of two symmet- rical parts coming together in a sharp edge. This edge divides the jet into two streams which follow the two parts of ! ^-i the vane, being gradually deflected and ~~vT finally discharged at its outer edges. The force exerted upon the vane by the jet. and the work done by this force per unit time may be computed by the methods explained in Chapter XIV. With the notation de- scribed in Art. 174, it is seen that U2 = ui = u, Ai=0, ai=0, Vi = Viu, while a 2 is determined by the tangent to the vane curve at the outer edge. The force constantly exerted upon the vane is equal to the change in the momentum of the water striking it in one second (Art. 158). If the two partial streams are equal, symmetry shows that the direction of the resultant force coincides with that in which the vane is moving; we therefore resolve the momentum in this direction. The momentum of W Ibs. of water before striking the vane is WVi/g, while that of an equal quantity leaving the vane is WVz/g. The component of the latter in the direction of the motion of the vane is W W V 2 cos A 2 = (u'+ v 2 cos a 2 ). y y 199 200 THE TANGENTIAL WATER WHEEL. Hence the force exerted upon the vane is W P = (Vi UV2 COS 02). y If frictional resistance be neglected, the relative velocity will remain constant during the flow over the vane, so that W and P= (1 -cos a 2 ) (7i - u). c/ The work done per second by the force P is L = Pu' } or on the assumption of no frictional loss, W L= (1 -cos a 2 ) (Vi u)u. y For a given value of W this work has its greatest value when u = Vi/2, the value being l-COSOa "" If a 2 = 180, this reduces to WV?/2g, showing that the kinetic energy of the water striking the vane is all given up to the vane. 206. General Features of Tangential Water Wheel. The tangential water wheel is designed to realize as nearly as pos- sible the ideal conditions assumed in the foregoing illustration. With this object buckets are mounted upon the rim of a wheel, and a cylindrical jet is thrown against these in such a way that each bucket, while receiving the jet, is moving as nearly as practicable in the same direction as the jet. If the water could be received and gradually deflected by each bucket without dis- sipation of energy until its relative velocity became equal and opposite to the velocity of the bucket, the energy of the jet would be wholly utilized. The practical limitations which pre- GENERAL FEATURES OF TANGENTIAL WATER WHEEL. 201 vent a close approximation to this ideal case are indicated in the following discussion. The general characteristics of wheels of this type are shown in Figs. 98 and 99. The fundamental features which distin- FIG. 98. Two views of a Doble water wheel generating 8000 H. P. under a working head of 1528 ft. guish these motors from those of the Girard type are the use of circular nozzles for " guide-passages/' and the double or " split " character of the buckets. Tangential wheel buckets have been made of various forms, the relative merits of which will not be discussed. Separate views of wheel-runners of two different makes are represented in Figs. 100 and 101. The relation of the buckets to the jet is shown in Fig. 102. Here the bucket A is receiving the jet centrally, while A' is just 202 THE TANGENTIAL WATER WHEEL. entering the jet and 'A" is nearly in the position where the jet ceases to strike it. While receiving the jet each bucket thus has a range of motion A' A" , the angle between its direction of motion and that of the jet varying by the amount A'CA". The following theoretical discussion will refer to conditions as FIG. 99. Two views of a Pelton unit consisting of two water wheels with a total capacity of 7500 H. P. under a working head of 872 ft. they exist for a mean position of the bucket. A horizontal section through the bucket and jet in this mean position is shown in Fig. 102 (B). The motion of the water relative to the bucket is not, however, in this plane, but has an upward com- ponent because of the downward component of the velocity of the bucket. This is shown by the vector triangle at (C). The angle a^ between the velocity of the bucket (u) and the relative BEST WHEEL SPEED. 203 velocity of the water leaving the wheel (v 2 ), is not shown in its true size in the figure, since neither u nor v 2 is parallel to the plane of the sectional view (B) . A mean value, depending upon the construction of the bucket, may be assumed for the angle a 2 , and the form of the bucket should be such as to make this angle as near 180 as will permit the water to clear the following bucket. 207. Best Wheel Speed. The theory of the tangential water- wheel may be expressed in substantially the same form as that of the Girard turbine, if the same notation is employed. The relative velocity v\ of the water just about to strike the bucket is computed, for a given value of u, from the vector triangle for u, i\, V\. This triangle lies in a vertical plane, and is shown at (C), Fig. 102. Thus (1) To compute v 2 the general formula (V) * is to be applied; and since practically u\ = u 2 and z\ = z 2 , the equation becomes The loss of head hf includes the loss in friction of the stream while flowing over the bucket ^surface, the loss due to impinge- ment of the jet against the sharp edge or "splitter," and the loss due to the interference of the edge of the bucket with the jet before it is in position to receive the whole stream. These losses all vary with the relative velocity. We may adopt the usual assumption that the loss is proportional to the square of this velocity, and express it in the same way as in the theory of the Girard turbine : *Art. 179. 204 THE TANGENTIAL WATER WHEEL. Equations (1), (2) and (3) then give from which v 2 may be found. FIG. 102. (4) The next step would be to substitute in the general formula (IV) for L cos cosa 2 , thus expressing L in terms of u. The resulting equation would be identical with (7) of Chapter XVII, with c = l. 208. Approximate Solution for Best Velocity. The appli- cation of the mathematical condition for maximum L leads to an equation of the fourth degree for determining u. As a simpler and sufficiently exact solution the same approximate BEST VELOCITY EFFICIENCY. 205 assumption may be made as in the case of the Girard turbine (Chapter XVII). The unutilized energy (per pound of water) is made up of the "frictional " loss k(v 2 2 /2g) and the kinetic energy possessed by the water as it leaves the wheel, V 2 2 /2g. While the former is the more important, the latter probably varies more rapidly as the speed varies from that giving greatest efficiency, so that the highest efficiency will be obtained when V 2 is small, and this will be when v 2 and u are nearly equal. Assuming v 2 = u, ........ (5) equation (4) becomes 0, .... (6) from which u/V\ may be computed. 209. Efficiency. Taking as the available energy the kinetic energy of the jet from the nozzle, the hydraulic efficiency is Putting V2 = u, s 2 = u + v 2 cos a 2 = w(l+cos a 2 ), the highest effi- ciency according to the foregoing solution is [u v? "1 y-cosAi-(l+cosa 2 )y-^, ... (8) in which u/Vi must have the value given by equation (6) . 210. Form of Bucket Surface. While a bucket is receiving the jet it moves through a certain distance in a direction oblique to that of the jet and turns through a small angle. The point at which the jet strikes it, and the direction of the relative motion of the impinging water, are therefore variable, and the form of the bucket surface should conform to this variation. Without entering into a detailed study, it may be said that the form should be such as to deflect the water as gradually as pos- sible throughout its entire passage over the bucket surface, and to make the direction of the relative velocity of outflow as 206 THE TANGENTIAL WATER WHEEL. nearly opposite to the velocity of the bucket as practicable. The varying aspect presented by the bucket to the stream seems to require a surface of double curvature if sudden changes of velocity are to be avoided. The dividing edge should, of course, be as sharp as practicable. The limitation of the direc- tion of outflow imposed by the necessity that the water leaving one bucket shall clear the next one has already been mentioned. 211. Comparison with Theory of Girard Turbine. The for- mulas above deduced for the tangential water wheel are seen to be identical in form with those obtained for the radial-flow Girard turbine except that c = l. They may also be obtained from the formulas applying to the axial-flow Girard wheel by putting Zi=z 2 . 212* Conditions Favorable for Use of Tangential Water Wheel. In general it may be said that wheels of this type may be used to advantage wherever the fall utilized is great and the supply of water relatively small, and are at a decided dis- advantage only when the fall is but a few feet, or when the supply of water is great. They are in efficient operation under falls as great as 2500 ft. 213. Actual Efficiencies. While accurate tests are rare, it is probable that the best tangential water wheels give effi- ciencies as great as are obtained from any type of turbine or water wheel. Hydraulic efficiencies as great as 80 to 85 per cent under heads up to 1500 or 2000 ft. are doubtless not uncommon. The extremely high efficiencies sometimes claimed by manufacturers must be regarded with suspicion. 214. Regulation. An important problem in connection with the practical working of a water wheel is the regulation of the discharge from the nozzle in order to vary the power output of the wheel, or to conform to variations in the supply of water. The partial closing of a valve of any ordinary form placed in the supply-pipe or nozzle would cause a serious loss of energy. In some cases two or more nozzles are placed at different points GOVERNING. 207 of the circumference of the wheel, any combination of which may be used as required. The needle nozzle (Fig. 103) is a regulating nozzle of peculiar FIG. 103. form, designed to vary the size of the jet at pleasure with little loss of efficiency. The conical valve or "needle " can be moved parallel to the axis of the jet, so as to leave any desired amount of opening between the needle and the nozzle tip. If carefully made, this device accomplishes the object of regulation with little loss of energy very satisfactorily.* 215. Governing. Another important problem is that of maintaining a uniform speed of rotation when the load fluctu- ates. This is usually accomplished by deflecting the jet so that a varying portion of it strikes the buckets, the deflection being permitted by a joint in the nozzle pipe, and the movement being controlled by some form of centrifugal governor. Governing has also been successfully accomplished by means of the needle nozzle, the needle being actuated by the governor so that the nozzle opening varies with any variation in the speed. EXAMPLES. 1. If Ai =20 and a 2 = 160, determine best value of u/Vi and highest efficiency assuming /c=0. Ans. u/V\ =.532. e = .966. 2. Solve Ex. 1 assuming k = .5. Ans. u/V = A73. e = .860. 3. In the same case, if the maximum efficiency is .80, what is the value of k, and what is the best wheel velocity? 4. If the effective head at the nozzle is 600 ft. and the wheel is to make 700 R.P.M., what should be its diameter? 5. If the nozzle throws a jet 1.5 inches in diameter, determine the power. * The needle nozzle is covered by U. S. patent. CHAPTER XIX. THEORY OF THE REACTION TURBINE. 216. General Features of Reaction Turbines. The funda- mental characteristic of a reaction turbine is the fact that the wheel passages are completely filled by the streams flowing through them. The pressure within these passages, therefore, is not determined by contact of the stream with air as in the impulse wheel, but in general varies continuously between the points of admission and discharge. The general arrangement of a reac- tion turbine with radial outward flow (the Fourneyron type) is shown in Fig. 92, while the case of axial flow (Jonval) is represented in Fig. 104, and that of inward flow (Francis type) in Fig. 105. The turbine shown in Fig. 105 does not closely resemble the original design of Francis, being adapted to a much greater fall. A design quite generally adopted in American practice is that of mixed flow (inward admission and axial discharge). This is represented in Fig. 106. There is no sharp distinction between this and the form represented in Fig. 105. Turbine runners of two forms are shown in Figs. 107 and 108, the former being of the inward-flow or Francis type, the latter of the mixed-flow type similar to those in Fig. 106. 208 FIG. 104. FIG. 100. PELTON WATER-WHEEL RUNNER. FIG. 101. DOBLE WATER-WHEEL RUNNER. FIG. 107. RUNNER OF FRANCIS TURBINE (PLATT IRON WORKS COMPANY). FIG. 108. RUNNER OF VICTOR TURBINE (PLATT IRON WORKS COMPANY). GENERAL FEATURES OF REACTION TURBINES. 209 FIG. 105. Modern Francis turbine, of capacity 5000 H. P., under a fall of 320 ft. Built by the Allis-Chalmers Company. FIG. 106. A pair of turbines installed by the I. P. Morris Co. The work, ing head is 82 ft., the speed 600 revolutions per minute, and the power gener- ated 750 H.P. 210 THEORY OF THE REACTION TURBINE. The following theory as at first presented refers to the Four. neyron or outward- flow turbine, as represented in Fig. 92. The main features of the theory hold for reaction turbines of other forms, the points of difference being indicated later. 217. Data and Notation. In the following discussion of reaction turbines the notation will for the most part be the same used in the foregoing theory of impulse wheels and ex- plained in Art. 174. In addition the following symbols will be used. Assuming the wheel to discharge into the atmosphere,* let h denote the total fall from surface of supply reservoir or head- race to the place of discharge from the wheel, and let h' denote the total loss of head (i.e., the part of h that is not utilized). In applying to the reaction turbine the general formulas deduced in Chapter XV, the special condition must be intro- duced that the cross-section of the stream within the wheel is everywhere fixed, not varying with the wheel speed nor with the velocity of flow. Thus in the equation of continuity (formula (I'), Art. 179) both FI and /2 are constant, so that v% and V\ are in a constant ratio. That is, ........ (1) in which c' = FI // 2 = constant .f The following data will- be taken as known: h, A\, a^ c? = Fi/f'2, c = r 2 /ri. 218. Relation between Wheel Speed and Rate of Flow through Wheel. The velocity of flow in every section of the stream is * If the wheel is submerged, h will mean the fall from surface of supply water to surface of waste water; the same formulas will then apply as in the case of discharge into the air. The effect of a suction-tube is considered in Art. 228. fit will be noticed that the condition Fi// a = constant in the reaction turbine replaces the condition p\ = p z in the impulse turbine. Formula (V), which was used in determining the flow through the wheel in the previous case, is of use in the present problem only in determining the relation between p l and p2. RATE OF FLOW THROUGH WHEEL. 211 determined if that in any one section is fixed. Let all such velocities be expressed in terms of FI, and let it be required to determine the relation between Vi and the wheel speed HI. From the above meanings of h and h' it is seen that h h' is the utilized head, and that the energy received by the wheel from the water per unit time is L = W(h-h') (2) Hence from formula (IV), Art. 179, W W(hh')= (t*i$j[ U2S 2 ) , tj or g(hh')=UiSiu 2 s 2 (3) The second member of this equation can be expressed in terms of YI and u\. Thus u 2 = cui ; Si = Vi cos AI ; S 2 = u 2 + v 2 cos a 2 = c so that (3) may be written g(h h') = (cosAicc? cosa 2 )ViU L c 2 u-f. . . (4) It is now necessary to express the value of h'. The most important losses of head may be expressed by two terms. One of these represents the "frictional " loss occurring throughout the stream, which by the usual rule of hydraulics is expressed as proportional to the square of the velocity of flow. Since the velocities in all sections vary in the same ratio, we may assume o k-~- = total frictional loss of head, ^9 k being a coefficient whose value depends upon the dimensions and character of the entire series of passages, and which will 212 THEORY OF THE REACTION TURBINE. be treated as constant. The other important loss of head is that represented by the kinetic energy of the water leaving the wheel. Combining this with the frictional loss, we may write From the vector triangle for u 2 , v 2 , and F 2 , V 2 2 = U 2 2 + V2 2 + 2U 2 V 2 COS a 2 = c 2 u l 2 + c' 2 V 1 2 +2cc'u l V l cos a 2 , so that (5) may be written in the form 2gh' = (1 + k) c'WJ + 2cc f cos a 2 V l u l + c 2 uj. . . (6) Eliminating h' between (4) and (6), . . (7) By solving this equation Vi may be determined for any value of MI. For a given wheel, FI being fixed in value, the rate of discharge varies as V\, being given (when V\ is known) by the equation W^wFtVi. ....... (8) 219. Power and Efficiency for any Wheel Velocity. Ex- pressing UiSi u 2 s 2 in terms of u\ and V\ as in Art. 218, the general formula (IV) may be written W L= [(cos AI -cc' cos a 2 )ViUi -c 2 Ui 2 ]. ... (9) From this, by using the value of FI given by (7), the power corresponding to any value of u\ may be computed, provided FI is known so that W can be determined. The efficiency, however, is independent of the value of FI. For W Ibs. of water the available energy is Wh foot-pounds, hence e = rr = ~[(cos AI - cc' cos a 2 ) ViUi - c%i 2 J, . (10) BEST SPEED AND HIGHEST EFFICIENCY. in which V\ must have the value given by (7) for any value f 220. Best Speed and Highest Efficiency. The speed giving maximum efficiency may be determined by applying the con- dition de/dui=Q. The direct method of procedure would be to solve equation (7) for V\ and substitute its value in (10), thus expressing e explicitly in terms of the single variable HI before differentiating. The following indirect method is less laborious. Differentiating (10), . de . . dV, gh ~-j = (cos AI cc cos a2)Uij + (cos AI -cc' cos a 2 )Vi 2c 2 Wi =0. Differentiating (7), + cos AI wj-r^-f cos ^i - 7i -c 2 wi =0. Eliminating dVi/dui between these two equations, and reduc- ing, (l+&)c' 2 (cos Ai-ccf cos a 2 )V l 2 -2(l+k)c 2 c f2 V l u 1 -c 2 (cos AI +cc r cos a 2 )wi 2 =0. (11) This equation must be satisfied when e is a maximum. Com- bining it with (7) , which is always true, the values of u\ and V\ corresponding to maximum efficiency may be found. The solution may be expressed as follows : Equation (11) determines * the value of Vi/u\. Calling this a, and substituting Vi = aui in (7), [(!+A;)c r2 a 2 + 2cosA 1 .a:-c 2 ]wi 2 =2^, . . . (12) from which may be computed the value of u\ giving maximum efficiency. * Two values of V\/u\ are given by (11); which value corresponds to the practical problem is seen in any particular case after substitution in (7). 214 THEORY OF THE REACTION TURBINE. 221. Best Angle of Wheel Vane. The direction of the wheel .vane at the point of inflow should agree with that of the rela- tive velocity of outflow from the guide passages, in order to prevent sudden deflection of the water and consequent loss of energy. The value of ai is fixed by that of VI/MI, and may be determined in the usual manner by solving the vector tri- angle whose sides are MI, Vi, Vi. Equation (12) of Art. 200 may be employed : i. . . . (13) Since u\/V\ varies with the wheel velocity, ai also varies with u\. The value corresponding to maximum efficiency should govern the design of the wheel. In the foregoing theory the loss of head due to sudden deflection of the stream entering the wheel is neglected, which is equivalent to assuming that the vane angle varies with the speed of rotation so as always to agree with ai. This is allow- able in solving for maximum efficiency, since the vane angle is to be made to agree with the value of ai which is finally found as the result of the solution. For a given wheel, however, there is only a single value of Ui/Vi which makes the assumption of no sudden deflection of the stream true, and the loss of head due to this cause should be taken into account in an accurate solution of the problem of flow through the wheel for any rotation speed. In spite of this, however, it is probable that equation (7) shows approxi- mately the way in which V\ varies with Ui in an actual wheel, for a considerable range of values of MI. 222. Pressure at Entrance to Wheel. If hi denotes the fall from head-race to point of outflow from guides, and H f the loss of 'head between these points, the equation of energy gives pi being reckoned from atmospheric pressure as zero. INTRODUCTION OF RATIOS OF VELOCITIES. 215 EXAMPLES. Given A^ =28, a, = 158, n =3.375 ft., r 2 =4.146 ft., F, =6.537 sq. ft., / 8 = 7.687 sq. ft., h = 12.8 ft. 1. Assuming fc = .5, determine best wheel velocity, best value of vane angle, and highest efficiency. Ans. w,=.50\ / 2^=28.7ft. per sec.; e = .71; a, =61 45'. 2. By trial of different values of k, determine what value gives a maximum efficiency of 80 per cent. Ans. A; = .25 very nearly. 223. Introduction of Ratios of Velocities. Let m denote the velocity equivalent to the fall h, i.e., then it is seen that the main equations reached by the above theory really involve the ratios of the three velocities Ui, V t , m, rather than their actual values. Thus, comparing different cases in which h has different values, if HI, Vi and m are changed in the same ratio, equations (7), (10), and (11) are unchanged* It is instructive to write the equations so as to involve the ratios of the three velocities explicitly. Let Ui/m=x, Vi/m = y. Then equations (7), (10), and (11) become osA l -yx-c 2 x 2 = l, ... (A) e = 2 (cos A i - cc' cos o 2 ) yx - 2c 2 x 2 , . . . (B) (1 +&X 2 (coB AI -cc' cos a 2 )y 2 -2(1 +k)c 2 c' 2 yx - c 2 (cos AI +cc* cos a 2 ).x 2 =0. (C) Of these equations (A) and (B) are general, while (C) holds only for maximum efficiency. The values of the rate of discharge and of the power depend not merely upon ratios of velocities, but upon their actual values. They may be written as follows: (D) ^ ey. . . . (E) 216 THEORY OF THE REACTION TURBINE. 224. Graphical Representation. If x and y are taken as rectangular coordinates, equation (A) represents a curve which shows graphically the way in which the rate of discharge varies with the wheel velocity for any constant fall; for when m in constant, Vi is proportional to y and HI to x. The value of e for any value of x and the corresponding value of y may be computed from (B). If a curve be drawn with e as ordinate and x as abscissa, it shows graphically the 1.00 .80 .60 .40 .20 (E) 20 Values .40 .60 FIG, 109. .80 1.00 variation of the efficiency with the wheel speed when h is con- stant. The variation of the power with the wheel velocity when h is constant is shown by a curve determined by taking as abscissas values of x and as ordinates values of ey, the variable factor in the value of L as given by (E) . These three curves, for the data of the examples after Art. 222, with A; =0.25, are shown in Fig. 109. APPROXIMATE SOLUTION FOR MAXIMUM EFFICIENCY. 217 Equation (A) represents an hyperbola, while (C) represents two straight lines, one of which intersects the hyperbola in a point whose coordinates are the values of x and y for maximum efficiency. These curves cannot, of course, be regarded as accurately representing the results actually to be expected from a turbine of given dimensions. The uncertain element in the theory is the value of the loss of head h' '; in particular the remarks made in Art. 221 regarding loss at entrance must be borne in mind. It should be noticed that the value of e given by equation (B) involves no constants except dimensions of wheel and guides, and that this equation is not affected by the uncer- tainty in the value of h f . If the true relation between x and y were known (as, for example, by experiment), the efficiency curve could be correctly drawn from equation (B). 225. Approximate Solution for Maximum Efficiency. Reasoning as in the case of the impulse turbine (Art. 198), it appears that the greatest efficiency will result when the absolute velocity of outflow from the wheel has a small value. The assumption ^2 = ^2 will therefore give an approximation to the solution for maximum efficiency. The more common assump- tion, however, is that ^4.2 = 90, or V 2 cosA 2 = u 2 + V2 cos a 2 -=0 ..... (14) Either of these assumptions leads to a simple equation in place of (C). Thus, from (14), or cx+c' cos a 2 -2/=0, ...... (C') which replaces (C). The solution is otherwise unchanged. In the case represented by the curves in Fig. 107 the ap- proximate solution for maximum efficiency gives practically the same result as the exact solution. The value of y/x or a given by (C') is almost identical with that given by (C); both these equations are represented by the straight line marked (C), and 218 THEORY OF THE REACTION TURBINE. the intersection of this line with the curve (A) gives the values of x and y corresponding to maximum efficiency. EXAMPLES. 1. Using the results of the exact solution of Ex. 1, Art. 222, deter mine the value of A 2 . Ans. 92 30'. 2. With same data, assume ^4.2=90, and determine x, y, e, ai. Ans. 3 = .51, 2/ = .79, e = .71, a,=Q2 50'. 226. Cases of Inward, Axial, and Mixed Flow. All these cases are covered by the foregoing theory, so long as the wheel discharges directly into the air or below the surface of the tail- race. With axial flow c = l, with inward flow or mixed inward and axial flow c, the pressure in the chamber X will vary according to the law already given. If r is the radius of the inner chamber (practically equal to that of the wheel), the pressure at any point of the cylindrical bounding surface of the chamber X will exceed that at a point in the axis of rotation at the same level by the equivalent of a water column of height r 2 a 2 /2g. Pressure is communicated to- the water in the outer chamber through the connecting orifices, and a condition of equilibrium will be reached in which the pressure within the chamber Y and the vertical pipe follows the hydrostatic law. At any point in this outer chamber the pressure will equal that at points on the same level at the outer boundary of the inner chamber. When equilibrium exists in the outer chamber the pressure conditions in the different parts of the apparatus are thus seen to be the following: (a) Throughout the reservoir R and the connecting pipe, and at all points of the chamber X which lie in the axis of rota- tion, the pressure increases with the depth below the surface of the reservoir according to the hydrostatic law. (6) Through- out the chamber Y and the connecting pipe the pressure varies according to the hydrostatic law; but the pressure at any point exceeds that at points on the same level in the reservoir 72 by the equivalent of a water column of height r 2 w 2 /2g, or u 2 /2g (u being the linear velocity of the outer ends of the wheel- blades) . Evidently, therefore, water will rise in the pipe to a point Q, such that PQ = u 2 /2g. Let the open end of the pipe be at a height h above P. If the velocity of rotation of the wheel be such that water will rise to the open end of the pipe; and if the velocity TRANSFORMATIONS OF ENERGY. 227 of rotation exceeds this value, water will flow out. In this way a continuous stream can be caused to flow from the reservoir, discharging at the open end of the pipe; the possible height of the lift being proportional to the square of the wheel velocity. We thus have a centrifugal pump. It is not necessary that the pump shall be lower than the surface of the supply reservoir; but if it is higher, some means must be provided for filling the chambers of the pump and the pipe leading from the reservoir, and to prevent the water from running back before the wheel velocity becomes great enough to sustain it. This may be accomplished by a foot- valve placed in the supply pipe, so arranged as to open only in the direction of flow. Neither is it necessary that the axis of rota- tion be vertical. 234. Transformations of Energy. In such an apparatus as that shown in Fig. 112 energy must constantly be supplied to the wheel in order to maintain the uniform rotational velocity. This energy is -given up to the water by the action of the wheel- blades. The particles of water next to the paddles, which receive energy directly from the wheel, pass it on to the neigh- boring particles, and through these it is transmitted to the particles remote from the moving blades. Since there will always be some relative motion of the particles of water among themselves, some energy is dissipated into heat by reason of internal friction in the water. Another portion of energy is dissipated by reason of the friction between the rotating body of water in the chamber X and the walls of the chamber. Such losses occur even if no flow takes place through the wheel. If the wheel velocity is such that flow takes place, other losses of energy occur. There is evidently a frictional loss such as always accompanies the flow of water. Also, since a particle flowing through the apparatus suffers sudden changes of velocity, there is a loss due to this cause. Following a particle from the point where it enters the wheel chamber to a point in the dis- charge-pipe, it will be seen in a general way how its velocity varies. In moving from the axis of rotation to the circum- 228 TURBINE PUMPS. ference of the chamber X, its velocity will be changed gradually from a small value to a value nearly equal to that of the outer ends of the paddles. As it enters one of the orifices leading into the outer chamber it is suddenly deflected; and on enter- ing this chamber its velocity is nearly destroyed. These losses of energy are so important that such a crude pump as that represented in Fig. 112 would have a very low efficiency. 235. Design for Efficient Pump. In Fig. 113* are shown the main features of a pump designed to reduce, as far as prac- FIG. 113. ticable, the losses of energy by dissipation into heat. The axis of rotation is horizon tal,t (A) being a section by a plane con- taining this axis, and (B) a section by a plane perpendicular to the axis. The rotating wheel is formed with continuous passages, bounded by curved surfaces in such a way as to cause a gradual deflection of the water instead of a sudden deflection. The vanes which separate the passages are curved "back- ward," i.e., in such a way that the relative velocity of a par- ticle of water when it reaches the outer circumference of the * This figure shows the essential hydraulic features of the Risdon-Sulzer pump. t The direction of the axis is not, however, essential. RELATION BETWEEN LIFT AND WHEEL SPEED. 229 wheel is directed as nearly as possible opposite to the velocity of the perimeter of the wheel at that point; the object being to make the absolute velocity of outflow as small as possible. If it were possible to save all the kinetic energy possessed by the water as it leaves the wheel, the direction of the absolute velocity of outflow would be immaterial, so far as efficient working is concerned; it might be advantageous to make the angle a 2 as small as 90, or even smaller, since by this means the rate at which the wheel would give energy to the water (when running at a given speed) would be increased. The reason for avoiding a high velocity of outflow from the wheel is that an important fraction of the kinetic energy due to this velocity is necessarily lost. The absolute velocity of outflow must, however, have a forward tangential component, since the head-equivalent of the energy imparted by the wheel to the water is proportioned to this component and to the speed of rotation.* It is important to conduct the water from the wheel to the receiving chamber in such a manner as to reduce the velocity of flow as gradually as possible. This is the aim of the design shown in Fig. 113. The water passes from the wheel into passages conforming to the direction of its absolute velocity just before leaving the wheel, and these passages enlarge very gradually to their place of discharge into the receiving chamber. 236. Relation between Lift and Wheel Speed. It was shown above (Art. 233) that the height of the column of water that can be sustained by the rotation of the wheel while no flow takes place is u 2 2 /2g, where u 2 is the velocity of the outer ends of the wheel-blades. When flow occurs, however, this i elation no longer holds. Energy is continually given up by the wheel to the water, and this results in a gain of effective head which may take the form of an increased lift. If h m is the head equivalent to the energy imparted to the water by the wheel (i.e., the energy given up by the wheel per pound of water discharged); h f the total loss of head by reason of dis- sipation of energy between supply reservoir and discharge * This follows from formula (IV), as will be shown below. 230 TURBINE PUMPS. reservoir; HI and H 2 the values of the effective head at sur- faces of supply reservoir and discharge reservoir respectively; the general equation of energy gives (Art. 96) H 2 -Hi=h'"-h'. The lift H 2 -Hi is thus equal to the head-equivalent of the energy supplied by the wheel diminished by the frictional losses of head. If a wheel is so designed and operated as to impart energy to the stream efficiently (i.e., with small loss by dissipa- tion), the lift maintained while flow occurs may be materially greater than that due to rotation when there is no flow. This is borne out by experience, some of the best pumps being found to give a lift greater than u 2 2 /2g. 237. Notation for Mathematical Theory. In the following mathematical theory the notation used in the theory of the reaction turbine will be employed, with the following modifica- tions. We now have to deal with a lift instead of a fall, and with energy given up by wheel instead of energy received by wheel from water. Hence we shall put L = energy given up by wheel per second; h = total lift from surface of supply reservoir to surface of discharge reservoir. No change is made in the notation for absolute and relative velocities and their direction angles. It is, however, found convenient to express the equations in terms of u 2 and v 2 , instead of ui and Vi as in the former theory. The discussion will refer to the case in which the admission and discharge occur substantially as in Fig. 113. The absolute velocity V\ thus means the velocity of flow in the suction-pipe close to the wheel, and FI means the cross-section at this point e Since V\ is parallel to the axis of ro- FIQ tation, ^i=90 and cos Ai=0. This greatly simplifies the main fundamental equations. ..1 FUNDAMENTAL EQUATIONS. ^ 231 The direction of flow in the passages leading from the wheel to the receiving chamber depends upon the form of the pas- sages. For the purpose of the mathematical theory it is necessary to idealize the conditions of flow within these pas- sages by assuming that the direction angle of the velocity of a particle immediately after leaving the wheel has the same value for all particles. This angle, measured as usual from the direc- tion of u 2 , will be called A 2 '. It must be remembered that A 2 is the direction angle of V 2 , which is the absolute velocity of a particle just before leaving the wheel. For a given pump A 2 and A 2 can be equal only for some one ratio of wheel velocity to lift velocity. It is desirable that they should be equal when this ratio has the value under which the pump is designed to operate, since if they are unequal a sudden change of velocity occurs which results in loss of energy. 238. Fundamental Equations. The energy given up by the wheel to the water is expended in two ways: in lifting the water a height h, and in overcoming the wasteful resistances represented by the lost head h'. Hence L = W(h+h') (1) The value of L may also be expressed by a formula similar to (IV) of Art. 179. In fact the reasoning leading to that formula (Art. 177) is strictly applicable to the case in which the energy given to the wheel by the water is negative. But with the present meaning of L we must change signs and write W L= y Or, since now i = Vi cos AI = 0, W we have L = u 2 S2 (2) y But s 2 = u 2 + v 2 cos a 2 ; 232 TURBINE PUMPS. W hence L= ^2(^2 + ^2 cos a 2 ) ...... (3) t7 Equations (1) and (3) now give g (h +h f ) =u 2 (u2~\-V2 cos c^) ..... (4) Equations (3) and (4) are the fundamental equations upon which the following theory is based. Equation (2) shows that if L has a positive value, s 2 must be positive; thus justifying the statement made in Art. 235 that the water must leave the wheel with a velocity having a forward tangential component. 239. Loss of Head. To complete the theory it is necessary to estimate the value of h'. Doubtless the most important losses of head, in a pump working at its best speed for the exist- ing lift, are of the kind included under the term friction losses. These depend upon the velocity of flow and the character of the passages, and may be expressed by means of a single term in accordance with the usual assumption regarding such losses. If there is a sudden change in the direction of the velocity of particles leaving the wheel, this causes a loss which cannot be expressed by a term of the above form. The passages into which discharge occurs should, if possible, be so formed that such sudden deflection of the water does not take place. This condition cannot be satisfied except for a particular relation between speed and lift, which should be that for which the pump is designed to work. In discussing the question of design for highest efficiency, it will be instructive to give first a solu- tion of the problem assuming that no sudden deflection of the outflowing stream occurs.* Although the resulting equations will not show the working of an actual pump for different rela- * This corresponds to the solution given in Art. 218 for turbine motors. DESIGN FOR HIGHEST EFFICIENCY. 233 tions of speed to lift, they should agree well with actual con- ditions for the case which is to govern the design. The question of efficiency curves and discharge curves for a given pump will be considered afterward. 240. Design for Highest Efficiency. Substituting in equa- tion (4) we obtain 2u 2 2 + 2u 2 v 2 cos a 2 - kv 2 2 = 2gh, .... (5) from which v 2 and the rate of discharge can be computed for any wheel velocity. The value of the efficiency may be expressed as follows: Since the energy utilized per second is Wh, while the energy supplied by the wheel to the water is W(h+h'), the hydraulic efficiency is Wh h ~W(h+h')~h+h'> . () or from (4) , e = , - -^ - , . (7) ^2(^2 + ^2 cos $2) The efficiency for any wheel speed is found by using in (7) the value of v 2 in terms of u 2 determined from equation (5). The mathematical condition for maximum efficiency is de/du 2 = Q. The solution may be carried out by the method employed in Art. 220. In the present problem, however, the following method is shorter. v 2 Since h f = k-^- , the value of e in (6) may be written *9 2gh This equation shows that, for any given value of h, e decreases continually as v 2 increases, having its greatest value when v 2 = 0. The maximum value of e is e m = l, ........ (9) 234 TURBINE PUMPS. while the corresponding value of u 2 , found by putting v 2 =Q in (5), is u 2 =Vgh, ....... (10) The angle of outflow of the water from the wheel is A 2 =0, . . . ..... (11) as may be seen from the relation between the three vectors representing u 2 , v 2 , V 2 . Thus, since always ], ...... (12) the condition v 2 =Q gives [V 2 ]=[u 2 ] ........ (13) The meaning of these results is more easily seen from a graphical representation. Let m=V2gh, x = u 2 /m, y = v 2 /m. Equations (5) and (7) may then be written (14) Taking x and y as rectangular coordinates, equation (14) represents an hyperbola whose center is at the origin. A par- ticular case of this curve is represented at (A) in Fig. 115, which shows only the part of the curve for which y is positive. One asymptote of the hyperbola is shown in the figure. Taking values of e as ordinates of a curve of which the abscissas are the corresponding values of x, the result is shown at (B), Fig. 115. For a constant value of the lift the rate of discharge varies as y, and the speed as x. Hence curves (A) and (B) show how DESIGN FOR HIGHEST EFFICIENCY. 235 the rate of discharge and the efficiency vary with the speed when the lift remains constant. It must be remembered that these results do not apply to a given pump working at different speeds, but show the com- parative results of an assumed series of pumps running at differ- ent speeds, assuming that in every case the absolute velocity 1.00 Values of x. 1.0 FIG. 115. 2.0 of outflow receives no sudden change, but that the flow from the wheel to the receiving chamber occurs with equal efficiency for all. S.trictly interpreted, the conclusion is that the case of maxi- mum efficiency is that in which the angle A 2 is zero, the cor- responding wheel velocity being Vgh or .707m, and the rate of discharge zero. As the ratio u 2 /m or x increases, the effi- ciency decreases and the rate of discharge increases, the angle A 2 also increasing. The practical conclusion is that the case of maximum effi- ciency is an ideal limit toward which the design should approxi- mate, but which it can never reach, and that the speed of work- ing should be such as to give a relatively small value of the dis- charge angle A 2. But in order to produce a given rate of discharge with as small a pump as practicable, it may be advis- able to let the working speed increase materially above the value corresponding to the ideal case of maximum efficiency. 236 TURBINE PUMPS. 241. Losses of Head in a Given Pump. In an actual pump wor Icing under different relations of speed to lift, the entire loss of head cannot be expressed by a term of the form assumed above. In addition to the " friction" losses there are losses due to sudden deflections of the stream as it enters and as it leaves the wheel, such deflections depending not upon the velocity of flow alone, .but also upon the speed of rotation. Since the water enters the wheel near the axis of rotation where the wheel speed is relatively small, the deflection of the stream due to lack of exact adjustment of vanes to stream velocity at this point is probably of small importance. Sudden deflection at the point of outflow from the wheel would appear, however, to be an important cause of loss of head. While no exact evaluation of this loss can be made on theoretical grounds alone, the following considerations lead to an expression which probably represents actual conditions at least approximately for a pump discharging in the way shown in Fig. 113. It may be noticed first that in the limiting case of no dis- charge (02 = 0) we ought to have h' = u 2 2 /2g. For in this case the gain of effective head due to the action of the wheel (which is always the value of L/W given by (3)) reduces to u 2 2 /g, while the actual lift has the value u 2 2 /2g by Art. 233, so that the lost head is the difference between these values. That u 2 2 /2g is a reasonable value for h f in this limiting case is seen also by considering what occurs when there is a very slight discharge. A particle just about to leave the wheel has absolute velocity u 2 , while just outside the wheel it enters a body of water at rest,* so that its kinetic energy is wholly lost. * Of course this is not an accurate statement of actual conditions, since the water in the passages just outside the wheel would be set in motion by the friction of the rotating body. With such a construction as that shown in Fig. 113, however, this water is prevented from taking up the motion of the wheel periphery, and the statement that the velocity of the particle leaving the wheel is wholly destroyed is practically true. The assumption above made as to the loss of head is in accordance with the usual rule for estimating loss of head due to a stream entering a body of water at rest. (Art. 82.) EQUATIONS FOR RATE OF DISCHARGE AND EFFICIENCY. 237 Considering now the general case in which u 2 and v 2 have any values, we will make the ideal assumption referred to in Art. 237, that the direction angle of the absolute velocity of a particle of water just after leaving the rotating wheel has the same value for all particles, this value being fixed by the form of the passages leading to the receiving chamber. In other 'words, it will be assumed that the absolute velocity changes from a value V 2 with direction angle A 2 to a value V 2 with direction angle A 2) and that the latter angle has a fixed value, while the former varies with u 2 and v 2 . In Fig. 116, AB, BC and AC represent u 2 , v 2 and V 2 respectively; while EAC' is the fixed angle A 2 , and V 2 is represented by some vector AC' '. The absolute velocity thus changes from AC to AC' as the particle leaves the wheel. It will be assumed that the loss of head due to this sudden change is (CC') 2 /2g. The value of CC r is to be expressed in terms of u 2 and v 2f Since with proper construction (the thickness of the wheel- passages being equal to that of the passages into which the discharge occurs) the component of V 2 and that of V 2 per- pendicular to u 2 are equal, CC' is parallel to AB, and its value is easily found to be sin(a 2 -A 2 ') sin A 2 = u 2 -k'v 2 , . . . (16) if k' is written for the constant sin (2 -A 2') The total loss of head may now be expressed by the equation 2g (17) 242. Equations for Rate of Discharge and Efficiency. Substituting the above value of h' in equation (4), the latter 238 TURBINE PUMPS. takes the form . . (18) The value of the hydraulic efficiency in terms of u 2 and v 2 , as given by equation (7), is not changed, but in applying it v 2 must have values satisfying (18). The equation is cos a 2 )' (19) Introducing ratios of velocities, x and y having meanings as in Art. 240, the above equations become y 2 = l; . . . (20) 243. Condition of Maximum Efficiency. The values of x and y for which e is a maximum may be found by applying the condition de/dx = 0. Writing (21) in the form cosa 2 4O and differentiating, 1 de , n v . d?/ . -^^- = (2x + ycosa 2 )+xcosa 2 --j-=Q. Differentiating (20), [x + (k f +cos a 2 )y] +[(k' +cos a 2 )x - Equating values of dy/dx given by these two equations, and reducing, = o. . . . (22) GRAPHICAL REPRESENTATION FOR CONSTANT LIFT. 239 This equation determines two values of y/x, one of which gives a real solution of the problem. Calling this value a, and putting y=ax in equation (20), the result is ]x 2 = l, . . . (23) which gives the value of x for maximum efficiency. 244. Graphical Representation for Case of Constant Lift. When the lift remains constant, x varies directly as the wheel speed, and y as the rate of discharge. Hence the curve repre- sented by (20), when x and y are made rectangular coordinates, shows how the rate of discharge varies when the wheel is run at different speeds under a constant lift. The relation between efficiency and wheel speed under the same conditions may be represented by a curve of which the ordinate and abscissa of any point are corresponding values of e and x. The value of e for any value of x is to be computed from (20) and (21). A third curve of importance is that showing the relation between wheel velocity and power. This relation is expressed by equation (3), or more conveniently by the equivalent equa- tion Wh Since W varies with varying speed, we substitute for it its value, W = wq = wf 2 v 2 = wf 2 my, with the result T w} 2 myh wj 2 m 3 y wf 2 m s , L = = -^ - = -75 I . . . (24) e 2g e 2g The variable factor in this value of L is y/e, which has been represented by 1. If I be made ordinate of a point whose ab- scissa is x, the locus of this point as x varies is a curve showing the way in which the power varies with the wheel velocity. 240 TURBINE PUMPS. The equations for the case of constant lift will for convenience be summarized below. With them will be included (22), which applies to the case of maximum efficiency, and which is seen to represent two straight lines, one of which intersects the general x-y curve in the point whose abscissa is the value of x for maximum efficiency.* l. . . f (A) 2x(x+ycosa 2 )' - a2 x 2 = <). . ^ . (Q (D) _ wf 2 m? y n , N 1 L= ' --2*2/(*+2/cosa 2 )=--Z. . (E) Special case. For a given wheel the angles a 2 and A 2 are known, and k' can be computed from them. The friction factor k cannot, however, be known apart from experiment. Consider a pump for which Equations (A), (B), and (C) become l', . . . (25) e== 2x(x-M6y)' ...... (26) 7 ^5 2f> -2.31*2, + ^3-7^=0 ..... (27) * It will be noticed that there is a correspondence between these equa- tions and those given in Art. 223 for the reaction turbine. GRAPHICAL REPRESENTATION FOR CONSTANT LIFT. 241 It appears that if k is small in comparison with 13.70, it has little influence on the form of the x-y curve. Fig. 117 shows the curves obtained by assuming* &=0. This reduces (25) and (27) to the forms (28) (29) while (26) is unchanged since it does not involve k explicitly. l.O IE), (A) (B) Values of x 1.0 FIG. 117. 2.0 Equation (28) represents an hyperbola whose center is at the origin of coordinates. The curve cuts the z-axis in the point x = l, and the asymptotes are the lines = .547z, The coordinates of the vertex are = .795, t/ = .148. A remarkable feature of the curve is that for a certain range of values of x there are double values of y\ i.e., to each value * It is evident that this assumption makes the maximum efficiency 1, since h' will reduce to when A 2 =A 2 ', 242 TURBINE PUMPS. .2 of the wheel velocity there correspond two values of the rate of discharge. Although this may at first sight seem anomalous-, it is in harmony with the general explanation given in Art. 236. If .70 the height of the column sustained by rotation without flow is u 2 2 /2g (in accordance with the theory of Art. 233), 2/ = should give ' x = l. But when flow occurs, the lift may 60 be greater than u 2 2 /2g because of the energy imparted by the rotating e wheel to the water flowing through the wheel passages; that is, values of .50 x less than 1 would give positive values of y. The soundness of this reasoning is corroborated by the re- sults of experiments. Fig. 118 shows the x-y curve and the effi- ciency curve * obtained from experiments on a pump similar in design to the one considered in the foregoing theory. The curves in Fig. 117 are marked with the same letters as the corresponding general equations. One solution of equation (29) is .1 x l.O FIG. 118. which represents a straight line intersecting the x-y hyperbola in the point whose coordinates are x = .808, y = .218, the values corresponding to maximum efficiency. The value of the maxi- mum efficiency is 1, as it must be if & = 0. 245. Effect of Friction Loss on Form of Curves. In order to show how the form of the curves and the solution for maxi- * It should be said that the efficiencies represented in the experimental curve are gross efficiencies computed from the actual work done in driving the pump, while the efficiencies given by equation (B) and represented by the corresponding curve in Fig. 117 are of course hydraulic efficiencies. The experimental curve is based upon data supplied by Mr. C. H. Stoddard, Chief Mechanical Engineer of the Risdon Iron Works, San Francisco. REMARK ON FRICTION LOSSES. 243 mum efficiency are affected by the friction factor k, which has been assumed zero in the foregoing example, the curves for k = 5 (all other data remaining unchanged) are shown in Fig. 119. Equations (25) and (27) now become 1.0 CE) Values 01 x 1.0 FIG. 119. 2.0 (30) (31) The last equation gives for maximum efficiency ?/ = .189x, which with (30) and (26) gives z = .843, ?/ = .160, e = M. 246. Remark on Friction Losses. In the above theory h was defined as the total lift from supply reservoir to discharge reservoir, and h' as the total head lost between supply and discharge reservoirs. The reasoning is not changed, however, if h = H 2 -Hi, where H l and H 2 are the values of the effective head at two sections A and B taken anywhere on the suction and discharge sides of the pump respectively, and h' is defined as the loss of head between these sections.* The value of v 2 for any *That is, H^z^p./w + vS/Zg, H 2 = z 2 + p 2 /w + v 2 2 /2g, If H, and H 2 are to be determined by experiment, it is advantageous to have the cross- sections at A and B equal, so that the velocities need not be considered in determining H, - H^ 244 TURBINE PUMPS. given value of u 2 will not be changed by changing the positions of the sections A and B, since the term k(v 2 2 /2g) expressing friction loss will increase or decrease exactly as H 2 -Hi de- creases or increases. The values of x and y will, however, change, since these are ratios of u 2 and v 2 to h. The compu- ted efficiencies will also depend upon the position of A and B, since the entire loss of head between these sections is charged against the pump when the efficiency is computed from equa- tion (B). EXAMPLES. 1. The following are dimensions of a centrifugal pump: r 2 =6.5" = .5417'; a 2 = 150; / 2 = .108 sq. ft.; A 2 ' = lo. Determine x and y for maximum efficiency, neglecting friction losses. Ans. k' =2.732, z = .857, y = .3U, e = l. 2. Solve with same data except assume fc=6. Ans. z = .903, ?/ = .168, e = .73. 3. With data as in Ex. 2, draw curves of discharge, efficiency and power under constant lift. Ans. x 2 + 3.732xy - 13.46y 2 = 1. 4. With same data, if the lift is 150 ft., what number of R.P.M. should give highest efficiency, and what would be the corresponding rate of discharge? If the speed were increased so as to double the rate of discharge, what would be the efficiency? Ans. For maximum efficiency .V = 1565 R.P.M. , q = 1.785 cu. ft. per sec. N = 1870 R.P.M. gives g =3.570 cu. ft. per sec., e = .59. 5. Solve examples 1 and 2 assuming ^4 2 '=30. 6. The following are dimensions of a pump: r 2 =8"; a 2 = 156; / 2 = 5.25 sq. in.; A 2 '=25. (a) Determine x and y for maximum efficiency if k=4. (b) Determine best speed of rotation and rate of discharge when the lift is 85 ft. 247. Curves for Rate of Discharge, Efficiency and Power when the Wheel Velocity Remains Constant. An important practical question relates to the effect of variations of the lift when the pump runs at constant speed. The equations apply- ing to such a case may be derived from those already given. It will be convenient to introduce the variable m/u 2 = I/x = x' instead of x, so that x' is proportional to V2gh or m. Further, the rate of discharge is not proportional simply to y when the CURVES FOR RATE OF DISCHARGE. 245 lift varies, since m in equation (D) is not constant. Since m = u 2 /x, (D) may be written We therefore introduce y' = y/x as a new variable to represent the rate of discharge. The five main equations, expressed in terms of x' and y', become = l>, . . . (A') x' 2 =0;. ... ((7) . . (D') In (E'), V is written for the variable factor in the value of L; it is equal to l/x 3 . If curves are constructed with x f as abscissa and y f , e, I', respectively, as ordinates, they will show the variation of rate of discharge, efficiency and power with the lift when the speed remains constant. Special case. For the particular data given above equa- tions (A') and (B') become 0/2 + (13.70 +%' 2 -5.672/' = l, .... (32) x' 2 2(1 -.8662/')' (33) Assuming as before & = 0, the curves are shown at (A'), (B'), (E'), Fig. 120. The equation 246 TURBINE PUMPS. represents an ellipse whose principal axes are parallel to the axes of coordinates, and whose center is at the point x' = 0, Of special interest is the form of curve (E'). This indicates that if the speed is constant, an increase in the lift causes a decrease in the power.* This explains what has sometimes been regarded as an anomaly in the practical working of cen- trifugal pumps. At first sight it might be inferred that an increased lift would increase the load upon the motor which drives the pump, but the opposite effect is often observed. This is in accordance with curve (E'), and is easily understood when it is considered that the power is proportional directly to the rate of discharge and to the lift, and inversely to the efficiency. An increase in the lift (the speed remaining con- stant) causes a decrease in the rate of discharge, and probably also an increase in the efficiency, so that the net result may be to decrease the power. EXAMPLE. With data as in Ex. 2, Art. 246, draw curves of discharge, efficiency and power for constant speed. * Except for very small values of the rate of discharge. , COMPOUNDING. 247 248. Compounding. A compound pump consists of two or more pumps arranged in series, i.e., so that the discharge of one is the supply of the next. Since the lift due to a wheel does not depend upon the actual value of the pressure upon it, the total lift due to such a series is the sum of the lifts which the several wheels would produce acting separately. For high lifts compounding is common, extremely high wheel velocities being thus avoided. The different wheels are commonly made alike in dimensions and mounted on the same shaft, so that they rotate together and produce equal lifts. See Fig. 121. FIG. 121. Worthington three-stage turbine pump. 249. Balancing. Since the impeller must have clearance space in order to rotate freely, it will be surrounded by water under pressure. Since friction will prevent this water from acquiring the full rotational velocity of the wheel, the intensity of pressure throughout the clearance space may approach the value existing at the periphery of the wheel. The intensity of pressure in the supply pipe being much less than this, the central portion of the wheel will experience a much greater total pressure from the back than from the front, thus subjecting the shaft to a thrust which must be provided for in the design. 248 TURBINE PUMPS. In order to balance this thrust a common practice is to mount upon the same shaft two wheels alike in all respects except that they are right- and left-handed. Fig. 122 shows FIG. 122. Risdon-Sulzer four-stage pump with balanced impellers. such a construction with two pairs of balanced runners, con- nected in series so as to form a four-stage pump. 250. Actual Lifts and Efficiencies. It was formerly sup- posed that turbine pumps could be efficiently used only with small lifts and large discharges. Experiment has shown, how- ever, that with proper design good efficiencies may be realized with very considerable lifts. Probably hydraulic efficiencies as high as 80 per cent can be obtained under a lift of more than 100 ft., with a single impeller wheel. By compounding, a lift of several hundred feet is entirely practicable. APPENDIX A. STEADY FLOW OF A GAS. A 1. Effect of Compressibility on Theory of Steady Flow. When the fluid is compressible the theory of steady flow requires important modification in two particulars. First, it is no longer necessary that equal volumes pass different sections in a given interval of time. Second, as the density of any definite portion of the fluid will in general change as it moves along the stream, it will do positive or negative work against the pres- sures acting on its bounding surface, resulting in the develop- ment of mechanical energy if the body expands, and the absorp- tion of mechanical energy if it contracts.* Thus both the equation of continuity and the general equation of energy will be changed. Equation of Continuity. If the flow remains steady, equal masses of fluid must pass all cross-sections of the stream during any given time. But the mass passing a section F where the velocity is v and the density w is wFv per unit time. Hence, comparing different sections, = wFv = constant ..... (1) A3. Energy Passing a Given Cross-section. The expression deduced in Art. 59 for the energy passing a given section of the stream is not changed by the fact that the fluid is com- pressible. But in applying it to different sections the appro- priate value of w must be used at each section. * In other words, there is a transformation of molecular energy into mechanical or the reverse. 249 250 STEADY FLOW OF A GAS. A 4. Energy Generated within a Given Portion of the Stream by Expansion. Let Fig. Al represent a portion of a steady stream of gas, and let it be assumed that the density decreases continuously in the direction of flow.* Let A and B be any two cross-sections, and consider the motion of the body of gas AB during a short time At, at the end of which it occupies the volume A'B' '. If W denotes the mass of fluid which passes any section per unit time, we have for the mass passing a section during the time At WAt = WiFiViM = w 2 F 2 v 2 At ; each of these expressions being equal to the mass of fluid in each of the volumes A A ', BB' '. During the small motion con- sidered, each elementary portion of gas expands slightly, doing work against pressure; it is desired to determine the total amount of such work done dur- ing the time At by all the ele- mentary portions of the body AB. Now in steady flow the condition at any given point of the volume A'B must be the same at the end of the time At as at the beginning. The whole work of expansion done by the body AB is therefore equal to that which would be done by the elementary portion AA' if it expanded into the volume BB' ', passing through all stages as to temperature, pressure, and density that actually exist in the stream from A to B. This work cannot, therefore, be com- puted, unless the condition of the gas is known at every point in the stream AB. Isothermal change of perfect gas. If the temperature of the gas is the same at every point and remains unchanged during the flow, the relation between pressure and density for the case of isothermal expansion is to be used in computing the work. * This is merely for convenience in stating the argument. The conclusion holds algebraically in any case. FIG. Al. WORK OF EXPANSION. 251 For a perfect gas this relation is 1-!*, (2) p 2 W 2 or = = = constant (3) w w\ w 2 The work done by unit mass in expanding from pressure p\ and density w\ to pressure p 2 and density w 2 is easily found to be The work of expansion within the volume AB per unit time is therefore TF^log^. (4) wi 5 p 2 Adidbatic change of perfect gas. If no heat is given out or received by any portion of the gas during the flow, the condi- tion of steady flow must be such that the temperature varies along the stream with the density in just the same way as it varies in a given portion of gas expanding or contracting adiabatically. For perfect gases the adiabatic law connecting pressure and density is M'2 in which k is the ratio of the specific heat at constant pressure to that at constant volume, and has the value 1.41 very nearly. The work done by unit mass in expanding from pressure pi and density w\ to pressure p 2 and density w 2 is easily shown to be _ k l\Wi w 2 Hence the work of expansion per unit time within the volume ABis w 252 STEADY FLOW OF A GAS. Expansion of steam. For saturated steam expanding adi- abatically the result given for perfect gases holds approxi- mately, except that a different number takes the place of k. Assuming the equation the work of expansion within the volume AB per unit time is in which the value of m is about 1.13 for dry saturated steam and somewhat less for steam mixed with water. A value com- monly used is - 1 7 -. The cases of isothermal and adiabatic change are the two extremes between which any actual case of flow of a gas will lie. In order to maintain the isothermal condition there must be free conduction of heat between the stream and sur- rounding bodies, which are kept at constant temperature Practically, the isothermal condition could be approximated to only in case of very slow flow. If the flow is very rapid, the quantity of heat gained or lost by conduction by any portion of the gas may be a small fraction of that absorbed or generated by reason of the work of expansion or contraction, so that the condition of adiabatic change may be nearly realized. In any case, let W-H" represent the mechanical energy gen- erated per unit time within the volume AB by reason of expan- sion. Then for isothermal flow # = a log & . . (9) Wi & p 2 For adiabatic flow fl !(>_>). . (10) k-l\Wi w 2 / GENERAL EQUATION OF ENERGY 253 A 5. General Equation of Energy for Steady Flow of a Gas. Recurring now to the reasoning used in Art. 60, consider the gains and losses of mechanical energy in the volume AB (Fig. A 1) per unit time. The volume receives across the section A the amount of energy It loses across the section B the amount It loses by dissipation an amount which may be called WH'. And finally it gains an amount WH" due to the work done by the gas in expanding. But the total energy gained must equal the total energy lost, since in steady flow the quantity of mechanical energy within the volume AB remains constant. Hence or Hi-H 2 = H'-H", ..... (12) in which HI and H 2 have meanings as in Art. 62. A 6. Equation of Energy for Isothermal Flow. Using equa- tion (9), and remembering that in isothermal change of volume PI/WI =p2/w 2 , the general equation of energy becomes +ff'. (13) A 7. Equation of Energy for Adiabatic Flow. In this case equation (10) is to be used, and the general equation of energy may be written in the form 254 STEADY FLOW OF A GAS. In applying this equation it must be remembered that p and w are connected by the relation (5). Equation (14) applies approximately to steam, k being given the proper value. A 8. Flow of Gas Through Small Orifice. If a chamber (X, Fig. A 2) containing gas under pressure be connected by a small orifice or short tube with a second chamber (Y) in which a less pressure exists, flow will take place under practically adiabatic condi- tions. Neglecting loss of energy by dissipation, a formula for the veloc- FIG A 2 ity of flow through the orifice may P & be deduced from equation (14). Let p' be the pressure within the chamber X, p" that within the chamber Y, and p that within the stream passing through the orifice.* In applying the equation of energy, let the point corresponding to A, Fig. A 1, be taken within the chamber X, and the point corresponding to B at the smallest cross-section of the stream passing the orifice. The terms z\ and z 2 may be neglected, and we have also Vi=0, PI/W!=P'/W', V 2 = v , so that the equation reduces to __ 2g~k-l\w' w Or, making use of the relation between pressures and densities in adiabatic change, we may write o 2 k The velocity of flow through the orifice thus depends upon the ratio of the pressure within the orifice to that within the chamber X. * It might seem that p could be assumed equal to p", but it will soon appear that this is not generally permissible. FLOW OF GAS THROUGH SMALL ORIFICE. 255 Mass discharged per unit time. If the cross-section of the jet is FQ, we have for the mass discharged per unit time W = w Q F v = j It will be seen that if p /p' be assumed to decrease from the value 1, W will increase up to a maximum and then decrease. The maximum occurs when + l (18) It appears, therefore, that po/p' will not fall below this value, however small the ratio p"/p'. Substituting this limiting value of PQ/P' in (16) and (17), the former becomes k p r 2g k + 1 w" and the latter takes the form (19) 5-7TTT. . (20) The limiting values of po/p' for two cases, with the corre- sponding formulas for V Q and W, are as follows : Substance. k Limiting Po/p'- Formula for VQ. Formula for W. f 1 rf Air 1 41 .527 V = 7Q5\*2o W- 485w'F n v |2<7 \ w' Steam. . . 1.111 .582 <^& W-. 446^^2^ 256 STEADY FLOW OF A GAS. The value of j//v/ for air at C. is 26,200 ft., while at t -C. it is 26,200(1 + .003660. For any given temperature, therefore, the velocity as given by (19) is independent of the value of p'. For C. it is v = 990 ft. per sec. For steam at an absolute pressure of 100 Ibs. per sq. in. the value of p' '/w' is about 63,300 ft.; at 200 Ibs. per sq. in. it is about 66,200 ft. The corresponding values of the velocity of efflux as given by (19) are 1,460 ft. per sec. and 1,500 ft. per sec. These results do not apply if p" is greater than the value of po given by (18); in that case equations (16) and (17) must be used, with po = p". A 9. Complete Solution of Problem of Adiabatic Flow. In the case of flow considered above, in which p" po require different treatment, as will be seen from the following examples. EXAMPLES. 1. A reservoir contains air at 20 C. under a pressure of 20 Ibs. per sq. in. (absolute). Compute the mass discharged per second into the atmosphere through an orifice of one square inch area. The rate of mass-discharge is W =wFv. In this case p">p,} we therefore take p" as the pressure at the orifice, and find the correspond- ing values of w and v. For 20 C. we have p' /w' = 26200(1 + .0732) = 28100 ft., and since p' =2880' Ibs. per sq. ft., w' =.01025 Ibs. per cu. ft. From the above table or diagram, w/w'=.8Q5, hence w = . 00825. Also ^ = .705; Vj = .765 V64.4 X 28100 = 1030; v = . 705X1030 =726 ft. per sec. Finally, W = wFv = . 00825 X X 726 .0416 Ibs. per sec. 2. Air is discharged from a reservoir in which the pressure is 100 Ibs. per sq. in. above atmospheric, and the temperature 20 C., through an orifice of one square inch area. Compute the mass discharged per second. In this case p"