H Y D R A U BY HORACE W. KING Professor of Hydraulic Engineering, University of Michigan AND CHESTER O. WISLER Associate Professor of Hydraulic Engineering, University of Michigan NEW YORK JOHN WILEY & SONS, INC LONDON: CHAPMAN & HALL, LIMITED 1922 -:..:/. Copyright, 1922, By HORACE W. KING AND CHESTER O. WISLER PRESS OF BRAUNWORTH & CO BOOK MANUFACTURERS BROOKLYN, N. Y. PREFACE THIS book deals with the fundamental principles of hydraulics and their application in engineering practice. Though many formulas applicable to different types of problems are given, it has been the aim of the authors to bring out clearly and logically, the underlying principles which form the basis of such formulas rather than to emphasize the importance of the formulas them- selves. Our present knowledge of fluid friction has been derived largely through experimental investigation and this has resulted in the development of a large number of empirical formulas. Many of these formulas have necessarily been included but, in so far as possible, the base formulas to which empirical coeffi- cients have been applied have been derived analytically from fundamental consideration of basic principles. The book is designed as a text for beginning courses in hydraulics and as a reference book for engineers who may be interested in the fundamental principles of the subject. Tables of coefficients are given which are sufficiently complete for class- room work, but the engineer in practice will need to supplement them with the results of his own experience and with data obtained from other published sources. Chapters I to VI inclusive and Chapter XI were written by Professor Wisler and Chapters VII to X inclusive were written by Professor King. Acknowldgement for material taken from many publications is made at the proper place in the text. H. W. K. C. O. W. University of Michigan, April, 1922. 5019i 3 ill CONTENTS CHAPTER I INTRODUCTION PAGE ART. 1. Fluids. 2. Definitions. 3. Units used in Hydraulics. 4. Weight of Water. 5. Compressibility of Water. 6. Viscosity. 7. Surface Tension. 8. Accuracy of Computations 1 CHAPTER II PRINCIPLES OF HYDROSTATIC PRESSURE ART. 9. Intensity of Pressure. 10. Direction of Resultant Pressure. 11. Pascal's Law. 12. Free Surface of a Liquid. 13. Atmospheric Pressure. 14. Vacuum. 15. Absolute and Gage Pressure. 16. Intensity of Pressure at any Point. 17. Pressure Head. 18. Trans- mission of Pressure. 19. Vapor Pressure. 20. The Mercury Barometer. 21. Piezometer Tubes. 22. Mercury Gage. 23. The Differential Gage, 24. Suction Pumps and Siphons , 7 CHAPTER III PRESSURE ON SURFACES ART. 25. Total Pressure on Plane Areas. 26. Center of Pressure on Plane Areas. 27. Graphical Method of Location of Center of Pressure. 28. Position of Center of Pressure with Respect to Center of Gravity. 29. Horizontal and Vertical Components of Pressure on any Surface 2o CHAPTER IV IMMERSED AND FLOATING BODIES ART. 30. Principle of Archimedes. 31. Center of Buoyancy. 32. Sta- bility of Floating Bodies. 33. Determination of Metacentric Height 38 Vl CONTENTS CHAPTER V RELATIVE EQUILIBRIUM OF LIQUIDS PAGE ART. 34. Relative Equilibrium Defined. 35. Vessel Moving with Constant Linear Acceleration. 36. Vessel Rotating about a Vertical Axis 45 CHAPTER VI PRINCIPLES OF HYDROKINETICS ART. 37. Introductory. 38. Friction. 39. Discharge. 40. Steady Flow and Uniform Flow. 41.' Continuity of Discharge. 42. Stream Line and Turbulent Motion. 43. Energy and Head. 44. Ber- noulli's Theorem. 45. Application of Bernoulli's Theorem to Hydrostatics. 46. Bernoulli's Theorem in Practice. 47. Venturi Meter, 48, Pitot Tube 51 CHAPTER VII FLOW OF WATER THROUGH ORIFICES AND TUBES ART. 49. Description and Definitions. 50. Characteristics of the Jet. 51. Fundamental Orifice Formula. 52. Orifice Coefficients. 53. Algebraic Transformation of Formula with Velocity of Approach Correction. 54. Head Lost in an Orifice. 55. Path of Jet. 56. Orifices under Low Heads. 57. Suppression of Contraction. 58. Standard Short Tube. 59. Converging Tubes. 60. Nozzles. 61. Diverging Tubes. 62. Borda's Mouthpiece. 63. Re-entrant Tubes. 64. Submerged Orifice. 65. Partially Submerged Orifices. 66. Gates. 67. Discharge under Falling Head 71 CHAPTER VIII FLOW OF WATER OVER WEIRS ART. 68. Description and Definitions. 69. Velocity at any Depth. 70. Theoretical Formulas for Discharge. 71. Theoretical Formula for Mean Velocity. 72. Weir Coefficients. 73. Weirs with End Contractions. 74. Modifications of Fundamental Formula. 75. Algebraic Transformation of Formula. 76. Weir Experiments. 77. Formulas for Sharp-crested Weirs. 78. Discussion of Weir Formulas. 79. Submerged Weirs. 80. Further Discussion of Sub- merged Weirs. 81. Triangular Weirs. 82. Trapezoidal Weirs. 83. The Cippoletti Weir. 84. Weirs not Sharp-crssted. 85. Broad- crested Weirs or Chutes. 86. Measurement of Head. 87. Con- ditions for Accurate Measurement over Sharp-crested Weirs 101 CONTENTS Vii CHAPTER IX FLOW OF WATER THROUGH PIPES PAGE ART. 88. Description and Definitions. 89. Wetted Perimeter and Hydraulic Radius. 90. Critical Velocities in Pipes. 91. Friction and Distribution of Velocities. 92. Energy of Water in a Pipe. 93. Continuity of Flow in Pipes. 94. Loss of Head. 95. Hydraulic Gradient. 96. Loss of Head Due to Friction in Pipes. 97. The Chezy Formula. 98. Hazen- Williams Formula. 99. King Formula. 100. General Discussion of Pipe Formulas. 101. Friction Formula for Non-turbulent Flow. 102. Detailed Study of Hydraulic Gra- dient and Minor Losses. 103. Part of Pipe above Hydraulic Gra- dient. 104. Special Problems. 105. Branching Pipe Connecting Reservoirs at Different Elevations. '106. Compound Pipe Con- necting Two Reservoirs. 107. Pipes of More than One Diameter Connected in Series 135 CHAPTER X FLOW OF WATER IN OPEN CHANNELS ART. 108. Description and Definition. 109. Wetted Perimeter and Hydraulic Radius. 110. Friction and Distribution of Velocities. 111. Energy Contained in Water in an Open Channel. 112. Con- tinuity of Flow in Open Channels. 113. Loss of Head. 114. Hy- draulic Gradient or Water Surface. 115. Loss of Head Due to Friction in Open Channels. 116. The Chezy Formula. 117. The Kutter Formula. 118. The Manning Formula. 119. Comparison of Manning and Kutter Formulas. 120. The Bazin Formula. 121. Open-channel Formulas in General. 122. Detailed Study of Hy- draulic Gradient or Water Surface. 123. Hydraulics of Rivers. 124. Irregular Sections. 125. Cross-section of Greatest Efficiency. 126. Circular Sections. 127. Non-uniform Flow. 128. Backwater. 129. Divided Flow 176 CHAPTER XI HYDRODYNAMICS ART. 130. Fundamental Principles. 131. Interpretation of Newton's Laws. 132. Relative and Absolute Velocities. 133. Jet Impinging Normally on a Fixed Flat Plate. 134. Jet Impinging Normally on a Moving Flat Plate. 135. Jet Deflected by a Fixed Curved Vane. 136. Jet Deflected by a Moving Curved Vane. 137. Work Done on Moving Vanes. 138. Forces Exerted upon Pipes. 139. Straight Pipe of Varying Diameter. 140. Pipe Bends. 141. Water Hammer in Pipe Lines. 142. Formulas for Water Hammer 211 HYDRAULICS CHAPTER I INTRODUCTION 1. Fluids. Fluids are substances which possess unlimited mobility and which offer practically no resistance to change of form. A perfect fluid yields to the slightest tangential stress, and can therefore have no tangential stress if it is at rest. Fluids may be divided into two classes, (a) liquids, or fluids that are practically incompressible, and (6) gases, or fluids that are highly compressible. 2. Definitions. Hydraulics is the science embodying the laws that relate to the behavior of liquids, and particularly of water. In its original sense the term hydraulics was applied only to the flow of water in conduits, but the scope of the word has been broadened by usage. Hydraulics may be divided conveniently into three branches : (a) hydrostatics, which deals with liquids at rest, (6) hydrokinetics, which treats of the laws governing the flow of liquids, and (c) hydrodynamics, which relates to the forces exerted upon other objects by liquids in motion or upon liquids by other objects in motion. The fundamental laws of hydraulics apply equally to all liquids, but in hydrokinetics empirical coefficients must be mod- ified to conform to the liquid considered. Water is the most common liquid and the only one that is of general interest to engineers. 3. Units Used in Hydraulics. It is common practice in the United States and Great Britain to base hydraulic computations on the foot-pound-second system of units. In practically all hydraulic formulas these units are used, and if not otherwise INTRODUCTION stated they are understood. Frequently the diameters of pipes or orifices are expressed in inches, pressures are usually stated in pounds per square inch, and volumes may be expressed in gallons. Before applying such data to problems, conversion to the foot- pound-second system of units should be made. Care in the conversion of units is essential. Errors in hydraulic computa- tions result more frequently from wrong use of units than from any other cause. 4. Weight of Water. Water has its maximum density at a temperature of 39.3 F. At this temperature pure water has been given a specific gravity of unity and it thus serves as a standard of density for all substances. The density of water decreases for temperatures above and below 39.3. It freezes at 32 and boils at 212 F. The weight of pure water at its temperature of maximum density is 62.424 Ibs. per cubic foot. The weights at various temperatures are given in the following table: WEIGHT OF PURE WATER Tempera- Pounds Tempera- Pounds Tempera- Pounds ture, per ture, per ture, per Fahrenheit cubic foot Fahrenheit cubic foot Fahrenheit cubic foot 32 62.416 90 62.118 160 61.006 39.3 62.424 100 61.998 170 60.799 40 62.423 110 61.865 180 60.586 50 62 . 408 120 61.719 190 60.365 60 62.366 130 61.555 200 60.135 70 62.300 140 61.386 210 59.893 80 62.217 150 61 . 203 212 59.843 For temperatures above the boiling point Rankine gives the following approximate formula: w being the weight of water in pounds per cubic foot and T the temperature in degrees Fahren- heit, 124.85 w = T+461 , 500 500 ' T+4G1 As water occurs in nature, it invariably contains a certain amount of salts and mineral matter in solution. Silt or other COMPRESSIBILITY OF WATER 3 impurities may also be carried in suspension. These substances are invariably heavier than water and they therefore increase its weight. The impurities contained in rivers, inland lakes and ordinary ground waters do not usually add more than one-tenth of a pound to the weight per cubic foot. Ocean water weighs about 64 Ibs. per cubic foot. After long-continued droughts the waters of Great Salt Lake and of the Dead Sea have been found to weigh as much as 75 Ibs. per cubic foot. Since the weight of inland water is not greatly affected by ordinary impurities nor changes of temperature, an average weight of water may be used which usually will be close enough for hydraulic computations. In this book the weight of a cubic foot of water is taken as 62.4 Ibs. Sea water will be assumed to weigh 64.0 Ibs. per cubic foot unless otherwise specified. In very precise work weights corresponding to different temperatures may be taken from the above table. 6. Compressibility of Water. Water is commonly assumed to be incompressible, but in reality it is slightly compressible. Upon release from pressure water immediately regains its original volume. For ordinary pressures the modulus of elasticity is constant that is, the amount of compression is directly proportional to the pressure applied. The modulus of elasticity, E, varies with the temperature as shown in the following table. Temperature, Fahrenheit Modulus of elasticity, pounds per square inch 35 77 212 288,000 327,000 360,000 These values hold only for pressures below 1000 Ibs. per square inch. Kite obtained a reduction in volume of 10 per cent for a pressure of 65,000 Ibs. per square inch, giving a value of E of 650,000 for this high intensity of pressure. The compressibility of water usually affects the solution of practical problems in hydraulics only by changing its unit weight. Since pressures commonly encountered are relatively small, in most cases water may be considered incompressible without introducing any appreciable error. 4 INTRODUCTION 6. Viscosity. One of the characteristic properties of a liquid is its ability to flow. A perfect liquid would be one in which every particle could move in contact with adjacent particles without friction. The pressures between all such particles would be normal to their respective surfaces at the points of contact since there could be no tangential stress without friction. All liquids are capable, however, of resisting a certain amount of tangential stress, and the extent to which they possess this ability is a measure of their viscosity. Water is one of the least viscous of all liquids. Oil, molasses and wax are examples of liquids having greater viscosity. 7. Surface Tension. At any point within a body of liquid the molecules are attracted towards each other by equal forces. The molecules forming the surface layer, however, are subjected to an attraction downward that is not balanced by an upward attraction. This causes a film or skin to form on the surface and results in many interesting phenomena. A needle may be made to float upon water so long as the surface film is not broken, but it will sink immediately when the film is broken. Surface tension causes the spherical shape of dewdrops or drops of rain. This phenomenon also makes possible the hook gage described in Art. 86. Where water flows in an open conduit, surface tension retards velocities at the surface, the maximum velocity ordinarily being below the surface (see Art. 110). Capillary action is also explained by the phenomenon of surface tension combined with that of adhesion. A (Fig, 1) illustrates an open tube of small diameter im- mersed in a liquid that wets the tube. Water rises in the tube higher than the level out - side, the meniscus being con- cave upward. The tube B is immersed in mercury or some FlG j other liquid which does not wet the tube. In this case the meniscus is convex upward and the level of the liquid in the tube is depressed. The effect of capillarity decreases as the size of tube increases. The water in a tube one-half inch in diameter is approximately at the same level as the outside water, but it is appreciably different for smaller tubes. For this reason, ACCURACY OF COMPUTATIONS 5 piezometer tubes (Art. 21) should not have diameters much smaller than one-half inch. 8. Accuracy of Computations. Accuracy of computations is most desirable, but results should not be carried out to a greater number of significant figures than the data justify. Doing this implies an accuracy which does not exist and may give results that are entirely misleading. Suppose, for example, that it be desired to determine the theoretical horse-power available in a stream where the discharge is 311 cu. ft. per second and the available head is 12.0 ft. The formula to be used is _wQH where w = the weight of a cubic foot of water; Q = the discharge in cubic feet per second; 77 = the available head. Substituting in the formula, 62.4X311X12.0 550 = 423.41 horse-power. Referring to t}ie table, page 2, it is seen that for the ordinary range of temperatures the weight of water may vary from 62.30 to 62.42 Ibs. per cubic foot. Furthermore, the statement that the discharge is 311 cu. ft. per second means merely that the exact value is more nearly 311 than 310 or 312. In other words, the true value may lie anywhere between 310.5 and 311.5. Likewise, the fact that the head is given as 12.0 merely indicates that the correct value lies somewhere between 11.95 and 12.05. Therefore substituting in the formula the lower of these values, 62.30X310 n 5Xll.95 = 42030 horse . power . 550 Again substituting in the formula the higher of the possible values, 62.42X311.5X12.05 550 = 426 . 00 horse-power. It is evident, therefore, that the decimal .41 in the original answer 423.41 is unjustified, and that the last whole number, 3, 6 INTRODUCTION merely represents the most probable value, since the correct value may lie anywhere between 420.30 and 426.00. The answer should, therefore, be given as 423. It may be stated in general that in any computation involving multiplication or division, in which one or more of the numbers is the result of observation, the answer should contain the same number of significant figures as is contained in the observed quan- tity having the fewest significant figures. In applying this rule it should be understood that the last significant figure in the answer is not necessarily correct, but represents merely the most probable value. To give in the answer a greater number of significant figures indicates a degree of accuracy that is unwarranted and misleading. CHAPTER II PRINCIPLES OF HYDROSTATIC PRESSURE 9. Intensity of Pressure. The intensity of pressure at any point in a liquid is the amount of pressure per unit area. If the intensity of pressure is the same at every point on any area, A, ,- = CD the symbol p representing the intensity of pressure and P the total pressure acting upon the area. If, however, the intensity of pressure is different at different points, the intensity of pressure at any point will be equal to the pressure on a small differential area surrounding the point divided by the differential area, or P-, > Intensities of pressure are commonly expressed in pounds per square inch and pounds per = ^^-^^^^ __^ square foot. Where there is no danger of ambiguity, the term pressure is often used as an abbre- viated expression for intensity of pressure. 10. Direction of Resultant Pressure. The resultant pressure on any plane in a liquid at rest is normal to that plane. Assume that the resultant pres- sure P, on any plane AB (Fig. 2), makes an angle other than 90 with the plane. Resolving P FIG. 2. into rectangular components PI 7 and P2, respectively parallel 8 PRINCIPLES OF HYDROSTATIC PRESSURE with and perpendicular to AB, gives a component PI which can be resisted only by a tangential stress. By definition, a liquid at rest can resist no tangential stress and therefore the pressure must be normal to the plane. This means that there can be no static friction in hydraulics. 11. Pascal's Law. At any point within a liquid at rest, the intensity of pressure is the same in all directions. This principle is known as Pascal's Law. Consider an infinitesimally small wedge-shaped volume, BCD (Fig, 3), in which the side BC is vertical, CD is horizontal and BD makes any angle 6 with the hori- zontal. Let AI, AZ and AS and pi, P2 and ps represent, respectively the areas of these sides and the intensities of pressure to which they are subjected. Assume that the ends of the wedge are "* vertical and parallel. Since the wedge is at the rest, the principles of equilibrium may be ap- plied to it. From Art. 10 it is known F that the pressures are normal to the faces of the wedge. Choosing the coordinate axes as indicated in Fig. 3 and setting up the equations of equilibrium, SY = and SF = 0, and neglecting the pressures on the ends of the wedge, since they are the only forces acting on the wedge which have components along the Z-axis and therefore balance each other, the following expressions result; in 6, m . . cos 0- But As sin 6= AI and As cos 6 Therefore Since BD represents a plane making any angle with the hori- zontal and the wedge is infinitesimally small so that the sides may be considered as bounding a point, it is evident that the intensity of pressure at any point must be the same in all directions. FREE SURFACE OF A LIQUID 9 12. Free Surface of a Liquid. Strictly speaking, a liquid having a free surface is one on whose surface there is absolutely no pressure. It will be shown later, however, that there is always some pressure on the surface of every liquid. In practice the free surface of a liquid is considered to be a surface that is not in contact with the cover of the containing vessel. Such a surface may or may not be subjected to the pres- sure of the atmosphere. It may be shown that the free surface of a liquid at rest is horizontal. Assume a liquid having a surface which is not hori- zontal, such as A BCDE (Fig. ^ n >N 4) . A plane MN, inclined to the horizontal, may be passed through any liquid having such a surface in such manner that a portion of the liquid BCD lies above the plane. Since the liquid is at rest, BCD must be in equilibrium, but the vertical force of gravity would necessarily have a component along the inclined plane which could be resisted only by a tangential stress. As liquids are in- capable of resisting tangential stress it follows that the free surface must be horizontal. 13. Atmospheric Pressure. All gases possess mass and con- sequently have weight. The atmosphere, being a fluid composed of a mixture of gases, exerts a pressure on every surface with which it comes in contact. At sea level under normal conditions atmospheric pressure amounts to 2116 Ibs. per square foot or about 14.7 Ibs. per square inch. VARIATION IN ATMOSPHERIC PRESSURE WITH ALTITUDE FIG. 4. Altitude above sea level in feet Pressure in pounds per square inch Altitude above sea level in feet Pressure in pounds per square inch 14.7 5,280 12.0 1000 14.15 6,000 11.7 2000 13.6 7,000 11.3 3000 13.1 8,000 10.9 4000 12.6 9,000 10.5 5000 12.1 10,000 10.1 10 PRINCIPLES OF HYDROSTATIC PRESSURE The intensity of atmospheric pressure decreases with the altitude. Owing to compressibility, the density of air also decreases with altitude, and therefore the intensity of pressure changes less rapidly as the altitude increases. The accompanying table gives approximate values of the atmospheric pressure corre- sponding to different elevations above sea level. 14. Vacuum. A perfect vacuum, that is, a space in which there is no matter either in the solid, liquid or gaseous form, has never been obtained. It is not difficult, however, to obtain a space containing a minute quantity of matter. A space in contact with a liquid, if it contains no other substance, always contains vapor from that liquid. In a perfect vacuum there could be no pressure. In practice, the word "vacuum" is used frequently in con- nection with any space having a pressure less than atmospheric pressure, and the term " amount of vacuum" means the amount the pressure is less than atmospheric pressure. The amount of vacuum is usually expressed in inches of mercury column or in pounds per square inch measured from atmospheric pressure as a base. For example, if the pressure within a vessel is reduced to 12 Ibs. per square inch, which is equivalent to 24.5 in. of mer- cury column, there is said to be a vacuum of 2.7 Ibs. per square inch or 5.5 inches of mercury. (Arts. 20 and 22.) 15. Absolute and Gage Pressure. The intensity of pressure above absolute zero is called absolute pressure. Obviously, a negative absolute pressure is impossible. Usually pressure gages are designed to measure the intensities of pressure, above or below atmospheric pressure as a base. Pres- sures so measured are called relative or gage pressures. Negative gage pressures indicate the amount of vacuum, and at sea level pressures as low as, but no lower than, 14.7 Ibs. per square inch are possible. Absolute pressure is always equal to gage pressure plus atmospheric pressure. Fig. 5 illustrates a gage dial, on the inner circle of which is shown the ordinary gage and vacuum scale. The outer scale indicates the corresponding absolute pressures. 16. Intensity of Pressure at any Point. To determine the intensity of pressure at any point in a liquid at rest or the variation in pressure in such a liquid, consider any two points such as 1 and 2 (Fig. 6) whose vertical depths below the free surface of the INTENSITY OF PRESSURE AT ANY POINT 11 liquid are hi and fa, respectively. Consider that these points lie in the ends of an elementary prism of the liquid, having a cross- sectional area dA and length I. Since this prism is at rest, all of the forces acting upon it must be in equilibrium. These forces consist of the fluid pressure on the sides and ends of the prism and the force of gravity, FIG. 5. Gage dial. Let X and Y, the coordinate axes, be respectively parallel with and perpendicular to the axis of the prism which makes an angle 6 with the vertical. Also let pi and p% be the intensities of pressure at points 1 and 2, respectively, and w be the unit weight of the liquid. Considering forces acting to the left along the J^-axis as negative and remembering that the pressures on the sides of the prism are normal to the .XT-axis and therefore have no X com- ponents, the following equation may be written: = p id A p2dA wldA cos 0=0. Since I cos B = hi fa, this reduces to (3) From this equation it is evident that in any liquid the difference in pressure between any two points is the product of the unit weight of the liquid and the difference in elevation of the points. If hi = fa, pi must equal p2 ; or, in other words, in any continuous homogeneous body of liquid at rest, the intensities of pressure at all points in a horizontal plane must be the same. Stated con- 12 PRINCIPLES OF HYDROSTATIC PRESSURE versely, in any homogeneous liquid at rest all points having equal intensities of pressure must lie in a horizontal plane. If in equation (3) hz is made equal to zero, p2 is the intensity of pressure on the liquid surface. In case that pressure is atmos- pheric, or p a , equation (3) becomes pi=whi+p a , (4) or, in general, p = wh+p a . . . . . . . . (5) In this equation p is evidently the absolute pressure at any point in the liquid at a depth h below the free surface. The correspond- ing gage pressure is p = wh (6) In the use of the above equations care must be taken to express all of the factors involved in their proper units. Unless otherwise stated p will always be understood to be intensity of pressure in pounds per square foot, w will be the weight of a cubic foot of the liquid and h will be measured in feet. At any point in a body of water at a depth h below the free surface, the absolute pressure in pounds per square foot is p = Q2. 4/1+2116. . (7) The relative or gage pressure in pounds per square foot is p = 62Ah (8) If, however, it is desired to express the pressure in pounds per square inch it is necessary only to divide through by 144. Hence if p' is used to express absolute pressure in pounds per square inch, j^G^ 2116 " 144 144 r 144 = .433/i+14.7, or, expressed as gage pressure in pounds per square inch, p'=. 433/1. PRESSURE HEAD 13 17. Pressure Head. Equation (6) may be written in the form, 2 = h. (9) W T) Here h, or its equivalent, , represents the height of a column of liquid of unit weight w that will produce an intensity of pressure, p. It is therefore called pressure head. In considering water pressures the pressure head, h, is expressed in feet of water column regardless of whether it is obtained by dividing the pressure in pounds per square foot by 62.4 or by dividing the pressure in pounds per square inch by 0.433. 18. Transmission of Pressure. Writing equation (3) in the form, h 2 ), ...... (10) it is evident that the pressure at any point, such as point 1 (Fig. 6 ), in a liquid at rest is equal to the pressure at any other point, such as point 2, plus the pressure produced by a column of the liquid whose height, h, is equal to the difference in elevation between the two points. Any change in the intensity of pressure at point 2 would cause an equal change at point 1. In other words, a pres- sure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point in the liquid. This principle is made use of in the hydraulic jack by means of which heavy weights are lifted by the application of relatively small forces. Example. In Fig. 7 assume that the piston and weight, W, are at the same elevation, the face of the piston having an area of 2 sq. in. and the face of the weight 20 sq. in. What weight W can be lifted by a force P of 100 Ibs. applied at the end of the lever as shown in the figure? Since atmospheric pressure FIG. 7. Hydraulic jack. is acting on both the piston and weight its resultant effect will be zero and it may therefore be 14 PRINCIPLES OF HYDROSTATIC PRESSURE neglected. Taking moments about the force F on the piston is F = 4X100 F = 600 Ibs. 600 = 300 Ibs. per square inch. & which is the intensity of pressure on the face of the piston, and since the two are at the same elevation in a homogeneous liquid at rest it is also the intensity of pressure on the weight. Therefore = 6000 Ibs. Evidently this is the value of W for equilibrium; any weight less than 6000 pounds could be lifted by the force of 100 Ibs. 19. Vapor Pressure. Whenever the free surface of any liquid is exposed to the atmosphere, evaporation is continually taking place. If, however, the surface is in contact with an enclosed space, evaporation takes place only until the space becomes saturated with vapor. This vapor produces a pressure, the amount of which depends only upon the temperature and is entirely independent of the presence or absence of air or other gas within the enclosed space. The pressure exerted by a vapor within a closed space is called vapor pressure. In Fig. 8, A represents a tube having its open end submerged in water and a stopcock at its upper end. Consider the air within A to be absolutely dry at the time the stopcock is closed. At the instant of closure the water surfaces in- side and outside the tube will stand at the same level. Evaporation within the tube, however, will soon saturate the space containing air and create a vapor pressure, p v , which will cause a depression of the Pa ~ FIG. 8. water surface within the vessel equal to . THE MERCURY BAROMETER 15 In the same figure, B represents a tube closed at the upper end. Assume a perfect vacuum in the space above the water in the tube. If this condition were possible the water level in B would stand at an elevation above the water surface outside. Vapor w pressure within the vessel, however, causes a depression exactly w equal to that produced within A, so that the maximum height of water column possible under conditions of equilibrium in such a tube is ~. Vapor pressures increase with the temperature, as is shown in the following table. VAPOR PRESSURES OF WATER IN FEET OF WATER COLUMN Tempera- ture, F. EL w Tempera- ture, F. Pv_ W Tempera- ture, F. EL w -20 0.02 60 0.59 140 6.63 -10 .03 70 0.83 150 8.54 .05 80 1.16 160 10.90 10 .08 90 1.59 170 13.78 20 .13 100 2.17 180 17.28 30 .19 110 2.91 190 21.49 40 .28 120 3.87 200 26.52 50 .41 130 5.09 212 33.84 20. The Mercury Barometer. The barometer is a device for measuring intensities of pressure exerted by the atmosphere. In 1643 Torricelli discovered that if a tube (Fig. 9) over 30 in. long and closed at one end, is rilled with mercury and then made to stand vertically with the open end submerged in a vessel of mercury, the column in the tube will stand approximately 30 in. above the sur- face of the mercury in the vessel. Such a device is known as a mercury barometer. Pascal proved that the height of the column of mercury depended upon the atmospheric pressure, when he carried a FlG - 9 - Mercury barometer to a higher elevation and found that barometer. the height of the column decreased as the altitude increased. 16 PRINCIPLES OF HYDROSTATIC PRESSURE Although theoretically water or any other liquid may be used for barometers, two difficulties arise in using water. First, the height of water column necessary to balance an atmospheric pres- sure of 14.7 Ibs. per square inch is about 34 feet at sea level, which height is too great for convenient use; and, second, as shown in Art. 19, water vapor collecting in the upper portion of the tube creates a pressure which partially balances the atmospheric pres- sure, so that the barometer does not indicate the total atmospheric pressure. ^ Since mercury is the heaviest known liquid and has a very low vapor pressure at ordinary air temperatures it is more satisfactory for use in barometers than any other liquid. 21. Piezometer Tubes. A piezometer tube is a tube tapped into the wall of a vessel or pipe for the purpose of measuring moderate pressures. Thus the height of water column in tube a (Fig. 10) is a measure of the pressure at J., the top of the pipe. Similarly the pressure at the elevation B is measured by the height of water column in tube 6, that is, p = wh. Piezometer tubes always measure gage pressures since the water surface in the tube is subjected to atmospheric pressure. Obviously, the level to which water will rise in a tube will be the same regardless of whether the connection is made in the side, bottom or cover of the containing vessel. Piezometer tubes are also used to measure pressure heads in pipes where the water is in motion. Such tubes should enter the pipe in a direction at right angles to the direction of flow and the connecting end should be flush with the inner surface of the pipe. If these precautions are not observed, the height of water column may be affected by the velocity of the water, the action being similar to that which occurs in Pitot tubes. (See Art. 48.) In order to avoid the effects of capillary action, piezometer tubes should be at least ^ in. in diameter. Pressures less than atmospheric pressure may be measured by either of the methods illustrated in Fig. 11 which shows a pipe section AB in which the water is flowing. The vertical distance, h t which the water surface, C, in the open tube drops below A is a measure of the pressure below atmospheric pressure, or, in FIG. 10. Cross- section of pipe with piezometer tubes. MERCURY GAGE 17 other words, it is a measure of the amount of vacuum existing at A. This is true since the pressure at C f is atmospheric, being at the same level as C in a homogeneous liquid at rest and the pressure at A must be less than at C' by the amount wh. (See Art. 17.) To the right in Fig. 11 is shown an inverted piezometer tube with the lower end immersed in an open vessel containing water. Atmospheric pressure acting on the water surface in the vessel forces water to rise in the tube to a height h which measures the vacuum existing at B. Air will be drawn into the pipe at B until the intensity of pressure on the water surface in the tube equals the pressure at B. Neglecting the weight of the air in the tube, it is evident that pn = p D = pa- w h. Here p B is expressed as absolute pressure since atmospheric pressure is included in the equation, FIG. 11. FIG. 12. Mercury gage. 22. Mercury Cage. In the measurement of pressures so great that the length of tube required for a water piezometer would be unwieldy, the mercury U-tube, illustrated in Fig. 12, is a con- venient substitute. Water under pressure fills the pipe, or vessel at A and the tube down to the level D. Mercury fills the tube from D around to B, above which level the tube is open to the atmosphere. The pressure at C equals the pressure at B which is atmos- pheric, or p a , plus the pressure produced by the mercury column, h. Hence, pc=pa+w'h. ,,,,,,, (11) If pc and p a are expressed in pounds per square foot and h in feet, w' is the weight of a cubic foot of mercury, or 13.6 (specific 18 PRINCIPLES OF HYDROSTATIC PRESSURE gravity of mercury) X 62. 4 = 848 Ibs. If, however, p c and p a are expressed in pounds per square inch and h in feet, 040 ' = (or 13. 6X0. 433) =5. 89. The pressure at D (Fig. 12) is the same as at C, being at the same level in a homogeneous liquid at rest. The pressure at A is equal to that at C minus the pressure produced by the water column a, or, PA = PCWCL ........ (12) Here again if p A and pc are expressed in pounds per square foot and a in feet, w equals 62.4, but if p A and p c are expressed in pounds per square inch and a in feet, w equals 0.433. Com- bining equations (11) and (12), pA = p a -\-w'hwa. . . . (13) Here PA is expressed as absolute pressure, since p a (atmospheric pressure) enters into the equation. If PA is less than atmospheric pressure by an amount greater than wa, h is negative, as in Fig. 13, and w'hwa. . . (14) FIG. 13. 23. The Differential Gage. The dif- ferential gage as the name indicates, is used only for measuring differences in pres- sure. A liquid heavier or lighter than water is used in the gage, depending upon whether the differences in pressure to be measured are great or small. In Fig. 14 is shown the form of dif- ferential gage usually employed for meas- uring large differences in pressure. M and N are two pipes containing water under different pressures which may be either greater or less than atmospheric pressure. The two pipes are connected by a bent tube, of which the portion BCD is filled with mercury, while all of the remaining space is filled with water. FIG. 14. Differential mercury gage. THE DIFFERENTIAL GAGE 19 If M and N were at the same elevation the difference in pres- sure in the two pipes would be measured by the pressure due to the mercury column DC minus that due to the water column EF or would equal w'hwh. If the two pipes are at different elevations, to the above difference in pressure must be added or subtracted the intensity of pressure produced by the water column whose height is equal to the difference in elevation of the pipes. Proof of these state- ments follows. The pressure at A equals that at M minus the pressure produced by the water column whose height is a. Evidently the pressure at B is the same as at A, being at the same elevation in a homo- geneous liquid at rest. For the same reason the pressures at B and C are also equal. Hence, pc=pB = pA = pMwa (15) The pressures at C and F are not equal, since these points are not connected by a homogeneous liquid. The pressure at D is equal to that at C minus the pressure produced by the mercury column h and is = pcw'h. (16) and Combining these equations or Obviously, the greater the difference between w f and w the greater the difference in pressure that can be measured for any given value of h. Fig. 15 illustrates a type of differential gage used when the difference in pressures to be measured is small. Usually a liquid, such as a light oil, whose specific gravity is slightly less than unity is used in the upper portion of the inverted U-tube, AC, the remainder of the tube being filled with water. FIG. 15. Differential oil gage. 20 PRINCIPLES OF HYDROSTATIC PRESSURE As with the mercury gage, if M and N are at the same elevation the difference in pressure will be equal to that produced by the oil column AB and the water column CD. If M and N are at different levels, then or .... (19) .... (20) the equations being the same as were obtained for the mercury differential gage. The use of a liquid whose unit weight w' is very nearly the same as that of water makes a very sensitive measuring device. With such a device small differences in pressure will produce relatively large values of h. The value of h produced by any particular difference in pressure is independent of the relative cross-sectional areas of the columns AB and CD. This is evident since it is the difference in intensities of pressure that is measured and not difference in total pressures. 24. Suction Pumps and Siphons. Suction pumps depend upon atmospheric pressure for their operation. The plunger creates a partial vacuum in the pump stock, and atmospheric pressure acting upon the outer water surface causes water to rise within the pump. The operation of siphons is also produced by atmospheric pressure. li\ Fig. 16 the two vessels, A and B, are connected by a tube. As long as the tube is filled with air there is no tendency for water to flow. If, however, air is exhausted from the tube at C, atmospheric pressure will cause water to rise in each leg of the tube an equal height above the water sur- FIG. 16. Siphon. faces. When water has been drawn up a distance d, equal to the height of the summit above the water surface in the upper vessel flow from A to B will begin. If the velocity of the water is high enough, any air entrapped in the tube will be carried out by the moving water, and the tube will flow full. If the sum- PROBLEMS 21 mit of the siphon is a distance greater than above the 10 water surface in the higher vessel, siphon action is impossible. It is not necessary that the discharge end of the tube be sub- merged to induce siphon action. If flow is started by suction at the free end of the tube or by other means, it will continue as long as the discharge end of the tube is lower than the water surface in the vessel or until the vacuum in the siphon is broken. PROBLEMS 1. Determine the intensity of pressure on the face of a.dam at a point 40 ft. below the water surface. *#}t*}^ (a) Expressed in pounds per square foot gage pressure. (6) Expressed in pounds per square inch gage pressure. (c) Expressed in pounds per square foot absolute pressure. (d) Expressed in pounds per square inch absolute pressure. 2. Determine the intensity of pressure in a vessel of mercury (sp. gr. = 13.6) at a point 8 in. below the surface, expressing the answer in the same units as in Problem 1. 3. A vertical pipe, 100 ft. long and 1 in. in diameter, has its lower end open and flush with the inner surface of the cover of a box, 2 ft. square and 6 in. high. The bottom of the box is horizontal. Neglecting the weight of the pipe and box, both of which are filled with water, determine: (a) The total hydrostatic pressure on the bottom of the box. (6). The total pressure exerted on the floor on which the box rests. 4. At what height will water stand in a water barometer at an altitude of 5000 ft. above sea level if the temperature of the water is 70 F.? Under similar conditions what would be the reading of a mercury barometer, neglecting the vapor pressure of mercury? 5. What are the absolute and gage pressures in pounds per square inch existing in the upper end of the water barometer under the conditions of Problem 4? 6. What height of mercury column will cause an intensity of pressure of 100 Ibs. per square inch? What is the equivalent height of water column? 7. A pipe 1 in. in diameter is connected with a cylinder 24 in. in diameter, each being horizontal and fitted with pistons. The space between the pistons is filled with water. Neglecting friction, what force will have to be applied to the larger piston to balance a force of 20 Ibs. applied to the smaller piston? 8. In Problem 7, one leg of a mercury U-tube is connected with the smaller cylinder. The mercury in this leg stands 30 in. below the center of the pipe, the intervening space being filled with water. What is the height of mercury in the other leg, the end of which is open to the air ? 9. A U-tube with both ends open to the atmosphere contains mercury in the lower portion. In one leg, water stands 30 in. above the surface of the mercury; in the other leg, oil (sp. gr.=0.80) stands 18 in. above the surface 22 PRINCIPLES OF HYDROSTATIC PRESSURE of the mercury. What is the difference in elevation between the surf: the oil and water columns? 10. Referring to Fig. 15, page 19, if the pressure at M is 20 Ibs. per inch, what is the corresponding pressure at N if a = 1 ft., b = 4 ft. and h - (Sp.gr. of oil = 0.80.) 11. In Fig. 15, page 19, determine the value of h, if a = 1 ft., b = 4 ft. a pressure at N is 1.4 Ibs. per square inch greater than at M. (Sp. gr 0.80.) 12. A vertical tube 10 ft. long, with its upper end closed and low open, has its lower end submerged 4 ft. in a tank of water. Neglecting pressure, how much will the water level in the tube be below the level tank? 13. In Fig. 7, page 13, if the diameters of the two cylinders are 3 : 24 in. and the face of the smaller piston is 20 ft. above the face of the W, what force P is required to maintain equilibrium if TF = 8000 Ibs.? 14. Referring to Fig. 12, page 17, if A = 20 in. and a = 12 in., what absolute pressure in pounds per square inch at A1 What is the gage pn 15. In Problem 14, if the surface of the mercury column in each leg U-tube stands at the same elevation as A when the pressure at A is pheric, determine the values of a and h when the gage pressure at A is per square inch, the diameter of the tube being the same throughout. 16. Referring to Fig. 13, page 18, determine the absolute press pounds per square inch at A when a =8 in. and h = 10 in. What is the spending gage pressure? 17. In Fig. 14, page 18, let c = 6 ft., and assume that A = wh< pressure at M is atmospheric. If the pressure at N remains constant, det the value of h when the gage pressure at If is increased to 8 Ibs. per inch. 18 In Fig. 14, page 18, if a = 24 in. and c=6 ft., what is the vali when the pressure at M is 10 Ibs. per square inch greater than at N. 19. A and B are, respectively, the closed and open ends of a I both being at the same elevation. For a distance of 18 in. below tube is filled with oil (sp. gr. =0.8); for a distance of 3 ft. below B, th is filled with water, on the surface of which atmospheric pressure is The remainder of the tube is filled with mercury. What is the al pressure at A expressed in pounds per square inch? 20. In Problem 19, if B were closed and A were open to the atmos what would be the gage pressure at B, expressed in pounds per square CHAPTER III PRESSURE ON SURFACES 25. Total Pressure on Plane Areas. The total pressure on any plane surface is equal to the product of its area and the intensity of pressure at its center of gravity. This may be proved as follows: Fig. 17 shows projections on two vertical planes normal to (a) (*) FIG. 17. each other, of any plane surface, MN, subjected to the full static pressure of a liquid with a free surface. Projection (6) is on a plane at right angles to MN. The surface MN makes any angle, 6, with the horizontal, and, extended upward, the plane of this surface intersects the surface of the liquid in the line 00, shown as the point in (b). Consider the surface MN to be made up of an infinite number of horizontal strips each having a width dy so small that the 24 PRESSURE ON SURFACES intensity of pressure on it may be considered constant. The area of any strip whose length is x, is dA = xdy. The liquid having a unit weight of w, the intensity of pressure on any strip at a depth h below the surface and at a distance y from the line 00 is p = wh = wy sin 6. The total pressure on the strip is dP = wy sin 6 dA and the total pressure on MN is (1) From the definition of center of gravity, JydA=Ay', , . , (2) where y' is the distance from the line 00 to the center of gravity of A. Hence, P = wsm S Ay' (3) Since the vertical depth of the center of gravity below the water surface is \ h' = y' sin0, . . .^ .... (4) it follows that P = wh'A, ' . Jj.< X ... (5) where wh' represents the intensity of pressure at the center of gravity of A. 26. Center of Pressure on Plane Areas. Any plane surface subjected to hydrostatic pressure is acted upon by an infinite number of parallel faNs%;vvhose magnitudes vary with the depth, below the free surface, of the various infinitesimal areas on which the respective forces act. Since these forces are parallel they may be replaced by a single resultant force P. The point on the surface at which this resultant force acts is called the center of pressure. In other words, if the total hydrostatic pressure on any area were applied at the center of pressure the same effect would be CENTER OF PRESSURE ON PLANE AREAS 25 produced on the area as is produced by the variable pressures distributed over the area. The position of the horizontal line containing the center of pressure of a plane surface subjected to hydrostatic pressure may be determined by taking moments of all the forces acting on the area about some horizontal axis in the plane of the surface. For the case described in the preceding article and illustrated in Fig 17, the line 00 may be taken as the axis of moments for the surface MN. Designating by y the distance to the center of pressure from the axis of moments, it follows from the defini- tion of center of pressure that, ........ (6) or fydP J y=p- ......... (7) It was shown in Art. 25 that dP = wy sin dA and P = wsin BAy' ..... ! . . (3) Substituting these values, equation (7) becomes w sin 6 I y 2 dA \ y 2 dA V = w sine Ay' ~~ Ay' ' ..... (8) in whichjy 2 dA is the moment of inertia, /o, of MN with respect to the axis OO, and Ay' is the statical moment, S, of MN with respect to the same axis. Therefore, Since the moment of inertia of an area about any axis equals the moment of inertia of the area about a parallel axis through its center of gravity plus the product of the area and the square of the distance between the two axes, equation (9) may be written, 26 or PRESSURE ON SURFACES y=y'+- y (ii) where k is the radius of gyration of the area. The above discussion refers only to the determination of 'the position of the horizontal line which contains the center of pressure that is, y gives only the distance from the horizontal axis of moments to the center of pressure. For any figure such that the locus of the midpoints of the horizontal strips is a straight line, as, for instance, a triangle or trapezoid with base horizontal, the center of pressure falls on that straight line. It is with such figures that the engineer is usually concerned. For other figures, the horizontal location of the center of pressure may be found in T t/V FIG. 18. a manner similar to that described above by taking moments about an axis, within the plane of the surface, at right angles to the horizontal axis of moments. Examples l : (a) Find the center of pressure on the vertical triangular gate shown in Fig. 18. 1 It is apparent that the solution of any problem involving the location of the center of pressure, for an area whose radius of gyration is known or can be readily found, may be accomplished by the simple substitution of values for k and y' in equation (11). This involves a mere mathematical process with no necessity on the part of the student for either thought or understanding of fundamental principles. It is therefore recommended that the beginner solve all such problems by the use of equation (6) which is really nothing more than a formulated expression of the definition of center of pressure. The examples given are solved by this method. CENTER OF PRESSURE ON PLANE AREAS 27 Using the equation Py = j'ydP (6) it will be necessary to express dP in terms of y. In this case it will be convenient to take moments about 0-0, a horizontal line through the vertex of the gate, lying in the plane of the gate. Moments could, however, be taken about any other horizontal axis lying within this plane. The total pressure dP on any thin horizontal strip at a distance y from the axis of moments equals the intensity of pressure, wh, times the area dA, or dP = whdA. Since w = 62A, h = 5+y and dA=x dy dP=62A(5+y)xdy. Since x varies with y it must be expressed in terms of y before integrating. From similar triangles, Substituting, x y 4 4 = 3 r x = &- and also therefore (5y+y 2 )dy fydP J wh'A = 62. 4X7X6, y 62.4X42 = A(*X27+iX81) = 2^ ft. below the vertex, or ^ ft. below the center of gravity of the gate. The horizontal location of the center of pressure in this case is on the median connecting the vertex with the base. 28 PRESSURE ON SURFACES (6) Find the center of pressure on the inclined rectangular gate shown in Fig. 19. Taking moments again about the top of the gate as an axis, where and j y dP \ ywh dA, dA=My 5' FIG. 19. Since P = wh f A =62.4(5+2 cos 30; 62.4X6 J Uy+ 62.4X24(5+V3) 1 5 2 V3 I 4 I 4 J ^58.47 26.93 = 2. 17 ft. below AB, measured along the plane of the gate. The horizontal location of the center of pressure is 3 ft. from either end of the gate. GRAPHICAL METHOD OF LOCATION 29 (c) In example (6) what force F applied normally to the gate at its lower edge will be required to open it? Knowing the total pressure, P, on the gate and the location of the center of pressure, by taking moments about the upper edge which is the center of rotation, 4F = 2.17P . 4X24(5+ \/3) = 5470lbs. The value of F can also be found directly without determining either the location of the center of pressure or the total pressure. Taking moments about the top of the gate, 4F= CydP JQ r = I ywh dA = 62.4X6 J (5y+-^y 2 }dy = 62.4X6 |i/ 2 +-?/ 3 V3 = 62.4X6^X16+X64) = 21890 F = 5470lbs. If this force were applied at the bottom of the gate, the gate would be in equilibrium and there would be no reaction on the supports along the lower edge or sides of the gate. Any force greater than 5470 Ibs. would open the gate. 27. Graphical Method of Location of Center of Pressure. Semigraphic methods may be used advantageously in locating the center of pressure on any plane area whose width is constant. 30 PRESSURE ON SURFACES The rectangular surface, A BCD, illustrated in Fig. 19 is shown in perspective in Fig. 20a. BC (Fig. 206) represents the projection of the rectangle on a vertical plane perpendicular to the plane of the surface. The vertical depths below the water surface of the top and bottom of the rectangle are, respectively, hi and h%. The intensity of pressure, whi, on the top of the rectangle is represented by the equal ordinates A A' and BB' (Fig. 20a), and on the bottom of the rectangle the equal ordinates CC' and DD' represent the intensity of pressure whz. BB 1 and CC' (Fig. 206) represent, to a reduced scale, the inten- sities of pressure at B and C. BB" and CC" are laid off equal (a) (6) FIG. 20. to 6BB ; and 6CC', respectively, and therefore represent the areas ABB' A' and DCC'D'. The total pressure acting on the surface ABCD is therefore represented by the area of the pressure diagram BCC"B" as it is similarly represented by the pressure volume ABCDA'B'C'D'. Also the resultant pressure on the surface acts through the center of gravity of the pressure area BCC"B" just as it acts through the center of gravity of the pressure volume ABCDA'B'C'D'. The trapezoid BCC"B" may be divided into the rectangle BB"EC and the triangle B"C"E, the locations of whose centers of gravity are known. By taking moments of each of these pressure areas about C"C and dividing the sum of these moments by the POSITION OF CENTER OF PRESSURE 31 area of the trapezoid, the distance of the center of pressure from C is determined. Thus and CC" therefore C"# Taking moments about C"C, - = 4Xjtu>X5X6+w(5+2V3)X6] = 1.83 ft. Example (c), page 29, can also be solved by taking moments about BB" as follows: 4X5X6X2+fX62. 4X2V3X6XI from which F = 54701bs. For areas having a variable width, OB"C" is not a straight line and the center of gravity of the pressure area is not so easily located. For such areas it will probably be easier to use the analytical method described in Art. 26. 28. Position of Center of Pressure with Respect to Center of Gravity. If the intensity of pressure varies over any surface, the center of pressure is below the center of gravity. Consider the equation (see Art. 26), 2 ~=y'+-, ......... (ii) k 2 Since must always be positive, y must be greater than y f . This may also be seen from Fig. 20. The center of pressure on A BCD is the normal projection on that plane of the center of gravity of the pressure volume ABCDA'B'C'D'. Evidently this projection must fall below the center of gravity of A BCD since it would fall at the center of gravity if the intensity of pressure on the surface were uniform, in which case the pressure volume would be ABCDA'B'EF. 32 PRESSURE ON SURFACES N* It also appears from the above discussion and from a study of Fig. 20 that for any area the greater its depth below the surface of the liquid the more nearly will the center of pressure approach the center of gravity. The two coincide at an infinite depth. In two cases the intensity of pressure is constant over the area and hence the center of pressure coincides with the center of gravity: (a) When the surface is horizontal. (6) When both sides of the area are completely submerged in liquids of the same density. As an illustration consider the gate AB (Fig. 21). Water stands hi feet above the top of the gate on one side and h?, feet above it on the other side. The distribution of pressure on the left is represented by the trapezoid ABMN and on the right by the trapezoid AHKB. The triangle GED is similar to CFG and equal to CE'P by construction. The trapezoid of pressure AHKB is therefore balanced by the trapezoid ON ML. The resultant in- tensity of pressure on the 7 surface is therefore constant, as represented by the rect- angle OABL, and the center of pressure must coincide with the center of gravity of the surface. This is true regardless of the shape of the surface. The resultant intensity of pressure equals wh } where h is the difference in elevation of water surfaces. In this latter case it should be observed that it is the center of the resultant pressure that coincides with the center of gravity of the gate, since the center of gravity of either of the trap- ezoidal areas of pressure, considered alone, falls below the center of gravity of the gate. 29. Horizontal and Vertical Components of Pressure on any Surface. It may be convenient to deal with the horizontal and vertical components of the pressure acting on a surface rather than with the resultant pressure. Consider, for example, the water pressure acting on the curved 7 FIG. 21. HORIZONTAL AND VERTICAL COMPONENTS 33 The dam may FIG. 22. face, AB, of the dam shown in section in Fig. 22. have any length normal to the plane of the paper. Choosing the coordinate axes as shown, let BF represent the trace of a vertical plane normal to the XY plane. Consider the equi- librium of the volume of liquid whose cross-section, as shown in the figure, is ABF and whose ends are parallel with the XY plane and separated by a distance equal to the length of the dam. Since this volume of liquid is assumed to be in equilibrium, 2X = and S7 = 0. The only forces that have any components parallel with the X-axis are the X-components of the normal pressures acting on the surface A B and the normal pressure on the vertical plan BF. Since BF is the projection of the face A B on a vertical plane nor- mal to the X-axis, it follows that the resultant of the X-com- ponents of the pressures on A B, or P x , equals the normal pressure on the projection of A B on a vertical plane normal to the X-axis. As the demonstration holds true independently of the manner in which the X-axis is chosen, it may be stated in general that the component, along any horizontal axis, of the pressure on any area is equal to the normal pressure on that vertical projection of the area which is normal to the chosen axis. In a similar manner consider the vertical forces acting on the volume of liquid whose cross-section is ABF (Fig. 22). The only vertical forces are the force of gravity, or the weight of the liquid, and the vertical components of the pressures on the surface AB, which forces must therefore be equal in magnitude. In other words, the vertical component of the pressure on any surface is equal to the weight of that volume of the liquid extending verti- cally from the surface to the free surface of the liquid. If the pres- sure were acting upward on the surface, its magnitude, as will be shown later (Art. 30), would be equal to the weight of that volume of the liquid that would extend from the surface to the free surface of the liquid. The pressures considered in this article are relative 34 PRESSURE ON SURFACES pressures, since obviously atmospheric pressure is acting on both sides of the dam and the resultant effect is zero. Examples: (a) What will be the resultant pressure on the base BC of the masonry dam subjected to water pres- sure, as shown in Fig. 23, and where will this resultant in- tersect the base? The following numerical values are given. Area of section ABCD = QQO sq. ft. Area of water section OAB = 200 sq. ft. Weight of ma- sonry = 150 Ibs. per cubic foot. Linear dimensions are shown in figure. A section of dam 1 ft. long will be considered to be in equi- FIG. 23. librium under the action of the following forces: TF = the total weight of the section, acting through the center of gravity of the cross-section A BCD; P = resultant hydrostatic pressure acting on the face AB; R = reaction between the earth and the base, BC, of the dam. This reaction must necessarily be equal, opposite and colinear with the resultant of W and P. Since the dam is in equilibrium when subjected to the above forces, the fundamental principles of equilibrium may be applied that is, 2X = 0, 27=0 and 2M = 0. > For 2X = 0; R X = P X and for the data given P x = 62 . 4 X4 X45 = 63,200 Ibs. \ For 27=0; R V = P V +W and for the data given P v = 200 X 62 . 4 = 12,480 Ibs. IF = 600X150 = 90,000 Ibs. R v = 12,480+90,000 = 102,480 Ibs. ^ HORIZONTAL AND VERTICAL COMPONENTS 35 The total resultant pressure on the base is 63,2002+ 102,4802 = 122,500 Ibs, For 2 Jlf = 0, taking moments about an axis through C, normal to the plane of the section, X being the distance from C at which the resultant, R, intersects the base of the dam, x - 25P y - 18W = 0. Substituting the numerical values of R v , P x , W and P y 102,480X+63,200X-V-~ 12,480X25-90,000X18 = and reducing Thus the resultant pressure of 122,500 Ibs. per linear foot of dam intersects the base 8.6 ft. from the toe of the dam. (6) Determine the tensile stress in the walls of a 24-in. pipe carrying water under a head of 100 ft. In a case like this where the head is relatively large compared to the diameter of the pipe, it is customary to consider that the intensity of pressure is uniform throughout the pipe. A cross-section of the pipe is shown in Fig. 24. Consider a semicircular segment, AB, of unit length, held in equilibrium by the two forces T. Evidently T is the tensile stress in the i --- >TTT^ T ' wall of the pipe, and if the intensity of pressure is 24" assumed to be constant, T * is constant at all points in __ I __ the section. The sum of . - * the horizontal components of the normal pressures acting on the semicircular segment is equal to the normal pressure on the vertical projection of this segment (Art. 29), Calling this normal pressure P, since 2# = 0, 2T=P = whA = 62. 4X100X2 = 12,480 Ibs. and 7 7 = 6,240 Ibs. 36 PRESSURE ON SURFACES The required thickness of a steel pipe, using a safe working stress of 16,000 Ibs. per square inch is or a little more than in. PROBLEMS 1. A vertical rectangular gate is 4 ft. wide and 6 ft. high. Its upper edge is horizontal and on the water surface. What is the total pressure on the gate and where is the center of pressure? 2. Solve Problem 1 if the water surface is 5 ft. above the top of the gate, other conditions remaining the same. 3. Solve Problem 2 if the plane of the gate makes an angle of 30 with the vertical, other conditions remaining unchanged. 4. A cubical box, 24 in. on each edge, has its base horizontal and is half filled with water. One of the sides is held in position by means of four screws, one at each corner. Find the tension in each screw due to the water pressure. 6. A vertical, triangular gate has a horizontal base 4 ft. long, 3 ft. below the vertex and 5 ft. below the water surface. What is the total pressure on the gate and where is the center of pressure? 6. A vertical, triangular gate has a horizontal base 3 ft. long and 2 ft. below the water surface. The vertex of the gate is 4 ft. below the base. What force normal to the gate must be applied at its vertex to open the gate? 7. A triangular gate having a horizontal base 4 ft. long and an altitude of 6 ft. is inclined 45 from the vertical with the vertex pointing upward. The base of the gate is 8 ft. below the water surface. What normal force must be applied at the vertex of the gate to open it? 8. A cylindrical tank, having a vertical axis, is 6 ft. in diameter and 10 ft. high. Its sides are held in position by means of two steel hoops, one at the top and one at the bottom. What is the tensile stress in each hoop when the tank is filled with water? 9. What is the greatest height, h, to which the water can rise without causing the dam shown in Fig. 25 to collapse? Assume b to be so great that water will not flow over the top of the dam. 10. If b in the figure is 20 ft., find the least value of h at which the dam will collapse. "FIG 25 ** ^ vert i ca l> triangular gate has a horizontal base 8 ft. long and 6 ft. below the water surface. Its vertex is 2 ft. above the water surface. What normal force must be applied at the vertex to open the gate? 12. A masonry dam of trapezoidal cross-section, with one face vertical, PROBLEMS 37 has a thickness of 2 ft. at the top and 10 ft. at the bottom. It is 22 ft. high and has a horizontal base. The vertical face is subjected to water pressure, the water standing 15 ft. above the base. The weight of the masonry is 150 Ibs. per cubic foot. Where will the resultant pressure intersect the base? 13. In Problem 12 what would be the depth of water when the resultant pressure intersects the base at the outer edge of the middle third, or If ft. from the middle of the base? 14. A vertical triangular surface has a horizontal base of 4 ft. and an altitude of 9 ft., the vertex being below the base. If the center of pressure is 6 in. below the center of gravity, how far is the base below the water surface? 16. Water stands 40 ft. above the top of a vertical gate which is 6 ft. square and weighs 3000 Ibs. What vertical lift will be required to open the gate if the coefficient of friction between the gate and the guides is 0.3? 16. On one side, water stands level with the top of a vertical, rectangular gate 4 ft. wide and 6 ft. high, hinged at the top; on the other side water stands 3 ft. below the top. What force applied at the bottom of the gate, at an angle of 45 with the vertical, is required to open the gate? 17. A vertical, trapezoidal gate in the face of a dam has a horizontal base 8 ft. below the water surface. The gate has a width of 6 ft. at the bottom and 3 ft. at the top, and is 4 ft. high. Determine the total pressure on the gate and the distance from the water surface to the center of pressure. 18. Determine the total pressure and the position of the center of pressure on a vertical, circular surface 3 ft. in diameter, the center of which is 4 ft. below the water surface. 19. A 6-in. pipe line in which there is a 90 bend contains water under a gage pressure of 450 Ibs. per square inch. Assuming that the pressure is uniform throughout the pipe and that the water is not in motion, find the total longitudinal stress in joints at either end of the bend. CHAPTER IV IMMERSED AND FLOATING BODIES 30. Principle of Archimedes. Any body immersed in a liquid is subjected to a buoyant force equal to the weight of liquid dis- placed. This is known as the principle of Archimedes. It may be proved in the following manner. The submerged body ABCD (Fig. 26) is referred to the coor- dinate axes X, Y and Z. Consider the small horizontal prism aia 2 , parallel to the X-axis, to have a cross-sectional area dA. The X-component of the normal force acting on a\ must be equal and opposite to the same force acting on 2, each being equal to whdA. There is, therefore, no tendency for this prism to move in a direction parallel to the X-axis. Since the same reasoning may be applied to every other prism parallel to a\ai it follows that there is no tendency for the body as a whole to move in this direc- tion. The same reasoning ap- plies to movement parallel to the Z-axis or to any other axis in a horizontal plane. If, therefore, there is any tendency for the body to move it must be in a vertical direction. Consider now the F-components of the hydrostatic pressures acting on the ends of any vertical prism 6162 having a cross- sectional area, dA, so small that the intensity of pressure on either end of the prism may be considered constant. The resultant of these pressures will be the difference between dP\, the vertical com- ponent of the normal pressure at 61 equal to whi dA, acting down- ward and dP2, the corresponding force acting at 62, equal to wli? dA, FIG. 26. CENTER OF BUOYANCY 39 acting upward. The resultant pressure will be upward and equal to w(h,2 hi) dA; but (h^ hi) dA is the volume of the elementary prism which, multiplied by w, gives the weight of the displaced liquid. Since the entire body, A BCD, is made up of an infinite number of such prisms, it follows that the resultant hydrostatic pressure on the body will be a buoyant force equal in magnitude to the weight of the displaced liquid. If the weight of the body is greater than the buoyant force of the liquid the body will sink. On the other hand, if the weight of the body is less than the buoyant force, the body will float on the surface, displacing a volume of liquid having a weight equal to that of the body. 31. Center of Buoyancy. A BCD (Fig. 27) represents a floating body. From the principles of the preceding article, the buoyant force acting on any elementary area of the submerged surface must be equal to the weight of the vertical prism of displaced liquid directly above it. Since the weight of each prism is directly proportional to its volume, the center of gravity of all these buoyant forces, or the center of buoyancy, must coincide with the center of gravity of the displaced liquid. 32. Stability of Floating Bodies. Any floating body is sub- jected to two systems of parallel forces; the downward force of gravity acting on each of the particles that goes to make up the body and the buoyant force of the liquid acting upward on the various elements of the submerged surface. In order that the body may be in equilibrium the resultants of these two systems of forces must be colinear, equal and oppo- site. Hence the center of buoyancy and the center of gravity of the floating body must lie in the same vertical line. Fig. 28 (a) shows the cross-section of a ship floating in an upright position, the axis of symmetry being vertical. For this position the center of buoj^ancy lies on the axis of symmetry at BO which is the center of gravity of the area ACL. The center of gravity of the ship is assumed to be at G. If, from any cause, such as wind or wave action, the ship is made to heel through an angle 6, as shown in Fig. 28 (6), the center of gravity of the ship and cargo remaining unchanged, the center of buoyancy will shift to a new position, B, which is the center of gravity of the area 40 IMMERSED AND FLOATING BODIES A'C'L. The buoyant force F, acting upward through B, and the weight of the ship W, acting downward through G, constitute a couple which resists further overturning and tends to restore the ship to its original upright position. In all cases, if the vertical line through the center of buoyancy intersects the inclined axis of symmetry at a point M above the center of gravity, the two forces F and W must produce a righting moment. If, however, M lies below G an overturning moment is produced. The point M is known as the metacenter and its distance, GM, from the center of gravity of the ship, is termed the metacentric height. The value of the metacentric height is a measure of the stability of the ship. For angles of inclination not greater than 10 or 15 the FIG. 28. position of M does not change materially, and for small angles of heel the metacentric height may be considered constant. For greater inclinations, however, the metacentric height varies to a greater extent with the angle of heel. 33. Determination of Metacentric Height. Fig. 29 illus- trates a ship having a displacement volume, V. When the ship is tilted through the angle 6 the wedge AOA' emerges from the water while the wedge C'OC is immersed. If the sides A A' and C'C are parallel, these wedges must be similar and of equal volume, t>, since the same volume of water is displaced by the ship whether in an inclined or upright position. The wedges therefore will have the same length and the water lines AC and A'C' must inter- sect on the axis of symmetry at 0. ^ When the ship floats in an upright position a buoyant force DETERMINATION OF METACENTRIC HEIGHT 41 F, equal to wv, acts upward through K, the center of gravity of the wedge AOA'. In the inclined position this force no longer acts, but an equal force F' acts at K', the center of gravity of the wedge C'OC. It may be considered that a downward force F", equal to F, has been introduced, the resultant of F" and F being zero. A righting couple has therefore been introduced equal to wvL, L being the horizontal distance between the centers of gravity of the wedges. Because of the shifting of the force F from K to K' the line of action of the buoyant force W acting on the entire ship is shifted from BQ to B, a horizontal distance S such that = wvL. (1) Consider now a small vertical prism of the wedge C'OC, at a distance x from 0, having a cross-sectional area dA. The buoyant force produced by this immersed prism is wx tan 6 dA, and the moment of this force about is wx 2 tan dA . The sum of all of these moments for both wedges must be equal to wvL or w tan 0J x 2 dA = wvL = wVS. But S = MB sin 0, and for small angles, since the sine is very nearly equal to the tangent, Jx 2 dA = V(MBo) = V(GMGB ), 42 IMMERSED AND FLOATING BODIES and since J x 2 dA is the moment of inertia, /, of the water-line section about the longitudinal axis through 0, f^TUf 1^7? fty} the sign in the last expression being positive if M falls below G and negative if above. Example. Find the me- tacentric height of the rect- angular scow shown in Fig. 30. The scow is 40 ft. long, 20 ft. wide and 8 ft. deep. It has a draft of 5 ft. when FIG. 30. floating in an upright posi- tion. The center of gravity of the scow is on the axis of symmetry, 1 ft. above the water surface. The angle of heel is 10. The problem may be solved by substituting values in formula (2), but the following method may be used conveniently for shapes such that the center of buoyancy can be readily found. Since the center of buoyancy, B, is at the center of gravity of HA'C'F, its position may be found by taking moments about the axes HF and EF. Before taking moments the distances KC f and KF must be determined. tan 10 = 3. 52, Taking moments about HF ' = 2.60ft. Taking moments about EF #=8.83 ft. PROBLEMS 43 The distance of B from the inclined axis of symmetry is ' = 10-8. 83 = 1.17 ft. From similar triangles MB' A'K ,. D/ 1.17X20 -BF=ra r MB^-^-^- =6.65 ft. The metacentric height is = 6. 65+2. 60-6 = 3. 25 ft. The righting moment is TFz = 5X20X40X62.4X3.25sinlO* = 140,900 ft.-lbs. PROBLEMS 1. A rectangular scow 15 ft. by 32 ft., having vertical sides, weighs 40 tons (80,000 Ibs.) . What is its draft? 2. If a rectangular scow 18 ft. by 40 ft. has a draft of 5 ft. what is its weight? 3. A cubic foot of ice (sp. gr.=0.90) floats freely in a vessel containing water whose temperature is 32 F. When the ice melts, will the water level in the vessel rise, lower or remain stationary? Explain why. 4. A ship of 4000 tons displacement floats with its axis of symmetry vertical when a weight of 50 tons is midship. Moving the weight 10 ft. towards one side of the deck causes a plumb bob, suspended at the end of a string 12 ft. long, to move 9 in. Find the metacentric height. 6. A rectangular scow 30 ft. wide, 50 ft. long, and 12 ft. high has a draft of 8 ft. Its center of gravity is 9 ft. above the bottom of the scow. If the scow is tilted until one side is just submerged, determine : (a) The position of the center of buoyancy. (6) The metacentric height. (c) The righting couple, or the overturning couple. 6. In Problem 5, what would be the height of the scow (all other data remaining unchanged) if, with one side just submerged, the scow would be in unstable equilibrium? 7. A box, 1 ft. square and 6 ft. high, has its upper end closed and lower end open. By submerging it vertically with the open end down what is the greatest weight the box can sustain without sinking? 8. In Problem 7, what weight would hold the box in equilibrium with the upper end submerged 10 ft. below the surface? 9. A solid block of wood (sp. gr. =0.6) in the shape of a right cone has a 44 IMMERSED AND FLOATING BODIES base whose diameter is 12 in. and an altitude of 18 in. In what position will this block float in water when it is in stable equilibrium? 10. A solid block of wood (sp. gr.=0.6) in the shape of a right cylinder has a diameter of 12 in. and a length of 15 in. Determine the position in which this block will float in water when in stable equilibrium. CHAPTER V RELATIVE EQUILIBRIUM OF LIQUIDS 34. Relative Equilibrium Defined. In the preceding chapters liquids have been assumed to be in equilibrium and at rest with respect both to the earth and to the containing vessel. The present chapter treats of the condition where every particle of a liquid is at rest with respect to every other particle and to the con- taining vessel, but the whole mass, including the vessel, has a uniformly accelerated motion with respect to the earth. The liquid is then in equilibrium and at rest with respect to the vessel, but it is neither in equilibrium nor at rest with respect to the earth. In this condition a liquid is said to be in relative equilibrium. Since there is no motion of the liquid with respect to the vessel and no movement between the water particles themselves there can be no friction. Hydrokinetics, which is treated in the following chapters, deals with the condition in which water particles are in motion with respect to the earth and also with respect to each other. In this case the retarding effects of friction must be considered. Relative equilibrium may be considered as an intermediate state between hydrostatics and hydrokinetics. Two cases of relative equilibrium will be dis- cussed. 35. Vessel Moving with Constant Linear Acceleration. If a vessel partly filled with any liquid moves horizontally along a straight line with a constant acceleration, j, the surface of the liquid will assume an angle with the horizontal as shown in Fig. 31. To determine the value of 6 for any value of j, consider the forces acting on a small mass of liquid, M, at any point on the surface. This mass is moving with a constant horizontal acceleration, j, and the force producing the acceleration is the resultant of all 45 FIG. 31. 46 RELATIVE EQUILIBRIUM OF LIQUIDS the other forces acting upon the mass. These forces are the force of gravity, W, acting vertically downward and the pres- sure of all the contiguous particles of the liquid. The resultant, F, of these forces must be normal to the free surface AB. Since force equals mass times acceleration, P-Mj-M, (1) and from the figure P = W tan0 (2) Solving these two equations simultaneously, tan0 = -?, (3) tf which gives the slope that the surface, AB, will assume for any constant acceleration of the vessel. Since was assumed to be anywhere on the surface and the values of j and g are the same for all points, it follows that tan 6 is constant at all points on the surface or, in other words, AB is a straight line. The same value of 8 will hold for a vessel moving to the right with a positive acceleration as for a vessel moving to the left with a negative acceleration or a retardation. To determine the intensity of pressure at any point 6, at a depth h below the free surface, consider the vertical forces acting on a vertical prism ab (Fig. 31). Since there is no acceleration vertically the only forces acting are atmospheric pressure at a, gravity, and the upward pressure on the base of the prism at 6. Hence, if the cross-sectional area is dA, pi,dA=whdA -\-p a dA, (4) or pi,=wh+pa, (5) or, neglecting atmospheric pressure which acts throughout, Pb = wh (6) Therefore, in a body of liquid moving with a horizontal accelera- tion the relative pressure at any point is that due to the head of liquid directly over the point, as in hydrostatics. In this case, however, it is evident that all points of equal pressure lie in an inclined plane parallel with the surface of the liquid. VESSEL ROTATING ABOUT A VERTICAL AXES 47 In equation (3) if j were zero, tan would equal zero; or, in other words, if the vessel were moving with a constant velocity the surface of the liquid would be horizontal. Also if the accelera- tion were vertically upward, the surface would obviously be horizontal. To determine the relative pressure at any point, 6, in a vessel with an acceleration upward, consider the forces acting on a vertical prism of liquid ab of height h and cross-sectional area dA (Fig. 32). The force, P, producing the acceleration is the resultant of all the forces acting on the prism, consisting of gravity equal to wh dA, acting downward and the static pres- sure on the lower end of the filament at 6, equal to pt dA, acting upward. There- fore FIG. 32. wh dA. From which = wh+wh j -. (7) 9 This shows that the intensity of pressure at any point within a liquid contained in a vessel hav- ing an upward acceleration, j, is greater than the static pressure by an amount equal to wh~ . Evidently, if the acceleration were downward, the sign of the last term in the above expression would become negative, and if j = g, pb = 0. In other words, if a vessel containing any liquid falls freely in a vacuum, so as not to be retarded by air friction, the pressure will be zero at all points throughout the vessel. 36. Vessel Rotating about a Vertical Axis. When the vessel shown in Fig. 33 is at rest, the FIG. 33. 48 RELATIVE EQUILIBRIUM OF LIQUIDS surface of the liquid is horizontal and at ran. ra'&V represents the form of surface resulting from rotating the vessel with a con- stant angular velocity co radians per second about its vertical axis OF. Consider the forces acting on a small mass of liquid, M, at a, distant r from the axis OF. Since this mass has a uniform circular motion it is subjected to a centripetal force, C = Ma>\ . . : (8) which force produces an acceleration directed toward the center of rotation and is the resultant of all the other forces acting on the mass. These other forces are the force of gravity, W = Mg, acting vertically downward, and the pressure exerted by the adjacent particles of the liquid. The resultant,'/' 1 , of this liquid pressure must be normal to the free surface of the liquid at a. Designating by 6 the angle between the tangent at a and the horizontal, _dh_ C_7ifco 2 r_co 2 r ~dr~W = Mg == g ' or co 2 r , ah = ar, \ which, when integrated becomes, The constant of integration equals zero, since when r equals zero h also equals zero. Since h and r are the only variables this is the equation of a parabola, and the liquid surface is a paraboloid of revolution about the F-axis. As the volume of a paraboloid is equal to one- half that of the circumscribed cylinder and since the volume of liquid within the vessel has not been changed, The linear velocity at a is v = ur. ... ..... (10) PROBLEMS 49 Substituting v for o>r in equation (9) Expressed in words, this means that any point on the surface of the liquid will rise above the elevation of greatest depression a height equal to the velocity head (see Art. 43) at that point. To determine the relative pressure at any point c afc a depth h' vertically below the surface at c' consider the' vertical forces acting on the prism cc', having a cross-sectional area dA. As this prism has no vertical acceleration, 2i/ = and * p c dA=wh'>dA or p e = wh'. ....... .'\ (12) * That is, the relative pressure at any point is that due to the hea*d of liquid directly over tjie point, as in hydrostatics. There- fore the distribution of pressure on the bottom of the vessel is represented graphically by the vertical ordinates to the curve m'b'n'r It also follows that the total pressure on the sides of the vessel is the same as though the vessel were filled to the level m'n 1 and were not rotating. 4 * ' PROBLEMS 1. A vessel containing water moves horizontally along a straight line with a constant velocity of 10 ft. per second. What is the form of its water surface? 2. A vessel partly filled with water moves horizontally with a constant, linear acceleration of 10 ft. per second per second. What is the form of its water surface? 3. An open cylindrical vessel, 2 ft. in diameter, 3 ft. high and two-thirds filled with water, rotates about its vertical axis with a constant speed of QOR.PfM. Determine: (a) The depth of water at the center of the vessel. (6) The total pressure on the cylindrical walls. (c) The total pressure on the bottom of the vessel. 4. In Problem 3, what is the greatest speed in revolutions per minute that the vessel can have without causing any water to spill over the sides? 5. In Problem 3, what speed in revolutions per minute must the vessel have in order that the depth at the center will be zero? 6. In Problem 3, what speed in revolutions per minute must the vessel have in order that there may be no water within 6 in. of the vertical axis? 50 RELATIVE EQUILIBRIUM OF LIQUIDS 7. If a closed cylindrical vessel, 2 ft. in diameter, 3 ft. high and completely filled with water, rotates about its vertical axis with a speed of 240 R.P.M., determine the intensity of pressure at the following points: (a) At the circumference, just under the cover. (6) At the axis, just under the cover. (c) At the circumference, on the bottom. (d) At the axis, on the bottom. CHAPTER VI PRINCIPLES OF HYDROKINETICS 37. Introductory. The principles relating to the behavior of water or other liquids at rest are based upon certain definite laws which hold rigidly in practice. In solving problems involving these principles it is possible to proceed by purely rational methods, the results obtained being free from any doubt or ambiguity. Calculations are based upon a few natural principles which are universally true and simple enough to permit of easy application. In all problems ordinarily encountered in hydrostatics, after the unit weight of the liquid has been determined, no other experi- mental data are required. A liquid in motion, however, presents an entirely different condition. Though the motion undoubtedly takes place in accordance with fixed laws, the nature of these laws and the influence of the surrounding conditions upon them are very complex and probably incapable of being expressed in any exact mathematical form. Friction and viscosity affect the laws of hydrokinetics in a varying degree for different liquids. Since water is the most common liquid with which the engineer has to deal and since, as a result, more is known about the laws relating to the flow of this liquid, the following treatise on hydrokinetics applies only to water. The fundamental principles discussed hold true for all liquids, but the working formulas would necessarily have to be modified for each different kind of liquid. A clearer conception of the underlying principles of hydro- kinetics is made possible by the assumption of certain ideal conditions. This also permits of the establishment of a few basic laws which may be expressed as fundamental formulas. These assumed conditions, however, vary widely from those which actually exist and working formulas based upon them must invari- ably be modified by experimental coefficients. 51 52 PRINCIPLES OF HYDROKINETICS A formula with its empirical coefficients included, which requires only that numerical values be affixed to the coefficients to make it adaptable to the solution of problems, is referred to as a base formula. Many formulas used in hydrokinetics differ so widely from the fundamental form that they have little if any claim to a rational basis. During the last two centuries many hundreds of experiments on flowing water have been performed. These experiments have covered a wide range of conditions, and the data obtained from them make possible the modern science of hydrokinetics. 38. Friction. There can be no motion between two substances in contact without friction. This principle applies to liquids and gases as well as solids. Water flowing in any conduit encounters friction with the surfaces with which it comes in contact. There is also friction between the moving particles of water themselves, commonly called viscosity (Art. 6). The free surface of water flowing in an open channel encounters the resistance of the air and also the greater resistance of the surface skin which results from surface tension (Art. 7). The amount of frictional resistance offered by any surface increases with the degree of roughness of the surface. The resistance which results from viscosity decreases as the tempera- ture of the water increases. The influence of friction and viscosity on the flow of water must be determined experimentally. To overcome frictional resistance requires an expenditure of energy. The expended energy is transformed into heat. After being so transformed it cannot, through the ordinary processes of nature, be reconverted into any of the useful forms of energy contained in flowing water and is therefore often referred to as lost energy. 39. Discharge. The rate of flow or the volume of water pa'ssing a cross-section of a stream in unit time is called the discharge. The symbol Q will be used to designate the discharge in cubic feet per second. Other units of discharge, such as cubic feet per minute, gallons per minute or gallons per day are sometimes employed for special purposes. If a uniform velocity at all points in the cross-section of a stream were possible there would be passing any cross-section every second a prism of water having a base equal to the cross-sectional area of the stream and a length equal to the velocity. Because, STEADY FLOW AND UNIFORM FLOW S3 however, of the varying effects of friction and viscosity, the different filaments of water move with different velocities. For this reason it is common in hydraulics to deal with mean velocities. If v is the mean velocity in feet per second past any cross-section, and a is the cross-sectional area in square feet, Q = av ....... . (1) and These simple formulas are of fundamental importance. 40. Steady Flow and Uniform Flow. If the same quantity of water passes any cross-section of a stream during equal successive intervals of time the flow is said to be steady. If the quantity of water passing any cross-section changes during successive intervals of time the flow is said to be unsteady. If not otherwise stated, the condition of steady flow will be assumed. The fundamental principles and formulas based upon steady flow do not generally hold for unsteady flow. ^Problems most commonly encountered in practice deal only with steady flow. If in any reach of a stream the velocity at every cross-section is the same at any instant the flow is said to be uniform. This condition requires a stream of uniform cross-section. If the cross- section is not uniform throughout the reach, in the portions of the reach where velocity changes occur, the flow is non-uniform. Thus, uniform flow implies instantaneous similarity of condi- tions at successive cross-sections, whereas steady flow involves permanency of conditions at any particular cross-section. 41. Continuity of Discharge. When, at any instant, the dis- charge is the same past every cross-section, it is said to be continu- ous, or there is continuity of discharge. The term continuity of flow is also used to express this condition. Letting Q, a and v represent, respectively, discharge, area and mean velocity with similar sub- scripts applying to the same cross-section, continuity of discharge exists when etc ..... . (3) Continuity of discharge may be illustrated by assuming water to be turned into a canal. At first there will be a greater volume #f water flowing near the entrance than at sections farther down. 54 PRINCIPLES OF HYDROKINETICS Under such conditions the discharge is not continuous. Ulti- mately, however, if the supply of water is constant and assuming no losses from seepage or evaporation, there will be the same quan- tity of water flowing past all sections of the canal, and the condi- tion of continuity of discharge will exist in the entire canal regard- less of whether or not all reaches of the canal have the same cross- sectional area. In a pipe flowing full, even though the pipe is made up of several diameters, the discharge is continuous. 42. Stream Line and Turbulent Motion. Flowing water is said to have stream-line motion if each particle follows the same path as was followed by every preceding particle that occupied the same position. If stream-line motion exists within a conduit having parallel sides the paths of the water particles are parallel to the sides of the conduit and to each other. Water flows with stream-line motion only at very low velocities, (a) FIG. 34. excepting in very small pipes where such motion may exist at quite high velocities (Art. 90). Under practically all conditions encoun- tered in the field of engineering, the motion is turbulent, the water particles moving without any regularity and not in accordance with any known laws. In Fig. 34 (a) and (6) represent, respect- ively, stream-line and turbulent motion. Friction and viscosity affect the flow of water whether the motion be stream line or turbulent, but the effects produced in the two cases are in accordance with different laws (Art. 90). 43. Energy and Head. Since the principles of energy are applied in the derivation of fundamental hydraulic formulas, an explanation of such principles as will be used is here introduced. Energy is defined as ability to do work. Where the English system of units is employed, both energy and work are measured in foot-pounds. The two forms of energy commonly recognized are kinetic energy and potential energy. Kinetic energy is the ability of a mass to do work by virtue of its ENERGY AND HEAD 55 velocity. Where v is the velocity in feet per second and M , the mass in gravitational units, is equal to W/g, the kinetic energy of any mass is expressed by the equation V 2 V 2 which reduces to for a weight of unity. The expression 4 is of the form feet per second X feet per second ,. , . - j - -- -5 - = feet, feet per second per second and it therefore represents a linear quantity expressed -in feet. It is the distance which a body must fall in a vacuum to acquire the velocity v. When applied to flowing water it is called the velocity head. Although representing a linear quantity, the velocity head is directly proportional to the kinetic energy of any mass having a velocity v and is equal to the kinetic energy of one pound of any substance moving with a velocity v. Potential energy is latent or potential ability to do work. Water manifests this ability in two ways: (a) By virtue of its position or elevation with respect to some arbitrarily selected horizontal datum plane, considered in con- nection with the action of gravity. This may be called energy of position, energy of elevation or gravitational energy. (6) By virtue of pressure produced by the action of gravity, or by the application of some external force, on the water. This may be called pressure energy. Energy of position may be explained by considering a mass having a weight of W pounds whose elevation above any horizontal datum plane is h feet. With respect to this plane the mass has Wh foot-pounds of energy. A mass weighing one pound will have h foot-pounds of energy. If a mass weighing one pound is placed h feet below the datum plane, its energy with respect to the plane will be h foot-pounds, being negative because this amount of energy will have to be exerted upon the mass to raise it to the datum plane against the action of gravity. Here again the expression for energy, in this case h, represents a linear quan- tity which is the elevation head of the mass, but it should be kept clearly in mind that it is also the energy expressed in foot-pounds 56 PRINCIPLES OF HYDROKINETICS contained in one pound of water by. virtue of its position with respect to the datum plane. It thus appears that the amount of energy of position possessed by a mass depends upon the elevation of the datum plane. In any particular problem, however, all masses should be referred to the same plane. This gives the relative amounts of energy contained in different masses or the relative amounts of energy in the same mass in different positions, which is all that is usually required. The action of pressure energy is illustrated by the piston and n cylinder arrangement shown R in Fig. 35, which is operated -p entirely by water under a gage pressure of p pounds per square foot. The area of the piston is A square feet. The cylinder is supplied with water through the valve R and may be emptied through the valve S. At the beginning of the stroke the piston is at CD, the valve S is closed and R is open. Water enters the cylinder and slowly drives the piston to the right against the force P. Neglecting friction, the amount of work done on the piston while moving through the distance I feet is Pl = pAl foot-pounds. If R is now closed and S opened, again neglecting friction, the piston may be moved back to its original position without any work being done upon it. The quantity of water required to do the work, pAl foot- pounds, is Al cubic feet and its weight is wAl pounds. The amount of work done per pound of water is therefore pAl p f r-j = foot-pounds. wAl w Since this work is done entirely at the expense of pressure energy and while the gage pressure is being reduced from p to zero, the p amount of pressure energy per pound of water is ~~ foot-pounds. It has been shown in Art. 17 that represents pressure head or a linear quantity. If pressure head is expressed .in feet of water column, it will also represent foot-pounds of energy per pound "of water as has been shown to be the case for velocity BERNOULLI'S THEOREM 57 head and elevation head. There are other forms of energy, such as heat energy and electrical energy, which have no direct bearing on the laws governing flowing water. The three forms of energy which water may have are illus- trated in Fig. 36. At A water is moving with a velocity v. The kinetic energy of a pound of water at A is v 2 /2g, the pressure energy is =h W and the energy of position referred to the datum plane MN is z. Thus, with respect to the plane MN the total energy per pound of water at any point A, expressed in foot-pounds, is (5) - FIG. 36. FIG. 37. 44. Bernoulli's Theorem. In 1738 Daniel Bernoulli, an Italian engineer, demonstrated that in any stream flowing steadily without friction the total energy contained in a given mass of water is the same at every point in its path of flow. In other words kinetic energy, pressure energy and energy of position may each be converted into either of the other two forms, theoretically without loss. Thus if there is a reduction in the amount of energy contained in any one form there must be an equal gain in the sum of the other two. In Fig. 37, bcde represents a filament of water flowing with steady stream-line motion surrounded by other water moving with the same velocity as that of the filament. Under these conditions 58 PRINCIPLES OF HYDROKINETICS the frictional loss that occurs will be extremely small and for the present it will be ignored. Every particle of water passing the section 6c will, a little later, pass the section de and no water will pass the section de which did not previously pass be. Consider now the forces acting on this filament of water. On the section be whose area is ai there is a normal pressure in the direc- tion of flow of intensity p\ producing motion. On the section de whose area is #2 there is a normal intensity of pressure p2 parallel with the direction of flow and resisting motion. On the lateral surface of the filament, indicated by the lines bd and ce } there is a system of forces acting normal to the direction of motion, which have no effect on the flow and can therefore be neglected. The force of gravity, equal to the weight of the filament, acts downward. The work performed on the filament by the three forces will now be investigated. Consider that in the time dt the particles of water at be move to b'c' with a velocity v\. In the same time interval the particles at de move to d'e' with a velocity v%. Since there is continuity of flow, The work, G\, done by the force acting on the section be in the time dt is the product of the total force and the distance through which it acts, or foot-pounds ...... (6) Similarly the work done on the section de is Gz = p2d2V2dt foot-pounds, .... (7) being negative because p2 is opposite in sense to p\ and resists motion. The work done by gravity on the entire mass in moving from the position bcde to b'c' d'e' is the same as though bcb'c 1 were moved to the position ded'e' and the mass b'c'de were left undisturbed. The force of gravity acting on the mass bcb'c' is equal to the volume aividt times the unit weight w. If z\ and 22 represent, respectively, the elevations of the centers of gravity of bcb'c' and ded'e' above the datum plane MN, the distance through which the force of BERNOULLI'S THEOREM 59 gravity would act on the mass bcb'c' in moving it to the position ded'e' is 21 22 and the work done by gravity is G3=waiV] L dt(zi 22) foot-pounds. . . . (8) The resultant gain in kinetic energy is Mv2 2 Mv\ 2 waiVidt, ~2 2 2i-W-V 0) From fundamental principles of mechanics, the total amount of work done on any mass by any number of forces is equal to the resultant gain in kinetic energy. Therefore from equations (6), (7), (8) and (9). wa\v\dt 2g Dividing through by waiVidt and transferring, and remembering that a\v i = a,2V2, there results w w This is known as Bernoulli's equation. It is the mathe- matical expression of Bernoulli's theorem which is in reality the law of conservation of energy applied to flowing water. It may be stated as follows: Neglecting friction, the total head, or the total amount of energy per pound of water, is the same at every point in the path of flow. Water invariably suffers a loss of energy through friction in flowing from one point to another. If the direction of flow is from point 1 to point 2, the total energy at 2 must be less than at 1. In order to make equation (11) balance, a quantity, h f , equal to the loss of energy, or what is equivalent, the loss of head due to friction between the two points, must be added to the right-hand side of the equation. Including the loss of head due to friction Bernoulli's equation becomes This equation is the basis of all rational formulas used in hydrokinetics. It is the foundation of the science. 60 PRINCIPLES OF HYDROKINETICS 45. Application of Bernoulli's Theorem to Hydrostatics. Although in hydrostatics it is not necessary to make use of Ber- noulli's equation, it is interesting to note that it applies to water at rest as well as GO water in motion. The points 1 and 2 (Fig. 38) are hi and fo, respectively, below the free surface of a liquid at rest and z\ and z 2 above the horizontal plane MN. The liquid being at rest, v\ and V2 equal zero, and, since without velocity there can be no friction, h f is zero. Therefore equation (12) reduces to or, transposing, From the figure w w w w (13) (14) FIG. 38. J FIG. 39. Substituting this value of 22 21, equation (14) may be written Pi-p2 = w(hi-h 2 ), (15) which is the same as equation (3), page 11. 46. Bernoulli's Theorem in Practice. Bernoulli's theorem (Art. 44) is based upon the assumptions of steady flow, stream-line motion and continuity of discharge. Under ordinary conditions water flows with turbulent motion (Art. 90) whereas stream-line motion is assumed in applying Bernoulli's theorem. The effect of turbulence is to increase the losses and therefore this additional loss is included with the loss of head due to friction and no further correction is necessary. It is permissible to write Bernoulli's equation between any two points on any assumed line of flow provided it is known that there is flow between the two points. Thus, in Fig. 39, which VENTURI METER 61 represents water discharging from a reservoir through a pipe, Bernoulli's equation may be written between points A and B. The relation obtained from this equation is then assumed to hold for each of the filaments in the pipe. 47. Venturi Meter. An illustration of the practical use of Bernoulli's equation is provided by the Venturi meter. This instrument, which is used for measuring the discharge through pipes, was invented by Clemens Herschel and named by him in honor of the original discoverer of the principle involved. A Venturi meter set in an inclined position is illustrated in Fig. 40, FIG. 40. Venturi meter. It consists of a converging section of pipe BC and a longer diverging section DE, the smaller ends being connected by a cylindrical section CD, called the throat. The larger ends B and E, termed the inlet and outlet, respectively, have the same diameter as the pipe line in which the meter is to be installed. Let ai,vij pi and z\ represent the area, velocity, pressure and elevation, respectively, at point 1 in the inlet. Also let a 2 , i2, P2 and Z2 represent the corresponding quantities at point 2 in the throat. Writing Bernoulli's equation between points 1 and 2, neglecting friction, vi 2 , Pi 62 PRINCIPLES OF HYDROKINETICS If piezometer tubes are connected at the inlet and throat, the heights of water in these tubes afford a measure of the pressure energy at the two points. Since the relative elevations of points 1 and 2 are also known, the only unknowns in equation (16) are vi and V2. For any given diameters of throat and inlet the corre- sponding areas can be found, and since V2 can be expressed in terms of v\, and substituting this equivalent value in equation (16) leaves v\ as the only unknown. Solving the equation for v\ and multiplying the result by ai gives the discharge through the pipe. For a given discharge the difference between elevations of water surfaces in the two piezometers will be the same regardless of FIG. 41. whether the meter is horizontal or inclined. If the meter is assumed to be rotated in a vertical plane about point 2 until it is in a horizontal position, the rate of discharge through the meter being unchanged, the total amount of energy contained in the water at the inlet must be the same as before the meter was rotated. Since v\ has remained constant +21 must also have w remained constant. The same reasoning applies to point 2, and hence the difference in elevation of the water levels in the two piezometers is constant for all angles of inclination. Venturi meters are usually installed in an approximately horizontal position. Example. A Venturi meter having a throat 4 in. in diameter is installed in a 12-in. pipe line. A mercury U-tube connected as shown in Fig. 41 shows a difference in height of mercury columns *&2x VENTURI METER 63 of 9 in., the remainder of the tube being filled with water. Find the rate of discharge, Q, in cubic feet per second, neglecting friction. Writing Bernoulli's equation between points 1 and 2 Since the angle of inclination will not affect the result, the meter can be assumed to be horizontal. Then z\ and 22 cancel and equation (16) becomes __ == _ 2g 2g w w From the given data ^ ^ E1_? = | X13 . 6 _ =9 .45. w w 12 12 The areas of circles being proportional to the squares of their diameters, ^-I 2 _!-Q a 2 ~4 2 ~ and since the flow is continuous and V2 a* Substituting these results in equation (17) and reducing 5^ = 9.45 9. vi = 2 . 76 ft. per second. 1 2 X7T Q = ai vi= ~ X 2. 76 = 2. 17 cu. ft. per second. The pressure in the throat of a Venturi meter is always much less than at the entrance. This may be seen from the following equation for a horizontal meter: vi 2 pi_v 2 2 p 2 64 PRINCIPLES OF HYDROKINETICS Since the velocity increases from point 1 to point 2 there must be a corresponding decrease in pressure, otherwise there would be an increase in the total amount of energy per pound of water. In the practical use of the Venturi meter the loss of head due to friction, though small, should not be neglected. Consider first the theoretical equation for the horizontal meter : v_l>Pi = v_l + P2 ( 2g^w 2g^w or il^l = ^^ =h (20) 20 w w h being the difference in pressure at points 1 and 2, measured in feet of water column. Since substituting, (20) becomes IjV-tfi 2 - = h (22) 20 This expression reduces to (23) Since equation (23) does not include the loss of energy (or head) resulting from friction, it gives a greater velocity than is ever obtained. In order to correct the formula for friction loss an empirical coefficient, K, is applied to it. The discharge through the meter is given by the formula (24) Substituting the value of v\ given in equation (23) and including the coefficient K, the formula for discharge through a Venturi meter becomes (25) PITOT TUBE 65 The value of K is affected by the design of the meter and also by the degree of roughness of its inner surface. It has been found from experiments that K usually lies between 0.97 and 0.99. 48. Pitot Tube. Fig. 42 illustrates several tubes immersed vertically in a stream of flowing water. The upper ends of the tubes are open and exposed to the atmosphere. At the same depth, ha, there is an opening in each tube which allows free communication between the tube and water in the stream. The velocity of the water at depth h* is v. Tubes (a), (b) and (c) are similar, being bent through an angle of 90, the tip of each tube being open. When the open end of such a tube is directed against the current as shown by (a), the velocity of the water causes water to rise in the tube a distance h above the free surface of the stream. It is shown later in this article that h is equal to the head due to the velocity v. If the same tube is placed with the open end directed down- stream as in (6) the pressure at the opening is less than h d and the water surface in the tube is a certain distance, hi, below the surface of the stream. A similar condition exists when the tube is placed with its lower leg transverse to the stream as shown in (c). In this case, according to experiments by Darcy, the vertical distance, h%, of the water surface in the tube below the water surface of the stream is a little greater than hi. Similarly for (d), which is a straight tube open at each end, there is a depression, /i3, of the water column in the tube. The tube (e) is the same as (a) except that the tip of the tube is closed and there is a small hole on each side of the lower leg. If this tube is held with the 66 PRINCIPLES OF HYDROKINETICS lower leg parallel to the direction of flow the water surface in the tube remains at about the same elevation as the water surface of the stream, hi, h 2 and h% are all less than the velocity head, but directly proportional to it. From experiments by Darcy the following approximate values of hi and h 2 were obtained : and fc 2 = The conditions of flow affecting the height of water column in tube (d) are similar to those encountered when piezometer tubes (Art. 21) extend through the conduit walls into the stream. Piezometer tubes are designed to measure pressure head only and in order that their readings may be affected a minimum amount h e FIG. 43. Pitot tube. FIG. 44. by the movement of the water, their ends should be set flush with the inner surface of the conduit and they should never project beyond this surface. A bent L-shaped tube with both ends open, similar to Fig. 42 (a), is called a Pilot tube, from the name of the French investigator who first used such a device for measuring the velocity of flowing water. When a Pitot tube is first placed in the position shown in Fig. 43, water enters the opening at e until the water surface within the tube rises a distance h above the surface of the stream. A condition of equilibrium is then established and the quantity of water contained within the tube will remain unchanged as long as the flow remains steady. Assuming stream-line motion, the conditions of flow near the entrance to the tube will be as shown in Fig. 44. There will be a volume of dead water in the tube, the PITOT TUBE 67 upstream limit of which is not definitely known, but which may be represented by some line such as abc or ab'c or by the interme- diate line ab"c. Since stream-line motion is assumed, there must be some such surface of quiescent water on the adjacent upstream side of which particles of water will be moving with an extremely low velocity. Consider a particle of water flowing from dtoe,d being on the axis of the tube far enough upstream so that the velocity is not affected by the presence of the tube, e being on the upstream sur- face of the quiescent water above referred to and at the same elevation as d. As this particle flows from d to e its velocity is gradually retarded from Vd to practically zero at e. The velocity head at e may therefore be called zero. Based on the above assumptions and neglecting friction, Bernoulli's equation between the points d and e may be written From Fig. 43 since from the figure h e h d = = h e and -*, w w w w Substituting in equation (26) (27) Since d is a point at any depth, the general expression may be written or v = V2gh ....... (29) Hence the velocity head at d is transformed into pressure head at e and because of this increased pressure inside the tube a column v 2 of water will be maintained whose height is above the water level outside. 68 PRINCIPLES OF HYDROKINETICS Pitot tubes of the type shown in Fig. 43 are not practicable for measuring velocities because of the difficulty of determining the height of the water surface in the tube above the surface of the stream. In order to overcome this difficulty Darcy used an instrument with two L-shaped tubes as shown in Fig. 45. One tube is directed upstream and the other downstream, the two tubes being joined at their upper ends to a single tube connected with ^ an air pump and provided with a stopcock at A. By opening the stopcock and pumping some of the air from the tubes both water columns are raised an equal amount, since the pressure on their surfaces is reduced equally. FIG. 45. FIG, 46. The stopcock can then be closed and the difference in height of water columns can be read. A tube of this kind requires rating, since the downstream leg, which is the same as Fig. 42 (6), does not measure the static pressure of the water. The difference in height of water columns, v 2 hi, is greater than the velocity head, , but the ratio between the t/ two is approximately constant. The value of this ratio must be determined experimentally for each instrument. The tubes shown in Fig. 46 give a difference in elevation of water columns practically equal to the velocity head. The leg measuring static head is similar to Fig. 42 (e). Pitot tubes may be rated by determining their readings when placed in water flowing with a known velocity or when they are moved at a known rate through still water. If the Pitot tube is used for measuring velocity in a closed PROBLEMS 69 conduit flowing under pressure, two tubes are absolutely essential, one measuring both pressure and velocity and the other measuring pressure only. PROBLEMS 1. The diameter of a pipe changes gradually from 6 in. at A to 18 in. at B. A is 15 ft. lower than B. If the pressure at A is 10 Ibs. per square inch and at B, 7 Ibs. per square inch when there are 5.0 cu. ft. per second flowing, determine : (a) The direction of flow. (6) The frictional loss between the two points. 2. If in Problem 1 the direction of flow is reversed, determine the pressure at A if all other factors, including the frictional loss, remain the same. 3. In Problem 1, determine the diameter of pipe at B in order that the pressure at that point will also be 10 Ibs. per square inch, all other factors remaining constant. 4. Determine the discharge in Problem 1, assuming no frictional loss, all other conditions remaining as stated. 5. What would be the difference in pressure in pounds per square inch between A and B, Problem 1, if there were 6.2 cu. ft. per second flowing, neglecting friction. 6. A siphon having a diameter of 6 in. throughout, discharges from a reservoir, A, into the air at B. The summit of the siphon is 6 ft. above the water surface in A and 20 ft. above B. If there is a loss of 3 ft. head between A and the summit and 2 ft. between the summit and B, what is the absolute pressure at the summit in pounds per square inch? Also determine the rate of discharge in cubic feet per second and in gallons per day. 7. In Problem 6 determine the absolute pressure in pounds per square inch at the summit and the discharge in cubic feet per second if the diameter at the summit is 5 in. and at the outlet, 6 in., all other data remaining the same. 8. A flaring tube discharges water from a reservoir at a depth of 36 ft. below the water surface. The diameter gradually increases from 6 in. at the throat to 9 in. at the outlet. Neglecting friction determine the maximum possible rate of discharge in cubic feet per second through this tube. What is the corresponding pressure at the throat? 9. In Problem 8 determine the maximum possible diameter at the outlet at which the tube will flow full. 10. A jet of water is directed vertically upward. At A its diameter is 3 in. and its velocity is 30 ft. per second. Neglecting air friction, determine its diameter at a point 10 ft. above A. 11. Water is delivered by a scoop from a track tank to a locomotive tender that has a speed of 20 mi. per hour. If the entrance to the tender is 7 ft. above the level of the track tank and 3 ft. of head is lost in friction at what velocity will the water enter the tender? 12. In Problem 11 what is the lowest possible speed of the train at which water will be delivered to the tender? 70 PRINCIPLES OF HYDROKINETICS 13. A Venturi meter having a diameter of 6 in. at the throat is installed in an 18 in. water main. In a differential gage partly filled with mercury (the remainder of the tube being filled with water) and connected with the meter at the throat and inlet, the mercury column stands 15 in. higher in one leg than in the other. Neglecting friction, what is the discharge through the meter in cubic feet per second? 14. In Problem 13 what is the discharge if there is 1 ft. of head lost between the inlet and the throat, all other conditions remaining the same. 15. In Problem 13 what would be the difference in the level of the mercury columns if the discharge is 5.0 cu. ft. per second and there is 1 ft. of head lost between the inlet and throat? 16. A Venturi meter is installed in a 12-in. water main. If the gage pressure at the meter inlet is 8 Ibs. per square inch when the discharge is 3.0 cu. ft. per second determine the diameter of the throat if the pressure at that point is atmospheric. Neglect friction. 17. A flaring tube discharges water from a vessel at a point 10 ft. below the surface on which the gage pressure is 8.5 Ibs. per square inch. If the diameter of the throat is 4 in., at which point the absolute pressure is 10 Ibs. per square inch, determine the discharge in cubic feet per second, neglecting friction. 18. In Problem 17 what is the diameter of the tube at the discharge end? CHAPTER VII FLOW OF WATER THROUGH ORIFICES AND TUBES 49. Description and Definitions. As commonly understood in hydraulics, an orifice is an opening with a closed perimeter and of regular form through which water flows. If the perimeter is not closed or if the opening flows only partially full the orifice becomes a weir (Art. 68). An orifice with prolonged sides, such as a piece of pipe two or three diameters in length set in the side of a reservoir, is called a tube. An orifice in a thick wall has the hydraulic proper- ties of a tube. Orifices may be circular, square, rectangular or of any other regular shape. The stream of water which issues from an orifice is termed the jet. An orifice with a sharp upstream edge so formed that water in passing touches only this edge is called a sharp-edged orifice. The term velocity of approach as applied to orifices means the mean velocity of the water in a channel leading up to an orifice, The portion of the channel where the velocity of approach is considered to occur is designated the channel of approach. An orifice is spoken of as a vertical or hori- zontal orifice depending upon whether it lies in a vertical or horizontal plane. 60. Characteristics of the Jet. Fig. 47 represents a sharp-edged, circular orifice. The water particles approach the orifice in converging directions as shown by the paths in the figure. Because of the inertia of those particles whose velocities have components par- allel to the plane of the orifice, it is not possible to make abrupt changes in their directions the instant they leave the orifice and they therefore continue to move in curvilinear 71 FIG. 47. Sharp-edged orifice. 72 FLOW OF WATER THROUGH ORIFICES AND TUBES paths, thus causing the jet to contract for some distance beyond the orifice. This phenomenon of contraction is referred to as the contraction of the jet and the section where contraction ceases is called the vena contracta. The vena contracta has been found to be at a distance equal to about one-half the diameter of the orifice from the plane of the orifice, at a in figure. Beyond the vena contracta the cross-sectional area of the jet does not undergo any change excepting insofar as it is affected by gravity. If the direction of the jet is vertically upward (Fig. 51), or if it has an upward component, the force of gravity retards its velocity and thus increases its cross-sectional area; and, con- versely, if the direction of the jet has a downward component, gravity increases its velocity and decreases its cross-sectional area. If a jet discharges into the air the pressure within the jet at its vena contracta and beyond is atmospheric pressure. This may be seen by investigating the conditions which will ^esult from assuming pressures greater or less than atmospheric pressure. If, for example, the pressure within a cross-section is assumed to be greater than atmospheric pressure, there will be an unbalanced pressure along every radius of the section that is, the pressure at the center will be greater than at the circumference. Since water is incapable of resisting tensile stress this would cause the jet to expand. In a similar manner if the internal pressure is assumed to be less than atmospheric, since water unconfined is incapable of resisting a compressive force, the unbalanced pressure will produce an acceleration and therefore a further contraction. Since neither expansion nor contraction occurs, it follows that the pressure throughout the vena contracta must be atmospheric pressure. Between the plane of the orifice and the vena contracta the pressure within the jet is greater than atmospheric pressure because of the centripetal force necessary to change the direction of motion of the particles. That this pressure must be greater than atmospheric can easily be proved by writing Bernoulli's equation between a water particle in the jet back of the vena contracta and another particle in the vena contracta. The form assumed by jets issuing from orifices of different shapes presents an interesting phenomenon. The cross-section of the jet is similar to the shape of the orifice until the vena con- FUNDAMENTAL ORIFICE FORMULA 73 o tracta is reached. Fig. 48 shows various cross-sections of jets issuing from square, triangular and elliptical orifices. The left- hand diagram in each case is a cross-section of the jet near the vena contracta. The following diagrams are cross-sections at suc- cessively greater distances from the orifice. This phenomenon, which is common to all shapes of orifices excepting circular orifices, is known as the inversion of the jet. After passing through the fourth stage A O Y O shown in the figure the jet reverts ^ ^ , . . . & FIG. 48. Form of jet from square, to its original form and continues triangular and elliptical orifices, to pass through the cycle of changes described above as long as it flows freely or is not broken up by wind or air friction. 51. Fundamental Orifice Formula. Fig. 49 represents the general case of water discharging through an orifice. In the , FIG. 49. Discharge from orifice. derivation of the fundamental formula it will be assumed that the water flows without friction and also that there is no contraction of the jet and therefore no pressure within the jet in the plane of the orifice. In order to write a general expression applicable to all filaments, it will be necessary to make the further assump- tion that all of the water particles in a cross-section of the channel of approach flow with the same velocity. There are two chambers, A and B, the gas pressure in chamber A being p A and in chamber B being p B , the relation of p A to p B 74 FLOW OF WATER THROUGH ORIFICES AND TUBES being such that water will flow from chamber A to chamber B. The flowing water may be considered to be made up of filaments of which mn is one, m being a point in the water in chamber A and n a point in the jet in the plane of the orifice. The filament passes through the orifice at a distance h below the surface of the water. The point m is at a distance h m below the water surface and at a distance z above n. v m is the velocity at m and v n is the velocity at n. Bernoulli's equation may be written between the points m and n as follows : and since h m -}-z = and These formulas are general expressions of relation between velocity and head for any filament. Since the filaments at different elevations discharge through a vertical orifice under different heads their velocities are not the same. Where, however, the head is large in comparison with the height of the opening, the mean velocity of the jet may be taken as the velocity due to the mean head. The theoretical mean velocity thus obtained may be represented by the symbol v t . Introducing also the assumption that all of the water particles in a cross-section of the channel of approach flow with the same velocity, V; v n and v m in formulas (2) and (3) may be replaced, respectively, by v t and V, which gives 2g 2g w w and From the definition (Art. 49) V is the velocity of approach. The condition most commonly encountered is that illustrated FUNDAMENTAL ORIFICE FORMULA 73 in Fig. 50, where the surface of the water and the jet are each exposed to the atmosphere. In this case p A = p B = atmospheric pressure and formulas (4) and (5) reduce, respectively, to (6) and If the cross-sectional area of the reservoir or chan- nel leading up to the orifice is large in com- parison with the area of the orifice the velocity of approach, V, may be called zero and equations (6) and (7) reduce, respect- ively, to and FIG. 50. Orifice with water surface and jet subjected to equal pressures. ,,2 (8) (9) These formulas express the theoretical relation between head and velocity for an orifice discharging from a relatively large body of water whose surface is subjected to the same pressure as the jet. It is under this condition that discharge from orifices ordinarily occurs and the above formulas are the ones most commonly used. Since these formulas also express the relation between potential head and velocity head (Art. 43) -they have a wide appli- cation in hydraulics. Formula (9) is also the formula for the velocity acquired by a body falling a distance h through space. The theoretical velocity of water flowing through an orifice is therefore the velocity acquired FIG. 51. Horizontal orifice. 76 FLOW OF WATER THROUGH ORIFICES AND TUBES by a body falling freely in vacuo through a distance equal to the head on the orifice. This principle, discovered by Torricelli in 1644, is known as Torricelli's theorem. Fig. 51 illustrates a hori- zontal orifice discharging under a head h. According to Torri- celli's theorem the jet should rise to a height h, but experiments show that the actual height to which the jet rises is slightly less than h. The discrepancy is due to the retarding effects of friction and viscosity. This matter is discussed more fully in the following article. 52. Orifice Coefficients. The assumptions which were made in the derivation of formula (5) may be summarized briefly as follows : (a) All water particles in a cross-section of the channel of approach flow with the same velocity. (6) There is no contraction of the jet. (c) The water flows without friction. Since these conditions do not in reality exist, it is necessary to modify the derived formulas to make them applicable to actual conditions. To accomplish this, three empirical coefficients are applied to formula (5), there being one coefficient to correct for the difference between the assumed conditions and the actual con- ditions for each of the above assumptions. The method of cor- recting for each assumption will be discussed in the order given above. (a) Correction for non-uniformity of velocity in cross-section of channel of approach. The effect of the variation in velocity in a cross-section of the channel of approach that is, the variation in the velocity with which the water particles in the different filaments approach the orifice, is similar to the effect of this condi- tion on the discharge over weirs. The matter being of relatively much greater importance in this connection is taken up under weirs and will not be discussed here. (See Art. 72 (a) .) The commonly accepted method of modifying formula (5) so as to have it include this correction is to apply a coefficient a to V 2 the term . The value of a has not been determined for orifices. It varies with the distribution of velocities in the channel of ap- proach and is always greater than unity. With the coefficient ORIFICE COEFFICIENTS 77 a included, calling v f the velocity after the correction has been applied, formula (5) becomes (b) Correction for contraction. The ratio of the cross-sectional area of the jet at the vena contracta to the area of the orifice is called the coefficient of contraction. Thus, if a' and a are, respect- ively, the cross-sectional area of the jet at the vena contracta and the area of the orifice and C c is the coefficient of contraction, . n o! , C c = or a' = C c a. If v is the actual mean velocity in the vena contracta the discharge through the orifice is en) The mean value of C c is about 0.62. It varies slightly with the head and size of orifice. (c) Correction for friction. The velocity of the jet suffers a retardation due to the combined effects of friction and viscosity. The ratio of the actual mean velocity, v, to the velocity, v', which would exist without friction, has been termed the coefficient of velocity, but it might more properly be called the coefficient of friction. Designating the coefficient of velocity by the sumbol C v C v = or v = C v v'. The average value of C v for a sharp-edged orifice is about 0.98. Substituting the value of v' given in formula (10), the general formula for mean velocity of a jet issuing from an orifice, with the two coefficients to correct, respectively, for friction and the assumption of uniform velocity in a cross-section of the channel of approach, becomes If the pressures p A and p B are equal . . (13) 78 FLOW OF WATER THROUGH ORIFICES AND TUBES If the velocity of approach is so small that it can be called zero without introducing an appreciable error v = C V2^h. . . (14) Substituting the value of v given by formula (12) in (11) the general formula for discharge through an orifice with the three corrective coefficients becomes . . .(15) It is usual to combine C C C V into a single coefficient, C, called the coefficient of discharge. Substituting C for C C C V , the general formula for discharge is .... (16) u/ / If the pressures p A and p B are equal \ (17) If the velocity of approach V is so small that it can be considered to equal zero (18) As orifices are ordinarily used, the pressures p A and p B are equal and the velocity of approach is so small that it may be neglected without appreciable error. Formula (18) is therefore recognized as the common discharge formula for an orifice. In the remainder of this chapter, if not otherwise specified, it will be assumed that the pressure on the water surface is the same as the pressure on the jet, and the velocity of approach will be considered to be so small as to be negligible. Formulas (14) and (18) then become the respective formulas for mean velocity and discharge. Numerical values of C c and C c may be obtained experimentally but an accurate determination is extremely difficult. C c may be obtained approximately by measuring the diameters of the vena contracta and orifice with calipers, the coefficient of contraction being equal to the ratio of the squares of their respective ORIFICE COEFFICIENTS 79 COEFFICIENTS OF DISCHARGE (C) FOR CIRCULAR ORIFICES Head h in feet Diameter of orifice in feet 0.02 0.04 0.07 0.1 0.2 0.6 1.0 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 1.0 .644 .623 .612 .608 .600 .595 .591 1.5 .637 .618 .608 .605 .600 .596 .593 2.0 .632 .614 .607 .604 .599 .597 .595 2.5 .629 .612 .605 .603 .599 .598 .596 3.0 .627 .611 .604 .603 .599 .598 .597 4.0 .623 .609 .603 .602 .599 .597 .596 6.0 .618 .607 .602 .600 .598 .597 .596 8.0 .614 .605 .601 .600 .598 .596 .596 10.0 .611 .603 .599 .598 .597 .596 .595 20.0 .601 .599 .597 .596 .596 .596 .594 50.0 .596 .595 .594 .594 .594 .594 .593 100.0 .593 .592 .592 .592 .592 .592 .592 COEFFICIENTS OF DISCHARGE (C) FOR SQUARE ORIFICES Head h in feet Side of square in feet 0.02 0.04 0.07 0.1 0.2 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 1.0 .648 .628 .618 .613 .605 .601 .599 1.5 .641 .622 .614 .610 .605 .602 .601 2.0 .637 .619 .612 .608 .605 .604 .602 2.5 .634 .617 .610 .607 .605 .604 .602 3.0 .632 .616 .609 .607 .605 .604 .603 4.0 .628 .614 .608 .606 .605 .603 .602 6.0 .623 .612 .607 .605 .604 .603 .602 8.0 .619 .610 .606 .605 .604 .603 .602 10.0 .616 .608 .605 .604 .603 .602 .601 20.0 .606 .604 .602 .602 .602 .601 .600 50.0 .602 .601 .601 .600 .600 .599 .599 100.0 .599 .598 .598 .598 .598 .598 .598 80 FLOW OF WATER THROUGH ORIFICES AND TUBES diameters. A Pitot tube may be used to determine approximately velocities in the vena contracta. The coefficient of discharge may be obtained with great accuracy by measuring the quantity of water flowing from an orifice of known dimensions in a given time and determining the ratio between this discharge and the theoretical discharge. Since in practice it is usually the discharge from orifices that is required, it is the coefficient of discharge that is of greatest value to en- gineers. An average value of the coefficient of discharge is about 0.60. It is not a constant, but varies with the head and also with the shape and size of the opening. On page 79 are tables of values of C for circular and square orifices taken from Hamilton Smith's Hydraulics. Sharp-edged orifices provide an accurate means of measuring small rates of discharge. 53. Algebraic Transformation of Formula with Velocity of Approach Correction. The fundamental orifice formula with velocity of approach correction as derived in Art. 52 is (17) By definition V = Q/A where A is the cross-sectional area of the stream in the channel of approach. Substituting this value of V and reducing, equation (17) becomes CaV2gh ^ I o* (19) Expanding the denominator by the binomial theorem gives a diminishing series and dropping all terms excepting the first two since they will be very small quantities, the formula may be expressed in the approximately equivalent form . (20) The term within the parenthesis is the velocity of approach corrective factor. It becomes unity when the ratio of the orifice to the cross-sectional area of the stream in the channel of approach is so small that it may be considered zero. Where a correction for velocity of approach is required, formula (20) will be found more convenient than formula (17). HEAD LOST IN AN ORIFICE 81 54. Head Lost in an Orifice. Consider water to be discharging from an orifice under a head h (Fig. 52). Because of friction, the velocity of discharge will be less than V2gh or, from formula (14), page 78 ...... (14) The head producing discharge is therefore, /= - C 2 2g' That is, h is the total head, including the lost head. The (19) FIG. 52. Sharp-edged orifice. FIG, 53. Path of jet. head that is not lost is the velocity head due to the actual velocity v. Therefore, Lost head = Total head velocity head or, placing the symbol ho for lost head, v 2 ' ' (22) (23) Formulas (22) and (23) are fundamental and are applicable to any orifice or tube whose coefficient of velocity is known. v 2 For a sharp-edged orifice, since C c = 0.98, h = 0.041 . Since v = C v v t , formula (22) reduces to and substituting the value of C t for a sharp-edged orifice h. 82 FLOW OF WATER THROUGH ORIFICES AND TUBES 55. Path of Jet. When water issues from an orifice the direction of the jet is at first normal to the plane of the orifice but, for orifices not in a horizontal plane, the force of gravity causes it immediately to begin to curve downward. Let x (Fig. 53) be the abscissa and y the ordinate of any point in the path of a jet discharging from a vertical orifice. The space x will be described uniformly in a certain time t and if v is the velocity with which water leaves the orifice x = vt. The jet has a downward acceleration which conforms to the law of falling bodies and therefore ^ y ~ 2' Eliminating t between the two equations (24) which is the equation of a parabola with its vertex at the orifice. Since by formula (14), page 78, v = C v V~2gh, (14) equation (24) may also be written x 2 = 4C*hy. . (25) This formula indicates an experimental method of obtaining C; x, y and h may be measured and substituted in the formula and C v may be computed 56. Orifices under Low Heads. Where the head on a vertical orifice is small in comparison with the height of the orifice there is theoretically an appreciable dif- ference between the discharge obtained by assuming the mean velocity to be that due to the mean head and the discharge ob- tained by taking into considera- tion the variation in head. The FIG. 54.-Rectangular orifice. 6XaCt formula f r octangular orifices is derived as follows: Fig. 54 shows & rectangular orifice of width L and height D, ORIFICES UNDER LOW HEADS 83 the water surface and jet being each subjected to atmospheric pressure, hi and hz are the respective heads on the upper and lower edges of the orifice. Neglecting velocity of approach, the theoretical discharge through any elementary strip of area Ldy at a distance y below the water surface, is given by the equation which integrated between the limits of h% and hi gives -hiK) ...... (26) This formula gives the theoretical discharge from an orifice where the pressures on the water surface and jet are equal and the velocity of approach is considered to be zero. When hi is zero that is, when the water surface does not touch the upper edge of the opening, the formula reduces to , ...... (27) which is the theoretical formula for discharge over a weir without velocity of approach correction (see Art. 70). To make formula (26) applicable to actual conditions a coeffi- cient of discharge must be introduced and the formula becomes -hiK) ...... (28) Values of C for this formula have not been well determined and it is seldom used in practice. Formula (18), which for rectangular orifices may be written ....... (29) h being the head on the center of the orifice and C the coefficient of discharge for rectangular orifices as given on page 79, may be used satisfactorily even for quite low heads since these values of C include corrections for the approximations contained in the formula. The theoretical difference between formulas (28) and (29) may be shown as follows : h being the head on the center of the rectangle, h 2 = h-\-^D and hi = h %D. Substituting these values in equation (26) and expanding them by the binomial theorem, . (30) 84 FLOW OF WATER THROUGH ORIFICES AND TUBES This shows that formula (29) always gives a greater discharge than formula (28) if the same value of C is used in each case. For h = D, the sum of the infinite series is 0.989 and for h = 2D, it is 0.997. The theoretical error introduced by using formula (29) is thus about 1 percent where h = D and 0.3 of 1 per cent where h = 2D. In a manner similar to that described above for rectangular orifices, the discharge for a circular orifice may be shown to be FIG. 55. Orifice with bottom con- traction partially suppressed. FIG. 56. Orifice with bottom con- traction completely suppressed. in which D is the diameter of orifice and h is the head on the center of orifice. This formula gives results differing from those obtained by the approximate formula similar to the corresponding formulas for rectangular orifices. If h = D the sum of the series is 0.992 and if h = 2D the sum is 0.998. Formula (31) is seldom, if ever, used in practice. 57. Suppression of Contraction. The effect of constructing an orifice so as to reduce the contraction is to increase the cross- sectional area of the jet and thus to increase the discharge. If an orifice is placed close to a side or the bottom of a reservoir the tendency of the filaments of water to approach the orifice from all directions (Fig. 55) is restricted and some of the filaments must approach in a direction more nearly parallel to the direction of the jet than they would otherwise. If the orifice is flush with one side or the bottom (Fig. 56) the contraction on that side of the orifice will be wholly suppressed. STANDARD SHORT TUBE 85 In a similar manner, rounding the inner edge of the orifice (Fig. 57) reduces contraction. An orifice constructed to con- form to the shape of the jet which issues from a sharp-edged orifice FIG. 57. Orifice with rounded entrance. FIG. 58. Bell-mouth orifice. FIG. 59. Sharp-edged orifice with extended sides. FIG. 60. Standard short tube. (Fig. 58) is called a bell-mouth orifice. The coefficient of con- traction of such an orifice approaches very close to unity. 58. Standard Short Tube. Extending the sides of an orifice does not affect the discharge so long as the jet springs clear. The 86 FLOW OF WATER THROUGH ORIFICES AND TUBES orifice illustrated in Fig. 59 has a sharp upstream corner and the conditions of flow are the same as for a sharp-edged orifice in a thin plate. When the jet touches the sides of the orifice the conditions of flow are changed. A circular orifice with a sharp edge having sides extended to about 1\ diameters is called a standard short tube. The jet is contracted by the edge of the orifice, as at m (Fig. 60), and for low heads it will expand and fill the tube. For high heads the jet may at first spring clear of the sides of the tube, but by temporarily stopping the tube at its discharging end and allowing the water to escape, the tube can be made to flow full. The moving water carries with it a portion of the air which is entrapped in the space, s, causing a pressure less than atmospheric pressure. The result is to increase the head under which water enters the orifice and therefore the discharge is greater than occurs from a sharp- edged orifice of the same diameter discharging freely into the air. Conditions at the outlet end of the tube will first be considered. By writing Ber- noulli's equation between a point in the reservoir where the velocity of approach may be considered zero and a point, n, in the outlet (Fig. 61) there is obtained, as for an orifice (Art. 51) the relation V,' (9) Since the tube flows full, the coefficient of contraction at the outlet equals unity. It has been found experimentally that the coefficient of dis- charge, C, and therefore the coefficient of velocity, C v , for the outlet equals approximately 0.82, the value of the coefficient varying slightly with the head and diameter of tube. Therefore from formula (14), page 78 FIG. 61. Standard short tube. (32), STANDARD SHORT TUBE 87 and since the coefficient of contraction is unity, Q = CaV2^h = O.S2aV2ih ...... (33) The discharge is thus about one-third greater than for a sharp- edged orifice of the same diameter. To investigate conditions at the contracted portion of the jet, Bernoulli's equation may be written between a point on the water surface, where the velocity is considered zero and the pressure is atmospheric, and a point m in the contracted portion of the jet. Thus = ++lost head ..... (34) Assuming the coefficient of contraction at m to be 0.62, the same as for a sharp-edged orifice discharging freely into the air, and writing the equation of continuity between m and n or v m =l.Glv ........ (35) v 2 The head lost between the reservoir and m (page 81) is 0.04-^-. 20 Substituting these values, equation (34) becomes , . . (36) and substituting v from equation (32) and reducing, ^ = 34-0.8ft ........ (37) There exists, therefore, a partial vacuum at m which will lift a water column 0.8ft (Fig. 61), the pressure being O.Swft less than atmospheric pressure. This has been confirmed experimentally. Evidently the relation does not hold when 0.8ft becomes greater than 34 ft., or when the head becomes greater than approxi- mately 42.5 ft., since this condition gives a negative value to w in equation (37) which is not possible. FLOW OF WATpR THROUGH ORIFICES AND TUBES Jiead in /the entire length of a standard short tube (see Art. 54) is given by the formula (22) v 2 'and since C p = 0.82, the formula gives h = Q.5Q . This case is important since the entrance to a pipe, where the end of the pipe is flush with a vertical wall, is usually considered as a standard short tube and the head lost at entrance to the pipe is taken as the head lost in a standard short tube (see page 156). FIG. 62. Converging tube with sharp-cornered entrance. FIG. 63. Converging tube with rounded entrance. 59. Converging Tubes. Converging tubes having a circular cross-section are frustums of cones with the larger end adjacent to the reservoir. They may have a sharp-cornered entrance as in Fig. 62 or a rounded entrance as in Fig. 63. The jet contracts slightly at a, just beyond the end of the tube. The coefficient of contraction, C c , decreases as the angle of convergence, 6, increases; becoming 0.62 for 0=180 when the tube becomes a sharp-edged orifice. The coefficient of velocity, C v , on the other hand, decreases as 6 decreases. As for any orifice (38) The following table gives coefficients for converging, conical tubes with sharp-cornered entrances, interpolated from experiments by ,1'Aubuisson and Castel. These results are interesting in that Lhcy show the general laws of variation of coefficients but, on NOZZLES 89 account of the small models used in the experiments, they should not be taken as generally applicable to all tubes of this type. COEFFICIENTS FOR CONICAL CONVERGING TUBES Coef- Angle of convergence, 6 (Fig. 62) ficient 5 10 15 20 25 30 40 50 c. 0.829 0.911 0.947 0.965 0.971 0.973 0.976 0.981 0.984 C c 1.000 .999 .992 .972 .952 .935 .918 .888 .859 C 0.829 .910 .939 .938 .924 .911 .896 .871 .845 The coefficient of velocity and therefore the coefficient of discharge is increased by rounding the entrance (Fig. 63), since this reduces the head lost in the tube. The coefficient of con- traction will not be materially changed. Exact values of coeffi- cients will depend upon the extent to which the corner is rounded. Maximum discharge is obtained when the shape of the entrance conforms to the shape of the contracted jet. 60. Nozzles. A nozzle is a converging tube attached to the end of a pipe or hose. The nozzle increases the velocity of the issuing jet, thus increasing the range of distance which it covers. Fig. 64 illustrates two types of nozzles in common use. Each of these has a cylindrical tip of such length that it will flow full, thus preventing contraction and increasing the discharge. The converging part of the tube may be the frustum of a cone as in Fig. 64 (a) or the inside may be convex as in (b). Each of these shapes gives an efficient stream. The following mean values of co- efficients of discharge for smooth nozzles, similar to Fig. 64 (a), having a diameter at the base of 1.55 in., have been determined from experiments by Freeman. 1 FIG. 64. Nozzles. Diameter in inches i 4 i 1 ii u If Coefficient of discharge. . 0.983 0.982 0.980 0.976 0.971 0.959 j 1 JOHN R. FREEMAN: Experiments Relating to Hydraulics of Fire Streams, Trans. Amer. Soc. Civ. Eng., vol. 21, pp. 303-482 (1889). 90 FLOW OF WATER THROUGH ORIFICES AND TUBES Since the coefficient of contraction is unity, the coefficients of discharge given above are also the coefficients of velocity. The nozzle being a form of tube to which formula (22), page 81, applies, the head lost in the nozzle is 20' or substituting values of C v given in the above table, v 2 h = (0.04 to 0.09)^, (22) (39) in which v is the mean velocity at the outlet of the nozzle. Expressed as a function of the velocity, vi, in the hose or pipe having a diameter D, the diameter of the nozzle being d, ..... (40) Bernoulli's equation, for a horizontal nozzle, may be written between a point at entrance to the nozzle and a point in the jet as follows: head > in which p\ is the gage pressure at the entrance, v\ is the velocity at entrance and v is the velocity in the jet. From this equation the pressure at the base of the nozzle, pi, may be determined if the discharge is known or the discharge may be de- termined if pi is known. 61. Diverging Tubes. Fig. 65 represents a coni- cal diverging tube, having FIG. 65. Diverging tube. rounded entrance corners, so that all changes in velocity occur gradually. Such a tube, provided the angle of flare is not too great nor the tube too long, will flow full. The theoretical velocity, v tj at the outlet of the tube, obtained in the same manner as for an orifice (Arts. 51 and 52), is (9) BORDA'S MOUTHPIECE 91 the actual mean velocity being v = C,V2gh> (14) where C v is the coefficient of velocity at the outlet. Experiments indicate that even under favorable conditions the value of C v is small. Venturi and Eytelwein, experimenting with a tube 8 in. long, 1 in. in diameter at the throat and 1.8 in. in diameter at the outer end, obtained results which give a value of C v of about 0.46. The lost head (formula (22), Art. 54) was, therefore, approximately 0.79/i. Even with this large loss of head the discharge through the tube was about two and one-half times the discharge from a sharp- edged orifice having the same diameter as the throat of the tube. The greater portion of the loss of head occurs between the throat and outlet of the tube where the stream is expanding and thus has a tendency to break up in eddies with a waste of energy. Experiments by Venturi indicate that an included angle, 6, of about 5 and a length of tube about nine times its least diameter give the most efficient discharge. A diverging tube, such as that shown in Fig. 65, is commonly called a Venturi tube. The pressure head at the throat is evidently less than atmos- pheric pressure. This may be shown by writing Bernoulli's equation between ra and n. When the throat is so small that Bernoulli's equation gives a negative absolute pressure at m, for- mula (14) no longer holds. The conditions are similar to those already described for a standard short tube, Art. 58. 62. Borda's Mouthpiece. Since the contraction of a jet issuing from an orifice is caused by the water entering the orifice from various directions inclined to the axis of the orifice, it follows that the greater the angle between the extreme directions the greater will be the contraction of the jet. The extreme case occurs in Borda's mouthpiece (Fig. 66), where the water approaches the orifice from all directions. This mouthpiece consists of a thin tube projecting into the reservoir about one diameter. The proportions are such that the jet springs clear of the walls of the tube. Borda's mouthpiece is of interest because it is possible to obtain its coefficient of contraction by rational methods. The cross-sectional area of the jet at the vena contracta, mn, 92 FLOW OF WATER THROUGH ORIFICES AND TUBES is a' and the velocity at this section is v. If a is the area of the opening, the coefficient of contraction C c = a'/a. The size of the reservoir is assumed to be so large in comparison with the area of the orifice that the velocity of the water within the reservoir may be neglected and that the pressure on the walls will, therefore, follow the laws of hydrostatics. Excepting the pressure act- ing on the horizontal pro- jection de of the mouthpiece on the opposite wall, the horizontal pressures on the walls will balance each other. The total pressure on de is wah, which is also FIG. 66. Borda's mouthpiece. the resultant horizontal accelerating force acting on the water entering the mouthpiece. Consider the mass of water xymn to move to the position x'y' m'n' in t seconds. The change in the momentum of the mass con- sidered is the difference in the momentum of the mass xx'yy' and mm'nn'. But the momentum of xx'yy' is entirely vertical, there- fore the change in momentum in a horizontal direction is equal to the momentum of mm'nn', which is produced by the action of the force wah. rp, ,. , , . a'vtw , . . a'v 2 tw The mass of mmnn is and its momentum is - . g . g The impulse of the force wah is waht. Equating impulse and change of momentum, a'v 2 tw what = therefore, and since g a' _gh v = a/_ = a 2C 2 ' RE-ENTRANT TUBES 93 Therefore, assuming the coefficient of velocity to be unity, the coefficient of contraction is theoretically 0.5, or calling the coeffi- cient of velocity 0.98, the same as for a sharp-edged orifice, the coefficient of contraction is 0.52. This value has been verified approximately by experiments. 63. Re-entrant Tubes. Tubes, having their ends project into a reservoir (Fig. 67), and having a length of about 2^ diameters, are called re-entrant or inward-projecting tubes. The action of water in such tubes is similar to that in standard short tubes (Art. 58), except that the contraction of the jet near the entrance is greater, At the discharge end the tube flows full and the FIG. 67. Re-entrant tube. FIG. 68. Submerged orifice. coefficient of velocity therefore equals the coefficient of discharge. Thus C c =l and C V = C. From experiments C p = 0.75. The head lost, from equation (22) (Art. 54) is, therefore, h Q = Q.7S ~. 2 9 This case is important since the entrance to a pipe, which projects into a body of water, may be considered as a re-entrant tube and the head lost at entrance GO the pipe is taken as the head lost in a re-entrant tube (Art. 102). 64. Submerged Orifice. An orifice discharging wholly under water (Fig. 68) is called a submerged orifice. The assumption is usually made that every filament of water passing through the orifice is being acted upon by a head, hi-h 2 = h, the difference in elevation of water surfaces. Based upon this assumption 94 FLOW OF WATER THROUGH ORIFICES AND TUBES and using the same nomenclature as for orifices with free dis- charge (9) and (14) (18) Coefficients of discharge for sharp-edged, submerged orifices are very nearly the same as for similar orifices discharging into the air. The assumption that h,2 is the pressure head on the center of the orifice at its lower side is not strictly true unless all of the velocity head, due to the velocity of the water leaving the orifice, is lost in friction and turbulence as the velocity is reduced to zero. It has been .shown experimentally that less than 90 per cent of this velocity head may be lost. Assuming a loss of 90 per cent, v 2 the pressure head at the center of the orifice is h 2 0. 12 5-. The effect of this condition on the discharge may be investigated by writing Bernoulli's equation between m and n and n and p (Fig. 68). This matter is not of great importance in connection with submerged orifices, since the discrepancy resulting from the use of formula (18) is relatively small and the coefficient of discharge which is determined from experiments eliminates this source of error. The loss of head sustained at the outlet of a pipe discharging into a body of still water is discussed in Art. 102. The conditions of discharge in this case are practically identical with those of the submerged orifice discussed above. 65. Partially Submerged Orifices. Fig. 69 represents a rectangular orifice, the bot- tom of which is submerged to a depth D. The upper and lower edges of the orifice are, respectively, hi and h 2 below the upper water surface. Z is the difference in elevation of water surfaces. L is the length of the FIG. 69. Partially submerged orifice. GATES 95 orifice. The total discharge through the orifice is evidently the combined discharge of the upper portion of the orifice discharging into the air and the lower portion discharging as a submerged orifice. The theoretical formula for discharge from this orifice is inter- esting because of its relation to the submerged weir (Art. 79). Let Qi and Q 2 be, respectively, the discharges from the free and submerged portions of the orifice. Then from Art. 56, if C" is the coefficient of discharge for the upper portion h,*), ......... (42) and by Art. 64, C" being the coefficient of discharge for the lower portion Z), ......... (43) and the total discharge, Q, for the orifice is Q = Qi +Q 2 = LVfy[%C'(Z 3A - hi*}+C"Vz(h 2 - Z)], (44) or since 1i2 Z = D, ~] ..... (45) Since the coefficient of discharge for an orifice with free discharge is very nearly equal to the coefficient for a submerged orifice the equation may be put in the approximately equivalent form (46) If /ii = the orifice is a submerged weir and equation (45) becomes (47) The submerged weir is discussed in Arts. 79 and 80. 66. Gates. As used in engineering practice gates are forms of orifices. They may discharge freely into the air or be partially or wholly submerged. Though the principles underlying the dis- charge through orifices have been discussed in the preceding pages they cannot be applied accurately to gates because of the fact that gates do not ordinarily conform to the regular sections for which coefficients are directly available. 96 FLOW OF WATER THROUGH ORIFICES AND TUBES Fig. 70 illustrates a cross-section of a head gate such as is commonly used in diverting water from a river into a canal. A curtain wall extends between two piers, having grooves in which the gate slides. The bottom of the opening is flush with the floor of the structure. Such an opening has suppressed contraction at the bottom, nearly complete contraction at the top and par- tially suppressed contractions at the sides. Other equally com- plex conditions arise. The selection of coefficients for gates is therefore a matter requiring mature judgment and an intelligent use of the few available experimental data. Even the most FIG. 70. Headgate. FIG. 71. Discharge under falling head. experienced engineers may expect errors of at least 10 per cent in the coefficients which they select and to provide for this uncer- tainty ample allowance should be made in designs. 67. Discharge under Falling Head. A vessel is filled with water to a depth hi (Fig. 71). It is desired to determine the time required to lower the water surface to a depth h,2 through a given orifice. A is the area of the water surface when the depth of water is y and a is the area of the orifice. The rate of discharge at any instant when the head is y, the coefficient of discharge being C, is CaV2gy, DISCHARGE UNDER FALLING HEAD 97 and in the infinitesimal time, dt, the corresponding volume of water which flows out is In the same infinitesimal time the head will drop dy and the volume of water discharged will be Equating the values of dV = CaV2gydt or dt= Ca\/2gy (48) From this expression, by integrating with respect to y between the limits hi and /i2, the time required to lower the water surface the amount (hi fo) may be determined or the time of emptying the vessel may be obtained by placing /i2 = 0, provided A can be expressed in terms of y. For a cylinder or prism the cross-sectional area, A, is constant and the formula after integration becomes t = 2A CaV2g (49) The above formulas apply also to vertical or inclined orifices, provided the water surface does not fall below the top of the orifice. The heads hi and h 2 are then measured to the center of the orifice. The time required to completely empty a vessel evidently can be determined only in the case of a horizontal orifice. Example. Two chambers, 1 and 2 (Fig. 72), with vertical sides, each chamber being 8 ft. wide, are separated by a par- tition. Chamber 1 is 25 ft. long and chamber 2 is 10 ft. long. At the bottom of the partition is an orifice 1 ft. by 2 ft. The orifice is at all times submerged. The coefficient of discharge is 0.85. At a certain instant the water surface is 10 ft. higher in chamber 1 than in chamber 2. After what interval 10' FIG. 72. 98 FLOW OF WATER THROUGH ORIFICES AND TUBES of time will the water surfaces in the two chambers be at the same elevation? Solution. Let y be the difference in elevation of water surfaces at any instant and dy be the change in the difference in elevation of water surfaces in time dt. The amount of water flowing into chamber 2 in time dt will be dV = CaVtyy dt = . 85 X 2 X 8 . Q2Vydt = 13 . 6 Vydt. Also in the same interval of time the head will drop ^| dy in chamber 1 and rise f f dy in chamber 2. Then , 10X25X8, 2000 dV= Equating values of dV or Civ , _ Vy Integrating between the limits 10 and and reducing, 2 = 26.5 seconds. PROBLEMS 1. A sharp-edged orifice, 2 in. in diameter, in the vertical side of a large tank, discharges under a head of 16 ft. If C c is 0.62 and C v is 0.98 determine the diameter and velocity of the jet at the vena contracta and the discharge in cubic feet per second. 2. In Problem 1 how far from the vertical plane containing the orifice will the jet strike a horizontal plane which is 6 ft. below the center of the orifice? 3. A sharp-edged orifice, 3 in. in diameter, lies in a horizontal plane, the jet being directed upward. If the jet rises to a height of 21.2 ft. and the coefficient of velocity is 0.98, what is the depth of the orifice below the water surface, neglecting air friction. The pressure in the jet and on the surface of the reservoir is atmospheric. 4. In Problem 3, if C c = 0.62, what is the diameter of the jet 16 ft. above the orifice? 6. If the orifice shown in Fig. 49, page 73, has a diameter of 2 in. and the diameter of the vena contracta is 1.6 in. determine the discharge if A = 3.6 ft., F = 0, p A =9 .7 Ibs. per square inch, p^ = 1.3 Ibs. per square inch and the head lost is 0.8 ft. 6. A sharp-edged orifice, 4 in. in diameter, in the vertical wall of a tank discharges under a constant head of 4 ft. The volume of water discharged PROBLEMS 99 in 2 minutes weighs 6352 Ibs. At a point 2.57 ft. below the orifice the center of the jet is 6.28 ft. distant horizontally from the orifice. Determine C c , C v andC. 7. Determine the theoretical discharge (neglecting velocity of approach) from a vertical rectangular orifice 3 ft. long and 1 ft. high, the head on the top of the orifice being 2 ft. 8. A standard short tube, 4 in. in diameter, discharges under a head of 20 ft. What is the discharge in cubic feet per second? In gallons per day? 9. If a s-in. hole is tapped into the standard short tube, referred to hi Problem 8, at a point 2 in. from the entrance, determine the discharge through the tube, assuming the friction losses to remain the same. 10. If the upper end of a piezometer tube is connected with the ^-in. hole referred to in Problem 9 and the lower end is submerged in a pan of mercury, to what height will the mercury rise in the tube? 11. A Borda's mouthpiece 6 in. in diameter discharges under a head of 10 ft. What is the discharge in cubic feet per second? What is the diameter of the jet at the vena contracta? 12. Water is discharging through a gate 18 in. square. On the upstream side the water surface is 5 ft. above the top of the gate and on the down- stream side it is 2 ft. above. If the coefficient of discharge is 0.82, what is the discharge in cubic feet per second? 13. A canal carrying 40 cu. ft. per second has a depth of water of 3 ft. A structure is built across the canal containing a gate 2 ft. square, the bottom of the gate being set flush with the bottom of the canal. If the coefficient of discharge is 0.85, what will be the depth of water on the upstream side of the gate? 14. If, in Problem 13, the gate has a width of 3 ft. and it is desired to increase the depth of water above the structure to 4 ft., what should be the height of the gate, all other conditions remaining the same? 15. A 3-in. fire hose discharges water through a nozzle having a diameter at the tip of 1 in. If there is no contraction of the jet and (7^ = 0.97, the gage pressure at the base of the nozzle being 60 Ibs. per square inch, what is the discharge in gallons per minute? 16. In Problem 15 to what vertical height can the stream be thrown, neglecting air friction? 17. In Problem 15, if it is desired to throw a stream to a vertical height of 100 ft., what must be the pressure at the base of the nozzle? 18. In Problem 15 what is the maximum horizontal range (in the plane of the nozzle) to which the stream can be thrown? 19. A fire pump delivers water through a 6-in. main to a hydrant to which is connected a 3-in. hose, terminating in a 1-in. nozzle. The nozzle, for which C c = l and C p = 0.97, is 10 ft. above the hydrant and the hydrant is 50 ft. above the pump. What gage pressure at the pump is necessary to throw a stream 80 ft. vertically above the nozzle? 20. A cylindrical vessel 4 ft. in diameter and 6 ft. high has a sharp-edged circular orifice 2 in. in diameter in the bottom. If the vessel is filled with water how long will it take to lower the water surface 4 ft.? 21. A tank, which is the frustum of a cone having its bases horizontal 100 FLOW OF WATER THROUGH ORIFICES AND TUBES and axis vertical, is 10 ft. high and filled with water. It has a diameter of 8 ft. at the top and 3 ft. at the bottom. What is the time required to empty the tank through a sharp-edged orifice 3 in. square? 22. A hemispherical shell, with base horizontal and uppermost, is filled with water. If the radius is 8 ft. determine, the time required to empty through a sharp-edged orifice 6 in. in diameter located at the lowest point. 23. A tank 12 ft. long has its ends vertical, top and bottom horizontal, and is 6 ft. high. The top and bottom are rectangular, having widths of 8 ft. and 5 ft., respectively. A standard short tube, 4 in. in diameter, is located in one end near the bottom. If at the beginning the tank is full, find the time necessary to lower the water surface 4 ft. 24. In the tank described in Problem 23 assume that there is a vertical partition parallel with the ends and 5 ft. distant from one end. Near the bottom of this partition there is a circular, sharp-edged orifice 4 in. in diameter. If at the beginning the larger chamber is filled and the smaller chamber con- tains water having a depth of 2 ft., find the time required for the water surfaces to come to the same level. CHAPTER VIII FLOW OF WATER OVER WEIRS 68. Description and Definitions. A weir may be described as any notch of regular form through which water flows. This notch may be in the side of a tank, reservoir or channel or it may be an overflow dam with retaining walls at its ends. In general any obstruction, having an approximately uniform cross-section, placed in a channel so that water must flow over it is a weir. The edge or surface over which the water flows is called the crest of the weir. The overfalling sheet of water has been termed the nappe. Weirs may be classified in two ways, (a) with reference to the shape of the notch and (6) with reference to the cross-sectional form of the crest. Rectangular weirs that is, weirs having a level crest and vertical sides, are the most generally used. Other weirs in more or less common use, named from the shape of the notch or opening, are triangular weirs, trapezoidal weirs and parabolic weirs. Weir crests are constructed of many cross-sectional forms, but they all come under one of the general headings, (a) sharp-crested weirs, which are used primarily for the measurement of flowing water and (6) weirs not sharp crested which are used pri- : 5^^? marily as a part of hydraulic structures. A sharp-crested weir is a weir with a sharp upstream edge so formed that water in passing touches only this edge. The nappe from such a weir FIG. 73. Sharp-crested weir. is contracted at its under side in the same way that the jet from a sharp-edged orifice is contracted. This is called crest contrac- tion. If the sides of the notch also have sharp upstream edges so 101 OF WATER OVER WEIRS that the nappe is contracted in width the weir is said to have end contractions. The nappe from a weir having a length equal to the width of the channel suffers no contraction in width and such a weir is said to have end contractions suppressed. Fig. 73 is a cross-section of a sharp-crested weir which illustrates crest con- traction. Figs. 74 and 75 are views of weirs with end contrac- tions. Fig, 76 shows a weir with end contractions suppressed. FIG. 74. W T eir with end contractions. There is a downward curvature to the water surface near the weir crest (Fig. 73). This is called the surface contraction. The head, H (Fig. 73), is the vertical distance from the water surface, back of the effects of surface contraction, to the crest of the weir. The curvature of the water surface is not perceptible beyond a distance of about 2H upstream from the weir. The head is usually measured at distances of 6 to 16 ft. upstream from the weir. DESCRIPTION AND DEFINITIONS 103 The vertical contraction of the nappe includes both the surface contraction and the crest contraction. The section where the effects of crest contraction disappear, corresponding to the vena contracta of the jet, will be referred to as the contracted section of the nappe. Incomplete contraction of the nappe occurs when the crest of a weir is so near the bottom, or the ends of a weir with end con- tractions are so near to the sides of the channel, as to interfere with the approach of the water filaments in directions parallel to the face of the weir. The conditions are similar to those causing par- FIG. 75. Weir with end contractions. tial suppression of the contraction of the jet issuing from an orifice, discussed in Art. 57. Weirs not sharp crested are constructed in a wide variety of cross-sectional forms as is exemplified in the many shapes of over- flow dams now in existence. Such weirs have surface contraction similar to sharp-crested weirs, but conditions at the crest are different and vary with the sectional form (see Figs. 84 to 87). A variety of cross-sections of weirs of this class are shown in Fig. 88. The term velocity of approach, as used in connection with weirs, means the mean velocity in the channel just upstream from the weir. The portion of the channel near where the head is measured is designated the channel of approach. The height of a weir, P (Fig. 73), is the vertical distance of the crest above the bottom of the channel of approach. 104 FLOW OF WATER OVER WEIRS 69. Velocity at any Depth. Consider water to be discharging over the weir crest A (Fig. 77). In the derivation of the funda- mental formula it will be assumed that the water flows without friction and also that there is no contraction of the nappe and therefore no pressure within the nappe. In order to write a gen- eral expression applicable to all filaments it will be necessary to make the further assumption that all of the water particles in FIG. 76. Weir with end contractions suppressed. a cross-section of the channel of approach flow with the same velocity. From the nature of the discharge over weirs it is evident that the water surface in the channel and the nappe must be subjected to the same pressure from surrounding gases, which is usually atmospheric pressure. All pressures excepting those resulting from the weight of the water may therefore be neglected. The flowing water may be considered to be made up of filaments of which mn is one, m being a point in the channel of approach THEORETICAL FORMULAS FOR DISCHARGE 105 and n a point in the nappe, in the plane of the weir. The filament passes over the weir at a distance y below the surface of the water. The point m is a distance h m below the water surface and a distance z below n. v m is the velocity at m and v n is the velocity at n. Bernoulli's equation may be written between the points m and n as follows: and since h m z = and (3) These formulas express the theoretical relation between depth and velocity for any point in the plane of the weir. Introducing the assumption that all of the water particles in a cross-section of the channel of approach flow with the same velocity, V; v m in formulas (2) and (3) may be replaced by V, which gives and (5) If the cross-sectional area of the channel of approach is very much larger than the area of the notch, the velocity of approach is small and V may be called zero. The depth, y, at which the velocity v n occurs is then from formula (4) and the theoretical velocity at a depth y, from formula (5) is v n = V2^. , . . . . ... . . (7) 70. Theoretical Formulas for Discharge. Referring again to Fig. 77, let an origin be assumed at 0, a distance H vertically 106 FLOW OF WATER OVER WEIRS above the weir crest A. Formula (5) as derived in the preceding article is This is also the equation of a parabola whose axis is the line OA V 2 and whose intersection with the axis is at M , a distance above *9 the origin. Assuming the curve MN to be the graph of the FIG. 77. equation, the abscissa at any depth, y, is the theoretical velocity at this depth. Considering a unit length of weir, the area of an elementary strip is dy and the theoretical discharge through this strip is dQi=v n dy Substituting the value of v n from equation (5) (8) dQi = and 5T * THEORETICAL FORMULA FOR MEAN VELOCITY 107 or -*] w This formula expresses the theoretical discharge over a weir 1 ft. long, assuming uniform velocity in a cross-section of the channel of approach, and neglecting the effects of friction and contraction of the nappe. It evidently is also the area of the surface OSNA (Fig. 77). As a matter of convenience the symbol h may be substituted V 2 for -^-. Making this substitution the theoretical discharge for a weir of length L becomes which formula may be transposed to the form ' TA *n . . (H) In this form the term within the brackets is the factor which corrects for velocity of approach. If the cross-sectional area of the channel of approach is large in comparison with the cross- sectional area of the nappe, the effect of velocity of approach will not be appreciable and may be considered to be zero. The above formulas then reduce to (12) which is the same as formula (27) (page 83). This formula may also be derived directly from Fig. 77. The area of the surface A OP which represents the discharge over a weir 1 ft. long, being half of a parabolic segment, is equal to two- thirds of the area of the circumscribed .rectangle ORPA or %HV2gH. The discharge for a weir of length L is therefore %L/V2gH % , which is the same as formula (12). 71. Theoretical Formula for Mean Velocity. Since formula (9) which is the theoretical formula for discharge over a weir 1 ft. long is also an expression for the area of the surface OSNA and since the abscissas to this curve at any depth are the velocities at the depth, the mean of the abscissas between and A gives the mean velocity of the water discharging over the weir. The 108 FLOW OF WATER OVER WEIRS mean velocity is therefore the area of the surface OSNA divided by H, and the expression for theoretical mean velocity is obtained by dividing formula (9) by H, which gives, after substituting h If the velocity of approach is considered to be zero, h also becomes zero and the above formula reduces to (14) Equating the right-hand members of equations (7) and (14) gives the theoretical depth at which the mean velocity occurs, or y = *H ......... (15) 72. Weir Coefficients. The assumptions which were made in the derivation of formula (10) may be summarized briefly as follows: (a) All water particles in a cross-section of the channel of approach flow with the same velocity. (6) There is no contraction of the nappe. (c) The water flows without friction. Since these conditions do not in reality exist, it is necessary to modify formula (10) and the formulas derived therefrom to make them applicable to actual conditions. To accomplish this, three empirical coefficients are applied to the formula, there being one coefficient to correct for the difference between assumed conditions and actual conditions for each of the above assumptions. The method of correcting for each assumption will be discussed in the order given above. (a) Correction for non-uniformity of velocity in cross-section of channel of approach. The velocity in any cross-section of a channel is never uniform. As a result of the combined effects of friction, viscosity and surface tension (Arts. 7 and 110) velocities are lowest near the sides and bottom of an open channel and, if the channel is straight and uniform, the maximum velocity is below the surface and near the center of the channel. If there are no obstructions, velocities in a vertical line (Art. 110) vary approxi- mately as the abscissas to a parabola. In the channel of approach where a weir obstructs the flow, the law of distribution WEIR COEFFICIENTS 109 of velocities is not well understood and in cases where these velocities have been measured they have been found to vary quite irregularly. It is not practicable therefore to determine by analysis the extent to which discharges over a weir may be affected by the distribution of velocities in a cross-section of the channel of approach, but the general effect may be seen by studying certain assumed conditions. Let the curve CMB (Fig. 78) represent any vertical distribu- tion of velocities in the channel of approach for a strip of water O' e= 3 < ^-!->| / ^^ d ^dy j y ^\ /I'M //'! V ' A ' /B X- X/ ! F \ FIG. 78. Velocities in channel of approach. 1 ft. wide. The velocity at any distance, y, above the bottom is v, the total depth being d. The kinetic energy for this strip of water -*- -* The kinetic energy for any distribution of velocities can be deter- mined from this formula where v can be expressed in terms of y. Three conditions of assumed velocities are illustrated in Fig. 78. Uniform velocities are indicated by the vertica. 1 line EF. Velocities decreasing uniformly downward with a bottom velocity of zero are illustrated by the line DA. The line CM B illustrates a parabolic distribution of velocities. As these three lines are drawn, the mean velocity, V, is the same for each case. For uniform velocities, v in equation (16) is constant and equal to V, Substituting this value and integrating, there results KE wdV* (17) 110 FLOW OF WATER OVER WEIRS For uniformly varying velocities, illustrated by the line DA (Fig. 78) v=2V^. Substituting this value in equation (16) and integrating which shows the kinetic energy for this distribution of velocities to be twice as great as for uniform velocity. It may be shown also by writing the equation of a curve similar to CMS, v being expressed as a function of y, and sub- stituted for v in equation (16), that the kinetic energy for this distribution of velocities is about 1.3 times the kinetic energy for uniformly distributed velocities. Similarly for any variation in velocities in the cross-section of a channel, it may be shown that the water contains more kinetic energy if the velocity is non-uniform than if it is uniform. In general, the kinetic energy contained in the water in the channel of approach may be written (19) in which a is an empirical coefficient always greater than unity, and since velocity head is the kinetic energy contained in 1 Ib. of water (Art. 43) the general expression for velocity head due to 72 velocity of approach is a -g- or vh. This expression should, therefore, be written for h in formula (13), and calling v' the velocity after this correction has been applied (6) Correction for contraction. The ratio of the thickness of the nappe at its contracted section to the head on the weir may be called the coefficient of contraction, C c . This includes only vertical contraction; a separate correction being required for weirs with end contractions (see Art. 73). If t is the thickness of the nappe and H the head C e = or t = C c H WEIR COEFFICIENTS 111 and if v is the actual mean velocity in the contracted section of the nappe, L being the length of the weir, the discharge over the weir is Q = tLv = C c LHv. ..'.., . . . (21) The average value of C e for .sharp-crested weirs is about 0.635. (c) Correction for friction. The velocity in the nappe suffers a a retardation by reason of the combined effects of friction and viscosity. The ratio of the actual mean velocity, v y to the velocity */, which would exist without friction, is called the coefficient of velocity, Designating the coefficient of velocity by the symbol C Vj C v = f or v = C v i/ 9 substituting this value of v in (21) Q = C c C v LHv' ......... (22) The average value of C P for a sharp-crested weir is probably about 0.98, the same as for a sharp-edged orifice. Substituting the value of v' given in formula (20), the formula for discharge over a weir with the three corrective coefficients becomes . . (23) It is usual to combine V2gC c C v into a single coefficient, C, called the weir coefficient, then (24) If Co = 0.635 and C, = 0.98, the values given above, (7 = 3.33, which is an average value of this coefficient. It is the value adopted by Francis as a result of his experiments on sharp-crested weirs. Later experiments by Fteley and Stearns, and Bazin con- sidered in connection with the Francis experiments, show quite conclusively that C is not a constant. Its value appears to be repre- sented quite closely by the expression 3.34 ~ ^0.03' 112 FLOW OF WATER OVER WEIRS An investigation by Bazin gave the following value of C: With C substituted, formula (23) becomes The expression within the brackets is the correction for velocity of approach. When the velocity of approach is so small that the head, h, due to this velocity may be considered zero, formula (25) becomes (26) Formula (25) includes coefficients which correct for all of the assumptions which were made in deriving the theoretical formula (10) for weirs with end contractions suppressed. Formula (25) is often written in the equivalent form )*-(, at bends or curves in pipes, in addition 144 FLOW OF WATER THROUGH PIPES to the loss which occurs in an equal length of straight pipe. This is designated the loss of head due to bends. If the symbol HI is used to designate all losses of head in a pipe line, the loss of head due to friction being represented by hf and all minor losses by #2, (9) in which 95. Hydraulic Gradient. The locus of the elevations to which water will rise in a series of piezometer tubes inserted in a pipe line is called the hydraulic gradient or hydraulic grade line. The hydraulic gradient of a straight pipe of uniform diameter having the same degree of roughness of interior surface through- out is a straight line. In Fig. 94 the line ac is the hydraulic gradient for the pipe. Where a pipe changes in diameter or where for any reason there is a change in velocity or direction of flow, there is a break in the hydraulic gradient, the change in elevation being the combined effects of the change in velocity head, where velocity changes occur, and the loss of head due to friction or turbulence. The broken line aiCte&ifoc^dife, Fig. 95, represents the hydraulic gradient for the system of pipes shown. The hydraulic grade line thus indicates all losses of head and changes in velocity head. FIG. 96. 96. Loss of Head Due to Friction in Pipes. Fig. 96 represents a straight pipe without obstructions or changes in diameter. The loss of head, h f , in the length I is a measure of the resistance to flow. The laws governing this loss are intricate and are not subject to exact analysis. There are, however, certain general laws LOSS OF HEAD DUE TO FRICTION IN PIPES 145 which are in the nature of conclusions resulting from observation and experiment, which appear to govern fluid friction in pipes and which are expressed in the most generally accepted pipe formulas now in use. These laws may be briefly stated as follows : (a) Frictional resistance is independent of the pressure within the pipe, and other things being equal: (&) Friction between moving water and the inner surface of the pipe increases with the roughness of the surface. This may be expressed as a coefficient K whose value increases with the degree of roughness of the pipe. (c) Friction between moving water and the inner surface of a pipe is directly proportional to the area of the wetted surface; that is, it is proportional to the product of the wetted perimeter and the length or irdl, d being the diameter and I the length of the pipe. (d) As the cross-sectional area of the pipe increases, the retarding influence of viscosity becomes less, and it usually is con- sidered to vary inversely as some power of the area, and there- fore of the diameter or as 1/d*. (e) Frictional resistance varies directly as some power of the velocity, or as v n . (/) Frictional resistance increases with the viscosity and therefore inversely with the temperature. This factor is usually omitted from pipe formulas, coefficients being selected which apply to average air temperatures. Combining the factors expressed in (6), (c), (d) and (e) above, the total head lost is represented by the equation, ...... (11) or substituting K f and m for KXn and xl respectively, the general expression for loss of head due to friction in pipes may be written, (12) since = s, formula (12) may be transposed to the form l_ n m 1_ r-f-^J ff, 146 FLOW OF WATER THROUGH PIPES or 1 substituting K" for ( -^r. ) , y f or - , and z for -, \A / n n v = K"d v s z , ......... (13) or since the hydraulic radius r, for a circular pipe flowing full, equals d/4, or d=4r, formula (13) may be written, or substituting K'" for 4?K", v = K"W .......... (14) Each of the above formulas, though expressed differently, con- tains all of the factors, excepting temperature, which are believed to affect fluid friction. The base formulas for friction losses in pipes are commonly written in any of the three forms expressed by equations (12), (13), and (14). The further consideration of loss of head due to friction in pipes must be purely empirical. The values of coefficients and exponents to be applied to the base formulas are determined from the available experimental data. Of the large number of published formulas for determining the loss of head due to friction in pipes, only a few are given. It should be kept in mind that in all of the following formulas, hf, Z, d and other linear quantities must be expressed in feet and v must be expressed in feet per second. 97. The Chezy Formula. This formula deserves a place of prominence among pipe formulas not only because it represents the first successful attempt to express friction losses in algebraic terms, but also because it embodies all of the laws of fluid friction as they are understood and applied at the present time, and with certain modifying factors that have been found necessary, its use is now more general than that of any other formula either for pipes or open channels. As written by Chezy in 1775 this formula is (15) = is the rate of slope of the hydraulic gradient. It will be observed in which v is the mean velocity, r is the hydraulic radius and s = THE CHEZY FORMULA 147 that formula (15) is of the same form as (14), y and z being each \ and C being substituted for K' n '. The coefficient C was supposed by Chezy to be constant, but it is now known to vary with the degree of roughness of the surface with which the water comes in contact as well as with the velocity and hydraulic radius (or diameter). Since C appears to be a function of v and r the Chezy formula evidently does not accurately express the law of fluid friction. In the ideal formula, the coefficient would vary only with the roughness of the channel, and many attempts have been made to obtain a formula with such a coefficient expressing v as a function of r and s. These attempts have met with rather indifferent success. Formula (15) is used with an accompanying table giving values of C for different velocities, diameters and kinds of pipe. The table on page 148 gives approximate average values of C for four different kinds of pipe, as obtained from the available experi- mental data. In an account of experiments on the flow of water in pipes, published by Darcy l in 1857, he expressed the Chezy formula in the form, the relations between C and / in formulas (15) and (16) being /=77o and C = 2-\ -r. It will be observed that formula (16) may be obtained from formula (12) by writing n = 2 and w=l, the two formulas being of the same general form. From his experiments Darcy deduced the following values of /, as representing the mean of his observations. For new, clean cast-iron pipes, d being the diameter of the pipe in feet, For old cast-iron pipes, 1 M. H. DARCY: Recherches Experimentales Relatives au Mouvement de 1'eau dans les Tuyaux. Paris, 1857. 148 THE CHEZY FORMULA p g H O & w PHCE s l CO CO O O5 to OO 03 CO OS iH CO t~ O CO to 00 to Os C^l to l>- TH -*HH CO OS t^OOOOOSOS OSOOOO T-HTHTH to . OSCOtOOOO C^tOoO 00 OS OS OS OS O O O TH TH I-H TH II O fq 8 |>tOrHtOOO THIOOSCOOOOOOO5 t>- TH HH t^ OS tH 55 W o ,- (Mt^iHCOtO t^OOQHH cocot^t^t^- i> oo oo oo TH HH CO 00 O CO tO t^. OS osososos ooooo OOGO GO 00 COOS OS CO GO OS OS (Nt^THCOlO l> O - O 00 GO 00 00 00 OS OS OS OS TH T-H T-H 1 1 T-H g w o |!S ?j iO >O iO CO CO oo co (N CO t^o co loos CO CO GO O to iO ;CO 00 TH HH CO O CO CO OS OS OS O O OTHTHTHTH GO O iH GO CO CO OS -rH IO GO O CO CO O CO CO OS iH HH l^ O CO 00 00 OS OS OS O O O TH TH TH CO >O lO Tj< CO CO O to Tfi -^ ooooo !>. OS lO CO TH c ooooo OOt^CO UTiiO ooooo lO T-H GO CO "* ^ CO CO o o o o ^ (M O O5t^ CO CO CO . O >O iO "^ 11 IO(N OCXDCO SSS88 CO CO CO . lO g 1 i-H O5 * O !> t^ lO >O lO ^t 1 ooooo O(M O 00 CO ^ ^ -* 1 CO CO OOOOO * CO CO (M (M (M o 88 01 oot^co 10 ooooo ooooo ooooo "* 00 O rt< (N CO - >O "^ CO CO OQ O5 O5 ooooo co I-H o 01 oo - CO O ooooo O"* CO COCM ooooo oT 02 1^1 S^ M OOTtno f-4 C. 159. KING FORMULA 153 only with v and not with d, a much simpler table of coefficients is required for formula (20) than is required for the Chezy formula. Average values of K to be used with formula (20) are given in the table below. These values are the same as the values of / on page 149 for d 12 inches. I v 2 VALUES OF K IN THE FORMULA hf=K 125 Velocity, Clean Old Clean Clean in cast-iron cast-iron wooden concrete feet per second pipe pipe pipe pipe 2 .021 .038 .025 .029 5 .019 .038 .020 .025 10 .018 .038 .017 .023 20 .016 .038 .015 .021 Formula (20) may also be transposed and written, ).625 in which h f /l = s is the slope of the hydraulic gradient. Sub- stituting this value and writing C' for A/-J, v=C f s y^- Q25 . (21) In this form the formula is more convenient for certain types of problems. Average values of C' are given in the following table : VALUES OF C' IN THE FORMULA v=C's lA d- 625 Velocity, Clean Old Clean Clean in cast-iron cast-iron wooden concrete feet per second pipe pipe pipe pipe 2 55 42 51 47 5 58 42 57 51 10 61 42 62 53 20 63 42 66 56; 154 FLOW OF WATER THROUGH PIPES Formulas (15), (16), (20) and (21), when used with their accompanying tables of coefficients, all give the same results. 100. General Discussion of Pipe Formulas. The foregoing formulas represent the more common types of formulas for deter- mining the loss of head due to friction in pipes. There are an indefinite number of formulas, many of which possess merit. The choice of one formula over another is not of so great impor- tance as the careful and intelligent use of the formula after it is selected. The engineer should select the formula for general use which he believes to be in the most convenient form, and after adopting it he should endeavor to become familiar with its coefficients. The tables contained in this volume are sufficient for class room exercises, but the practicing engineer should extend his knowledge of coefficients by study and observation, and obtain values from actual measurements whenever the opportunity offers. 101. Friction Formula for Non-turbulent Flow. If v' is the velocity in feet per second, d\ the diameter of the pipe in inches, hf the friction loss in a length /, and P the viscosity coefficient (formula (3), Art. 90), the velocity in a pipe where stream-line flow exists according to Reynolds is ....... (22) If a case be assumed where dt=l in., hf= 1 ft. and 1= 100 ft., v' for a temperature of zero degrees Centigrade is 3.61 ft. per second, and for higher temperatures the velocity would be greater. The table on page 138 shows this velocity to be above the higher critical velocity and the flow must be turbulent. Formula (22) therefore does not apply and one of the formulas for turbulent flow should be used. 102. Detailed Study of Hydraulic Gradient and Minor Losses. In the discussion of loss of head due to friction (Art. 96), it has been shown that, other things being equal, the loss of head varies as v n and that usually n is less than 2 but does not vary greatly from this value. In some formulas, therefore, this loss of head is expressed as a function of the velocity head and coeffi- cients varying in value with v are applied to the formulas to make them represent average friction losses as given by experiments. In a similar manner it has been found that minor losses (Art. DETAILED STUDY OF HYDRAULIC GRADIENT 155 94) vary roughly with the square of the velocity and they are commonly expressed in formulas as functions of the velocity head. Variable coefficients are then applied to these formulas so as to make them give losses in accordance with the available experimental data. These losses (see Art. 94) expressed alge- braically are V 2 V 2 V 2 ho = K ^-, h d = K d -^ } h c = K c , etc., and formula (10), page 144, may be written, +K - (23) In the above equation, v is a general expression for velocity. It is the velocity in the pipe where the loss of head occurs and in case of enlargement or contraction it is the velocity in the smaller pipe. KQ, K d , K c , etc., are variable coefficients whose values must be determined from experiments. Fig. 97 illustrates entrance and discharge conditions for a pipe leading from one reservoir into another reservoir at a lower elevation. The water starts with zero velocity in the upper reservoir, finally coming to rest in the lower reservoir, and all of the energy represented by the difference in elevation of water surfaces is utilized in overcoming resistance. In Fig. 97, a\a% represents the hydraulic grade line which results from changing the velocity of the water from zero to the velocity which it attains in the pipe. The vertical distance between a\ and #2, that is, the distance which 2 is below the surface of the water, is the velocity head or v 2 /2g, where v is the mean velocity in the pipe. The line 0,10,2, must be considered as the hydraulic gradient of some particular filament of water, such as xy, since points in other filaments which are the same horizontal distance from the entrance to the pipe may have different veloci- ties and therefore different hydraulic gradients. It may appear that the pressure at any point in the filament should be that due to the weight of the water column above it. This would be true if the laws of hydrostatics might be applied. The laws of hydro- statics do not, however, apply to water in motion, the pressure being less than it would be at the same depth for water at rest. That this is true has been proved experimentally. It also follows from writing Bernoulli's equation between a point x where the 156 FLOW OF WATER THROUGH PIPES velocity is practically zero and a point y at the entrance to the pipe where the velocity equals v, the velocity in the pipe. Assum- ing the points to be of the same elevation the equation becomes or since v x is practically zero, . (24) The head lost at entrance to a pipe takes place within a distance of about two or three diameters from the entrance and is similar to the loss of head in a short tube. The line c^as, Fig. 97, is the portion of the hydraulic gradient which shows this loss of Fig. 97. Pipe connecting two reservoirs. head. There is a depression in the hydraulic gradient at o! because of the contraction of the jet. Vertically below as, the jet has expanded and fills the tube. The head lost at entrance is the vertical distance between az and as, or ho. Since the first two or three diameters of a pipe are similar to a short tube, entrance losses for pipes may be considered to be the same as for short tubes. The general formula for loss of head at entrance to a pipe is then (formula (22), page 81), (25) in which the coefficient of discharge, C, depends for its value upon tl^e conditions at entrance, and Ko = -~l. For convenience of DETAILED STUDY OF HYDRAULIC GRADIENT 157 reference, values of C and KQ given in Chapter VII, are repeated in the following table : COEFFICIENTS FOR DETERMINING Loss OF HEAD AT ENTRANCE TO PIPES Entrance to pipe Reference C K Inward projecting .... Art 63 75 78 Sharp cornered Art 58 82 50 Slightly rounded Art. 57 0.90 23 Bell mouth Art. 57 98 04 Since the effect of entrance conditions can not be determined accurately the selection of a proper value of KQ is to some extent a matter of judgment. Unless the entrance is known to be other than sharp cornered, a value of 0.5 is commonly used. Conditions at the outlet of a pipe may be illustrated in a similar manner. If there were no loss of head where water enters the lower reservoir the hydraulic grade line would connect as and e\ (Fig. 97), the latter point being v 2 /2g below the surface of the water. The distance 61^2 represents the portion of the velocity head lost through shock and turbulence. This may be illustrated by writing Bernoulli's equation between a point s at the outlet of the pipe where the velocity equals v, the velocity in the pipe, and a point u where the velocity v u is practically zero, ha equals the loss of head due to turbulence. If the two points are at the same elevation, ^-f^ = > T i M +^+^, (26) or since t> = 0, v 2 (27) huhs represents the portion of the velocity head which is not lost but which is reconverted into pressure head. The rate at which this reconversion takes place is represented in the figure by the line 62^3, the end of the hydraulic gradient. Expressing the head lost at discharge, ha, as a function of the velocity head, ^=^ (28) 158 FLOW OF WATER THROUGH PIPES The loss at discharge is the special case of loss of head due to sudden enlargement in which the ratio of smaller to larger dia- meter is practically zero. Values of Kd may therefore be taken from column 2 of the table on page 161. Since these values are nearly unity for the ordinary velocities encountered in pipes it is commonly considered that the entire velocity head is lost. Change in gradient resulting from sudden contraction in pipes is illustrated in Fig. 98. If there were no loss of head between any two points, on opposite sides of the contracted section, the difference in heights of water columns in two piezometer tubes as b and e } above these points would measure the gain in velocity head. If the two points are considered so close together that pipe friction may be neglected, the difference in height of water Fig. 98. Sudden contraction in pipe. columns 6 and e measures the gain in velocity head plus the loss of head due to sudden contraction. The hydraulic gradient as determined experimentally is illustrated by the line abcdef. There is a depression at d, due to contraction of the jet, similar to the depression at a' in the hydraulic grade line of Fig. 97. The piezometer tube c measures the pressure in the corner where there is little or no velocity. If piezometer tubes, c and d, were arranged to measure pressures near the axis of the pipe where the velocities are higher, the hydraulic gradient would be below bcde and would resemble Wd'e. It is important to note that the ordinary piezometer tube, which is set flush with the inner surface of the pipe, measures the pressure at the surface of the pipe but does not necessarily measure the pressure at points in the same cross-section at some DETAILED STUDY OF HYDRAULIC GRADIENT 159 distance from the surface. In smooth straight pipes the differ- ence between pressures at the surface and interior points is prob- ably not great but the difference may be quite large near sections where changes in diameter occur. The loss of head due to sudden contraction expressed as a function of the velocity head is (29) in which K c is an empirical coefficient, and v is the velocity in the smaller pipe. The following table gives experimental values of JC* VALUES OF THE COEFFICENT K c , FOR SUDDEN CONTRACTION Velocity in RATIO OF SMALLER TO LARGER DIAMETER smaller pipe, V 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 2 0.49 0.49 0.48 0.45 0.42 0.38 0.28 0.18 0.07 0.03 5 .48 .48 .47 .44 .41 .37 .28 .18 .09 .04 10 .47 .46 .45 .43 .40 .36 .28 .18 .10 .04 20 .44 .43 .42 .40 .37 .33 .27 .19 .11 .05 40 .38 .36 .35 .33 .31 .29 .25 .20 .13 .06 The loss of head at entrance to pipes is a special case of loss of head due to contraction. If the body of water is large the conditions conform approximately to a ratio of diameters of zero, and for -a square-cornered entrance, where the end of the pipe is flush with a wall having a plane surface, the values of KQ are comparable with the values of K c in the second column of the above table. If the change to a smaller diameter takes place gradually, as is the case with a gradually tapering section connecting the two pipes, or if the corners of the smaller pipe are rounded so as to reduce contractions, values of K c will be much smaller than those given for a sudden reduction of diameter. If the change is made as gradually as in a Venturi meter or if a bell-mouth connection 160 FLOW OF WATER THROUGH PIPES (page 85) between the two pipes is used, K c may become prac- tically negligible. Change in gradient resulting from sudden enlargement is shown in Fig. 99. In this case the change in height of water columns in piezometer tubes b and e, before and after enlargement, measures the gain in pressure head and this gain is equal to the loss in velocity head minus the loss of head due to sudden enlargement, the loss of head due to pipe friction being assumed so small as to be negligible. This may be verified by writing Bernoulli's equation between points on either side of the enlarge- ment. The hydraulic gradient as plotted from experiments by Gibson is abcde. The pressures shown by tubes c and d being 2 g Fig. 99. Sudden enlargement in pipe. measured at the surface of the pipe are undoubtedly less than if they were measured near the center of the pipe. In this case the piezometer tube c would read practically the same as the tube b and the pressure would be greater than that indicated. The portion of the hydraulic gradient bcde would then be similar to bc'd'e. The loss of head due to sudden enlargement expressed as a function of the velocity head is (30) Archer l has shown from an investigation of his own experi- 1 W. H. ARCHER: Loss of Head Due to Enlargements in Pipes. Trans. Amer. Soc. Civ. Eng., vol. 76, pp. 999-1026 (1913). DETAILED STUDY OF HYDRAULIC GRADIENT 161 ments and the experiments of others that h c is quite accurately represented by the formula, A, = 1.098 (31) in which v is the velocity in the smaller pipe and v\ is the velocity in the larger pipe. Experiments at the University of Michigan indicate that Archer's formula holds quite accurately in the limit where v\ is zero. In this case the conditions become those of loss of head at discharge, discussed on page 157. By equating the values of h e given by equations (30) and (31) and transposing, the following value of K e is obtained: K e 1.098 ,,67osi d 1.919 (32) d/di being the ratio of the smaller to the larger diameter. The following table of values of K e are computed from formula (32) : VALUES OF THE COEFFCIENT K e , for SUDDEN ENLARGEMENT Velocity in RATIO OF SMALLER TO LARGER DIAMETER smaller pipe, V 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 2 1.00 1.00 0.96 0.86 0.74 0.60 0.44 0.29 0.15 0.04 5 .96 .95 .89 .80 .69 ,56 .41 .27 .14 .04 10 .93 .91 .86 .77 .67 .54 .40 .26 .13 .04 20 .86 .84 .80 .72 .62 .50 .37 .24 .12 .04 40 .81 .80 .75 .68 .58 .47 .35 .22 .11 .03 The loss of head due to enlargement in pipes may be reduced by changing the diameters gradually. If the diameter increases at a uniform rate, the amount of loss increases as the angle between the axis and surface of the pipe increases and is practi- cally negligible for very small angles. Experimental values of K e have not been well determined for gradual enlargements, but those given in the following table are the approximate mean of such data as are available. There are not sufficient experimental data to determine the extent to which K e varies with the velocity. 162 FLOW OF WATER THROUGH PIPES VALUES OF THE COEFFCIENT K e , FOR GRADUAL ENLARGEMENT RATIO OF SMALLER TO LARGER DIAMETER Angle between axis and surface of pipe 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 5 0.04 0.04 0.04 0.04 0.04 0.04 0.03 0.02 0.01 15 .16 .16 .16 .16 .16 .15 .13 .10 .06 30 .49 .49 .48 .48 .46 .43 .37 .27 .16 45 .64 .63 .63 .62 .60 .55 .49 ,38 .20 60 .72 .72 .71 .70 .67 .62 .54 .43 .24 Gates or Valves when partially closed obstruct the flow and cause a loss of head and consequent drop in the hydraulic gradient. The difference in elevation of the hydraulic gradient on opposite sides of the obstruction measures the loss of head due to the obstruction. Following the form used for other losses, the loss of head in pipes due to gates, valves, or other obstructions may be written, lh=K,, (33) v being the mean velocity in the pipe. Experiments indicate that Kg does not vary appreciably with the velocity but increases with the amount of restriction. The following values of K g are the average values obtained from the best experimental data available. VALUES OF THE COEFFICIENT K g , FOR OBSTRUCTIONS IN PIPES Ratio of area of opening to cross-sectional area K. Ratio of area of opening to cross-sectional area of pipe of pipe 0.1 13.5 0.5 2.7 0.15 10.0 0.6 1.8 0.2 8.0 0.7 1.1 0.3 5.7 0.8 0.6 0.4 4.0 0.9 0.2 Bends or curves in pipes cause a gradual drop in the hydraulic gradient which is in excess of the drop that would occur in an DETAILED STUDY OF HYDRAULIC GRADIENT 163 equal length of the same kind of straight pipe. From a careful investigation of the available experimental data Fuller x deduced the following empirical formula, h b = cu 2 - 25 , .' '. . V . .' . (34) hd being the loss of head due to bends in excess of the loss which would occur in an equal length of straight pipe, v being the mean velocity in the pipe and c being a coefficient whose value varies with the radius, R, of the axis of the pipe expressed in feet. Fuller gives the following values of c for bends of 90: For R = Q,c = 0.0135; for R = 0.5, c = 0.0040; for R = l, c = 0.00275; for R = 3, c = 0.0024; for # = 6, c = 0.0023; for #=10, c = 0.00335; for R = 20, c = 0.0060; for # = 30, c = 0.0070; for # = 40, c = 0.0075; for # = 60, c = 0.0086. Following the form used for other losses, v 2 h b = K b , (35) and equating the right-hand members of equations (34) and (35) and reducing, ........ (36) From this formula the values of K b given in the following table have been computed: VALUES OF THE COEFFICIENT, K b . FOR Loss OF HEAD DUE TO BENDS OF 90 Mean RADIUS OF BEND IN FEET velocity in pipe, v 0.5 1 2 6 8 10 20 30 40 50 2 1.03 0.31 0.21 0.19 0.18 0.21 0.26 0.45 0.53 0.57 0.61 5 1.30 .38 .26 .23 .22 .26 .32 .57 .67 .72 .77 10 1.54 .46 .31 .28 .26 .31 .38 .68 .79 .86 .92 20 1.84 .54 .37 .33 .31 .37 .46 .81 .95 1.02 1.08 40 2.18 .65 .44 .39 .37 .44 .54 .97 1.12 1.21 1.30 From the above table it will be seen that the minimum loss of head from bends occurs when the radius of the axis of the 1 W. E. FULLER: Loss of Head in Bends. Journal of New England Water Works Association, December, 1913. 164 FLOW OF WATER THROUGH PIPES pipe is from 2 to 6 ft. For bends of 45 the coefficients will be about three-fourths and for 22 J about one-half of the values given in the table. 103. Part of Pipe Above Hydraulic Gradient. Fig. 100 shows a pipe of uniform diameter leading from a reservoir and discharg- ing under the head H. The summit, M , is a distance y above the straight line BeC but at a lower elevation than the water surface in the reservoir. Two conditions will be considered: first, Pa~ PC Pa pv where y< , and second, where y> , p a being atmos- pheric pressure and p, being the vapor pressure corresponding to the temperature of the water in the pipe. Assume the pipe AMSC, Fig. 100, to be empty when water Fig. 100. Part of pipe above hydraulic gradient. is turned into it at A. Water will first rise to the summit M and begin to flow down the decline toward the depression S, at which point it will collect and seal the pipe entrapping air between M and S. Eventually water will discharge from the outlet C. If the velocity is high enough the air entrapped between M and S will be removed by the flowing water, other- wise it will remain there and obstruct the flow. In such cases the air may be removed by a suction pump at the summit. If there is no air in the pipe and y< , assuming the loss of head to be uniform, the hydraulic gradient will be the straight line BeC and the flow will be the same as though all of the pipe were below the hydraulic gradient. ty\ - /y\ If y> , the flow of water will be restricted, even though all air is exhausted from the pipe, and the hydraulic gradient PART OF PIPE ABOVE HYDRAULIC GRADIENT 165 will no longer be a straight line. This condition is illustrated in Fig. 100. The hydraulic gradient is a straight line to a point, nr\ nr\ D, which is a distance d = below the summit. M. At M w (assuming no air in the pipe) the absolute pressure in the p ; pe is the vapor pressure corresponding to the temperature within the pipe and this pressure continues on down to N, the pipe flowing partially full between M and N. Throughout all por- tions of the pipe flowing full the velocity must necessarily be the same since the discharge is constant and therefore, assuming a uniform degree of roughness for the pipe, the slope of hydraulic gradient in such portions must be uniform. In other words, the slope of EC must be the same as the slope of BD. Through- out the distance where the pipe is not flowing full, the hydraulic gradient, represented by the line DE, is the same vertical distance, d, below the water surface in the pipe. The point E is the intersection of the line CE, parallel to BD, and the line DE. The section at N where the pipe begins to flow full is vertically above E. The conditions of flow, especially at low velocities, are not usually as favorable as those described above, because of the tendency of air to collect at a summit. Water flowing at low velocities will not remove air and may even liberate it, and cause air to collect at high places such as M, Fig. 100. The condition will be worse at summits above the hydraulic gradient if the pipe leaks, since the movement of air will be inward. In such cases the occasional operation of an air pump at the summit will be necessary to remove the air. At a summit below the hydraulic gradient, where the pressure within the pipe is greater than atmospheric, the air which collects may be removed through a valve. Air at a summit which is below the elevation of the water surface will not stop the flow of water entirely but will cause a portion of the pipe to flow partially full. Summits in pipe lines are always objectionable, and especially so are summits above the hydraulic gradient. Where they cannot be avoided special pro- vision should be made for removing the air which collects. 104. Special Problems. Pipe lines may be composed of pipes of several diameters connected in series, or they may branch in different ways so as to divide the flow, thus presenting a 166 FLOW OF WATER THROUGH PIPES variety of problems. Oftentimes such problems may be solved more readily by trial solutions, though some formulas may be derived which are of assistance. A few special cases are given in the following pages. Problems of this type are encountered frequently in designing mains for city water supplies. If the pipes are long (1000 diameters or more) the minor losses will ordinarily be comparatively small and are usually neglected. If, however, it is desired to include these losses, a solution should be made first neglecting them and then correcting the results to include them. 105. Branching Pipe Connecting Reservoirs at Different Elevations. A, B and C are three reservoirs connected by pipes 1, 2 and 3, as shown in Fig. 101. Let l\, di, Qi and v\ FIG. 101. Branching pipe connecting three reservoirs. represent, respectively, the length, diameter, discharge and mean velocity for pipe 1, and the same symbols with subscripts 2 and 3, the corresponding terms for pipes 2 and 3. If a piezom- eter is assumed to be at the junction P, the water surface in the tube will be a certain distance, hi } below the surface in reser- voir A . The surface of reservoir B is a distance H B = hi +7*2 below that of reservoir A and the surface of reservoir C is Hc=hi-\-h^ below the surface of reservoir A. If hiHB, reservoirs A and B will supply reservoir C. There are many problems suggested by this figure, in which certain quantities are given with others to be determined. Methods of solving three of these problems are given. Case L Having given the lengths and diameters of all pipes, and elevations of the three reservoirs; to determine Qi, Q% and Q 3 . BRANCHING PIPE CONNECTING RESERVOIRS 167 This problem is most conveniently solved by .trial. Assume 61 and solve for hi. Then using H B minus this computed value of h\ for h2, the loss of head due to friction in pipe 2, solve for Q 2 . Similarly, using H c minus the computed value of hi for friction loss, hs, in pipe 3, solve for 63. Evidently 61 = 62+63, 62 being negative if the direction of flow is from B toward P. The correct value of Qi will lie between the assumed value and the computed value of 62+63. Continue to assume new values of Qi, between these limits, and repeat computations until 61 = 02+63. It may be found helpful in making assumptions to plot com- puted values of 61? Fig. 102, against the error made in each assumption, that is, against 61 ~ (62+63). The resulting differ- ence may be either plus or minus. If the assumed values of 61 are well selected they will define a curve whose intersection with the 6i-axis will give the discharge as accurately as is usually required. The points should be on both sides of the 6i-axis and preferably one of the points should be quite close to it. Usually not more than three trial solutions will be necessary. This problem may also be solved analytically. Assuming any formula for pipe friction, as, for example, the Chezy formula, 10 9 3 7 6 4 3 2 1 1 2 3 4 5^6 7 FIG. 102. 910 h also and From Fig. (101), and substituting the above values of hi and h 2 , (37) 168 FLOW OF WATER THROUGH PIPES and in a similar manner, TT f h V l 2 _Lf k ^3 2 . ,oox Hc=fl di2J+ f *Z2ii> (38) also since Qi = Q2+Qs di 2 v 1 =d 2 2 V2+d 3 2 v 3 (39) By solving equations (37), (38) and (39) simultaneously, vi, V2 and vz may be determined if the other quantities are given. Since fi, /2 and /a are dependent upon vi, V2 and v$, respectively for their values, the equations must first be solved with assumed friction coefficients, to be corrected after the first solution for velocities has been completed. With these corrected values of fit /2 and /3, another solution of the equations for more accurate values of vi, v% and v% may be made. Case 2. Having given the lengths and diameters of all pipes, Qi, and the elevations of water surfaces in reservoir A and one of the other reservoirs as B] to determine the elevation of water surface in reservoir C. Using Qi, determine the lost head, hi, in pipe 1. Then h2 = Hshi is the lost head in pipe 2, using which, $2 may be computed. $2 will be plus or minus depending upon whether the direction of flow in pipe 2 is towards B or P. Then Qs = Qi $2. With $3 determined, the head lost in pipe 3 may be computed, and the elevation of water surface in reservoir C obtained. Case 3. Having given the lengths of all pipes, the elevations of water surfaces in all reservoirs, Qi, and the diameters of two pipes as d\ and d2', to determine cfe. Determine hi, $2 and Qs as for Case 2. Then with Qs and hz = Hchi known, compute d%. 106. Compound Pipe Connecting Two Reservoirs. The reser- voirs A and B are connected by a system of pipes as shown in Fig. 103. Pipe 1 leading from reservoir A divides at S into pipes 2 and 3 which join again at T. Pipe 4 leads from the junction T to a reservoir B. Let li, d\ and vi be respectively the length, diameter and mean velocity for pipe 1, and the same symbols with subscripts 2, 3 and 4 the corresponding quanitites for pipes 2, 3 and 4. $2 and Qs are the respective discharges for pipes 2 and 3, the sum of which discharges equals Q, the total discharge through pipes 1 and 4. Assuming piezometer tubes COMPOUND PIPE CONNECTING TWO RESERVOIRS 169 at and T, H is the total head lost in the system of pipes, hi is the head lost in pipe 1, h 2 = hs is the head lost in pipes 2 or 3, and h* is the head lost in pipe 4. Before taking up any of the special problems suggested by this figure a general analysis will be given. Any of the formulas for determining loss of head due to friction may be employed. It will be found advantageous to use a formula whose coefficient does not vary with the diameter, FIG. 103. Compound pipe. but a fractional exponent for v will be objectionable. Formula (20) has, therefore, been selected. Since h 2 = ^3, from formula (20), 1 2 k (40) also = ^ = ^|= and ,3=^ = Ci 2 TTC1 2 $3 writing these values of v 2 and v% in equation (40), k 16Q 2 2 1 h 16< % (4D K 2 Since in formula (20) K varies only with the velocity and not with the diameter, the error introduced by assuming K 2 = K% will not be important unless v\ and v 2 are widely different. Assum- ing them equal and canceling, or (42) (43) 170 FLOW OF WATER THROUGH PIPES Placing - ...... <> Q-3 = FQ 2 , ........ (45) and since Q, ....... (46) Q 2 +FQ 2 = Q, ...... (47) and Therefore to determine approximately (within the limits of the error introduced in assuming K to be a constant) the quantity of water passing through pipe 2 divide the total discharge by 1+F and the total discharge minus Q 2 gives Q 3 . The expression for loss of head may be written, H = h 1 +h 2 +h 4 ........... (49) The losses of head in the pipes 1, 2 and 4 may be expressed by either formula (16) or (20). Using (20), the expression for lost head becomes V 2 2 U V4 2 ~ ' (50) But V2= Q2 = *Q2 a 2 ird 2 2 and Q 4Q --- ............ Substituting these values of v\, v 2 and ^i in equation (50) and reducing ICQ"/ Z, fe Z* \ ff = 2^? V d? 38 (H-^)A" di^j ' ( If formula (16) in place of (20) had been used in writing equation (50) the above formula would be COMPOUND PIPE CONNECTING TWO RESERVOIRS 171 In formulas (54) and (55) the coefficients are not constants. Ki, K2, and K vary with the velocity and fi, /2, and /4 vary both with the velocity and diameter. Formula (54) will be more convenient in solving problems where all diameters are not known since KI, K^ and K do not depend upon the diameter for their value. Three types of problems are explained below. Case 1. Having given the lengths and diameters of all pipes and the total lost head; to determine Q. Determine F by formula (44), then determine an approxi- mate value of Q from formula (54) or (55) estimating the velocities in pipes 1, 2 and 4 for obtaining trial values of KI, K^ and K, or /i, /2 and /4. If the estimated velocities were not too much in error this solution may give the value of Q as accurate as is desired; otherwise with the value of Q obtained by the first solution determine the velocities and corresponding values of coefficients in the three pipes and again solve equation (54) or (55) for Q. The second solution should always give Q within the desired degree of accuracy. To check the result, from the computed value of Q determine hi and 7i4, then using H(hi-\-h4) as the friction loss in pipes 2 and 3 compute 62 and 63- The results obtained should show Q approximately equal to 62+63. If closer agreement between the computed values of Q and 62+63 is required than may be obtained readily by this method the results may be adjusted by trial solutions. Since there is always uncertainty as to the proper value of coefficients to use, it is not usually desirable to work for too close an agreement. If preferred this problem may be solved entirely by trial but it will save time in trial solutions to determine first by formulas (44) and (48) the portion of the total flow that passes through one of the branching pipes. Then successive values of 6 may be assumed and the lost head in each pipe computed until the sum of the losses in the three pipes equals the total lost head. A final check should be made to see that 6 equals approximately 62+63. Case 2. Having given the discharge, diameters and lengths of all pipes; to determine the total lost head. The lost head, H, may be determined from equations (44) and (54). If preferred 62 may be first obtained from formulas (44) and 172 FLOW OF WATER THROUGH PIPES (48) and Q% from formula (45). Then using these discharges, compute the head lost in pipes 2 and 3. These should be equal. If the computations do not show them equal adjust the dis- charges by trial (reducing the discharge in the pipe showing the greater loss of head and increasing by the same amount the discharge in the other pipe) , and again compute the head lost in each pipe. Repeat the assumption and computations until the losses of head in each of the pipes become the same or until the agreement is close enough for the purpose. This loss of head plus the loss of head in pipes 1 and 4, which may be computed in the usual manner, gives, H, the total lost head. Case 8. Having given the discharge, the total lost head, the lengths of all pipes and the diameters of three pipes; to deter- mine the other diameter. Assume that the diameter of pipe 2 is to be determined. Compute the head lost in pipes 1 and 4 by one of the formulas for determining loss of head due to friction. Deduct from the total lost head the sum of these computed losses. With this difference, which is the head lost in each of pipes 2 and 3, deter- mine Qs. Then, Qz = Q Q%. With Q 2 known, and the lost head determined as described above, compute the diameter of pipe 2. If the diameter of one of the single pipes, as for example, pipe 4, is to be determined, compute the head lost in pipe 1, as described in the preceding paragraph and also the head lost in the branching pipes 2 and 3 as described under Case 2. The difference between the total lost head and the sum of the above losses is the head lost in pipe 4, from which the diameter of this pipe may be computed. 107. Pipes of More than One Diameter Connected in Series. Fig. 104 represents a pipe of three diameters with lengths Zi, 2 and Is', diameters d\ t d 2 and cfa; and velocities 01, V2 and ^3. The total loss of head, H, assuming formula (19) for friction loss is u^iT' ' ' (56) also _xJ 2 xJ_2 .rJ-2 .... (57) PROBLEMS 173 1 substituting the values of vi, V2 and v% obtained from (57) in equa- tion (56) and transposing the following formula is obtained: 16Q 2 , . (58) From this formula with all quantities but one given, the unknown quantity may be obtained. If the velocity is not known, values of Ki t solution. FIG. 104. Pipe of three diameters. and K$ must be assumed and corrected from a trial A second solution of the problem may then be made. PROBLEMS 1. A new cast-iron pipe 1200 ft. long and 6 in. in diameter carries 1.3 cu. ft. per second. Determine the frictional loss. 2. Determine the discharge of the pipe described in Problem 1 if it dis- charges under a head of 80 ft. 3. What diameter of new cast-iron pipe, 1 mile long is required to dis- charge 4.5 cu. ft. per second under a head of 50 ft.? 4. What diameter of concrete pipe 8000 ft. long is required to discharge 40 cu. ft. per second under a head of 8 ft.? 5. Determine the loss of head due to sudden enlargement if a pipe carrying 2.0 cu. ft. per second suddenly changes from a diameter of (a) 6 in. to 8 in., (&) 6 in. to 12 in., and (c) 6 in. to 18 in. Also determine the difference in pressure resulting from these changes. 6. Solve Problem 5 if the direction of flow is reversed in each case. 7. A cast-iron pipe 12 in. in diameter and 100 ft. long having a sharp- cornered entrance draws water from a reservoir and discharges into the air. What is the difference in elevation between the water surface in the reservoir and the discharge end of the pipe if the rate of discharge is 16.0 cu. ft. per second? 8. If the pipe described in Problem 7 connects two reservoirs, both ends beinsc sharp-cornered and submerged, other conditions remaining the same, 174 FLOW OF WATER THROUGH PIPES determine the difference in elevation of the water surfaces in the two reser- voirs. 9. A wood-stave pipe having a diameter of 48 in. is laid on a down-grade of 6 ft. per mile. Determine the difference in pressure between two points 1 mile apart if the discharge is 45 cu. ft. per second. 10. A wood-stave pipe, 500 ft. long, is to be designed to carry 60 cu. ft. per second across a ravine so as to connect the ends of an open flume. If the difference in elevation between the water surfaces in the two ends of the flume is to be 5 ft., determine the necessary diameter of pipe, assuming abrupt changes in section, and neglecting the effect of velocity in the flume. 11. A concrete-pipe culvert 90 ft. long and 3 ft. in diameter is built through a road embankment. The culvert is laid on a grade of 1 ft. per 100 ft. Water is backed up to a depth of 5 ft. above the top of the pipe at the entrance and at the outlet the top of the pipe is submerged to a depth of 2 ft. Assume a sharp-cornered entrance. What is the discharge? 12. In Problem 11 (assuming all other conditions to be the same), what diameter of pipe will be required to discharge 100 cu. ft. per second? 13. In Problem 11 (assuming all other conditions to be the same) what will be the depth of water above the top of the pipe at its entrance when the culvert is discharging 50 cu. ft. per second? 14. A pipe line is to be laid connecting two tangents which intersect at an angle of 90. Between two points on these tangents, each distant 100 ft. from their point of intersection, will the total loss in head be less if a bend having a radius of 6 ft. or one having a radius of 50 ft. is used? 15. Three new cast-iron pipes are connected in series as shown in Fig. 104. The first has a diameter of 12 in. and a length of 1200 ft.; the second has a diameter of 24 in. and a length of 2000 ft.; and the third has a diameter of 18 in. and a length of 1500 ft. If the discharge is 8 cu. ft. per second, determine the lost head neglecting the minor losses. 16. If, in Problem 15, the total lost head in the three pipes is 45 ft., neglecting the minor losses, determine the discharge. 17. If the three pipes, described in Problem 15, have lengths of 500 ft. each, the entrance and all changes in section being sharp-cornered, deter- mine the total lost head when the discharge is 8 cu. ft. per second. 18. Referring to Fig. 101, page 166, if pipes 1, 2 and 3 have diameters of 24 in., 12 in. and 18 in., and lengths of 1200 ft., 500 ft. and 1000 ft., respect- ively, determine the discharge through pipe 1, if ## = 12 ft. and Hc = 2Q ft. Neglect minor losses. 19. In Problem 18, determine HC, if the discharge through pipe 1 is 20 cu. ft. per second, all other conditions remaining the same. 20. In Problem 18, determine the diameter of pipe 3 if the discharge through pipe 1 is 24 cu. ft. per second, all other conditions remaining constant. 21. Referring to Fig. 103, page 169, if pipes 1, 2, 3 and 4 have diameters of 36 in., 18 in., 24 in. and 30 in., and lengths of 3000 ft,, 2000 ft., 2400 ft. and 1500 ft., respectively, determine H, if the discharge through the system is 60 cu. ft. per second. Neglect the minor losses. 22. In Problem 21, determine the discharge if H is 12 ft., other con- ditions remaining the same. PROBLEMS 175 23. In Problem 21, determine the necessary diameter of pipe 3 if H is 15 ft., other conditions remaining the same. 24. Three smooth rubber-lined fire hose, each 200 ft. long and 1\ in. in diameter and having 1 in. nozzles, are connected to a 6-in. fire hydrant If, for the nozzles C c = l and C p = 0.97, determine the necessary pressure in the hydrant in order to throw streams 100 ft. high, the nozzles being 10 ft. above the hydrant. 25. Determine the height of streams that can be thrown if the pressure in the hydrant is 70 Ibs. per square inch, all other conditions remaining as stated in Problem 24. CHAPTER X FLOW OF WATER IN OPEN CHANNELS 108. Description and Definition. An open channel is a con- duit which conveys water without exerting any pressure, above atmospheric pressure, other than that due to the actual weight of water carried. Water therefore ordinarily flows in an open channel with a free water surface, though for an enclosed conduit, like a sewer flowing full, water may touch the top surface without exert- ing pressure. In this case it is classed as an open channel. Exam- ples of open channels are rivers, canals, flumes, and sewers and aqueducts when carrying water not under pressure. Open channels have various forms of cross-section. Artificial channels are commonly of rectangular, trapezoidal, or circular cross-section, while natural streams have irregular channels. Though friction losses in open channels follow the same general laws as in pipes and pipe formulas could be adapted to them, special friction formulas for open channels are usually employed. 109. Wetted Perimeter and Hydraulic Radius. The wetted perimeter of any conduit is the line of intersection of its wetted surface with a cross-sectional plane. In Fig. 105 the wetted perimeter is FIG. 105.-Cross-section of trape- the length of the broken Hne abcd zoidal channel. T , , ., n In a circular conduit flowing part full, as a sewer, the wetted perimeter is the arc of a circular seg- ment and in a natural stream, Fig. 116, it is the irregular line abode. The hydraulic radius of any section of a channel is its area divided by the wetted perimeter. All open channel formulas express the velocity as a function of the hydraulic radius. 110. Friction and Distribution of Velocities. As described under pipes (Art. 91) there is friction between the moving water and the surfaces of any conduit. If there were no other influences 176 FRICTION AND DISTRIBUTION OF VELOCITIES 177 the maximum velocity would ordinarily occur at places most remote from the surfaces of the conduit which produce friction, and consequently at the water surface. Owing to surface tension, however, there is a resistance to flow at the surface of the water, and the maximum velocity occurs at some distance below the sur- face. Under ideal conditions, where there are no disturbing influ- ences of any kind, the distribution of velocities in a regular channel will be uniform and similar on either side of the center. There are always, however, some irregularities in every channel sufficient to prevent a uniform distribution of velocities. The lines of equal velocity plotted from a large number of velocity measurements for the Sudbury conduit near Boston, Fig. 106, shows a more regular distribution of velocities than will be found in most channels. The distribution of velocities in a river of irregular cross-section, as FIG. 106. Distribution of velocities in Sudbury conduit. determined from velocity measurements with a current meter, is shown in the upper portion of Fig. 107. The figures show the velocities obtained at the points, where measurements were made and the irregular lines are interpolated equal velocity lines. The curves in the lower portion of Fig. 107 show the distribu- tion of velocities in vertical lines. These curves are called vertical velocity curves and the velocities from which they are plotted are called velocities in the vertical. The following properties of ver- tical velocity curves have been determined from the measurement of velocities of a large number of streams and a study of the curves plotted from them. (a) Vertical velocity curves have approximately the form of parabolas with horizontal axes passing through the thread of maximum velocity. In general the maximum velocity occurs somewhere between the water surface and one-third of the depth, 178 FLOW OF WATER IN OPEN CHANNELS the distance from the surface to the point of maximum velocity being at a greater proportional depth for greater depths of water. For shallow streams the maximum velocity is very near to the surface while for very deep streams it may lie at about one-third Distances in feet 10 20 30 40 50 60 70 80 90 100 UO 120 130 140 150 , CURVES OF EQUAL VELOCITY VERTICAL VELOCITY CURVES Note; Figures at top of curves indicate measuring points Figures at bottom of curves indicate mean velocity in the vertical 20 30 40 50 60 70 80 90 100 UO 120 130 140 150 1.12 -1.91 1.99 1.87 2.22 2.27 -/ 1.58 2.28- 2.25 2.11 1.88 Horizontal divisions represent 1 foot per second velocity FIG. 107. Velocities in natural stream. of the depth. A strong wind blowing either upstream or down- stream will affect the distribution of velocities in the vertical. (6) The mean velocity in the vertical is ordinarily found at a distance below the surface varying from 0.55 to 0.65 of the depth. The velocity at 0.6 depth is usually within 5 per cent of the mean velocity. FRICTION AND DISTRIBUTION OF VELOCITIES 179 (c) The mean of the velocities at 0.2 and 0.8 depth usually gives the mean velocity in the vertical within 2 per cent. (d) The mean velocity in the vertical is ordinarily from 0.80 to 0.95 of the surface velocity. The smaller percentage applies to the shallower streams. The above properties of vertical velocity curves are made use of in measuring the discharges of streams. Mean velocities in successive verticals are first obtained by measuring the velocity at 0.6 of the depth in each vertical or, where greater accuracy is required, by taking the mean of the velocities at 0.2 and 0.8 of the depth. The mean of the velocities in any two adjacent ver- ticals is considered as the mean velocity of the water between these verticals. The area between the verticals having been determined, the discharge through this portion of the cross-section of the stream is the product of this area and the mean velocity. The sum of all discharges between successive verticals is the total discharge. The mean velocity in the vertical may be obtained by taking the mean of several velocity measurements. This method is more laborious, however, and does not give the mean velocity appreciably more accurately than that obtained by taking the mean of velocities at 0.2 and 0.8 of the depth. The distribution of velocities in an ice-covered stream, Fig. 108, is modified by the effect of friction between the water and the ice. The amount of this friction exceeds the skin friction of a free water surface and the maximum velocity therefore occurs nearer mid- depth. The mean velocity in the vertical for an ice-covered stream is not at 0.6 depth but the mean of velocities at 0.2 and 0.8 depth gives approximately the mean velocity the same as for a stream with a free-water surface. 111. Energy Contained in Water in an Open Channel. This subject has been discussed (Art. 72) in connection with the velocity of approach for weirs, and the energy of water in a pipe is discussed in Art. 92. In open channels where velocities in different parts of a cross-section are not the same, the total kinetic energy contained in the water flowing past any cross-section is (1) W being the total weight of water passing the cross-section in one second, v the mean velocity, and a an empirical coefficient depend- 180 FLOW OF WATER IN OPEN CHANNELS ing for its value upon the distribution of velocities in the channel but always greater than unity. The range of variation in a has not been determined but in channels with unobstructed flow it probably lies between 1.1 and 1.2. This matter is not of great importance in ordinary hydraulic problems. Velocity head in open channels is commonly taken as the head due to the mean Distances in feet CURVES OF EQUAL VELOCITY ft 8 2 s I 4 1 5 ! VERTICAL VELOCITY CURVES Note: Figures at top of curves indicate measuring points Figures at bottom of curves indicate mean velocity in the vertical Horizontal divisions represent 1 foot per second velocity FIG. 108. Velocities in ice-covered stream. velocity, that is, a is assumed to equal unity. The slight error introduced by this assumption may be partially eliminated by a proper selection of coefficients. 112. Continuity of Flow in Open Channels. In any open channel fed by a constant supply of water there is the same rate of flow past every cross-section. In pipes flowing full, Art. 93, since water is practically incompressible, it is not necessary that the supply be constant in order to have the same rate of flow past LOSS OF HEAD 181 every cross-section at the same instant; but in open channels the water is not confined and a variation in the rate of supply will cause unequal rates of flow past different sections of the channel. In other words, in order to have continuity of flow in an open channel it is necessary at the same time to have steady flow, but in pipes flowing full there is always continuity of flow regardless of whether the flow is steady or variable. When continuity of flow exists in an open channel, the mean velocities are equal at all cross-sections having equal areas but if the areas are not equal the velocities are inversely proportional to the areas of the respective cross-sections. Thus, if a\ and vi, and 2 and v are respectively areas and mean FIG. 109. Open channel with accelerating velocity. velocities at any two cross-sections in an open channel where con- tinuity of flow exists, and 113. Loss of Head. Figs. 109 and 110 represent channels of constant cross-sections receiving water at uniform rates from reservoirs. Potential energy changes to kinetic energy as move- ment of the water takes place. As the water flows down these channels a portion of the energy is lost by friction between the water and the surfaces of the channels. If there were no friction ^sses of any kind the velocity, v t at any section, s, would be v t =\/2gh, in which h is the vertical distance between the water surface in the reservoir and the water surface at the section of the channel. The velocity of the water would thus continue to accel- erate as long as the downward slope of the channel continued. 182 FLOW OF WATER IN OPEN CHANNELS The actual conditions of flow as modified by friction for the case illustrated in Fig. 109 are as follows. The water receives a certain initial velocity at a, (Art. 85) where the channel leaves the reservoir. This channel, as indicated in the figure, has a slope steeper than is required to carry the water with the initial velocity which it receives at a. The velocity therefore accelerates for some distance, but a portion of the energy which the water contains is used in overcoming friction. As the water proceeds down the channel at a continually increasing velocity, a point is finally reached (since frictional resistance increases with the velocity, as has already been shown for pipes and will be shown for open channels) beyond which the energy used in overcoming friction in any reach exactly equals the potential energy contained in the water within the reach by reason of the slope of the channel. After this point has been reached, approximately at b in the figure, FIG. 110. Open channel with constant velocity. the water will flow to c and beyond with a constant velocity as long as channel conditions remain unchanged. As water moves from a to 6 a portion of its potential energy is continually being changed to kinetic energy and the remainder is used in overcoming friction. Between 6 and c the kinetic energy remains constant and all of the potential energy is used in overcoming friction. Fig. 110 illustrates the case where the slope is no greater than is required to carry the water at the initial velocity which it receives at a. Under these circumstances, as long as the channel conditions remain constant, the velocity in the channel will be the same at all sections. In other words the potential energy of the water will all be used in overcoming friction and the kinetic energy will remain constant. This case is the one most commonly encountered in open channel problems. In this connection it is important to bear in mind that the velocity remains constant only so long as the channel conditions LOSS OF HEAD 183 remain constant. Any change in the size of the channel changes the velocity. An increase or decrease in the slope of the channel causes a corresponding increase or decrease in velocity. The velocity is also modified by any conditions affecting frictional resistance. It will be noted that velocity conditions for open channels are different than for pipes. In a pipe flowing full, under a constant head, the water being confined and considered incom- pressible, the mean velocity can change only when the diameter of the pipe changes. In an open channel, however, the water being unconfined, the velocity changes with every change in channel conditions. Losses of head in open channels are in every respect analogous to losses of head in pipes. In addition to the loss of head due to friction between the moving water and the surface of the channel there is a loss of head wherever the velocity of water or direction of flow is changed. The same symbols will be used to represent losses of head in open channels that are used to designate the cor- responding losses in pipes (Art. 94). These losses for open channels are as follows: (a) A continuous loss of head throughout the channel due to friction between the moving water and the surface of the channel and to viscosity. This loss is commonly referred to as the loss of head due to friction. (6) A loss of head at entrance to the channel, that is, where a channel takes water from a reservoir or other body of compara- tively still water. (c) A loss of head at discharge, that is, where a channel dis- charges into a reservoir or other body of comparatively still water. (d) A loss of head due to contraction where a channel changes to a smaller cross-sectional area causing an increase in velocity. The loss of head at entrance to a channel (referred to under (6) above) is a special case of this loss. (e) A loss of head due to enlargement where a channel changes to a larger cross-sectional area causing a decrease in velocity. The loss of head at discharge (referred to under (c) above) is a special case of this loss. (/) A loss of head due to obstructions of any kind in a channel, such as gates, bridge piers or submerged weirs. (g) A loss of head due to curves in a channel in addition to the loss which occurs in an equal length of straight channel. 184 FLOW OF WATER IN OPEN CHANNELS 114. Hydraulic Gradient or Water Surface. The hydraulic gradient (Art. 95) of an open channel coincides with the water sur- face. In the case of steady, uniform flow, the water surface is parallel to the bottom of the channel. Any changes in channel conditions which will cause either an increase or decrease in velocity will cause a change in the elevation 3f the water surface the same as a change in velocity in a pipe will cause a change in the hydraulic gradient. Changes in the hydraulic gradient of a pipe line resulting from changes in section are frequently of minor importance and need be considered only insofar as they affect the total loss of head, while for an open channel the effect of any changes in the cross-section should be thoroughly understood and designs for the transition of the water from one velocity to another, should be worked out with great care. FIG. 111. Failure to provide properly for changes in elevation of water sur- face at the place where such changes occur may make it necessary for the channel to carry water at a depth other than that for which it was designed and thus interfere with its satisfactory operation. 115. Loss of Head Due to Friction in Open Channels. Fig. Ill represents the condition of steady, uniform flow in a straight channel. Since all of the head, h f , is used in overcoming friction in the distance, Z, this lost head is a measure of the resistance to flow. The ratio h//l is called the slope, and is represented by the symbol s. Since friction losses in open channels and pipes are of the same character they are governed by the same laws. To make the general laws as stated for pipes on page 145 apply to open channels it is necessary only to substitute, respectively, the words channel and hydraulic radius for pipe and diameter. It is evident, therefore, that the base formulas for pipes apply equally to open channels. Formula (14), page 146, is in the form generally used THE CHEZY FORMULA 185 for open channels. For convenience of reference it is here repeated. v = K"'r v s z ...... . . : (2) The further consideration of friction losses in open channels must be purely empirical. Numerical values of empirical coef- ficients and exponents and, if necessary, modifications in the form of the base formula must be derived from experimental data. A few of the more commonly used open channel formulas are given in the following pages. 116. The Chezy Formula. This formula as stated in the pre- ceding chapter (Art. 97) is, with proper modification, applicable either to open channels or to pipes, though it was originally designed for open channels. The formula as written by Chezy is ....... (3) in which v is the mean velocity, r is the hydraulic radius, and s is the slope of water surface. The Chezy formula is of the same form as formula (2) ; y and z each being equal to J and C being written for K'". The value of the coefficient C varies with the character- istics of the channel. In the form given it is not therefore readily adaptable to open channels but with modifications it forms the basis of most of the formulas in common use. 117. The Kutter Formula. An elaborate investigation of all available records of measurements of flow in open channels was made by Ganguillet and Kutter, 1 Swiss engineers, in 1869. As a result of their study they deduced the following empirical formula, commonly called the Kutter formula, for determining the value of C in the Chezy formula. 000281 ^ (A) Vr In the above formula, C is expressed as a function of the hydraulic radius, r, and slope, s, as well as the coefficient of roughness, n, whose value increases with the degree of roughness of the channel. 1 GANGUILLET and KUTTER: Flow of Water in Rivers and Other Channels. Translation by HERRING and TRAUTWINE, John Wiley & Sons, Publishers. 186 FLOW OF WATER IN OPEN CHANNELS VALUES OF C FROM KUTTER'S FORMULA HYDRAULIC RADIUS r IN FEET Slope n 0.2 0.3 0.4 0.6 0.8 1.0 1.5 2.0 2.5 3.0 4.0 6.0 8.0 0.0 5.0 .00005 0.010 87 98 109 123 133 140 154 164 172 177 187 199 207 213 220 .012 68 78 88 98 107 113 126 135 142 148 157 168 176 182 189 .015 52 58 66 76 83 89 99 107 113 118 126 138 145 150 159 .017 43 50 57 65 72 77 86 93 98 103 112 122 129 134 142 .020 35 41 45 53 59 64 72 80 84 88 95 105 111 116 125 .025 26 30 35 41 45 49 57 62 66 70 78 85 92 96 104 .030 22 25 28 33 37 40 47 51 55 58 65 74 78 83 90 .0001 0.010 98 108 118 131 140 147 158 167 173 178 186 196 202 206 212 .012 76 86 95 105 113 119 130 138 144 148 155 165 170 174 180 .015 57 64 72 81 88 93 103 109 114 118 125 134 140 143 150 .017 48 55 62 70 75 80 88 95 99 104 111 118 125 128 135 .020 38 45 50 57 .63 67 75 81 85 88 95 102 107 111 118 .025 28 34 38 43 48 51 59 64 67 70 77 84 89 93 98 .030 23 27 30 35 39 42 48 52 55 59 64 72 75 80 85 .0002 0.010 105 115 125 137 145 150 162 169 174 178 185 193 198 202 206 .012 83 92 100 110 117 123 133 139 144 148 154 162 167 170 175 .015 61 69 76 84 91 96 105 110 114 118 124 132 137 140 145 .017 52 59 65 73 78 83 90 97 100 104 110 117 122 125 130 .020 42 48 53 60 65 68 76 82 85 88 94 100 105 108 113 .025 30 35 40 45 50 54 60 65 68 70 76 83 86 90 95 .030 25 28 32 37 40 43 49 53 56 59 63 69 74 77 82 .0004 0.010 110 121 128 140 148 153 164 171 174 178 184 192 197 198 203 .012 87 95 103 113 120 125 134 141 145 149 153 161 165 168 172 .015 64 73 78 87 93 98 106 112 115 118 123 130 134 137 142 .017 54 62 68 75 80 84 92 98 101 104 110 116 120 123 128 .020 43 50 55 61 67 70 77 83 86 88 94 99 104 106 110 .025 32 37 42 47 51 55 60 65 68 70 75 82 85 88 92 .030 26 30 33 38 41 44 50 54 57 59 63 68 73 75 80 .001 0.010 113 124 132 143 150 155 165 172 175 178 184 190 195 197 201 .012 88 97 105 115 121 127 135 142 145 149 154 160 164 167 171 .015 66 75 80 88 94 98 107 112 116 119 123 130 133 135 141 .017 55 63 68 76 81 85 92 98 102 105 110 115 119 122 127 .020 45 51 56 62 68 71 78 84 87 89 93 98 103 105 109 .025 33 38 43 48 52 55 61 65 68 70 75 81 84 87 91 .030 27 30 34 38 42 45 50 54 57 59 63 68 72 74 78 .01 0.010 114 125 133 143 151 156 165 172 175 178 184 190 194 196 200 .012 89 99 106 116 122 128 136 142 145 149 154 159 163 166 170 .015 67 76 81 89 95 99 107 113 116 119 123 129 133 135 140 .017 56 64 69 77 82 86 93 99 103 105 109 115 118 121 126 .020 46 52 57 63 68 72 78 84 87 89 93 98 102 105 108 .025 34 39 44 49 52 56 62 65 68 70 75 80 83 86 90 .030 27 31 35 39 43 45 51 55 58 59 63 67 71 73 > 77 THE KUTTER FORMULA 187 Some of the values of n as published by the authors of the formula are as follows : n = 0.010 for well-planed timber or neat cement; n = 0.012 for common boards; n = 0.013 for ashlar or neatly joined brickwork; n= 0.017 for rubble masonry; n = 0.020 for canals in firm gravel; n = 0.025 for canals and rivers in good condition; n = 0.030 for canals and rivers with stones and weeds; n = 0.035 for canals and rivers in bad order. The above values do not include all present construction mate- rials, and later experimental data have shown the need of revising MORTON'S VALUES OF THE COEFFICIENT OF ROUGHNESS, n, FOR KUTTER'S AND MANNING'S FORMULAS RANGE o] p VALUES Commonly Surface From To used values Vitrified sewer pipe . . . 010 017 013 Common, clay drain tile Oil 017 014 Glazed brickwork Oil 015 013 Brick in cement mortar . . . 012 017 015 Neat-cement surfaces 010 013 Cement-mortar surfaces Oil 016 015 Concrete pipe . . ... 012 016 015 Plank flumes planed 010 014 012 Plank flumes, unplaned Oil 015 013 Plank flumes with battens 012 016 015 Concrete-lined channels 012 018 015 Rubble masonry 0.017 030 Dry rubble . . . . 025 035 Ashlar masonry 013 017 Smooth metal flumes .... Oil 015 Corrugated metal flumes 022 030 Earth canals in good condition 0.017 025 0225 Earth canals with weeds and rocks Canals excavated in rock 0.025 025 0.040 035 0.035 033 Natural streams in good condition 0.025 033 Natural streams with weeds and rocks .... gluggish rivers very weedy 0.035 0.050 0.060 150 188 FLOW OF WATER IN OPEN CHANNELS them. A more complete list of values of "Kutter's n" based upon later data and practice, and showing the probable ranges of variation has been prepared by Horton, extracts from which are tabulated on page 187. The solution of the Kutter formula may be obtained from tables which usually accompany the formula. The use of the Chezy formula with the Kutter coefficient thus becomes much simplified. A short table of values of C corresponding to different values of r, s, and n is given on page 186. Interpolations are necessary in using this table. 118. The Manning Formula. This formula was first suggested by Manning 1 in 1890. His study of the experimental data then available led to the conclusion that the values of the exponents y and z (formula (2)) which best represented the law of flow in open channels were, respectively f and J. Expressed in English units the Manning formula is 1.486 (5) This may be considered as the Chezy formula with The coefficient of roughness, n, is to be given the same value as n in the Kutter formula. The values of n applicable to different channel conditions are tabulated on page 187. Expressed in metric units the Manning formula is (6) 119. Comparison of Manning and Kutter Formulas. Using the same value of n in each case, the Kutter and Manning formulas give identical results for r=l meter = 3. 28 feet. This may be proved by substituting 3.28 for r in each formula. It will be found that each formula then reduces to 1 ROBERT MANNING : Flow of Water in Open Channels and Pipes. Trans. lust. Civ. Eng. of Ireland, 1890, vol. 20. COMPARISON OF MANNING AND KUTTER FORMULAS 189 Since s does not appear in this equation, it follows that when r equals 1 meter, the Kutter formula gives the same value of C for all slopes. Further investigation shows that for hydraulic radii less than 1 meter, with the same value of n used in each formula, the Kutter formula gives somewhat higher values of C than the Manning formula. For hydraulic radii greater than 1 meter the values of C obtained by the Kutter formula are in some cases slightly less and in others slightly greater than the values obtained by the Manning formula. It has been found, however, from several hundred gagings of open channels which were made under a wide range of conditions as regards shape, size and variation in roughness that the proper values of n to be used in the two formulas are so nearly identical that for all practical purposes the same values may be used. In other words, with the same value of n, problems solved by means of one of the formulas will give results agreeing very closely with those obtained by using the other formula. The Kutter formula has for many years been the most widely used of all of the open-channel formulas. It has been used almost exclusively in the United States and England and more commonly than any other formula in other parts of the world. The Manning formula has been used for a number of years, in Egypt, India, and Australia and quite recently many American engineers have come to see its advantages over the more cumbersome Kutter formula. It is because of the established use of the Kutter formula and the general familiarity of engineers with the type of channel represented by different values of " Kutter 's n " that there is an advantage in including n in the Manning coefficient. Expressed as it is, engineers familiar with the Kutter formula may adopt the Manning formula without the necessity of familiarizing themselves with a new coefficient, and at the same time get prac- tically the same results as with the formula with which they are familiar. Very evidently the Manning formula could be written ....... (7) and values of K could be selected for different types of channels the same as values of n are now selected. It is to be hoped that this form of the formula will eventually come into general use, and 190 FLOW OF WATER IN OPEN CHANNELS that the present form will be simply a step in the transition from the use of the Kutter formula to the use of the simplified form of the Manning formula. In order to assist in determining K in terms of n the following table of values is given : MANNING'S K IN TERMS OF n. K = 1.486 n K n K n K n K n K n K 0.009 165 0.016 93 0.023 65 0.030 50 0.044 34 0.070 21 .010 149 .017 87 .024 62 .032 46 .046 32 .075 20 .011 135 .018 83 .025 59 .034 44 .048 31 .080 19 .012 124 .019 78 .026 57 .036 41 .050 30 .085 18 .013 114 .020 74 .027 55 .038 39 .055 27 .090 17 .014 106 .021 71 .028 53 .040 37 .060 25 .095 16 .015 99 .022 68 .029 51 .042 35 .065 23 .100 15 The Kutter formula shows C to be a function of the slope, s, while the Manning formula does not. The terms involving s in the Kutter formula were introduced to make the formula agree with the measurements of flow of the Mississippi river by Humphreys and Abbott. These measurements have since been shown to be in error by at least 10 per cent and to this extent the Kutter formula is based upon incorrect data. Later experiments do not verify the conclusions of Ganguillet and Kutter regarding this matter. The value of C in the Kutter formula is not materially affected by the s terms unless the slope is very small, much smaller than is ordinarily used in designing channels, and so the formula has been satisfactorily used for the conditions ordinarily encountered in practice. It is probable that the Kutter formula would have been more satisfactory for all channels, including those with very small slopes, with the " s " terms omitted. That this is so is shown quite conclusively by some recent experiments on flow in the Chicago drainage canal. 1 The foregoing discussion may be summed up briefly with the statement that the Manning formula is much simpler to use than the Kutter formula and that, with the same value of n, it gives practically the same results as the Kutter formula except for flat 1 MURRAY BLANCHARD : Hydraulics of the Chicago Sanitary Districts Main Channel. Journal of the Western Society of Engineers, Sept., 1920. THE BAZIN FORMULA 191 slopes. In the latter case the Manning formula appears to give more accurate results than the Kutter formula. 120. The Bazin Formula. This formula, first published l in 1897, like the Kutter formula, determines the value of C in the Chezy formula. It considers C to be a function of r but not of s. Expressed in English units the formula is 157.6 ~ ....... 8 in which m is a coefficient of roughness. Values of m given by Bazin are m = 0.109 for smooth cement or planed wood; m = 0.290 for planks, ashlar and brick; m = 0.833 for rubble masonry; ra= 1.540 for earth channels of very regular surface; m = 2.360 for ordinary earth channel; m = 3.170 for exceptionally rough channels. The Bazin formula is used extensively in France but it has not been generally adopted in other countries. The value of m is subject to fully as wide a variation as n in the Kutter or Manning formulas. 121. Open-channel Formulas in General. The Kutter, Manning, and Bazin formulas are the best known and most widely used of the open-channel formulas. There are a large number of other formulas which have been published, and many of these doubtless possess merit. It is not ordinarily advisable, however, to use any except the commonly accepted formulas unless there is very good reason for so doing. The successful use of any open- channel formula requires an accurate knowledge of conditions, and judgment in the selection of coefficients. Even the most expe- rienced engineers may expect errors of at least 10 per cent in select- ing coefficients with corresponding errors in their results. 122. Detailed Study of Hydraulic Gradient or Water Surface. In the investigation of minor losses in pipes, Art. 102, it has been shown that these losses may be added collectively to the loss of head due to friction and that this sum, which represents the total 1 Annales des Fonts et Chaussees, 1897. 192 FLOW OF WATER IN OPEN CHANNELS lost head, may be considered as a unit. The losses of head for open channels are similar to those for pipes with the exception that they must be provided for at the places where they occur. Losses of head in open channels which result from changes in velocity may be expressed as functions of the velocity head the same as for pipes. Thus, V 2 . V 2 V 2 ho = K , h d = K d , h c = K c , etc. in which v is the mean velocity in the channel having the smaller cross-sectional area. While these losses of head for open channels are frequently of much greater importance than the similar losses for pipes, the values of coefficients for determining them are not so well established. More experimental data in this field are needed. FIG. 112. Change of channel to smaller section. Fig. 112 shows the change in water surface resulting from con- tracting the cross-sectional area of a channel. The mean velocity in the larger channel is v\ and in the smaller channel v. It is assumed that the grades of the two channels are just sufficient to maintain these velocities; that is, there is uniform flow in each channel. Kinetic energy is always obtained at the expense of potential energy. In any open channel the drop in water surface, h in Fig. 112, occurring at any change in section measures the loss in potential energy resulting from the change. A portion of h, h v in figure, is used in producing kinetic energy, that is in increasing the velocity of the water. The remainder, h c , is the head used in overcoming friction losses at the place where the change in velocity occurs. Referring to the figure, *-!-!' '<' DETAILED STUDY OF WATER SURFACE 193 in which v and Vi are the mean velocities in the smaller and larger channels respectively. This expression is also obtained by writing Bernoulli's equation between points in a filament on either side of the change in section when equal velocities at all points in a cross-section are assumed for each channel. From the figure h c +- V , ..... (10) and in a manner similar to that employed for losses due to contrac- tion in pipes hc = K c .......... (11) and substituting this value in equation (10) V 2 V 2 Vi 2 ( . There are no experimental data from which K c for open channels can be determined but it appears reasonable that it may have a value corresponding to that for contractions in pipes. The max- imum value for a sharp-cornered entrance may thus be taken as 0.5 with a smaller value for a rounded or tapered entrance. With care in design the value of K c may be reduced very nearly to zero. Under the most favorable conditions where K c is zero, the differ- ence in elevation of water surfaces will be h v . This drop in water surface should always be provided for when a canal changes to a smaller section. Example. Assume entrance conditions, such that ^ = 0.25; the velocity in the upper channel, vi = 2.0 ft. per second; in the lower channel z; = 8.0 ft. per second. Determine the drop in water surface. From equation (12) If a canal discharges from a reservoir or other body of com- paratively still water the conditions are the same as above except that vi may be considered zero. In the above problem v\ could have been considered zero without materially affecting the result. Fig. 113 shows the change in water surface resulting from enlarg- ing the cross-sectional area of a channel. The mean velocity in the 194 FLOW OF WATER IN OPEN CHANNELS smaller channel is v and in the larger channel vi. It is assumed that the slopes of the two channels are just sufficient to maintain these velocities. The velocity v being greater than v\ there is a loss in kinetic energy with a resultant gain in potential energy. If there were no loss of energy from friction, all of the kinetic energy in the smaller channel in excess of the kinetic energy in the larger channel would be converted into potential energy, and the water surface in the larger channel would be at a distance h above the water surface in the smaller channel. Since some energy is required to overcome friction and turbulence losses where the transition in velocities occurs, the actual elevation of water surface in the larger channel is lower than it would be if there were no friction losses, or, as indicated in the figure, at a distance, h, above the water surface in the smaller channel. The vertical distance h e = h v h thus rep- FIG. 113. Change of channel to larger section. resents the loss of head due to enlargement. This may be expressed algebraically as follows : also h = h,-h e = f g - V -^-h e , .... (14) V 2 and substituting K t for h e the formula may be written, v 2 v 2 vi 2 There are no satisfactory experimental data giving values of K e . In general, however, it is known that for abrupt changes in velocity, very little of the kinetic energy in the smaller channel is con- verted into potential energy. In other words nearly all of this energy is lost in friction and turbulence and there is little differ- DETAILED STUDY OF WATER SURFACE 195 ence in elevation in the water surfaces in the two channels. In this case K e approaches very near to unity. By exercising care in design and construction and making velocity changes gradual so as to produce a minimum of turbulence the value of K e may be greatly reduced. It should be remembered that equations (9) to (15) are approx- imations in that they assume the kinetic energy in each channel to be that due to the mean velocity, that is, they assume equal velocities at all points in a cross-section for each channel.- This is equivalent to assuming a in equation (1) to be equal to unity. The error introduced, thereby is however, of no great importance, especially in view of the fact that a correction for this error is necessarily included in the empirical values selected for K c or K e . The most common types of obstructions in open channels are submerged weirs, gates, and bridge piers. Losses of head resulting from weirs of all kinds and gates, which are types of orifices, are treated in earlier chapters. These structures are commonly employed to deflect water from main canals to secondary channels. Bridge piers, Fig. 114, restrict the cross-sectional area of a chan- nel and therefore obstruct the flow. The loss of head, h g , or what amounts to the same thing, the amount that the water will be backed up by piers is not sufficient to be of any im- portance except for com- paratively high velocities. The most important case arises in determining the backing-up effect of bridge piers during flood stages of FlG 114. Bridge pier obstructing flow, streams. The total loss of head is made up of three parts; a loss of head due to contraction of the channel at the upstream end of the piers, a loss of head due to enlargement of the channel at the downstream end, and an increase in loss of head due to friction resulting from the increase in velocity in the contracted portion of the channel. On account of the higher velocity, the surface of the water between the piers is depressed, the vertical distance, h, measuring the increase in velocity head plus the loss of head. The distance h h g is a measure of the velocity head reconverted into static head. 196 FLOW OF WATER IN OPEN CHANNELS The quantitative determination of losses of head from piers is entirely empirical. Though many experimental data are available, no satisfactory general formula for the solution of this problem is known. In general, however, piers that are so designed as to allow the changes in velocity to occur gradually with a minimum amount of turbulence, cause the smallest loss of head. Fig. 115 represents two horizontal sections of piers. Sec- tion A will cause less turbulence and consequently less loss of head than section B> Curves or bends in the alignment of a channel cause a loss of head. For low velocities such as occur in earth canals the loss in head is slight and ordinarily no allowance is made for it unless the curves are sharp and fre- quent. For sharp curves in concrete- lined canals or flumes designed to carry water at high velocities, an increase in the slope should be pro- vided. There are few experiments for determining the loss of head in curves, but the data for bends in pipes (page 163) may be used as a guide. Some engineers prefer to correct for loss of head at curves by using a higher coefficient of roughness. 123. Hydraulics of Rivers. Open-channel formulas do not apply accurately to natural streams since the channel sections and slope of water surface vary and the flow is non-uniform. At the lower stages, streams usually contain alternating reaches of riffles and slack water. During high stages this condition largely dis- appears and the water surface becomes approximately parallel to the average slope of the bottom of the channel. The degree of roughness of natural streams varies greatly within short reaches and even within different parts of the same cross-section. This may be seen from Fig. 116, which illustrates a stream in flood stage. The channel of normal flow, abc, will probably have an entirely different coefficient of roughness than the flood plain cde. Also the portion of the left-hand bank, Fig. 116, lying above ordinary high water may be covered with trees or other vegetation and have a FIG. 115. Effect of shape of bridge piers in causing tur- bulence. IRREGULAR SECTIONS 197 higher coefficient than the lower portion. Rocks and other channel irregularities are frequent and cause varying conditions of turbu- lence, the effect of which on the coefficient of roughness is difficult to estimate. There are, however, times when the engineer must estimate as well as he can, the carrying capacity of a natural channel. This is done by making a survey of the stream, from which cross-sections may be plotted and the slope of water surface may be determined. A certain reach is selected for which are obtained an average cross- section and slope of water surface, the computations being based upon these average values. It is apparent that results obtained in this manner will be very approximate, and that the degree of accuracy obtained will depend largely upon the ability of the engineer to judge the effect of these varying conditions upon the coefficient of roughness. Bet- ter results will be obtained for natural streams that have fairly straight and uni- form channels and are free FIG. 116. Natural stream in flood stage, from conditions causing tur- bulence. At the higher stages channel irregularities have less effect upon the slope of water surface, and open-channel formulas then apply more accurately. 124. Irregular Sections. Open-channel formulas should not be applied directly to sections having a break or pronounced irregularity in the wetted perimeter. Fig. 116, which illustrates a stream in flood stage shows a break in the wetted perimeter at c. That the open-channel formulas do not apply directly to the entire cross-section in such a case may be shown by the following example. The Manning formula is used, though any of the other open-channel formulas will show substantially the same results. Assume s = 0.00.1; n = 0.035; the length of the portion, abc, of the wetted perimeter = 200 ft., and of the portion cde = 300 ft. ; the area of the portion, abcm of the cross-section = 3000 sq. ft. and of the portion mode = 900 sq. ft. Then for the entire cross-section, a = 3900, p = 500, r = - = ~^^ =7.8, and by J) oUU the Manning formula, Q = 20,600 cu. ft. per second. The cross- section may now be divided into the two portions, abcm and 198 FLOW OF WATER IN OPEN CHANNELS mcde. Assuming the depth of water over the flood plain to be 3 ft., the wetted perimeter of the portion of the cross-section, a&cwwill be about 203 ft. Then r = - = =U.78 and from the Manning formula, Q = 24,250 cu. ft. per second. This indi- cates that a portion of the channel discharges more water than all of it, which is clearly impossible. Where an open-channel formula must be applied to an irregular section such as that indicated in Fig. 116, it is necessary to divide the cross-section into two portions and compute the discharges for each portion separately. As the two portions of the channel will differ in roughness, different coefficients should be selected for each. 125. Cross-section of Greatest Efficiency. The most efficient channel cross-section, from a hydraulic standpoint, is the one which, with a given slope and area, will have the maximum capacity. This cross-section is the one having the smallest wetted perimeter, since frictional resistance increases directly with the wetted perimeter. This also may be seen from an examination of one of the open-channel formulas. Take for example the equation for Q as given by the Manning formula Under the assumptions made a, n, and s are constant. Q there- fore increases with r. Since r=, r increases as p decreases and since Q varies only with r it is a maximum when p is a minimum. It should be borne in mind in 'this connection that there are usually practical objections to using cross- sections of minimum area but the dimen- sions of such cross-sections should be known and adhered to as closely as conditions appear to justify. FIG. I17.-Semicircular Of a11 cross-sections, having a given area channel. ^ ne semicircle, Fig. 117, has the smallest wetted perimeter and it is therefore the cross-section of highest hydraulic efficiency. Semicircular cross- sections are sometimes used for concrete or brick channels but they are not used for earth channels. CROSS-SECTION OF GREATEST EFFICIENCY 199 Trapezoidal cross-sections, Fig. 118, are used more commonly than any others. They are the only practical sections for earth canals, and masonry and wooden conduits are usually of this form. The rectangular section, usually used for wooden flumes, may be considered as a special case of the trapezoidal section. Properties of trapezoidal sections and methods of determining sections of greatest efficiency are shown in the following analysis. e b From Fig. 118, = z and = y, r e = Dz and b = Dy. Then and or p=(y+2Vl+z 2 )D, (17) a = D 2 (z+y), (18) D = (19) () (6) FIG. 118. Trapezoidal channels. Substituting this value of D in equation (17), p=(y+2Vl+*). ' a (20) Equating the first derivative with respect to y to zero and reducing, y = 2(VT+^-z), (21) b = 2D(Vl+z 2 -z) (22) or From which may be obtained the relation between depth of water and bottom width of canal of the most efficient cross- section for any values of z. From equations (17) and (18), P D(y+2Vl+z 2 )' (23) 200 FLOW OF WATER IN OPEN CHANNELS Substituting y from equation (21) and reducing r = D/2, .......... (24) or, the cross-section of greatest efficiency has a hydraulic radius equal to one-half the depth of water. By substituting y from equation (21) in equation (20) and reducing, the following expression is obtained -z, ..... (25) equating the first derivative with respect to z to zero and reducing z = 4= = tan30 - \/3 It may be seen from an examination of equation (22) that when z = tan 30 the length of each side is equal to 6, Fig. 118 (c) and the section becomes a half hexagon. Thus, of all the trape- zoidal sections (including the rectangle), for a given area, the half hexagon has the smallest perimeter and it is therefore the most efficient trapezoidal cross-section. From equation (22) are obtained the following relations between bottom width and depth (for different side slopes) for trapezoidal cross-sections of maximum efficiency, e/D = z i i 1 1J 2 3 4 b 2D 1.567) 1.247) 0.837) 0.617) 0.477) 0.327) 0.257) A semicircle having its center in the middle of the water surface may always be inscribed within a cross-section of maximum efficiency. This is illustrated for a trapezoidal cross-section in Fig. 119. OA, OB, and OC are drawn from a point on the center line of the water surface perpendicular to the sides of the channel EF, FG, and GH respectively. Let EF = GH=x] FG = b ; OA = OC = R ; and OB = D. As before a = area of section and p = wetted perimeter. Then from the figure and p = 2x+b. CIRCULAR SECTIONS 201 And since (equation (24)) the hydraulic radius equals one-half the depth of water D From which p~ 2x+b R = D. That is, OA, OB, and OC are all equal and a semicircle with center at is tangent to the three sides. 126. Circular Sections. The maximum discharge from a channel of circular cross-section occurs at a little less than full depth. This may be seen from an examination of open-channel formulas. The discharge by the Manning formula is Q= (16) a being the cross-sectional area. In the investigation of a particular channel n and / s will be constant. From Fig. 120, R being / the radius of the circle, and sin 6. FIG. 120.-Circular channel. With these equations an expression for ar % may be written, dif- ferentiating which and equating to zero, the value of 6 which makes Q a maximum is found to be 57 40'. The corresponding depth of water is D = 0.938d. Other open-channel formulas will give substantially the same result. This means that a pipe carry- ing water not under pressure, when free from obstructions and laid on a true grade, will not flow full. 127. Non-uniform Flow. In uniform flow the velocity past all cross-sections in the channel is constant. This condition obtains in ordinary conduits where successive cross-sections are uniform in size and shape and the slope of water surface is parallel to the bed of the channel. There are, however, certain cases Where the velocities are being accelerated or retarded, that is, the flow is non-uniform, although the same quantity of water passes all 202 FLOW OF WATER IN OPEN CHANNELS cross-sections so that continuity of flow exists, or expressed by symbols, , etc. An example of non-uniform flow is illustrated in Fig. 121, which rep- resents a canal supplied by another canal or reservoir. Water enters the canal at a certain initial velocity which may be computed by the method described in Art. 85. There is a sudden drop in the water surface at the entrance. The slope of the canal is greater than that required to carry the water at its initial velocity and the velocity therefore continues to accelerate until it becomes equal to the velocity at which the channel will carry water under condi- tions of uniform flow. Fig. 122 which represents a canal connecting two reservoirs is FIG. 121. Non-uniform flow in channel with steep slope. another example of non-uniform flow. The bottom of the canal is on a slope different from that which the water surface will attain. The total head producing flow is H. As before, there is a drop in the water surface at the place where the water receives its initial velocity. The same analysis applies to the two cases illus- trated in Figs. 121 and 122. This problem may best be analyzed by considering the channel to be divided up into reaches A, B, C, etc., Figs. 121 and 122, the computations being made for one reach at a time. There is a certain degree of approximation introduced in doing this as com- putations are based upon an average cross-sectional area, but by reducing the length of reach considered any desired degree of accuracy may be obtained. The following nomenclature is used: NON-UNIFORM FLOW 203 1 = si s hi d\ 60 61 VQ v\ v r z n length of reach considered; slope of bottom of canal; average slope of water surface in reach; fall of water surface in reach; loss of lead due to friction in reach; depth of water in upper end of reach; depth of water in lower end of reach; bottom width of trapezoidal section at upper end of reach; bottom width of trapezoidal section at lower end of reach; mean velocity at upper end of reach ; mean velocity at lower end of reach; mean velocity at middle of reach; hydraulic radius of section at middle of reach; slope of sides of canal for trapezoidal section ; coefficient of roughness in Manning's formula. FIG. 122. Non-uniform flow in channel with flat slope. From Bernoulli's equation, assuming equal velocities at all points in a cross-section. (26) From Manning's formula, 1.486 . . . (27) and approximately, putting V = (28) Substituting this value of h f in equation (26) and transposing, ^Irl^ Vsit* (29) This is the general equation for non-uniform flow. 204 FLOW OF WATER IN OPEN CHANNELS If the areas and wetted perimeters of the cross-sections at the upper and lower ends of a reach and also the drop in water surface, hi, are measured, the velocity at one end of the reach may be expressed in terms of the velocity at the other end, as, for example, VQ = -- , and the other velocity, vi, may be computed from 0*0 equation (29). Similarly if Q and the cross-section at the upper end of a reach are known, the cross-section at the lower end of a reach may be computed. Assume for example a trapezoidal cross-section. Then hi=do+8il-di ..... ^ . . . (30) zd y ......... ......... (32) and for the average section, In the right-hand members of the above equations 61 and di are the only unknown quantities, and one of these must be assumed. The known quantities and the assumed value of 61 or d\ are then substituted in these equations and the expressions for hi; v , vi, and r thus obtained are substituted in equation (29) which equation may be solved for 61 or di depending upon which has been assumed. Suppose for example that di, the depth of water at the lower end of the reach, has been assumed; 61 is then computed. If the pro- portions of canal section thus obtained are not satisfactory, a new value of di may be assumed and 61 may be recomputed. Ordina- rily the general form of the canal will be well enough known so that recomputations will be unnecessary. To get the dimensions of other cross-sections the above process will be repeated, the section at the lower end of one reach becoming the upper cross-section of the next reach below. 128. Backwater. A common problem in non-uniform flow occurs where water is backed up by a dam, weir or other obstruc- tion. Usually it is required to determine the amount that the BACKWATER 205 water surface will be raised at certain specified distances upstream from the obstruction. Fig. 123 indicates a channel whose water surface without the obstruction would be the line mn; with the obstruction the water surface assumes the curved line abode. This latter line is called the backwater curve. It is required to determine the position of sufficient points on the backwater curve so that it may be plotted. In this case the velocity is retarded but the general principles are the same as for non-uniform flow with accelerating velocity, and the same general method of solution may be followed. The channel is divided up into reaches A, B, C, etc., as before, but as the elevation of water surface at the obstruction is usually given the computations are begun at this section and continued upstream. FIG. 123. Backwater. This is not necessary, however, as the computations may be begun at any section with a known elevation of water surface and be carried either upstream or downstream. The nomenclature is that given on page 203. The general equation for non-uniform flow as given on page 203 , ^vi _vo 2g 2g 8.83r* (29) do, the depth of water at the upper end of the reach is the quantity sought. With this determined the depth at the upper end of the next reach may be obtained in the same manner. This problem may be solved by equation (29) in a manner practically identical with that described in Art. 126, the only difference being that the computations proceed upstream instead of downstream. One of the commonest applications of the backwater curve 206 FLOW OF WATER IN OPEN CHANNELS problem is to the determination of the elevations of points in the backwater curve above a dam. This is necessary when the damage which will result from submerging property during floods or the effect of backwater on some power plant farther upstream, is to be determined. Since in natural streams the channels are irregular, average sections in each reach and also average velocities are used. It is usual, therefore, to put the average velocity in the reach, v, in formula (29) in the place of v and vi, from which the follow- ing simplified expression is obtained : In 2 v 2 This is simply a transposed form of the Manning formula, hi in this case being equal to h f . In solving backwater problems by this formula, Q and the elevation of water surface at the lower end of reach A are usually known. Then hi is assumed, from which a trial value of elevation of water surface in the middle of the reach may be obtained. Based upon this assumed elevation an average channel cross-section for the reach may be plotted and the area and hydraulic radius determined. Then hi may be computed and if this computed value of hi differs from the assumed value sufficiently to materially affect the results of the computations, a new assumption for hi may be made and the computations repeated until the assumed value of hi is as near as is desired to the assumed value. With a little experience it will be found that the first assumed value of hi will give the computed value close enough without repeating the computations. As considerable uncertainty exists in the selection of a proper value of n and since the error thus introduced into the result is raised to the second power any solution of this problem is neces- sarily approximate. 129. Divided Flow. Fig. 124 represents a channel divided by an island. The total discharge, Q, is given and it is required to determine Qi and $2, the portion of discharge going respectively to channels 1 and 2, and also the total lost head hi } that is, the drop in water surface from m to n. From the Manning formula <> DIVIDED FLOW 207 and since v = , using subscripts 1 and 2 to refer to the respective channels, hi and la 2.21r 2 K' (36) (37) FIG. 124. Channel divided by island. Equating values of hi and reducing Putting and since k (38) F = nia 2 r 2 < Q2 = T+F> (40) (41) (42) h\ may now be obtained by substituting Qi and Q% in formulas (36) and (37). In computing backwater curves where the channel is divided by an island of considerable length it may be necessary to make computations for separate reaches, as described on page 205. In this case, Qi and Q 2 may be determined approximately from formulas (40) and (42) using an average cross-section for each channel and with these discharges determined the slope, or rise in water surface, hi, may be computed. If the values Qi and Q 2 are correct, the computations should show hi the same for each 208 FLOW OF WATER IN OPEN CHANNELS channel. If the computed values of hi do not agree, the com- putations should be repeated, reducing the discharge of the channel for which the computations give the greatest value of hi and increas- ing the discharge of the other channel by an equal amount. The computations are repeated until they give approximately the same values of hi for both channels. It may be helpful, in order to reduce the number of trial solutions, to plot values of Q against the error made in each assump- tion. The method is similar to that for divided flow in pipes, described on page 167. The problem presented by channels having a flood plain, Fig. 116 is similar to that of the channel divided by an island. The flow over the plain should be computed separately from that for the main channel, as discussed on page 207. If the total dis- charge is known, Qi and $2 being respectively the portions of the flow for the main channel and flood plain, formulas (40) and (42) may be used to determine approximate values of Qi and $2. With these approximate discharges determined, the two values of hi may be computed, using the proper value of n for each channel. If these values of hi do not agree the computations should be repeated, correcting Qi and 62 and continuing the computations in the same manner as that described above for channels divided by an island. PROBLEMS 1. An earth canal in good condition having a bottom width of 12 ft. and side slopes of 2 horizontal to 1 vertical is designed to carry 180 cu. ft. per second at a mean velocity of 2.25 ft. per second. What is the necessary grade of the canal? 2. What is the capacity of the canal in Problem 1, if the grade is 2 ft. per mile, all other conditions remaining the same? 3. Determine the depth of water in the canal, described in Problem 1, if the grade is 2 ft. per mile, other conditions remaining as stated. 4. Determine the bottom width of the canal having the same capacity and side slopes as the canal described in Problem 1, that will give the most efficient section. 6. A circular concrete sewer 5 ft. in diameter and flowing half full has a grade of 4 ft. per mile. Determine the discharge. 6. In Problem 5, what slope in feet per mile must the sewer have if the mean velocity is to be 8 ft. per second when flowing full capacity? 7. A smooth-metal flume of semicircular cross-section, has a diameter of 6 ft. and a grade of 0.005. What diameter of corrugated metal flume will be required to have the same capacity. PROBLEMS 2|)9 8. An earth canal carries a depth of water of 6 ft. The canal is 20 ft. wide on the bottom and has side slopes of 1.5 horizontal to 1 vertical. s = 0.0002. Using a value of n of 0.025 compute the discharge by the Manning formula and with this discharge determine the value of n in the Kutter formula and also the value of m in the Bazin formula. 9. An earth canal is to be designed to carry 400 cu. ft. per second at a mean velocity of 2.2 ft. per second. The sides of canal have a slope of 2 horizontal to 1 vertical. The depth of water is to be one-fourth of the bot- tom width. Assuming that the canal will be maintained in good condition find the necessary grade. 10. An earth canal in good condition carries 200 cu. ft. per second at a velocity of 2 ft. per second. Side slopes of canal are 2 horizontal to 1 vertical. The depth of water is one-third of bottom width of canal. This canal discharges into a flume with a tapered entrance, the conditions being such that the loss of head at entrance may be considered to be one-half of what it would be for an abrupt change in section. The flume is 7 ft. wide and has vertical sides. The slope of the bottom of the flume is such that it carries a depth of water of 3.5 ft. Determine how much the bottom of the flume should be above or below the bottom of the canal. 11. An earth canal containing weeds and grass has a bottom width of 15 ft. and side slopes of 2 horizontal to 1 vertical. The depth of water is 4 ft. and the slope is 2.75 ft. per mile. It is desired to change the section to a semicircular concrete-lined channel having a slope of 1.5 ft. in 1000 ft. Determine the radius of the semicircular channel if it flows full. If the change in section is abrupt and sharp-cornered, what will be the drop in water surface where the change in section occurs? 12. An earth canal in good condition has a bottom width of 10 ft., side slopes of 1.75 horizontal to 1 vertical, a grade of 0.00025, and carries 140 cu. ft. per second. A gate is constructed at the lower end of the canal which dis- charges freely into the air. The coefficient of discharge of the gate is 0.83. The gate is to be 3.0 ft. high. How wide should the gate be to maintain a constant depth of water in the canal? 13. A rectangular flume of unplaned timber connects two reservoirs 300 ft. apart. The flume is 16 ft. wide and both entrance and exit are sharp- cornered. The bottom of the flume, which is on a level grade, is 5 ft. below the water surface in the upper reservoir and 2 ft. below the level in the lower reservoir. Determine the discharge. 14. A rectangular flume of unplaned timber carries water from a reser- voir. The width of flume is 20 ft., the length is 1000 ft. and the slope is 1 ft. per 100 ft. If the entrance is sharp-cornered and the bottom of the flume at the entrance is 4 ft. lower than the water surface in the reservoir, determine the rate of discharge. 15. An earth canal in good condition is 60 ft. wide on the bottom and has side slopes of 2 horizontal to 1 vertical. One side slope extends to an elevation of 20 ft. above the bottom of the canal. The other bank, which is a practically level meadow at an elevation of 6 ft. above the bottom of the canal, extends back 500 ft. from the canal and then rises abruptly. The meadow is covered with short grass and weeds. If the slope of the canal 210 FLOW OF WATER IN OPEN CHANNELS is 2.2 ft. per mile, determine the discharge when the water is 8 ft. deep in the canal. 16. A flume built of planed lumber, with vertical sides, 8 ft. wide has a grade of 2 ft. per 1000 ft. A sharp-crested weir 3.5 ft. high is constructed across the flume. When the head of water over the weir is 3.1 ft., what is the elevation of water surface at a section 200 ft. upstream from the weir? How much higher is the water surface at this section than it would be with the same quantity of water flowing but with the weir removed? 17. A canal carries 300 cu. ft. per second of water at a depth of 5.5 ft. The water in this canal is to be dropped to a lower elevation through a concrete chute of rectangular cross-section, having a grade of 1 ft. in 10 ft. The chute is to carry a uniform depth of water of 3.0 ft., the width to vary as required to maintain this depth. Determine the width of chute at entrance, at 100 ft. below the entrance, and at 200 ft. below the entrance. Also determine the minimum width possible for this depth of water. 18. A canal, 58,000 ft. long (580 Sta.'s), is to be constructed with a capacity of 300 cu. ft. per second. The canal diverts from a river and terminates at a reservoir into which it discharges. The water surface in the river at the point of diversion is to be maintained at an elevation of 770 ft. (a) Water is to be diverted through six head gates, having rectangular orifices each 2 ft. by 5 ft. Determine the head required to force the water through these openings, assuming a coefficient of discharge of 0.80. (6) From Sta. to Sta. 425 the canal is in earth section, having side slopes of 2 horizontal to 1 vertical, and a depth of water of 0.3 of the bottom width of the canal. Velocity of water is to be 2.1 ft. per second. Assume a coefficient of roughness of 0.0225. Determine grade or slope of canal. (c) Between Sta. 425 and 500 the canal is in rock and is to have a semi- circular section lined with concrete. The grade of the canal is to be 2 ft. per 1000 ft. Coefficient of roughness, 0.014. Determine the head lost at entrance, using a coefficient of velocity of 0.92. Also determine diameter of canal section. (d) From Sta. 500 to Sta. 580 the section of canal is the same as from Sta. to Sta. 425. At the reservoir end of the canal (Sta. 580) a weir is to be constructed in order that a uniform depth of water may be maintained throughout the entire length of earth section. Length of this weir is to be equal to the bottom width of the earth canal. The weir has a rectangular section, with horizontal crest 2 ft. 8 in. wide. Determine height of crest above bottom of the canal. (e) Tabulate the elevations of the water surface, to nearest 0.1 ft., at the following stations: Sta. 0+10, Sta. 424+90, Sta. 425+10, Sta. 500 and Sta. 579+90. (Assume K e -l at Sta. 500.) CHAPTER XI HYDRODYNAMICS 130. Fundamental Principles. Newton's laws of motion form the basic principles of the subject of hydrodynamics. These laws are clear and definite and lead to results that agree exactly with experiment. Briefly stated they are as follows: I. Any body at rest or in motion with a uniform velocity along a straight line will continue in that same condition of rest or motion until acted upon by some external force. II. Any change in the momentum of a moving body is propor- tional to the force producing that change and occurs along the same straight line in which the force acts. III. To every action there is always an equal and opposite reaction. These three laws of Newton's are frequently referred to as the Laws of Inertia, Force, and Stress, respectively. The solution of practically any problem in hydrodynamics may be accomplished by the direct application of these laws. It is therefore essential that a clear conception be had of their full significance. As an aid in acquiring this conception the following discussion is presented. 131. Interpretation of Newton's Laws. Newton's first law of motion is merely a statement of the now well-known fact that matter is inert; that is, it possesses no ability, per se, to change its condition of rest, or motion, and that any such change must be brought about through the action of some external force, as for instance, friction of the air in retarding the velocity of a bullet. Since change in motion results from the application of a force it may be assumed that the magnitude of the change in motion will depend upon the magnitude of the force producing that change ; in other words it may be assumed that there is, as usual, a direct relation between cause and effect. Making this assump- tion, the second law follows naturally from the first. Momentum is by definition quantity of motion, and is equal to the product of 211 212 HYDRODYNAMICS the mass and velocity. Designating by Z\, the momentum of a mass of water, M, having a velocity v\. Zi=Mvi. If a force P acting upon this mass for a time A changes the velocity to v 2 . Z 2 = Mv 2 , and Z 2 -Z 1 =M(v 2 -v 1 ), (1) or &Z and since = acceleration, and mass times acceleration equals force. Equation (2) may be written in the form, P =^l^ (3) or substituting for Av its equivalent (v 2 vi) and letting At equal 1 second, equation (3) becomes, P = M(v 2 -vi) (4) This means that when a force acts upon a mass and thereby changes its velocity from v\ to v 2 , the force is equal to the product of the mass whose velocity is changed each second from v\ to v 2 , and the change in velocity. This may be demonstrated in another manner. The amount of work done upon any mass is equal to the gain in kinetic energy, or Pl = ^ ^ =-^-(v 2 vi)(v 2 -\-vi), ... (5) Zi Zi Zi in which I is the distance through which the force P acts upon the mass M. But "Y ... . . (6) Therefore, from equations (1) and (2) ),.. .->... (7) INTERPRETATION OF NEWTON'S LAWS 213 or letting t = 1 second, (4) It should be kept clearly in mind that P is the force acting upon the mass whose velocity is changed from v\ to V2. Newton's third law of motion states that the mass reacts with a force equal in magnitude but opposite in direction to the force P. To avoid confusion the reacting force that water exerts upon an object will be designated by F. Hence F=-P=-M(v 2 -vi)=M(vi-v 2 ). . . . (8) Since force and velocity are vector quantities, it follows that if a jet of water impinges against a vane which is either moving or at rest and thereby has its velocity in any direction changed, a force F is exerted upon the vane whose magnitude in any direction is equal to the change in momentum per second that the jet undergoes in the same direction. In other words the force F is equal to the mass impinging per second times the change in velocity in the direction of the force. The X- and F-components of the force exerted by a jet whose path lies in the XY plane will therefore be, ^2 = Mass impinging per second X change in velocity along the X-axis. Fi, = Mass impinging per second X change in velocity along the F-axis. F = \/F X 2 +F V 2 , and the tangent of the angle which this resultant makes with the v . . F y A-axis is -=T. f x The change in velocity may be either positive or negative, the only difference being that in the case of a decrease in velocity the dynamic force exerted by the water on the vane is in the same direction as flow, whereas in the case of an increase in the velocity the dynamic force exerted on the vane is opposed to the direction of flow. For instance, referring to Fig. 127, the flow being from the left and the X-component of the velocity being decreased, F x is directed toward the right, whereas the F-component of the velocity being increased and the flow being directed upward, F v is directed downward. 214 HYDRODYNAMICS 132. Relative and Absolute Velocities. Strictly speaking, all motion is relative. No object in the universe is known to be fixed in space. An airplane is said to be flying one hundred miles per hour but this is its velocity only with respect to the surface of the earth beneath it. The earth's surface itself is moving at a tre- mendous speed both with respect to its axis and to the sun, each of which are whirling through space at a still greater rate. It is nevertheless convenient in connection with this subject to consider all motion with respect to the earth's surface as abso- TIG. 125. lute motion. The airplane above referred to has therefore an abso- lute velocity of 100 miles per hour. Another plane in pursuit may have an absolute velocity of 120 miles per hour but its relative velocity with respect to the first plane is only 20 miles per hour. If the two planes were to fly in opposite directions, each retaining its same absolute velocity, the relative velocity between them would be 220 miles per hour. If they were to fly at right angles to each other their relative ve- locities would be Vl00 2 +120 2 = 156.2 miles per hour. Since velocities are vector quantities, these results may be obtained graphically as in Fig. 125, in which, V A is the absolute velocity of A, V B is the absolute velocity of B, and u is the relative velocity of either with respect to the other. 133. Jet Impinging Nor- mally on a Fixed Flat Plate. Fig. 126 shows a jet impinging at right angles against a fixed flat FIG. 126. Jet impinging against a flat plate. JET IMPINGING ON A MOVING FLAT PLATE 215 plate. It is assumed that the plate is large enough with respect to the size of the jet so that the jet is deflected through a full 90. The pressure exerted on the plate varies from a maximum at the axis of the jet where it is very nearly equal to that due to the full velocity head, to zero at a distance approximately equal to the diameter of the jet from its axis. The total pressure exerted is equal to the product of the mass impinging per second and the orig- inal velocity of the jet, since the final velocity of the water as it leaves the plate has no component in its original direction. Hence, the force exerted on the plate is, , ,,. wQ wav 2 p = Mv = v = -- , ...... (9) Q 9 where M and Q represent respectively the mass and quantity striking the plate per second, and a and v are the cross-sectional area and mean velocity of the jet, all terms being expressed in the foot-pound-second system. 134. Jet Impinging Normally on a Moving Flat Plate. Con- sider the case of a jet impinging normally against a flat plate mov- ing in the same direction as the jet or at least having a component of its motion in that direction. It is assumed that the plate has a uniform velocity, being restrained from accelerating by some external agency. The mass impinging per second is ....... (10) where Q' is the quantity of water striking the plate in cubic feet per second, a is the cross-sectional area of the jet in square feet and u is the relative velocity of the jet with respect to the plate. The change in velocity is v v' = u, since the velocity in the direction of the force is changed from v to v' and the jet leaves the plate tangentially with a velocity whose X-component is equal to the velocity of the moving plate. The force acting on the plate is (11) 9 9 135. Jet Deflected by a Fixed Curved Vane. The jet, shown in Fig. 127, is deflected through an angle 6 by a fixed, curved, trough-shaped vane A B. It is assumed that the vane is so smooth that friction may be neglected so that the velocity with which the 216 HYDRODYNAMICS jet leaves at B may be considered the same as that with which it strikes at A. It is also assumed that the vertical height of the vane is so small that gravity will not appreciably retard the velocity of the jet. Considering the horizontal and vertical components of the force acting on the vane, = ~ (l-cos0), 9 sinfl. (12) (13) FIG. 127. Jet impinging on a fixed curved vane. FIG. 128. Jet impinging on a curved, moving vane. The negative sign in equation (13) means that the force, F y , is exerted in a direction opposite to v v , or in other words, down- ward. The resultant force is y sn (14) If the angle of deflection is greater than 90, cos 0, in equation (12) becomes negative. If the jet is deflected through a full 180, 0= 1, and sin = 0, and the equations become 2wav 2 2Mv = y (15) (16) 136. Jet Deflected by a Moving Curved Vane. Consider the vane shown in Fig. 128 to be moving with a uniform velocity v' JET DEFLECTED BY A MOVING CURVED VANE 217 in the original direction of the jet. The absolute velocity of the jet as it impinges at A is v and its relative velocity with respect to the vane is v v' = u. The mass impinging per second is there- wau fore 9 Neglecting friction the relative velocity of the jet with respect to the vane remains unchanged while flowing from A to B so that the jet leaves the vane at B with a relative velocity u in a tangential direction, the X-component of which is u cos 6. The change in velocity along the X-axis is therefore u w cos or u(l cos B) and (a) In a similar manner the change in velocity in a vertical direc- tion is from zero to u sin B and therefore wau 2 9 sin 6 (18) The absolute velocity and direction of the jet as it leaves the vane are shown in Fig. 128 by the vector v a which is the result- ant of the relative velocity u and the velocity, v', of the vane. If a jet is directed against a double-cusped vane as shown in Fig. 129 (a) so that half the jet is deflected by each cusp through equal angles, F x will be determined by equation (17) but F y is zero 218 HYDRODYNAMICS since the two F-components balance being equal and opposite in direction. If a jet is deflected through a full 180 either by a single- or double-cusped vane as shown in Fig. 129 (6) and (c), obviously, for a stationary vane, (19) y and for a moving vane (20) If a series of vanes are so arranged on the periphery of a wheel that the entire jet, directed tangentiallyjto the circumference is striking either one vane or another successively, the mass imping- ing again becomes M= and the force exerted is, y F* = M(l-cos0), ...... (21) l7 F v = uswe ........ (22) y It should be noted that when F v is radial, F x is the only com- ponent of the force tending to produce rotation. 137. Work Done on Moving Vanes. Since work is equal to force times distance it is apparent that for a jet to do any work upon a vane, the vane must be moving with a velocity between zero and the velocity of the jet since at these limiting velocities either the distance or the force is equal to zero. The question then arises as to what velocity the vane should have, for any given velocity of jet, to perform the maximum amount of work. The amount of work done per second is the product of the force acting in the direction of motion and the distance through which it acts. Assuming that the direction of motion of the vane is parallel with the direction of the jet, the force acting is (Art. 136), (23) g and the distance through which it acts per second is equal to the WORK DONE ON MOVING VANES 219 velocity of the vane, v'. Representing the work done per second, expressed in foot-pounds, by G, (24) Considering v' as the only variable in this expression and equat- ing the first derivative to zero, the relation between v and v' may be determined for which G is a maximum. from which v' = v and z/ = ..... (25) o When v 1 ' = v, no work is done since the force exerted is then zero and this value represents a condition of minimum work. For maximum work therefore v' = ^ . o In the case of a series of vanes so arranged that the entire jet strikes either one vane or another successively, the force exerted in the direction of motion, which is assumed parallel with the direction of the jet is (Art. 136), F x = (v-v')(l- C osd) ...... (26) i/ The distance through which this force acts in one second is v', and therefore, G =(v-v')(l-cose)v' ..... (27) J/ Differentiating, and equating to zero, dG wav(l cos 6) dv' g and for maximum work v (0-200=0, (28) 220 HYDRODYNAMICS Substituting this value of v f in equation (27), G-^d-oos*), ....... (29) or G = ^(l-cos0), ....... (30) which is - - ~ - - times the total kinetic energy available in the 2i jet. For = 180 this expression equals unity and the total kinetic energy of the jet being converted into work. This also appears from considering that the relative velocity of the jet as it leaves the vane is , which, is also the velocity of the vane. These two velocities being equal and opposite in direction have a resultant of zero. The water thus leaves the vane with zero velocity, signifying that all of its original energy has been utilized in performing work. The above principles are made use of in the design of impulse turbines, which consist of a series of vanes attached to the per- iphery of a wheel. The angle 6 must be somewhat less than 180 so that the jet in leaving a vane will not interfere with the suc- ceeding vane. Making the angle 6 equal to 170 in place of 180 reduces the force applied to the wheel by less than 1 per cent. 138. Forces Exerted upon Pipes. In the preceding articles of this chapter the discussion has been restricted to forces exerted by jets impinging against flat and curved surfaces. As it was always considered that the flow was free and unconfined the only forces acting were dynamic. Consideration will now be given to the longitudinal thrust exerted upon a section of pipe by water flowing through it under pressure. This thrust will usually be found to be the resultant of both static and dynamic forces. The transverse forces which determine the necessary thickness of pipe were discussed in Art. 29. 139. Straight Pipe of Varying Diameter. Under conditions of steady flow through a straight pipe of varying diameter there is a STRAIGHT PIPE OF VARYING DIAMETER 221 longitudinal thrust exerted upon the pipe. This thrust is the resultant of a dynamic force, a static pressure, and frictional resistance. Fig. 130 shows a straight section of converging pipe. Let p\, ai t an d v\ represent respectively the pressure, area, and mean veloc- ity at A B and ^2, 2, and V2 the corresponding values at CD. In flowing from A B to CD the water is accelerated from vi to V2 and the force, P, producing this acceleration is the resultant of all the component forces acting on the mass A BCD. These forces consist of the pressures on the sections AB and CD, the pressure exerted by the pipe walls ACBD and the force of gravity, the last FIG. 130. of which can be neglected since it acts vertically and has no com- ponent in the direction of acceleration. The pressures on AB and CD are a\,p\ and a^pz respectively, a\p\ acting in the direction of acceleration and a^p^ being opposed to it. The pressure, dR, exerted on the water by any differential area, da, of the pipe walls will be inclined slightly from the normal toward the direction of flow on account of friction. The vertical component of dR, being normal to the direction of acceleration may be neglected, leaving dR x as the only component to be considered. All the values of dR x for the various elementary areas of pipe wall, being parallel and acting in the same direction, may be combined into the resultant, R x , whose magnitude is as yet unknown but whose 222 HYDRODYNAMICS direction is opposed to acceleration. Therefore the force pro- ducing acceleration, R x ........ (32) But in Art. 131 it was shown that and therefore, from equations (32) and (33), y (33) (34) Since R x is the X-component of the forces exerted upon the water by the pipe it follows that the thrust exerted upon the section of pipe by the water must be equal and opposite to R x , or in other words the thrust will act toward the right in Fig. 130. Considering a straight section of pipe of constant diameter throughout, equation (34) reduces to (35) since 0,1 = 0,2 = a, and Vi = v 2 . In equation (35) pi p 2 is the drop in pressure resulting from friction between sections A B and CD. 140. Pipe Bends. The thrust exerted upon a curved section of pipe of either constant or varying diameter is the re- sultant of component forces similar to those discussed in the preceding article. The chief difference lies in the fact that the resultant thrust on a curved section of pipe has both X- and F-components since there is a change in ve- locity along both of these axes. Fig. 131 shows a pipe bend, having a diameter de- creasing from AB to CD and FlG 131 a deflection angle, 6. Let R x and R y represent the X- and F-components of the forces exerted by the pipe upon the water. WATER HAMMER IN PIPE LINES 223 The X- and F-components of the thrust exerted by the water upon the pipe are equal in magnitude but opposite in direction to R z and R y respectively. Assuming that the bend lies in a horizontal plane so that the action of gravity is normal to the direction of acceleration and therefore may be ignored, the resultant X and Y forces producing acceleration are, C12P2 cos 6 R x = (v2 cos vi), . (36) R y = v 2 sin 0, ..... (37) J the right-hand members in these equations representing the in- crease in momentum along the X- and F-axes resulting from the accelerating forces. From the above equations, U2P2 cos 0-f (vi v 2 cos 0), . . . (38) t/ sin0+ v 2 sme ........ (39) \J If the pipe bend is one of constant diameter throughout, ai=a2, vi=V2 t and pi=p2 (approximately), and the equations reduce to -cos*), ....... (40) h* .......... (41) If the angle equals 90 these equations become (42) y 141. Water Hammer in Pipe Lines. In Fig. 132 is shown a pipe line leading from a reservoir, A, and discharging into the air at B near which is located a gate valve. If the valve is suddenly closed a dynamic pressure is at once exerted in the pipe in excess of the normal static pressure. The magnitude of this pressure is frequently much greater than that of any static pressure to which the pipe may ever be subjected and the possibility of the occurrence 224 HYDRODYNAMICS of such pressure must therefore be investigated in connection with the design of any pipe line of importance. This dynamic pressure, commonly called water hammer, is the result of a sudden transformation of the kinetic energy of the moving mass of water within the pipe into pressure energy. Since force equals mass times acceleration, or M- M ' (43) it follows that if the velocity of the mass M could be reduced from v to zero instantaneously, this equation would become P = M 0' (44) or in other words the pressure resulting from the change would be infinite. Such an instantaneous change is, however, impossible. FIG. 132. Consider the conditions within the pipe immediately following the closure of the valve. Let l\, 1%, Is, ... In represent infmitesi- mally short sections of pipe as shown in Fig. 132. The instant the valve is closed, the water in contact with it in section l\ is brought to rest, its kinetic energy is transformed into pressure energy, the water is somewhat compressed and the pipe wall with which it is in contact expands slightly as a result of the increased stress to which it is subjected. Because of the enlarged cross-sectional area of l\ and the compressed condition of the water within it, a greater mass of water is now contained within this sec- tion than before the closure. It is evident then that a small volume of water flowed into section h after the valve was closed. An WATER HAMMER IN PIPE LINES 225 instant later a similar procedure takes place in 1 2 and then in k, so that evidently a wave of increased pressure travels up the pipe to the reservoir. The instant this wave reaches the reservoir the entire pipe is expanded and the water within it is compressed by a pressure greater than that due to the normal static head. There is now no longer any moving mass of water within the pipe, the conversion of whose kinetic energy into pressure energy serves to maintain this high pressure and therefore the pipe begins to contract and the water to expand with a consequent return to normal static pressure. This process starts at the reservoir and travels as a wave to the lower end. During this second period some of the water stored within the pipe flows back into the reservoir but on account of the inertia of this moving mass an amount flows back greater than the excess amount stored at the end of the first period so that the instant this second wave reaches the valve the pressure at that point drops not only to the normal static pressure but below it. A third period now follows during which a wave of pressure less than static sweeps up the pipe to the reservoir. When it reaches the reservoir the entire pipe is under pressure less than static but since all the water is again at rest the pressure in / immediately returns to the normal static pressure due to the head of water in the reservoir. This starts a fourth period marked by a wave of normal static pressure moving down the pipe. -When the valve is reached the pressure there is normal and for an instant the conditions throughout the pipe are similar to what they were when the valve was first closed. The velocity of the water (and the resultant water hammer) is now, however, somewhat less than it was at the time of closure because of friction and the imperfect elasticity of the pipe and the water. Instantly another cycle begins similar to the one above described, and then another, and so on, each set of waves succes- sively diminishing, until finally the waves die out from the influ- ences above mentioned. Equation (44) shows that for instantaneous closure of valve the pressure created would be infinite if the water were incom- pressible and the pipe were inelastic. Instantaneous closure is, however, physically impossible. To determine the amount of excess pressure actually resulting from water hammer it is neces- sary to take into consideration the elasticity of the pipe and the compressibility of water. This leads to a rather lengthy mathe- 226 HYDRODYNAMICS matical analysis which will here be avoided and there will be given only the resulting workable equations. The following nomen- clature will ,be used, all units being expressed in feet and seconds except E and E f , which are in pounds per square inch: b = thickness of pipe walls; D = inside diameter of pipe; E = modulus of elasticity of pipe walls in tension; E' = modulus of elasticity of water in compression; g = acceleration of gravity; h = head due to water hammer (in excess of static head) ; H = normal static head in pipe; L = length of pipe line; 7 7 = time of closing valve; v = mean velocity of water in pipe before closure of valve; v w = velocity of pressure wave along pipe. 142. Formulas for Water Hammer. In the following dis- cussion whenever the term " pressure " is used it is understood to mean " pressure due to water hammer " and is the amount of pressure in excess of that due to the normal static head. If the valve is closed instantaneously the pressure in l\ imme- diately rises to p max and remains at this value while the pressure wave travels to the reservoir and returns. The time required nr for the wave to travel to the reservoir and back to the valve is -. v w The pressure in l n reaches this same p max but remains at that value only for an instant. At any intermediate section, p m&K is maintained only until the wave of reduced pressure reaches that section. If the time taken to completely close the valve is or exactly , p max will occur only in section l\, and will last only Vw for an instant, being immediately lowered by the return of the static pressure wave. If the time of closing the valve is greater 2L than , p max will never be attained since the wave of reduced Vw pressure will then have reached li before the valve is completely closed or p max is reached. Evidently two formulas are necessary; one to determine the maximum water hammer, when the time of closure is less than FORMULAS FOR WATER HAMMER 227 Q7" , and the other to determine the ordinary water hammer that v w or occurs when the time of closure is greater than , as is usually Vw the case. The same formula for the determination of maximum water hammer has been quite generally adopted. The general expres- sion is, *m = ^, ......... (45) y where 4660 (46) Eb Substituting this value in equation (45), * f ( } Eb For steel pipe this reduces to 1 A.ZV (48) Frequently these expressions appear in different forms but they may all be reduced to the above forms. There is no such general agreement as to the proper formula to be used for the determination of ordinary water hammer when T 2L is greater than . Many formulas have been derived, some ^w giving results twice as great as others. Certain assumptions as to the manner of valve closure, the effect of friction and the manner in which the waves are reflected, etc., must be made before any theoretical formula can be derived. It appears that the main reason for the wide discrepancy in results lies in the difference in these fundamental assumptions. Assuming that the valve is closed in such manner that the rate of rise in pressure will be constant throughout the entire closure, OA, Fig. 133, represents the variation in pressure at the 228 HYDRODYNAMICS valve during the time T, provided that no return pressure wave 2Z/ interferes. After a time, , however, the returning wave will v w reach the valve and assuming that its intensity has been undi- minished by friction or other cause, it will exactly annul the tendency for the pressure to increase, due to continued closing, and as a result the pressure remains constant during the remainder FIG. 133. of the time T, as shown by the horizontal line BA'. similar triangles, From From which g v w 2Lv (49) This formula was first proposed by Professor Joukovsky 1 of Moscow, Russia, in 1898 and, it is claimed, was substantiated by a series of experiments which he conducted. Other commonly used formulas 2 are as follows : NH (Allievi) 1 N. JOUKOVSKY: Trans, of Prof. N. Joukovsky's paper on Water Ham- mer, by Boris Simin; Journal American Waterworks Association (1904). 2 MILTON M. WARREN: Penstock and Surge-Tank Problems. Trans. Amer. Soc. Civ. Eng., vol. 79 (1915). Contains discussions of all commonly used water hammer formulas. PROBLEMS 229 where I, . . (Johnson) -iV - where Lv and N = 2gHT, Lv (Warren) h = m (Mead, et al.) Joukovsky's formula gives results twice as great as Mead's formula. Johnson's formula can readily be reduced to Allievi's. It may be noted that these two formulas make h vary with H, which the other formulas do not do. Justification for this varia- tion is not apparent. Eliminating Johnson's formula (which reduces to Allievi's), no two of the above formulas give results that are at all similar under all conditions. Discrepancies of from 100 to 200 per cent are possible. A comprehensive and carefully conducted series of experiments are necessary before any formula for ordinary water hammer can be relied upon to give trustworthy results. PROBLEMS 1. A jet 1 in. in diameter and having a velocity of 25 ft. per second strikes normally against a fixed, flat plate. Determine the pressure on the plate. 2. In Problem 1, what would be the pressure on the plate if it were moving with a uniform velocity of 10 ft. per second in the same direction as the jet? 3. What should be the velocity of the plate, in Problem 2, if the jet is to perform the maximum amount of work? Determine the corresponding amount of work in foot-pounds per second. 4. A jet having a diameter of 2 in. and a velocity of 40 ft. per second is deflected through an angle of 60 by a fixed, curved vane. Determine the X- and F-components of the force exerted. 6. Solve Problem 4 if the vane is moving with a velocity of 25 ft. per second in the same direction as the jet. 6. A H-in. nozzle has a coefficient of velocity of 0.97 and a coefficient of contraction of unity. The base of the nozzle has a diameter of 4 in., at which point the gage pressure is 80 Ibs. per square inch. The jet is 230 HYDRODYNAMICS deflected through an angle of 150 by a double-cusped vane that has a velocity, in the direction of the jet of 30 ft. per second. What is the pressure exerted on the vane arid what is the amount of work done expressed in foot-pounds? 7. In Problem 6 determine the velocity the vane must have if the jet is to perform the maximum amount of work. What is the maximum work in foot-pounds? 8. If the jet, in Problem 6, strikes a series of vanes so arranged on the periphery of a wheel that the entire jet is deflected through an angle of 170, what is the maximum amount of work that can be done? 9. A horizontal straight pipe gradually reduces in diameter from 12 in. to 6 in. If, at the larger end, the gage pressure is 40 Ibs. per square inch and the velocity is 10 ft. per second, what is the total longitudinal thrust exerted on the pipe? Neglect friction. 10. A bend in a pipe line gradually reduces from 24 in. to 12 in. The deflection angle is 60. If at the larger end the gage pressure is 25 Ibs. per square inch and the velocity is 8 ft. per second, determine the X- and F-com- ponents of the dynamic thrust exerted on the bend. Also determine the X- and F-components of the total thrust exerted on the bend, neglecting friction. 11. A 24-in. cast-iron pipe | in. thick and 6000 ft. long discharges water from a reservoir under a head of 80 ft. What is the pressure due to water hammer resulting from the instantaneous closure of a valve at the discharge end? 12. If the time of closing the valve, in Problem 11, is 6 sec., determine the resulting pressure due to water hammer, comparing results obtained by use of formulas by Joukovsky, Johnson, Warren and Mead. INDEX Absolute and gage pressure, 10 Absolute velocity, 214 Acceleration, 45, 48, 181, 202 Accuracy of computations, 5 Air in pipes, 164, 165 Algebraic transformation of orifice formula, 80 of weir formula, 113 Allievi formula for water hammer, 228 Angle of heel, 40, 42 Angular velocity, 48 Approach, channel of, 71, 76, 103, 108 Approach, velocity of, 71, 76, 80, 103, 108, 112, 120, 128 Archimedes, principle of, 38 Atmospheric pressure, 9, 15, 17 Axis of moments, 25 Backwater, 204, 207 Barometer, mercury, 15 water, 16 Barr's experiments, 123 Base formula, 52 Bazin experiments, 111 open-channel formula, 191 submerged-weir formula, 121 weir formula, 116 Bends in pipes, loss of head. 143, 162 coefficients for, 163 thrust at, 222 Bernoulli's equation, 59, 63, 74, 87, 141 Bernoulli's theorem, 57 application to hydrostatics, 60 in practice, 60 Boiling point of water, 2 Borda's mouthpiece, 91 Bridge piers, 195 Broad-crested weirs, 129 Buoyancy, center of, 39 Buoyant force, 39, 40 Canals, 199 Capillary action, 4, 16 Cast-iron pipes, coefficients, 148, 149, 152, 153 Center of buoyancy, 39, 42 Center of gravity, 31, 39, 42 Center of pressure, 24, 29, 31, 32 and center of gravity, 31, 32 Channel entrance, 131, 202, 203 Channel of approach, 71, 76, 103, 108 Characteristics of jet, 71 Chezy formula, open channels, 185 pipes, 146 Chutes, 129 Cippoletti weir, 124 Circular channels, 176, 201 orifice, 71 Coefficient of contraction, nozzles, 89 orifices, 77, 78 tubes, 86, 88, 92, 93 weirs, 110 Coefficient of discharge, nozzles, 89 orifices, 78, 80 submerged orifice, 94 tubes, 86, 93 Coefficient of roughness, 185 Coefficient of velocity, nozzles, 90 orifices, 77, 78 tubes, 86, 93 weirs, 111 Coefficients of discharge, nozzles, 90 orifices, 79 tubes, 86 231 232 INDEX Coefficients, hose, 150 open channels, 187, 190, 191 pipes, 148, 149, 153. 161, 162, 163 weirs, 108 Coefficient, weir, 111 Components of pressure, 33, 35 Compound pipes, 168 Compressibility of water, 3 Computations, accuracy of, 5 Concrete pipes, coefficients, 148, 149, 153 Conduits, 135, 199 Conical tubes, 88 Conservation of energy, 59 Converging tubes, 88 Continuity of discharge, 53 of flow, 53, 140, 180, 202 Contraction, coefficient of, 77, 78, 89, 92, 110 crest, weirs, 101 end, weirs, 102 gradual, pipes, 143, 159 in channels, 183, 192 of jet, 72, 76, 77, 88, 92 of nappe, 103, 108, 110 sudden, pipes, 143, 158 suppressed, 84, 102 surface, weirs, 102 vertical, weirs, 103 Converging tubes, 88 Crest contraction, 101 Critical velocity, 136 Cross-section, most efficient, 198 Curves, backwater, 205 in open channels, 183, 196 in pipes, 143, 162 Curved surface, pressure on, 33 Dams, coefficients for, 127 pressure against, 33, 34 Darcy, experiments on Pitot tube, 65, 66 modification of Pitot tube, 68 pipe formula, 147 Density of water, 2 Depth of flotation, 39 Differential gage, 18 Direction of resultant pressure, 7 Discharge, 52 continuity of, 53, 140, 180 head lost at, 143, 155, 157, 183 under falling head, 96 Discharge coefficient, nozzles, 89 orifices, 78 tubes, 86, 93 Discussion of open-channel formulas, 191 of pipe formulas, 154 of weir formulas, 118 Distance, unit of, 1 Distilled water, properties of, 2 Distribution of velocities, 108, 138, 176 Diverging tube, 90 Divided flow, 169, 206 Dynamic force, 214-229 Efficient channel section, 198 Elasticity of water, 3 Elevation and air pressure, 9 Emptying vessel, 97 End contractions, weirs, 102 suppressed, 102 Energy and head, 54 kinetic, 54, 57 of position, 55, 57 of water in channel, 109, 179 of water in pipe, 139 per pound of water, 57, 140 potential, 55, 57 pressure, 56 English system of units, 1 Enlargement of section, channels, 193 pipes, 143 Entrance losses, channels, 183 pipes, 143, 155, 156 Entrance to channels, 131 Equation of continuity, 53, 181, 202 Falling head, discharge under, 96 Falls, 117 Floating bodies, 38 stability of, 39 Flow, continuity of, 53, 140, 180, 202 in open channels, 176-209 in rivers, 196 INDEX 233 Flow, non-uniform, 53, 201 over weirs, 101-134 steady, 53 stream line, 54, 136 through nozzles, 89 through orifices, 71-99 through pipes, 135-175 through tubes, 85-93 turbulent, 54, 136 uniform, 53 Fluids, definition, 1 Flumes, 199 Force, acting on mass, 212, 213 exerted on pipe, 220-223 exerted on vane, 213-218 Francis coefficient for weirs, 111 experiments, 111 formula for end contractions, 112 weir formula, 115 Free surface of liquid, 9 Freezing temperature of water, 2 Friction, 52 and distribution of velocities, 108, 138, 176 Bernoulli's equation with, 59, 141 coefficients for hose, 150 coefficients for open channels, 187, 190, 191 coefficients for pipes, 148, 149, 153 formulas for open channels, 185- 191 formulas for pipes, 146-154 independent of pressure, 145 in open channels, 176, 184 in orifices, 76, 77 in pipes, 138-146 in Venturi meter, 64 of flowing water, 59, 138, 176 over weirs, 108 Fteley and Stearns, experiments, 111 weir formula, 116 Fundamental orifice formula, 73 Gage, differential, 18 hook, 131 mercury, 17 oil, 19 water stage, 132 Gage pressure, 10, 16 Ganguillet and Kutter, 185 Gases, 1, 9 Gates, 95, 195 and valves, 143, 162 coefficients for, 96 pressure on, 26, 28, 32 Grade line, hydraulic, 144, 154, 164, 184, 191 Gradient, hydraulic, 144, 154, 164, 184, 191 Hazen- Williams formula, 150 Head, definition of, 54 energy and, 54 lost at bends or curves, 143, 162, 183, 196 lost at discharge, 143, 155, 157, 194 lost at entrance, 143, 155, 156, 193 lost at obstructions, 143. 162, 195 lost by contractions, 143, 158, 192 lost by enlargements, 143, 160, 193 lost by friction, 59 lost by friction in open channels, 181, 183, 184 lost by friction in pipes, 140, 'l43, 144 lost in nozzles, 90 lost in orifices, 81 lost in tubes, 87 measurement of, 131 pressure, 13, 16, 56 velocity, 55 Height of weir, 103 Herschel, Venturi meter, 61 Hook gage, 4, 131 Horizontal and vertical components of pressure, 32, 34 Horizontal orifice, 71, 75 Horton's values of "n," 187 weir coefficients, 127 Hose, 150 coefficients, 150 Hydraulic grade line, 144, 154, 164, 184, 191 gradient, 144, 154, 164, 184, 191 jack, 13 radius. 135, 176 234 INDEX Hydraulics, definition, J or rivers, 196 Hydrodynamics, 211-230 definition, 1 fundamental principles, 211 Hydrokinetics, 1, 51-210 Hydrostatic pressure, 7-37 Hydrostatics, 1, 7-44 Immersed and floating bodies, 38-44 Impurities in water, 2 Incomplete contractions, 84, 103 Intensity of pressure, 7, 8, 10, 30, 35 Inversion of jet, 73 Inward-projecting tubes, 93 Irregular sections of channels, 197 Jack, hydraulic, 13 Jet, characteristics of, 71 contraction of, 72, 76, 77, 88, 92 definition, 71 deflected by curved vane, 216 force of, 213-218 forms assumed by, 72 impinging on fixed plate, 214 inTpinging on moving plate, 215 inversion of, 73 path of, 82 pressure in, 72 velocity with respect to vane, 217 work done by, 218 Johnson water-hammer formula, 229 Joukovsky, water-hammer experi- ments, 228 formula, 228 Kinetic energy, 54, 57 in channel of approach, 109 in open channels, 109, 179 in pipes, 139 King pipe formula, 152 submerged-weir formula, 121 weir formula, 117 Kutter formula, 185 table for solving, 186 Lea, investigation of pipes, 152 Linear acceleration, 45 Liquid, buoyant force of, 38 definition, 1 free surface of, 9 in motion, 51 intensity of pressure within, 7, 8 relative equilibrium, 45 rotating, 47 surface tension, 4 uniformly accelerated, 45 water most common, 51 Long pipes, 166 Loss of head, at discharge, 143, 155, 157, 183, 194 at entrance, 143, 155, 156, 183, 194 due to bends or curves, 143, 162, 183, 196 due to contractions, 143, 158, 192 due to enlargements, 143, 160, 193 due to friction in channels, 181, 184 due to friction in pipes, 140-146 due to obstructions, 143, 162, 195 in nozzles, 90 in open channels, 181 in orifices, 81 hi pipes, 140 in tubes, 87 Manning formula, 188 Mead water-hammer formula, 229 Mean velocity over weirs, 107 Measurement of head, 131 Mercury barometer, 15 gage, 17 Metacenter, 40 Metacentric height, 40 Meter, Venturi, 61 Modifications of fundamental weir formula, 112 Modulus of elasticity, 3 Moment of inertia, 25 Momentum, definition, 211 Motion, stream line, 54, 136 turbulent, 54, 136 Mouthpiece, Borda's, 91 Nappe, 101 contracted section of, 103 contraction of, 108, 110 depth of mean velocity in, 108 INDEX 235 Nappe, pressure in, 104 velocities in, 104, 107, 111 Negative pressure, 10 Newton's laws of motion, 211 Non-uniform flow, 53, 201 Nozzles, 89 Numerical computations, 5 Obstructions in channels, 183, 195 in pipes, 143, 162 Oil-differential gage, 19 Open-channel friction formulas, 185- 191 Bazin formula, 191 Chezy formula, 185 Kutter formula, 185, 188 Manning formula, 188 Open channels, backwater in, 204 bends or curves in, 183, 196 contractions in section, 183, 192 description and definition, 176 discussion of formulas, 191 divided flow in, 206 efficient section for, 198 energy of water in, 109, 179 enlargement in section, 183, 193 entrance to, 131, 183 friction in, 176 hydraulic gradient of, 184, 191 hydraulic radius of, 176 loss of head in, 181 non-uniform flow, 201 obstructions in, 183, 195 water surface of, 184, 191 wetted perimeter of, 176 Orifice, circular, 71 coefficients, 76 definition and description, 71 discharge under falling head, 96 flow through, 71-85, 93-98 fundamental formula, 73 head lost in, 81 horizontal, 71, 75 partially submerged, 94 rectangular, 71, 82 square, 71 submerged, 93 suppressed contraction, 84 Orifice, under low heads, 82 vertical, 71 with velocity of approach, 80 Parabolic weir, 101 Partially submerged orifices, 94 Pascal's law, 8 Path of jet, 82 Perfect liquid, 4 Piezometer tubes, 5, 16 Pipe-friction formulas, 146-154 Chezy formula, 146 Dracy formula, 147 Hazen- Williams formula, 150 King formula, 152 non-turbulent flow, 154 Pipes, bends or curves in, 143, 162 branching, 166 coefficients, 148, 149, 151, 153 compound, 168 continuity of flow, 140 contractions in, 143, 158 critical velocity in, 136 description and definition, 135 discharge from, 143, 155, 157 divided flow in, 169 energy of water in, 139 enlargements in, 143, 160 entrance to, 143, 155, 156 flow through, 135-175 force exerted upon, 220 friction coefficients, 148, 149, 153 friction in, 138, 144 hydraulic gradient, 144, 154, 164 hydraulic radius of, 135 loss of head in, 140 more than one diameter, 142, 172 obstructions in, 143, 162 special problems, 165 tensile stress in walls, 35 thrust at bends, 222 viscosity, effect on friction, 145 water hammer in, 223 wetted perimeter of, 135 Piston, 13, 56 Pitot tube, 65-69 Plate, jet impinging against, 214, 215 Potential energy, 55, 57 236 INDEX Pressure, absolute and gage, 10 atmospheric, 9, 15 center of, 24, 29, 31 energy, 56, 62 head, 13, 16, 56 intensity of, 7, 8, 10, 30, 32, 35 negative, 10 on curved surfaces, 33, 35 on plane areas, 23 on surfaces, 23-37 relative, 10 transmission of, 13 vapor, 14, 164 Principle of Archimedes, 38 Principles of equilibrium, 8 Principles of hydrokinetics, 51-70 Principles of hydrostatic pressure, 7- 22 Pumps, suction, 20 Radius, hydraulic, 135, 176 of bends in pipes, 163 Reaction of jet, 213 Rectangular channels, 176, 199 orifice, 71, 82 weir, 101 Relative and absolute velocities, 214 Relative equilibrium of liquids, 45-50 Relative pressure, 10 Resultant pressure, direction of, 7, 33 position of, 32 Rivers, hydraulics of, 196 Rotating vessel, 47 Salt water, 3 Scow, stability of, 42 Section of greatest efficiency, 198 Semicircular channel, 198 Sewers, 135, 176 Ships, stability of, 40 Siphon, 20 Skin of water surface, 4 Slope, "s," 145, 184 Special pipe problems, 165 Specific gravity of mercury, 17 of water, 2 Square orifice, 71 Stability of scow, 42 Stability of ship, 40 Standard short tube, 85 Steady flow, 53 Stream line motion, 54, 136 Submerged bodies, 38 orifices, 93 weirs, 95, 118, 120, 195 Suction pumps, 20 Sudbury conduit, 177 Summit of pipe, 164, 165 of siphon, 20 Suppressed contraction, orifices, 84 Suppressed weirs, 102 Suppression of contraction, 84, 102 Surface contraction, 102 Surfaces, pressure on, 23-37 Surface tension, 4, 177 Tangential stress of water, 1, 4, 8 Temperature, 2 effect on critical velocity, 138 effect on viscosity, 138 effect on weight of water, 2 Tensile stress in pipe walls, 35 Theoretical discharge over weirs, 105 velocity over weirs, 107 velocity through orifices, 74 Thickness of pipes required, 35 Thompson's experiments, 123 Time of emptying vessel, 97 Time, unit of, 1 Torricelli's theorem, 76 Transmission of pressure, 13 Trapezoidal channels, 176, 199 weirs, 123 Triangular weirs, 121 Tube, definition of, 71 Tubes, capillary action in, 4 converging, 88 diverging, 90 immersed in flowing water, 65 standard short, 85 Turbulent motion, 54, 136 Uniform flow, 53 Units in hydraulics, 1 Unsteady flow, 53 U-tube, 17 INDEX 237 Vacuum, 10, 17 amount of, 10 perfect, 10 Vane, jet deflected by, 215, 216 fixed curved, 215 moving curved, 216 work done on, 218 Vapor pressure, 14, 164 variation with temperature, 15 Velocities, distribution of, 138, 176 in vertical, 177 Velocity, absolute and relative, 214 at any depth, over weirs, 104 critical, 136 curves of, 177 from orifices, 74 head, 55, 81 of approach, orifice, 71, 75, 76, 78, 80 weir, 103, 108, 112, 114, 120, 128 Velocity coefficient, nozzles, 90 orifices, 77, 78 tubes, 86, 93 weirs, 111 Vena contracta, 72 Venturi meter, 61 Vertical contraction, 103 Vertical jets, 75 Vertical orifice, 71 Vertical velocity curves, 177 Vessel, rotating, 47 time of emptying, 96 with constant acceleration, 45 Viscosity, 4, 137, 143 Warren water-hammer formula, 229 Water barometer, 16 boiling point of, 2 compressibility of, 3 dynamic force, 214-229 freezing point of, 2 hammer, 223 maximum density of, 2 physical properties of, 2-5 Water, weight of, 2 Weight of submerged bodies, 38 of mercury, 17 of water, 2 unit of, 1 Weir, broad crested, 129 channel of approach, 71, 76, 103, 108 Cippoletti, 124 coefficient, 111, 127 coefficients, 108, 110, 111 conditions for accurate use, 132 definition and description, 101 discharge formulas for sharp crest, 115 discussion of formulas, 118 experiments, 111, 115 formula for mean velocity, 105 fundamental formulas, 83, 107, 112 head measurement, 131 height of, 103 modification of fundamental for- mula, 112 not sharp-crested, 101, 103, 125 range of accuracy, 132 rectangular, 101 sections, 126 sharp-crested, 101 submerged, 95, 118 theoretical formula for discharge, 105 transformation of formula, 113 trapezoidal, 101, 123 triangular, 101, 121 velocity of approach to, 103 with end contractions, 102, 112 with end contractions suppressed, 102 Wetted perimeter, 135, 176 Wooden pipes, coefficients, 148, 149, 153 Work done on vane, 218 on mass of water, 58 THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. <)CT 23 1940 M AUG 31 1943 MOV 9 1940M Sp 23 194 , Vty ,j H^R^S&I^ I94j to * PSN/l^> 1J fjf. ?9 ?&:- **^ *2 i /I AU6 26 M42 n tlS^ MOV SI ) 1040 o M n v o ^C |,, CCD *l 4 4O44 J^^n rto 11 iy4d n AT ir . . . . UOI Ib J LD 21-100m-7,'40 (6936s) UNIVERSITY OF CALIFORNIA LIBRARY - - .C