ELEMENTS OF HYDRAULICS McGraw-Hill DookCompany Electrical World The Engineering and Mining Journal En5ineering Record Engineering News Railway Age Gazette American Machinist Signal Lngin."..=.'. Selection of type Action and reaction wheels Speed 181 criterion Capacity criterion Characteristic speed Spe- cific discharge Specific power Specific speed Relation between characteristic speed and specific speed Classi- fication of reaction turbines Classification of impulse wheels Numerical applications Normal operating range Selection of stock runner. 37. POWER TRANSMITTED THROUGH PIPE LINE AND NOZZLE . . . 195 Effective head at nozzle Velocity of flow for maximum power Maximum efficiency Diameter of nozzle for maximum output of power Graphical relation between power and efficiency Method of determining nozzle diameter. 38. EFFECT OF TRANSLATION AND ROTATION 198 Equilibrium under horizontal linear acceleration Equi- librium under vertical linear acceleration Free surface of liquid in rotation Depression of cup below original level in open vessel Depression of cup below original level in closed vessel Practical applications. 39. WATER HAMMER IN PIPES 202 Increase in pressure due to suddenly checking flow Bulk modulus of elasticity of water Pressure waves in pipe produced by suddenly checking flow Velocity of com- pression wave Period of compression wave Increase in pressure produced by instantaneous stoppage of flow Joukovsky's experiments Gibson's experiments. 40. HYDRAULIC RAM .;.......... 205 Principle of operation Efficiency of ram. CONTENTS xi ART. PAGE 41. DISPLACEMENT PTJMPS 207 Suction pump Maximum suction lift Force pump Stress in pump rod Direct driven steam pump Calcula- tion of pump sizes Power required for operation Diam- eter of pump cylinder. 42. CENTRIFUGAL PUMPS 215 Historical development Principle of operation Impeller types Conversion of kinetic energy into pressure Volute casing Vortex chamber Diffusion vanes Stage pumps. 43. PRESSURE DEVELOPED IN CENTRIFUGAL PUMP 224 Pressure developed in impeller Pressure developed in diffusor General formula for pressure head developed. 44. CENTRIFUGAL PUMP CHARACTERISTICS 227 Effect of impeller design on operation Rising and droop- ing characteristics Head developed by pump Effect of throttling the discharge Numerical illustration. 45. EFFICIENCY AND DESIGN OF CENTRIFUGAL PUMPS 234 Essential features of design Hydraulic and commercial efficiency. 46. CENTRIFUGAL PUMP APPLICATIONS 236 Floating dry docks Deep wells Mine drainage Fire pumps Hydraulic dredging Hydraulic mining. APPLICATIONS 242 SECTION IV HYDRAULIC DATA AND TABLES TABLE 1. Properties of water 260 2. Head and pressure equivalents . ............. 261 3. Discharge equivalents . '. . . . 262 4. Weights and measures .;....., 263 5. Specific weights of various substances . . 264 6. Dimensions of steam, gas, and water pipe . . ... . . . 265 7. Capacity of reciprocating pumps 266 8. Circumferences and areas of circles 268 9. Efflux coefficients for circular orifice .... . . . . . . 273 10. Efflux coefficients for square orifice . . . . . 272 11. Fire streams 275 12. Coefficients of pipe friction 277 13. Friction head in pipes 278 14. Bazin's values of Chezy's coefficient 282 15. Kutter's values of Chezy's coefficient 283 16. Discharge coefficients for rectangular notch weirs 285 17. Discharge per inch of length over rectangular notch weirs . 286 18. Discharge per foot of length over rectangular notch weirs . 287 19. Principles of mechanics 288 20. Discharge per foot of length over suppressed weirs .... 290 INDEX . 291 ELEMENTS OF HYDRAULICS SECTION I HYDROSTATICS 1. PROPERTIES OF A PERFECT FLUID Definition. -A fluid is defined, in general, as a substance which offers no resistance to change in form provided this deformation is not accompanied by change in volume. The fundamental property of a fluid is the perfect mobility of all its parts. Most of the applications of the mechanics of fluids relate to water, and this domain of mechanics is therefore usually called hydraulics or hydromechanics. It is convenient to subdivide the subject into hydrostatics, relating to water at rest; hydro- kinetics, relating to water in motion; and hydrodynamics, relating to the inertia forces exerted by fluids in motion, and the energy available from them. Distinction between Liquid and Gas. A liquid such as water has a certain degree of cohesion, causing it to form in drops, whereas a gaseous fluid tends to expand indefinitely. A gas is therefore only in equilibrium when it is entirely enclosed. In considering elastic fluids such as gas and steam, it is always necessary to take account of the relation between volume and pressure. For a constant pressure the volume also changes greatly with the temperature. For this reason the mechanics of gases is concerned chiefly with heat phenomena, and forms a separate field called thermodynamics, lying outside, and sup- plementary to, the domain of ordinary mechanics. Elasticity of Water. Water, like other fluids, is elastic, and under heavy pressure its volume is slightly diminished. How- ever, since a pressure of one atmosphere, or 14.7 Ib. per square inch, exerted on all sides only decreases its volume about 20 QQQ* and a pressure of 3000 Ib. per square inch, only 1 per cent., it is customary in all practical calculations, and sufficiently accurate 1 2 , i t/j > t v .-ELEMENTS OF HYDRAULICS . for ordinary purposes, to assume that it is incompressible. The volume of an ideal liquid is therefore assumed to remain constant. Fluid Pressure Normal to Surface. Since a perfect fluid is one which offers no resistance to change in form, it follows that the pressure on any element of surface of the fluid is everywhere normal to the surface: To prove this proposition, consider any small portion of a fluid at rest, say a small cube. Since this cube is assumed to be at rest, the forces acting on it must be in equi- librium. In the case of a fluid, however, the general conditions of equilibrium are necessary but not sufficient, since they take no account of the fact that the fluid offers no resistance to change in form. Suppose, therefore, that the small cube under con- sideration undergoes a change in form and position without any change in volume. Since the fluid offers no resistance to this deformation, the total work done on the elementary cube in producing the given change must be zero. In particular, suppose that the cube is separated into two parts by a plane section, and that the deformation consists in sliding one of these parts on the other, or a shear as it is called. Then in addition to the forces acting on the outside of each part, it is necessary to consider those acting across the plane section. But the total work done on each part separately must be zero inde- pendently of the other part, and also the total work done on the entire cube must be zero. Therefore, by subtraction, the work done by the forces acting across the plane section must be zero. But when any force is displaced it does work equal in amount to the product of this displacement by the component of the force in the direction of the displacement. Therefore if the work done by the force acting across the plane section of the cube is zero, this force can have no component in the plane of the section, and must therefore be normal, i.e., perpendicular, to the section. Viscosity. This absence of shear is only rigorously true for an ideal fluid. For water there is a certain amount of shear due to internal friction, or viscosity, but it is so small as to be practi- cally negligible. The greater the shear the more viscous the fluid is said to be, and its amount may be taken as a measure of viscosity. It is found by experiment that the internal friction depends on the difference in velocity between adjacent particles, and for a given difference in velocity, on the nature of the fluid. The viscosity of fluids is therefore of great importance in consider- ing their motion, but does not affect their static equilibrium. HYDROSTATICS 3 For any fluid at rest, the pressure is always normal to any element of surface. Density of Water. In hydraulic calculations the unit of weight may be taken as the weight of a cubic foot of water at its tempera- ture of greatest density, namely, 39 F. or 4 C. It is found by accurate measurement that a cubic foot of water at 39 F. weighs 62.42 Ib. This constant will be denoted in what follows by the Greek letter 7. In all numerical calculations it must be remem- bered therefore that 7 = 62.4 Ib. per cubic foot. (i) The density and volume of water at various temperatures are given in Table 1. Specific Weight. The weights of all substances, whether liquids or solids, may be expressed in terms of the weight of an equal volume of water. This ratio of the weight of a given volume of any substance to that of an equal volume of water is called the specific weight of the substance, and will be denoted in what follows by s. For instance, a cubic foot of mercury weighs 848.7 Ib., and its specific weight is therefore 848 7 Its exact value at C. is s = 13.596, as may be found in Table 1. The weight of 1 cu. ft. of any substance in terms of its specific weight is then given by the relation Weight = 75 = 62.45 Ib. per cubic foot. (2) 2. HYDROSTATIC PRESSURE Equal Transmission of Pressure. The fundamental principle of hydrostatics is that when a fluid at rest has pressure applied to any portion of its surface, this pressure is transmitted equally to all parts of the fluid. To prove this principle consider any portion of the fluid limited by a bounding surface of any form, and suppose that a small cylindrical portion is forced in at one point and out at another, the rest of the boundary remaining unchanged (Fig. 1). Then if AA denotes the cross-sectional area of one cylinder and An its height, its volume is AA-An. Similarly, the volume of the 4 ELEMENTS OF HYDRAULICS other cylinder is AA'-An', and, since the fluid is assumed to be incompressible, AA-An = AA'-An'. Now let p denote the unit pressure on the end of the first cylinder, i.e., the intensity of pressure, or its amount in pounds per square inch. Then the total pressure normal to the end is pAA, and the work done by this force in moving the distance An is pAA-An. Sim- ilarly, the work done on the other cylinder is p'AA'-An'. Also, if 7 denotes the heaviness of the fluid per unit volume, the work done by gravity in moving this weight 7AA-An through the distance h, F - where h denotes the difference in level between the two elements considered, is yAA-An-h. Therefore, equating the work done on the fluid to that done by it, we have n + 7AA-An-/i = p'-AA'-An'. Since AA-An = AA'-An', this reduces to p' = p + 7h. (3) If h 0, then p' = p. Therefore the pressure at any point in a perfect fluid is the same in every direction. Also the pressure at the same level is everywhere the same. Moreover, if the intensity of pressure p at any point is increased by an amount w, so that it becomes p + w, then by Eq. (3) the intensity of pressure at any other point at a difference of level h becomes P" = (p + w But since p' = p + yh, we have by subtraction, P tf = p' + w, that is, the intensity of pressure at any other point is increased by the same amount w. A pressure applied at any point is there- fore transmitted equally to all parts of the fluid. HYDROSTATICS For fluids such as gas and steam the term yh is negligible, and consequently for such fluids the intensity of pressure may be assumed to be everywhere the same. Pressure Proportional to Area. To illustrate the application of this principle consider a closed vessel or tank, filled with water, having two cylindrical openings at the same level, closed by movable pistons (Fig. 2). If a load P is applied to one piston, then in accordance with the result just proved there is an increase of pressure throughout the vessel of amount p = where A denotes the area of the piston. The force P' exerted on the second piston of area A' is therefore PA ' P' = PA' = ^f, whence p; p FIG. 2. FIG. 3. The two forces considered are therefore in the same ratio as their respective areas. This relation remains true whatever shape the ends of the pistons may have, the areas A and A' in any case being the cross-sectional areas of the openings. For instance, an enlargement of the end of the piston, such as shown in Fig. 3, has no effect on the force transmitted, since the upward and '' downward pressures on the ring of area ird 2 as the effective area. . cancel, leaving 6 ELEMENTS OF HYDRAULICS Hydraulic Press. An important practical application of the law of hydrostatic pressure is found in the hydraulic press. In its essential features this consists of two cylinders, one large and one small, each fitted with a piston or plunger, and connected by a pipe through which water can pass from one cylinder to the other (Fig. 4). Let p denote the intensity of pressure within the fluid, D, d the diameters of the two plungers, P the load applied FIG. 4. to one and W the load supported by the other, as indicated in the figure. Then and consequently W D , s If the small plunger moves inward a distance h y the large one will be forced out a distance H such that each will displace the same volume, or whence 4 Neglecting friction, the work done by the force P in moving the distance h is then HYDROSTATICS and is therefore equal to the work done in raising W the dis- tance H. Frictional Resistance of Packing. Usually, however, there is considerable frictional resistance to be overcome, since for high pressures, common in hydraulic presses, heavy packing is necessary to prevent leakage. One form of packing ex- tensively used is the U leather packing shown in Fig. 5. In this form of packing the water leaking past the plunger, or ram as it is often called, enters the leather cup pressing one side against the cylinder and the other against the ram, the pres- sure preventing leakage being proportional to the pressure of the water. To take into account the fric- tional resistance in this case, let ju denote the coefficient of fric- tion between leather and ram, C, c the depths of the packing on the large and small rams (Fig. 4), and p the intensity of water pressure. Then the area of leather in contact with the large ram is irDC and its frictional resistance is therefore irDCp^. Similarly, the frictional resistance for the small ram is irdcpfjL. Consequently and whence W TV/ 1 ~ 4M D\ (5) Efficiency of Hydraulic Press. The efficiency of any ap- paratus or machine for transforming energy is defined as . _ Useful work or effort Efficiency - Total ava ilable work or effort' 8 ELEMENTS OF HYDRAULICS and is therefore always less than unity. In the present case if there were no frictional resistance the relation between W and P would be given by Eq. (4). The efficiency for this type of press is therefore the ratio of the two equations (5) and (4), or Efficiency = -4M) (6) 3. SIMPLE PRESSURE MACHINES Hydraulic Intensifier.- in Art. 2 there are a High Pressure Outlet -Besides the hydraulic press described number of simple pressure machines based on the principle of equal distribution of pressure through- ".- out a liquid. Four types are here illustrated and described, as well as their combination in a hydraulic installation. When a hydraulic machine such as a punch or riveter is finishing the operation, it is re- quired to exert a much greater force than at the beginning of the stroke. To provide this in- crease in pressure, an intensifier is used (Fig. 6). This consists of several cylinders telescoped one inside another. Thus in Fig. 6, which shows a simple form of intensifier, the largest cylinder A is fitted with a ram B. This ram is hollowed out to form another cylinder C, fitted with a smaller ram D, which is fixed to the yoke at the top. In operation, water at the ordinary pump pressure enters the cylinder A through the intake, thereby forcing the ram B up- ward. This has the effect of forcing the ram D into the cylinder C, and the water in C is thereby forced out through D, which is hollow, at an increased pressure. Let p\ denote the pressure Low Pressure Intake FlG. 6. HYDROSTATICS 9 of the feed water, p 2 the intensified pressure in C and di, d 2 the diameters of the cylinders B and C respectively, as indicated in the figure. Then Pi = whence en I I I I The intensity of pressure in the cylinders is therefore inversely proportional to the areas of the n n rams. When a greater .intensification of pressure is required a compound intensifier is used, consisting of three or four cylinders and rams, nested in telescopic form, the general arrangement and principle of operation being the same as in the simple intensifier shown in Fig. 6. Hydraulic Accumulator. A hy- draulic accumulator is a pressure regulator or governor, and bears somewhat the same relation to a hydraulic system that the flywheel does to an engine; that is, it stores up the excess pump delivery when the pumps are delivering more than is being used, and delivers it again under pressure when the demand is greater than the supply. There are two principal types of hydraulic accumulator, in one of which the ram is fixed or stationary, and in the other the cylinder. The latter type is shown in Fig. 7. When the delivery of the pumps is greater or less than required by the machine, water enters or leaves the cylinder of the accumulator through the pipe A. The ram is thereby raised or lowered, and with it the weights suspended by the yoke from its upper end. The pressure in Weights tr A ^3 Water FlG. 7. 10 ELEMENTS OF HYDRAULICS the system is thereby maintained constant and free from the pulsations of the pump. The capacity of the accumulator is equal to the volume of the ram displacement, and should be equal to the delivery from the pump in five or six revolutions. The diameter of the ram should be large enough to prevent a high speed in descent, so as to avoid the inertia forces set up by sudden changes in speed. FIG. 8. FIG. 9. Hydraulic Jack. The hydraulic jack is a lifting apparatus operated by the pressure of a liquid under the action of a force pump. Thus in Fig. 8 the hand lever operates the pump piston B, which forces water from the reservoir A in the top of the ram through the valve at C into the pressure chamber D under the ram. The force exerted is thereby increased in the direct ratio of the areas of the two pistons. Thus if the diameter of the pump piston is 1 in. and the diameter of the lifting piston or ram is 4 in., HYDROSTATICS 11 the area of the ram will be sixteen times that of the pump piston. If then a load of, say, 3 tons is applied to the pump piston by means of the lever, the ram will exert an upward lifting force of 48 tons. Pump to Return Valve Discharge to Main System FIG. 10. Hydraulic Crane. The hydraulic crane, shown in Fig. 9, con- sists essentially of a ram and cylinder, each carrying a set of 12 ELEMENTS OF HYDRAULICS pulleys. A chain or rope is passed continuously over the two sets of pulleys as in the case of an ordinary block and tackle, the free end passing over guide pulleys to the load to be lifted. When water is pumped into the cylinder under pressure, the two pulley blocks are forced apart, thereby lifting the load at the free end of the chain or rope. Hydraulic Elevator. In hydraulic installations two or more of these simple pressure machines are often combined, as in the hy- draulic elevator shown in Fig. 10. In this case the accumulator serves to equalize the pump pressure, making the operation of the system smooth and uniform. The main valve for starting and stopping is operated, in the type shown, by the discharge pressure, maintained by means of an elevated discharge reservoir. A pilot valve, operated from the elevator cab, admits this low-pressure discharge water to opposite sides of the main valve piston as desired, thereby either admitting high-pressure water from the pump and accumulator, or opening the outlet valve into the discharge. The other details of the in- stallation are indicated on the diagram. 4. PRESSURE ON SUBMERGED SURFACES Change of Pressure with Depth. For a liquid at rest in an open vessel or tank, the free upper surface is perfectly level. Let the atmospheric pressure on this surface be denoted by p. Then, from Eq. (3), the pressure p' at a depth h below the surface is given by P f = P + yh. Since the atmospheric pressure is practically constant, the free surface of the liquid may be assumed as a surface of zero pressure when considering only the pressure due to the weight of the liquid. In this case p = 0, and the pressure p' at any depth h, due to the weight of the liquid, becomes P' = 7h. (7) Hence the pressure at any point in a liquid due to its own weight is directly proportional to the depth of this point below the free upper surface. Moreover, let A^l denote any element of a submerged surface. Then the pressure on it is p'AA HYDROSTATICS 13 Therefore the pressure on any element of area of a submerged surface is equal to the weight of a column of water of cross section equal to the element considered, and of height equal to the depth of this element below the surface. Pressure on Submerged Area. Consider the pressure on any finite area A in the side of a tank a reservoir containing a liquid at rest (Fig. 11). Let A A denote any element of this area and x its distance below the surface of the liquid. Then, by what precedes, the pressure on this elementary area is FIG. 11. and consequently the total pressure P on the entire area A is given by the summation But from the ordinary formula for finding the center of gravity of an area, the distance XQ of the center of gravity of A below the surface is given by Ax Q = Zx&A and consequently P = 7 Ax . (8) Therefore, the pressure of a liquid on any submerged plane surface is equal to the weight of a column of the liquid of cross section equal to the given area and of height equal to the depth of the center of gravity of this area below the free surface of the liquid. Center of Pressure. The point of application of the resultant pressure on any submerged area is called the center of pressure, and for any plane area which is not horizontal, lies deeper than the 14 ELEMENTS OF HYDRAULICS center of gravity of this area. For instance, consider the water pressure against a masonry dam with plane vertical face (Fig. 12). By Eq. (7) the pressure at any point A is proportional to the depth of A below the water surface. If, then, a length AB is laid off perpendicular to the wall and equal to the depth of A below the surface, that is, AB = AO, then AB will represent to a certain scale the normal pressure at A. If the same is done at various other points of the wall, their ends, B, D, etc., will lie in a straight line inclined at 45 to the horizontal. For a portion of FIG. the wall of length b, the pressure acting on it will then be equal to the weight of the water prism OEF, namely, P=7y- 2 ; (9) and the center of pressure will coincide with the center of gravity of this prism. It therefore lies at a distance of f h below the water surface, which is below the center of gravity of the rectan- gular area under pressure since the latter is at a distance of -^ from the surface. General Formula for Center of Pressure. To obtain a general expression for the location of the center of pressure, consider any plane area inclined at an angle a to the horizontal (or water sur- face) and subjected to a hydrostatic pressure on one side. Let 00' denote the line of intersection of the plane in which the given area lies with the water surface (Fig. 13). Also, let AA denote any element of the given area, h the depth of this element below the HYDROSTATICS 15 surface, and x its distance from 00', as indicated in the figure. Then from Eq. (7), the pressure AP acting on this element is AP = yhAA, and the moment of this force with respect to the line 00' is xAP = yhx AA. Now let P denote the total resultant pressure on the area A and x c the distance of the point of application of this resultant from 00', i.e., x c represents the x coordinate of the center of pressure. Then since the sum of the moments of all the elements of pressure FIG. 13. with respect to any axis 00' is equal to the moment of their re- sultant with respect to this axis, we have 2zAP = Px c . But AP = yhAA and P = ZyhAA. Consequently this becomes Zyhx&A = XcZyhAA. Also, since h = x sin a, this may be written 7 sin a2x 2 AA = 7 sin ax c '2xAA or, cancelling the common factor 7 sin a, The left member of this expression is by definition the moment of inertia, 7, of the area A with respect to the line 00' } that is 16 ELEMENTS OF HYDRAULICS while by the formula for the center of gravity of any area A we also have x A where x denotes the x coordinate of the center of gravity of A . The x coordinate of the center of pressure is therefore determined by the general formula I = Ax * Application. In applying this formula it is convenient to use the familiar relation I = I, + Ad 2 where 7 = moment of inertia of A with respect to the axis 00' ; I g = moment of inertia of A with respect to a gravity axis parallel to 00'; d = distance between these two parallel axes. For example, in the case of the vertical dam under a hydrostatic head h, considered above, we have for a rectangle of breadth 6 and height h, _ bh* lg ~'' 12' Consequently the moment of inertia / with respect to its upper edge is T ^' : and therefore the depth of the center of pressure below the sur- face is bh* I 3 2 which, of course, agrees with the result obtained geometrically. 5. EQUILIBRIUM OF TWO FLUIDS IN CONTACT Head Inversely Proportional to Specific Weight. If two open vessels containing the same fluid, say water, are connected by a tube, the fluid will stand at the same level in both vessels (Fig. 14). If the two vessels contain different fluids which are of HYDROSTATICS 17 different weights per unit of volume, that is to say, of different specific gravities, then since the fluid in the connecting tube must exert the same pressure in either direction, the surface of the lighter fluid will be higher than that of the heavier. For instance, let i and s 2 denote the specific gravities of the two fluids in the apparatus shown in Fig. 15, and let A denote the area of their surface of contact. Then for equilibrium ys 2 AH whence Ji _ 82 H~SI* (10) The ratio of the heights of the two fluids above their surface of separation is therefore inversely proportional to the ratio of their specific gravities. FIG. 14. FIG. 15. Water Barometer. If one of the fluids is air and the other water, we have what is called a water barometer. For example, suppose that a long tube closed at one end is filled with water and the open end corked. Then if it is placed cork downward in a vessel of water and the cork removed, the water in the tube will fall until it stands at a certain height h above the surface of the water in the open vessel, thus leaving a vacuum in the upper end of the tube. The absolute pressure in the top of the tube, A (Fig. 16), is therefore zero, and at the surface, B, is equal to the pressure of the atmosphere, or approximately 14.7 pounds per square inch. But from Eq. (3) we have P = P where in the present case 2 18 ELEMENTS OF HYDRAULICS PB = 14.7 Ib. per square inch; PA = 0; 7 = 62.4 Ib. per cubic foot and by substitution of these values we find that h = 34 ft. approximately. This is the height, therefore, at which a water column may be maintained by ordinary atmospheric pressure. It is therefore also the theoretical height to which water may be raised by means of an ordinary suction pump. As it is impossible in practice to secure a perfect vacuum, however, the actual working lift for a suction pump does not exceed 20 or 25 ft. Mercury Barometer. If mercury is used instead of water, since the specific gravity of mercury is s = 13.5956, we have 14.7 X 144 == 13.6 X 62.4 = m *> a PP roximatel y> FIG. 16. FIG. 17. which is accordingly the approximate length of an ordinary mercury barometer. Piezometer. When a vessel contains liquid under pressure, this pressure is conveniently measured by a simple device called a piezometer. In its simplest form this consists merely of a tube inserted in the side of the vessel, of sufficient height to prevent overflow and large enough in diameter to avoid capillary action, say over 1/4 in. inside diameter. The height of the free surface of the liquid in the tube above any point B in the vessel then measures the pressure at B (Fig. 17). Since the top of the tube is open to the atmosphere, the absolute pressure at B is that due to a head of h + 34 ft. HYDROSTATICS 19 Mercury Pressure Gage. In general it is convenient to use a mercury column instead of a water column, and change the form of the apparatus slightly. Thus Fig. 18 shows a simple form of mercury pressure gage, the difference in level, h, of the two ends FIG. 18. of the mercury column measuring the pressure at B. Let s denote the specific weight of mercury and s' the specific weight of the fluid in the vessel. The pressure at any point C in the vessel is then p c = 7 sh - 7s'h'. (u) FIG. 19. For example, if the fluid in the vessel is water, then 7 = 62.4, s' = 1, s = 13.596, and consequently p c = 62.4(13.596/i - h'). 20 ELEMENTS OF HYDRAULICS In case there is a partial vacuum in the vessel, the gage may be of the form shown in Fig. 19. The pressure on the free surface AB in the reservoir is then the same as at the top of the barometer column, C, namely, p c = 14.7 ysh. 6. EQUILIBRIUM OF FLOATING BODIES Buoyancy. When a solid body floats on the water partially submerged, as in the case of a piece of timber or the hull of a ship, each element of the wetted surface experiences a unit normal pressure of amount p = yh where h denotes the depth of the element in question below the water surface. Since the body is at rest, the total pressure acting = W-B FIG. 20. on the wetted surface together with the weight of the body, which in this case is the only other external force, must then form a sys- tem in equilibrium. Since the weight of the body acts vertically downward, the water must therefore exert an upward pressure of the same amount. This resultant upward pressure of the water is called the buoyant effort, or buoyancy, and the point of applica- tion of this upward force is called the center of buoyancy. For equilibrium, therefore, the buoyancy must be equal to the weight of the body and act vertically upward along the same line, since otherwise these two forces would form a couple tending to tip or rotate the body (Fig. 20). HYDROSTATICS 21 Floating Equilibrium. To calculate the buoyancy, suppose that the solid body is removed and the space it occupied below the water line refilled with water. Then since the lateral pressure of the water in every direction must be exactly the same as before, the buoyancy must be equal to the weight of this volume of water. The buoyancy is therefore equal to the weight of the volume of water displaced by the floating body, and the center of buoyancy coincides with the center of gravity of the displacement. For equilibrium, therefore, a solid body must sink until the weight of the water it displaces is equal to the weight of the body, and the centers of gravity of the body and its displacement must lie in the same vertical. These conditions also apply to the case when a body is entirely submerged. As the density of water increases with the depth, if a solid is slightly heavier than the water it displaces, it will sink until it reaches a depth at which the density is such that the weight of the water it displaces is exactly equal to its own weight. Theorem of Archimedes. If a solid is heavier than the weight of water it displaces, equilibrium may be maintained by suspend- ing the body in water by a cord (Fig. 20), in which case the ten- sion, T, in the cord is equal to the difference between the weight of the body and its buoyancy, that is, the weight of the water it displaces. A solid immersed in a liquid therefore loses in weight an amount equal to the weight of the liquid displaced. This is known as the Theorem of Archimedes, and was discovered by him about the year 250 B. C. Physical Definition of Specific Weight. Consider a solid com- pletely immersed in a liquid, and let V denote the volume of the solid, and 7 the weight of a cubic unit of the liquid, say 1 cu. ft. Then the buoyancy, B, of the body is B = yV. Also, if 71 denotes the weight of a cubic unit of the solid, regarded as uniform and homogeneous, its weight is W =7iF. The ratio = ^ = s () B 7 is called the specific weight of the solid with respect to the liquid in which it is immersed (compare Art. 1). In general, the liquid to which the specific weight refers is assumed to be water at a 22 ELEMENTS OF HYDRAULICS. temperature of 39 F. The specific weight of any substance is then that abstract number which expresses how many times heavier it is than an equal volume of water at 39 F. The specific weight of water is therefore unity; for lighter substances such as wood or oil it is less than unity; and for heavier substances like lead and mercury it is greater than unity. Determination of Specific Weight by Experiment. The spe- cific weight of a body may be determined by first weighing it in air and again when immersed in water. The actual weight of the body in air is then W = 71 V = ysV where s denotes its specific weight, and its apparent weight T when immersed (Fig. 20) is T = W - B = ysV - yV = yV(s - 1), that is, T = 7 V(s - i). (13) Therefore, by division, W s T ~ s -1 whence W S = The specific weight of a body is therefore equal to its weight in aif divided by its loss in weight when immersed in water. Application to Alloy. If a body is an alloy or mixture of two different substances whose specific weights are known, the volume of each substance may be determined by weighing the body in air and in water. Thus let V\ denote the volume, and s\ the spe- cific weight, of one substance, and F 2 , s 2 of the other. Then the weight of the body in air is W = and its apparent weight T when immersed is, from equation (13), T = 7 V 1 (s 1 ~ 1) + T F 2 (s 2 - 1). Solving these two equations simultaneously for V\ and F 2 , the result is T - (l - I) W sj -17 HYDROSTATICS 23 en Hfl This method of determining relative volumes was invented by Archimedes in order to solve a practical problem. Hiero, King of Syracuse, had furnished a quantity of gold to a goldsmith to be made into a crown. When the work was completed the crown was found to be of full weight, but it was suspected that the gold- smith had kept out a considerable amount of gold and substi- tuted an equal weight of silver. To test the truth of this sus- picion Archimedes first balanced the crown in air against an equal weight of gold, and then immersed both in water, when the gold was found to outweigh the crown, proving the goldsmith to be dis- honest. Zero Buoyancy. When a body lies flat against the bottom of a vessel filled with water, fitting the bottom so closely that no water can get under it, its buoyancy is zero. In this case if W denotes the weight of the body, A the area of its horizontal cross section, and h the depth of water on it, the force T required to lift it is (Fig. 21) T = W + yAh. That is to say, the force T is the same as would be necessary to lift the body itself and the entire column of water vertically over it. This same principle underlies the action of a leather sucker or vacuum tipped arrow, the fluid in that case being air. FIG. 21. 7. METACENTER Stability of Floating Body. When a floating body is shoved to one side it remains in this position and is therefore in neutral equilibrium as regards lateral translation. In determining the stability of a floating body it is therefore only necessary to con- sider its equilibrium as regards rotation. 24 ELEMENTS OF HYDRAULICS After a floating body has been tipped or rotated a small amount from its position of equilibrium, the buoyancy, in general, no longer passes through the center of gravity of the body. Conse- quently the weight and buoyancy together form a couple tending to produce rotation or tip the body. If this couple tends to' right the body the equilibrium is stable, whereas if it tends to tip it over it is unstable. This evidently depends on the form of the FIG. 22. wetted surface, and also on the form of the part immersed by the rotation. Metacenter. For example, consider a floating box of rectangu- lar cross section, immersed to a depth d below the surface A A (Fig. 22), and suppose it is tipped by an external couple until the water line becomes A' A'. In this new position the displace- ment is trapezoidal, and the cen- ter of buoyancy B is the center of gravity of this trapezoid. But since the buoyancy is of the same amount as before the box was tipped and the triangle of immer- sion mno is equal to the triangle of emersion opq, the lines AA and A' A' intersect on the vertical axis CC. The intersection M of the line of action of the buoyancy with the vertical axis CC is called the metacenter. Evidently the location of the metacenter depends on the angle of tip and is different for each position. It is also apparent that the equilibrium is stable if the metacenter M lies above the center of gravity G of the body, and unstable if M lies below (7. It is also shown in what follows that the metacenter moves higher as FIG. 23. HYDROSTATICS 25 the angle of tip, a, increases. Its lowest position is called the true metacenter. Coordinates of Metacenter. For the special case of the rec- tangular cross section shown in Fig. 22, let x, y denote the co- ordinates of the center of gravity of the trapezoid, and a, 6, c the lengths of three sides (Fig. 23). Then from geometry, _ c(2a + 6) a 2 + ab + & 2 = 3 (a + 6) ' y ~ 3 (a + 6) From Fig. 21 the sides a and b of the trapezoid expressed in terms of d and a are c c a = d o tan a; b = d + ~ tan a. Inserting these values of a and b in the expressions for x and ?/, the result is c (3d - I tan a) 3d 2 + ^ tan 2 a Also, from Fig. 22, the total height H of the metacenter above the bottom of the vessel or box is / c \ H = y + ( ^ ~ x I cot a. Hence by inserting the above values for x and y in this expression for H and reducing, we obtain the relation H = 2 The height H therefore increases with a; that is, the greater the angle of tip, the higher the metacenter M . Moreover, by substi- tuting a = in Eq. (15) the position of the true metacenter, or limiting position of M, is found to be at a height H f above the bottom given by To prevent a ship from capsizing, it is necessary to so design and load it that the height of its center of gravity above the bottom shall be less than H'. Metacentric Height. To consider the general ease of equi-- librium of a floating body, take a vertical cross section through 26 ELEMENTS OF HYDRAULICS the center of gravity G of the body (Fig. 24), and suppose that by the application of an external couple it is slightly tipped or rotated about an axis OY, drawn through perpendicular to the plane of the paper. Then the volume displaced remains un- changed, but the center of buoyancy B is moved to some other point B r . To find the metacentric height h y , or distance from the center of gravity G of the body to the metacenter M, let V = volume of liquid displaced, A = cross-sectional area of body in plane of flotation, 6 = distance from center of gravity G to center of buoyancy B, k v = radius of gyration of area of flotation A about the axis OY. Then it can be shown that 1 h y = -y^--b. (17) Similarly, for rotation about the axis OX, the metacentric height h x is given by where k x denotes the radius of gyration of the area of flotation A about the axis OX. Evidently the metacentric height is greater for a displacement about the shorter principal axis of the section A. For instance, it is easier to make a ship roll than to cause it to tip endwise or pitch. The locus of the centers of buoyancy for all possible displace- 1 Webster, Dynamics of Particles, p. 474. (Teubner). HYDROSTATICS 27 ments is called the surface of buoyancy, and the two metacenters given by Eqs. (17) and (18) are the centers of curvature of its principal sections. Period of Oscillation. When a floating body is tipped and then released, it will oscillate, or roll, with a simple harmonic motion. To find the period of the oscillation, the general ex- pression for the period of oscillation of a solid body rotating about a fixed axis may be applied, namely 1 *> where P = period or time of a complete oscillation, W = weight of the body, / = moment of inertia of the body with respect to the axis of rotation, h = distance from the center of gravity of the body to the axis of rotation. Since I = MK 2 , where M denotes the mass of the body and K its radius of gyration, and also W = Mg, Eq. (19) for the period may be written P = 2 ,JM^ = ^. (20) V Mgh Vgh Rolling and Pitching. In the present case, consider rotation about the two principal axes OX and OF of the section .A in the plane of floatation, and let K x , K y denote the radii of gyration of the solid with respect to these axes, and P x , P y the correspond- ing periods, or times of performing a complete oscillation about these axes. Then from Eq. (20), ftrg. ~ r Vgh, Substituting in these expressions the values of h x and h y given by Eqs. (17) and (18), they become r 2irK x .p 271-Kj, * ~ k , ,- b \ t ././A b\ For a body shaped like a ship, K and k increase together, and consequently the larger value of k corresponds to the smaller period P. A ship therefore pitches more rapidly than it rolls. 1 Slocum, Theory and Practice of Mechanics, p. 302. (Holt & Co.) 28 ELEMENTS OF HYDRAULICS For further applications of the metacenter the student is re- ferred to works on naval architecture. APPLICATIONS 1. The ram of a hydraulic press is 10 in. in diameter and the plunger is 2 in. in diameter. If the plunger is operated by a handle having a leverage of 8 to 1, find the pressure exerted by the ram, neglecting friction, when a force of 150 Ib. is applied to the handle. 2. In a hydraulic press the diameter of the ram is 15 m. and of the plunger is 3/4 in. The coefficient of friction may be assumed as 0.12 and the width of the packing on ram and plunger is 0.2 of their re- spective diameters. What pressure will be exerted by the ram when a force of 200 Ib. is applied to the plunger? , 3. Water in a pipe AB is to be kept at a constant pressure of 1200 Ib. per square inch by forcing in a plunger of diameter d (Fig. 25). This is operated by a piston of di- ameter Z), whose lower surface is sub- jected to the pressure of a column of water 75 ft. high. Find the ratio of the two diameters d and D. 4. In a hydraulic pivot bearing, a vertical shaft carrying a total load W is supported by hydraulic pressure (Fig. 26). The pivot is of diameter Z), and is surrounded by a U leather packing of width c. Show that the frictional moment, or resistance to rotation, is given by the relation J} FIG. 25. M = 2 where /* denotes the coefficient of friction. 5. For an ordinary flat pivot bearing of the same diameter D and for the same coefficient of friction /z as in the preceding prob- lem, the frictional moment is given by the relation 1 M = WD. o Slocum, Theory and Practice of Mechanics (Holt) p. 194. HYDROSTATICS 29 Show that the hydraulic pivot bearing is the more efficient of the two provided that D c<-. Calculate their relative efficiency when c = 0.2D. 6. An instrument for measuring the depth of the sea consists of a strong steel flask, divided into two compartments which are connected by a valve. The upper compartment is filled with 920 grams of distilled water and the lower compartment with mer- cury (Fig. 27). When lowered to the bottom, the outside pres- _I FIG. 26. FIG. 27. sure forces the sea water through a small opening in the side of the flask and thereby forces the mercury through the valve into the upper compartment. Assuming that the depth of the sea in certain parts of the Pacific ocean is 9429 meters, and that the ratio of the densities of distilled and salt water is 35:36, find how many grams of mercury enter the upper compartment. 1 The modulus of compressibility of water is 0.000047, that is, an in- crease in pressure of one atmosphere produces this decrease in volume. 7. A hydraulic jack has a 3-in. ram and a 3/4-in. plunger. If the leverage of the handle is 10 to 1, find what force must be applied to the handle to lift a weight of 5 tons, assuming the effi- ciency of the jack to be 75 per cent. 1 Wittenbauer, Aufgaben aus der Technischen Mechanik, Bd. III. 30 ELEMENTS OF HYDRAULICS 8. A hydraulic intensifier is required to raise the pressure from 600 Ib. per square inch to 2500 Ib. per square inch with a stroke of 3 ft. and a capacity of 4 gal. Find the required diameters of the rams. 9. In a hydraulic intensifier like that shown in Fig. 6, the diameters are 2 in., 5 in. and 8 in., respectively. If water is sup- plied to the large cylinder at a pressure of 500 Ib. per square inch, find the pressure at the high pressure outlet. 10. How would the results of the preceding problem be modi- fied if the frictional resistance of the glands, or packing, is taken into account, assuming that the frictional resistance of one stuff- ing box is 0.05 pd, where p denotes the water pressure in pounds per square inch, and d is the diameter of the ram in inches? 11. A hydraulic crane has % ram 10 in. in diameter and a velocity ratio of 1 : 12, that is, the speed of the lift is twelve times the speed of the ram. Assuming the efficiency of the crane to be 50 per cent., find what load it will lift with a water pressure of 1500 Ib. per square inch. 12. A hydraulic crane has a velocity ratio of 1:9 and is required to lift a load of 4 tons. Find the required size of the ram for a pressure in the mains of 750 Ib. per square inch, a loss of head due to friction of 75 Ib. per square inch, and a mechanical efficiency of 70 per cent. 13. How many foot pounds of work can be stored up in a hydraulic accumulator having a ram 10 in. in diameter and a lift of 12 ft., with a water pressure of 800 Ib. per square inch? 14. Find the energy stored in an accumulator which has a ram 10 in. in diameter, loaded to a pressure of 1000 Ib. per square inch, and having a stroke of 25 ft. If the full stroke is made in 1 minute find the horsepower available during this time. 15. The stroke of a hydraulic accumulator is fifteen times the diameter of the ram and the water pressure is 1200 Ib. per square inch. Find the diameter of the ram for a capacity of 125 horse- power minutes. 16. The ram of a hydraulic accumulator is 20 in. in diameter, the stroke 25 ft., and the water pressure 1050 pounds per square inch. If the work during one full downward stroke is utilized to operate a hydraulic crane which has an efficiency of 50 per cent, and a lift of 35 ft., find the load raised. 17. An accumulator is balanced by means of a chain of length I passing over two pulleys A and B (Fig. 28) and carrying a count- HYDROSTATICS 31 erweight W equal to the total weight of the chain. Find the distance apart of the pulleys and the required weight of chain per unit of length in order that this arrangement may balance the difference in pressure during motion. B FIG. 28. Hint. Let A denote the area of the ram and w the weight of the chain per unit of length. Then for the dimensions shown in the figure, we have the relations wx wz = yAc, x + y + z = I, x + h = b + c, yAc whence w = 2c + 2b - 2h - y 18. A hydraulic accumulator has a ram 15 in. in diameter and carries a load of 60 tons. Assuming the total frictional resistance to be 3 tons, find the required water pressure when the load is being raised and when it is being lowered. 19. Show that the depth of the center of pressure below the sur- face for a vertical rectangle of breadth b and depth d, with upper edge immersed to a depth hi and lower edge to a depth h 2 (Fig. 29) is given by the equation 2A 2 3 - x c =- 32 ELEMENTS OF HYDRAULICS 20. Show that the center of pressure for a vertical plane tri- angle with base horizontal and vertex at a distance hi below the surface (Fig. 30) is given by the equation X c u T i i i 1 i FIG. 29. 21. From the results of the preceding problem show that if the vertex of the triangle lies in the surface, the depth of the center of pressure is x, = - d, and if the base of the triangle lies in the surface FIG. 30. 22. Show that the depth of the center of pressure below the sur- face for a vertical circular area of radius r, immersed so that its center lies at a depth h below the surface is given by HYDROSTATICS 33 FIG. 31. 23. A circular opening, 2 ft. in diameter in the vertical side of a tank is closed by a circular cover held on by two bolts, one 14 in. above the center of the cover and the other 14 in. below its center. When water stands in the tank at a level of 20 ft. above the center of the opening, find the stress in each bolt. 24. A pipe of 4 ft. inside diameter flows just full, and is closed by a valve in the form of a flat circular plate balanced on a horizontal axis. At what distance from the center should the axis be' placed in order that the valve may bal- ance about it? 25. An automatic movable flood dam, or flashboard, is made of timber and pivoted to a back stay at a certain point C, as shown in Fig. 31. The point C is so located that the dam is stable provided the water does not rise above a certain point A, but when it rises above this point the dam automatic- ally tips over. Determine where the point C should be located. 26. An opening in a reservoir wall is closed by a plate 2-1/2 ft. square, hinged at the upper edge, and inclined at 60 to the hori- zontal. The plate weighs 250 lb., and is raised by a vertical chain attached to the middle point of its lower edge. If the center of the plate is 15 ft. below the surface, find the pull on the chain required to open it. 27. A rectangular cast-iron sluice gate in the bottom of a dam is 3 ft. high, 4 ft. wide and 3 in. thick. The head of water on the center of the gate is 35 ft. Assuming the coeffi- cient of friction of the gate on the slides to be 1/4, and that there is no water on the lower side of the gate, find the force required to lift it. Weight of cast iron is 450 lb. per cubic foot. 28. Flow from a reservoir into a pipe is shut off by a flap valve, as shown in Fig. 32. The pivot A is so placed that the weight of FIG. 32. 34 ELEMENTS OF HYDRAULICS the valve and arm balance about this point. Calculate the pull P in the chain required to open the valve for the dimensions given in the figure. 29. The waste gate of a power canal is 8 ft. high and 5 ft. wide, and when closed there is a head of 10 ft. of water on its center. If the gate weighs 1000 Ib. and the coefficient of friction between gate and seat is 0.4, find the force required to raise it. 30. A lock gate is 30 ft. wide and the depth of water on the two sides is 28 ft. and 14 ft. respectively. Find the total pressure on the gate and its point of application. 31. A lock is 20 ft. wide and is closed by two gates, each 10 ft. wide. If the depth of water on the two sides is 16 ft. and 4 ft. respectively, find the resultant pressure on each gate and its point of application. 32. A dry dock is 60 ft. wide at water level and 52 ft. wide at floor, which is 40 ft. below water level. The side walls have a straight batter. Find the total pressure on the gates and its point of application when the gates are closed and the dock empty. 33. A concrete dam is 6 ft. thick at the bottom, 2 ft. thick at top and 20 ft. high. The inside face is vertical and the outside face has a straight batter. How high may the water rise without causing the resultant pressure on the base to pass more than 6 in. outside the center of the base? Note. A dam may fail either by overturning or by sliding. In general, however, if a well-laid masonry dam is stable against overturning it will not fail by sliding on a horizontal joint. To prevent sliding on the base, an anchorage should be provided by cutting steps or trenches in the foundation if it is of rock, or in the case of clay or similar material by making the dam so massive that the angle which the resultant pressure on the base makes with the vertical is less than the angle of friction. In designing dams it is customary to proportion the section so that the resultant pressure on any horizontal joint shall fall within the middle third of the joint. If this condition is satisfied there will be no danger of tensile stresses developing in the face of the dam. 1 If water is allowed to seep under a dam, it will exert a lifting effort equal to the weight of a column of water of height equal to static head at this point. To secure stability it is therefore 1 Slocum and Hancock, Strength of Materials (Ginn), Revised Edition, p. 220. HYDROSTATICS 35 WASTE WEIR --i-x^r MAXIMUM SECTION 3?3?5 i 5f&'HES? * 'i* 6 "'P" Copper strip 1 ^Steel bor^Inspection-uiell Vyclopecm HORIZONTAL SECTION AT EXPANSION-JOINT >lssumjrace: of&re \ KENSICO DAM FIG. 33. Catskill Aqueduct System. 36 ELEMENTS OF HYDRAULICS essential to prevent seepage by means of a cut-off wall, as indi- cated in Figs. 33 and 34. In investigating the stability of a dam, however, the best prac- tice provides for accidental seepage by making allowance for an upward pressure on the base due to a hydrostatic head of two- thirds the actual depth of water back of the dam. 34. Figure 33 shows a typical section of the Kensico Dam, form- ing part of the Catskill Water System of the City of New York. The Kensico Reservoir covers 2218 acres, with a shore line 40 120 ft. above flow ' berm Drainage u/eff- OLIVE BRIDGE DAM MAXIMUM MASONRY SECTION FIG. 34. Catskill Aqueduct System. x miles in length, and has a storage capacity of 38,000,000,000 gal. The dimensions of the main dam are length 1843 ft.; height 300 ft. ; thickness at base 230 ft. ; thickness at top 28 ft. Investigate the stability of this dam in accordance with the conditions stated in the note to Problem 33. 35. Figure 34 shows a section of the Olive Bridge Dam and typi- cal dyke section of the Ashokan Reservoir, which forms part of the Catskill Water System of the City of New York. This reservoir covers' 8 180 acres, with a shore line 40 miles in length and a stor- age capacity of 132,000,000,000 gal. The principal dimensions HYDROSTATICS 37 of the main dam are, length 4650 ft. ; height 220 ft. ; thickness at base 190 ft. ; thickness at top 23 ft. Investigate the stability of this dam as in the preceding problem. 36. In the $25,000,000 hydraulic power development on the Mississippi river at Keokuk, Iowa, the dam proper is 4650 ft. long, with a spillway length of 4278 ft. The power plant is designed for an ultimate development of 300,000 h.p., and consists of vertical shaft turbines and generators in units of 10,000 h.p. each. Transmission lines convey the current at 110,000 volts to St. Louis, 137 miles distant, and to other points. 1 Operating Mechanism''' H.W.EI.S2S .eGage Lamp FIG. 35. Section of side walls of lock at Keokuk, Iowa, showing valves for intake and discharge conduits. A notable feature of the plant is the ship lock which is of unus- ual size for river navigation, the lock chamber being 400 ft. long by 110 ft. wide with a single lift of from 30 to 40 ft., the total water content of the lock when full being about 2,200,000 cu. ft. The locks at Panama are the same width but the maximum lift on the Isthmus is 32 ft., the average lift being about 28 ft. Find the maximum pressure on the lock gates at Keokuk and its point of application. (See frontispiece.) 37. The side walls of the Keokuk lock are monolithic masses of concrete, with a base width of 33 ft., a top width of 8 ft., and 1 Eng. News, Sept. 28, 1911. 38 ELEMENTS OF HYDRAULICS an outside batter of 1 :1.5, as shown in Fig. 35. If the water stands 48 ft. above the floor of the lock on the inside and 8 ft. on the outside, find the point where the resultant pressure on the side walls intersects the base, neglecting the weight of the road- way on top and the arches which support it. 38. The lower lock gates at Keokuk are of the mitering type, as shown in Figs. 36 and 37, and are very similar to those in the Panama canal locks. The gates are 49 ft. high and each leaf consists of 13 horizontal ribs curved to a radius of 66 ft. 4-3/4 in. on the center line, framed together at the ends by the quoin and FIG. 36. Steel miter gates for lower end of lock, showing buoyancy chamber. miter posts, and also having seven lines of intermediate framing. The chord length over the posts is 66 ft. 4-3/4 in. and the rise of the curve is 10 ft. 8-1/2 in. 1 Each leaf contains a buoyancy chamber to relieve the weight on the top hinge. This consists of a tank of about 3840 cu. ft. capacity, placed between the curve of the face and the chord line of the bracing. The total weight of each gate in air is about 240 tons. Find how much the buoyancy chamber relieves the weight on the top hinge. 39. The upper gates of the Keokuk lock are of a floating type never before used, and consist essentially of floating tanks moving in vertical guides and sinking below the level of the sill (Fig. 38). 1 Eng. News, Nov. 13, 1913. HYDROSTATICS 39 To close the lock, compressed air is admitted to an open-bottom chamber in the gate, which forces out the water and causes the gate to rise. To open the lock, the air in this chamber is al- Section of Gate erf Center (E-F) e "'- Hsw * Sectional Plan C'D FIG. 37. Detail of steel miter gates for lower end of lock. lowed to escape, when the weight of the gate sinks it to its lower position. The flotation of the gate is controlled by two closed displace- 40 IT tj ELEMENTS OF HYDRAULICS \ ^ HYDROSTATICS 41 ment chambers, one at each end, and one open buoyancy cham- ber. Each of the former is 42 ft. long, 4 ft. deep and 16 ft. wide. The buoyancy chamber is 2-1/2 ft. high beneath the displacement chambers and 6-1/2 ft. high in the 28-ft. space between them, its capacity being 6000 cu. ft. With the gate floating and its bottom just clear of the sill, the weight of the part above water is 190 tons, which is increased by the ballast in the displacement chambers to 210 tons. The dis- placement of the submerged part of the gate is 12 tons so that the buoyant effort required is 198 tons. Find the equivalent displacement in cubic feet, from this result subtract the volume of the displacement chambers, and then find the required air pressure in the buoyancy chamber. In raising the gate it is actually found that this pressure varies from 2 Ib. per square inch to as high as 12 Ib. per square inch when the gate is leaving its lower seat. 40. A gas tank is fitted with a mercury gage as shown in Fig. 18. The height h of the mercury column is 20 in. Find the excess of pressure in the tank above atmospheric. 41. A piece of lead weighs 20 Ib. in air. What will be its apparent weight when suspended in water, assuming the specific weight of lead to be 11.4? 42. A pail of water is placed on a platform scales and found to weigh 12 Ib. A 6-lb. iron weight is then suspended by a light cord from a spring balance and lowered into the water in the pail until completely immersed. Find the reading on the spring balance and on the platform scales., 43. A brass casting (alloy of copper and zinc) weighs 200 Ib. in air and 175 Ib. in water. If the specific weight of copper is 8.8 and of zinc is 7, how many pounds of each metal does the casting contain? 44. One end of a wooden pole 12 ft. long, floats on the water and the other end rests on a wall so that 2 ft. project inward beyond the point of support (Fig. 39). If the point of support is 18 in. above the water surface, find how much of the pole is immersed. 45. A floating platform is constructed of two square wooden beams each 16 ft. long, one 18 in. square and the other 1 ft. square. On these is laid a platform of 2-in. plank, 10 ft. wide. Find~ where a man weighing 160 Ib. must stand on the platform to make it float level, and how high its surface will then be above 42 ELEMENTS OF HYDRAULICS the water (Fig. 40). The weight of timber may be assumed as 50 Ib. per cubic foot. 46. A piece of timber 4 ft. long and 4 in. square has a weight W attached to its lower end so that it floats in water at an angle of 45 (Fig. 41). Find W. 47. A rectangular wooden barge is 30 ft. long, 12 ft. wide and 4 ft. deep, outside measurement, and is sheathed with plank FIG. 39. 3 in. thick, the frame weighing half as much as the planking. Find the position of the water line when the barge floats empty, and also the load in tons it carries when the water line is 1 ft. from the top. Assume the weight of wood as 50 Ib. per cubic foot. 48. A prismatic wooden beam 10 ft. long, 1 ft. wide and 6 in. thick floats flat on the water with 4 in. submerged and 2 in. above water. Find its specific weight. FIG. 40. 49. A dipper dredge weighs 1200 tons and floats on an even keel with bucket extended and empty. When the bucket carries a load of 3 tons at a distance of 50 ft. from the center line of the scow, a plumb line 15 ft. long, suspended from a vertical mast, swings out 5 in. Find the metacentric height. HYDROSTATICS 43 50. A steamer is of 14,000 tons displacement. When its life boats on one side are filled with water, a plumb line 20 ft. long suspended from a mast is found to swing out 9-1/2 in. If the FIG. 41. total weight of water in the boats is 75 tons and their distance from the center line of the vessel is 25 ft., find the period with which the ship will roll. SECTION II HYDROKINETICS 8. FLOW OF WATER FROM RESERVOIRS AND TANKS Stream Line. In the case of a flowing liquid, the path fol- lowed by any particle of the liquid in its course is called a stream line. In particular, if a reservoir or tank is filled with water and a small opening is made in one side at a depth h below the surface, the water flows out with a certain velocity depending on the depth, or head, h. Since the par- ticles of water flowing out con- verge at the opening, the stream lines inside the vessel are, in general, comparatively far apart, but become crowded more closely together at the orifice. Liquid Vein. Under the condi- tions just considered, suppose that a closed curve is drawn in any horizontal cross section of the vessel and through each point of the closed curve draw a stream line. The totality of all these stream lines will then form a tube, called a liquid vein (Fig. 42). From the definition of a stream line it is evident that the flow through such a tube or vein is the same as though it were an actual material tube. In particu- lar, the same amount of liquid will flow through each cross section of the vein and therefore the velocity of flow will be greatest where the cross section of the vein is least, and vice versa. Ideal Velocity Head. In any particular vein let v denote the velocity of flow at a distance h below the surface, and Q the quantity of water per second flowing through a cross section of the vein at this depth. Then the weight of water flowing through the cross section per second is yQ and its potential energy at the 44 1: FIG. 42. HYDROKINETICS 45 height h is yQh. The kinetic energy of this quantity of water jQv 2 flowing at the velocity v is ~ Therefore by equating the potential energy lost to the kinetic energy gained and neglecting all frictional and other losses we have whence V = V2gE (22) This relation may also be written in the form k= 2g The quantity h is therefore called the ideal velocity head, since it is the theoretical head required to produce a velocity of flow v. Torricelli's Theorem. The relation v = V2gh is known as Torricelli's Theorem. Expressed in words, it says that the ideal velocity of flow under a static head h is the same as would be acquired by a solid body falling in a vacuum from a height equal to the depth of the opening below the free surface of the liquid. Actual Velocity of Flow. The viscosity of the liquid, as well as the form and dimensions of the opening, have an important effect in modifying the discharge. Considering viscosity first, its effect is to reduce the velocity of the issuing liquid below the ideal velocity given by the relation v = ^2gh. It is therefore necessary to modify this relation so as to conform to experiment by introducing an empirical constant called a velocity coefficient. Denoting this coefficient by C v , the expression for the velocity becomes v = C v V2gh. (23) For water the value of the velocity (or viscosity) coefficient for an orifice or a nozzle is approximately C v = 0.97. Contraction Coefficient. In the case of flow through an orifice or over a weir, the oblique pressure of the water approaching from various directions causes a contraction of the jet or stream so that the cross section of the jet just outside the orifice is some- 46 ELEMENTS OF HYDRAULICS what less than the area of the opening. Consequently the dis- charge is also less than it would be if the jet were the full size of the opening. If the area of the orifice is denoted by A, the area of the jet at the contracted section will be some fraction of this amount, say C C A , where C c is an empirical constant called a contraction coeffi- cient, which must be determined experimentally for openings of various forms and dimensions. Efflux Coefficient. Taking into account both the viscosity of the liquid and the contraction of the jet, the formula for dis- charge becomes Q = actual velocity X area of jet = (C v V2 + >?*s o S ^ t' S CO CO CO CO CO i 1 ^2 ZQ ^^ 8 O5 rH tf O II II ii n II SH 'S r 0- 0> o> ^ 0>fS H t I NfM ^ ^ 1 : Si3 o o << II 1 2 ^ GO o -3 * ^ >H * ^ *5 m I-J-I r __^ II * ^ c3 . "i E 02 $} " II % CO ^" '"^ F** ti a *^ ^*" C*c^ /^ r 1 ^ ^ ^4 ^ s Qi> ^ O & PH 1 ^ . c^ & o O D O o3 0) rt_) -^ H fe S 2 ^0 S o ^ ^ J .s H ft . i 1" Q^ ^ S <5 u 3 ^ .'S -H Experimenter rH 'S cq 23 ^ ^ CC *+-i z g CO IN ^ O o o -^ _ p r* O CO rH OO n GO ^3 O S" ^ '"I *S ^ c5 Hamilton Smith, , 1886 4 Resume of work of > ous experimenters 1 Lowell Hydraulic 2 Flow of Water o 3 Annales des Pon 4 Hamilton Smith; HYDROKINETICS 53 For a sharp-edged opening the mean value of the efflux coefficient is K = 0.62, as stated in Art. 9. In the present case, therefore, KA = 0.626A, and if b and h are expressed in feet, the above formula becomes Q = {o.62bh)\/2gih, = 3.3bh 3 ^ cu. ft. per sec. (29) It is often convenient to express b and h in inches, and the dis- charge Q in cubic feet per minute. Expressed in these units, the formula becomes or, reducing and simplifying, Q = o.4bh^ cu. ft. per min. (30) where b and h are both expressed in inches. These formulas are the basis of many of the weir tables used in practical work, such as Tables 17 and 18 in this book. 11. STANDARD WEIR MEASUREMENTS Construction of Weir. From the experiments summarized in the preceding article it was found that any empirical weir formula could only be relied upon to give accurate results when the con- ditions under which the measurement was made were approxi- mately the same as those under which the formula was deduced. To obtain accurate results from weir measurements it is therefore customary to construct the weir according to certain standard specifications, as follows: 1. A rectangular notch weir is constructed with its edges~ sharply beveled toward the intake, as shown in Fig. 49. The bottom of the notch, called the crest of the weir, must be per- fectly level and the sides vertical. 2. The length, or width, of the weir should be between four and eight times the depth of water flowing over the crest of the weir. 3. The channel or pond back of the weir should be at least 50 per cent, wider than the notch, and of sufficient depth so that the velocity of approach shall not be over 1 ft. per second. In general it is sufficient if the area bh is not over one-sixth the area 54 ELEMENTS OF HYDRAULICS of the channel section where 6 denotes the width of the notch and h ,the head of water on the crest. 4. To make -the end contractions complete there must be a clearance of from 2h to 3h between each side of the notch and the corresponding side of the channel. 5. The head h must be accurately measured. This is usually accomplished by means of an instrument called a hook gage (Fig. 50), located as explained below. For rough work, however, FIG. 49. the head may be measured by a graduated rod or scale, set back of the weir at a distance not less than the length of the notch, with its zero on a level with the crest of the weir (Fig. 49). Hook Gage. As usually constructed, the hook gage consists of a wooden or metal frame carrying in a groove a metallic sliding scale graduated to feet and hundredths, which is raised and low- ered by means of a milled head nut at the top (Fig. 50). By means of a vernier attached to the frame, the scale may be read to thou- sandths of a foot. The lower end of the frame carries a sharp- pointed brass hook, from which the instrument gets its name. In use, the hook gage is set up in the channel abave the weir and leveled by means of a leveling instrument so that the scale HYDROKINETICS 55 reads zero when the point of the hook is at the exact level of the crest of the weir. The hook is then raised until its point just reaches the surface, causing a distortion in the reflection of light from the surface of the water. If slightly lowered the distortion disappears, thus indicating the surface level with precision. The reading of the vernier on the scale then gives the head on the crest to thousandths of a foot. Location of Hook Gage. To avoid surface oscillations, and thereby obtain more precise readings, the hook gage should be set up in a still box com- municating with the channel. The channel end of the opening or pipe leading into the still box must be flush with a flat surface set parallel to the direction of flow, and the pipe itself must be normal to this direction. The channel end of the pipe must be set far enough above the weir to avoid the slope of the surface curve, but not so far as to increase the head by the natural slope of the stream. If the formula of any particular ex- perimenter is to be used, his location for the still box should be duplicated. Proportioning Weirs. To illustrate the method of proportioning a weir, suppose that the stream to be measured is 5-1/2 ft. wide and 1-1/2 ft. deep, and that its average velocity, deter- mined by timing a float over a meas- ured distance or by using a current meter or a Pitot tube (Arts. 26 and 27), is approximately 4 ft. per second. The flow is then approximately 1980 cu. ft. per minute. To determine the size of weir which will flow approximately this amount, try first a depth of say 10 in. From Table 17 it is found that each inch of length for this depth will deliver 12.64 cu. ft. per minute. The required length of weir would FIG. 50. Hook gage. 56 ELEMENTS OF HYDRAULICS then be T^QA = 156.6, which is fifteen and two-thirds times the depth and therefore too long by Rule 2 of the specifications. Since the weir must evidently be deeper, try 18 in. From the table the discharge per linear inch for this depth is 30.54 cu. ft. per minute, and consequently the required length would be 1980 3Q 54 = 64.8 in., which is now only 3.6 times the depth and therefore too short. By further trial it is found that a depth of 15 in. gives a 1980 length of 23~23 = ^ 5 ' 2 * n *' wm ^ h * s 5 -^ times the depth and therefore comes within the limits required by Rule 2. Suppose then that the notch is made 7 ft. long and say 20 in. deep, so that the depth may be increased over the calcu- lated amount if necessary. If then the width of the pond back of the weir is not 50 per cent, greater than the width of the notch, or if the velocity of flow should be in excess of 1 ft. per second, the pond should, if possible, be enlarged or deepened to give the desired result. With the weir so constructed suppose that the depth of water over the stake back of the weir is found to be 15-1/8 in. From the table the discharge per linear inch cor- responding to this head is found to be 23.52 cu. ft. per minute, and this multiplied by 84, the length of the weir in inches, gives 1975.7 cu. ft. per minute for the actual measured discharge. 12. TIME REQUIRED FOR FILLING AND EMPTYING TANKS Change in Level under Constant Head. To find the time required to raise or lower the water level in a tank, reservoir, or lock, let A denote the area of the orifice through which the flow takes place and K its coefficient of discharge or efflux. Several simple cases will be considered. The simplest case is that in which the water level in a tank is raised, say from AB to CD (Fig. 51), by water flowing in under a constant head h. Let V denote the total volume of water flow- ing in, represented in cross section by the area ABCD in the figure. Then since the discharge Q through the orifice per second is Q = KA V20&, the time t in seconds required to raise the surface to the level CD is HYDROKINETICS 57 Varying Head. It is often necessary to find the time required to empty a tank or reservoir, or raise or lower its level a certain amount. A common case is that in which the level is to be raised or lowered from AB to CD (Fig. 52) by flow through a sub- merged orifice, the head on one side, EF, of the orifice being constant. If the cross section of the tank is variable, let Y denote its area at any section mn. In the time dt the level changes from the height y to y dy, and consequently the volume changes by the amount dV = Ydy. But by considering the flow through the orifice, of area A t the volume of flow in the time dt is FIG. 51. dV Hence, by equating these values of dV, we obtain the relation dt = Ydy whence t = KA (32) Canal Lock. A practical application of Eq. (32) is in find- ing the time required to fill or empty a canal lock. For an 58 ELEMENTS OF HYDRAULICS ordinary rectangular lock of breadth 6 and length I, the cross sec- tion is constant, namely Y = bl, and consequently the expression for the time integrates into t = bl KAV2g! Vy KAVg - ( VH - Vh). (33) Rise and Fall in Connected Tanks. When one tank discharges into another without any additional supply from outside, the Level Raised T Level Lowered FIG. 52. level in one tank falls as that in the other rises. If both tanks are of constant cross section, then when the level in one tank has been lowered a distance y, that in the other tank will have been raised a distance y' (Fig. 53), such that if M and N denote their sec- tional areas, respectively, My = Ny'. HYDROKINETICS 59 In the interval of time dt suppose y changes to y + dy. Then con- sidering the flow through the orifice of area A, as in the preceding case, we have Mdy = KA^2g[H - (y + y f )]dt, or, since y' = -TT-, this may be written MdyilN dt = - y(M Simplifying this expression and integrating, the resulting expres- sion for the time t is found to be r 1. FIG. 53. KA r Jo dy + N) KA-iYg - y(M + N) ] -I o Substituting the given limits, the time t required to lower the level a distance D is VNH - VNH - D(M + N) I (34) When the level becomes the same in both tanks, since the volume discharged by one is received by the other, we have or MD = N(H - D), NH D M + N 60 ELEMENTS OF HYDRAULICS Substituting this value of D in Eq. (34), it becomes 2MNVH KAV2g(M+ N)' t = (35) which is therefore the length of time required for the water in the tanks to reach a common level. Marietta's Flask. It is sometimes desirable in measuring flow to keep the head constant. It is difficult to accomplish this by keeping the supply constant, a more convenient method being by the arrangement shown in Fig. 54, which is known as Mariotte's Flask. This consists of putting an air-tight cover on the tank, having a corked orifice holding a vertical pipe open to the atmos- FIG. 54. phere. Since the pressure at the lower end A of the tube is always atmospheric, the flow is the same as though the water level was constantly maintained at this height. Therefore as long as the water level does not sink below the bottom of the pipe, the effective head on the orifice is its distance h below the bottom of the pipe, and the discharge is given by the formula Q = KAV 2 gh. (36) 13. FLOW THROUGH SHORT TUBES AND NOZZLES Standard Mouthpiece. When a short tube (adjutage, mouth- piece or nozzle) is added to an orifice, the flow through the open- ing is changed both in velocity and in amount. In general the velocity is diminished by the mouthpiece, due to increased fric- HYDROKINETICS 61 tional resistance, whereas the quantity discharged may be either increased or diminished, depending on the form of the mouth- piece. What is called the standard mouthpiece consists of a circular tube projecting outward from a circular orifice, and of length equal to two or three diameters of the orifice (Fig. 56). At the inner end of the tube the jet is contracted as in the case of a standard orifice, but farther out it expands and fills the tube. The velocity of the jet is reduced by this form of mouthpiece to v = 0.82 V2J/S, which is considerably less than for a standard orifice, but since there is no contraction, the quantity discharged is Q = 0.824 V2p, where A denotes the area of the orifice. The discharge is there- fore nearly one-third larger than for a standard orifice of the same area with complete contraction (Fig. 55). , Stream -line Mouthpiece. By rounding the inner edge of the mouthpiece so that its contour approximates the form of a stream line, the velocity of the jet is greatly increased, its value for the relative dimensions shown in Fig. 57 being about v = 0.96 and since the jet suffers no contraction, the quantity discharged is Q = 0.964 the area A, as before, referring to the area of the orifice. Borda Mouthpiece. A mouthpiece projecting inward and having a length of only half a diameter is called a Borda mouth- piece (Fig. 58) . The velocity is greatly increased by this form of mouthpiece, its value being about v = 0.99 V2p, but the contraction of the jet is more than for a standard orifice, so that the discharge is only Q = 0.534 where A denotes the area of the orifice. 62 ELEMENTS OF HYDRAULICS FIG. 57. FIG. 58. Conical Diverging Tube Re-entrant Tube : " :::f -- ^ F!G. 59. FIG. 60. Venturi Adjutage Angle a = 5 to 8 Q=1.5 A\/2gh Area A Measured on A Section AB _ ^ii 1 1 1 - Conical Converging Tube o Angle a=5to 10 FIG. 61. FIG. 62. HYDROKINETICS 63 Fire Hose; Smooth Cone Nozzle Q=.WA~V 2gh FIG. 63. Fire Hose; Smooth Convex Nozzle Q=.Q7A\/ FIG. 64. Fire Hose; Square Ring Nozzle A\rZrtT FIG. 65. Fire Hose; Undercut Ring Nozzle Q=.1lAV2 FIG. 66. If, however, the length of the mouthpiece is increased to two or three diameters (Fig. 59) the discharge is increased nearly 50 per cent., becoming Q = 0.724 ^J2gh. Diverging Conical Mouthpiece. For a conical diverging tube with sharp edge at entrance (Fig. 60) the jet contracts at the inner end as for an orifice, but farther on expands so as to fill the tube at outlet provided the angle of divergence is not over 8. The discharge is therefore greater than for a standard mouth- piece, its amount referred to the area A at the smallest section being Q = 0.95 A V20S. Venturi Adjutage. If the entrance to a diverging conical mouthpiece has a stream-line contour, it is called a Venturi adjutage (Fig. 61). In experiments by Venturi and Eytelwein with diverging mouthpieces of the relative dimensions shown in Fig. 61, a discharge was obtained nearly two and one-half times as great as for a standard orifice of the same diameter as that at the smallest section, or about twice that for a standard short tube 64 ELEMENTS OF HYDRAULICS of this diameter, the formula for discharge referred to the area A at the smallest section being Q = 1.55AV20S. Converging Conical Mouthpiece. In the case of a conical converging tube with sharp corners at entrance (Fig. 62) the jet contracts on entering and then expands again until it. fills the tube, the most contracted section being just beyond the tip, and the greatest discharge occurring for an angle of convergence of approxi- mately 13. Fire Nozzles. The fire nozzles shown in Figs. 63, 64, 65 and 66 are practical examples of converging mouthpieces. The smooth cone nozzle with gradually tapering bore has been found to be the most efficient, the coefficient of discharge for the best specimen being 0.977 with an average coefficient for this type of 0.97. For a square ring nozzle like that shown in Fig. 65 the coefficient of discharge is 0.74; and for the undercut type shown in Fig. 66 the coefficient of discharge is 0.71. 14. KINETIC PRESSURE IN A FLOWING LIQUID Kinetic Pressure. For a liquid at rest, the normal pressure exerted by it on any bounding surface is called the hydrostatic pressure and is given by the expression deduced in Art. 2, namely, p = p' + yh. If a liquid is in motion, however, the normal pressure it exerts on the walls of the vessel containing it, or on the bounding surface of a liquid vein or filament, follows an entirely different law, as shown below. To distinguish the hydrostatic pressure from the normal pres- sure exerted on any bounding surface by a liquid in motion, the latter will be called the kinetic pressure. Bernoulli's Theorem. To determine the kinetic pressure at any point in a flowing liquid, consider a small tube or vein of the liquid bounded by stream lines, as explained in Art. 8, and follow the motion of the liquid through this tube for a brief interval of time. Let A and A' denote the areas of two normal cross sections of the vein (Fig. 67). Then since the liquid is assumed to be in- compressible, the volume Ad displaced at one end of the tube HYDROKINETICS 65 must equal the volume A'd! displaced at the other end. If p denotes the average unit pressure on A, and p' on A', the work done by the pressure on the upper cap, A, is + pAd, and that on^the lower cap, A', is - p'A'd'. Also, if h denotes their difference in static head, as indicated in Fig. 67, the work done by gravity in the displacement of the vol- ume Ad a distance h is IP "T vAdh. Since the forces acting on the lateral surface of the vein are normal to this surface they do no work. Assuming, then, the case of steady flow, that is to say, assuming that each particle arriving at a given cross section experiences the same velocity and pressure as that experienced by the preceding particle at this point, so that the velocities v and v' through the caps A and A' are constant, the change in kinetic energy between these two positions is FIG. 67. Therefore, equating the total work done to the change in energy, the result is pAd - p'A'd' + 7 Adh = or, since Ad = A f d f , this reduces to v"* - z; 2 ), P' yy^ 2 2g TV (37) This result is known as Bernoulli's Theorem, and shows that in the case of steady parallel flow of an ideal liquid, an increase in velocity at any point is accompanied by a corresponding decrease 5 66 ELEMENTS OF HYDRAULICS in kinetic pressure, or vice versa, in accordance with the relation just obtained. Kinetic Pressure Head. If the theoretic heads corresponding to the velocities v and v' are denoted by H and H', respectively, then in accordance with Torricelli's theorem (Art. 8) we have v 2 v' 2 ^ TT TTf _ . 2 VB = b' Substituting these values of V A and V B in the preceding equation and solving for Q, the result is n ab Q = 68 ELEMENTS OF HYDRAULICS If h A and h B denote the static piezometer heads corresponding to the kinetic pressures p A and PB, respectively, this formula may be written ab (40) FIG. 70. Venturi meter and recording gage, manufactured by the Builders Iron Foundry. Ordinarily the throat diameter in this type of meter is made one-third the diameter of the main pipe, in which case a = 96. HYDROKINETICS 69 70 ELEMENTS OF HYDRAULICS If then, h denotes the difference in piezometer head between the up-stream end and the throat, the formula for discharge, ignoring frictional losses, becomes Q = 1.0062 b\/2gh. (41) By experiment it has been found that ordinarily for all sizes of Venturi meters and actual velocities through them, the actual discharge through the meter is given by the empirical formula Q = (0.97 o.03)b\4gh. (42) Commercial Meter. A typical arrangement of meter tube and recording apparatus is shown in Fig. 70, the lower dial indicating the rate of flow, and the upper dial making a continuous auto- graphic record of this rate on a circular chart. Catskill Aqueduct Meter. The Venturi meter affords the most accurate method yet devised for measuring the flow in pipe lines. Fig. 71 shows one of the three large Venturi meters built on FIG. 72. Venturi rate of flow controller manufactured by the Simplex Valve and Meter Co. the line of the Catskill Aqueduct, which is part of the water supply system of the City of New York. Each of these meters is 410 ft. long and is built entirely of reinforced concrete except for the throat castings and piezometer ring, which are of cast bronze. Provision is also made in connection with the City aqueduct for the installation of a Venturi meter upon each connection between the aqueduct and the street distribution pipes. Rate of Flow Controller. Fig. 72 illustrates a rate of flow con- troller operated by the difference in pressure in a Venturi tube. HYDROKINETICS 71 This apparatus is designed for use in a water pipe or conduit through which a constant discharge must be maintained regard- less of the head on the valve. It consists of a perfectly balanced valve operated by a diaphragm which is actuated by the differ- ence in pressure between the full and contracted sections of a Venturi tube. The valve and diaphragm are balanced by an adjustable counter- weight, which when set for any required rate of flow will hold the valve discs in the proper position for that flow. 16. FLOW OF WATER IN PIPES Critical Velocity. Innumerable experiments and investiga- tions have been made to determine the laws governing the flow of water in pipes, but so far with only partial success, as no general and universal law has yet been discovered. Experiments made by Professor Osborne Reynolds have shown that for a pipe of a given diameter there is a certain critical velocity, such that if the velocity of flow is less than this critical value, the flow proceeds in parallel filaments with true stream- line motion; whereas if this critical value is exceeded, the flow becomes turbulent, that is, broken by whirls and eddies and similar disturbances. The results of Professor Reynold's experi- ments showed that at a temperature of 60 F. this critical velocity occurred when Dv a = 0.02 where D denotes the diameter of the pipe in feet and v a is the average velocity of flow in feet per second. For parallel, or non-sinuous, flow it is possible to give a theoret- ical explanation of what occurs and deduce the mathematical law governing it, as shown below. No one, however, has yet ex- plained why the flow suddenly becomes turbulent at the critical velocity, or what law governs it subsequently. Viscosity Coefficient. The loss of energy accompanying pipe flow is due to the internal resistance arising from the viscosity of the liquid. This shear or drag between adjacent filaments is analogous to ordinary friction but follows entirely different laws. Unlike friction between the surfaces of solids, fluid friction has been found by experiment to be dependent on the temperature and the nature of the liquid; independent of the pressure; and, for ordinary velocities at least, approximately proportional to the 72 ELEMENTS OF HYDRAULICS difference in velocity between adjacent filaments. When this difference in velocity disappears, the frictional resistance also disappears. The constant of proportionality required to give a definite numerical value to fluid friction is called the viscosity coefficient and will be denoted by ju. This coefficient M is an empirical con- stant determined by experiment, the values tabulated below being the result of experiments made by O. E. Meyer. Temperature in degrees Fahr. 50 60 65 70' Viscosity coefficient // in Ib. sec. 32 X 10~ 6 28 X 10- 6 26 X 10~ 6 24 X 10~ 6 ft. 2 FlG 73 The dimensions of /* are, of course, such as to make the equa- tion in which it appears homogeneous in the^units involved, as will appear in what follows. Parallel (non-sinuous) Flow. 1 Consider non-sinuous flow in a straight pipe of uniform circular cross section, that is, at a veloc- ity less than the critical ve- locity and therefore such that the filaments or stream lines are all parallel to the axis of the pipe. By reason of symmetry the velocity of any particle depends only on its distance from the axis of the pipe. Let v denote the velocity of any particle and x its distance from the center (Fig. 73) . Then if x changes by an amount dx, the velocity changes by a corresponding amount dv, and since the velocity is least near the pipe walls, v decreases as x increases and consequently the ratio -r- is negative. Since the pipe is assumed to be of constant cross section and the flow uniform and parallel, the forces acting on any element of volume must be in equilibrium. Considering therefore a small water cylinder of radius x and length dy r in order to equilibrate the frictional resistance acting on the convex surface of this cylinder there must be a difference in pressure on its ends. 1 The following derivation is substantially that given by Foppl in his Dynamik. HYDROKINETICS 73 This explains the fall in pressure along a pipe, well known by experiment. Let dp denote the difference in pressure in a length dy. Then the difference in pressure on the ends of a cylinder of radius x is (7rx 2 )dp and the shear on its convex surface is (27rxdy) JJL-T- Equat- ing these two forces and remembering that -y- is negative, we obtain the relation Also, since the difference in pressure on the ends of any cylinder is proportional to its length, we have dp _ pi -p 2 _ dy~ I where pi and p 2 denote the unit pressures at two sections at a distance I apart, and the constant ratio is denoted by k for convenience. Substituting dp = kdy from the second equation in the first and cancelling common factors, we have finally dv k dx = ~ 2ju x ' whence, by integration, kx 2 ' 4^ + c ' where c denotes a constant of integration. To determine c assume that the frictional resistance between the pipe wall and the liquid follows the same law as that be- tween adjacent filaments of the liquid. Then it follows that the liquid in contact with the pipe must have zero velocity, as otherwise it would experience an infinite resistance. This seems also to be confirmed by the experiments of Professor Hele-Shaw, who showed that in the case of turbulent flow there was always a thin film of liquid adjoining the pipe walls which showed true stream-line motion, proving that its velocity was certainly less than the critical velocity and therefore small. Furthermore, the walls of commercial pipes are comparatively rough and conse- quently a thin skin or layer of liquid must be caught in these roughnesses and held practically stationary. Assuming then that v = when x = r, and substituting this 74 ELEMENTS OF HYDRAULICS pair of simultaneous values in the above equation, the value of c is found to be fcr 2 and consequently (r - x"). (43) This is the equation of a parabola, and therefore the velocity dia- gram is a parabolic arc with its vertex in the axis of the pipe; that is, the velocity is a maximum at the center where x = 0, its value being fcr 2 V V max = Average Velocity of Flow in Small Pipes. Let the discharge through any cross section of the pipe be denoted by Q. Then if the velocity at any radius x is denoted by v, we have -f Q = I 2irxdxv, )o k or, since v = j- (r 2 z 2 ), this becomes But if v a denotes the average velocity of flow we also have Q = tfa&rr 2 ) whence by substituting the above value for Q, we have Q Comparing this expression with that previously obtained for the maximum velocity, it is evident that the maximum velocity is twice the average velocity of flow. Loss of Head in Small Pipes. The loss in pressure in a length I is given by the relation obtained above, namely, HYDROKINETICS 75 or, if the difference in head corresponding to this difference in pressure is denoted by h, then, since p = yh, we have i u j z, PI - P* kl loss in head, h = - - = Substituting in this relation the value of k in terms of the average velocity of flow, the result is 8Mlv a _ 2Mly_ . ~ kr 2 T " kd 2 T For small pipes, therefore, the loss of head is proportional to the first power of the average velocity, and inversely proportional to the square of the diameter of the pipe. This result has been verified experimentally for small pipes by the experiments carried out by Poiseuille. Ordinary Pipe Flow. Under the conditions usually found in practice the velocity of flow exceeds the critical velocity and con- sequently the flow is turbulent and a greater amount of energy is dissipated in overcoming internal resistance than in the case of parallel flow. The result of Professor Reynold's experiments indicated that the loss of head in turbulent flow was given by the relation In commercial pipes the degree of roughness is a variable and un- certain quantity, so that the exact loss of head cannot be pre- dicted with accuracy. Practical experiments have shown, how- ever, that ordinarily the loss in head is proportional to the square of the average velocity, so that the relation becomes Since the theoretical head corresponding to a velocity v is v 2 A = ?r> the expression for the loss in head for a circular pipe *Q running full may in general be written or, denoting the constant of proportionality by/, this becomes 76 ELEMENTS OF HYDRAULICS Here/ is an empirical constant, depending on the condition of the inner surface of the pipe, and is determined by experiment. Eq. (45) is identical with Chezy's well-known formula as will be shown in Art. 19. 17. PRACTICAL FORMULAS FOR LOSS OF HEAD IN PIPE FLOW Effective and Lost Head. In the case of steady flow through long pipes, much of the available pressure head disappears in frictional and other losses, so that the velocity is greatly dimin- ished. Thus if h denotes the static head at the outlet and hi the head lost in overcoming frictional and other resistances to flow, the velocity v at the outlet is given by the relation or its equivalent, h = 5 + h " (46) The lost head hi is the sum of a number of terms, which will be considered separately. Loss at Entrance. A certain amount of head is lost at the entrance to the pipe, as in the case of a standard adjutage. If v denotes the velocity due to the head h with no losses, then v* . ^ whereas if V A denotes the actual velocity of flow the head corre- sponding to this velocity is v - v ^- 20 The head, hi, lost at entrance, is therefore If C v denotes the velocity coefficient for the entrance, then V A = C v v, HYDROKINETICS 77 and consequently the expression for the head lost at entrance may be written = _ ~ 2g 2g 2g 2g 2g \(V For the standard short tube C v = 0.82 (Art. 13) and therefore 1 1 C^ 2 ~ ^ = (0 82) 2 ~~ ^ = ' ^e nea ^ l s ^ a ^ enhance is therefore v 2 h i = 0.5 (47) If the pipe projects into the reservoir, C v 0.72 (Art. 13), and the head lost at entrance is thereby increased to hi = 0.93 |~- For ordinary service taps on water mains it may be assumed as hi = 0.62 ~ Friction Loss. In flow through long pipes the greatest loss in head is that due to the friction between the liquid and the walls of the pipe. Let d denote the internal diameter of the pipe and I its length. Then it has been found by experiment (Art. 16) that the head lost in internal friction, or friction head as it is called, is given by the formula h '=' f a'$' ( * 8) the quantity/ being an empirical constant determined by experi- ment. The average values of this coefficient for cast iron pipes are For new smooth pipes, f = 0.024 ; . . For old rusty pipes, f = 0.03. For more exact values of this coefficient, refer to Table 12. Bends and Elbows. Bends and elbows in a pipe also greatly diminish the effective head. From experiments by Weisbach 78 ELEMENTS OF HYDRAULICS it has been found that the lost head due to a sharp elbow of angle a (Fig. 74) is given by the formula (So) where m is a function of the angle a, given by the equation m = 0.9457 sin 2 - + 2.047 sin 4 Values of m, calculated from this formula for various values of the angle a, are tabulated as follows: a = 20 \ 30 \ 40 50 60 70 80 90 m = .046 .073 .139 .234 .364 .533 .740 .984 FIG. 74. FIG. 75. For a curved elbow of radius R and central angle a (Fig. 75) the lost head is given by the formula where the coefficient n has the value n = 0.131 +0.163(^ Values of n calculated from this formula for various values of the ratio ^ are tabulated below for convenience in substitution. it d R~ .2 .3 .4 .5 .6 .8 1.0 1.2 1.25 1.3 1.4 1.6 1.8 2.0 n = .1311. 1331.138 .145i. 1581. 206| .2941 .4401 .487| .539! .66l| .977|l.40|l.98 Enlargement of Section. A sudden enlargement in the cross section of a pipe decreases the velocity of flow and causes a loss of head due to eddying in the corners, etc. (Fig. 76). If the HYDROKINETICS 79 velocity is decreased by the enlargement from v\ to vz, it has been found by experiment, and can also be proved theoretically, that the head lost in this way is given by the formula 111 Ju-k^ (S') To obtain a more convenient expression for h 4 , let a denote the %?. ' " FIG. 76. area of cross section of the smaller pipe and A of the larger. Then via = v 2 A, whence and consequently the expression for h* may be written V 2 2 /A 2 (53) I - FIG. 77. Contraction of Section. A sudden contraction in section also causes a loss in head, similar to that due to a standard orifice or adjutage (Fig. 77). The lost head in this case has been found by experiment to be given by the equation (54) 80 ELEMENTS OF HYDRAULICS where q denotes an empirical constant, determined experimentally. The following tabulated values of the coefficient q are based on experiments by Weisbach, A being the cross-sectional area of the larger pipe and a of the smaller. 1 a I .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 q |.362|.338|.308 1.267 |.22l|.164 . 105 |. 053 .015 .000 Gate Valve in Circular Pipe. The loss in head due to a partly closed gate valve (Fig. 78) has been determined by experiment for different ratios of height of opening to diameter of pipe with the following results. 2 In this Table, x denotes the height of the opening, d the diameter of the pipe, h 6 the loss in head and v z f the empirical coefficient in the formula h& = f~-- X d i i 4 I \ 1 * 7 I i 97.8 17.0 5.52 | 2.06 0.81 | 0.26 0.07 FIG. 78. FIG. 79. FIG. 80. Cock in Circular Pipe. For a cock in a cylindrical pipe (Fig. 79) the coefficient f has been determined in terms of the angle of closure with the following results. e 5 I 10 15 20 25 30 | 35 40 45 | 50 55 | 60 65 82 f .05 .29 .75 1.56 3.1 5.47 9.68 17.3 31.2 52.6 106 206 486 Valve closed 1 Hoskins, Text-book on Hydraulics, page 74. 2 The coefficient for losses at valves are based on experiments by Weisbach and are given in most standard texts on Hydraulics. See for example Wittenbauer, Aufgabensammlung, Bd. Ill, S. 318; Gibson, Hydraulics and its Applications, pp. 249, 250. HYDROKINETICS 81 Throttle Valve in Circular Pipe. The coefficient { in the for- v 2 mula/ie = ^ has been determined experimentally for a throttle valve of the butterfly type (Fig. 80) for various angles of closure with results as follows: e 5 10 20 | 30 | 40 | 45 | 50 60 | 70 f 0.24 0.52 1.54 | 3.91 | 10.80 18.70 32.6 118 | 751 Summary of Losses. The total head, hi, lost in flow through a pipe line is then the sum of the six partial losses in head mentioned above, namely hi = hi + h 2 + hz + &4 + hs + h 6 . The values of these six terms may be tabulated as follows: Loss of head in pipe flow Head lost at entrance * - 4 Coefficient modified by nature of entrance and varies from .5 to .9 Friction head *-'/() ' New pipe, / = .024 Old pipe, / = .03 See table 12 Head lost at bends and elbows fi3 = m,n" (Sharp bend) * 3 = n (9"d) (Curved elbow) TO = .9457 sin' + 2 - 047 sin * / d\ 3 ' 5 n = .131 + .168(2) (f) Head lost at sudden enlargement (i - t>>* * 4 - -*-* Head lost at sudden contraction 715 = % For values of coefficient, see Table page 80 Head lost at parti- ally closed valve *' = 4 See Tabular values of f , pag and 81 es 80 From Eq. (36) we have v 2 v 2 h = T- + hi = -f- -f and inserting in this relation the values tabulated above and solv- ing for the velocity of flow, v, we have 2gh (55) 82 ELEMENTS OF HYDRAULICS Application. To give a simple illustration of the application of Formula (55), suppose it is required to find the velocity of flow fora straight new cast iron pipe, 1 ft. in diameter and 5000 ft. long, with no valve obstructions, which conducts water from a reservoir the surface of which is 150 ft. above the outlet of the pipe. In this case 2gh 2(32.2)150 1+0.5+ 0.024 = 8.9 ft. per sec. and the discharge is Q = Av = \ X 8.9 X 60 = 419.4 cu. ft. per min. 18. HYDRAULIC GRADIENT Kinetic Pressure Head. In the case of steady flow through a long pipe, if open piezometer tubes are inserted at different points of its length and at right angles to the pipe, the height at FIG. 81. which the water stands in any tube represents the kinetic pressure head at this point. Assuming that the pipe is straight and of uniform cross section, the velocity head is constant throughout, and therefore as the frictional head increases the pressure head decreases. The head lost in friction between any two points m and n (Fig. 81) as given by Eq. (48), Art. 17, is and is therefore proportional to the distance I between these points. Consequently, the drop in the piezometer column be- tween any two points is proportional to their distance apart, and HYDROKINETICS 83 therefore the tops of these columns must lie in a straight line. This line is called the hydraulic gradient, or virtual slope of the pipe. Evidently the vertical ordinate between any point in the pipe and the hydraulic gradient measures the kinetic pressure head at the point in question. Slope of Hydraulic Gradient. When a pipe is not straight, successive points on the hydraulic gradient may be determined by computing the loss of head between these points from the relation taking as successive values of I the length of pipe between the points considered. In water mains the vertical curvature of the pipe line is gen- erally small, and its effect on the hydraulic gradient is usually A- FIG. 82. neglected. When, however, a valve or other obstruction occurs in a pipe there is a sudden drop in the hydraulic gradient at the obstruction, due to the loss of head caused by it. It should be noted that the upper end of the hydraulic gradient lies below the water level in the reservoir a distance equal to the head lost at entrance plus the velocity head. The slope of the hydraulic gradient is usually defined, however, as , , .. static head Slope of hydraulic gradient = , TT r : > length of pipe which is equivalent to neglecting the velocity head and head lost at entrance, thereby making the assumed hydraulic gradient slightly steeper than it actually is. Peaks above Hydraulic Gradient. When part of the pipe line rises above the hydraulic gradient (Fig. 82), the pressure in this portion must be less than atmospheric since the pressure head h' becomes negative. If the pipe is air-tight and filled be- 84 ELEMENTS OF HYDRAULICS fore the flow is started this will not affect the discharge. If the pipe is not air-tight, air will collect at the summit above the hydraulic gradient, changing the slope of the latter from AB to AC as indicated in Fig. 82, thereby reducing the head to h" with a corresponding diminution of the flow. Before laying a long pipe line the hydraulic gradient should therefore be plotted on the profile to make sure there are no summits projecting above the gradient. In case such summits are unavoidable, provision should be made for exhausting the air which may collect at these points, so as to maintain full flow. 19. HYDRAULIC RADIUS Definition. That part of the boundary of the cross section of a channel or pipe which is in contact with the water in it is called the wetted perimeter, and the area of the cross section of the stream divided by the wetted perimeter is called the hydraulic radius, or hydraulic mean depth. lii what follows the hydraulic radius will be denoted by r, defined as Area of flow Hydraulic radius, r = =^r r Wetted perimeter Some writers apply the term hydraulic radius only to circular pipes, and use the term hydraulic mean depth for flow in channels. For a channel of rectangular cross section having a breadth 6 and depth of water h, the hydraulic radius is bh ~ b + 2h In a circular pipe of diameter d, running full, the hydraulic radius is r For the same pipe running half full, ird 4 8 d r - = ^d == -*' 2 and is therefore the same as when the pipe is full. HYDROKINETICS 85 Other examples of the hydraulic radius are shown in Figs. 98-104. Chezy's Formula. The formula proposed by Chezy for the velocity of flow in a long pipe is v = C Vrs^ (56) where s denotes the slope of the hydraulic gradient, defined in the preceding article; r is the hydraulic radius, defined above; and C is an empirical constant which depends on the velocity of flow, diameter of pipe, and roughness of its lining. For a circular pipe flowing full Chezy 's formula is identical with the formula for friction loss in a pipe, given by Eq. (48), Art. 17, namely, To show this identity, substitute in Chezy 's formula the values d . h r = -7 and s = T- Then it becomes whence, by squaring and solving for h, it takes the form 4 il\ Sgil Consequently, if the constant term is denoted by /, that is, Chezy 's formula assumes the standard form, The most important application of Chezy's formula is to flow in open channels, as explained in Art. 24. It has been found by experiment that the coefficient C in Chezy 's formula is not strictly constant for any particular chan- nel and dependent only on the roughness of the channel lining, but that it also varies with the slope and the hydraulic radius. 86 ELEMENTS OF HYDRAULICS This variation may be taken into account by writing the formula in the exponential form v = Cr m s n where the exponents m and n are also empirical constants. Ex- perimental data for this formula are not yet sufficient, however, to make its use practical. 20. DIVIDED FLOW Compound Pipes. In water works calculations the problem often arises of determining the flow through a compound system of branching mains. To illustrate the method of finding the discharge through the various branches, consider first the simple case of a main tapped f M B N Plan FIG. 83. by a branch pipe which later returns to the main, as indicated in Fig. 83. The solution in this case is based on the fundamental relation deduced in Art. 17, namely, where h denotes the static head, and hi the head lost in friction. Using the notation indicated on the figure and considering the HYDROKINETICS 87 d2g two branches separately, we obtain the following equations :- For line ABMCD, For line ABNCD, By subtraction of these two equations we have which shows that the frictional head lost in the branch BMC is equal to that lost in BNC. a nf FIG. 84. Since the total discharge through the branches is the same as that through the main before dividing and after uniting, we also have the two relations i ft\f\\ QiVi G 2 V 2 ~T~ ttsVs 4. WW By assuming an average value for the frictional coefficient /, the four Eq. (57), (58), (59) and (60) may then be solved for the four unknowns Vi, v 2 , v 3j v. Having found approximate values of the velocities, corresponding values of / may be sub- stituted in these equations and the solution repeated, thus giving more accurate values of the velocities. Having found the velocities, the discharge through the various pipes may be obtained from the relations r\ f\ Icgi == t^/4 88 ELEMENTS OF HYDRAULICS The solution for more complicated cases is identical with the above, except that more equations are involved. Branching Pipes. Another simple case of divided flow which is often met is that in which a pipe AB of diameter d divides at some point B into two other pipes, BC and BD, of diameters di and d 2 respectively, which discharge into reservoirs or into the air (Fig. 84). If any outlet, as C, is higher than the junction B, then in order for flow to take place in the direction BC, the hy- draulic gradient must slope in this direction; that is to say there must be a drop in pressure between the junction B and the level of the outlet reservoir C, or, in the notation of the figure, the con- dition for flow in the direction BC is hi > h. Assuming this to be the case, the solution is obtained from the same fundamental relation as above, namely, Using the notation indicated on the figure for length, diameter and velocity in the various pipes and considering one line at a time, we thus obtain the following equations: ForlineABC, ^ = + / +/' (*> Forline^Z), *,- Also, from the condition that the discharge through the main pipe must equal the sum of the discharges through the branches, denoting the cross-sectional areas by a, ai, 2 respectively, we have av = aiVi + a 2 v 2 . (49) By assuming an average value for the frictional coefficient /, these three equations may then be solved for the three unknowns v, Vi and v%. Having thus found approximate values of the veloc- ities, the exact value of / corresponding to each velocity may be substituted in the above equations and the solution repeated, giving more accurate values of the velocities. Having found the velocities, the discharge from each pipe is obtained at once from the relations Q = av; Qi = a^i; Q 2 = a 2 v 2 . HYDROKINETICS 89 The method of solution is the same for any number of branches, there being as many equations in any given case as there are un- known velocities to be determined. Other simple cases of divided flow are illustrated in the numer- ical examples at the end of the chapter. 21. FIRE STREAMS Freeman's Experiments. Extensive and accurate experi- ments on discharge through fire hose and nozzles were made by John R. Freeman at Lawrence, Mass., in 1888 and 1890. 1 From these experiments it was found that the smooth cone nozzle with simple play pipe is the most efficient for fire streams, the coefficient of discharge being nearly constant for the various types tried and having an average value of 0.974 for smooth cone nozzles and 0.74 for square ring nozzles. The friction losses for fire hose were found to depend chiefly on the nature of the interior surface of the hose, the loss in unlined linen hose being about 2-1/2 times as great as in smooth rubber hose of the same diameter. The friction loss was also found to vary approximately as the square of the velocity of flow. For fire hose laid in ordinary smooth curves but not cramped or kinked, the friction loss was found to be about 6 per cent, greater than in perfectly straight hose. Formulas for Discharge. The following formulas for discharge were deduced by Freeman from these experiments. Notation: Q = discharge in cubic feet per second, G = discharge in gallons per minute = 448.83Q, h = piezometer reading at base of nozzle in feet of water, p = pressure at base of nozzle in pounds per square inch = 0.434/i, K = coefficient of discharge, C c = coefficient of contraction, d = diameter of nozzle orifice in inches, D = diameter of channel, where pressure is measured, in inches, H = total hydrostatic head in feet = - 1 Trans. Am. Soc. C. E. Vol. 21, pp. 303-482; Vol. 24, pp. 492-527. 90 Then ELEMENTS OF HYDRAULICS Q = o.04374Kd 2 and o.o66 4 5Kd (63) G= i -K' d\ 4 = 29.83 i -K 2 ^ (64) Height of Effective Fire Stream. It was also found that the height, y, of extreme drops in still air from nozzles ranging in size from 3/4 in. to 1-3/8 in. in diameter was given by the formula H 2 y = H - 0.00135^' (65) The height of a first class fire stream will then be a certain frac- tion of y as indicated in the following table: When y = 50ft. 75ft. 100 ft. 125 ft. 150 ft. Height of first class fire stream = 0.82 y 0.79 y 0.73 y 0.67 y 0.63 y Table 11 is abridged from a similar table computed by Free- man from these and other formulas, not here given, and will be found convenient to use in solving fire-stream problems. 22. EXPERIMENTS ON THE FLOW OF WATER Verification of Theory by Experiment. The subject of hydrau- lics as presented in an elementary text book is necessarily limited to simple demonstrations of the fundamental principles. It should not be inferred from this that the subject is largely experi- mental and not susceptible of mathematical analysis. As a mat- ter of fact, hydrodynamics is one of the most difficult branches of applied mathematics, and its development has absorbed the HYDROKINETICS 91 best efforts of such eminent mathematical physicists as Poinsot, Kirch hoff, Helmholz, Maxwell, Kelvin, Stokes and Lamb. Natur- ally the results are too technical to be generally appreciated, but afford a rich field for study to those with sufficient mathematical preparation. Sudden contraction. Sudden enlargement. FIG. 85. Some of the results concerning the flow of liquids derived by mathematical analysis have been verified experimentally by the English engineers, Professor H. S. Hele-Shaw and Professor Osborne Reynolds. The chief importance of these experiments is that they serve to visualize difficult theoretical results. Contraction. Enlargement. FIG. 86. Method of Conducting Experiments. In Art. 8 a stream line was defined as the path followed by a particle of liquid in its motion. A set of stream lines distributed through a flowing liquid therefore completely determines the nature of the flow. To make such stream lines visible, so as to make it possible to actually trace the motion of the particles of a clear fluid, both experimenters named above allowed small bubbles of air to enter 92 ELEMENTS OF HYDRAULICS a flowing stream. These bubbles do not make the motion directly visible to the eye, but by making the pipe or channel of glass and projecting a portion of it on a screen by means of a lantern, its image on the screen as viewed in this transmitted light clearly shows certain characteristic features. Effect of Sudden Contraction or Enlargement. Figs. 85 and 86, reproduced by permission of Professor Hele-Shaw, show the effect of a sudden contraction or enlargement of the channel sec- tion. It is noteworthy that the disturbance or eddying is much FIG. 87. greater for a sudden enlargement than for a sudden contraction. This is due to the inertia of the fluid which prevents it from imme- diately filling the channel after passing through the orifice. This also confirms what has already been observed in practice, namely, that the loss of energy due to a sudden enlargement in a pipe is much greater than that due to a corresponding contraction. Disturbance Produced by Obstacle in Current. If the channel is of considerable extent and a small obstacle is placed in it, the stream lines curve around the obstacle, leaving a small space behind it, as shown in Fig. 87. If the object is a square block or flat plate this effect is greatly magnified, as shown in Fig. 88. The water is prevented from closing at once behind the obstacle by reason of its inertia. This indicates why the design of the HYDROKINETICS 93 stern of a ship is so much more important than that of the bow, since if there are eddys in the wake of "a ship, the pressure of the water at the stern is decreased, thereby increasing by just this much the effective resistance to motion at the bow. Stream -line Motion in Thin Film. In these experiments it was also observed that there was always a clear film of liquid, or border line, on the sides of the channel and around the obstacle. This observed fact was accounted for on the ground that by reason of the friction between a viscous liquid and the sides of the FIG. 88. channel or obstacle, the thin film of liquid affected was not mov- ing with turbulent motion but with true stream line motion, as in an ideal fluid. To isolate this film so as to observe its motion, water was allowed to flow between two plates of glass in a sheet so thin that its depth corresponded to the thickness of the border line previously observed. When this was done it immediately became apparent that the flow was no longer turbulent but a steady stream-line motion. The flow of a viscous fluid like gly- cerine in a thin film thus not only eliminates turbulent flow, but also to a certain extent the inertia effects, thereby resulting in true stream-line flow. Cylinder and Flat Plate. To make the stream lines visible, colored liquid was injected through a series of small openings, the result being to produce an equal number of colored bands or stream lines in the liquid. Fig. 89 shows these stream-lines for 94 ELEMENTS OF HYDRAULICS FIG. 89. FIG. 90. Theory Experiment HYDROKINETICS 95 a cylinder, and Fig. 90 for a flat plate placed directly across the current, while Fig. 91 shows a comparison of theory and experi- ment for a flat plate inclined to the current. Velocity and Pressure. The variation in thickness of the bands is due to the difference in velocity in various parts of the channel, the bands of course being thinnest where the velocity is greatest. Since a decrease in velocity is accompanied by a cer- tain increase in pressure, the wide bands before and behind the obstacle indicate a region of higher pressure. This accounts for the standing bow and stern waves of a ship, whereas the narrow- ing of the bands at the sides indicates an increase of velocity and reduction of pressure, and accounts for the depression of the water level at this part of a ship. FIG. 92. In the case of a sudden contraction or enlargement of the chan- nel section, the true stream-line nature of the flow was clearly apparent, as shown in Fig. 92, the stream lines following closely the form derived by mathematical analysis for a perfect fluid. 23. MODERN SIPHONS Principle of Operation. In its simplest form, a siphon is merely an inverted U-shaped tube, with one leg longer than the other, which is used for emptying tanks from the top when no outlet is available below the surface. In use, the tube is filled with liquid and the ends corked, or otherwise closed. The short end of the tube is then placed in the reservoir to be lowered, so that the level of the end outside is lower than the surface of the reservoir (Fig. 93). When the ends of the tube are opened, the liquid in the reservoir begins to flow through the tube with a head, 96 ELEMENTS OF HYDRAULICS T h, equal to the difference in level between the surface of the reservoir and the lower, or outer, end of the tube. If the inner end of the siphon is placed close to the bottom of the reservoir it can be practically emptied in this manner. For emptying small tanks a siphon can conveniently be made of a piece of ordi- nary tubing or hose. Siphon Spillway. Re- cent commercial applica- tions of the siphon on a large scale are the siphon spillways and siphon lock 93 on the New York State barge canal. At three lo- calities on the Champlain division of the Barge canal the siphon principle is being used for the first time to create a spillway of any considerable size. At the place shown in Fig. 94, it was necessary _L PIG. 94. Siphon Spillway, Champlain division, New York State barge canal. to provide for a maximum outflow of about 700 cu. ft. per second and to limit the fluctuation of water surface to about 1 ft. The ordinary waste weir of a capacity sufficient to take care of this HYDROKINETICS 97 98 ELEMENTS OF HYDRAULICS HYDROKINETICS 99 flow, with a depth of only 1 ft. of water on the crest, would re- quire a spillway 200 ft. long. The siphon spillway measures only 57 ft. between abutments and accomplishes the same results. This particular structure consists of four siphons, each having a 100 ELEMENTS OF HYDRAULICS cross-sectional area of 7-3/4 sq. ft. and working under a 10-1/2-ft. head. There is also a 20-ft. drift gap to carry off floating debris. The main features of construction are shown in Fig. 95. The siphon spillway was designed and patented by Mr. George F. Stickney, one of the Barge canal engineers. Siphon Lock. The siphon lock on the New York State Barge Canal is located in the city of Oswego, and is the only lock of this type in this country and the largest ever built on this principle. It consists of two siphons, as shown in Fig. 96, the crown of each being connected by a 4-in. pipe to an air tank in which a partial vacuum is maintained. To start the flow, the air valve is opened, the vacuum in the tank drawing the air from the siphon and there- by starting the flow. When the siphon is discharging fully, its draft is such that the air is sucked out of the tank, thus restoring the partial vacuum. To stop the flow, outer air is admitted to the crest of the siphon by another valve, thereby breaking the flow, as indicated in Fig. 97. The operating power is thus self renewing, and ; except for air leakage, lockages can be conducted by merely manipulating the 4-in. air valves. However, to avoid the necessity of refilling the tank when traffic is infrequent, it is customary to close the 20-in. outlet valve, thus holding the water in the tank. Using both siphons, the lock chamber can be filled in 4-1/2 to 5 min., and emptied in 5-1/2 to 6 min. 24. FLOW IN OPEN CHANNELS Open and Closed Conduits. Conduits for conveying water are usually classified as open and closed. By a closed conduit is. meant one flowing full under pressure, as in the case of ordinary pipe flow discussed in Art. 15. Water mains, penstocks, draft tubes and fire hose are all examples of closed conduits. Open channels, or conduits, are those in which the upper sur- face of the liquid is exposed to atmospheric pressure only, the pressure at any point in the stream depending merely on the depth of this point below the free surface. Rivers, canals, flumes, aqueducts and sewers are ordinarily open channels. A river or canal, however, may temporarily become a closed chan- nel when covered with ice, and an aqueduct or sewer may also become a closed channel if flowing full under pressure. Steady Uniform Flow. The fundamental laws applying to flow in open and closed channels are probably identical, and in HYDROKINETICS 101 the case of steady, uniform flow the same formulas apply to both. For steady flow in an open channel the quantity of water passing any transverse section of the stream is constant, and for uniform flow the mean velocity is also constant. Under these conditions the cross-sectional area of the stream is constant throughout its length, and the hydraulic gradient is the slope of the surface of the stream. The formula for velocity of flow is then the one given in Art. 19 under the name of Chezy's formula, namely v = C Vrs. (66) Kutter's Formula. Numerous experiments have been made to determine the value of the coefficient C for open channels. In 1869, E. Ganguillet and W. R. Kutter, two Swiss engineers, made a careful determination of this constant, the result being expressed in the following form : v = 0.00281 1.811 41-65 + - + i + 41.65 + Vrs (67) in which s = hydraulic gradient, or slope of channel, area of flow r = hydraulic radius = - , -- ~. > wetted perimeter n = coefficient of roughness. The coefficient of roughness, n, depends on the nature of the channel lining. Approximate values of n for various surfaces are given in the following table: Nature of Channel Lining Planed timber carefully joined, glazed or enameled surfaces Smooth clean cement Cement mortar, one-third sand Unplaned timber or good new brickwork Smooth stonework, vitrified sewer pipe and ordinary brickwork . . Rough ashlar and good rubble masonry Firm gravel Ordinary earth Earth with stones, weeds, etc Earth or gravel in bad condition 0.009 0.010 0.011 0.012 0.013 0.017 0.020 0.025 0.030 0.035 Limitations to Kutter 's Formula. Kutter's formula, Eq. (67), is widely used and is reliable when applied to steady, uniform 102 ELEMENTS OF HYDRAULICS flow under normal conditions. From a study of the data on which this formula is based, its use has been found to be subject to the following limitations: It is not reliable for hydraulic radii greater than 10 ft., or ve- locities greater than 10 ft. per sec., or slopes flatter than 1 in 10,000. Within these limits a variation of about 5 per cent, may be expected between actual results and those computed from the formula. Table 15 gives numerical values of the coefficient C calculated from Formula (67). Bazin's Formula. In 1897, H. Bazin also made a careful determination of the coefficient C from all the experimental data then available, as the result of which he proposed the following formula: Vr where r = hydraulic radius, m = coefficient of roughness. Bazin' s formula has the advantage of being simpler than Kutter's, and is independent of the slope s. Values of the coeffi- cient of roughness, m, for use with this formula are given in the following table: Nature of Channel Lining Planed timber or smooth cement Unplaned timber, well-laid brick or concrete Ashlar, good rubble masonry or poor brickwork Earth in good condition Earth in ordinary condition Earth in bad condition . . 0.06 0.16 0.46 0.85 1.30 1.75 Table 14 gives numerical values of the coefficient C calculated from formula (68). Kutter's Simplified Formula. A simplified form of Kutter's formula which is also widely used is the following: f 100 Vr 1 , v = - -- -=. V Lb + Vr J rs, where b is a coefficient of roughness which varies from 0.12 to 2.44. For ordinary sewer work the value of this coefficient may be assumed as b = 0.35. HYDROKINETICS 103 26. CHANNEL CROSS SECTION Condition for Maximum Discharge. From the Chezy formula for flow in open channels, namely, Q = Av = ACVrs, it is evident that for a given stream section A and given slope s, the maximum discharge will be obtained for that form of cross section for which the hydraulic radius r is a maximum. Since area of flow wetted perimeter this condition means that for constant area the radius r, and therefore the discharge, is a maximum when the wetted perimeter is a minimum. The reason for this is simply that by making the area of contact between channel lining and water as small as possible, the frictional resistance is reduced to a minimum, thus giving the maximum discharge. Maximum Hydraulic Efficiency. In consequence of this, it follows that the maximum hydraulic efficiency is obtained from a semicircular cross section, since for a given area its wetted perimeter is less than for any other form (Fig. 98) . For rectangu- lar sections the half square has the least perimeter for a given area, and consequently is most efficient (Fig. 99). Similarly, for a trapezoidal section the half hexigon is the most efficient (Fig. 100). In each case the hydraulic radius is half the water depth, as proved below. In the case of unlined open channels it is necessary to use the trapezoidal section, the slope of the sides being determined by the nature of the soil forming the sides. This angle having been determined, the best proportions for the section are obtained by making the sides and bottom of the channel tangent to a semi- circle drawn with center in the water surface (Fig. 101). Regular Circumscribed Polygon. Any section which forms half of a regular polygon of an even number of sides, and has each of its faces tangent to a semi-circle having its center in the water surface, will have its hydraulic radius equal to half the radius of this inscribed circle (Figs. 98-103). To prove this, draw radii from the center of the inscribed circle to each angle of the polygon. Then since the area of each of the triangles so formed is equal to 104 ELEMENTS OF HYDRAULICS Semicircle FIG. 98. Half Square FIG. 99. 4=3#tan 30 w.p. = 6 R tan 30 r _8Jg 2 tan 30 _R QR tan 30 2 4/2 tan 30 P= Wetted Perimeter A=!/2 PR =y?R = R '2/2 tan 30 ol Half Hexagon FIG. 100/ Trapezoid FIG. 101. P= Wetted Perimeter A= V 2 PR Half Octagon FIG. 102. Triangular Section FIG. 103. HYDROKINETICS 105 II- a- (N a go 04 04 04 .a -g c3 ""cu 04 04 a SD . -3 '1 .1 I 00 O 04 | CO O O M 106 ELEMENTS OF HYDRAULICS one-half its base times its altitude, and since the altitude in each case is a radius of the inscribed circle, the total area is D Area = ^ X perimeter. Consequently the hydraulic radius r is area of flow * X perimeter R wetted perimeter perimeter 2 Properties of Circular and Oval Sections. For circular and oval cross sections, the maximum velocity and maximum dis- Angle in Radians Circular Section FIG. 104. i Standard Oval Section FIG. 105. charge are obtained when the conduit is flowing partly full, as apparent from the table on page 105, which is a collection of the most important data for circular and oval sections, 1 as shown in Figs. 104 and 105. Theoretically, the maximum discharge for a circular pipe oc- curs when the pipe is filled to a depth of 0.949 D, but if it is attempted to maintain flow at this depth, the waves formed in the pipe strike against the top, filling it at periodic intervals and thus producing impact losses. To obtain the maximum discharge' without danger of impact, the actual depth of flow should not exceed 5/6 D. 26. FLOW IN NATURAL CHANNELS Stream Gaging. In the case of a stream flowing in a natural channel the conditions determining the flow are so variable that 1 Weyrauch, Hydrauliches Rechnen, S. 51. HYDROKINETICS 107 no formula for computing the discharge has been devised that can claim to give results even approximately correct. To obtain accurate results, direct measurements of cross sections and veloc- ities must be made in the field. The two methods of direct measurement in general use are as follows : Either FIG. 106. Electric current meter. 1. The construction of a weir across the stream, and the calcula- tion of the discharge from a weir formula; or, if this is not feasible, 2. The measurement of cross sections of the stream by means of soundings taken at intervals, and the determination of average velocities by a current meter or floats. The first of these methods is explained in Art. 11. 108 ELEMENTS OF HYDRAULICS Current Meter Measurements. The current meter, one type of which is shown in Fig. 106, consists essentially of a bucket wheel with a heavy weight suspended from it to keep its axis horizontal, and a vane to keep it directed against the cur- rent, together with some form of counter to indicate the speed at which the wheel revolves. The meter is first rated by towing it through still water at various known velocities and tabulating the corresponding wheel speeds. From these results a table, or chart, is constructed giving the velocity of the current corre- sponding to any given speed of the wheel as indicated by the counter. This method of calibration, however, is more or less inaccurate, as apparent from Du Buat's paradox, explained in Art. 27. Float Measurements. When floats are used to determine the velocity, a uniform stretch of the stream is selected, and two cross sections chosen at a known distance apart. Floats are then put Water Surface t Average Velocity Bed ofyStream into the stream above the upper section and their times of transit from one section to the other observed by means of a stop watch. A sub-surface float is commonly used, so arranged that it can be run at any desired depth, its position being located by means of a small surface float attached to it. If the cross section of the stream is fairly uniform, rod floats may be used. These consist of hollow tubes, so weighted as to float upright and extend nearly to the bottom. The velocity of the float may then be assumed to be equal to the mean velocity of the vertical strip through which it runs. . Variation of Velocity with Depth. The results of such measure- ments show, in general, that the velocity of a stream is greatest midway between the banks and just beneath the surface. In HYDROKINETICS 109 particular, the velocities at different depths along any vertical are found to vary as the ordinates to a parabola, the axis of the parabola being vertical and its vertex just beneath the surface, as indicated in Fig. 107. From this relation it follows that if a float is adjusted to run at about 0.6 of the depth in any vertical strip, it will move with approximately the average velocity of all the particles in the vertical strip through which it runs. Calculation of Discharge. In order to calculate the discharge it is necessary to measure the area of a cross section as well as the average velocities at various points of this section. The total cross section is therefore subdivided into parts, say A\ t A 2 , As, etc. (Fig. 108), the area of each being determined by measuring the ordinates by means of soundings. The average velocity for each division is then measured by one of the methods explained above, and finally the discharge is computed from the relation Q = + A 2 v 2 + A 3 v 3 + 27. THE PITOT TUBE Description of Instrument. An important device for measur- ing the velocity of flow is the instrument known as the Pilot tube. In 1732 Pitot observed that if a small vertical tube, open. at both ends, with one end bent at a right angle, was dipped in a current so that the horizontal arm was directed against the current as indicated in Fig. 109, A, the liquid rises in the vertical arm to a height proportional to the velocity head. The height of the col- umn sustained in this way, or hydrostatic head, is not exactly equal to the velocity head on account of the disturbance created by the presence of the tube. No matter how small the tube may be, its dimensions are never negligible, and its presence has the effect of causing the filaments of liquid, or stream lines, to curve 110 ELEMENTS OF HYDRAULICS around it, thereby considerably modifying the pressure. Since the column of liquid in the tube is sustained by the impact of the current, this arrangement is called an impact tube. If a straight vertical tube is submerged, or a bent tube having its horizontal arm directed transversely, that is, perpendicularly, to the current, the presence of the tube causes the stream lines to turn their concavity toward the orifice, thereby producing a suc- tion which is made apparent by a lowering of the water level in this tube, as shown in Fig. 109, B. In the case of the bent tube, if the horizontal arm is directed with the current, as shown in Fig. Impact Tube Suctionor Pressure Tube Tube Direction of Flow FIG. 109. 109, C, the suction effect is more pronounced, and the level in the tube is still further lowered. When the horizontal arm of a bent tube is directed with the current, the arrangement is called a suction or trailing tube. It is practically impossible, however, to obtain satisfactory numerical results with this simple type of Pitot tube, as in the case of flow in open channels the free surface of the liquid is usu- ally disturbed by waves and ripples and other variations in level, which are often of the same order of magnitude as the quantities to be measured; while in the case of pipe flow under pressure there are other conditions which strongly affect the result, as will appear in what follows. Darcy's Modification of Pitot's Tube. In 1850 Darcy modified the Pitot tube so as to adapt it to general current measurements. This modification consisted in combining two Pitot tubes, as shown in Fig. 110, the orifice of the impact tube being directed HYDROKINETICS 111 up stream, and the orifice of the suction tube transverse to the current. In some forms of this apparatus, the suction tube is of the trailing type, that is, the horizontal arm is turned directly down stream. To make the readings more accurate, the difference in eleva- tion of the water in the two tubes is magnified by means of a differential gage, as shown in Fig. 110. Here A denotes the impact tube and B the suction tube (often called the pressure tube), connected with the tubes C and D. between which is a graduated scale. After placing the apparatus in the stream to be gaged, the air in both tubes is equally rarified by suction at F, thereby causing the water level in both to rise propor- tional amounts. The valve at F is then closed, also the valve at E, and the appara- tus is lifted from the water and the read- ing on the scale taken. It was assumed by Pitot and Darcy that the difference in level in the tubes was pro- portional to the velocity head ^-t where v denotes the velocity of the current. Call- H~[ ing hi and h% the difference in level, that is, the elevation or depression of the g -^J water in the impact and suction tubes re- spectively, and mi, m% the constants of proportionality, we have therefore T D FIG. 110. If, then, h denotes the difference in elevation in the two tubes (Fig. 110), we have The velocity v is therefore given in terms of h by the equation v = m Vigh (69) where m = mi + w 2 112 ELEMENTS OF HYDRAULICS The coefficient m depends, like mi and m 2 , on the form and dimen- sions of the apparatus, and when properly determined is a con- stant for each instrument, provided that the conditions under which the instrument is used are the same as those for which m was determined. The value of m in this formula has been found to vary from 1 to as low as 0.7; the value m = 1 corresponding to h = ~-; and y V 2 the value m = 0.7 to h = 9 The explanation of this ap- parent discrepancy is given below under the theory of the impact tube. In the case of variable ve- locity of flow it has been shown by Rateau 1 that the Pitot, or Darcy, tube meas- _j_ ures not the mean velocity but the mean of the squares of the velocities at the point where it is placed during the experiment. To obtain the mean velocity it is necessary v 2 to multiply <=>- by a coeffi- cient which varies according to the rate of change of the velocity with respect to -the time. From Rateau's experi- ments this coefficient was found to vary from 1.012 to 1.37, having a mean value of 1.15. This corresponds to a mean value for m of 0.93. Pitometer. A recent modification of the Pitot tube is an instrument called the Pitometer (Fig. 111). The mouth-piece of this apparatus consists of two small orifices pointing in oppo- site directions and each provided with a cut- water, as shown in the figure. When in use, these are set in line parallel to the current, Cutwater FIG. 111. 1 Annales des Mines, Mars, 1898. HYDROKINETICS 113 so that one points directly against the current and the other with it. The differential gage used with this instrument consists of a U-tube, one arm of which is connected with one mouthpiece and the other arm with the other mouthpiece, and which is about half filled with a mixture of gasoline and carbon tetrachloride, colored dark red. The formula for velocity as measured by this instrument is given in the form v = k[2g(s - where k = empirical constant = 0.84 for the instrument as manufactured and calibrated, s = specific weight of the tetrachloride mixture = 1.25, d = difference in elevation in feet between the tops of the two columns of tetrachloride. Inserting these numerical values, the formula reduces to v = 3.368 VS. It is claimed that velocities as low as 1/2 ft. per second can be measured with this instrument. Pitot Recorders. The Pitot meter is used in power houses, pumping stations and other places where a Venturi tube cannot be installed, and is invaluable as a water-works instrument to determine the pipe flow in any pipe of the system. A recent portable type, especially adapted to this purpose is shown in Fig. 112. This instrument is 34 in. high, weighs 75 lb., and furnishes a chart of the Bristol type which are averaged with a special planimeter furnished with the instrument. A 1-in. tap in the water main is required for inserting the Pitot mouth- piece. It is claimed that these instruments have a range from 1/2 ft. per second to any desired maximum. Theory of the Impact Tube. The wide variation in the range of coefficients recommended by hydraulic engineers for use with the Pitot tube can be accounted for only on the ground of a faulty understanding of the hydraulic principles on which its action is based. The most important of these are indicated below, with- out presuming to be a complete exposition of its action. It will be shown in Art. 29 that the force produced by the im- pact of a jet on a flat plate is twice as great as that due to the hy- 114 ELEMENTS OF HYDRAULICS drostatic head causing the flow. That is to say, if the theoretical velocity of a jet is that due to a head h, where FIG. 112. Pitot recording meter, Simplex Valve and Meter Co. the force exerted on a fixed plate by the impact of this jet is equal to that due to a hydrostatic head of h f = 2h, in which case The orifice in a Pitot tube is essentially a flat plate subjected to the impact of the current. Considering only the impact effect, HYDROKINETICS 115 therefore, the head which it is theoretically possible to attain in a Pitot tube is h ' = v -, 9 which corresponds to a value of m of 0.7 in the formula v m D' There are other considerations, however, which often modify this result considerably. The effect of immersing a circular plate in a uniform parallel current has been fully analyzed theo- retically and the results confirmed experimentally. The re- sults of such an analysis made by Professor Prasil, as pre- sented in a paper by Mr. N. W. Akimoff, 1 are shown in Fig. 113. The diagram here shown represents a vertical section of J Jour. Amer. Water Works Assoc., May, 1914. 116 ELEMENTS OF HYDRAULICS a current flowing vertically downward against a horizontal cir- cular plate. The stream lines S, shown by the full lines in the figure, are curves of the third degree, possessing the property that the volumes of the cylinders inscribed in the surface of revolution generated by each stream line are equal. For instance, the volume of the circular cylinder shown in section by AA'BB' is equal to that of the cylinder CC'DD', etc. It may also be noted that the size of the plate does not affect the general shape or properties of the curves shown in the diagram. The surfaces of equal velocity are ellipsoids of revolution having the center of the plate as center, and are shown in sec- tion in the figure by the ellipses marked EV. In general, each of these ellipses intersects any stream line in two points, such as F and G. Therefore somewhere between F and G there must be a point of minimum velocity, this being obviously the point of contact of the corresponding ellipse with the stream line. The locus of these points of minimum velocity is a straight line OH in section, inclined to the plate at an angle of approxi- mately 20. The surface of minimum velocity is therefore a cone of revolution with center at 0, of which OH is an element. The surfaces of equal pressure are also ellipsoids of revolution with common center below 0, and are shown in section by the ellipses marked PP in the figure. The surface of maximum pres- sure is a hyperboloid of revolution of one sheet, shown in section by the hyperbola YOY. It should be especially noted that the cone of minimum ve- locity is distinct from the hyperboloid of maximum pressure so that in this case minimum velocity does not necessarily imply maximum pressure, as might be assumed from a careless applica- tion of Bernoulli's theorem. This analysis shows the reason for the wide variation in the results obtained by different experimenters with the Pitot tube, and makes it plain that they will continue to differ until the hy- draulic principles underlying the action of the impact tube are generally recognized and taken into account. Construction and Calibration of Pitot Tubes. The impact end of a Pitot tube is usually drawn to a fine point with a very small orifice, whereas the vertical arm is given a much larger diame- ter in order to avoid the effect of capillarity. The tubes used by Darcy had an orifice about 0.06 in. in diameter which was HYDROKINETICS 117 enlarged in the vertical arm to an inside diameter of about 0.4 in. In his well-known experiments for determining the velocity of fire streams (Art. 21), Freeman used for the mouthpiece of his impact tube the tip of a stylographic pen, having an aperture 0.006 in. in diameter. With this apparatus and for the high velocities used in the tests, the head was found to be almost v 2 exactly equal to ^ corresponding to a value of m = 1.0 in the formula v = m ^2gh. It is also important that the impact arm should be long enough so that its orifice is clear of the standing wave produced by the current flowing against the vertical arm. The cutwater used with some forms of apparatus (see Fig. Ill) is intended to eliminate this effect but it is doubtful just how far it accomplishes its purpose. The most prolific source of error in Pitot-tube measurements is in the calibration of the apparatus. The fundamental principle of calibration is that the tube must be calibrated under the same conditions as those for which it is to be used. Thus it has been shown in Art. 16 that flow below the critical velocity follows an entirely different law from that above this velocity. Flow in a pipe under pressure is also essentially different from flow in an open channel. Du Buat's Paradox. Furthermore, the method of calibration is of especial importance. This is apparent from the well-known hydraulic principle known as Du Buat's paradox. By experiment Du Buat has proved that the resistance, or pressure, offered by a body moving with a velocity v through a stationary liquid is quite different from that due to the liquid flowing with the same velocity v past a stationary object. ' The pressure of the moving liquid on the stationary object was found by him to be greater than the resistance experienced by the moving object in a station- ary liquid in the ratio of 13 to 10. All methods of calibration which depend on towing the instrument through a liquid at rest therefore necessarily lead to erroneous and misleading results. Since the Pitot tube is so widely used for measuring velocity of flow, its construction and calibration should be standardized, so that results obtained by different experimenters may be sub- ject to comparison, and utilized for a more accurate and scientific construction of the instrument. 118 ELEMENTS OF HYDRAULICS 28. NON-UNIFORM FLOW; BACKWATER Energy Equation for Stream of Variable Depth. In Art. 24 it is shown that steady, uniform flow implies constant slope and cross section of the channel, in which case the velocity and depth of water are also constant. In the case of natural channels, however, the cross section is never uniform but varies from point to point, in consequence of which the flow is non-uniform, both the velocity and hydraulic radius changing with the depth. For a stream with free upper surface, the hydraulic gradient coincides with the water surface, and consequently the drop in the water surface in any distance I measures the head lost in this distance. FIG. 114. Now consider two adjacent cross sections of a stream of vari- able depth (Fig. 114). Then by Bernoulli's theorem we have Pa . *>1 2 _ p a V 2 2 where z if 2 2 denote the elevations of the surface at the two sections considered above a horizontal datum plane, and hi is the frictional head lost in the given distance I. Since the atmospheric pressure p a is constant, this expression simplifies into (22 - 21) + + hi = 0. An expression for the lost head hi in terms of the average velocity v may be obtained from Chezy's formula, namely, v = C V, for by squaring and transposing, it becomes C 2 r HYDROKINETICS 119 and since y = s, these two relations combine to give to fe = cv" Substituting this value of hi in the above expression it takes the form (22 + (70) Differential Equation of Surface Profile. 1 To obtain the equation of a longitudinal profile of the surface, or surface curve FIG. 115. as it is called, let the distance between the two cross sections considered be infinitesimal, and denote this distance by dx. Then the difference in elevation z will also be infinitesimal, that is, 2 2 Zi = dz, and also v 2 2 Vi 2 = d(v 2 ) = 2vdv, where d(v z ) is the ordinary differential notation indicating an infinitesimal change in v 2 . Substituting these differential values in Eq. (70), it becomes , vdv , v 2 dx , x - dz + + TJ^T = o, (71) which is therefore the differential equation of the surface curve. Now take the origin on the down-stream side of the section at a distance x from it, and suppose the adjacent section to be at a distance x + dx from the origin, as indicated in Fig. 115. If s denotes the slope of the bed of the stream and y the depth of the 1 The method here given for determining the equation of the surface curve is substantially that followed in all standard texts on hydraulics. For exam- ple, see Foppl, Vorlesungen iiber Technische Mechanik, Bd. 1. >i 120 ELEMENTS OF HYDRAULICS water, the fall of the bed m the distance dx is sdx and the change in depth is dy. The/ latter, however, is negative since the slope of the surface is less than that of the bed, so that dz = sdx dy. Also, if A denotes the area of the cross section with abscissa x, b its average width, and Q the quantity of water flowing past this section per unit of time, then A = by, dA = bdy, , = | = |, *=-|Jf- Substituting these values for dz and dv in Eq. (71), it becomes Q 2 dy , Q 2 dx dy - sdx - ^ 8 + jfc = o. (72) Back-water Curve for Broad Shallow Stream. The case of most practical importance is that in which the level of a stream is raised by means of a dam or weir, and it is required to determine the new elevation of the surface back of the dam or weir. In this case let h denote the original depth of the stream before the weir was built, or the depth below the weir after it is constructed (Fig. 116). Then assuming that the breadth of the stream and the total discharge remains unchanged, we have Q = v bh = C Vr's bh where r 1 denotes the hydraulic radius for the original channel section. Substituting this value of Q in Eq. (72) it becomes C 2 r'sbWdy , Wr'sWdx dy-sdx- ^^T-+-l^c^- > which simplifies into 2 ' 2 (73) If the stream is assumed to be shallow in comparison with its breadth, this expression can be greatly simplified. Since the expressions for the cross sections and hydraulic radii of the new channel and the original channel are HYDROKINETICS 121 if y and h are small in comparison with the breadth b, the values of the hydraulic radii are approximately r = y, r' = h. Inserting these values of r, r' and A in Eq. (73) and multiplying l*< through by r^, it becomes (74) which is the required differential equation of the backwater curve for the special case considered. FIG. 116. Integration of Backwater Function. To integrate Eq. (74) let r- = u. Then dy = hdu, and Eq. (74) becomes u 3 _ g hdu = u z which may be written in the form u 3 _ By ordinary division, u* 1 - n 1 + du. g u* - 1 w 3 - 1 ' and consequently the above relation becomes SI i* 2 o \ slii I O 6 \ liM 5*rr* + (i- )^i 122 ELEMENTS OF HYDRAULICS Now integrating between limits x\ and x 2 for x, and corresponding limits ui and u 2 for u, we have The values of the first two integrals are simply fxi r ui I dx = Xi Xz = I, I du = HI u z , Jxi Juz where I denotes the distance between the cross sections con- sidered. For convenience let the remaining integral be denoted by

; - J urm e log (u - i) vi Substituting the limits, we have in this notation , 2u + 1 cot and therefore Eq. (75) becomes or, since T = u, we have y\ = hui, y% = hu z and consequently this equation may be written - ^)l^(ui) - <^(u 2 )], ( 7 6) g / which is the required equation of the backwater curve. Values of Integral. To avoid the labor of calculating the numerical value of the integral (u] u 250 approximately, there will be a sudden ver- tical drop in the backwater curve, whereas if s is less than this critical value, -r- is always negative and the curve contains no such singularity. Numerical Application. To illustrate the application of Eq. (76), suppose that on a certain river it is proposed to create a waterfall for a hydraulic power station by building a dam. Before proceeding with the construction it is required to know among other things whether the water can be raised 5 ft. at the dam without interfering with the operation of an existing power plant 7 miles up stream. Suppose that the stream at the proposed site is 10 ft. deep and 500 ft. wide, the slope of its bed is 2 ft. per mile, and the maximum annual discharge is 25,000 cu. ft. per second. Then its mean velocity is 25,000 , v " = 500 X 10 = 5 ft ' per sec "> the hydraulic radius for the given section is 500 X 10 500 + 20 and the value of the constant C in Chezy's formula is 5 = 82.9. 5280 HYDROKINETICS 125 First, let it be required to find how far up stream the increase in depth will amount to 2 ft. From the given data, h = 10, s = g|g, C = 82.9, yi --= 15, 2/2 = 12, 15 12 Ul == 10 =: L5j U2 = 10 = L2j and from the table for v 2 Head lost at entrance, hi = 0.62 H~ Head lost at stopcock =1/2 head lost in tap, Head lost in pipe by Prony's formula, where, in this last formula, d = diameter of pipe in inches, H = head in feet, L = length in yards, G = discharge in U. S. gal. per minute. 1 Discharge of Water through Street Taps and House Service Pipes, Cassier's Mag., Nov., 1905. 130 ELEMENTS OF HYDRAULICS Using these formulas we obtain in the present case the following numerical results : Pressure lost in tap = 2. 24 Ib. per square inch. Pressure lost at stopcock = 1 . 12 Ib. per square inch. Pressure lost at entrance = 3 . 08 Ib. per square inch. Pressure lost in 72.5 ft. of 1-in. pipe = 17.53 Ib. per square inch. Total pressure required in main = 23 . 97 Ib. per square inch. 74. A building is to be supplied with 2500 gal. of water per hour through 180 ft. of service pipe at a pressure at the building line of 15 Ib. per square inch. The pressure in the main is 35 Ib. per square inch. Find the required size of service pipes and taps. Solution. The total drop in pressure in this case is 20 Ib. per square inch. Therefore using the formulas given in the pre- ceding problem and assuming different sizes of service pipes, the results are as follows: One 1.25-in. full-size pipe 180 ft. long discharges 1715 gal. per hour. Two 1-in. full-size pipes discharge together 1920 gal. per hour. Two 1.25-in. pipes with 5/8-in. taps discharge together 2880 gal. pe*r hour. One 1.5-in. pipe with 1-in. tap discharges 2519 gal. per hour. The last has sufficient capacity and is cheapest to install, and is therefore the one to be chosen. 75. A pipe 1 ft. in diameter connects two reservoirs 3 miles apart and has a slope of 1 per cent. Assuming the coefficient of friction as 0.024, find the discharge and the slope of the hydrau- lic gradient when the water stands 30 ft. above the inlet end and 10 ft. above the outlet end. 76. Two reservoirs 5 miles apart are connected by a pipe line 1 ft. in diameter, the difference in water level of the two reservoirs being 40 ft. Assuming the value of Chezy's constant in feet and second units to be 125, find the discharge in gallons per hour. 77. A 12-in. main 5000 ft. long divides into three other mains, one 6 in. in diameter and 6000 ft. long, one 10 in. in diameter and 7000 ft. long, and one 8 in. in diameter and 4000 ft. long. The total static head lost in each line between reservoir and outlet is the same and equal to 100 ft. Find the discharge in gallons per 24 hours at each of the three outlets. HYDROKINETICS 131 Solution. The head lost in friction in the length I is given by the relation and the discharge by Eliminating v between these relations, we have 16/ZQ 2 Q = 10S h t rd 2 whence Q = 8/7 Assuming/ = 0.02 and / = 1000 ft., the discharge Q for pipes of various sizes in terms of the head lost per 1000 ft. is given by the following relations : Diameter of pipe in inches Discharge in gal. per 24 hours in terms of head lost per 1000 ft. 4 Q = 58,430 Vhi 6 161,OOOVfe 8 330,500Vfo 10 = 577,500Vfo 12 = 911,000V/?! 16 = l,870,OOOVfo 20 = 3,266,OOOVfo 24 = 5,147,OOOVfo 30 = 9,002,OOOVfo 36 = 14,200,OOOVfo 48 = 29, 150,000 Vhi 56 = 42,850,OOOVfo 60 = 50,920,OOOV/ii 66 ' = 64,600,OOOVfe 72 = 80,320,000 Vhi In the present noted by Q with Also if h with the pipe per 1000 ft., case let the flow in gallons per 24 hours be de- a subscript indicating the size of pipe. Then 612 = Qe + Qs + Qio. proper subscript denotes the head lost in each we have from the above relations Q = 911,000 = 161,000 = 330,500 = 577,500 132 ELEMENTS OF HYDRAULICS and since from the conditions of the problem the head lost in each line amounts to 100 ft., we also have the relations 5Ai2 + 6A 6 = 100, 5/*i2 + 7hi = 100, 5&i2 + 4/* 8 = 100. From these relations we find 100 - and substituting these values in the first equation, the result is F E D C B A 1 9. ^ Q, , 1 M N R FIG. 118. 161,000 V/i 6 + 330,500 + 577,500 = 91 1,000 J 100 - whence ^6 = 7.52 and consequently h 8 = 12.28; h lo = 6.446; h 12 = 10.976. Substituting these values of h in the formulas for discharge, the results are Q 6 = 444,360 gal. per 24 hr. Q 8 = 1,110,480 gal. per 241ir. Qio = 1,465,700 gal. per 24 hr. Oe + Qs + Qio = 3,020,540 gal. per 24 hr. The actual calculated value of Qi% is Qi 2 = 3,015,400 gal. per 24 hr., HYDROKINETICS 133 the discrepancy between these results being due to slight inac- curacy in extracting the square roots. 78. A pipe of constant diameter d discharges through a number of laterals, each of area A and spaced at equal distances I apart (e.g., street main and house service connections). Find the rela- tion between the volume of flow in three successive segments of the main 1 (Fig. 118). Solution. The discharge at A is Oi = KA where vi denotes the velocity of flow at this point. Also the head lost in friction in the segment AB is l f(\ Vl =f (d) 2 and consequently where 8/7 ^ = 2^75* At B the pressure head is h + hi and the discharge is Q 2 - Q, = KA Similarly, for the discharge at C and D we obtain the relations O, - Q 2 = KA Qt Q 3 = KA V20 V/i + a(Qi< whence by elimination where 6 2 = The general relation is therefore (Q ~ Qn-l) 2 ~ (Qn-l - Qn-t) 2 = The following geometrical construction may be used for deter- mining Q n . Determine an angle such that 6 = tan 6 1 J. P. Frizell, Jour. Franklin Inst., 1878. 134 ELEMENTS OF HYDRAULICS and lay off Q n -i and Q n _ 2 on a straight line so that OM = Q n _ 2 and ON = Q n --\ as shown in Fig. 118. At N erect a perpendicular NP to ON, and then lay off NR = MP. Then OR = Q n . 79. A reservoir discharges through a pipe line made up of pipes of different sizes, the first section being 4000 ft. of 24-in. pipe, followed by 5000 ft. of 20-in. pipe, 6000 ft. of 16-in. pipe and 7000 ft. of 12-in. pipe. The outlet is 100 ft. below the level of the reservoir. Find the discharge in gallons per 24 hours. Solution. Using the same notation as in Problem 77, we have in the present case 4/> 24 + 5/*2o + 6Ai6 + 7h 12 = 100. Also, since the loss in head per 1000 ft. varies inversely as the fifth power of the diameter, and consequently 24 = ^4= 7.594A 24; Solving these three equations simultaneously with the first one, the results are ^24 = 0.35; /*2o = 0.871; A 16 = 2.658; h 12 = 11.20. As a check on the correctness of these results we have 4X 0.35 = 1.400 5 X 0.871 = 4.355 6 X 2.658 = 15.948 7 X 11.20 = 78.400 100.103 The discharge may be found from the formulas given in Problem 77, the results being as follows: Qi2 = Qie = Q 2 o = Q 24 = 3,046,000 gal. per 24 hr. 80. A pipe AB, 1000 ft. in length, divides at B into two pipes, BC which is 600 ft. long and BD which is 900 ft. long. The HYDROKINETICS 135 fall for AB is 25 ft., for BC is 10 ft. and for BD is 20 ft. Find the required diameters of the three pipes to deliver 500 gal. per minute at C and 300 gal. per minute at D. 81. A reservoir empties through a pipe AB (Fig. 119) which branches at B into two pipes BC and BD, one of which discharges FIG. 119. 20,000 gal. per hour at C and the other 30,000 gal. per hour at D. The lengths of the pipes are AB = 1200 ft., BC = 900 ft., BD = 600 ft., and the depths of the outlets below the surface of the reservoir are hi = 25 ft., A 2 = 60 ft. The pipes are of cast iron, r FIG. 120. and the velocity of flow in A B is to be 3 ft. per second. Calculate the diameters of all three, and the velocity of flow in BC and BD. 82. Two reservoirs empty through pipes which unite at C (Fig. 120) into a single pipe which discharges at D. The lengths 136 ELEMENTS OF HYDRAULICS of the pipes are h = 1500 ft., 1 2 = 900 ft. and I = 2400 ft. The diameters of the pipes are d = 6 in., d 2 = 4 in., and d = 9 in., and the depths of the outlet below the levels of the reservoirs are hi = 75 ft., A 2 = 100 ft. Find the velocity of flow in each pipe and the total discharge in gallons per hour. 83. A water main 3 ft. in diameter divides into two smaller mains of the same diameter and whose combined area equals that of the large main. If the velocity of flow is 3 ft. per second, compare the heads lost per mile in the large and small mains. Old Croton. Aqua Claudia. FIG. 121. Comparison of ancient and modern aqueducts. 84. The head on a fire hydrant is 300 ft. Find its discharge in gallons per minute through 400 ft. of inferior rubber-lined cotton hose 2.5 in. in diameter and a 1.5-in. smooth nozzle. 85. What head is required at a fire hydrant to discharge 250 gal. per minute through a 1.25-in. ring nozzle and 500 ft. of 2.5-in. best rubber-lined cotton hose? HYDROKINETICS 137 86. A fire stream is delivered through 100 ft. of 2-in. rubber- lined cotton hose and a nozzle 1-1/8 in. in diameter. The hy- drant pressure is 75 Ib. per square inch. Find pressure at nozzle, discharge in gallons per minute and height of effective fire stream. 87. Two reservoirs are connected by a siphon 16 in. in diameter and 50 ft. long. If the difference in level in the reservoirs is 25 ft., calculate the discharge, assuming the coefficient of pipe friction to be 0.02 and considering only friction losses. 88. A cast-iron pipe 2 ft. in diameter has a longitudinal slope of 1 in 2500. If the depth of water in the pipe is 18 in., calculate the discharge. 89. A rectangular flume 6 ft. wide, 3 ft. deep and 1 mile long is constructed of unplaned lumber and is required to deliver 120 cu. ft. per second. Determine the necessary gradient and the total head lost. 90. The Aqua Claudia, shown in Fig. 121, was one of the nine principal aqueducts in use in the First Century, A. D., for supply- ing Rome with water. The lengths and capacities of these nine aqueducts were as follows: Name Date of construc- tion P Length in feet Length in miles Discharge per day in cu. ft. Alti- tude of springs above sea level in feet Level in Rome in feet Aqua Appia Anio Vetus Aqua Marcia. . . Aqua Tepula . . . Aqua Julia . . 312 B. C. 272-269 144-140 125 33 53,950 209,000 299,960 58,200 74,980 10.2 39.6 56.8 11.0 14 2 4,072,500 9,814,200 10,465,800 993,000 2,691,200 98 918 1043 495 1148 65 157 192 199 209 Aqua Virgo Aqua Absietina. 19 67,900 107,775 12.9 20.4 5,587,700 874,800 79 685 65 54 Aqua Claudia. . . Anio Novus .... 38-52 A. D. 38-52 A. D. 225,570 285,330 42.7 54.0 7,390,800 10,572,900 1050 1312 221 231 The construction of the earliest aqueducts was the simplest, most of them being underground. In the Aqua Appia only 300 ft. were above ground, and in the Anio Vetus only 1100 ft. were above ground. In the Aqua Marcia 7.5 miles were supported on arches; in the Aqua Claudia 10 miles were on arches, and in the Anio Novus 9.5 miles were on arches. The con- struction of the last shows the greatest engineering skill, as it follows a winding course, at certain points tunnelling through hills and at others cross- ing ravines 300 ft. deep. 138 ELEMENTS OF' HYDRAULICS The cross section of the channels (specus) varied at different points of the course, that of the largest, the Anio Novus, being 3 to 4 ft. wide and 9 ft. high to the top, which was of pointed shape. The channels were lined with hard cement (opus signinum) containing fragments of broken brick. The water was so hard that it was necessary to clean out the calcareous deposits frequently, and for this purpose shafts or openings were constructed at intervals of 240 ft. Filtering and settling tanks (pisdnce limarioe, or "purgatories") were constructed on the line of the aqueduct just outside the city, and within the city the aqueducts ended in huge distributing reservoirs (Castella) from which the water was conducted to smaller reservoirs for distribution to the various baths and fountains. Supposing the population of Rome and suburbs to have then numbered one million, there was a daily water supply of nearly 400 gal. per capita. Modern Rome with a population of half a million has a supply of about 200 gal. per capita. The volume of water may also be compared with that of the Tiber which discharges 342,395,000 gal. per day, whereas in the First Cen- tury, A. D., the aqueducts carried not less than 392,422,500 gal. per day, which by the Fourth Century had been increased by additional supplies to 461,628,200 gal. per day. i Assuming that the Aqua Claudia had an average width of 3 ft. with 6 ft. depth of water, and that the grade was uniform and the difference in head lost in friction, calculate from the values tabu- lated above the velocity of flow and Chezy's constant C in the formula v = C Vrs. 91. A channel of trapezoidal section with side slopes of two horizontal to one vertical is required to discharge 100 cu. ft. per second with a velocity of flow of 3 ft. per second. Assuming Chezy's constant as 115, compute the required bottom width of channel and its longitudinal slope. 92. A channel of trapezoidal cross section has a bottom width of 25 ft. and side slopes of 1 to 1. If the depth of water is 6 ft. and the longitudinal inclination of the bed is 1 in 5000, find the discharge, assuming the coefficient of roughness, n, in Kutter's formula to be 0.02. 93. A channel of rectangular section has a bottom width of 20 ft., depth of water 6 ft. and longitudinal slope of 1 in 1000. Calculate the discharge, assuming the coefficient of roughness, n, in Kutter's formula to be 0.01. 94. A reservoir A supplies another reservoir B with 400 cu. ft. of water per second through a ditch of trapezoidal section, with earth banks, 5 miles long. To avoid erosion, the flow in this channel must not exceed 2 ft. per second. HYDROKINETICS 139 From reservoir B the water flows to three other reservoirs, C, D, E. From B to C the channel is to be rectangular in section and 4 miles long, constructed of unplaned lumber, with a 10-ft. fall and a discharge of 150 cu. ft. per second. From B to D the channel is to be 5 miles long, semicircular in section and constructed of concrete, with 12-ft. fall and a discharge of 120 cu. ft. per second. From B to E the channel is to be 3 miles long, rectangular in section and constructed of rubble masonry, with 15 ft. fall and a discharge of 130 cu. ft. per second. Find the proper dimensions for each channel section. 95. The flow through a circular pipe when completely filled is 25 cu. ft. per second at a velocity of 9 ft. per second. How much 0.2 0.4 0.6 0.8 1.0 1.2 [ Ratio of Depth to Diameter | ^ s^ Y 1.0 0.8 O.G 0.4 0.2 \ / I \ ^ ^ i \ Q x / A / ^ / . X / / ' / ^ / / / / r / / ^ S f ^ ^ ^ 0.2 0.4 0.6 0.8 1.0 1.2 Ratio of Q and V to their Values when Pipe is Full FIG. 122. would it discharge if rilled to 0.8 of its depth, and with what velocity? Solution. Fig. 122 shows a convenient diagram for solving a problem of this kind graphically. 1 The curve marked v (velocity) is plotted from Kutter's simplified formula 100Vr\ ,- p I Vrs + Vr/ for a value of b of 0.35, and the discharge Q from the formula Q = Av, the ordinates to the curves shown in the figure being the Imhoff. Taschenbuch fur Kanalisations Ingenieure. 140 ELEMENTS OF HYDRAULICS ratio of the depth of the stream to the diameter of the pipe, and the abscissas the ratios of Q and v respectively to their values when the pipe flows full. To apply the diagram to the problem under consideration, observe that for a depth of 0,8d the abscissa of the discharge curve is unity, and consequently the discharge for this depth is the same as when the pipe is completely filled. The abscissa of the velocity curve corresponding to this depth 0.8d (i.e., with abscissa 0.8) is 1.13, and consequently the velocity at this depth is 1.13 X 9 = 10.17 ft. per second. Similar diagrams have been prepared by Imhoff for a large variety of standard cross sections and are supplemented by other .diagrams or charts which greatly simplify ordinary sewer calculations. Problem 96. In the Catskill Aqueduct, which forms part of the water supply system of the City of New York, there are four FIG. 123. Cut and cover conduit, Catskill aqueduct. distinct types of conduit; the cut-and-cover type, grade tunnel, pressure tunnel, and steel pipe siphon. The cut-and-cover type, shown in section in Fig. 123, is 55 miles in length, and is constructed of concrete and covered with an earth embankment. This is the least expensive type, and is used wherever the eleva- tion and nature of the ground permits. The hydraulic data for the standard type in open cut is as follows : HYDROKINETICS 141 s = 0.00021 Depth of Area of Wetted Hydraulic flow flow perimeter . radius in feet in sq. ft. in feet in feet 17.0 241.0 57.4 4.20 Full, 16.2 237.7 50.8 4.67 Max. cap. 15.3 230.9 47.7 4.84 14.0 217.6 44.1 4.92 12.0 192.6 39.4 4.88 10.0 163.2 35;0 4.65 8.0 130.7 30.9 4.24 6.0 97.1 26.8 3.61 4.0 62.2 22.8 2.72 2.0 27.0 18.8 1.47 In the preliminary calculations the relative value of Chezy's coefficient for this type was assumed to be C = 125. Using this value, calculate the maximum daily discharge. Problem 97. Where hills or mountains cross the line of the Aqueduct, tunnels are driven through them at the natural eleva- tion of the Aqueduct (Fig. 124). There are 24 of these grade tunnels, aggregating 14 miles. The hydraulic data for the stand- ard type of grade tunnel is as follows : s = 0.00037 Depth of flow in feet Area of flow in sq. ft. Wetted perimeter in feet Hydraulic radius in feet 17.0 16.25 15.3 .198.6 195.6 188.5 52.2 46.0 42.7 3.80 4.25 4.41 Full, Max. cap. 14.0 175.7 39.3 4.46 12.0 152.4 35.0 4.35 10.0 126.8 31.0 4.10 8.0 100.2 26.9 3.72 6.0 73.8 22.9 3.22 4.0 47.6 18.9 2.51 - 2.0 21.0 14.9 1.49 The relative value of Chezy's coefficient for this type was assumed in the preliminary calculations to be C = 120. Using this value, calculate the maximum daily discharge and the corresponding velocity of flow. Problem 98. Where the line of the Aqueduct crosses broad and deep valleys and there is suitable rock beneath them, circular 142 ELEMENTS OF HYDRAULICS tunnels are driven deep in the rock and lined with concrete (Fig. 125). There are seven of these pressure tunnels, with an FIG. 124. Grade tunnel, Catskill FIG. 125. Pressure tunnel, Catskill aqueduct. aqueduct. aggregate length of 17 miles. The hydraulic data for these pressure tunnels is as follows: Slope Diameter Area of waterway Wetted perimeter Hydraulic radius . 00059 14 ft. 6 in. 165.1 sq. ft. 45.55 ft. 3.625 ft. Assuming the relative value of Chezy's coefficient to be C = 120, calculate the velocity of flow and the daily discharge. \*Width j. 5--*< On steep hillside ~-Hcif sect/at at joint -holes shown dotted FIG. 126. Steel pipe siphon, Catskill aqueduct. Problem 99. In valleys where the rock is not sound, or where for other reasons pressure tunnels are impracticable, steel pipe siphons are used (Fig. 126). These are made of steel plates riveted together, from 7/16 to 3/4 of an inch in thickness, and HYDROKINETICS 143 are 9 ft. and 11 ft. in diameter respectively. These pipes are embedded in concrete and covered with an earth embankment, and are lined with 2 in. of cement mortar as a protection to the steel and also for the sake of smoothness. There are 14 of these siphons aggregating 6 miles in length, and three pipes are required for the full capacity of the Aqueduct. Assuming three mortar lined 11-ft. pipes, having a relative coefficient of = 120 and a slope s = . 00059, calculate the velocity of flow through them and the maximum daily discharge. Problem 100. A broad shallow stream has naturally a depth of 3 ft. and a longitudinal slope of 5 ft. per mile. If a dam 8 ft. high is erected across the stream, determine the rise in level one mile up stream assuming the value of the constant C in Chezy's formula as 75. SECTION III HYDRODYNAMICS 29. PRESSURE OF JET AGAINST STATIONARY DEFLECTING SURFACE Normal Impact on Plane Surface. When a jet of water strikes a stationary flat plate or plane surface at right angles, the water spreads out equally in all directions and flows along this plane surface, as indicated in Fig. 127. The momentum of the water after striking the surface is equal to the sum of the momenta of its separate parti- cles, but since these flow off in opposite directions their alge- braic sum is zero. Consequently the entire momentum of a jet is destroyed by normal impact against a stationary plane sur- face. To find the pressure, P, ex- erted by the jet on the surface, let A denote the cross section of the jet and v its velocity. Then the mass of water flowing per unit of time is yAv FIG. 127. M = (J and, consequently, from the principle of impulse and momentum, Pdt = Mv = iA^_ For uniform or steady flow, P is constant, and if M denotes the quantity flowing per unit of time, then t is unity. In this case the above expression for the hydrodynamic pressure P of the jet on the surface becomes p = (77) 144 HYDRODYNAMICS 145 If h denotes the velocity head, then h =o~> an< ^ ^^ (^ may be written P = 27 Ah. (78) Relation of Static to Dynamic Pressure. If the orifice is closed by a cover or stopper, then the hydrostatic pressure P' on this cover is approximately equal to the weight of a column of water of height h and cross section A ; and consequently P' = 7 Ah. (79) Comparing Eq. (78) and (79), it is apparent that the normal hydrodynamic pressure of a jet on an external plane surface is twice as great as the hydrostatic pressure on this surface would be if it was shoved up against the opening so as to entirely close the orifice. In deriving this relation, the coefficient of efflux is assumed to be unity; that is, the area A of the jet is assumed to be the same as that of the orifice, and the velocity v to be the full value corresponding 'to the head h. Since the coefficient is actually less than unity, the hydrodynamic pressure never attains the value given by Eq. (78). For instance, in the case of flow from a standard orifice, if A denotes the area of the orifice and a the cross section of the jet, then from Arts. 8 and 9 a = 0.62^1, and v = 0.97 Therefore the expression for P becomes vav 2 , , p = L - _ 27(0.62^.) (0.97 2 A) 9 instead of 2yAh, as given by Eq. (78). Note, however, that this apparently large discrepancy is due chiefly to the fact that the area A in Eq. (78) denotes the cross section of the jet, whereas in Eq. (79) it denotes the area of the orifice. If the area A in both expressions denotes the cross section of the jet, Eq. (78) is practically true, and the hydrodynamic pressure is approxi- mately twice the hydrostatic pressure on an equal area. Oblique Impact on Plane Surface. If a jet strikes a stationary plane surface obliquely, at an angle a (Fig. 128), the axial velocity v of the jet may be resolved into two components, v sin a normal to the surface, and v cos a tangential to the surface. If the surface is perfectly smooth, the water flowing along the surface experiences no resistance to motion, and the pressure, P, exerted on the surface is that corresponding to the normal velocity 10 146 ELEMENTS OF HYDRAULICS component v' = v sin a. The area to be considered, however, is not a right section, A, of the jet, but a section A' normal to the component v sin a, as *A ' /'/'W * m ^icated in Fig. 128. The total pressure, P, exerted on the surface, is then or, since v' = v sin a and A' = , this may be written in sin the form P = g sin a. (80) If a = 90, this reduces to Eq. (77). Axial Impact on Surface of Revolution. If the surface on which the jet impinges is a surface of revolution, coaxial with the jet (Fig. 129), then in this case also the particles spread out FIG. 129. equally in all directions, and consequently the sum of the momenta of the particles in the direction perpendicular to the axis of the jet is zero. The velocity of any particle in a direction parallel to the axis of the jet, however, becomes v cos a, where a denotes the angle which the final direction taken by the particles makes with their initial direction, as indicated in Fig. 129. The total initial momentum is then Mv = ^f, HYDRODYNAMICS 147 and the total final momentum is yAv 2 Mv cos a = - - cos a. g Therefore, equating the impulse to the change in momentum, we obtain the relation p = (i _ COS a). (81) For a jet impinging normally on a plane surface, a = 90, and this expression reduces to Eq. (77). Complete Reversal of Jet. If a is greater than 90, then cos a becomes negative and the pressure P is correspondingly FIG. 130. increased. ' For example, if the direction of flow is completely reversed, as shown in Fig. 130, then a = 180, cos a = 1, and hence 27 Av 2 P = (82) The hydrodynamic pressure in this case is therefore twice as great as the normal pressure on a flat surface, and four times as great as the hydrostatic pressure on a cover over an orifice of the same area as the cross section of the jet. Deflection of Jet. When a jet is deflected in an oblique direction, the final velocity v may be resolved into components v cos a and v sin a, as indicated in Fig. 131. The component of the final momentum parallel to the initial direction of the jet is then Mv(l cos a) = yAv' (1 COS a), 148 ELEMENTS OF HYDRAULICS and the horizontal component, H, exerted in this direction is H = g (l COS a). (83) Similarly, the component of the final momentum perpendicular to the initial direction of the jet is ,, . jAv 2 . Mv sm a = - sin a, V cosef FIG. 131. and the vertical component, V, exerted in this direction is V = g , sin a. (84) The total pressure of the jet on the deflecting surface, or reaction of the surface on the jet, is, then, P = V# 2 +F 2 = 1^- V(l - cos a) 2 + sin 2 a tJ which simplifies into P = - V2(l - COS a). (85) o A more convenient expression for P may be obtained by using the trigonometric relation +1 p^ - = sin ^a, by means of which Eq. (85) may be written in the form (86) HYDRODYNAMICS 149 Dynamic Pressure in Pipe Bends and Elbows. When a bend or elbow occurs in a pipe through which water is flowing, the change in directon of flow produces a thrust in the elbow, as in the case of the deflection of a jet by a curved vane, considered in the preceding article. From Eq. (85) and (86), the amount of this thrust P is g V2(l - cos a) = a sin H, and the direction of the thrust evidently bisects the angle a, as indicated in Fig. 132. FIG. 132. In the case of jointed pipe lines if the angle of deflection is large or the velocity of flow considerable, this thrust may be sufficient to disjoint the pipe unless provision is made for taking up the thrust by some form of anchorage, as, for example, by filling in with concrete on the outside of the elbow. 30. PRESSURE EXERTED BY JET ON MOVING VANE Relative Velocity of Jet and Vane. In the preceding article it was assumed that the surface on which the jet impinged was fixed or stationary. The results obtained, however, remain valid if the surface moves parallel to the jet in the same or opposite direction, provided the velocity, v, refers to the relative velocity between jet and surface. Thus if the surface moves in an opposite direction to the jet with a velocity v', the relative velocity *of jet and surface is v + v' and the pressure is correspondingly increased, whereas if they move in the same direction, their relative velocity is v v', and the pressure is diminished. 150 ELEMENTS OF HYDRAULICS Work Done on Moving Vane. Consider, for example, the case of a jet striking a deflecting surface and assume first that this surface moves in the same direction as the jet with velocity v r (Fig. 133). Since the surface, or vane, is in motion, the mass of water, M' } reaching the vane per second is not the same FIG. 133. as the mass of water, M f passing a given cross section of the jet per second. That is, the mass, M, issuing from the jet per second is g whereas the mass, M', flowing over the vane per second is M , = yA(v -^ g Therefore the components of the force acting on the vane, given by Eq. (83) and (84), become in this case H = M'(v - ')(!- cos a) = (v - v')\l - cos a), yA Q V = M'(v - */) sin a = J ~(v - v'Y sin a. Since the motion of the vane is assumed to be in the direction of the component H, the component V, perpendicular to this direction, does no work. The total work, W, done on the vane by the jet is therefore W = Hv' = TAv/(v ~ V/)2 (i - cos a). (88) Speed at which Work Becomes a Maximum. The condition that the work done shall be a maximum is ^r = = (1 - cos a)[(v - v'Y - 2v'(v - v')], whence v' = - (89) HYDRODYNAMICS 151 % Substituting this value of v r in Eq. (88), the maximum amount of work that can be realized under the given conditions is found to be A- TT7 3 / *V/1 *yAt; Wmax = - (v gj (1 cos a) = 27 (1 cos a). Maximum Efficiency for Single Vane. The efficiency of a, motor or machine is denned in general as Useful work Efflciency = Total energy available" ' . (9o) Since, in the present case, the total kinetic energy of the jet is the efficiency, E, becomes 4jAv 3 -277 ( ~~ E = - - = (l _ cos a). yAv 3 27 The maximum efficiency occurs when a 180, in which case Emax = ~_ = 59-2 per cent. (91) Maximum Efficiency for Continuous Succession of Vanes. If there is a series of vanes following each other in succession so tthat each receives only a portion of the water, allowing this portion to expend its energy completely on this vane before leaving it, then the mass M' in Eq. (87) is replaced by M, and the component H becomes H == M(v - v')(l - cosa)= yAv(v ~^~ (I - cos a). The work done on the series of vanes is therefore W . HT' = . (I - cos a). (92) *** . The condition for a maximum in this case is dW Nr/ /N ., -^7- = = l (1 - cos a)[(v - v f ) - v'], whence v (93) 152 ELEMENTS OF HYDRAULICS Substituting this value of v' inEq. (92), the maximum work which can be realized from a series of vanes moving parallel to the jet is yA v 2 2 w '' (-I) /i \ /-, \ (1 COS a) = 7 (1 COS a). Hence the efficiency in this case is ! (1 COS a) E 2(7 The maximum efficiency therefore occurs when a 180, its value being x =- (2) = ioo per cent. (94) The actual efficiency of course can never reach this upper limit, as the conditions assumed are ideal, and no account is taken of frictional and other losses. Impulse Wheel; Direction of Vanes at Entrance and Exit. In general, it is not practicable to arrange a series of vanes so Shaft FIG. 134. as to move continuously in a direction parallel to the jet. As usually constructed, the vanes are attached to the circumference of a wheel revolving about a fixed axis (Fig. 134). Let co denote the angular velocity of the wheel about its axis, and n, r 2 the radii HYDRODYNAMICS 153 of the inner and outer edges of the vanes. Then the tangential or linear velocities at these points, say HI and u^, are u\ rico; Uz = 7*200. Now let Vi denote the absolute velocity of the jet at entrance to the vane, and v% its absolute velocity at exit. Then by forming a parallelogram of velocities on u\ and Vi as sides, the relative velocity, Wi, between jet and vane at entrance is determined, as indicated in Fig. 134. Similarly, the parallelogram on Uz, Wz as sides determines the absolute velocity, v 2 , at exit. In order that the water may glide on the vane without shock, the tip of the vane at entrance must coincide in direction with the vector wi. Work Absorbed by Impulse Wheel. Let M denote the mass of water passing over the vane per second. At entrance the velocity of this mass in the direction of motion (i.e., its tangential velocity) is v\ cos a, and at exit is Vz cos 0, where a and ]8 are the angles indicated in Fig. 134. The linear momentum of the mass M at entrance is then Mvi cos a, and its angular momentum is MV&I cos a. Similarly, its linear momentum at exit is Mvz cos and its angular momentum is MV^TZ cos /3. The total change in angular momentum per second, that is, the amount given up by the water or imparted to the wheel, is then MviTi cos a MvzTz cos |8. For a continuous succession of vanes, as in the case of an ordinary impulse wheel, the mass M is the total amount of water supplied by the jet per second. Hence, if T denotes the total torque exerted on the wheel, by the principle of angular impulse and momentum, remembering that M is the mass of water flowing per unit of time, and consequently that the time is unity, T = M(viii cos a-v 2 r 2 cos 0). (95) The total work imparted to the wheel is W= Tu, or, since M = - > u\ riw, u 2 = r 2 co, the expression for the y work becomes W = Tco = 1 (UiVi COS a - U 2 V 2 COS |8). (96) These relations will be applied in Art. 34 to calculating the power and efficiency of certain types of impulse wheels. 154 ELEMENTS OF HYDRAULICS 31. REACTION OF A JET Effect of Issuing Jet on Equilibrium of Tank. Consider a closed tank containing water or other liquid, and having an orifice in one side closed by a cover. When the cover is removed the equilibrium of water and tank will be destroyed. At the instant of removal this is due to the disappearance of the pressure pre- viously exerted on the cover considered as part of the tank. After the jet has formed and a steady flow has been set up, as- suming that the depth of water is maintained constant by supply- ing an amount equal to that flowing out, as indicated in Fig. 135, the pressure within the fluid and on the walls of the tank will not regain its original static value, since, in accordance with Ber- noulli's theorem, an increase in velocity must be accompanied by a corresponding decrease in pressure. Energy of Flow Absorbed by Work on Tank. To calculate the effect of the flow on the equilibrium of the tank, suppose that the FIG. 135. tank is moved in the direction opposite to that of the jet and with the same velocity, v, as that of the jet. Then the relative velocity of the jet with respect to the tank is still v, but the absolute velocity of the jet is zero and consequently its kinetic energy is also zero. If h denotes the head of water on the orifice (Fig. 135) and Q the quantity of water flowing per second, then its loss in potential energy per second is yQh. Moreover, while this volume of water Q moved with the tank, it had a velocity v, and therefore possessed kinetic energy of amount -~ The total energy given up by the water in flowing from the tank is then HYDRODYNAMICS 155 v 2 or, since h = ~ approximately, these terms are equal, and the Z Q total energy lost by the water becomes yQv 2 Energy given up = - i/ Now let P denote the reaction of the jet, that is, the resultant of all the pressure exerted on the tank by the water except that due to its weight. Then, since the distance traversed by the force P in a unit of time is the velocity v of the tank, by equating the work done by P to the energy given up by the water, we have whence = = = g g The reaction P is therefore twice the hydrostatic pressure due to the head h. This is also apparent from the results of Art. 27, since the pres- sure of a jet on a fixed surface close to the orifice must be equal to its reaction on the vessel from which the jet issues. The actual reaction of the jet is of course somewhat less than its theoretical value, as given by the relation P = 2yAh, since there are various losses, due to internal friction, etc. Principle of Reaction Turbine. In order for the tank to retain its uniform velocity, v, a resistance of amount P must constantly be overcome, for if the resistance is less than this amount the motion will be accelerated. It is apparent, therefore, that by a proper choice of the velocity, v, of the tank it is possible to utilize almost the entire energy of the jet in overcoming a resistance coupled up with the tank. This is the principle on which the reaction turbine is based, as explained in Art. 35. It should be noted that if the water flowing out is continually replaced from above, half of the available energy must be used in giving the supply water the same velocity as the tank. The use- ful work is therefore reduced to one-half the previous amount, and the available energy is only that due to the velocity head h. Barker's Mill. The simplest practical application of the reaction of a jet is the apparatus known as Barker's mill (Fig. 136). In this apparatus water flows from a tank into a hollow vertical arm, or spindle, pivoted at the lower end, and from this into a 156 ELEMENTS OF HYDRAULICS SIDE ELEVATION o horizontal tubular arm, having two orifices near the ends on opposite sides. The jets issue from these orifices, and their reactions cause the horizontal arm to rotate, driving the central spindle from which the power is taken off by a belt and pulley. The steam turbine invented by Hero of Alexandria in the first cen- tury B. C., is an almost identical arrangement, the motive power in this case being due to the reaction of a jet of steam instead of a jet of water. 32. TYPES OF HYDRAULIC MOTORS Current Wheels. There are three general types of hydraulic motors, namely 1. Current and gravity wheels, 2. Impulse wheels and turbines, 3. Reaction turbines. The current wheel is the oldest type of prime mover, and in its primitive form consisted of a large vertical wheel, with a set of paddles or buckets attached to its circumference, and so placed in a running stream that the current acting on the lower, or im- mersed, portion produced revolution of the wheel. A later im- provement consisted in placing the wheel at the foot of a water fall and conducting the water by a flume to the top of the wheel, the action of the water in this case being due almost entirely to its weight (Art. 33). Impulse Wheels. The impulse wheel, as its name indicates, is designed to utilize the impulsive force exerted by a jet moving with a high velocity and striking the wheel tangentially. The wheel, or runner, in this case carries a series of curved buckets or vanes which discharge into the atmosphere. A feature of this type is that the runner rotates at a high velocity and can there- fore be made of comparatively small diameter. The two prin- cipal types of impulse wheel are the Girard impulse turbine, which originated in Europe, and the Pelton wheel, which was developed in the United States (Art. 34). PLAN F!G. 136. HYDRODYNAMICS 157 Reaction Turbines. The reaction turbine depends chiefly on the reaction exerted by a jet on the vessel from which it flows, which in this case is the passage between the vanes on the runner. In an impulse wheel the energy of the water as it enters the wheel is entirely kinetic, and as there is free circulation of air between the vanes and they discharge into the atmosphere, the velocity of the water is that due to the actual head. In a reaction turbine the energy of the water as it enters the wheel is partly kinetic and partly pressure energy, and as the water completely fills the passages between the vanes, its velocity at entrance may be either greater or less than that due to the static head at that point. A feature of the reaction turbine is that it will operate when completely submerged. Classification of Reaction Turbines. Reaction turbines are subdivided into four classes, according to the direction in which the water flows through the wheel. These are (1) Radial outward flow turbines, (2) Radial inward flow turbines, (3) Parallel or axial flow turbines, (4) Mixed flow turbines, the direction of flow being partly radial and partly axial, changing from one to the other in passing over the vanes (Art. 35). Classification of Hydraulic Motors. The following tabulated classification is useful as a basis for the description of the various types of hydraulic motors given in Arts. 33, 34 and 35. Current and Gravity Wheels: Utilizes impact of current or weight of the water. Impulse Wheels and Turbines: Utilizes kinetic energy of jet at high velocity. Suitable for limited amount of water under high head. Ordinarily used for heads from 300 ft. to 3000 ft. Reaction Turbines: Utilizes both kinetic and pressure en- ergy of water. Suitable for large quantities of water under low or me- dium head. Ordinarily used for heads from 5 to 500 ft. Current wheel, Undershot wheel (Poncelet), Breast wheel, Overshot wheel. Girard turbine (European), Pelton wheel (American). Radial inward flow (Francis type), Radial outward flow (Fourneyron type), Parallel or axial flow (Jonval type), Mixed flow (American type). 158 ELEMENTS OF HYDRAULICS 33. CURRENT AND GRAVITY WHEELS Current Wheels. The vertical current wheel, mentioned in Art. 32, was the earliest type of hydraulic motor, dating from prehistoric times, although they are still in use in China and Syria. Undershot Wheels. The first improvement consisted in confining the water in a sluice and delivering it directly on the vanes. This type was known as the Undershot wheel, and was in common use until about the year 1800 A. D. (Fig. 137). Flat radial vanes were used with this type, for which the maximum theoretical efficiency was 50 per cent., the velocity of the vanes to realize this efficiency being one-half the velocity of the stream, as explained in Art. 29. The actual efficiency of such wheels was much lower, being only from 20 to 30 per cent. ^^^^^^^^^^^^^^%^^% FIG. 137. FIG. 138. Poncelet Wheels. Undershot wheels were greatly improved by Poncelet who curved the vanes, so that the water entered without shock and was discharged in a nearly vertical direction (Fig. 138). The water thus exerted an impulse on the vanes during the entire time it remained in the wheel, thereby raising the actual efficiency to about 60 per cent. Undershot wheels of the Poncelet type are adapted to low falls, not exceeding 7 ft. in height. Breast Wheels. A modification of the undershot wheel is the Breast wheel, the water being delivered higher up than for an ordinary undershot wheel, and retained in the buckets during the descent by means of a breast, or casing, which fits the wheel as closely as practicable (Fig. 139). Wheels of this type are known as high breast, breast, and low breast according as the water is delivered to the wheel above, at, or below the level of the center HYDRODYNAMICS 159 of the wheel. The high breast wheel operates almost entirely by gravity, that is, by the unbalanced weight of the water in the buckets, its efficiency being from 70 to 80 per cent. Breast and low breast wheels operate partly by gravity and partly by impulse, the efficiency varying from about 50 per cent, for small wheels to 80 per cent, for large wheels. This type was in use until about 1850. Overshot Wheels. A more recent type is the Overshot wheel, the characteristic of this type being that the water is delivered at the top of the wheel by a sluice, as indicated in Fig. 140. For I Head Race, Head Breast FIG. 139. FIG. 140. maximum efficiency the diameter of the wheel should be nearly equal to the height of the fall, the efficiency for well-designed overshot wheels ranging from 70 to 85 per cent., which is nearly as high as for a modern turbine. An overshot wheel at Troy, N. Y., is 62 ft. in diameter, 22 ft. wide, weighs 230 tons, and develops 550 H.P. Another on the Isle of Man is 72 ft. in diameter and develops 150 H.P. 34. IMPULSE WHEELS AND TURBINES Pelton Wheel. The intermediate link between the old type of water wheel and the modern impulse wheel was the Hurdy Gurdy, which was introduced into the mining districts of Cali- fornia about 1865. This somewhat resembled the old current wheel, being vertical with flat radial vanes, but differed from it in that it was operated by a jet impinging on the vanes at high velocity. The maximum theoretical efficiency of the Hurdy Gurdy was 50 per cent. (Art. 29), while its actual efficiency varied from 25 to 35 per cent. 160 ELEMENTS OF HYDRAULICS The substitution of curved buckets for the flat radial vanes was the great improvement which converted the Hurdy Gurdy into the Pelton wheel. The construction of the bucket is shown in Fig. 141, the jet being divided by the central ridge and each half deflected through nearly 180. Evidently the angle of deflection must be slightly less than 180, so that the discharge from one bucket may clear the one following. A later improve- ment is the Doble bucket, also shown in Fig. 141, each half of Pelton Bucket Doble Bucket FIG. 141. which is ellipsoidal in form, with part of the outer lip cut away so as to clear the jet when coming into action. The relation of the jet to the wheel is shown in Fig. 142, the type there shown being arranged with a deflecting nozzle for eco- nomic regulation. A more recent type of Pelton wheel is shown in Fig. 143, the features of this type being the Doble buckets and the so-called chain type of attachment of the buckets. One of the most important features of construction in this type of impulse wheel is the needle valve for regulating the flow. The HYDRODYNAMICS 161 T FIG. 142. FIG. 143. Pelton-Doble runner, diameter 153 in., weight 28,000 Ibs. 162 ELEMENTS OF HYDRAULICS cross section shown in Fig. 144 indicates the location of the needle valve with respect to the nozzle. The methods of operating the valve and of elevating and depressing the nozzle are shown in Fig. 145. This form of nozzle under the high heads ordinarily used gives a very smooth and compact jet, as shown by the instantaneous photograph reproduced in Fig. 146. Efficiency of Pelton Wheel. If the jet was completely reversed in direction and the speed of the buckets was one-half that of the jet, the theoretical efficiency of the Pelton wheel would be unity, or 100 per cent., as shown by Art. 30, Eq. (94). This is FIG. 144. also apparent from other considerations; for if the velocity of the jet is v and that of the buckets is -~ , then the velocity of the water relative to the lowest bucket is v ~-, or -~ (Fig. 147). Zi Z Therefore at exit the water is moving with velocity -^ relative to the bucket while the bucket itself is moving in the opposite direction with velocity -~ * Hence the absolute velocity of the water at exit is -^ -H , or zero, and therefore, since the total kinetic energy of the water has been utilized, the theoretical efficiency of the wheel is unity. As a matter of fact there are hydraulic friction losses to be taken into account and also the direction of flow is not completely reversed. The efficiency of the Pelton wheel has been found in a number of authentic tests to exceed 86 per cent. The actual efficiency in operation de- HYDRODYNAMICS 163 ^L FIG. 145. Pelton regulating needle nozzle. FIG. 146. 164 ELEMENTS OF HYDRAULICS pends of course on the particular hydraulic conditions under which the wheel operates. A good idea of what may be ex- pected in practice, however, is given by the following data, ob- tained from a unit of 4000 KW normal capacity, operating under a head of 1300 feet: Load in KW Percentage of normal capacity Wheel efficiency 5000 4000 3000 2000 125% 100% 75% 50% 81% 83% 82% 70% FIG. 147. Since the Pelton wheel operates by utilizing the kinetic energy of the water, it is best adapted to a small discharge under a high head. Girard Impulse Turbine. A somewhat different type of impulse wheel has been developed in Europe, known as the Girard impulse turbine. In this type the shaft may be mounted either vertically or horizontally, and the flow may be either radial or axial. The type shown in Fig. 148 is arranged for radial flow, with vertical shaft. The construction is practically the same in all cases, the water entering through a pipe B, as shown in Fig. 148, and proceeding through one or more guide passages C, which direct the water onto the vanes. The quantity of water admitted to the vanes is regulated by some kind of gate, that indicated in Fig. 148 being a sliding gate operated by a rack and pinion not shown in the figure. HYDRODYNAMICS 165 As the vanes are more oblique at exit than at entrance, they are necessarily closer together at exit. To prevent choking, it is therefore necessary to widen the vanes laterally at exit, as shown in elevation in Fig. 148. As the water discharges under atmospheric pressure, ventilating holes are made through the sides of the vanes at the back to allow free admission of air. FIG. 148. Power and Efficiency of Girard Turbine. The power and efficiency of a Girard turbine may easily be calculated from the results of Art. 30. Using the same notation, as indicated on Fig. 134, from Eq. (96), Art. 30, the work done per second on the wheel is given by the relation Work per second = - - (u\ Vi cos a u 2 Vz cos )* 166 ELEMENTS OF HYDRAULICS Since the water is under atmospheric pressure, the absolute velocity Vi of the water at entrance is calculated from the effective head h by means of the relation vi = ^J2gh. From Fig. 149 we have by geometry Vi cos a = Ui + Wi cos 6, v% cos /3 = u% Wz cos tf>, and from the law of cosines v^ = u^ + Wi 2 + 2u iWi cos 6. COS . FIG. 149. Now the total energy imparted to the wheel per second is the difference in kinetic energy at entrance and exit, namely Therefore, equating this to the expression for the work done per second, as given above, we have COS a COS Substituting in this equation the values given above for Vi cos a, Vz cos j3, Vi 2 and v 2 2 and reducing, the result is finally .2 _ (97) It is evident that the efficiency will be greater the more nearly the jet is reversed in direction, that is, the smaller

. v 2 = 2u 2 sin TT, and Ui = cos a Now the peripheral velocity of the inner and outer ends of the vanes in terms of the angular velocity co of the runner is given by the relations u\ = TI , u 2 = r 2 co, whence u 2 r 2 - = ' or u 2 = Substituting this value of u 2 in the expression given above for 02, we have 2uir 2 sin ^ Vir 2 sin ^ .

the efficiency, E, of the wheel, as defined by Eq. (90), Art. 30, is a 168 ELEMENTS OF HYDRAULICS Since E is less than unity, it is evident that the maximum theoretical efficiency is always less than 100 per cent., and also that the efficiency is greater the smaller the angles a and V2gh (lOO) where

Ul == 7 == C and consequently Ui is about 3 per cent, more than )c 2 V/i. It is customary, however, to express the height of a runner in terms of its diameter as b = c s d, where the coefficient c 3 is a constant for homologous runners of a given type. For American reaction turbines c 3 varies from about 0.10 to 0.30. Substituting this value for b in the expression for the discharge, it becomes Q = (7rCiC 2 c 3 )d 2 V/i . Therefore if the constant part of this expression is denoted by k q , it may be written Q = M 2 Vh: (102) The coefficient k q may be called the capacity constant of the run- ner, and for any given runner may be computed from the relation d'Vh For American reaction turbines the capacity constant ranges in value from about 2 to 4. Since k q has approximately the same HYDRODYNAMICS 185 value for all runners of a given type, it serves as a criterion for comparing the capacities of runners of different types. Characteristic Speed. The speed constant and capacity constant taken separately are not sufficient to fix the require- ments of combined speed and capacity. That is to say, two runners may have different values of k v and k q and yet be equivalent in operation. To fix the type, therefore, another criterion must be introduced which shall include both k v and k q . The most convenient combination of these constants is that introduced by Professor Camerer of Munich and the well- known turbine designer, Mr. N. Baashuus of Toronto, Ontario. This criterion may be obtained as follows. 1 The horse power of a turbine is given by the expression 62.37QA6 550 where e denotes the hydraulic efficiency of the turbine. If the horse power, discharge Q and head h are given, the efficiency may be calculated from this relation by writing it in the form 550ff.P. ~ 62.37 Qh If the efficiency is known, the constants in the above expression for the horse power may be combined into a single constant k, and the equation written in the form H.P. = kQh. (104) When the efficiency is not known it is usually assumed as 80 per cent., in which case k = yj- From Eq. (101) the speed in r.p.m. is given by the relation 6ok v Vh and from Eq. (103) the nominal diameter of the runner is given as Eliminating d between these two relations, we have therefore 60 /b v V 1 S. G. Zowski : A Comparison of American Highspeed Runners for Water Turbines. Engineering News, Jan. 28, 1909, pp. 99-102. 186 ELEMENTS OF HYDRAULICS Moreover, from Eq. (104) we have H.P. Q kh ' and substituting this value of Q in the preceding expression for n, we have finally n = ( - - ) 7~=- (106) The expression in parenthesis is a constant for any given type and may be denoted by N a , in which case we have For any given type of turbine this constant N a may be calculated from the relation n VH.P. (I07) Various names have been proposed for this constant N s such as "type constant" and "type characteristic." In Germany, where its importance as determining the type and performance of a turbine seems to have first been recognized, it is called the specific speed (spezifische Geschwindigkeit, or spezifische Um- laufzahl). This term, however, is not entirely satisfactory to American practice, as it seems desirable to use the term specific speed in another connection, as explained in what follows. The term for the constant N 8 favored by the best authorities as more fully describing its meaning is characteristic speed, which is there- fore the name adopted in this book. 1 For impulse wheels the characteristic speed ranges in value from about 1 to 5, while for radial, inward flow turbines its value lies between 10 and 100. Specific Discharge. It is convenient to express the discharge, power, speed, etc., in terms of their values under a 1 ft. head. The discharge under a 1 ft. head is called the specific dis- charge, and its value is found by substituting h = I in Eq. (102). 1 The use of the term characteristic speed has been recommended to the author by the well known hydraulic engineer Mr. W. M. White, who is using this term in preparing the American edition of the German handbook "de Hutte," and strongly advocates its general adoption in American practice. HYDRODYNAMICS 187 Consequently if the specific discharge is denoted by Qi, its value is Oi = M 2 , and therefore Q = QiVh. (108) For reaction turbines the specific discharge ranges in value from 0.302d 2 for the slowest speeds to 2.866d 2 for the highest speeds, the diameter d being expressed in feet. Specific Power. Similarly, the power developed under 1 ft. head is called the specific power and will be denoted in what fol- lows by H.P.i. From Eq. (104) we have H.P. = kQh and since from Eq. (102) Q = M 2 VJf , by eliminating Q between these two relations we have H.P. = kk g d 2 h^h. Substituting h = 1 in this equation, the specific power is there- fore given as H.P.! = kk q d\ and consequently H.P. = ILP.ihVh. (109) Specific Speed. By analogy with what precedes, the speed under 1 ft. head will be called the specific speed and denoted by HI. Substituting h = 1 in Eq. (105), we have therefore 60/b, HI = -T-' ird and consequently n = niVh. (no) For reaction turbines the specific speed ranges in value from 78 147 r for the lowest speeds, to j- for the highest speeds, the diameter d being expressed in feet. Relation between Characteristic Speed and Specific Speed. From the relation 188 ELEMENTS OF HYDRAULICS the characteristic speed N s may be defined in terms of the quan- tities defined above as specific. Thus, assuming h = I and H.P. = 1, we have N s = n, expressed in r.p.m. Therefore, the characteristic speed is the speed in r.p.m. of a turbine dimin- ished in all its dimensions to such an extent as to develop 1 H.P. when working under a head of 1 ft. Since it is apparent from Eq. (106) that N B stands for the combination where A: is a function of the efficiency e, the characteristic speed N s is an absolute criterion of turbine performance as regards speed, capacity and efficiency. From Eq. (107), however, it is evident that N s may be calculated directly from the speed, power and head without knowing the actual dimensions of the runner, its discharge, or its efficiency. Classification of Reaction Turbines. The characteristic speed N 8 may be used as a means of classifying both the various types of impulse wheels and of reaction, or pressure, turbines. In the following table practically all the different kinds of pressure turbines of the radial inward flow type are classified by their characteristic speeds, the corresponding efficiencies being also given in each case. 1 Type of pressure Characteristic Efficiency turbine speed, N Maximum At half power Low speed 10 - 20 82 per cent, at 3/4 power 76 per cent. Medium speed. . . High speed 30 - 50 60 - 80 82 per cent, at 3/4 power 80 per cent, at .8 power 75 per cent. 70 per cent. Very high speed. 90 -100 73 per cent, at . 9 power 53 per cent. The values of the constant N 8 in this table refer to the maximum power of one runner only. In case the characteristic speed is higher than 100, it is necessary to use a multiple unit. At maximum power, the efficiencies are slightly lower than the maximum efficiencies given above. From this table it is apparent that low speed turbines show a favorable efficiency over a wide range of loads but are prac- tically limited to high heads, whereas high speed turbines are efficient at about 0.8 load but show a notable decrease in effi- 1 Water Power Plants and the Type Characteristics of Turbines. N. Baashuus, Engineering News, Mar. 2, 1911, pp. 248-250. HYDRODYNAMICS 189 ciency at half load. The use of the latter is therefore indicated for low heads where the water supply is ample at all seasons. Classification of Impulse Wheels. In a way similar to the preceding, the characteristic speed may be used to classify the various types of impulse wheels, as indicated in the following table. 1 Impulse wheels N g | 1 | 2 | 3 | 4 | 5 Efficiency at 3/4 load. 80 per cent. 79 per cent. 78 per cent. 77 per cent. 76 per cent. The numerical values of N 8 in this table refer to the maximum power of one nozzle only. In case the characteristic speed lies between 5 and 10 it is therefore necessary to use more than one nozzle. Numerical Applications. To illustrate the use of the preceding numerical data, suppose that it is required to determine the proper system of hydraulic development for a power site with an available flow of 310 cu. ft. per second under an effective head of 324 ft. The power capacity in this case is Hp= 324X310 =9m | Of this amount about 100 H.P. will be required for exciter and lighting purposes. There would therefore be installed two. exciter units running at 550 r.p.m., one of which would be a re- serve unit. The characteristic speed for these units would then be 550 /100 N * = 3lWl = Since this lies between 1 and 5, an impulse wheel would be used for driving the exciter generators. The main development of 9000 H.P. would be divided into three units of 3000 H.P. each, running at 500 r.p.m., with a fourth unit as a reserve. The characteristic speed for these main units would then be 500 /3000 N * ~ 310 \17.6 = As N s lies between 10 and 100, a pressure turbine would be used for driving the main generators. 1 Water Power Plants and the Type Characteristics of Turbines. N. Baashuus, Engineering News, Mar. 2, 1911, pp. 248-250. 190 ELEMENTS OF HYDRAULICS As a second illustration, suppose that an impulse wheel is required to develop 1300 H.P. under a 400 ft. head at an efficiency of not less than 78 per cent. From the preceding table for im- pulse wheels it is apparent that it is necessary to use a wheel having a characteristic speed of about 3. If a single wheel and nozzle is used, the speed in r.p.m. at which it must run is found from Eq. (107) to be n H.P. 3 X 400 150 r.p.m. If two nozzles are used, each furnishes half of the power and the corresponding speed is 150 V 2 = 212 r.p.m. perating Rai ; ! Overload 10 20 30 40 50 60 70 80 90 100 Per Cent Turbine Load FlG. 164. With four nozzles acting on two runners the required speed would be 150V4 = 300 r.p.m., and for 6 nozzles acting on 3 runners, n = 150V6 = 367 r.p.m. Since the value of N, is the same in each case, the efficiency is practically 78 per cent, in each case although there is a wide difference in the speed and setting. Normal Operating Range. Having determined the proper type of development, it is necessary, in case a reaction turbine is used, to determine the required size and type of runner to HYDRODYNAMICS 191 develop maximum efficiency under the given conditions of operation. For a turbine direct connected to a generator, the capacity of the turbine, in general, should be such as to permit the full over- load capacity of the generator to be developed and at the same time place the normal operating range of the unit at the point of maximum efficiency of the turbine, as indicated in Fig. 164. The normal horse power, or full-load, here means the power at which the max- imum efficiency is attained, any excess power being regarded as an overload. When the supply of water is ample but the head is low, efficiency may to a certain extent be sacrificed to speed and capacity in order that the greatest power may be developed from each runner, thereby reducing the investment per horse power of the installation. On the other hand, when the flow of water is insufficient to meet all power requirements, an increase in efficiency shows a direct financial return in the increased output of the plant. FIG. 165. FIG. 166. Selection of Stock Runner. Ordinarily it is required to select a stock runner which will operate most favorably under the given conditions. To explain how an intelligent selection of size and type of runner may be made from the commercial constants given by manufacturers, the following runner table of a standard make of turbine is introduced. 1 (Page 193.) 1 The Allis-Chalmers Company, Milwaukee. 192 ELEMENTS OF HYDRAULICS The cut accompanying each of the six types given in the table shows the outline of runner vane for this type. To indicate its relation to the runner and to the turbine unit as a whole, Fig. FIG. 168. 165 shows a typical cross section of runner; Fig. 166 shows how this is related to the casing; and Fig. 167 shows a cross section of the entire turbine unit. The runner is also shown in per- spective in Fig. 168. HYDRODYNAMICS 193 TYPE "A" RUNNER nk TYPE"B" RUNNER r' TYPE V RUNNER -j-w- * -" l> \ v ^/ I IliV X Z JJL^S N 5 v II 4 N a '= 13.55 N s = 20.3 N a = 29.4 Ui - 0.585 Ui = 625 Ui = 0.665 Diam. | R.P.M.i H.P.i | Qi R.P.M.i H.P.i Ql RP.M.i H.P .i Qi 15 ^71.7 0.0358 0.394 76.6 0.0705 0.776 81.4 0.130 1.43 18 "59.8 0.0514 0.565 63.8 0.105 1.155 67.8 0.187. 2.06 21 51.2 0.0705 0.776 54.7 0.138 1.523 58.2 0.225 2.48 24 44.8 0.0915 1.007 47.8 0.182 2.00 51.0 0.333 3.66 27 39 . 8 0.116 1.276 42.5 0.229 2.52 45.2 0.423 4.65 30 35.8 0.142 1.562 38.3 0.284 3.12 40.7 0.520 5.72 34 31 _6 0.184 2.024 33.8 0.363 3.99 35.9 0.668 7.35 38 28.3 0.230 2.53 30.2 0.453 4.98 32.2 0.835 9.19 42 25.6 0.280 3.08 27.4 0.551 6.06 29.1 1.016 11.18 46 23.4 0.336 3.69 25.0 0.665 7.32 26.6 1.225 13.48 50 21.5 0.398 4.38 23.0 0.7b 8.69 24.4 1.450 15.95 55 19.5 0.480 5.28 20.9 0.95 10.45 22.2 1.745 19.20 60 17.9 0.573 6.30 19.1 1.13 12.43 20.4 2. OS \ 22.88 65 16.5 0.672 7.39 17.7 1.33 14.63 18.8 2.44 26.84 70 15.4 0.785 8.64 16.4 1.53 16.83 17.5 2.82 31.00 TYPE" D" RUNNER j . j_ TYPE"E" RUNNER TYPE" F" RUNNER ^ V \ h" > \ f) r ' a kr \ s ^9- ^ Q v ^ ' N a = 40.7 Ns = 51.760 5 N 8 = 71 . 479 Ui - 0.70 Ui = 0.75 - Ui = 85 Diam.| R.P.M.i |HJ>.i Qi Diam. R.P.M.i H.P.i I Qi R.P.M.i H.P.i Qi 14 98.4 0.277 3.05 111.5 0.410 4.51 16 86.1 0.367 4.04 97.7 0.541 5.95 18 76.5 0.471 5.18 86.8 0.704 7.74 15 85 7 0.226 2.49 20 69.0 0.597 6.57 78.1 0.912 10.03 18 71 4 0.324 3.56 22 62.6 0.731 8.04 71.0 1.133 12.46 21 61.3 0.442 4.86 24 57.4 0.883 9.70 65.1 1.375 15.13 24 53.6 0.577 6.35 26 53.0 .055 11.60 60.1 1.62 17.85 27 47 6 0.731 8.04 28 49.2 .243 13.67 55.8 1.93 21.25 30 42 8 0.902 9.92 30 46.0 .436 15.80 SS.l 2.20 24.20 34 37 8 1. 158 12.74 32 43.0 .65 18.15 48.8 2.55 28.10 38 33 9 1.444 15.88 34 40.5 .8 9 20.80 46.0 2.82 31.10 42 30 6 1.765 19.4 36 38.3 2.15 23.65 te.5 3.14 34.55 46 28 2.12 23.3 38 36.3 2.42 26.60 41.1 3.52 38.70 50 25 7 2.50 27.5 40 34.4 2.75 30.25 39.1 3.93 43.20 55 23 4 3.04 33.4 421/2 32.4 3.09 34.0 36.8 4.33 47.60 60 21 4 3.61 39.7 45 30.6 3.53 38.8 34.7 4.92 54.10 65 19.8 4. 22 46.4 47 1/2 29.0 4.01 44.1 32.9 5.66" 62.25 70 18 4 4.90 53.9 50 27.6 4.45 49.0 31.2 6.13 67.40 52 1/2 26.3 4.95 54.5 29.8 6.75 74.25 55 25.1 5.52 60.7 28.4 7.50 82.50 57 1/2 24.0 6.10 67.1 27.2 8.16 89.75 60 23.0 6.80 74.8 26.0 8.94 98.30 13 194 ELEMENTS OF HYDRAULICS From Eq. (107) it is evident that, other things being equal, the characteristic speed for high heads will be relatively small whereas for low heads it will be large. Thus in the runner table above, type "A." with a characteristic speed of 13.55 is adapted to high heads, running up to 600 ft., while at the other end of the series, type "F" with a characteristic speed of about 75, is adapted to effective heads as low as 10 ft. To give a numerical illustration of the use of the runner table, suppose it is required to determine the type of runner and the speed in r.p.m. to develop 750 H.P. under a head of 49 ft. In this case h^h = 49V49 = 343, and consequently H.P. 750 H - Rl -/*V/i = 343 = 2 ' 2 > which corresponds to a 30-in. type "F" runner. Referring to the table for this type and size, we have n\ = 52.1, from which the required speed is found to be n = 52. lV^ = 364.7, say 360 r.p.m. If twin turbines were used, we would have 750 which corresponds to a 22-in. type "F" runner, having a speed of n = 71.0Vft~ = 497, say 500 r.p.m. As a second illustration, let it be required to find from the table the type of runner and speed to develop 4000 H.P. under an effective head of 300 ft. In this case h-Jh = 300V300 = 5190, and consequently the specific power is H.P. 4000 H - Rl == AV/T := 5190 = ' 77 ' which corresponds to a 50-in. type "B" runner. Referring to the table for this type and size we have H.P.i = 0.79, and m = 23, and consequently the power and speed for this type and size is H.P. = 0.79ft V/T = 4100, and n = 23Y/T= 395, say 400 r.p.m. HYDRODYNAMICS 195 37. POWER TRANSMITTED THROUGH PIPE LINE AND NOZZLE Effective Head at Nozzle. Hydraulic power is frequently delivered through a pipe line and nozzle (as, for instance, when power is developed by an impulse wheel), in which case there is considerable loss due to pipe friction and other causes. To investigate the amount of this loss and the condition for maximum efficiency, let D = inside diameter of pipe in feet, I = length of pipe in feet, h static head at nozzle in feet, v = velocity of flow in feet per second. To simplify the solution it is customary to neglect the slight loss of head at entrance to the pipe and in the nozzle, as both of these terms are small in comparison with the head lost in pipe fric- tion. From Eq. (48), Art. 17, the head lost in friction in a pipe of length I is given by the expression I v 2 Lost friction head = / IV o~~ where / denotes an empirical constant (Eq. (49), Art. 17, and Table 12). Therefore the effective head at the nozzle is Effective head = h - f^- (in) Velocity of Flow for Maximum Power. Since the discharge Q is irD 2 Q = ~-A~vy Ib. per second, the horse power delivered at the nozzle is TrDVy /, ,1 H.P. at nozzle ==^ T56 (h ~ f The value of v for which the horse power is a maximum is found from the calculus condition d(H.P.) dv =0 ' namety, dv ~ 4X550 V" J D2g whence 2ghD 3fl ' 196 , ELEMENTS OF HYDRAULICS Maximum Efficiency. Substituting this value of v in the expression for the horse power, its maximum value is found to be If there was no friction, the total available horse power would be Total H.P. - g^- The efficiency E is therefore _ actual H.P. J_ v^_ "total H.P. " ' .X2) V the maximum value of which, when v = -* / ^ , is I 2 " wuzz. power ~ I ~ ~~ = ~T~' (,^^4/ 3 3 Diameter of Nozzle for Maximum Output of Power.--To investigate conditions at the nozzle, let p = pressure in pounds per square foot before entering the nozzle, A = area of cross section of pipe, a = area of cross section of nozzle, v = velocity of water in pipe, V = velocity of water through nozzle. Then since all the water in the pipe discharges through the nozzle, we have Av = aV, or - -, a ~~ v and therefore, from Bernoulli's theorem, p v 2 V 2 (A\ 2 v 2 ~y Jr 2~g == ~2g = ' \a) 2g and also 2 J 4 i2 Therefore, by subtraction, we have /A\ 2 v 2 ,1 v 2 (a) 2g =h ~ f D2g whence a = Av 2ghD -fW HYDRODYNAMICS 197 For maximum power the velocity v, as given by Eq. (112), must be v = +1 9 , and inserting this value of v in the expression just \ oft obtained for the nozzle area a, the condition for maximum output of power is ^ (115) For a circular pipe and nozzle we have a Trd 2 , , irD 2 - and A = - -. > where d denotes the nozzle diameter; hence for maximum output we must have (116) 30,000 25,000 20,000 15,000 10,000 5,000 100 2 60 .3 40 20 10 15 20 25 30 Velocity in Feet per Second FlG. 169. 40 45 Graphical Relation between Power and Efficiency. The rela- tion between horse power and efficiency is shown graphically in Fig. 169 for the installation of the Vancouver Power Co. The length of pipe line in this case was I = 6678 ft., diameter D = 4 ft., and head h = 1214 ft. Assuming / =0.024, the horse power at the nozzle is found from the equation H.P. at nozzle = ~^ Q (h - f 5-^) = 1.42570(1214 - 0.62220 2 ). 198 ELEMENTS OF HYDRAULICS This is a maximum when " 25.5, its value being 29,423 H.P. The efficiency in per cent, is found from the relation E = (l - /^- = (1 - 0.0005124^)100. The horse power and efficiency, plotted on a velocity base, are shown in Fig. 169. Method of Determining Nozzle Diameter. In finding the nozzle diameter, d, a tentative value of / may be assumed as in the example just cited (see Eq. (48), Art. 17), and the cor- responding value of v found from Eq. (112). The value of / corresponding to this velocity may then be determined from Table 12, and with this value of/, the nozzle diameter may be determined from Eq. (116). Note that the nozzle diameter so determined gives the maximum output of power for the size of pipe specified, but that more power may be developed from the same nozzle by using a larger supply pipe. 38. EFFECT OF TRANSLATION AND ROTATION Equilibrium under Horizontal Linear Acceleration. Consider the equilibrium of a body of water having a motion of translation as a whole but with its particles at rest relatively to one another, such, for example, as the water in the tank of a locomotive tender when in motion on a straight level track. If the speed is constant, the forces acting on any particle of the liquid are in equilibrium, and conditions are the same as when the tank is at rest. If the motion is accelerated, how- ever, every particle of the liquid must experience an inertia force proportional to the acceleration. Thus, if the acceleration is denoted by a, the inertia force F acting on any particle of mass m, according to Newton's law of motion, is given by the relation F = ma. a FIG. 170. HYDRODYNAMICS 199 For a particle on the free surface of the liquid (Fig. 170), the inertia force F acting on this particle must combine with its weight W into a resultant R having a direction normal to the free surface of the liquid. From the vector triangle shown in the figure we have F = W tan a, and by Newton's law whence by division tan a = (117) Equilibrium under Vertical Linear Acceleration. If the tank is moving vertically upward or downward, the surface of the liquid will remain horizontal. If the motion is uniform, that is, with constant velocity, the conditions will be the same as though the tank was at rest. If it is moving upward with acceleration a, the surface will still remain horizontal but the pressure on the bottom of the tank will be increased by the W amount ma = a ) where W denotes the weight of a column of y water of unit cross section and height equal to the depth of water in the tank. Thus if p denotes the pressure on the bottom W of the tank in pounds per square inch, then since ma = a = yh a, we have ("8) If the acceleration is vertically downward, the pressure on the bottom of the tank is diminished by the amount yh () , its */ value being (119) Free Surface of Liquid in Rotation. If the tank is in the form of a circular cylinder of radius r, and revolves with angular velocity co about its vertical axis Y Y (Fig. 171), the free surface of the liquid will become curved or dished. To find the form assumed by the surface, let P denote any particle on the free 200 ELEMENTS OF HYDRAULICS surface at a distance x from the axis of rotation. Then if m denotes the mass of this particle, the centrifugal force C acting on it is n 9 W C = mx co 2 = #co 2 . g From the vector triangle shown in the figure we have W fj n I and since the slope of the surface curve is ~ = tan 0, we have as its differential equation ~ = tan = do; g whence, by integration, its algebraic equation is found to be y = co ^- (120) The surface curve cut out by a diametral section is therefore a parabola with vertex in the axis of rotation, and the free surface is a paraboloid of revolution. Y E 10 FIG. 172. Depression of Cup below Original Level in Open Vessel. Since the volume of a paraboloid is half the volume of the circumscribing cylinder, the volume of liquid above the level OX of the vertex (Fig. 172) is Vol. OCDEF = HYDRODYNAMICS But if AB is the level of the liquid when at rest, then 201 where k denotes the depth of the cup below the original level, and therefore 2,.,2 r'co 4g (121) Consequently the depth of the cup below the original level is proportional to the square of the angular velocity. Depression of Cup below Original Level in Closed Ves- sel. If the top of the tank is closed and the angular velocity increased until the liquid presses against the top, as shown in J? Fig. 173, the surface will still j c remain a paraboloid. If the total depth of the cup is denoted by H and its greatest radius by R, then since its volume must be the same as that of the cup of depth h, we have the relation FIG. 173. whence hr 2 = HR* ^ tt '' H' ,2-2 But from the equation of the surface curve, y = ~^~, by substituting the simultaneous values y H, x = R, we have H = ^-, and substituting in this relation the value of R z from the previous equation, the result is H = ^> whence H = o>r A /^. (122) 202 ELEMENTS OF HYDRAULICS Therefore after the liquid touches the top cover of the tank, the total depth of the cup is proportional to the first power of the angular velocity. Practical Applications. An important physical application of these results consists in the formation of a true parabolic mirror by placing mercury in a circular vessel which is then rotated with uniform angular velocity, the focus of the mirror depending on the speed of rotation. Another practical application has been found in the con- struction of a speed indicator. A glass cylinder containing a colored liquid is mounted on a vertical spindle which is geared to the shaft whose speed is required. The required speed is then obtained by noting the position of the vertex of the para- boloid on a vertical scale. From the level AB to the level CD (Fig. 173) the graduations on the scale are at unequal distances apart, as apparent from Eq. (121), but below this point they are equidistant, as shown by Eq. (122). 39. WATER HAMMER IN PIPES Increase in Pressure due to Suddenly Checking Flow. If water is flowing in a pipe with uniform velocity and the flow is suddenly stopped, as by the closure of a valve, the pressure in the pipe is greatly increased. If the velocity of flow is denoted by v and the mass of water in the pipe by m, the momentum of this mass will be mv. Assuming that the pipe is rigid and the water incompressible, if the flow is stopped in a length of time t, the increased pressure F near the valve may be obtained from the principle of impulse and momentum, namely Ft = mv. (123) From this relation it appears that if the valve is closed instan- taneously, that is, t = 0, the pressure F is infinite since the right member, mv, is finite and different from zero. Bulk Modulus of Elasticity of Water. In reality, however, this result is not valid, since the pipe walls are elastic and the water compressible. Thus it has been found by experiment that a pres- sure of one atmosphere, or 14.7 Ib. per square inch, on each face of a cube of water at 32 F. causes it to lose about 0.00005 of its original volume. Consequently the bulk modulus of elasticity of water, B, defined as _ _ unit Stress unit volume deformation HYDRODYNAMICS 203 has for its numerical value B = - = 294,000 Ib. per square inch. (124) 0.00005 Pressure Waves in Pipe Produced by Suddenly Checking Flow. What actually happens when the flow in a pipe is suddenly shut off, is an increase in pressure, beginning at the valve, which compresses the water and distends the pipe. Beginning at the valve, this effect travels back toward the reservoir or supply, producing a wave of compression in the water and a wave of distortion in the pipe. When all the water in the pipe has been brought to rest, the total kinetic energy originally possessed by the flowing water is stored up in the elastic deformation of the water and pipe walls. Since this condition cannot be maintained under the actual head in the reservoir, the pipe then begins to contract and the water to expand, thereby forcing the water back into the reservoir until it acquires a velocity approximately equal to its original velocity but in the opposite direction, that is, back toward the reservoir. After this wave has traversed the pipe, the water again comes to rest, but the kinetic energy ac- quired by the flow toward the reservoir will have reduced the pressure below normal. Consequently water again enters the pipe from the reservoir and flows toward the valve, beginning a new cycle of operations. Velocity of Compression Wave. An expression for the velocity of the wave of compression and distortion has been deduced by Professor I. P. Church, 1 and is given by the formula , '--VicSrid) ' <'*> where v c = velocity of pressure wave, in feet per second. B = bulk modulus of elasticity of water = 294,000 Ib. per square inch, E = modulus of elasticity in tension of pipe material, c = thickness of pipe wall in inches, d = internal diameter of pipe in inches, 8 = -^r-r = 0.4333. 144 Period of Compression Wave. The period of the pressure Church: Hydraulic Motors, p. 203. 204 ELEMENTS OF HYDRAULICS wave, or time T required for a wave to make a complete round trip from one end to the other and back again, is evidently T - f- 1 , (126) Vc where I denotes the length of the pipe in feet. Increase in Pressure Produced by Instantaneous Stoppage of Flow. Professor Church has also derived a formula for the in- crease in pressure due to an instantaneous stop, namely, <5vv c / 5BEa ( , P~ = -y - v Vg(Ec+ BdT (I27) g(Ec+ where v = velocity of flow before closure, p max = increase in pressure in pounds per square inch in excess of original hydrostatic pressure. Joukovsky 's Experiments. Elaborate experiments on water hammer in pipes, made by Professor Joukovsky 1 of Moscow, Russia, in 1897-98, confirm -these theoretical results, and also show that if the time t of closure is less than the period T (Eq. (126)), the full force of the maximum pressure (p max , Eq. (127)) will be felt in the pipe. But if the time t of closure is greater than the period T, the intensity of excess pressure, p, is reduced ac- cording to the relation that is, the excess pressure is inversely proportional to the time of closure. Gibson's Experiments. From more recent experiments, made by A. H. Gibson 2 of Manchester, England, it has been found that when the time of stoppage of flow is greater than 2 T 4:1 or , the excess pressure p is given by the formula <- where a = cross section of pipe, i = area through valve when closing begins, h = difference in static head in feet between the two sides of valve when there is no flow, t = time of uniform closing of valve. 1 Joukovsky: Abstract by Simin; Jour. Amer. Water Works Assn., 1904, p. 335. 2 Gibson: Water Hammer in Hydraulic Pipe Lines, Van Nostrand, 1909. HYDRODYNAMICS 205 The best precaution against hydraulic shock of this nature has been found to be the use of slow closing valves. Air chambers placed near the valves have also been found effective if kept filled with air, and safety valves of course reduce the shock to a pres- sure corresponding to the strength of spring used. 40. HYDRAULIC RAM Principle of Operation. A, useful application of water hammer is made in the hydraulic ram. In principle, a hydraulic ram is an automatic pump by which the water hammer produced by sud- denly checking a stream of running water is used to force a portion of that water to a higher elevation. Drive Pipe Connection A Discharge Pipe F Air Chamber E Escape Valve C Delivery Valve D Air Feeder H FIG. 174. Rife hydraulic ram. To illustrate the method of operation, a cross section of a ram is shown in Fig. 174. The ram is located below the level of the supply water in order to obtain a flow in the drive pipe. If located some distance from the supply, the water is first conducted to a short standpipe, as shown in Fig. 175, and from here a drive pipe of smaller diameter than the supply pipe conducts the water to the ram. The object of this arrangement is to utilize the full 206 ELEMENTS OF HYDRAULICS head of water available without making the drive pipe too long for the capacity of the ram. Referring to Fig. 174, the water flowing in the drive pipe A at first escapes around the valve C, which is open or down. This permits the velocity of flow to increase until the pressure against C becomes sufficient to raise it against its seat B. Since the water can then no longer escape through the valve C, it enters the Spring Strainer FIG. 175. air chamber E through the valve D, thereby increasing the pres- sure precisely as in the case of water hammer discussed in the preceding article. When the pressure in E attains a certain maxi- mum value, the flow is checked and the valve D falls back into place, closing the opening and trapping the water which has already entered the chamber E. The pressure in E then forces this water into the supply pipe F, which delivers it at an elevation proportional to the pressure in E. Hydraulic rams are also so built that they can be operated from one source of supply and pump water from a different source (Fig. 176). Muddy or impure water from a creek or stream may thus be used to drive a ram, and the water pumped from a pure spring to the delivery tank. Efficiency of Ram. The mechanical efficiency of a ram depends on the ratio of fall to pumping head, ranging from 20 per cent, for a ratio of 1 to 30, up to 75 per cent, for a ratio of 1 to 4. HYDRODYNAMICS 207 Its efficiency as a pump is of course very small, as only a small fraction of the water flowing in the drive pipe reaches the delivery pipe. The advantages of the hydraulic ram are its small first cost, simplicity of operation, and continuous service day and night without any attention. To obtain an expression for the mechanical efficiency of a ram, let H = supply head, h = effective delivery head including friction, q = quantity delivered, Q = quantity wasted at valve. Then the total input of energy to the ram is (Q + q)H, and the total output is qh. Consequently the mechanical efficiency is given by the ratio v - ~ This is known as d'Aubuisson's efficiency ratio. The hydraulic efficiency, however, is the ratio of the energy required for delivery to the energy of the supply. Consequently its value is QH The latter expression is known as Rankine's formula. 41. DISPLACEMENT PUMPS There are two types of pumps in general use; the displacement, or reciprocating type, and the centrifugal type. In the displace- ment pump the liquid is raised by means of a bucket, piston, or plunger, which reciprocates backward and forward inside a cylindrical tube called the pump barrel or cylinder. In the centrifugal pump, as its name indicates, the operation depends on the centrifugal force produced by rotation of the liquid. Suction Pump. One of the simplest forms of displacement pump is the ordinary suction pump shown in Fig. 177. Here the essential parts are a cylinder or barrel C containing a bucket jB, which is simply a piston provided with a movable valve, permitting the water to pass through in one direction only. This bucket is made to reciprocate up and down inside the barrel by means of a rod E. A suction pipe S leads from the lower 208 ELEMENTS OF HYDRAULICS end of the barrel to the liquid to be raised, and a delivery pipe D discharges the liquid at the desired elevation. In operation, the bucket starts from its lowest position, and as it rises, the valve m closes of its own weight. The closing of this valve prevents the air from entering the space below the bucket, and consequently as the bucket rises the increase in volume below it causes the air confined in this space to expand and thereby lose in pressure. As the pressure inside the suction pipe S thus becomes less than atmospheric, I j- ^^5=^ the pressure outside forces some of jo E3-t=l r "D ? the liquid up into the lower end of the pipe. When the bucket reaches the top of its stroke and starts to descend, the valve n closes, trapping the liquid already in the suction pipe S and also that in the barrel, thereby lifting the valve m as the bucket descends. When the bucket reaches its lowest position, it again rises, repeating the whole cycle of operations. At each repetition the water rises higher as it replaces the air, until finally it fills the pump and a continuous flow is set up through the delivery pipe. Maximum Suction Lift. Since atmospheric pressure at sea level is 14.7 Ib. per square inch, a pump operating by suction alone cannot raise water to a height greater than the head cor- responding to this pressure. Since a cubic foot of water weighs 62.4 Ib., the head corresponding to a pressure of one atmosphere is FIG. 177. h = 62.4 " 0.434 144 ft '> which is therefore the maximum theoretical height to which water can be lifted by suction alone. As there are frictional and other losses to be considered, the actual suction lift of pumps is only about two- thirds of this amount, the practical lift for differ- ent attitudes and pressures being as given in the following table. HYDRODYNAMICS 209 Altitude Barometric pressure Equivalent head of water Practical suction lift of pumps Sea level .... 1/4 mile. . . 1/2 mile... 3/4 mile... 1 mile 14.70 Ib. per sq. in. 14.02 Ib. per sq. in. 13.33 Ib. per sq. in. 12.661b. per sq. in. 12.02 Ib. per sq. in. 33.95ft. 32.38ft. 30.79ft. 29.24ft. 27.76ft. 22ft. 21ft. 20ft. 18ft. 17ft. 1-1/4 miles. 1-1/2 miles. 2 miles. . . . 11.42 Ib. per sq. in. 10.88 Ib. per sq. in. 9.881b. per sq. in. 26.38ft. 25.13ft. 22 . 82 ft. 16ft. 15 ft. 14ft. Force Pump. When it is necessary to pump a liquid to a height greater than the suction lift, or when it is desired to equalize the work between the up and down strokes, a combination suction and force pump may be used, as shown in Fig. 178. In the simple type here illustrated, the bucket is replaced by a solid piston, the movable valves being at m and n as shown. On the up stroke of the piston the valve m closes and the pump operates like a simple suction pump, filling the barrel with liquid. When the piston starts to descend, the valve n closes, and the liquid in the barrel is there- fore forced out through the valve m into the delivery pipe D. By making the suction and pressure heads equal, the piston can therefore be made to do the same amount of work on the down as on the up stroke; or the entire suction head may be utilized and the pressure head made whatever may be necessary. Stress in Pump Rod. To find the pull P on the pump rod E for the type shown in Fig. 177, let A denote the area of the bucket and hi, h 2 , the heads above and below the bucket, as indicated in the figure. Then the downward pressure PI on top of the bucket is FIG. 178. Pi = 14.74 + 62.4A 2 A, 14 210 ELEMENTS OF HYDRAULICS and the upward pressure P 2 on the bottom of the bucket is P 2 = 14.7 A - 62.4/hA. Therefore the total pull P in the rod is P = Pi - P* = 62.4A(Ai + h*) = Q2AAh. If I denotes the length of the stroke, the work done per stroke is then work per stroke = PI = 62AAhl. For the combined suction and pressure type shown in Fig. 178, the pressure in the rod on the down stroke is P = 62.4AA 2 , and the tension in the rod on the up stroke is P = Q2AAhi. Direct Driven Steam Pump. The modern form of recipro- cating power pump of the suction and pressure type is the FIG. 179. direct driven steam pump, illustrated in Figs. 179 and 180. In this type the steam and water pistons are on opposite ends of the same piston rod and therefore both have the same stroke, although their diameters are usually different. Until recently this was the standard type of general service pump, being used HYDRODYNAMICS 211 for all pressures and capacities, from boiler feed pumps to muni- cipal pumping plants. Although the centrifugal type is rapidly taking its place for all classes of service, the displacement pump is the most efficient where conditions demand small capacity at a high pressure, as in the operation of hydraulic machinery. Fig. 181 illustrates the use of a displacement pump in connection with a hydraulic press. The best layout in this case would be FIG. 180. to use a high-pressure pump and place an accumulator (Art. 3) in the discharge line between pump and press. The press cylinder can then be filled immediately at the maximum pressure and the ram raised at its greatest speed, the pump running meanwhile at a normal speed and storing excess power in the accumulator. Calculation of Pump Sizes. To illustrate the calculation of pump sizes, suppose it is required to find the proper size for a duplex (i.e., two cylinder, Fig. 180) boiler feed pump to supply a 100-H.P. boiler. 212 ELEMENTS OF HYDRAULICS For large boilers the required capacity may be figured as 34-1/4 Ib. of water evaporated per hour per horse power. For small boilers it is customary to take a larger figure, a safe practical rule being to assume 1/10 of a gallon per minute per boiler horse power. ^ C @ C ( ) <8 ) <2 5 < ^ L- . 1 || HI rti ,Kevcrse Valve To Pressure Governor Steam Inlet Supply to I'russ .^ Relief Valve o7scharge|| jil ,Exhnnst FIG. 181. In the present case, therefore, a 100-H.P. boiler would require a supply of 10 gal. per minute. Assuming 40 strokes per minute as the limit for boiler feed pumps, the required capacity is ^Q = 0.25 gal. per stroke. Therefore, assuming the efficiency of the pump as 50 per cent., the total capacity of the pump per stroke should be 25 Q~ = 0.5 gal. per stroke. Since we are figuring on a duplex, or two cylinder, pump, the required capacity per cylinder is -~ = 0.25 gal. per stroke per cylinder, z and consequently the displacement per stroke on each side of the piston must be 25 -~ = 0.125 gal. piston displacement. A HYDRODYNAMICS 213 Referring to Table 7 it is found that a pump having a water cylinder 2-3/4 in. in diameter, with a 5-in. stroke, will have the required capacity. Power Required for Operation. To find expressions for the H.P. and steam pressure required to operate a displacement pump, let Q = discharge of pump in gal. per minute, h = total pumping head in feet (including friction and suction head if any), D = diameter of steam piston in inches, d = diameter of water piston in inches, p = steam pressure in pounds per square inch, w water pressure in pounds per square inch, n = number of full strokes (i.e., round trips) per minute, c = number of pump cylinders (e.g., for duplex pump, Fig. 180, c = 2), I = length of stroke in inches, E = efficiency of pump. Since a gallon of water weighs 8.328 lb., the total work per minute required to raise the given amount Q to the height h is work = 8.328Q/i ft.-lb. per minute. Taking into account the efficiency of the pump, the actual horse power required is therefore * <> Diameter of Pump Cylinder. If the pump makes n full strokes per minute, the piston displacement per minute for each cylinder is 2n /7Td 2 \ far 1 ) and the actual effective displacement of the pump per minute is / fJ2 \ effective displacement =2ncE l^TH cu. in. per minute. Equating this to the required discharge Q, expressed in cubic inches per minute, we have ~l = 231Q, 214 ELEMENTS OF HYDRAULICS whence the required diameter of the pump cylinder in terms of the speed is found to be I2 - 13 Vl^E (13I) Steam Pressure Required for Operation. Since the total pressure on the steam piston cannot be less than that on the water piston, the minimum required steam pressure, p, is given by the relation /7rZ> 2 \ p(- 4 -) = o, whence (132) Numerical Application. To illustrate the application of these results, suppose it is required to determine the indicated horse power to operate a fire engine which delivers two streams of 250 gal. per minute each, to an effective height of 60 ft. Since the height of an effective fire stream is approximately four-fifths that of the highest drops in still air, the required head at the nozzle is 4/& = -^ X 60 = 75 ft. To this must be added the friction head h f lost in the hose between pump and nozzle, which is given by the relation (Art. 17) where I is the length of the hose and d its diameter, both expressed in inches, v is the velocity of flow through the hose, and/ is an empirical constant. For the best rubber lined hose, / = 0.02 for the first 100 ft. of hose and 0.0025 for each additional 100 ft. whereas for unlined hose/ = 0.04 for the first 100 ft. and 0.005 for each additional 100 ft. In the present case, assuming 100 ft. of the best 2-1/2-inch rubber-lined hose, we have / = 0.02, and since the quantity of water delivered is 231 Q = 250 X Tyno cu * ^' P er m i nu ^ e > and the area of the hose is ird 2 7r(2.5) 2 A = -j- = = 4.908 sq. in., HYDRODYNAMICS 215 ^ the velocity of flow in the hose is v = -- = 16.3 ft. per second. Consequently the friction head h f is and therefore the total pumping head H is H = 75 + 40 = 115 ft. From Eq. (130) the total horse power required, assuming a pump efficiency of 50 per cent., is then found to be H.P. = 0.00025 5 X 50 115 = 28.75. Assuming the efficiency of the engine to be 60 per cent., the total indicated horse power required would be 28.75 LILP - == "060" : 18 ' 42. CENTRIFUGAL PUMPS Historical Development. The centrifugal pump in its modern form is a development of the last 15 years although as a type it is by no means new. The inventor of the centrifugal pump was the celebrated French engineer Denis Papin, who brought out the first pump of this type in Hesse, Germany, in 1703. Another was designed by Euler in 1754. These were regarded as curiosi- ties rather than practical machines until the type known as the Massachusetts pump was produced in the United States in 1818. From this time on, gradual improvements were made in the centrifugal pump, the most important being due to Andrews in 1839, Bessemer in 1845, Appold in 1848, and John and Henry Gynne in England in 1851. Experiments seemed to show that the best efficiency obtainable from pumps of this type ranged from 46 to 64 per cent, under heads varying from 4-1/2 to 15 ft., and 40 ft. was considered the maximum head for practical operation. About the year 1901 it was shown that the centrifugal pump was simply a water turbine reversed, and when designed on similar lines was capable of handling heads as large, with an efficiency as 216 ELEMENTS OF HYDRAULICS high, as can be obtained from the turbines themselves. Since this date, great progress has been made in both design and construction, the efficiency of centrifugal pumps now ranging from 55 to over 90 per cent., and it being possible to handle heads as high as 300 ft. with a single stage-turbine pump and prac- tically any head with a multi-stage type. 1 The advantages of the centrifugal over the displacement type are its greater smoothness of operation, freedom from water hammer or shock, absence of valves, simplicity and compactness, and its adaptability for driving by belt or by direct connection to modern high-speed prime movers, such as steam turbines and electric motors. Under favorable conditions the first cost of a high-lift centrifugal pump may be as low as one-third that of a displacement pump, and the floor space occupied one-fourth that required by the latter. However, for small quantities of water discharged under a high head the displacement pump is preferable to the centrifugal type, as the latter requires too much compound- ing under such conditions. Principle of Operation. The principle on which the original centrifugal pumps of Papin and Euler operated was simply that when water is set in rotation by a paddle wheel, the centrifu- gal force created forces the water outward from the center of rotation. Appold discovered that the efficiency depended chiefly on the form of the blade of the rotary paddle wheel, or impeller, and the shape of the enveloping case, and that the best form for the blade was a curved surface opening in the opposite direction to that in which the impeller revolved, and for the case was a spiral form or volute. The first engineer to discover the value of compounding, that is, leading the discharge of one centrifugal pump into the suction of another similar pump, was the Swiss engineer Sulzer of Winterthur, who was closely followed by A. C. E. Rateau of Paris, France, and John Richards and Byron Jack- son of San Francisco, California. In its modern form, the power applied to the shaft of a cen- trifugal pump by the prime mover is transmitted to the water by means of a series of curved vanes radiating outward from the center and mounted together so as to form a single member called 1 Rateau found by experiment that with a single impeller 3.15 in. in diameter, rotating at a speed of 18,000 r.p.m., it was possible to attain a head of 863 ft. with an efficiency of approximately 60 per cent. Engineer, Mar. 7, 1902. HYDRODYNAMICS 217 Hollow arm impeller. Concave arm impeller. Sand pump impeller. Open impeller used in sewage pumps. Enclosed side suction impeller. Enclosed double suction impeller. FIG. 182. Impeller types. (Courtesy Morris Machine Works.) 218 ELEMENTS OF HYDRAULICS the impeller (Fig. 182). The water is picked up at the inner edges of the impeller vanes and rapidly accelerated as it flows between them, until when it reaches the outer circumference of the impeller it has absorbed practically all the energy applied to the shaft. Impeller Types. There are two general forms of impeller, the open and the closed types. In the former the vanes are attached to a central hub but are open at the sides, revolving between two stationary side plates. In the closed type, the FIG. 183. vanes are formed between two circular disks forming part of the impeller, thus forming closed passages between the vanes, extending from the inlet opening to the outer periphery of the impeller. The friction loss with an open impeller is considerably more than with one of the closed type, and consequently the design of pumps of high efficiency is limited to the latter. Conversion of Kinetic Energy into Pressure. As the water leaves the impeller with a high velocity, its kinetic energy forms a considerable part of the total energy and the efficiency of the pump therefore depends largely on the extent to which this kinetic energy is converted into pressure in the pump casing. In some forms of pump no attempt is made to utilize this kinetic energy, the water simply discharging into a concentric HYDRODYNAMICS 219 chamber surrounding the impeller, from which it flows into a discharge pipe. The result of such an arrangement is that only the pressure generated in the impeller is utilized and all the ki- netic energy of the discharge is dissipated in shock and eddy formation. Volute Casing. This loss of kinetic energy may be partially avoided by making the casing spiral in section, so that the sectional area of the discharge passage increases uniformly, FIG. 184. Double suction volute pump, built by the Platt Iron Works Co. making the velocity of flow constant (Fig. 183). This type of casing is called a volute chamber (Fig. 184). When the volute is properly designed, a high efficiency may be obtained with this type of casing. 1 Vortex Chamber. An improvement on the simple volute chamber is that known as the whirlpool chamber, or vortex chamber, suggested by Professor James Thomson. In this type the impeller discharges into a concentric chamber considerably larger than the impeller, outside of and encircling which is a volute chamber. In its original form this necessitated exces- 1 With the De Laval volute type of centrifugal pump shown in Fig. 185, efficiencies as high as 85 per cent, have been obtained under favorable conditions. 220 ELEMENTS OF HYDRAULICS FIG. 185. Longitudinal section of De Laval single-stage double-suction volute pump. FIG. 186. Longitudinal section of Alberger volute pump. HYDRODYNAMICS 221 sively large dimensions, but in a modified form it is now very generally used (Figs. 185 and 186). The effectiveness of this arrangement depends on the principle of the conservation of angular momentum. Thus, after the water leaves the impeller no turning moment is exerted on it (neglecting frictional resistance) and consequently as a given mass of water moves outward, its speed decreases to such an extent as to keep its angular momentum constant. For a well- designed vortex chamber, the velocity of the water at the outside FIG. 187. Diffusion vanes. of the diffusion space is less than the velocity of the water as it leaves the impeller in the inverse ratio of the radii of these points, and if this ratio is large, a large part of the kinetic energy of the discharge may therefore be converted into pressure head in this manner. This method of diffusion is therefore well adapted to the small impellers of high speed pumps, since the ratio of the outer radius of the diffusion chamber to the outer radius of the impeller may be made large without unduly increasing the size of the casing. Diffusion Vanes. Another method for converting the kinetic energy of discharge into pressure head consists in an application 222 ELEMENTS OF HYDRAULICS of Bernoulli's law as illustrated in the Venturi tube; namely, that if a stream flows through a diverging pipe the initial velocity head is gradually converted into pressure head without appreciable END SECTIONAL VIEW SIDE SECTJONAL VIEW FIG. 188. Alberger turbine pump. loss. To apply this principle to a centrifugal pump, the impeller is surrounded by stationary guide vanes, or diffusion vanes (Fig. 187), so designed as to receive the water without shock on leaving the impeller and conduct it by gradually diverging passages into a vortex chamber or volute casing. This type of HYDRODYNAMICS 223 construction is therefore essentially a reversed turbine, and is commonly known as a turbine pump (Fig. 188). The angle which the inner tips of the diffusion vanes make with the tangents to the discharge circle is calculated exactly as in the case of the inlet vanes of a turbine, that is, so that they shall be parallel to the path of the water as it leaves the impeller. As this angle changes with the speed, the angle which is correct for one speed is incorrect for any other and may actually obstruct the discharge. A turbine pump must therefore be designed for a particular speed and discharge, and when required to work under variable conditions loses considerably in efficiency. If the conditions are very variable, the vortex chamber type is preferable, both by reason of its greater average efficiency under such conditions and also on account of its greater simplicity and cheapness of construction. Discharge Suction FIG. 189. Worthington multi-stage turbine pump. Stage Pumps. Single impellers can operate efficiently against heads of several hundred feet, but for practical reasons it is desirable that the head generated by a single impeller should not exceed about 200 feet. When high heads are to be handled, therefore, it is customary to mount two or more impellers on the same shaft within a casing so constructed that the water flows successively from the discharge of one impeller into the suction 224 ELEMENTS OF HYDRAULICS of the next. Such an arrangement is called a stage pump, and each impeller, or stage, raises the pressure an equal amount. Fig. 189 shows a multi-stage pump of the turbine type and Fig. 190 one of the volute type. A single impeller pump may be either of the side suction or double suction type. In the latter, half of the flow is received on each side of the impeller which is therefore perfectly balanced im m iHi FIG. 190. De Laval multi-stage volute pump. against end thrust (Fig. 185). A side suction pump, however, is simpler in construction, and it is also possible to balance them hydraulically against end thrust (Fig. 186). In stage pumps the device sometimes used for balancing is to arrange the impellers in pairs so that the end thrust of one impeller is balanced by the equal and opposite end thrust of its mate. 43. PRESSURE DEVELOPED IN CENTRIFUGAL PUMP Pressure Developed in Impeller. The pressure produced in a centrifugal pump must be sufficient to balance the static and frictional heads. When there is no volute, vortex chamber or diffusor, the kinetic energy of the discharge is all dissipated and the entire change in pressure is produced in the impeller. If, however, the velocity of discharge is gradually reduced by means of one of these devices, a further increase in pressure is produced in the casing or diffusion space, and if a diverging discharge pipe is used the pressure is still further increased. The change in pressure which is produced in passing through the impeller may be deduced by applying Bernoulli's theorem. For this purpose it is convenient to separate the total difference HYDRODYNAMICS 225 in pressure between the inlet and discharge circles into two com- ponents; one due to the rotation of the water in a forced vortex with angular velocity co, and the other due to the outward flow, FIG. 191. Detail of labyrinth rings in pump shown in Fig. 190. ^ FIG. 192. i.e.,the relative motion of the water with respect to the vanes of the impeller. Let the subscripts 1 and 2 refer to points on the inlet and discharge circles respectively. Then the radii of 15 226 ELEMENTS OF HYDRAULICS these circles will be denoted by r\, r 2 ; the pressure at any point on these circles by pi, p 2 , etc. Also let co denote the angular velocity of rotation of the impeller, and ui, u 2 the tangential velocities of the vanes at their inner and outer ends (Fig. 192), in which case Ui = r^u and Ui = r 2 o>. Applying Bernoulli theorem to the change in pressure produced by rotation alone, we have therefore pi w 2 n 2 _ ^2 co 2 r 2 2 ~ ~ ~~ Consequently the total change in pressure due to rotation, say p r where p r = pz PI, is given by the relation y- j 20 V 2g This expression is often called the centrifugal head. By similar reasoning the change in pressure produced by the outward flow is given by the relation p\ wi 2 _ p^ wi = 90 and consequently w^ = v\ 2 + Ui 2 . In this case, denoting the difference in pressure at inlet and exit due to the flow by p f , where p f = p' 2 p'i, we have Pf = P' Z ~ P' 1 = ^i 2 + ^i 2 ~ w-2 2 T " T 2gr The total increase in pressure in the impeller between the inlet and discharge ends of the vanes is therefore given by the relation p r + p f _ vi 2 + ui 2 - w z 2 + u 2 2 - ui 2 vi 2 + uz 2 - w 2 2 ~ 29 ^ Pressure Developed in Diffusor. Besides the increase in pressure produced in the impeller, the use of a suitable diffusion chamber permits part of the kinetic energy at exit, due to the absolute velocity v 2 of the discharge from the impeller, to be converted into pressure. Thus if k denotes the fraction of this kinetic energy which is converted into pressure in the diffusor, HYDRODYNAMICS 227 2 the expression derived above is increased by the term k^ When diffusion vanes are used, as in a turbine pump, the value of k may be as high as 0.75, and for a vortex chamber it may reach 0.60. General Formula for Pressure Head Developed. Combining the terms derived above, the total pressure head H developed by the pump is given by the simple formula kv 2 2 + Vi 2 + u 2 2 w 2 2 . H = - -^- (133) In applying this formula it is convenient to note that the total head H developed in the pump consists of three terms, as follows: ^ = head at eye (entrance) of impeller; = head developed in impeller; ^ = head developed in casing or diffusor. 44. CENTRIFUGAL PUMP CHARACTERISTICS Effect of Impeller Design on Operation. The greatest source of loss in a centrifugal pump is that due to the loss of the kinetic energy of the discharge. As only part of this kinetic energy can be recovered at most, it is desirable to reduce the velocity of discharge to as low a value as is compatible with efficiency in other directions. This may be accomplished by curving the outer tips of the impeller vanes backward so as to make the discharge angle less than 90. The relative velocity of water and vane at exit has then a tangential component acting in the opposite direc- tion to the peripheral velocity of the impeller, which therefore reduces the absolute velocity of discharge. This is apparent from Fig. 193 in which the parallelogram of velocities in each of the three cases is drawn for the same peripheral velocity u 2 and radial velocity at exit w 2 sin 6%. A comparison of these diagrams indicates" how the absolute velocity at exit v 2 increases as the angle 62 increases. The backward curvature of the vanes also gives the passages a more uniform cross section, which is favorable to efficiency. The effect which the design of the impeller has on the operation of the pump is most easily illustrated and understood by plotting 228 ELEMENTS OF HYDRAULICS curves showing the relations between the variables under con- sideration. Assuming the speed to be constant, which is the usual condition of operation, three curves are necessary to com- pletely illustrate the operation of the pump; one showing the relation between capacity and head, one between capacity and Wz FIG. 193. power, and one between capacity and efficiency. The first of these curves is usually termed the characteristic. Rising and Drooping Characteristics. The principal factor influencing the shape of the characteristic is the direction of the HYDRODYNAMICS 229 CAPACITY FIG. 195. Characteristics and efficiency curves obtained with DeLaval pumps. 230 ELEMENTS OF HYDRAULICS tips of the impeller blades at exit, although there are other fac- tors which affect this somewhat. If the tips are curved forward in the direction of rotation the characteristic tends to be of the rising type, whereas if they curve backward the characteristic tends to be of the drooping type (Figs. 194 and 195) . For a rising characteristic the head increases as the delivery increases and con- sequently the power curve also rises, since a greater discharge against a higher head necessarily requires more power (Fig. 196). A drooping power curve may be obtained by throttling at the eye of the impeller, but a greater efficiency results from designing the impeller so as to give this form of curve normally. 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 Percentage of Normal Capacity FIG. 196. For a high lift pump under an approximately constant head, as in the case of elevator work, a pump with radial vanes is most suitable as the discharge may be varied with a small altera- tion in delivery head. This is also true for a pump working under a falling head, as in the case of emptying a lock or dry dock, as it makes it possible to obtain a large increase in the discharge as the head diminishes, thereby saving time although at a loss of efficiency. One of the most important advantages of a drooping char- acteristic (Fig. 197) is that it is favorable to a drooping power delivery curve, making it impossible for the pump to overload the driving motor. For an electrically driven pump, in which the overload is limited to 20 per cent., or at most 25 per cent., of the normal power, backward curved vanes are therefore essential. Moreover, with a pump designed initially to work against a HYDRODYNAMICS 231 certain head, if the vanes at exit are radial, or curved forward, the possible diminution in speed is very small, the discharge ceasing altogether when the speed falls slightly below normal. As the backward curvature of the vanes increases the range of speed also increases, and consequently when the actual working head is not constant, as in irrigation at different levels, or in delivering cooling water to jet condensers in low head work, where the level of the intake varies considerably, a pump with drooping characteristic is much better adapted to meet varying conditions without serious loss of efficiency. 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 Percentage of Normal Capacity FIG. 197. Head Developed by Pump. These facts may be made more apparent by the use of the expression for the head developed by the pump, derived in the preceding article. Considering only the head developed in the impeller and casing, and omitting that due to the velocity of flow at entrance, t>i, which does not depend on the design of the pump, the expression for the head developed s H = Since v z is the geometric resultant of u 2 and w 2 , we have by the law of cosines, COS For an ideal pump, that is, one in which all the velocity head ~ is converted into pressure head in the diffusor, k is unity. As- 232 ELEMENTS OF HYDRAULICS suming k = 1 and substituting the expression for # 2 2 in the equa- tion for H, the result is W 2 2 UvWz COS 62 Jn =~ g For constant speed of rotation, u 2 is constant. For forward curved vanes 2 is greater than 90 and therefore cos B 2 is negative. In this case as w z increases H also increases; i.e., the greater the delivery the greater the head developed. For radial tipped vanes, 62 = 90 and cos 62 = 0. In this case H = , which is constant for all deliveries. y For backward curved vanes 62 is less than 90 and cos 62 is positive. Consequently in this case as the delivery increases the head diminishes. Although these relations are based on the assumption of a perfect pump, they serve to approximately indicate actual condi- tions, as is evident by inspection of the three types of charac- teristic. Effect of Throttling the Discharge. It is always necessary to make sure that the maximum static head is less than the head developed by the pump at no discharge. This is self-evident for the drooping characteristic, but the rising characteristic is misleading in this respect as the head rises above that at shut- off. Since for a certain range of head two different outputs are possible, it might seem that the operation of the pump under such conditions would be unstable. This instability, however, is counteracted by the frictional resistance in the suction and delivery pipes, which usually amounts to a considerable part of the total head. Any centrifugal pump with rising characteristic will therefore work satisfactorily if the maximum static head is less than the head produced at shut-off. If the frictional resistance is small it may be increased by throttling the discharge, so that by adjusting the throttle it is possible to operate the pump at any point of the curve with absolute stability. Numerical Illustration. The particular curves shown in Fig. 198 were plotted for an 8-in., 3-stage turbine fire pump built by the Alberger Co., New York, and designed to deliver 750 gal. per minute against an effective head of 290 ft., the pump being direct connected to a 75 H.P. 60-cycle induction motor operating at a synchronous speed of 1200 r.p.m. HYDRODYNAMICS 233 The head curve shows that this pump would deliver two fire streams of 250 gal. per minute each at a pressure of 143 Ib. per square inch; three streams of 250 gal. per minute each at a pressure of 125 Ib. per square inch; four streams of 250 gal. per minute each at a pressure of nearly 100 Ib. per square inch; or even five fairly good streams at a pressure of 80 Ib. per square inch. With the discharge valve closed the pump delivers no water but produces a pressure equivalent to a head of 308 ft. 420 396 ti $ fe 3GO ,vr ( it d \ ^v* \\ ^ \^~ - ti-P 3 _____ ___ ^J ^-- ***' n ^ >* "- 3 -^ ^ ^ ao & o 300 X ^ <, 7ft ?" /i x ^ \ 270 x' ^ V en W X* / \ ^ X ,e^ cS c. _^_^ \ \ X /- \ ^ s ^\ 50 pq X x ^ \ / / / \ N \ f /I \ k \ OA b ^ / ^ \ & / \ \ o / \ 1 2 30 120 / \ 3 / 1 / i 10 30 / / / 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 Gallonsiper Minute FIG. 198. If the head against which the pump operates exceeds this amount, it is of course impossible to start the discharge. The head for which this particular pump was designed was 290 ft., which cor- responds to the point of maximum efficiency. It is therefore apparent that the operating head must be carefully ascertained in advance, for if it is higher than that for which the pump was de- signed, both the efficiency and the capacity are diminished, whereas if it is lower, the capacity is increased but the efficiency is diminished. 234 ELEMENTS OF HYDRAULICS The power curve shows that under low heads the power rises. Also that the overload in the present case is confined to about 12 per cent, of the normal power. Consequently the motor could only be overloaded 12 per cent, if all the hose lines should burst, whereas the head curve shows that if all the nozzles were shut off no injurious pressure would result. The efficiency curve always starts at zero with zero capacity, as the pump does no useful work until it begins to discharge. The desirable features of an efficiency curve are steepness at the two ends, a flat top and a large area. Steepness at the beginning shows that the efficiency rises rapidly as the capacity increases, whereas a flat top and a steep ending show that it is maintained at a high value over a wide range. Since the average efficiency is obtained by dividing the area enclosed by the length of the base, it is apparent that the greater the area for a given length, the greater will be the average efficiency. 45. EFFICIENCY AND DESIGN OF CENTRIFUGAL PUMPS Essential Features of Design. The design of centrifugal pumps like that of hydraulic turbines requires practical ex- perience as well as detailed mathematical analysis. The general principles of design, however, are simple and readily understood, as will be apparent from what follows: Three quantities are predetermined at the outset. The inner radius of the impeller, r iy is ordinarily the same as the radius of the suction pipe; the outer radius, r^ is usually made twice rijand the angular speed co at which the impeller is designed to run is fixed by the particular type of prime mover by which the pump is to be operated. The chief requirement of the design is to avoid impact losses. In order therefore that the water shall glide on the blades of the impeller without shock, the relative velocity of water at entrance must be tangential to the tips of the vanes. Assuming the direction of flow at entrance to be radial, which is the assumption usually made although only approximately realized in practice, the necessary condition for entrance with- out shock is (Fig. 192) Vi = HI tan 0i, HYDRODYNAMICS 235 which determines the angle 0i. The relative velocity of water and vane at entrance is then The direction of the outer tips of the vanes, or angle 0%, Fig. 192, is determined in practice by the purpose for which the pump is designed, as indicated in Art. 44. For an assigned value of 62, the absolute velocity of the water at exit is and consequently as 2 increases, the absolute velocity at exit, t>2, also increases. Let Sij 2, Fig. 192, denote the radial velocity of flow at en- trance and exit, respectively, and AI, A 2 the circumferential areas of the impeller at these points. Then for continuous flow SiAi = SzA 2 . Usually si = $2, in which case A i = A 2 . If 61 and 62 denote the breadth of the impeller at inlet and outlet respectively, then AI = 2wr 1 bi and A 2 = 27rr 2 6 2 , and conse- quently for AI = Az we have btfi = b 2 r 2 . Assuming the radial velocity of flow throughout the impeller to be constant, the breadth b at any radius r is given by the relation br = bjTi. Hydraulic and Commercial Efficiency. Let H' denote the total effective head against which the pump operates, including suction, friction, delivery and velocity heads. Then if w denotes the velocity of the water as it leaves the delivery pipe, h the total lift including suction and delivery heads, and h/ the friction head, we have H' = h + h, + |j. The total theoretical head H developed by the pump, as derived in Art. 43, is H= -. Consequently the hydraulic efficiency of the pump is the ratio of these two quantities, that is, jj/ Hydraulic efficiency = =- n The commercial efficiency of the pump is the ratio of the work actually done in lifting the water through the height h to the 236 ELEMENTS OF HYDRAULICS total work expended in driving the impeller shaft, and is of course less than the hydraulic efficiency. 46. CENTRIFUGAL PUMP APPLICATIONS Floating Dry Docks. To illustrate the wide range of ap- plications to which centrifugal pumps are adapted, a few typical examples of their use will be given. The rapid extension of the world's commerce in recent years has created a demand for docking facilities in comparatively isolated ports, which has given rise to the modern floating dry dock (Fig. 199). In docks of this type the various compartments into FIG. 199. which they are divided are provided with separate pumps so that they may be emptied in accordance with the distribution of weight on the dock. Provision is usually made for handling one short vessel, two short vessels, or one extremely long ship, the balancing of the dock on an even keel being accomplished by emptying the various compartments in proportion to the weight sustained. The number of pumps in docks of this type varies from 6 to 20, depending on the number of compartments. The centrifugal pump is widely used and particularly suitable for this class of work, where a large quantity of water has to be discharged in a short time against a changing head which varies from zero H YD ROD YNA MICS 237 when the pumping begins to 30 or 40 feet when the dock is nearly dry (Fig. 200). l Deep Wells. In ob- taining a water supply from deep wells, the problem is to secure a pump which will handle a large quantity of water efficiently in a drilled well of moderate diam- eter, the standard di- ameters of such wells being 12 and 15 in. To meet this demand, cen- trifugal pumps are now built which will deliver from 300 to 800 gal. per minute from a 12-in. well, and from 800 to 1500 gal. per minute from a 15-in. well, with efficiencies ranging from 55 to 75 per cent. The depth from which the water is pumped may be 300 ft. or more, the pumps being built in several stages according to the depth (Fig. 201). Mine Drainage. The extensive use of electric power for operating mining machinery has led to the employment of centrifugal pumps for mine drainage. The advantages of this type of pump when direct connected to a high FIG. 200. 1 Figs. 199-203 are reproduced by permission of the Platt Iron Works Co., Dayton, Ohio. 238 ELEMENTS OF HYDRAULICS Deep | Well Centrif- t ugal Pump FiG. 201. FIG. 202. HYDRODYNAMICS 239 speed motor are its compactness, simplicity and low first cost. Fig. 202 illustrates a mine sinking turbine pump which operates against a 1250 ft. head in a single lift. Pumps of similar design are in operation in nearly all the important mining regions of the United States and Mexico. The turbine pump is used to best advantage where it is required to unwater a flooded mine shaft. For actual sinking work a displacement pump is preferable unless an ample sump is provided in order to keep the turbine pump well supplied with water so that it will not take air. Fire Pumps. The use of centrifugal pumps for fire protection has been formally approved by the Fire Insurance Underwriters, FIG. 203. who have issued specifications covering the essential features of a pump of this type to comply with their requirements. In the case of fire boats the centrifugal pump has been found to fully meet all demands. The New York fire boats " James Duane" and "Thomas Willett" are equipped with turbine pumps, each of which has a capacity of 4500 gal. per minute against 150 Ib. per square inch pressure. For automobile fire en- gines, the great range of speed for gas engines gives the centrifugal pump a great advantage, making it possible to throw streams to a great height by merely increasing the speed of the motor. This type can also be readily mounted on a light chassis and driven from the driving shaft of the machine, making a light, compact, flexible and efficient unit (Fig. 203). Hydraulic Dredging. The rapid development and improve- ment of internal waterways in the United States has demonstrated the efficiency of the hydraulic or suction dredge. The advantage of the hydraulic dredge over the dipper and ladder types is that it not only dredges the material but also delivers it at the desired 240 ELEMENTS OF HYDRAULICS point with one operation. Its cost for a given capacity is also less than for any other type of dredge, while its capacity is enormous, some of the Government dredges on the Mississippi handling over 3000 cu. yd. of material per hour. In operation the dredging pump creates a partial vacuum in the suction pipe, sufficient to draw in the material and keep it moving, and also produces the pressure necessary to force the discharge to the required height and distance. Hundreds of such pumps, ranging from 6 to 20 in. in diameter, are used on western rivers for dredging sand and gravel for building and other purposes. The dredge for this class of service is very simple, consisting principally of the dredging pump with its driving equipment mounted on a scow, the suction pipe being of sufficient length to reach to the bottom, and the material being delivered into a flat deck scow with raised sides, so that the sand is re- tained and the water flows overboard. For general dredging service where hard material is handled, it is necessary to use an agitator or cutter to loosen the material so that it can be drawn into the suction pipe. In this case the suction pipe is mounted within a structural steel ladder of heavy proportions to stand the strain of dredging in hard material, and of sufficient length to reach to the depth required. The cutter is provided with a series of cutting blades, and is mounted on a heavy shaft supported on the ladder, and driven through gearing by a separate engine (Fig. 204). Usually two spuds are arranged in the stern of the dredge to act as anchors and hold the dredge in position. The dredge is then swung from side to side on the spuds as pivots by 'means of lines on each side controlled by a hoisting engine, thus controlling the operation of the dredge. Suction dredges are usually equipped with either 12-, 15-, 18- or 20-in. dredging pumps, the last named being the standard size. For most economical operation as regards power, the velocity through the pipe line should not be greater than just sufficient to carry the material satisfactorily. With easily handled material the delivery pipe may be a mile or more in length, but with heavy mateiial requiring high velocity the length should not exceed 4000 ft. The practical maximum discharge pressure is about 50 Ib. per square inch. For long pipe lines it therefore becomes necessary to use relay pumps, the dredging pump delivering through a certain length of pipe HYDRODYNAMICS 241 10 242 ELEMENTS OF HYDRAULICS into the suction of the relay pump, and the latter delivering it through the remainder of the line. For high elevations or very long lines, several relay pumps may have to be used. The efficiency of a dredging pump is usually only 40 or 50 per cent., a high efficiency in this case not being so important as the ability to keep going. Hydraulic Mining. The centrifugal pump is also successfully used in hydraulic mining, where a high pressure jet is used to wash down a hill. A number of centrifugal pumps are used for this purpose in the phosphate mines of Florida. Other uses for centrifugal pumps besides those described above are found in municipal water works, sewage and drainage plants, sugar re- fineries, paper mills and irrigation works. APPLICATIONS 101. A jet 2 in. in diameter discharges 5 cu. ft. of water per second which impinges on a flat vane moving in the same direc- tion as the jet with a velocity of 12 ft. per second. Find the horse power expended on the vane. 102. A fireman holds a hose from which a jet of water 1 in. in diameter issues at a velocity of 80 ft. per second. What force will the fireman have to exert to support the jet? 103. A small vessel is propelled by two jets each 9 in. in diam- eter. The water is taken from the sea through a vertical inlet pipe with scoop facing forward, and driven astern by a centrifugal pump 2 ft. 6 in. in diameter running at 428 r.p.m. and delivering approximately 2250 cu. ft. of water per second. If the speed of the boat is 12.6 knots (1 knot = 6080 ft. per hour), calculate the hydraulic efficiency of the jet. 104. In the preceding problem, the efficiency of the pump was 48 per cent, and efficiency of engine and shafting may be assumed as 80 per cent. Using these values, calculate the total hydraulic efficiency of this system of propulsion. NOTE. The jet propeller is more efficient than the screw pro- peller, the obstacle preventing the adoption of this system in the past being the low efficiency obtainable from centrifugal pumps. 105. A locomotive moving at 60 miles per hour scoops up water from a trough between the rails by means of an L-shaped pipe with the horizontal arm projecting forward. If the trough is 2000 ft. long, the pipe 10 in. in diameter, the opening into the HYDRODYNAMICS 243 tank 8 ft. above the mouth of the scoop, and half the available head is lost at entrance, find how many gallons of water are lifted into the tank in going a distance of 1600 ft. Also find the slowest speed at which water will be delivered into the tank. 106. A tangential wheel is driven by two jets each 2 in. in diameter and having a velocity of 75 ft. per second. Assuming the wheel efficiency to be 85 per cent, and generator efficiency 90 per cent., find the power of the motor in kilowatts (1 H.P. = 746 watts = 0.746 kilowatt). 107. In a commercial test of a Pelton wheel the diameter of the jet was found to be 1.89 in., static head on runner 386.5 ft., head lost in pipe friction 1.8 ft., and discharge 2.819 cu. ft. per second. The power developed was found by measurement to be 107.4 H.P. Calculate the efficiency of the wheel. 108. A nozzle having an efflux coefficient of 0.8 delivers a jet 1-1/2 in. in diameter. Find the amount and velocity of the dis- charge if the jet exerts a pressure of 200 Ib. on a flat surface normal to the flow. 109. A jet 2 in. in diameter is deflected through 120 by striking a stationary vane. Find the pressure exerted on the vane when the nozzle is discharging 10 cu. ft. per second. 110. A power canal is 50 ft. wide and 9 ft. deep, with a velocity of flow of 1-1/2 ft. per second. It supplies water to the turbines under a head of 30 ft. If the efficiency of the turbines is 80 per cent., find the horse power available. 111. It is proposed to supply 1200 electrical horse power to a city 25 miles from a hydraulic plant. The various losses are estimated as follows: Generating machinery, 10 per cent.; line, 8 per cent.; trans- formers at load end, 9 per cent. ; turbine efficiency, 80 per cent. The average velocity of the stream is 3 ft. per second, available width 90 ft., and depth 6 ft. Find the net fall required at the dam. 112. The head race of a vertical water wheel is 6 ft. wide and v the water 9 in. deep, flowing with a velocity of 5 ft. per second. If the total fall is 20 ft. and the efficiency of the wheel is 70 per cent., calculate the horse power available from it. 113. A stream is 150 ft. wide with an average depth of 4 ft. and a velocity of flow of 1 ft. per second. If the net fall at the dam is 20 ft. and the efficiency of the wheel is 75 per cent., find the horse power available. 244 ELEMENTS OF HYDRAULICS 114. Eighty gallons of water per minute are to be pumped from a well 12 ft. deep by a pump situated 50 ft. from the well, and delivered to a tank 400 ft. from the pump and at 80 ft. elevation. The suction pipe is 3 in. in diameter and has two 3-in. elbows. The discharge pipe is 2-1/2 in. in diameter and has three 2-1/2-in. elbows. Find the size of engine required. NOTE. The lift is 92 ft. and the friction head in pipe and elbows amounts to about 25 ft., giving a total pumping head of 117 ft. The pump friction varies greatly, but for a maximum may be assumed as 50 per cent, of the total head, or, in the present case, 58-1/2 ft. 115. A single acting displacement pump raises water 60 ft. through a pipe line 1 mile long. The inside diameter of the pump barrel is 18 in., the stroke is 4 ft., and the piston rod is driven by a connecting rod coupled to a crank which makes 30 r.p.m. The velocity of flow in the pipe line is 3 ft. per second. Assuming the mechanical efficiency of the pump to be 75 per cent., and the slip 5 per cent., find the horse power required to drive the pump and the quantity of water delivered. 116. A 6-in. centrifugal pump delivers 1050 gal. per minute, elevating 20 ft. The suction and discharge pipes are each 6 in. in diameter and have a combined length of 100 ft. Find the friction head, total horse power required, and speed of pump for 50 per cent, efficiency. NOTE. The velocity of flow in this case is 12 ft. per second and the corresponding friction head for 100 ft. of 6-in. pipe is 8.8 ft. The total effective head is therefore 28.8 ft., requiring 15.26 H.P. at a speed of 410 r.p.m. 117. In the preceding problem show that if an 8-in. pipe is used instead of 6-in. there will be a saving in power of over 22 per cent. 118. A hydraulic ram uses 1000 gal. of water per minute under a 4-ft. head to pump 40 gal. per minute through 300 ft. of 2-in. pipe into a reservoir at an elevation of 50 ft. above the ram. Calculate the mechanical and hydraulic efficiencies of the ram, assuming the coefficient of pipe friction as 0.024. 119. An automobile booster fire pump, used for making a quick initial attack on a fire, is required to deliver two streams through 3/8-in. nozzles and 250 ft. of 1-in. hose. The pump is of the centrifugal type and is geared up to a speed of 3500 r.p.m. from the gas engine which drives the machine. Calculate the HYDRODYNAMICS 245 discharge in gallons per minute and the horse power required to drive the pump, assuming 50 per cent, efficiency. NOTE. For this size nozzle, the maximum discharge is reached with a nozzle pressure of about 68 Ib. per square inch corre- sponding to a velocity of about 100 ft. per second. 120. Feed water is pumped into a boiler from a round vertical tank 2-1/2 ft. in diameter. Before starting the pump the water level in the boiler is 38 in., and in the tank 22 in., above the floor level, and when the pump is stopped these levels are 40 in. and 15 in. respectively. If the steam pressure in the boiler while the pump is at work is 100 Ib. per square inch find the number of foot-pounds of work done by the pump. 121. A fire pump delivers three fire streams, each discharging 250 gal. per minute under 80 Ib. per square inch pressure. Find the horse power of the engine driving the pump if the efficiency of the engine is 70 per cent, and of the pump is 60 per cent. 122. A mine shaft 580 ft. deep and 8 ft. in diameter is full of water. How long will it take a 6-H.P. engine to unwater the shaft if the efficiencies of pump and engine are each 75 per cent.? 123. A fire engine pumps at the rate of 500 gal. per minute against a pressure of 100 Ib. per square inch. Assuming the overall efficiency to be 50 per cent., calculate the indicated horse power of the engine. 124. A water power plant is equipped with tangential wheels having an efficiency of 80 per cent. The water is delivered to the wheels through a cylindrical riveted steel penstock 5 miles long with a total fall of 900 ft., practically the entire penstock being under this head. The cost of power house and equipment is estimated at $50,000, penstock 6 cents per pound, operating expenses $5000 per annum, and interest on total investment 4 per cent, per annum. The income is to be derived from the sale of power at $12 per horse power per annum. A constant supply of water of 100 cu. ft. per minute is available. (A) Plot a curve with diameter of penstock as abscissa and yearly gross income as ordinate. (B) Plot a curve with diameter of penstock as abscissa and yearly gross expenses as ordinate. (C) From these two curves determine the two values of the diameter of penstock for either of which the net income is zero, 246 ELEMENTS OF HYDRAULICS and the one value for which the net income is a maximum, and find the amount of the latter. 125. A hydraulic pipe line is required to transmit 150 H.P. with a velocity of flow not greater than 3 ft. per second and a delivery pressure of 900 Ib. per square inch. Assuming that the most economical size of pipe is one which allows a pres- sure drop of about 10 Ib. per square inch per mile, determine the required size of pipe. 126. Find the maximum horse power which can be transmitted through a 6-in. pipe 4 miles long assuming the inlet pressure to be 800 Ib. per square inch and the coefficient of pipe friction to be 0.024. Also determine the velocity of flow and outlet pressure. 127. A 6-in. pipe half a mile long leads from a reservoir to a nozzle located 350 ft. below the level of the reservoir and discharging into the air. Assuming the coefficient of friction to be 0.03, determine the diameter of nozzle for maximum power. 128. A 10-in. water main 900 ft. long is discharging 1000 gal. of water per minute. If water is shut off in 2 sec. by closing a valve, how much is the pressure in the pipe increased? 129. In a series of experiments made by Joukowsky on cast-iron pipes, the time of valve closure in each case being 0.03 sec., the following rises in pressures were observed. 1 Show that these results give the straight line formula, p = 57v. CAST-IRON PIPE, DIAM. 4 IN., LENGTH 1050 FT. Vel. in ft. /sec 0.5 2.0 3.0 4.0 9.0 Observed pressure in lb./in. 2 . . . 31 119 172 228 511 Cast-iron pipe, diameter 6 in., length 1066 ft. Vel in ft /sec . . . 0.6 2.0 3.0 7.5 Observed rise in pressure .in lb./in. 2 . . . . 43 113 173 426 130. It is customary in practice to make allowance for possible water hammer by designing the pipes to withstand a pressure of 100 Ib. per square inch in excess of that due to the static head. Show that this virtually allows for an instantaneous stoppage at a velocity of 1.6 ft. per second. 131. A bowl in the form of a hemisphere, with horizontal rim, Gibson: Hydraulics and Its Applications, p. 239. HYDRODYNAMICS 247 is filled with liquid and then given an angular velocity co about its vertical axis. How much liquid flows over the rim (Fig. 205)? 132. A closed cylindrical vessel of height H is three-fourths full of water. With what angular velocity co must it revolve around its vertical axis in order that the surface paraboloid shall just touch the bottom of the vessel (Fig. 206). 133. A closed cylindrical vessel of diameter 3 ft. and height 6 in. contains water to a depth of 2 in. Find the speed in r.p.m. at which it must revolve about its vertical axis in order that the water shall assume the form of a hollow truncated paraboloid for which the radius of the upper base is 1 per cent, greater than the radius of the lower base; or, referring to Fig. 207, such that r\ = 1.01r 2 . 134. The test data for a 19-in. New American turbine runner are as follows: Head 25 ft.; speed 339 r.p.m.; discharge 2128 cu. ft. per min.; power developed 80 H.P. Calculate the turbine constants including the characteristic speed. FIG. 205. FIG. 206. FIG. 207. Solution. In this case, from Article 36, 60V/i " 60V25 k, 5.62 V64.4 = 0.7; 248 ELEMENTS OF HYDRAULICS 2128 Qi _ 7.0933 19 /\ V12/ = __ 5 K/S " 25^25 " 135. Two types of turbine runner, A and 5, are to be compared. From tests it is known that runner A will develop a maximum of 2080 H.P. at 500 r.p.m. under 100 ft. head, and runner B will develop 4590 H.P. at 580 r.p.m. under 150 ft. head. Deter- mine which of these types is the higher speed. 136. Show that to transform the characteristic speed N 8 from the English to the metric system it is necessary to multiply by the coefficient 4.46; that is to say, if the horse power and head are expressed in foot-pound units, and N 8 in the metric system, we have the relation A7 VlLR Ns = 137. Five two-runner Francis turbines installed in the power house of the Pennsylvania Water and Power Co. at McCalPs Ferry on the Susquehanna River are rated at 13,500 H.P. each under a head of 53 ft. at a speed of 94 r.p.m. The quantity of water required per turbine is 2800 cu. ft. per second. Calculate from this rating the characteristic speed, efficiency, and other turbine constants. ' 138. Four two-runner Francis turbines operating in the Little Falls plant of the Washington Water Power Co. have a nominal power capacity of 9000 H.P. each under a head of 66 ft. at a speed of 150 r.p.m. The quantity of water required per turbine is 1500 cu. ft. per second. From this rating calculate the charac- teristic speed, efficiency, and specific constants for these units. 139. The upper curve shown in Fig. 208 is the official effi- ciency test curve of the 9000 H.P. turbines, built by the I. P. HYDRODYNAMICS 249 Morris Co. for the Washington Water Power Co. These wheels are of the horizontal shaft, two-runner, central discharge type, with volute casings. Head 66 ft., speed 150 r.p.m., and rated runner diameter 6 ft. 2 in. The lower curve shown in the figure is derived from a test at Holyoke of a homologous experimental runner having a rated diameter of 2 ft. 8-13/14 in. These curves are almost identical in shape, the efficiency of the large units exceeding by a small margin that of the experimental runner. 70 GO t 50 10 90 50 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000 7500 8000 3500 9000 9500 Horse Power FIG. 208. Calculate the discharge and characteristic speed at maximum efficiency, and from these results compute the specific constants, 140. In testing a hydraulic turbine it was found by measure- ment that the amount of water entering the turbine was 8000 cu. ft. per minute with a net fall of 10.6 ft. The power devel- oped was measured by a friction brake clamped to a pulley. The length of brake arm was 12 ft., reading on scales 400 lb., and speed of pulley 100 r.p.m. Calculate the efficiency of the turbine. 141. One of a series of 65 tests of a 31-in. Wellman-Seaver- Morgan turbine runner gave the following data: 1 Gate opening 75 per cent.; head on runner 17.25 ft.; speed 186.25 r.p.m.; discharge 63.12 cu. ft. per second; power developed 111.66 H.P. Calculate the efficiency and the various turbine constants. 142. One of a series of 82 tests of a 30-in. Wellman-Seaver- Morgan turbine runner gave the following data: 2 1 Characteristics of Modern Hyd. Turbines, C. W. Larner, Trans. Am. Soc. C. E., Vol. LXVI (1910), pp. 306-386. 2 Ibid. 250 ELEMENTS OF HYDRAULICS Gate opening 80.8 per cent.; head on runner 17.19 ft.; speed 206 r.p.m.; discharge 85.73 cu. ft. per second; power developed 146.05 H.P. Calculate the efficiency and the other turbine constants. 143. Four of the turbines of the Toronto Power Co. at Niagara Falls are of the two-runner Francis type, with a nominal develop- ment of 13,000 H.P. each under a head of 133 ft. at a speed of 250 r.p.m. The quantity of water required per turbine is 1060 cu. ft. per second. Calculate the efficiency, characteristic speed and specific turbine coefficients for these units. 90 70 CO SO 10 400 800 1200 1COO 2000 2400 2800 3200 3GOO 4000 4400 4800 5200 5600 GOOD 6400 Horse Power FIG. 209. 144. The upper curve shown in Fig. 209 is the official test curve of the 6000 H.P. turbines designed by the I. P. Morris Co. for the Appalachian Power Co. The rated runner diameter is 7 ft. 6-1/4 in., head 49 ft., and speed 116 r.p.m. These turbines are of the single runner, vertical shaft type. The lower curve is derived from a test at Holyoke of the small, homologous, experimental runner, having a rated diameter of 27-3/8 in. The curves are identical in shape, but owing to the better arrangement of water passages in the large plant, its efficiency considerably exceeds that of the experimental runner. It may also be noted that the efficiency shown on this diagram is the highest ever recorded in a well-authenticated test. Calculate the discharge and characteristic speed at maximum efficiency, and" from these results compute the specific turbine constants. HYDRODYNAMICS 251 145. The following data, taken from the official Holyoke test reports, give the results of tests made on a 35-in. vertical Samson turbine built by the James Leffel Co. of Springfield, Ohio. Calculate the turbine constants and characteristic speeds. TESTS OF 35-iN. VERTICAL SAMSON TURBINE . Gate opening Head on wheel in feet Speed in rev. per min. Discharge in. cu. ft. per sec. . Horse power developed Efficiency in per cent. Full gate 0.9 gate 0.8 gate 75 gate. . 16.57 16.69 16.78 16 86 187 191 189 187 120.61 114.35 105.10 100 29 188.27 188.88 179.87 172 57 83.06 87.26 89.93 89 99 0.7 gate 0.6 gate...... 0.5 gate 17.08 17.23 17.47 188 185 188 92.83 77.15 66.89 160 . 03 128.22 108.72 88.99 85.05 82.03 146. The speed and water consumption of a turbine vary as the square root of the head ( V h), and the power varies as the square root of the cube of the head ( V h 3 ) . Thus if the head on a wheel is multiplied by 4, the speed and discharge will be multiplied by 2 and the power by 8. Given that a 12-in. turbine under 12-ft. head develops 14 H.P. at 480 r.p.m. using 762 cu. ft. of water per minute, find the power, speed and discharge for the same turbine under 48-ft. head. 147. On page 252 is given a rating table of turbines manu- factured by the S. Morgan Smith Co. of York, Pa., computed from actual tests of each size turbine under the dynamometer at the Holyoke testing flume. Calculate the nominal efficiency and characteristic speed for each size runner, and determine whether it is of the low, medium or high speed type. NOTE. Data of this kind may be used by the instructor as problem material for an entire class without duplicating results, the final results being collected and tabulated, thus serving as a check on the calculations and also showing the range of the constants involved. 148. On pages 253, 254 and 255 is given a rating table of Victor Turbines manufactured by the Platt Iron Works Co., Dayton, Ohio. Calculate the nominal efficiency, characteristic speed, and speed and capacity constants for each diameter and head. 252 ELEMENTS OF HYDRAULICS HYDRODYNAMICS 253 all Hor cub .s M fib! S3 ^ o TH r^ co (M OS i l TH O OS 1> TH TH IO CO CO TH TH CO OS rH O 00 TH TH to o co t^. rH 10 00 CO to rH IO CO Ol CO OS 00 TH TH 00 CO OS O t CO OS TH W TH 00 CO oo O CO co TH TH 00 oo 00 CO Ofl O1 00 1> (N l> to to t^ TH rH CO OS O TH CO 1> 01 TH OS CO CO CO OS rH CO CO IO rH 10 to TH 00 THCOCO CO to oo os TH OO CO rH OQ CO OQ 1> OS r- I O1 to CO CO eo CO tO CO OS 10 (N OS O5 to iO co b- tO TH TH 00 OS O* TH to CO O 00 t^ TH TH GO lO CO iO i i 10 CO O OS OS 10 TjH CO (N 1> (N 33 CO O O OS TtH 00 to to N- 00 CO J> CO TH to t CO I> pe Cubi Revo tO 0? eet ions Horse power. Cubic Revolu oo srH CO o TH 00 CO rH O co powe feet o ub H C R 1^ CO l> (N rJH rjn >O OS CO iO CO IO TH 00 OS (N 00 (M TH CO Horse power Cubic feet .. Revolutions . (N O 00 (M t^ TH t^ TH TH CO TH OS >O (N CO l> t^ CO TH CO TH (N OQ TH to TH TH CO IS tO 01 TH O to to TH O CO Os CO 00 00 CO TH CO OS CO CO CO iO TH b- CO TH co oo co Horse Cubic Revolu 254 ELEMENTS OF HYDRAULICS S po 9? += "d *Sg 2 -3 S 3 g w ;3 revolutio per minu ^ rH iO to OS CO I> OS CO OS 00 i i oo 10 CO CO ^ l> TjH CO Tj< CO CO T^ co co co rH tO OS O CO CO CO (N CO rH CO Tt< oo to co !> r- i CO CO O O 00 oo O >O co 00 o -^ 10 GO OS r-H rt< >O CO " rH O O CO l>- i i Tt< Tt< CO 00 Horse powe Cubic feet Revolutions CO O5 O 10 rj< (M 00 OO CO ^ 00 iO CO O OS ^ 00 CO TH l> Tt< CO oo TH d CO O "tf O 00 O 1 > * CO CO O5 1> O 00 "tf OS OS OO 00 TJH (M l> CO CO -^ 00 CO CO CO J> CO 00 00 l> TH !> CO b l> 00 (N TH to (N OS 1> CO CO CO (N (N to 00 O to (N (M (N Horse powe Cubic feet CO rt^ OS t^ 00 l>- (M 00 (N CO TH t^ (M I> (M TH CO 8 CO O Tt< t^ (N OS TH TH TH CO OS 00 O 00 rH CO CO OS (N to co co 00 IO (N "* CO 00 l> iO OS TJH (N I> O Tt< T* OS O OS (N (M 00 OS o (M 00 OS 00 iO CO rH (M OS CO CO TfH 00 IO CO OS (M rH 00 JO ^ OS O OO ^T* CO 1>* C^l rH 00 rH co CO~I>~ CO 1> Tt^ (N iO (N (N oo co ^ "^ O^ CO O t^ (M rH rH (N 00 O (N OS CO CO rH rH CO CO OS 00 - rH CO CO (M co o CO CO 1> (M CO O (N CO (M os co co tO rH CO O iO to (N CO CO CO iO T^ 1 I O O5 O (N s co co os CO O5 OS GO CO TH O GO OS GO O TH ^ (M CO OS TH CO '" CO co os t^ CO ^ rH CO rH (N co CO t^ CO 1> <*! OS CO rH . (N O OS 00 * (M CO GO OS 1> CO O TH TH GO O t^ * CO (N rH CO IO rJH T*H O5 OS CO CO ^ GO OS UO O GO CO t^ TH CO ^ TH CO GO t^ GO ^ OS TH l>- CO TH (M TH 00 CO C<1 TH CO rH |>. TH^TH GO OS Os to CO 00 |>. IO CO GO rH rH CO O (N O O CO CO 00 TJH rH Tt< O CO co co co b- O rH CO l>. 00 TH TH CO CO i-i GO OS OS co CO CO O rH co to CO rH .a , xj g revolutions per minute ee io we t po ee io Ho Cu Re Hor Cub Hor Cub Re Horse po Cubic fee Revolutions II! ee io Horse Cubic 256 ELEMENTS OF HYDRAULICS 149. Fig. 210 shows a vertical section of the 10,800-H.P. turbines designed by the I. P. Morris Co. for the Cedar Rapids Mfg. and Power Co. The rated diameter of these tur- bines is 11 ft. 10-1/2 in., head 30 ft., and speed 55.6 r.p.m. These turbines are at present the largest in the world, and it may be noted that all the latest features have been incorporated in the design, namely, volute casings and draft tubes molded in FIG. 210. the concrete; cast-iron speed rings supporting the concrete, genera- tor and thrust bearing loads from above; lignum vitse turbine guide bearing; thrust bearing support located above the generator ; Kingsbury thrust bearing with roller auxiliary; and pneumatic brakes acting on the rotor of the generator. Calculate the characteristic speed from the rating given above, and from the table on page 188 determine to which speed type it belongs. 150. The following table gives the results of 20 tests out of a total of 66 made Sept. 3-4, 1912, at the testing flume of the HYDRODYNAMICS 257 Holyoke Water Power Co. on a 24-in. Morris turbine type "O" runner. Calculate the characteristic speed for each test, and note its high value in test number 10. NOTE. Two of the wheels for the Keokuk installation and nine wheels for the Cedar Rapids plant are built with this type of runner, the large wheels being geometrically similar to the experimental wheel tested at Holyoke. REPORT OF TESTS OF A24-IN. MORRIS TURBINE, TYPE "O" RUNNER MADE IN THE TESTING FLUME OF THE HOLYOKE WATER POWER Co. Number of experi- ment Open- ing of speed gate in inches Per cent. of full dis- charge of wheel Head on wheel in feet Speed in rev. per min. Dis- charge in cu. ft. per sec. Horse power devel- oped Effi- ciency in per cent. 1 3.0 0.792 17.39 344.25 84.66 103.98 62.31 2 3.0 0.779 17.40 298.00 83.35 126.01 76.66 3 3.0 0.775 17.39 275.75 82.91 133. 2& 81.55 4 3.0 0.778 17.36 257.50 83.13 139.99 85.59 5 3.0 0.779 17.33 249.20 83.13 143.01 87.58 6 3.0 0.782 17.30 241.25 83.35 145.73 89.17 7 3.0 0.779 17.30 235.50 83.13 146.53 89.89 8 3.0 0.781 17.28 238.25 83.24 146 . 80 90.05 9 3.0 0.783 17.27 240.20 83.46 146.55 89.71 10 3.0 0.781 17.23 236.80 83.13 146.62 90.32 11 3.0 0.762 17.31 214.00 81.28 138.97 87.15 12 3.0 0.782 17.24 237.20 83.24 146.51 90.08 13 3.0 0.744 17.34 185.50 79.44 128.86 82.54 14 3.5 0.902 17.15 357.20 95.73 107.89 57.98 15 3.5 0.900 17.10 320.50 95.38 135.52 73.31 16 3.5 0.903 17.06 287.75 95.61 156.44 84.62 17 3.5 0.892 17.14 265.75 94.70 160.53 87.26 18 3.5 0.871 17.17 243.25 92.55 154.29 85.67 19 3.5 0.903 17.15 280.20 95.84 160.80 86.32 20 3.5 0.904 17.19 278.60 96.07 164.09 87.67 17 HYDRAULIC DATA AND TABLES 260 ELEMENTS OF HYDRAULICS TABLE 1. PROPERTIES OF WATER Density and Volume of Water Temp, in degrees Centigrade Density Volume of 1 gram in cu. cm. Temp, in degrees Centigrade Density Volume of 1 gram in cu. cm. 0.999874 1.00013 24 0.997349 .00266 1 0.999930 1.00007 26 . 996837 .00317 2 0.999970 1 . 00003 28 0.996288 .00373 2 0.999993 1.00001 30 0.995705 .00381 4 1.000000 1.00000 32 0.995087 .00394 5 0.999992 1.00001 35 0.995098 .00394 6 0.999970 1 . 00003 40 0.99233 .00773 7 0.999932 1.00007 45 0.99035 1.00974 8 0.999881 1.00012 50 0.98813 1.01201 9 0.999815 1.00018 55 0.98579 1.01442 10 0.999736 1 . 00026 60 0.98331 1.01697 11 0.999643 1.00036 65 0.98067 1.01971 12 0.999537 1.00046 70 0.97790 1.02260 13 0.999418 1 . 00058 75 0.97495 1.02569 14 0.999287 1.00071 80 0.97191 1.02890 16 0.998988 1.00101 85 0.96876 1.03224 18 0.998642 1.00136 90 0.96550 1.03574 20 0.998252 1.00175 95 0.96212 1.03938 22 0.997821 1.00218 100 0.95863 1.04315 Weight of Water Temp, in degrees Fahrenheit Weight in pounds per cu. ft. Temp, in degrees Fahrenheit Weight in pounds per cu. ft. Temp, in degrees Fahrenheit Weight in pounds per cu. ft. 32 40 50 60 70 80 90 62.42 62.42 62.41 62.37 62.31 62.23 62.13 100 110 120 130 140 150 160 62.02 61.89 61.74 61.56 61.37 61.18 60.98 170 180 190 200 210 212 60.77 60.55 60.32 60.07 59.82 59.56 TABLES 261 TABLE 2. HEAD AND PRESSURE EQUIVALENTS Head of Water in Feet and Equivalent Pressure in Pounds per Sq. In. Feet head Pounds per sq. in. Feet head Pounds per sq. in. Feet head Pounds per sq. in. 1 0.43 55 23.82 190 82.29 2 0.87 60 25.99 200 86.62 3 1.30 65 28.15 225 97.45 4 1.73 70 30.32 250 108.27 5 2.17 75 32.48 275 119.10 6 2.60 80 34.65 300 129.93 7 3.03 85 36.81 325 140.75 8 3.40 90 38.98 350 151.58 9 3.90 95 41.14 375 162.41 10 4.33 100 43.31 400 173.24 15 6.50 110 47.64 500 216.55 20 8.66 120 51.97 600 259.85 25 10.83 130 56.30 700 303 . 16 30 12.99 140 60.63 800 346.47 35 15.16 150 64.96 900 389.78 40 17.32 160 69.29 1000 433.09 45 19.49 170 73.63 50 21.65 180 77.96 Pressure in Pounds per Sq. In. and Equivalent Head of Water in Feet Pounds per sq. in. Feet head Pounds per sq. in. Feet head Pounds per sq. in. Feet head 1 2.31 55 126.99 180 415.61 2 4.62 60 138 . 54 190 438.90 3 6.93 65 150.08 200 461 . 78 4 9.24 70 161.63 225 519.51 5 11.54 75 173.17 250 577.24 6 13.85 80 184.72 275 643.03 7 16.16 85 196.26 300 692 . 69 8 18.47 90 207.81 325 750.41 9 20.78 95 219.35 350 808.13 10 23.09 100 230.90 375 865.89 15 34.63 110 253.98 400 922.58 20 46.18 120 277.07 500 1154.48 25 57.72 125 288 . 62 30 69.27 130 300 . 16 35 80.81 140 323.25 40 92.36 150 346 . 34 45 103 . 90 160 369 . 43 50 115.45 170 392.52 262 ELEMENTS OF HYDRAULICS TABLE 3. DISCHARGE EQUIVALENTS Gallons per inin. Cubic feet per sec. Cubic feet per min. Gallons per hour Gallons per 24 hours Bbls. per minute, 42 gal. bbl. Bbls. per hour, 42 gal. bbl. Bbls. per 24 hours, 42 gal. bbl. 10 1 3368 600 14,400 0.24 14 28 342 8 12 1 . 6042 720 17,280 0.29 17. 14 411 4 15 2.0052 900 21,600 0.36 21.43 514 3 18 2 4063 1,080 25,920 0.43 25.71 617 1 20 2 6733 1,200 28,800 0.48 28.57 685 7 25 3.342 1,500 36,000 0.59 35.71 857 27 3.609 1,620 38,880 0.64 38.57 925 30 4.001 1,800 43,200 0.71 42.85 1,028 35 4 678 2 100 50,400 83 50.0 1 200 36 4 812 2 160 51,840 86 51 43 1 234 40 45 50 60 70 0.1 5.348 6.015 6.684 8.021 9.357 2,400 2,700 3,000 3,600 4,200 57,600 64,800 72,000 86,400 100,800 0.95 1.07 1.19 1.43 1.66 57.14 64.28 71.43 85.71 100.0 1,371.0 1,543.0 1,714.0 2,057.0 2 400 75 10 026 4,500 108,000 1 78 107 14 2 570 80 10 694 4,800 115,200 1 90 114 28 2 742 90 100 0.2 12.031 13 368 5,400 6,000 129,600 144,000 2.14 2 39 128.5 142 8 3,085.0 3 428 125 16.710 7,500 180,000 2 98 178.6 4 286 135 150 0.3 18.046 20 052 8,100 9 000 194,400 216 000 3.21 3 57 192.8 214 3 4,628.0 5 143 175 23 394- 10 500 252 000 4 16 250 6 000 180 200 0.4 24.062 26 736 10,800 12 000 259,200 288,000 4.28 4 76 257.0 285 7 6,171.0 6 857 225 250 0.5 30.079 33.421 13,500 15,000 324,000 360,000 5.35 5 95 321.4 357.1 7,714.0 8 570 270 300 315 360 400 0.6 0.8 36.093 40.104 42.109 48.125 53 472 16,200 18,000 18,900 21,600 24 000 388,800 432,000 453,600 518,400 576,000 6.43 7.14 7.5 8.57 9 52 385.7 428.5 450.0 514.3 571 8 9,257.0 10,284.0 10,800.0 12,342.0 13 723*0 450 500 1.0 60.158 66 842 27,000 30 000 648,000 720,000 10.7 11 9 642.8 714 3 15,428.0 17 143 540 600 1.2 72.186 80.208 32,400 36,000 777,600 864,000 12.8 14.3 771.3 857.1 18,512.0 20,570 630 675 720 800 900 ,000 1.4 1.5 1.6 2.0 84.218 90.234 96.25 106.94 120.31 133.68 37,800 40,500 43,200 48,000 54,000 60,000 907,200 972,000 1,036,800 1,152,000 1,296,000 1,440,000 15.0 16.0 17.0 19.05 21.43 23 8 900.0 964.0 1,028.0 1,142.0 1,285.0 1,428 21,600.0 23,143.0 24,685.0 27,387.0 30,857.0 34 284 ,125 ,200 ,350 ,500 2.5 3.0 150.39 160.42 180.46 200.52 67,500 72,000 81,000 90,000 1,620,000 1,728,000 1,944,000 2,160,000 26.78 28.57 32.14 35 71 1,607.0 1,714.0 1,928.0 2 142 38,571.0 41,143.0 46,085.0 51 427 ,575 ,800 2,000 3.5 4.0 210.54 240.62 267 36 94,500 108,000 120 000 2,268,000 2,592,000 2 880 000 37.5 42.85 47 64 2,250.0 2,571.0 2 857 54,000.0 61,710.0 68 568 2,025 2,250 4.5 270.70 300 78 121,500 135 000 2,916,000 3 240 000 48.21 53 57 2,892.0 3 214 69,425.0 77 143 2,500 334 21 150 000 3 600 000 59 52 3 571 85 704 2,700 3,000 6.0 360.93 401.04 162,000 180,000 3,880,000 4,320,000 64.3 71.43 3,857.0 4,285.0 92,572.0 102,840.0 TABLES 263 o o- 1 1 1 HI 1 111 Bill Cubic meter 00 Tf t^ lO -CO co * -oo O O C co ^-t t^- 00 lO O CO > "s & 8-a I-;-- p H II II II II ^11 ? || || II 5 CO Tt* O 00 N : :8"8 o o *^ 111 III r-H.-l.-H ft O< ,0 a CT M a 3 (M O I> M< CO T}< -GO O O =2 a fe-3 | -2 n CD 5 ill | 1 O* ^2 d o o >c o o o O O O -3 II II II * i * i J ^ i fl 2 g S3 -5 3 K .2 i & s^^ i s M 1 S 1 -Q n 3 2 | GO CO O CO 1C CO CO O i-l OO ^H < e 2 o> 2 H S * 3 3 i-Hi-lOOOOOCSOCO dddrHO'-IiccJic CO CO r EIGH ee e I i 4-3 -*^> O B 3 S (Q t/3 -^ II g s 2 2 S ,jg s O (N O O 1> O O ^ [ * ^ ^ ^ O 5 5 S ^ M 'D W el -^ CJ Q> m 1 -gill 1 1 1 1 - : | ^ 8 g S a s s | i | > u &< 1 U 1M CM t~ O '. '. ce K J pq < EH 1 g>-S5S s 1 s 1 ^^ ,a "^SSSi 2^5^ >o M -eigd^ ss-ss H H g M ^ .3 g 3 a % 2 a ^ t^ CO CO O CO O CM OOOOCMrHCMOCMO ^ II II II II co d d d -g ii ii ii ii ja 43 "^ *" o o "- 1 -73 -ri a ^ 1 S M o-ocoo-^oo CM o C o> CU **-< i-iO (N (N'-it^ OOO5O 01>t^ OO^iO t^lNrH o o o o 1-1 r* N e eo *Q t>o(NTj OOCOOOMCS "oc M . OOWOOOiN Ol>00(NOO c a i-3 o tH OO lO t>- t CO t^ rt< * * t^ Tfi rt* lO CO **< I s - CO O -^ IN tO Tt< CO ddddd O l> CO I-HI-H NIMCOCO-* ujcooooi-i co-^ CO <* 1-1 OS OOCO l>- rt< i 1 T^ CO C^COOOCD lOOCsOCO CDOiCOC Or-I^HCO'C OO^OCO OOO'OO O IH N CO COt^ i^ 0000 OS COCO OOOOCOOOS OOCOTjOt^OO OO 1>O5(NCOS GoVoddlNOO >OCO (NCOS GoolN ^H^HI-H cO O ^ i-J 1H (N I^SSoo OOOO^j iOt>. i-HlN-^iO C5(MtOOOi-l MOCO COO I-II-IT-HI-II-I (NNC4COCO CO -OO 05 O 00 oo d 1-4 ci i Nominal internal "5 W5W5 ^HCO t~t^ COC005COOO CO1OO'OO o o t^ 06 os d 1-4 12-1/8 12-1/4 37.699 38.091 38.484 113.097 115.466 117.859 2 6 . 2832 3.1416 7-1/8 22 . 383 39.871 12-3/8 38 . 877 120.276 2-1/8 6.6759 3 . 5465 7-1/4 22.776 41.282 12-1/2 39 . 270 122.718 2-1/4 7.0686 3.9760 7-3/8 23.169 42.718 12-5/8 39 . 662 125.184 2-3/8 7.4613 4 .'4302 7-1/2 23.562 44.178 12-3/4 40 . 055 127.676 2-1/2 7.8540 4.9087 7-5/8 23.954 45.663 12-7/8 40.448 130.192 2-5/8 2-3/4 8.2467 8.6394 5.4119 5.9395 7-3/4 7-7/8 24.347 24.740 47.173 48.707 13 13-1/8 40.840 41.233 132.732 135.297 2-7/8 9.0321 6.4918 13-1/4 41.626 137.886 8 25.132 50.265 13-3/8 42.018 140.500 3 9.4248 7.0686 8-1/8 25.515 51.848 13-1/2 42.411 143.139 3-1/8 9.8175 7.6699 8-1/4 25.918 53 . 456 13-5/8 42.804 145.802 3-1/4 10.210 8.2957 8-3/8 26.310 55.088 13-3/4 43.197 148.489 3-3/8 10.602 8.9462 8-1/2 26.703 56.745 13-7/8 43.589 151.201 3-1/2 10.995 9.6211 8-5/8 27.096 58.426 3-5/8 11.388 10.320 8-3/4 27.489 60.132 14 43.982 153.938 3-3/4 11.781 11.044 8-7/8 27.881 61.862 14-1/8 44.375 156.699 3-7/8 12.173 11.793 14-1/4 44 . 767 159.485 9 28.274 63.617 14-3/8 45.160 162.295 4 12.566 12.566 9-1/8 28.667 65 . 396 14-1/2 45.553 165.130 4-1/8 12.959 13.364 9-1/4 29 . 059 67 . 200 14-5/8 45.945 167.989 4-1/4 13.351 14.186 9-3/8 29.452 69 . 029 14-3/4 46.338 170.873 4-3/8 13.744 15.033 9-1/2 29.845 70.882 14-7/8 46.731 173.782 TABLES 269 TABLE 8. CIRCUMFERENCES AND AREAS OF CIRCLES (Continued) Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches 15 47.124 176.715 21 65.973 346.361 27 84.823 572.556 15-1/8 47.516 179.672 21-1/8 66.366 350.497 27-1/8 85.215 577.870 15-1/4 47.909 182.654 21-1/4 66.759 354.657 27-1/4 85.608 583.208 15-3/8 48.302 185.661 21-3/8 67.151 358.841 27-3/8 86.001 588.571 15-1/2 48.694 188.692 21-1/2 67.544 363.061 27-1/2 86.394 593.958 15-5/8 49.087 191.748 21-5/8 67.937 367.284 27-5/8 86.786 599 . 370 15-3/4 49.480 194.828 21-3/4 68.329 371.543 27-3/4 87.179 604 . 807 15-7/8 49.872 197.933 21-7/8 68.722 375.826 27-7/8 87.572 610.268 16 50.205 201.062 22 69.115 380 . 133 28 87.964 615.753 16-1/8 50.658 204.216 22-1/8 69 . 507 384.465 28-1/8 88.357 621 . 263 16-1/4 51.051 207.394 22-1/4 69 . 900 388.822 28-1/4 88.750 626.798 16-3/8 51.443 210.597 22-3/8 70.293 393.203 28-3/8 89.142 632.357 16-1/2 51.836 213.825 22-1/2 70.686 397.608 28-1/2 89 . 535 637.941 16-5/8 52.229 217.077 22-5/8 71.078 402.038 28-5/8 89 . 928 643 . 594 16-3/4 52.621 220.353 22-3/4 71.471 406.493 28-3/4 90.321 649 . 182 16-7/8 53.014 223.654 22-7/8 71 . 864 410.972 28-7/8 90.713 654.837 17 53.407 226.980 23 72.256 415.476 29 91.106 660.521 17-1/8 53.799 230.330 23-1/8 72.649 420.004 29-1/8 91.499 666.277 17-1/4 54.192 233.705 23-1/4 73.042 424.557 29-1/4 91.891 671.958 17-3/8 54 . 585 237.104 23-3/8 73.434 429.135 29-3/8 92.284 677.714 17-1/2 54 . 978 240.528 23-1/2 73.827 433.731 29-1/2 92.677 683 . 494 17-5/8 55.370 243.977 23-5/8 74.220 438.363 29-5/8 93.069 689.298 17-3/4 55.763 247.450 23-3/4 74.613 443.014 29-3/4 93 . 462 695.128 17-7/8 56.156 250.947 23-7/8 75.005 447.699 29-7/8 93.855 700.981 18 56.548 254.469 24 75.398 452.390 30 94.248 706.860 18-1/8 56.941 258.016 24-1/8 75.791 457.115 30-1/8 94.640 712.762 18-1/4 57.334 261.586 24-1/4 76.183 461.864 30-1/4 95.033 718.690 18-3/8 57.726 265.182 24-3/8 76.576 466.638 30-3/8 95.426 724.641 18-1/2 58.119 268.803 24-1/2 76.969 471.436 30-1/2 95.818 730.618 18-5/8 58.512 272.447 24-5/8 77.361 476.259 30-5/8 96.211 736.619 18-3/4 58.905 276.117 24-3/4 77.754 481.106 30-3/4 96.604 742.644 18-7/8 59 . 297 279.811 24-7/8 78.147 485.978 30-7/8 96.996 748.694 19 59 . 690 283 . 529 25 78.540 490.875 31 97.389 754 . 769 19-1/8 60.083 287.272 25-1/8 78.932 495.796 31-1/8 97.782 760 . 868 19-1/4 60.475 291.039 25-1/4 79 . 325 500.741 31-1/4 98.175 766.992 19-3/8 60.868 294.831 25-3/8 79.718 505.711 31-3/8 98.567 773 . 140 19-1/2 61.261 298.648 25-1/2 80.110 510.706 31-1/2 98.968 779.313 19-5/8 61.653 302.489 25-5/8 80 . 503 515.725 31-5/8 99 . 353 785.510 19-3/4 62.046 306.355 25-3/4 80.896 520.769 31-3/4 99.745 791.732 19-7/8 62.439 310.245 25-7/8 81.288 525.837 31-7/8 100.138 797.978 20 62 . 832 314.160 26 81.681 530.930 32 100.531 804.249 20-1/8 63 . 224 318.099 26-1/8 82.074 536.047 32-1/8 100.924 810.545 20-1/4 63.617 322.063 26-1/4 82.467 541.189 32-1/4 101.316 816.865 20-3/8 64.010 326.051 26-3/8 82.859 546.356 32-378 101.709 823 . 209 20-1/2 64 . 402 330 . 064 26-1/2 83.252 551 . 547 32-1/2 102.102 829.578 20-5/8 64 . 795 334.101 26-5/8 83.645 556.762 32-5/8 102.494 835.972 20-3/4 65.188 338.163 26-3/4 84.037 562.002 32-3/4 102.887 842.390 20-7/8 65.580 1342.250! 26-7/8 84.430 567.267: 32-7/8 103.280 848.833 270 ELEMENTS OF HYDRAULICS CIRCUMFERENCES AND AREAS OF CIRCLES (Continued} Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches 33 103.672 855.30 39 122.522 1194.59 45 141.372 1590.43 33-1/8 104.055 861 . 79 39-1/8 122.915 1202.26 45-1/8 141.764 1599.28 33-1/4 104 . 458 868.30 39-1/4 123.307 1209.95 45-1/4 142.157 1608.15 33-3/8 104.850 874.84 39-3/8 123.700 1217.67 45-3/8 142.550 1617.04 33-1/2 105.243 881.41 39-1/2 124 . 093 1225.42 45-1/2 142.942 1625.97 33-5/8 105.636 888.00 39-5/8 .124.485 1233.18 45-5/8 143.335 1634.92 33-3/4 106.029 894.61 39-3/4 124.878 1240.98 45-3/4 143.728 1643.89 33-7/8 106.421 901.25 39-7/8 125.271 1248.79 45-7/8 144.120 1652.88 34 106.814 907.92 40 125.664 1256.64 46 144.513 1661.90 34-1/8 107.207 914.61 40-1/8 126.056 1264.50 46-1/8 144.906 1670.95 34-1/4 107.599 921.32 40-1/4 126.449 1272.39 46-1/4 145.299 1680.01 34-3/8 107.992 928.06 40-3/8 126.842 1280.31 46-3/8 145.691 1689.10 34-1/2 108.385 934.82 40-1/2 127.234 1288.25 46-1/2 146.084 1698.23 34-5/8 108.777 941 . 60 40-5/8 127.627 1296.21 46-5/8 146.477 1707.37 34-3/4 109.170 948.41 40-3/4 128.020 1304 . 20 46-3/4 146.869 1716.54 34-7/8 109.563 955.25 40-7/8 128.412 1312.21 46-7/8 147.262 1725.73 35 109.956 962.11 41 128.805 1320.25 47 147.655 1734.94 35-1/8 110.348 968.99 41-1/8 129.198 1328.32 47-1/8 148.047 1744.18 35-1/4 110.741 975.90 41-1/4 129.591 1336.40 47-1/4 148.440 1753.45 35-3/8 111.134 982.84 41-3/8 129.983 1344.51 47-3/8 148.833 1762.73 35-1/2 111.526 989 . 80 41-1/2 130.376 1352.65 47-1/2 149 . 226 1772.05 35-5/8 111.919 996.78 41-5/8 130.769 1360.81 47-5/8 149.618 1781.39 35-3/4 112.312 1003.78 41-3/4 131.161 1369.00 47-3/4 150.011 1790.76 35-7/8 112.704 1010.82 41-7/8 131.554 1377.21 47-7/8 150.404 1800.14 36 113.097 1017.88 42 131.947 1385.44 48 150.796 1809.56 36-1/8 113.490 1024.95 42-1/8 132.339 1393.70 48-1/8 151.189 1818.99 36-1/4 113.883 1032.06 42-1/4 132.732 1401.98 48-1/4 151.582 1828.46 36-3/8 114.275 1039.19 42-3/8 133.125 1410.29 48-3/8 151.974 1837.93 36-1/2 114.668 1046.35 42-1/2 133.518 1418.62 48-1/2 152.367 1847.45 36-5/8 115.061 1053.52 42-5/8 133.910 1426.98 48-5/8 152.760 1856.99 36-3/4 115.453 1060.73 42-3/4 134.303 1435.36 48-3/4 153.153 1866.55 36-7/8 115.846 1067.95 42-7/8 134.696 1443.77 48-7/8 153.545 1876.13 37 116.239 1075.21 43 135.088 1452.20 49 153.938 1885.74 37-1/8 116.631 1082.48 43-1/8 135.481 1460.65 49-1/8 154.331 1895.37 37- 1/4 117.024 1089 . 79 43-1/4 135.874 1469.13 49-1/4 154.723 1905.03 37-3/8 117.417 1097.11 43-3/8 136.266 1477.63 49-3/8 155.116 1914.70 37-1/2 117.810 1104.46 43-1/2 136.659 1486.17 49-1/2 155.509 1924.42 37-5/8 118.202 1111.84 43-5/8 137.052 1494.72 49-5/8 155.901 1934.15 37-3/4 118.595 1119.24 43-3/4 137.445 1503.30 49-3/4 156.294 1943.91 37-7/8 118.988 1126.66 43-7/8 137.837 1511.90 49-7/8 156.687 1953.69 38 119.380 1134.11 44 138.230 1520.53 50 157.080 1963.50 38-1/8 119.773 1141.59 44-1/8 138.623 1529.18 50-1/4 157.865 1983.18 38-1/4 120.166 1149.08 44-1/4 139.015 1537.86 50-1/2 158.650 2002.96 38-3/8 120.558 1156.61 44-3/8 139.408 1546.55 50-3/4 159.436 2022 . 84 38-1/2 120.951 1164.15 44-1/2 139.801 1555.28 51 160.221 2042 . 82 38-5/8 121.344 1171.73 44-5/8 140.193 1564.03 51-1/4 161.007 2062 . 90 38-3/4 121.737 1179.32 1| 44-3/4 140.586 1572.81 51-1/2 161.792 2083 . 07 38-7/8 122.129 1186. 94! | 44-7/8 140.979 1581.61 51-3/4 162.577 2103.35 TABLES 271 CIRCUMFERENCES AND AREAS OF CIRCLES (Continued) Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches 52 163.363 2123.72 63 197.920 3117.25 74 232.478 4300 . 84 52-1/4 164.148 2144.19 63-1/4 198.706 3142.04 74-1/4 233.263 4329.95 52-1/2 164.934 2164.75 63-1/2 199.491 3166.92 74-1/2 234.049 4359.16 52-3/4 165.719 2185.42 63-3/4 200.277 3191.91 74-3/4 234 . 834 4388 . 47 53 166.504 2206.18 64 201.062 3216.99 75 235.620 4417.86 53-1/4 167.490 2227.05 64-1/4 201 . 847 3242.17 75-1/4 236.405 4447.37 53-1/2 168.075 2248.01 64-1/2 202.633 3267.46 75-1/2 237.190 4476.97 53-3/4 168.861 2269.06 64-3/4 203.418 3292.83 75-3/4 237.976 4506.67 54- 169.646 2290 . 22 65 204.204 3318.31 76 238.761 4536.46 54-1/4 170.431 2311.48 65-1/4 204.989 3343.88 76-1/4 239.547 4566.36 54-1/2 171.217 2332.83 65-1/2 205.774 3369.56 76-1/2 240.332 4596.35 54-3/4 172.002 2354.28 64-3/4 206.560 3395.33 76-3/4 241.117 4626.44 55 172.788 2375.83 66 207.345 3421 . 19 77 241.903 4656.63 55-1/4 173.573 2397.48 66-1/4 208.131 3447.16 77-1/4 242.688 4686.92 55-1/2 174.358 2419.22 66-1/2 208.916 3473.33 77-1/2 243.474 4717.30 55-3/4 175.144 2441.07 66-3/4 209.701 3499.39 77-3/4 244.259 4747.79 56 175.929 2463.01 67 210.487 3525 . 66 78 245.044 4778.36 56-1/4 176.715 2485.05 67-1/4 211.272 3552.01 78-1/4 245 . 830 4809.05 56-1/2 177.500 2507.19 67-1/2 212.058 3578.47 78-1/2 246.615 4839.83 56-3/4 178.285 2529.42 67-3/4 212.843 3605.03 78-3/4 247.401 4870.70 57 179.071 2551 . 76 68 213.628 3631.68 79 248.186 4901 . 68 57-1/4 179.856 2574.19 68-1/4 214.414 3658.44 79-1/4 248.971 4932.75 57-1/2 180.642 2596.72 68-1/2 215.199 3685.29 79-1/2 249 . 757 4963.92 57-3/4 181.427 2619.35 68-3/4 215.985 3712.24 79-3/4 250.542 4995.19 58 182.212 2642.08 69 216.770 3739.28 80 251 . 328 5026.55 58-1/4 182.998 2664.91 69-1/4 217.555 3766.43 80-1/2 252.898 5089.58 58-1/2 183.783 2687.83 69-1/2 218.341 3793.67 58-3/4 184.569 2710.85 69-3/4 219.126 3821.02 81 254.469 5153.00 81-1/2 256 . 040 5216.82 59 185.354 2733.97 70 219.912 3848.45 82 257.611 5281.02 59-1/4 59-1/2 186.139 186.925 2757.19 2780.51 70-1/4 70-1/2 220.697 221.482 3875.99 3903.63 82-1/2 259.182 5345.62 59-3/4 187.710 2803.92 70-3/4 222.268 3931.36 83 260.752 5410.61 83-1/2 262 . 323 5476.00 60 188.496 2827.43 71 223.053 3959.19 60-1/4 189.281 2851.05 71-1/4 223.839 3987.13 84 263.894 5541.77 60-1/2 190.066 2874 . 76 71-1/2 224 . 624 4015.16 84-1/2 265.465 5607 . 95 60-3/4 190.852 2898.56 71-3/4 225 . 409 4043 . 28 85 267.035 5674.51 61 191.637 2922.47 72 226.195 4071 . 50 85-1/2 268.606 5741.47 61-1/4 192.423 2946.47 72-1/4 226.980 4099 . 83 86 270. 177 5808 . 80 61-1/2 61-3/4 193 . 208 193.993 2970.57 2994.77 72-1/2 72-3/4 227.766 228.551 4128.25 4156.77 86-1/2 271.748 5876.55 87 273.319 5944 . 68 62 194 . 779 3019.07 73 229.336 4185.39 87-1/2 274.890 6013.21 62-1/4 195.564 3043.47 73-1/4 230.122 4214.11 62-1/2 196.350 3067 . 96 73-1/2 230.907 4242.92 88 276.460 6082.12 62-3/4 197.135 3092.56 73-3/4 231.693 4271.83 II 88-1/2 278.031 6151.44 272 ELEMENTS OF HYDRAULICS CIRCUMFERENCES AND AREAS OF CIRCLES (Continued} Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches Diame- ter inches Circum- ference inches Area square inches 89 89-1/2 279.602 281.173 6221.14 6291.25 100 100-1/2 314.159 315.730 7853.98 7938.72 111 111-1/2 348.717 350.288 9766.89 9674 . 28 90 90-1/2 282 . 744 284.314 6361 . 73 6432.62 101 101-1/2 317.301 318.872 8011.85 8091 . 36 112 112-1/2 351.858 353.430 9852.03 9940 . 20 91 91-1/2 285.885 287.456 6503.88 6573.56 102 102-1/2 320.442 322.014 8171.28 8251.60 113 113-1/2 355 . 000 356.570 10028.75 10117.68 92 92-1/2 289 . 027 290 . 598 6647.61 6720.07 103 103-1/2 323 . 584 325.154 8332.29 8413.40 114 114-1/2 358.142 359.712 10207.03 10296.76 93 93-1/2 292.168 293 . 739 6792.91 6866.16 104 104-1/2 326.726 328.296 8494.87 8576.76 115 115-1/2 361 . 283 362.854 10386 . 89 10477 . 40 94 94-1/2 295.310 296.881 6939 . 78 7013.81 105 105-1/2 329.867 331.438 8659.01 8741.68 116 116-1/2 364.425 365.996 10568.32 10659.64 95 95-1/2 298.452 300.022 7088.22 7163.04 106 106-1/2 333.009 334 . 580 8824.73 8908.20 117 117-1/2 367.566 369 . 138 10751.32 10843 . 40 96 96-1/2 301 . 593 302.164 7238. 23 7313.84 107 107-1/2 336.150 337.722 8992.02 9076.24 118 118-1/2 370.708 372.278 10935.88 11028.76 97 97-1/2 304.734 306.306 7389.81 7474.20 108 108-1/2 339 . 292 340 . 862 9160.88 9245.92 119 119-1/2 373.849 375.420 11122.02 11215.68 98 98-1/2 307.876 309.446 7542.96 7620. 12 109 109-1/2 342.434 344 . 004 9331.32 9417.12 120 376.991 11309.73 99 311.018 7697.69 110 345.575 9503.32 99-1/2 312.588 7775.64 110-1/2 347.146 9589.92 TABLES 273 TABLE 9. 1 EFFLUX COEFFICIENTS FOR CIRCULAR ORIFICE Values of efflux coefficient K in Eq. (25), Art. 8, Q = 2/3Kb\/2g(H 3 / 2 - fc 3 / 2 ), for circular, vertical orifices, with sharp edges, full contraction and free discharge in air. For heads over 100 ft., use K = 0.592. Head on cen- ter of orifice in feet Diameter of orifice in feet 0.02 0.03 0.04 0.05 0.07 0.10 0.12 0.15 0.20 0.40 0.60 0.80 1.0 0.3 0.4 0.5 0.6 0.7 0.637 6280.621 613 608 0.655 0.651 0.643 0.640 0.637 0.637 0.633 0.630 0.628 0.631 0.627 0.624 0.622 0.624 0.621 0.618 0.616 0.618 0.615 0.613 0.611 0.612 0.610 0.609 0.607 0.606 0.605 0.605 0.604 0.600 0.601 0.601 0.596 0.596 0.597 0.592 0.593 0.594 0.590 0.591 0.590 0.8 0.9 1.0 1.2 1.4 0.648 0.646 0.644 0.641 0.638 0.634 0.632 0.631 0.628 0.625 0.626 0.624 0.623 0.620 0.618 0.620 0.618 0.617 0.615 0.613 0.615 0.613 0.612 0.610 0.609 0.610 0.609 0.608 0.606 0.605 0.606 0.605 0.605 0.604 0.603 0.603 0.603 0.603 0.602 0.601 0.601 0.601 0.600 0.600 0.600 0.597 0.598 0.598 0.598 0.599 0.594 0.595 0.595 0.596 0.596 0.592 0.593 0.593 0.594 0.594 0.591 0.591 0.591 0.592 0.593 1.6 1.8 2.0 2.5 3.0 0.636 0.634 0.632 0.629 0.627 0.624 0.622 0.621 0.619 0.617 0.617 0.615 0.614 0.612 0.611 0.612 0.611 0.610 0.608 0.606 0.608 0.607 0.607 0.605 0.604 0.605 0.604 0.604 0.603 0.603 0.602 0.602 0.601 0.601 0.601 0.601 0.601 0.600 0.600 0.600 0.600 0.599 0.599 0.599 0.599 0.599 0.599 0.599 0.599 0.599 0.5970.595 0.5970.595 0.5970.596 0.5980.597 0.5980.597 0.594 0.595 0.595 0.596 0.597 3.5 4.0 5.0 6.0 7.0 0.625 0.623 0.621 0.618 0.616 0.616 0.614 0.613 0.611 0.609 0.610 0.609 0.608 0.607 0.606 0.606 0.605 0.605 0.604 0.603 0.604 0.603 0.603 0.602 0.601 0.602 0.602 0.601 0.600 0.600 0.601 0.600 0.599 0.599 0.599 0.600 0.599 0.599 0.599 0.599 0.599 0.599 0.598 0.598 0.598 0.599 0.598 0.598 0.598 0.958 0.598 0.597 0.597 0.597 0.597 0.597 0.597 0.596 0.596 0.596 0.596 0.596 0.596 0.596 0.596 8.0 9.0 10.0 20.0 50.0 100.0 0.614 0.613 0.611 0.601 0.596 0.593 0.608 0.607 0.606 0.600 0.596 0.593 0.605 0.604 0.603 0.599 0.595 0.592 0.603 0.602 0.601 0.598 0.595 0.592 0.601 0.600 0.599 0.597 0.594 0.592 0.600 0.599 0.598 0.596 0.594 0.592 0.599 0.599 0.598 0.596 0.594 0.592 0.598 0.598 0.597 0.596 0.594 0.592 0.5980.597 0.5970.597 0.5970.597 0.5960.596 0.5940.594 0.5920.592 0.596 0.596 0.596 0.596 0.594 0.592 0.596 0.596 0.596 0.595 0.593 0.592 0.596 0.595 0.595 0.594 0.593 0.592 1 From Hamilton Smith's Hydraulics. 18 274 ' ELEMENTS OF HYDRAULICS TABLE 10. l EFFLUX COEFFICIENTS FOR SQUARE ORIFICE Values of efflux coefficient K in Eq. (25), Art. 8, Q = 2/3 Kb\/2-j(H z ' 2 - h 3/2 ), for square, vertical orifices, with sharp edges, full contraction, and free discharge in air. For heads over 100 ft., use K = 0.598 Head on cen- ter of orifice in feet Side of square in feet 0.02 0.03 0.04 0.50 0.07 0.10 0.12 0.15 0.20 0.40 0.60 0.80 1.0 0.3 0.4 0.5 0.6 0.7 0.660 0.656 . 648 0.645 0.642 0.643 0.639 0.636 0.633 0.642 0.637 0.633 0.630 0.628 0.632 0.628 0.625 0.623 0.621 0.624 0.621 0.619 0.617 0.616 0.617 0.616 0.614 0.613 0.612 0.612 0.611 0.610 0.610 0.609 0.605 0.605 0.605 0.601 0.597 0.601 0.598 0.6020.599 0.596 0.598 0.596 0.8 0.9 .0 .2 .4 0.652 0.650 0.648 0.644 0.642 0.639 0.637 0.636 0.623 0.630 0.631 0.629 0.628 0.625 0.623 0.625 0.623 0.622 0.620 0.618 0.620 0.619 0.618 0.616 0.614 0.615 0.614 0.613 0.611 0.610 0.611 0.610 0.610 0.609 0.608 0.608 0.608 0.608 0.607 0.606 0.605 0.605 0.605 0.605 0.605 0.6020.6000.598 0.603 ! 0.601 0.599 0.6030.601j0.600 0.6040.6020.601 0.6040.6020.601 0.597 0.598 0.599 0.600 0.601 .6 .8 2.0 2.5 3.0 0.640 0.638 0.637 0.634 0.632 0.628 0.627 0.626 0.624 0.622 0.621 0.620 0.619 0.617 0.616 0.617 0.616 0.615 0.613 0.612 0.613 0.612 0.612 0.610 0.609 0.609 0.609 0.608 0.607 0.607 0.607 0.607 0.606 0.606 0.606 0.606 0.606 0.605 0.605 0.605 0.606 0.606 0.606 0.606 0.606 0.605 0.605 0.605 0.605 0.605 0.6050.6030.602 0.6050.603!0.602 0.6050.6040.602 0.605^.6040.603 0.605'Q. 604 0.603 0.601 0.602 0.602 0.602 0.603 3.5 4.0 5.0 6.0 7.0 0.630 0.628 0.626 0.623 0.621 0.621 0.619 0.617 0.616 0.615 0.615 0.614 0.613 0.612 0.611 0.611 0.610 0.610 0.609 0.608 0.609 0.608 0.607 0.607 0.607 0.607 0.606 0.606 0.605 0.605 0.605 0.605 0.605 0.605 0.604 0.605 0.605 0.604 0.604 0.604 0.605 0.605 0.604 0.604 0.604 0.604 0.603 0.603 0.603 0.603 0.6030.602 0.6030.602 0.6020.602 0.6020.602 0.6020.602 8.0 9.0 10.0 20.0 50.0 100.0 0.619 0.618 0.616 0.606 0.602 0.599 0.613 0.612 0.611 0.605 0.601 0.598 0.610 0.609 0.608 0.604 0.601 0.598 0.608 0.607 0.606 0.603 0.601 0.598 0.606 0.606 0.605 0.602 0.601 0.598 0.605 0.604 0.604 0.602 0.600 0.598 0.604 0.604 0.604 0.602 0.600 0.598 0.604 0.604 0.603 0.602 0.600 0.598 0.604 0.603 0.603 0.602 0.600 0.598 0.603 0.603 0.603 0.601 0.600 0.598 0.603 0.602 0.602 0.601 0.599 0.598 0.602 0.602 0.602 0.601 0.599 0.598 0.602 0.601 0.601 0.600 0.599 0.598 1 From Hamilton Smith's Hydraulics. TABLES 275 TABLE 11. FIRE STREAMS From Tables Published by John R. Freeman 3/4-in. Smooth Nozzle Pressure in pounds per sq. in. re- Pressure at nozzle in pounds per Discharge in gallons per min. Height of effective fire stream Horizontal distance of stream quired at hydrant or pump to main- tain pressure at nozzle through vari- ous lengths of 2-1/2-in. smooth, rubber-lined hose. sq. in. 50 100 200 300400 500 600 800 1000 ft. ft. ft. ft. ft. ft. ft. ft. ft. 35 97 55 41 37 38 40 42 44 46 48 53 57 40 104 60 44 42 43 46 48 50 53 55 60 65 45 110 64 47 47 48 51 54 57 59 62 68 73 50 116 67 50 52 54 57 60 63 66 69 75 81 55 122 70 52 58 59 63 66 69 73 76 83 89 60 127 72 54 63 65 68 72 76 79 83 90 97 65 132 74 56 68 70 74 78 82 86 90 98 106 70 137 76 58 73 75 80 84 88 92 97 105 114 75 142 78 60 79 81 85 90 94 99 104 113 122 80 147 79 62 84 86 91 96 101 106 1.11 120 130 85 151 80 64 89 92 97 102 107 112 117 128 138 90 156 81 65 94 97 102 108 113 119 124 135 146 95 160 82 66 99 102 108 114 120 125 131 143 154 100 164 83 68 105 108 114 120 126 132 138 150 163 7/8-in. Smooth Nozzle 35 133 56 46 38 40 44 48 52 56 60 68 76 40 142 62 49 43 46 50 55 59 64 68 78 87 45 150 67 52 49 51 57 62 67 72 77 87 97 50 159 71 55 54 57 63 69 74 80 86 97 108 55 166 74 58 60 63 69 75 82 88 94 107 119 60 174 77 61 65 69 75 82 89 96 103 116 130 65 181 79 64 71 74 82 89 96 104 111 126 141 70 188 81 66 76 80 88 96 104 112 120 136 152 75 194 83 68 82 86 94 103 111 120 128 145 162 80 201 85 70 87 91 101 110 119 128 137 155 173 85 207 87 72 92 97 107 116 126 136 145 165 184 90 213 88 74 98 103 113 123 134 144 154 174 195 95 219 89 75 103 109 119 130 141 152 163 184 206 100 224 90 76 109 114 126 137 148 160 171 194 216 1-in. Smooth Nozzle 35 174 58 51 40 44 51 57 64 71 78 92 105 40 186 64 55 46 50 58 66 73 81 89 105 120 45 198 69 58 52 56 65 74 83 91 100 118 135 ' 50 208 73 61 57 62 72 82 92 102 111 131 151 55 218 76 64 63 69 79 90 101 112 122 144 166 60 228 79 67 67 75 87 98 110 122 134 157 181 65 237 82 70 75 81 94 107 119 132 145 170 196 70 246 85 72 80 87 101 115 128 142 156 183 211 75 255 87 74 86 94 110 123 138 152 167 196 226 80 263 89 76 92 100 115 131 147 162 178 209 241 85 274 91 78 98 106 123 139 156 173 189 222 90 279 92 80 103 112 130 147 165 183 200 236 95 287 94 82 109 118 137 156 174 193 211 249 100 295 96 83 115 125 144 164 183 203 223 276 ELEMENTS OF HYDRAULICS FIRE STREAMS (Continued} 1-1/8-inch Smooth Nozzle % Pressure in pounds per sq. in. required Pressure at nozzle in pound per sq. in. Discharge in gallons per min. Height of effective fire stream Horizontal distance of stream at hydrant or pump to maintain pres- sure at nozzle through various lengths of 2-1/2-in. smooth, rubber-lined hose 50 100 200 300 400 500 600 800 jll 000 ft. ft. ft. ft. ft. ft. ft. ft. ft. 35 222 59 54 43 49 60 71 82 94 105 127 149 40 238 65 59 50 56 69 81 94 107 120 145 171 45 252 70 63 56 63 77 92 106 120 135 163 192 50 266 75 66 62 70 86 102 118 134 150 181 213 55 279 80 69 68 77 95 112 130 147 165200 235 60 291 83 72 74 84 103 122 141 160 180 218 256 65 303 86 75 81 91 112 132 153 174 195 236 70 314 88 77 87 98 120 143 165 187 209 254 75 325 90 79 93 105 12,9 153 177 201 224 80 336 92 81 99 112 138 163 188 214 239 85 346 94 83 106 119 146 173 200 227 254 90 356 96 85 112 126 15& 183 212 241 95 366 98 87 118 133 163 194 224 254 ... 100 376 99 89 124 140 172 204 236 1-1/4-inch Smooth Nozzle 35 277 60 59 48 57 74 91 109 126 142 178 212 40 296 67 63 55 65 84 104 124 144 164 203 243 45 314 72 67 62 73 95 117 140 162 184 229 50 331 77 70 68 81 106 130 155 180 204 254 55 347 81 73 75 89 116 143 170 198 225 60 363 85 76 82 97 127 156 186 216 245 65 377 88 79 89 105 137 169 201 234 70 392 91 81 96 113 148 182 217 252 75 405 93 83 103 121 158 195 232 80 419 95 85 110 129 169 208 248 85 432 97 88 116 137 179 221 90 444 99 90 123 145 190 234 95 456 100 92 130 154 210 247 100 468 101 93 137 162 211 261 1-3/8-inch Smooth Nozzle 35 340 62 62 54 67 94 120 146 172 198 250 40 363 69 66 62 77 107 137 166 196 226 45 385 74 70 70 87 120 154 187 221 454 50 406 79 73 78 96 134 171 208 245 55 426 83 76 86 106 147 188 229 270 60 445 87 79 93 116 160 205 250 65 463 90 82 101 125 174 222 70 480 92 84 109 135 187 239 75 497 95 86 117 145 201 256 on 514 97 88 124 154 214 oU 85 529 99 90 132 164 227 90 545 100 92 140 173 240 95 560 101 94 148 183 254 100 574 103 96 156 193 TABLES 277 TABLE 12. l COEFFICIENTS OF PIPE FRICTION Values of the friction coefficient, /, in the formula Computed from the exponential formulas of Thrupp, Tutton and Unwin Diameter ' /elocity of flow in feet per second in inches 2 4 6 8 10 Lead pipe 1 2 3 4 0.032 0.030 0.029 0.028 0.026 0.025 0.024 0.023 0.024 0.023 0.022 0.021 0.022 0.021 0.020 0.020 0.021 0.020 0.019 0.019 6 0.034 0.033 0.032 0.032 12 18 0.027 024 0.027 024 0.026 023 0.026 023 24 36 48 0.022 0.020 018 0.022 0.019 018 0.021 0.019 017 0.021 0.019 017 Asphalted pipe 6 9 12 18 24 36 48 0.026 0.025 0.024 0.023 0.022 0.021 0.020 0.023 0.022 0.021 0.020 0.020 0.019 0.018 0.022 0.021 0.020 0.019 0.018 0.017 0.017 0.021 0.020 0.019 0.018 0.017 0.017 0.016 0.020 0.019 0.019 a. 018 0.017 0.016 0.015 Bare wrought iron pipe 3 6 12 24 36 48 60 0.024 0.022 0.019 0.017 0.016 0.015 0.015 0.021 0.019 0.017 0.015 0.014 0.013 0.013 0.019 0.017 0.015 0.014 0.013 0.012 0.012 0.018 0.016 0.014 0.013 0.012 0.011 0.011 0.017 0.016 0.014 0.012 0.011 0.011 0.010 Riveted wrought iron or steel pipe 12 24 36 48 60 72 0.025 0.020 0.017 0.016 0.015 0.014 0.022 0.018 0.016 0.014 0.013 0.013 0.021 0.017 0.015 0.014 0.013 0.012 0.020 0.016 0.014 0.013 0.012 0.011 0.019 0.016 0.014 0.013 0.012 0.011 3 6 0.028 024 0.026 022 0.025 022 0.025 021 9 0.021 0.020 0.020 0.019 New cast-iron pipe 12 18 24 36 0.020 0.018 0.017 015 0.019 0.017 0.016 015 0.018 0.017 0.016 014 0.018 0.016 0.015 014 3 6 0.059 050 0.058 050 0.058 050 0.058 049 Old cast iron pipe 9 12 18 24 36 0.046 0.043 0.039 0.037 0.033 0.045 0.042 0.039 0.036 0.033 0.045 0.042 0.038 0.036 0.033 0.044 0.042 0.038 0.036 0.032 1 Compiled from data in Gibson's Hydraulics 278 ELEMENTS OF HYDRAULICS TABLE 13. FRICTION HEAD IN PIPES Friction head in feet for each 100 ft. of straight, clean, cast-iron pipe. For old pipes the tabular values of the friction head should be doubled. Computed from Williams and Hazen's formula, = Cr-ss- 54 0.001 -' 04 : = velocity in feet per sec., s = slope; r = hydraulic radius in feet, C = 100. c 'a t k II Inside diameter of pipe 1/2 in. 3/4 in. 1 in. 1-1/2 in. 2 in. Velocity in feet per sec. Friction head in feet per 100 it. Velocity in ft. per sec. ll Velocity in feet per seo. || .3 g en fel !.] Id 1*8 a d .-a a O "^ 'c c 2 fc . > * fe If ^ i 2 3 4 5 1.05 2.10 3.16 4.21 5.26 2.1 7.4 15.8 27.0 41.0 i| 1.20 1.80 2.41 3.01 1.9 4.1 7.0 10.5 1.12 1.49 1.86 1.26 2.14 3.25 0.63 0.79 0.26 0.40 i 0.61 0.82 1.02 1.23 1.53 6 8 10 12 15 6.31 8.42 10.52 57.0 98.0 147.0 3.61 4.81 6.02 7.22 9.02 14.7 25.0 38.0 53.0 80.0 2.23 2.98 3.72 4.46 5.57 4.55 7.8 11.7 16.4 25.0 0.94 1.26 1.57 1.89 2.36 0.56 0.95 1.43 2.01 3.05 0.20 0.33 0.50 0.70 1.07 20 25 30 35 40 12.03 136.0 7.44 9.30 11.15 13.02 14.88 42.0 64.0 89.0 119.0 152.0 3.15 3.93 4.72 5.51 6.30 5.2 7.8 11.0 14.7 18.8 2.04 2.55 3.06 3.57 4.08 1.82 2.73 3.84 5.1 6.6 50 60 70 80 90 7.87 9.44 11.02 12.59' 14.17 28.4 39.6 53.0 68.0 84.0 5.11 6.13 7.15 8.17 9.19 9.9 13.9 18.4 23.7 29.4 100 120 140 160 180 ! 15.74 18.89 22.04 102.0 143.0 190.0 10.21 12.25 14.30 16.34 18.38 35.8 50.0 67.0 86.0 107.0 | 200 250 1 i 20.42 25.53 129.0 196.0 1 TAKLKS FRICTION HEAD IN PIPES (Continued) Discharge in gallons per min. Insult' ili:imt'tt>r of pipe 2-1/2 in. 3 in. 4 in. 5 in. 6 in. a I gS 1.1 Friction head in feet per 100 ft. Velocity in feet per sec. Jl JJU !* a l 11 I Jlfi i" Velocity in feet per sec. Friction head in feet per 100ft. s i t* |l |i |J* f. S 10 15 20 25 30 0.65 0.98 1.31 1.63 1.96 0.17 0.37 0.61 ' 0.92 1.29 0.45 0.68 0.91 1.13 1.36 0.07 0.15 0.25 0.38 0.54 0.51 0.64 0.77 0.06 0.09 0.13 0.49 0.04 35 40 50 60 70 2.29 2.61 3.27 3.92 4.58 1.72 2.20 3.32 4.65 6.2 1.59 1.82 2.27 2.72 3.18 0.71 0.91 1.38 1.92 2.57 0.89 1.02 1.28 1.53 1.79 2.04 2.30 2.55 3.06 3.57 0.17 0.22 0.34 0.47 0.63 0.57 0.65 0.82 0.98 1.14 0.06 0.08 0.11 0.16 0.21 0.45 0.57 0.68 0.79 0.03 0.05 0.07 0.09 80 90 100 120 140 5.23 5.88 6.54 7.84 9.15 7.9 9.8 12.0 16.8 22.3 3.63 4.09 4.54 5.45 6.35 3.28 4.08 4.96 7.0 9.2 0.81 1.0 1.22 1.71 2.28 1.31 1.47 1.63 1.96 2.29 0.27 0.34 0.41 0.58 0.76 0.91 1.02 1.13 1.36 1.58 0.11 0.14 0.17 0.24 0.31 160 180 200 250 300 10.46 11.76 13.07 16.34 19.61 29.0 35.7 43.1 65.5 92.0 7.26 8.17 9.08 11.35 13.62 11.8 14.8 17.8 27.1 38.0 4.08 4.60 5.11 6.38 7.66 2.91 3.61 4.4 6.7 9.3 2.61 2.94 3.27 4.08 4.90 0.98 1.22 1.48 2.24 3.14 1.82 2.05 2.27 2.84 3.40 0.41 0.53 0.61 0.93 1.29 1.73 2.21 2.75 3.35 3.98 350 400 450 500 550 22.87 26.14 29.41 122.0 156.0 196.0 15.89 18.16 20.43 22.70 24.96 50.5 65.0 81.0 98.0 117.0 8.93 10.21 11.49 12.77 14.04 12.4 16.0 19.8 24.0 28.7 5.72 6.54 7.35 8.17 8.99 4.19 5.4 6.7 8.1 9.6 3.98 4.54 5.11 5.68 6.24 600 700 800 900 1000 27.23 137.0 15.32 17.87 20.42 22.98 33.7 44.9 57.0 71.0 9.80 11.44 13.07 14.71 16.34 11.3 15.1 19.4 24.0 29.2 6.81 7.95 .9.08 10.22 11.35 4.68 6.24 7.98 9.93 12.04 1100 1200 1300 1400 1500 1 17.97 19.61 34.9 40.9 12.49 13.62 14. 7 15.89 ! 17.03 14.4 16.9 19.6 22.5 25.6 280 ELEMENTS OF HYDRAULICS FRICTION HEAD IN PIPES (Continued) Discharge in thousands of gallons per 24 hourd 3 S3 6 * 8 I 1 S 15 .2 Q Inside diameter of pipe 8 in. 10 in. 12 in. 16 in. 20 in. Velocity in feet per sec. Friction head in feet per 100 ft. Velocity in feet per sec. Friction head in feet per 100 ft. Velocity in feet per sec. Friction head in feet per 100 ft. Velocity in feet per sec. Friction head in feet per 100 ft. Velocity in feet per sec. a v II I* II 200 250 300 350 400 0.309 0.386 0.464 0.541 0.619 0.89 1.11 1.33 1.56 1.77 0.08 0.12 0.14 0.22 0.28 0.57 0.71 0.85 0.99 1.13 0.04 0.05 0.06 0.07 0.09 0.39 0.49 0.59 0.69 0.79 0.01 0.02 0.02 0.03 0.04 0.22 0.28 0.33 0.39 0.44 0.003 0.004 0.006 0.008 0.010 0.28 0.003 450 500 550 600 700 0.696 0.774 0.851 0.928 1.083 1.99 2.22 2.44 2.66 3.10 0.34 0.42 0.50 0.59 0.78 1.28 1.42 1.56 1.70 1.99 0.12 0.14 0.17 0.20 0.26 0.89 0.99 .09 .18 .38 0.05 0.06 0.07 0.08 0.11 0.50 0.55 0.61 0.66 0.77 0.012 0.015 0.017 0.02 0.03 0.31 0.35 0.39 0.43 0.50 0.004 0.005 0.006 0.007 0.009 0.012 0.014 0.017 0.020 0.024 800 900 1,000 1,100 1,200 .238 .392 .547 .702 .857 3.55 3.99 4.43 4.88 5.37 0.99 1.24 1.51 1.80 2.11 2.27 2.55 2.84 3.12 3.40 0.34 0.42 0.51 0.61 0.71 .58 .77 .97 2.17 2.36 2.96 3.94 4.92 5.91 6.89 0.14 0.17 0.21 0.25 0.29 0.89 1.00 1.11 1.22 1.33 0.03 0.04 0.05 0.06 0.07 0.57 0.64 0.71 0.78 0.85 1,500 2,000 2,500 3,000 3,500 2.321 3.094 3.868 4.642 5.41 6.65 8.86 11.08 13.30 3.18 5.4 8.4 11.6 4.26 5.67 7.10 8.51 9.93 1.C8 1.84 2.78 3.86 5.19 0.44 0.76 1.15 1.60 2.13 1.66 2.22 2.77 3.32 3.88 0.11 0.19 0.28 0.40 0.53 1.06 1.42 1.77 2.13 2.48 0.04 0.06 0.09 0.13 0.18 4,000 5,000 6,000 7,000 8,000 6.19 7.74 9.28 10.83 12.38 11.35 14.19 17.03 6.65 10.05 14.09 7.88 9.85 11.82 13.79 15.76 2.70 4.10 5.8 7.7 9.9 4.43 5.54 6.65 7.76 8.86 0.68 1.02 1.43 1.90 2.42 2.84 3.55 4.26 4.96 5.67 0.23 0.34 0.48 0.64 0.82 9,000 10,000 11,000 12,000 15,000 13.92 15.47 17.02 18.57 23.21 17.73 19. 7C 12.2 15.0 9.97 11.08 12.19 13.30 16.62 3.02 3.68 4.40 5.2 7.8 6.38 7.09 7.80 8.51 10.64 1.02 1.24 1.48 1.74 2.62 16,000 17,000 18,000 19,000 20,000 24.76 26.30 27.85 29.40 30.94 11.35 12.06 12.77 13.47 14.18 2.96 3.31 3.68 4.07 4.48 TABLES 281 FRICTION HEAD IN PIPES (Continued) C 3 Inside diameter of pipe 3 ,jg 3 3 24 in. 30 in. 36 in. 42 in. 48 in. is a . "S d 4$ (3 o la ss \ tn d c5 V t, .a li 2 a li 45 .3 |i * a li Jj C IS .32 ?? fl M fe >> o d ** >> o rl M >> g >_ >> .: a ** >> ^ (3 *" a |1 5 "s I: 13 li II 1 a .s - B| o " |s Frictioi feet pe 3 ,2 <3 .ft ~ ft 11 Velocit per se( 1! fj I S. T^ 1! li 1 1 5471 1 n 4Ql n on? 32 002 22 002 1 5 2 321 74 015 47 005 33 003 2 3 094 98 026 63 009 44 004 2 5 3 868 1 23 039 79 013 55 005 3.0 4.642 1.48 0.055 0.95 0.018 0.66 0.008 0.48 0.004 3 5 5 41 1 72 07 1 10 025 77 010 56 005 4.0 6.19 1.97 0.09 1.26 0.032 0.88 0.013 0.64 0.006 0.49 0.003 4.5 6.96 2.22 0.12 1.42 0.039 0.99 0.72 0.007 0.55 0.004 5.0 7.74 2.46 0.14 1.58 0.048 1.09 0.020 0.80 0.009 0.62 0.005 6.0 9.28 2.96 0.20 1.89 0.067 1.31 0.027 0.96 0.013 0.74 0.007 7.0 10.83 3.45 0.26 2.21 0.09 1.53 0.036 .13 0.017 0.86 0.009 8.0 12.38 3.94 0.34 2.52 0.11 1.75 0.047 .29 0.022 0.98 0.012 9.0 13.92 4.43 0.42 2.84 0.14 1.97 0.058 .45 0.027 .10 0.014 10.0 15.47 4.92 0.51 3.15 0.17 2.19 0.071 .61 0.033 .23 0.017 12.0 18.57 5.91 0.71 3.78 0.24 2.63 0.099 .93 0.047 .48 0.024 14.0 21.66 6.89 0.95 4.41 0.32 3.06 0.13 2.25 0.06 .72 0.032 16.0 24.76 7.88 1.22 5.04 0.41 3.50 0.17 2.57 0.08 .97 0.042 18.0 27.85 8.86 1.52 5.67 0.51 3.94 0.21 2.89 0.10 2.22 0.052 20.0 30.94 9.85 1.83 6.30 0.62 4.38 0.25 3.22 0.12 2.46 0.063 22.0 34.04 10.83 2.19 6.93 0.74 4.82 0.30 3.53 0.14 2.71 0.075 24.0 37.13 11.82 2.59 7.56 0.87 5.25 0.36 3.86 0.17 2.96 0.09 26.0 40.23 12.80 2.99 8.20 1.01 5.69 0.41 4.18 0.20 3.20 0.10 28.0 43.32 13.79 3.42 8.83 1.16 6.13 0.48 4.50 0.22 3.45 0.12 30.0 46.42 14.77 3.90 9.46 1.32 6.57 0.54 4.82 0.26 3.69 0.13 32.0 49.51 10.09 1.48 7.00 0.61 5.15 0.29 3.94 0.15 34.0 52.6 10.72 1.66 7.44 0.68 5.47 0.32 4.19 0.17 36.0 55.7 11.35 1.84 7.88 0.76 5.79 0.36 4.43 0.19 38.0 58.8 11.98 2.04 8.32 0.84 6.11 0.40 4.68 0.21 40.0 61.9 8.76 0.92 6.45 0.44 4.92 0.23 50.0 77.4 10.95 1.39 8.04 0.66 6.16 0.34 60.0 92.8 13.13 1.96 9.65 0.92 7.39 0.48 70.0 108.3 11.26 1.22 8.62 0.64 80.0 123.8 12.86 1.57 9.85 0.82 90.0 139.2 11.08 1.02 100.0 154.7 12.31 1.24 ' 282 ELEMENTS OF HYDRAULICS TABLE 14. BAZIN'S VALUES OF CHEZY'S COEFFICIENT Values of the coefficient C in Chezy's formula v = C\/Vs according to Bazin's formula (Art. 24): C 87 m 0.552 + - Vr Coefficient of roughness, m Hydraulic radius r, in feet Planed tim- ber or smooth cement Unplaned timber, well laid brick, or concrete Ashlar, good rubble mas- onry, or poor brick- work Earth in good condition Earth in ordinary condition Earth in bad condition TO = 0.06 TO = 0.16 TO = 0.46 TO = 0.85 TO = 1.30 TO = 1.75 0.1 117 82 43 27 19 14 0.2 127 96 55 35 25 19 0.3 131 103 63 41 30 23 0.4 135 108 68 46 33 26 0.5 136 112 71 50 3a 29 0.6 138 115 . 76 53 39 31 0.7 139 117 79 55 41 33 0.8 141 119 82 58 43 35 0.9 141 121 84 60 45 36 1.0 142 122 86 62 47 38 1.25 143 125 90 66 51 41 1.50 145 127 94 70 54 44 1.75 145 129 97 73 57 47 2.00 146 131 99 75 59 49 2.5 147 133 104 80 63 53 3.0 . 148 135 106 83 67 57 4.0 150 138 111 89 72 61 5.0 150 140 115 93 77 65 6.0 151 141 118 97 80 69 7.0 152 142 120 100 83 72 8.0 152 143 122 102 86 74 9.0 152 144 123 104 88 77 10.0 152 145 125 106 90 79 12.0 153 145 127 109 94 82 15.0 153 147 130 113 98 86 20.0 154 148 133 117 103 92 30.0 155 150 137 123 110 100 40.0 155 151 139 127 115 105 50.0 155 151 141 129 118 109 TABLES 283 TABLE 15. KUTTER'S VALUES OF CHEZY'S COEFFICIENT Values of the coefficient C in Chezy's formula v (Eq. (67), Art. 24): 41. 65 + according to Kutter's formula Slope, Coefficient of Hydraulic radius r, in feet s roughness, n 0.1 0.20.4 0.6|0.8 1 |l.5| 2 | 3 | 4 | 6 | 8 10 15 20 0.009 65 87 111J127 138 148 166 179 197 209 226 t 238,246 262J271 -H --" II II II 0.025 24 31 36 40 46 50 54 60 64 72 76 82 89 95 98 00 0.030 18 25 29 32 37 41 44 49 54 59 63 69 76 82 85 0.035 15 21 24 27 31 34 37 42 45 51 55 60 67 72 76 0.009 104 126 138 148 157 166 172 183 190 199 204 211 219 224 227 JH 0.010 89 110 120 129 140 148 154 164 170 179 184 191 199 203 207 H 0.011 78 97 107 115 126 133 138 148 154 162 168 175 183 187 190 S 0.012 69 87 96 104 113 121 125 135 141 149 154 161 168 172 176 ^ 0.013 62 78 87 94 103 110 115 124 130 138 142 149 157 162 164 3f 35 ~ fl 2 0.017 43 54 62 68 75 81 85 93 98 105 110 116 123 128 131 . " ""! 0.020 34 44 50 55 62 67 70 78 83 89 94 99 107 110 115 ii ii ii 0.025 25 32 37 42 47 51 55 61 65 71 76 81 88 92 96 00 0.030 19 25 30 33 38 42 45 50 54 59 63 69 75 80 83 0.035 16 21 24 27 31 35 37 42 45 51 55 60 66 70 73 0.009 110 129 141 150 161 169 175 184 191 199 204 211 218 222 225 0.010 94 113 124 1'31 142 150 155 165 171 179 184 190 197 202205 jg 0.011 83 99 109 117 127 134 139 149 155 163 168 174 181 186 188 b 0.012 73 89 98 105 115 122 127 136 142 149 154 160 167 171 175 & 0.013 65 81 89 96 104 111 116 124 130 138 142 149 155 160 163 c 22 0.017 45 57 63 69 76 82 86 93 98 105 110 116 122 127 129 0.020 36 45 51 56 63 68 71 78 83 89 93 99 105 110 113 II II II 0.025 27 34 39 43 48 52 56 62 66 71 75 81 87 91 94 0.030 21 27 30 34 39 42 45 50 54 59 63 68 74 78 81 0.035 17 22 25 28 32 35 38 43 46 51 54 59 65 68 72 0.009 110 130 143 151 162 170 175 185 191 199 204 210 217 222 225 0.010 95 114 125 133 143 151 156 165 171 179 184 190 196 200 204 J! 0.011 83 100 111 119 129 135 141 149 155 162 167 173 180 184 187 S 0.012 74 90 100 107 116 123 128 136 142 149 154 160 166 170 173 8. 0.013 66 81 90 98 106 112 117 125 130 138 142 148 154 159 161 ,-H "~ rH 00 0.017 46 57 64 70 77 82 87 94 99 105 109 115 121 126 128 o .g 0.020 36 46 52 57 64 68 72 79 83 89 93 99 105 108 112 II II II 0.025 27 34 39 44 49 53 56 62 66 71 76 81 86 90 93 00 0.030 21 27 31 35 39 43 45 51 55 59 63 68 74 77 80 0.035 17 22 25 29 33 35 38 43 1 46 51 55 59 65 68| 71 TABLES 285 :;j s I sc co 2 1 S -s H O S < S 3 + o> ! + ! ^ A {fi C CM CD CO I-H t^ CO C 1C >C CO O CO CD CO d d d d d ^H CO iC C CO CO ooooo CM 1C i-l >C * * O5 CM GO CO <* Tj( CO CO CD CO CO CO d d d d ooooo o o o o o 00000 O O O O 05 10 00 -* i-l 1C Tfi CO CO CO o d d d d d o o o o o GO * l>- CM 00 1C * CO CO CM CO CO CO CO CO o d d d d o o o o o 00000 T}< CO ^* 1C CM CM CM CM o o o o o o o o o o 00000 o o o o o d d d d d 00000 Sco co co CO CO CO d d d d 1-H CM co co co cc iC r*< d d i-H l> CO (N o d o o o o o o o o CO CM *O CD CM Tt< O5 C * CO CM CN CO CD CO CO o 00000 O O $3 O O O O O CO CO CO CO CO d d d d o" O i-H CM CM CM CO * I 1C 1C CO CD CM CM CM I CM CM CM CM CO CO CO i o T I OS GO GO 00 CO 3 CO CO CO d o o d d rJH C CO t- CO d d d d d OOO I O O O O ooooo O CO * CO (N CO CD CD CD CO d d d d d 00 "C CO CM i-H CO CO CD CO CO ooooo ooooo 1C I-H 00 CO Tt* T-I I-H O O O CO CD CO CD CO d d o d d 1-H 1C i-H CO (N CM CD CO CO ooooo i o o ooo o o o t~- >C CO O * os CO CM .-H CD CO CO odd Ol CO 00 CM CM <-l CO CO CO CO <** CO 1-1 CO TJ< CO CM CM TH CO CO CO O CO d d d d d CO CO CD CD CD d o o d o odd o' d I-H O O O O CO CO CO CO 1C d d odd CM ^ O O 05 I CO CO CO CO c O O O O O CO CO ooooo ooooo ooooo o o o o o o o o ooooo ooooo ooooo 2? S8 odd ooooo d d o d d CM O t> C CM Oi Oi 00 00 OO 1C 1C 1C 1C 1C d d d d d I o 1-4 CO CO O CD C 1C 1C dodo 1C O I*- C SOi 00 00 1C 1C C ddddd I dddd CM OS 1-1 >C iH CO l-H i-H O O CO CO CO CO CO 286 ELEMENTS OF HYDRAULICS TABLE 17. DISCHARGE PER INCH OF LENGTH OVER RECTANGULAR NOTCH WEIRS Discharge over sharp-crested, vertical, rectangular notch weirs in cubic feet per minute per inch of length. Computed from Eq. (30), Art. 10: Q = 0.4 bh^ for b = 1 in. Depth on crest in inches 1/8 1/4 3/8 1/2 5/8 3/4 7/8 0.00 0.01 0.05 0.09 0.14 0.19 0.26 0.32 1 0.40 0.47 0.55 0.64 0.73 0.82 0.92 1.02 2 1.13 1.23 1.35 1.46 1.58 1.70 1.82 1.95 3 2.07 2.21 2.34 2.48 2.61 2.76 2.90 3.05 4 3.20 3.35 3.50 3.66 3.81 3.97 4.14 4.30 5 4.47 4.64 4.81 4.98 5.15 5.33 5.51 5.69 6 5.87 6.06 6.25 6.44 6.62 6.82 7.01 7.21 7 7.40 7.60 7.80 8.01 8.21 8.42 8.63 8.83 8 9.05 9.26 9.47 9.69 9.91 10.13 10.35 10.57 9 10.80 11.02 11.25 11.48 11.71 11.94 12.17 12.41 10 12.64 12.88 13.12 13.36 13.60 13.85 14.09 14.34 11 14.59 14.84 15.09 15.34 15.59 15.85 16.11 16.36 12 16.62 16.88 17.15 17.41 17.67 17.94 18.21 18.47 13 18.74 19.01' 19.29 19.56 19.84 20.11 20.39 20.67 14 20.95 21.23 21.51 21.80 22.08 22.37 22.65 22.94 15 23.23 23.52 23.82 24.11 24.40 24.70 25.00 25.30 16 25.60 25.90 26.20 26.50 26.80 27.11 27.42 27.72 17 28.03 28.34 28.65 28.97 29.28 29.59 29.91 30.22 18 30.54 30.86 31.18 31.50 31.82 32.15 32.47 32.80 19 33.12 33.45 33.78 34.11 34.44 34.77 35.10 35.44 20 35.77 36.11 36.45 36.78 37.12 37.46 37.80 38.15 21 38.49 38.84 39.18 39.53 39.87 40.24 40.60 40.96 22 41.28 41.64 41.98 42.36 42.68 43.04 43.44 43.76 23 44.12 44.48 44.84 45.20 45.56 45.96 46.32 46.68 24 47.04 47.40 47.76 48.12 48.52 48.88 49.28 49.64 25 50.00 50.40 50.76 51.08 51.52 51.88 52.28 52.64 26 53.04 53.40 53.80 54.16 54.56 54.96 55.36 55.72 27 56.12 56.52 56.92 57.32 57.68 58.08 58.48 58.88 28 59.28 59.68 60.08 60.48 60.84 61.28 61.68 62.08 29 62.48 62.88 63.28 63.68 64.08 64.52 64.92 65.32 30 65 . 72 66.16 66.56 66.96 67.36 67.80 68.20 68.64 TABLES 287 TABLE 18. DISCHARGE PER FOOT OF LENGTH OVER RECTANGULAR NOTCH WEIRS Discharge over sharp crested, vertical, rectangular notch weirs in cubic feet per second per foot of length. Computed from Eq. (29), Art. 10: Q = 3. 3bh 3 / 2 for 6 = 1 ft. Depth on crest in feat 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.000 0.003 0.009 0.017 0.026 0.037 0.049 0.061 0.075 0.089 0.1 0.104 0.120 0.137 0.155 0.173 0.192 0.211 0.231 0.252 0.273 0.2 0.295 0.317 0.341 0.364 0.388 0.413 0.438 0.463 0.489 0.515 0.3 0.542 0.570 0.597 0.626 0.654 0.683 0.713 0.743 0.773 0.804 0.4 0.835 0.866 0.898 0.931 0.963 0.996 1.030 1.063 1.098 1.458 1.132 0.5 1.167 1.202 1.238 1.273 1.309 1.346 1.383 1.420 1.496 0.6 1.538 1.572 . 1.611 1.650 1.690 1.729 1.769 1.810 1.850 1.892 0.7 1.933 1.974 2.016 2.058 2.101 2.143 2.187 2.230 2.273 2.317 0.8 2.361 2.406 2.450 2.495 2.541 2.586 2.632 2.678 2.724 2.768 0.9 2.818 2.865 2.912 2.960 3.008 3.055 3.104 3.152 3.202 3.251 1.0 3.300 3.350 3.399 3.449 3.501 3.551 3.600 3.653 3.703 3.755 1.1 3.808 3.858 3.911 3.963 4.016 4.069 4.122 4.178 4.231 4.283 1.2 4.340 4.392 4.448 4.501 4.557 4.613 4.666 4.722 4.778 4.834 1.3 4.891 4.947 5.006 5.062 5.118 5.178 5.234 5.293 5.349 5.409 1.4 5.468 5.524 5.584 5.643 5.702 5.762 5.82'l 5.881 5.940 6.003 1.5 6.062 6.125 6.184 6.247 6.306 6.369 6.428 6.491 6.554 6.617 1.6 6.679 6.742 6.805 6.867 6.930 6.993 7.059 7.121 7.187 7.250 1.7 7.316 7.379 7.445 7.508 7.573 7.639 7.706 7.772 7.838 7.904 1.8 7.970 8.036 8.102 8.171 8.237 8.303 8.372 8.438 8.507 8.573 1.9 8.643 8.712 8.778 8.847 8.917 8.986 9.055 9.125 9.194 9.263 2.0 9.332 9.405 9.474 9.544 9.616 9.686 9.758 9.827 9.900 9.969 2.1 10.042 2.2 10.768 2.3 11.510 10.115 10.841 11.586 10.187 10.916 11.662 10.260 10.989 11.738 10.332 11.065 11.814 10.405 11.138 11.887 10.478 11.213 11.966 10.550 11.286 12.042 10.623 11.362 12.118 10.695 11.435 12.194 2.4 12.269 12.345 12.425 12.500 12.576 12.656 12.731 12.811 12.890 12.966 2.5 12.935 13.124 13 . 200 13.279 13.358 13.438 13.517 13.596 13.675 13 . 754 2.6 2.7 13 . 834 14.642 13.916 14.721 13.995 14 . 804 14.075 14.886 14.154 14.969 14.236 15.048 14.315 15.131 14.398 15.213 14.477 15.296 14 . 560 15.378 2.8 15.461 15.543 15.629 15.711 15.794 15.876 15.962 16.045 16.130 16.213 2.9 16.299 16.381 16.467 16.550 16.635 16.721 16.807 16.889 16.975 17.061 3.0 17.147 17.233 17.318 17.404 17.490 17.579 17.665 17.751 17.837 17.926 3.1 18.011 18.101 18.186 18.275 18.361 18.450 18.536 18.625 18.714 18.803 3.2 18.889 18.978 19.067 19.157 19 . 246 19.335 19.424 19.513 19 . 602 19.694 3.3 19.784 19.873 19.962 20.054 20.143 20.236 20.325 20.414 20.506 20.599 3.4 20.688 20.780 20.873 20.962 21.054 21.146 21.239 21.331 21.424 21.516 22.447 3.5 21 . 608 21.701 21.793 21.886 21.978 22.074 22.166 22.259 22.354 3.6 22.542 22.635 22.730 22.823 22.919 23.011 23.107 23 . 202 23.295 23.390 3.7 23.486 23.582 23.678 23.773 23.869 23.965 24.060 24.156 24.252 24.347 3.8 24.446 24.542 24.638 24.734 24.833 24.928 25.027 25.123 25.222 25.318 3.9 25.417 25.516 25.611 26.598 25.710 25.809 25.905 26.004 26.103 26-. 202 26.301 4.0 26.400 26.499 26.697 26.796 26.895 26.997 27.096 27. 195 27.298 288 ELEMENTS OF HYDRAULICS TABLE 19. PRINCIPLES Kinematics (motion) Linear motion Angular motion s = displacement 6 = displacement . v = velocity w = velocity a = acceleration a = acceleration vo = initial velocity coo = initial velocity F = force M = torque about fixed axis Notation W = Fs = work W = MB = work ~v~ at -\- Ci Derivation of above formulas s = i a < 2 + Cit + Cz if t = o, v = uo :.Ci = vo If t = 0, co = coo .'.Ci = coo If t = 0, s = .'.C 2 = o if t = o, e = o .'.C 2 = o Relation between v = rco linear and at = ra at tang. comp. of accel. angular motion a n = v 2 /r = rco 2 a n = normal comp. of accel. v 7? Arc AB ds = rdO v = rco Derivation of first .\ ds dO dv dta two formulas T/' \ds dt ~ T dt d? = r dT ..- \A. v rco at = ra A v B If body at A were free, it would proceed in direction of tangent AB and in time t would reach B where AB = vt. Derivation of Since it is found at C instead of B it normal accel. must have experienced a central for uniform circular motion acceleration. Let on denote this central acceleration. Then BC = %a n t 2 - By geometry BC X BD = A B 2 and in the limit BD approaches 2r. Hence |a n i 2 X 2r = u 2 i 2 , from which a n v 2 /r = co 2 r. TABLES 289 OF MECHANICS Dynamics (force) | Linear motion Angular motion Fundamental law F = ma M = la Discussion and derivation By experiment it is found that F ex a (Newton's 2nd Law) .'. F/a = const., say m, whence F = ma. m = intrinsic prop- erty of body called its mass. Mass = measured inertia. Consider rota- Fk. tion of rigid O r body about a \m fixed axis. 1 Then for a par- ticle of mass m at distance r from axis of rota- tion, law F = ma becomes Fr = mar, or since a = ra, Fr = mr 2 a. By summation 2Fr = Smr 2 a But ZFr = M , and Smr 2 = I . .M = la. If F = then a = and hence v = or constant, which ex- presses Newton's 1st Law. F = impressed force, ma = kine- tic reaction or inertia force. Equality F ma is dynamical expression of Newton's 3rd Law. Principle of work and energy w = Fs = m _m .,.!-! Derivation F = TOO, 2 = t>o 2 + 2as 1,2 - VO* M = la, w 2 = wo 2 + 2a0 a, 2 - wo 2 a TOU 2 TWO 2 TT MO I J " 2 " /W 2 W - Fs - mas - 2 2 Principle of impulse and momentum Fl = mv mvo Mt = Iw - Iwo Derivation F = ma, v = vo + at .'. at = v vo, and Ft mat = mv mvo M = la, u = wo ~\- at .' . at = w wo, and Mt = lat = lu luo Power Fv Power = Fv, h.p. = ^7: Mw Power = Mct>, h.p. = ^7: OoU Centrifugal force w w v 2 w F, = - a n = - = w 2 r c g g r g Derivation w v^ Special case of F = ma where m = and a = D'Alembert's principle ,-0 Explanation and use F = ma where F = external impressed force and a = accel. produced. Introduce another force P, given by P = ma. Then by addition, F + P = 0; i.e., the body is in equilibrium under the action of F and P. P is called the kinetic reaction, or reversed effective force, since P = F. By introducing this idea of the kinetic reactions equili- brating the impressed forces, all problems in dynamics are reduced to statical problems. This is called d'Alembert's Principle, and is usually expressed in the form ,-,g-0. 290 ELEMENTS OF HYDRAULICS TABLE 20. l DISCHARGE PER FOOT OP LENGTH OVER SUPPRESSED WEIRS Discharge over sharp-crested, vertical, suppressed weirs in cubic feet per second per foot of length. Computed by Bazin's formula (Art. 10) : Q = (0.405 . 00984 )(l +0.55 bhVZffh, for b = 1 ft. Head on crest, h, Height of weir, d, in feet in feet 2 4 6 8 10 | 20 30 0.1 0.13 0.13 0.13 0.13 0.13 0.13 0.13 0.2 0.33 0.33 0.33 0.33 0.33 0.33 0.33 0.3 0.58 0.58 0.58 0.58 0.58 0.58 0.58 0.4 0.88 0.88 0.87 0.87 0.87 0.87 0.87 0.5 1.23 1.21 1.21 1.21 1.21 1.20 1.20 0.6 1.62 1.59 1.58 1.58 1.57 1.57 1.57 0.7 2.04 1.99 1.98 1.98 1.97 1.97 1.97 0.8 2.50 2.43 2.41 2.41 2.40 2.40 2.40 0.9 3.00 2.90 2.88 2.86 2.86 2.85 2.85 1.0 3.53 3.40 3.36 3.35 3.34 3.33 3.33 .1 4.09 3.92 3.87 3.86 3.85 3.84 3.84 .2 4.68 4.48 4.42 4.40 4.38 4.36 4.36 .3 5.31 5.07 4.99 4.96 4.94 4.91 4.91 .4 5.99 5.68 5.58 5.54 5.52 5.49 5.48 .5 6.68 6.30 6.20 6.16 6.13 6.10 6.09 .6 7.40 6.97 6.84 6.78 6.74 6.69 6.69 .7 8.14 7.66 7.49 7.42 7.39 7.33 7.32 .8 8.93 8.37 8.18 8.09 8.05 7.98 7.96 1.9 9.75 9.11 8.89 8.79 8.74 8.65 8.63 2.0 10.58 9.87 9.62 9.51 9.44 9.34 9.32 2.1 11.45 10.65 10.37 10.25 10.17 10.05 10.02 2.2 12.34 11.46 11.14 10.99 10.91 10.78 10.75 2.3 13.24 12.29 11.93 11.77 11.66 11.52 11.48 2.4 14.20 13.15 12.75 12.56 12.45 12.28 12.24 2.5 15.17 14.03 13.59 13.38 13.26 13.06 13.01 2.6 16.16 14.92 14.44 14.20 14.07 13.85 13.80 2.7 17.18 15.83 15.31 15.04 14.92 14.65 14.60 2.8 18.23 16.79 16.21 15.92 15.76 15.48 15.42 2.9 19.29 17.77 17.11 16.79 16.63 16.33 16.25 3.0 20.39 18.74 18.06 17.71 17.52 17.18 17.10 3.1 21.50 19.74 19.02 18.64 18.42 18.04 17.96 3.2 22.64 20.77 19.98 19.58 19.34 18.93 18.83 3.3 23.81 21.80 20.98 20.55 20.27 19.82 19.73 3.4 24.98 22.89 21.99 21.52 21.24 20.75 20.63 3.5 26.20 24.00 23.01 22.48 22.22 21.69 21.60 3.6 27.41 25.09 24.06 23.52 23.20 22.62 22.48 3.7 28.64 26.22 25.14 24.56 24.20 23.59 23.43 3.8 29.94 27.38 26.22 25.60 25.23 24.56 24.39 3.9 31.21 28.53 27.33 26.65 26.26 25.53 25.34 4.0 32.54 29.74 28.45 27.74 27.32 26.55 26.35 4.1 33.85 30.95 29.59 28.83 28.36 27.55 27.33 4.2 35.22 32.18 30.75 29.96 29.48 28.59 28.36 4.3 36.59 33.43 31.93 31.10 30.58 29.62 29.37 4.4 37.99 34.70 33.12 32.24 31.70 30.66 30.42 4.5 39.40 35.98 34.33 33.39 32.83 31.74 31.47 4.6 40.83 37.29 35.56 34.58 33.98 32.84 32.53 4.7 42.29 38.62 36.82 35.75 35.13 33.93 33.61 4.8 43.75 39.96 38.07 37.00 36.33 35.05 34.70 4.9 45.22 41.30 39.35 38.20 37.49 36.15 35.77 5.0 46.71 42.67 40.62 39.44 38.70 37.28 36.88 Compiled from extensive hydraulic tables by Williams and Hazen. INDEX Accumulator, hydraulic, 9 Adjutage, Venturi, 63 American type of turbine, 170 Aqua Claudia, 137 Aqueducts, comparison of, 136 Archimedes, theorem of, 21 B Backwater, 118 Barge canal, N. Y. State, 96 Barker's mill, 155 Barometer, mercury, 18 water, 17 Bazin's coefficient, 280 formula, 102 Bends and elbows, head lost at, 77 Bernoulli's theorem, 64 Borda mouthpiece, 61 Branching pipes, 88, 133 Breast wheel, 158 Buoyancy, 20 chamber, Keokuk lock, 39 Canal lock, 58 Capacity criterion, 184 Catskill aqueduct, 36, 140 Venturi meter, 69, 70 Center of pressure, 13 Centrifugal pumps, 215 characteristics, 227 efficiency and design of, 234 Channel cross section, 103 Characteristic for centrifugal pumps, 227 speed, 185 Characteristics of impulse wheels and turbines, 181 Chezy's formula, 85 Church's formula for water hammer, 204 Circles, circumferences and areas of, 268 Circular orifice, efflux coefficients for, 273 sections, hydraulic properties of, 106 Cock in circular pipe, 80 Complete contraction, 49 Compound pipes, 86 Conduits, flow in open, 100 Conical mouthpiece, 63 Contracted weir, 51 Contraction coefficient, 45 of jet, 48 of section, 79 effect of, 92 partial and complete, 49 Crane, hydraulic, 11 Critical velocity of water in pipes, 71 Cross section of channel, 103 Current meter, 108 wheel, 156, 158 D Darcy's modification of Pitot's tube, 110 Deflection of jet, 147 Density of water, 3 Design of centrifugal pumps, 234 Diffusion vanes, 221 Diffusor, pressure developed in, 226 Discharge, conditions for maximum, 103 equivalents, 260 from rectangular orifice, 47 of rectangular notch weir, 48 through sharp-edged orifice, 48 Divided flow, 86 Doble bucket, 160 Draft tube, theory of, 173 Dry dock, floating, 236 DuBuat's paradox, 117 291 292 INDEX Dynamic pressure, 145 in bends and elbows, 149 E Effective fire stream, height of, 90 head, 46 Efficiency curves for turbines, 249, 250 . of hydraulic press, 7 maximum, for vanes, 151 Efflux coefficients, 46 for circular orifice, 273 for square orifice, 274 Elasticity, bulk modulus of, for water, 202 of water, 1 Elevator, hydraulic, 12 Empirical weir formula, 51, 52 Enlargement of section, 78 effect of, 92 Equilibrium of floating bodies, 20 of fluids in contact, 16 Experiments on flow of water, 90 F Fire nozzles, 64 pumps, 239 streams, 89 stream table, 275 Flat plate in current, 93 Float measurements, 108 Floating equilibrium, 21 gate, Keokuk lock, 40 Flow in small pipes, velocity of, 74 Force pump, 209 Fourneyron type of turbine, 169 Francis type of turbine, 170 Freeman's experiments on fire streams, 89 Friction head in pipes, 278 loss in pipe flow, 77 Gate valve in circular pipe, 80 Gibson's experiments on water hammer, 204 Girard impulse turbine, 164 Gradient, hydraulic, 82 H Head and pressure equivalents, 47, 259 developed by centrifugal pump, 231 effective, on orifice, 46 lost in pipe flow, 76 Hele-Shaw's experiments, 92 Hook gage, 54 House service pipes, 129 Hydraulic accumulator, 9 crane, 11 dredge, 239 efficiency of centrifugal pumps, 235 elevator, 12 gradient, 82 slope of, 83 intensifier, 8 jack, 10 mining, 242 motors, types of, 156 press, 5 radius, 84 Hydrostatic pressure, 3 Ideal velocity head, 44 Impact on plane surface, 144 on surface of revolution, 146 tube, theory of, 113 Impeller, pressure developed in, 224 Impellers for centrifugal pumps, 218 Impulse wheel, 152, 156 wheels, classification of, 189 Intensifier, hydraulic, 8 J Jack, hydraulic, 10 Jonval type of turbine, 169 Joukovsky's experiments on water hammer, 204 K Kensico dam, 35, 36 Keokuk turbines, 179 power plant, 37 Kinetic pressure, 64 head, 66, 82 INDEX 293 Ku tier's coefficients, 283 formula, 101, 102 Liquid vein, 44 Lock, Keokuk, 38 Loss of head in small pipes, 74 Lost head in pipe flow, 76 summary of, 81 M Marietta's flask, 60 Mechanics, principles of, 288 Mercury barometer, 18 pressure gage, 19 Metacenter, 23 coordinates of, 25 Metacentric height, 25 Mine drainage, 237 Mississippi River Power Co., 37, 179 Modulus of elasticity for water, 202 Motors, classification of hydraulic, 157 Moving vanes, 149 N Natural channels, flow in, 106 New York State barge canal, 96 Niagara Falls power plant, 172 Non-sinuous flow, 72 Nozzle diameter, determination of, 198 for maximum power, 196 Nozzles, flow through, 60 O Ontario Power Co., 172 Operating range, normal, 190 Oval sections, hydraulic properties of, 106 Overshot wheel, 159 Packing, frictional resistance of, 7 Parallel flow, 72 Partial contraction, 49 Pelton wheel, 159 efficiency of, 162 Period of oscillation of ship, 27 Piezometer, 18 Pipe flow, ordinary, 75 friction, coefficient of, 277 lines, power transmitted through, 195 Pipes, dimensions of, 265 friction head in, 278 Pitotmeter, 112 Pitot recorder, 113 tube, 109, 116 Poncelet wheel, 158 Power and efficiency, relation be- tween, 197 Pressure gage, mercury, 19 hydraulic, 5 machines, 8 static and dynamic, 145 Properties of water, 260 Pump sizes, calculation of, 211 Pumps, capacity of reciprocating, 266 centrifugal, 215 displacement, 207 steam, 210 suction, 207 R Ram, efficiency of hydraulic, 206 hydraulic, 205 Rate of flow controller, 70 Rational weir formulas, 51 Reaction of jet, 154 Reaction turbine, principle of oper- ation, 155, 157 turbines, 169 classification of, 188 Rectangular notch weir, 51 discharge from, 48 weirs, coefficients for, 285-287 orifice, discharge from, 47 Relative velocity of jet and vane, 149 Reversal of jet, 147 Rolling and pitching of ships, 27 Rotation of liquids, 198 S Siphon lock, 100 spillway, 96 steel pipe, 142 294 INDEX Siphons, 95 Specific discharge, 186 power, 187 speed, 186, 187 weight, 3 determination of, 22 physical definition of, 21 weights of various substances, 264 Speed criterion, 182 Square orifice, efflux coefficients for, 274 Stage pumps, 223 Standard mouthpiece, 60 Steam pump, 210 Stock runner, selection of, 191 Stream gaging, 106 line, 44 motion, 93 mouthpiece, 61 Submerged surfaces, pressure on, 12 Suction lift, 208 Suppressed weir, 51 weirs, discharge coefficients for, 290 Surface of liquid in rotation, 199 Tanks, connected, rise and fall in, 58 filling and emptying, 56 Throttle valve in circular pipe, 81 Torricelli's theorem, 45 Translation, effect of, 198 Tubes, short, flow through, 60 Turbine pumps, 223 setting, 177 test data, 251-255 U Undershot wheel, 158 Vanes, pressure of jet on, 149 Varying head, 58 Velocity head, ideal, 44 of approach, 49 of flow, actual, 45 variation with depth, 108 Venturi adjutage, 63 meter, 67 Catskill aqueduct, 69, 70 Viscosity, 2 coefficient, 71 Volute casing, 219 Vortex chamber, 219 W Water barometer, 17 hammer in pipes, 202 properties of, 260 Waves in pipes, pressure, 203 period of compression, 203 velocity of compression, 203 Weights and measures, 263 Weir formulas, 51, 52 measurements, 53 Weirs, construction of, 53 proportioning, 55 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. a* r* **' 1 '* - - .. FEB 19 1935 FEB 24W37 ,r ^ 12 1 ' '&<*. \lf\, . *** 181940 LD 21-100m-8,'34 ^X> X 300315 UNIVERSITY OF CALIFORNIA LIBRARY