Chapter 20

Hydraulic Machines

§ 152. Hydraulic machines may be broadly divided into two classes: (1)Motors, in which water descending from a higher to a lower level, or from a higher to a lower pressure, gives up energy which is available for mechanical operations; (2)Pumps, in which the energy of a steam engine or other motor is expended in raising water from a lower to a higher level. A few machines such as the ram and jet pump combine the functions of motor and pump. It may be noted that constructively pumps are essentially reversed motors. The reciprocating pump is a reversed pressure engine, and the centrifugal pump a reversed turbine. Hydraulic machine tools are in principle motors combined with tools, and they now form an important special class.

Water under pressure conveyed in pipes is a convenient and economical means of transmitting energy and distributing it to many scattered working points. Hence large and important hydraulic systems are adopted in which at a central station water is pumped at high pressure into distributing mains, which convey it to various points where it actuates hydraulic motors operating cranes, lifts, dock gates, and in some cases riveting and shearing machines. In this case the head driving the hydraulic machinery is artificially created, and it is the convenience of distributing power in an easily applied form to distant points which makes the system advantageous. As there is some unavoidable loss in creating an artificial head this system is most suitable for driving machines which work intermittently(seePower Transmission). The development of electrical methods of transmitting and distributing energy has led to the utilization of many natural waterfalls so situated as to be useless without such a means of transferring the power to points where it can be conveniently applied. In some cases, as at Niagara, the hydraulic power can only be economically developed in very large units, and it can be most conveniently subdivided and distributed by transformation into electrical energy. Partly from the development of new industries such as paper-making from wood pulp and electro-metallurgical processes, which require large amounts of cheap power, partly from the facility with which energy can now be transmitted to great distances electrically, there has been a great increase in the utilization of water-power in countries having natural waterfalls. According to the twelfth census of the United States the total amount of water-power reported as used in manufacturing establishments in that country was 1,130,431 h.p. in 1870; 1,263,343 h.p. in 1890; and 1,727,258 h.p. in 1900. The increase was 8.4% in the decade 1870-1880, 3.1% in 1880-1890, and no less than 36.7% in 1890-1900. The increase is the more striking because in this census the large amounts of hydraulic power which are transmitted electrically are not included.

XII. IMPACT AND REACTION OF WATER

§ 153. When a stream of fluid in steady motion impinges on a solid surface, it presses on the surface with a force equal and opposite to that by which the velocity and direction of motion of the fluid are changed. Generally, in problems on the impact of fluids, it is necessary to neglect the effect of friction between the fluid and the surface on which it moves.During Impact the Velocity of the Fluid relatively to the Surface on which it impinges remains unchanged in Magnitude.—Consider a mass of fluid flowing in contact with a solid surface also in motion, the motion of both fluid and solid being estimated relatively to the earth. Then the motion of the fluid may be resolved into two parts, one a motion equal to that of the solid, and in the same direction, the other a motion relatively to the solid. The motion which the fluid has in common with the solid cannot at all be influenced by the contact. The relative component of the motion of the fluid can only be altered in direction, but not in magnitude. The fluid moving in contact with the surface can only have a relative motion parallel to the surface, while the pressure between the fluid and solid, if friction is neglected, is normal to the surface. The pressure therefore can only deviate the fluid, without altering the magnitude of the relative velocity. The unchanged common component and, combined with it, the deviated relative component give the resultant final velocity, which may differ greatly in magnitude and direction from the initial velocity.From the principle of momentum, the impulse of any mass of fluid reaching the surface in any given time is equal to the change of momentum estimated in the same direction. The pressure between the fluid and surface, in any direction, is equal to the change of momentum in that direction of so much fluid as reaches the surface in one second. If Pais the pressure in any direction, m the mass of fluid impinging per second, vathe change of velocity in the direction of Padue to impact, thenPa= mva.Fig. 152.If v1(fig. 152) is the velocity and direction of motion before impact, v2that after impact, then v is the total change of motion due to impact. The resultant pressure of the fluid on the surface is in the direction of v, and is equal to v multiplied by the mass impinging per second. That is, putting P for the resultant pressure,P = mv.Let P be resolved into two components, N and T, normal and tangential to the direction of motion of the solid on which the fluid impinges. Then N is a lateral force producing a pressure on the supports of the solid, T is an effort which does work on the solid. If u is the velocity of the solid, Tu is the work done per second by the fluid in moving the solid surface.Let Q be the volume, and GQ the weight of the fluid impinging per second, and let v1be the initial velocity of the fluid before striking the surface. Then GQv12/2g is the original kinetic energy of Q cub. ft. of fluid, and the efficiency of the stream considered as an arrangement for moving the solid surface isη = Tu / (GQv12/ 2g).§ 154.Jet deviated entirely in one Direction.—Geometrical Solution(fig. 153).—Suppose a jet of water impinges on a surface ac with a velocity ab, and let it be wholly deviated in planes parallel to the figure. Also let ae be the velocity and direction of motion of the surface. Join eb; then the water moves with respect to the surface in the direction and with the velocity eb. As this relative velocity is unaltered by contact with the surface, take cd = eb, tangent to the surface at c, then cd is the relative motion of the water with respect to the surface at c. Take df equal and parallel to ae. Then fc (obtained by compounding the relative motion of water to surface and common velocity of water and surface) is the absolute velocity and direction of the water leaving the surface. Take ag equal and parallel to fc. Then, since ab is the initial and ag the final velocity and direction of motion, gb is the total change of motion of the water. The resultant pressure on the plane is in the direction gb. Join eg. In the triangle gae, ae is equal and parallel to df, and ag to fc. Hence eg is equal and parallel to cd. But cd = eb = relative motion of water and surface. Hence the change of motion of the water is represented in magnitude and direction by the third side of an isosceles triangle, of which the other sides are equal to the relative velocity of the water and surface, and parallel to the initial and final directions of relative motion.Fig. 153.Special CasesFig. 154.§ 155. (1)A Jet impinges on a plane surface at rest, in a direction normal to the plane(fig. 154).—Let a jet whose section is ω impinge with a velocity v on a plane surface at rest, in a direction normal to the plane. The particles approach the plane, are gradually deviated, and finally flow away parallel to the plane, having then no velocity in the original direction of the jet. The quantity of water impinging per second is ωv. The pressure on the plane, which is equal to the change of momentum per second, is P = (G/g) ωv2.(2)If the plane is moving in the direction of the jet with the velocity±u, the quantity impinging per second is ω(v ± u). The momentum of this quantity before impact is (G/g)ω(v ± u)v. After impact, the water still possesses the velocity ±u in the direction of the jet; and the momentum, in that direction, of so much water as impinges in one second, after impact, is ±(G/g) ω (v ± u)u. The pressure on the plane, which is the change of momentum per second, is the difference of these quantities or P = (G/g) ω (v ± u)2. This differs from the expression obtained in the previous case, in that the relative velocity of the water and plane v ± u is substituted for v. The expression may be written P = 2 × G × ω (v ± u)2/2g, where the last two terms are the volume of a prism of water whose section is the area of the jet and whose length is the head due to the relative velocity. The pressure on the plane is twice the weight of that prism of water. The work done when the planeis moving in the same direction as the jet is Pu = (G/g) ω (v − u)2u foot-pounds per second. There issue from the jet ωv cub. ft. per second, and the energy of this quantity before impact is (G/2g) ωv3. The efficiency of the jet is therefore η = 2(v − u)2u/v3. The value of u which makes this a maximum is found by differentiating and equating the differential coefficient to zero:—dη / du = 2 (v2− 4vu + 3u2) / v3= 0;∴ u = v or1⁄3v.The former gives a minimum, the latter a maximum efficiency.Putting u =1⁄3v in the expression above,η max. =8⁄27.(3) If, instead of one plane moving before the jet, a series of planes are introduced at short intervals at the same point, the quantity of water impinging on the series will be ωv instead of ω(v − u), and the whole pressure = (G/g) ωv (v − u). The work done is (G/g)ωvu (v − u). The efficiency η = (G/g) ωvu (v − u) ÷ (G/2g) ωv3= 2u(v-u)/v2. This becomes a maximum for dη/du = 2(v − 2u) = 0, or u =1⁄2v, and the η =1⁄2. This result is often used as an approximate expression for the velocity of greatest efficiency when a jet of water strikes the floats of a water wheel. The work wasted in this case is half the whole energy of the jet when the floats run at the best speed.§ 156. (4)Case of a Jet impinging on a Concave Cup Vane, velocity of water v, velocity of vane in the same direction u (fig. 155), weight impinging per second = Gw (v − u).Fig. 155.If the cup is hemispherical, the water leaves the cup in a direction parallel to the jet. Its relative velocity is v − u when approaching the cup, and −(v − u) when leaving it. Hence its absolute velocity when leaving the cup is u − (v − u) = 2u − v. The change of momentum per second = (G/g) ω (v − u) {v − (2u − v)} = 2(G/g) ω (v − u)2. Comparing this with case 2, it is seen that the pressure on a hemispherical cup is double that on a flat plane. The work done on the cup = 2(G/g) ω (v − u)2u foot-pounds per second. The efficiency of the jet is greatest when v = 3u; in that case the efficiency =16⁄27.If a series of cup vanes are introduced in front of the jet, so that the quantity of water acted upon is ωv instead of ω(v − u), then the whole pressure on the chain of cups is (G/g) ωv {v − (2u − v)} = 2(G/g)ωv (v − u). In this case the efficiency is greatest when v = 2u, and the maximum efficiency is unity, or all the energy of the water is expended on the cups.Fig. 156.§ 157. (5)Case of a Flat Vane oblique to the Jet(fig. 156).—This case presents some difficulty. The water spreading on the plane in all directions from the point of impact, different particles leave the plane with different absolute velocities. Let AB = v = velocity of water, AC = u = velocity of plane. Then, completing the parallelogram, AD represents in magnitude and direction the relative velocity of water and plane. Draw AE normal to the plane and DE parallel to the plane. Then the relative velocity AD may be regarded as consisting of two components, one AE normal, the other DE parallel to the plane. On the assumption that friction is insensible, DE is unaffected by impact, but AE is destroyed. Hence AE represents the entire change of velocity due to impact and the direction of that change. The pressure on the plane is in the direction AE, and its amount is = mass of water impinging per second × AE.Let DAE = θ, and let AD = vr. Then AE = vrcos θ; DE = vrsin θ. If Q is the volume of water impinging on the plane per second, the change of momentum is (G/g) Qvrcos θ. Let AC = u = velocity of the plane, and let AC make the angle CAE = δ with the normal to the plane. The velocity of the plane in the direction AE = u cos δ. The work of the jet on the plane = (G/g) Qvrcos θ u cos δ. The same problem may be thus treated algebraically (fig. 157). Let BAF = α, and CAF = δ. The velocity v of the water may be decomposed into AF = v cos α normal to the plane, and FB = v sin α parallel to the plane. Similarly the velocity of the plane = u = AC = BD can be decomposed into BG = FE = u cos δ normal to the plane, and DG = u sin δ parallel to the plane. As friction is neglected, the velocity of the water parallel to the plane is unaffected by the impact, but its component v cos α normal to the plane becomes after impact the same as that of the plane, that is, u cos δ. Hence the change of velocity during impact = AE = v cos α − u cos δ. The change of momentum per second, and consequently the normal pressure on the plane is N = (G/g) Q(v cos α − u cos δ). The pressure in the direction in which the plane is moving is P = N cos δ = (G/g)Q (v cos α − u cos δ) cos δ, and the work done on the plane is Pu = (G/g)Q(v cos α − u cos δ) u cos δ, which is the same expression as before, since AE = vrcos θ = v cos α − u cos δ.Fig. 157.Fig. 158.In one second the plane moves so that the point A (fig. 158) comes to C, or from the position shown in full lines to the position shown in dotted lines. If the plane remained stationary, a length AB = v of the jet would impinge on the plane, but, since the plane moves in the same direction as the jet, only the length HB = AB − AH impinges on the plane.But AH = AC cos δ / cos α = u cos δ / cos α, and therefore HB = v − u cos δ / cos α. Let ω = sectional area of jet; volume impinging on plane per second = Q = ω(v − u cos δ / cos α) = ω (v cos α − u cos δ) / cos α. Inserting this in the formulae above, we getN =Gω(v cos α − u cos δ)2;gcos α(1)P =Gω cos δ(v cos α − u cos δ)2;gcos α(2)Pu =Gωucos δ(v cos α − u cos δ)2;gcos α(3)Three cases may be distinguished:—(a) The plane is at rest. Then u = 0, N = (G/g) ωv2cos α; and the work done on the plane and the efficiency of the jet are zero.(b) The plane moves parallel to the jet. Then δ = α, and Pu = (G/g)ωu cos2α (v − u)2, which is a maximum when u =1⁄3v.When u =1⁄3v then Pu max. =4⁄27(G/g)ωv3cos2α, and the efficiency = η =4⁄9cos2α.(c) The plane moves perpendicularly to the jet. Then δ = 90° − α; cos δ = sin α; and Pu = G/g ωu (sin α / cos α) (v cos α − u sin α)2. This is a maximum when u =1⁄3v cos α.When u =1⁄3v cos α, the maximum work and the efficiency are the same as in the last case.Fig. 159.§ 158.Best Form of Vane to receive Water.—When water impinges normally or obliquely on a plane, it is scattered in all directions after impact, and the work carried away by the water is then generally lost, from the impossibility of dealing afterwards with streams of water deviated in so many directions. By suitably forming the vane, however, the water may be entirely deviated in one direction, and the loss of energy from agitation of the water is entirely avoided.Let AB (fig. 159) be a vane, on which a jet of water impinges at the point A and in the direction AC. Take AC = v = velocity ofwater, and let AD represent in magnitude and direction the velocity of the vane. Completing the parallelogram, DC or AE represents the direction in which the water is moving relatively to the vane. If the lip of the vane at A is tangential to AE, the water will not have its direction suddenly changed when it impinges on the vane, and will therefore have no tendency to spread laterally. On the contrary it will be so gradually deviated that it will glide up the vane in the direction AB. This is sometimes expressed by saying that the vanereceives the water without shock.Fig. 160.§ 159.Floats of Poncelet Water Wheels.—Let AC (fig. 160) represent the direction of a thin horizontal stream of water having the velocity v. Let AB be a curved float moving horizontally with velocity u. The relative motion of water and float is then initially horizontal, and equal to v − u.In order that the float may receive the water without shock, it is necessary and sufficient that the lip of the float at A should be tangential to the direction AC of relative motion. At the end of (v − u)/g seconds the float moving with the velocity u comes to the position A1B1, and during this time a particle of water received at A and gliding up the float with the relative velocity v − u, attains a height DE = (v − u)2/2g. At E the water comes to relative rest. It then descends along the float, and when after 2(v − u)/g seconds the float has come to A2B2the water will again have reached the lip at A2and will quit it tangentially, that is, in the direction CA2, with a relative velocity −(v − u) = −√ (2gDE) acquired under the influence of gravity. The absolute velocity of the water leaving the float is therefore u − (v − u) = 2u − v. If u =1⁄2v, the water will drop off the bucket deprived of all energy of motion. The whole of the work of the jet must therefore have been expended in driving the float. The water will have been received without shock and discharged without velocity. This is the principle of the Poncelet wheel, but in that case the floats move over an arc of a large circle; the stream of water has considerable thickness (about 8 in.); in order to get the water into and out of the wheel, it is then necessary that the lip of the float should make a small angle (about 15°) with the direction of its motion. The water quits the wheel with a little of its energy of motion remaining.§ 160.Pressure on a Curved Surface when the Water is deviated wholly in one Direction.—When a jet of water impinges on a curved surface in such a direction that it is received without shock, the pressure on the surface is due to its gradual deviation from its first direction. On any portion of the area the pressure is equal and opposite to the force required to cause the deviation of so much water as rests on that surface. In common language, it is equal to the centrifugal force of that quantity of water.Fig. 161.Case1.Surface Cylindrical and Stationary.—Let AB (fig. 161) be the surface, having its axis at O and its radius = r. Let the water impinge at A tangentially, and quit the surface tangentially at B. Since the surface is at rest, v is both the absolute velocity of the water and the velocity relatively to the surface, and this remains unchanged during contact with the surface, because the deviating force is at each point perpendicular to the direction of motion. The water is deviated through an angle BCD = AOB = φ. Each particle of water of weight p exerts radially a centrifugal force pv2/rg. Let the thickness of the stream = t ft. Then the weight of water resting on unit of surface = Gt ℔; and the normal pressure per unit of surface = n = Gtv2/gr. The resultant of the radial pressures uniformly distributed from A to B will be a force acting in the direction OC bisecting AOB, and its magnitude will equal that of a force of intensity = n, acting on the projection of AB on a plane perpendicular to the direction OC. The length of the chord AB = 2r sin1⁄2φ; let b = breadth of the surface perpendicular to the plane of the figure. The resultant pressure on surface= R = 2rb sinφ×Gt·v2= 2Gbtv2sinφ,2grg2which is independent of the radius of curvature. It may be inferred that the resultant pressure is the same for any curved surface of the same projected area, which deviates the water through the same angle.Case2.Cylindrical Surface moving in the Direction AC with Velocity u.—The relative velocity = v − u. The final velocity BF (fig. 162) is found by combining the relative velocity BD = v − u tangential to the surface with the velocity BE = u of the surface. The intensity of normal pressure, as in the last case, is (G/g) t (v − u)2/r. The resultant normal pressure R = 2(G/g) bt (v − u)2sin1⁄2φ. This resultant pressure may be resolved into two components P and L, one parallel and the other perpendicular to the direction of the vane’s motion. The former is an effort doing work on the vane. The latter is a lateral force which does no work.P = R sin1⁄2φ = (G/g) bt (v − u)2(1 − cos φ);L = R cos1⁄2φ = (G/g) bt (v − u)2sin φ.Fig. 162.Fig. 163.The work done by the jet on the vane is Pu = (G/g) btu (v − u)2(1 − cos φ), which is a maximum when u =1⁄3v. This result can also be obtained by considering that the work done on the plane must be equal to the energy lost by the water, when friction is neglected.If φ = 180°, cos φ = −1, 1 − cos φ = 2; then P = 2(G/g) bt (v − u)2, the same result as for a concave cup.§ 161.Position which a Movable Plane takes in Flowing Water.—When a rectangular plane, movable about an axis parallel to one of its sides, is placed in an indefinite current of fluid, it takes a position such that the resultant of the normal pressures on the two sides of the axis passes through the axis. If, therefore, planes pivoted so that the ratio a/b (fig. 163) is varied are placed in water, and the angle they make with the direction of the stream is observed, the position of the resultant of the pressures on the plane is determined for different angular positions. Experiments of this kind have been made by Hagen. Some of his results are given in the following table:—Larger plane.Smaller Plane.a/b = 1.0φ = ...φ = 90°0.975°721⁄2°0.860°57°0.748°43°0.625°29°0.513°13°0.48°61⁄2°0.36°..0.24°..§ 162.Direct Action distinguished from Reaction(Rankine,Steam Engine, § 147).The pressure which a jet exerts on a vane can be distinguished into two parts, viz∴—(1) The pressure arising from changing the direct component of the velocity of the water into the velocity of the vane. In fig. 153, § 154, ab cos bae is the direct component of the water’s velocity, or component in the direction of motion of vane. This is changed into the velocity ae of the vane. The pressure due to direct impulse is thenP1= GQ (ab cos bae − ae) / g.For a flat vane moving normally, this direct action is the only action producing pressure on the vane.(2) The term reaction is applied to the additional action due to the direction and velocity with which the water glances off the vane. It is this which is diminished by the friction between the water and the vane. In Case 2, § 160, the direct pressure isP1= Gbt (v − u)2/ g.That due to reaction isP2= −Gbt (v − u)2cos φ / g.If φ < 90°, the direct component of the water’s motion is not wholly converted into the velocity of the vane, and the wholepressure due to direct impulse is not obtained. If φ > 90°, cos φ is negative and an additional pressure due to reaction is obtained.Fig. 164.§ 163.Jet Propeller.—In the case of vessels propelled by a jet of water (fig. 164), driven sternwards from orifices at the side of the vessel, the water, originally at rest outside the vessel, is drawn into the ship and caused to move with the forward velocity V of the ship. Afterwards it is projected sternwards from the jets with a velocity v relatively to the ship, or v − V relatively to the earth. If Ω is the total sectional area of the jets, Ωv is the quantity of water discharged per second. The momentum generated per second in a sternward direction is (G/g) Ωv (v − V), and this is equal to the forward acting reaction P which propels the ship.The energy carried away by the water=1⁄2(G/g) Ωv (v − V)2.(1)The useful work done on the shipPV = (G/g) Ωv (v − V) V.(2)Adding (1) and (2), we get the whole work expended on the water, neglecting friction:—W =1⁄2(G/g) Ωv (v2− V2).Hence the efficiency of the jet propeller isPV/W = 2V / (v + V).(3)This increases towards unity as v approaches V. In other words, the less the velocity of the jets exceeds that of the ship, and therefore the greater the area of the orifice of discharge, the greater is the efficiency of the propeller.In the “Waterwitch” v was about twice V. Hence in this case the theoretical efficiency of the propeller, friction neglected, was about2⁄3.Fig. 165.§ 164.Pressure of a Steady Stream in a Uniform Pipe on a Plane normal to the Direction of Motion.—Let CD (fig. 165) be a plane placed normally to the stream which, for simplicity, may be supposed to flow horizontally. The fluid filaments are deviated in front of the plane, form a contraction at A1A1, and converge again, leaving a mass of eddying water behind the plane. Suppose the section A0A0taken at a point where the parallel motion has not begun to be disturbed, and A2A2where the parallel motion is re-established. Then since the same quantity of water with the same velocity passes A0A0, A2A2in any given time, the external forces produce no change of momentum on the mass A0A0A2A2, and must therefore be in equilibrium. If Ω is the section of the stream at A0A0or A2A2, and ω the area of the plate CD, the area of the contracted section of the stream at A1A1will be cc(Ω − ω), where ccis the coefficient of contraction. Hence, if v is the velocity at A0A0or A2A2, and v1the velocity at A1A1,vΩ = ccv (Ω − ω);∴ v1= vΩ / cc(Ω − ω).(1)Let p0, p1, p2be the pressures at the three sections. Applying Bernoulli’s theorem to the sections A0A0and A1A1,p0+v2=p1+v12.G2gG2gAlso, for the sections A1A1and A2A2, allowing that the head due to the relative velocity v1− v is lost in shock:—p1+v12=p2+v2+(v1− v)2;G2gG2g2g∴ p0− p2= G (v1− v)2/ 2g;(2)or, introducing the value in (1),p0− p2=G(Ω− 1)2v22gcc(Ω − ω)(3)Now the external forces in the direction of motion acting on the mass A0A0A2A2are the pressures p0Ω1− p2Ω at the ends, and the reaction −R of the plane on the water, which is equal and opposite to the pressure of the water on the plane. As these are in equilibrium,(p0− p2) Ω − R = 0;∴ R = GΩ(Ω− 1)2v2;cc(Ω − ω)2g(4)an expression like that for the pressure of an isolated jet on an indefinitely extended plane, with the addition of the term in brackets, which depends only on the areas of the stream and the plane. For a given plane the expression in brackets diminishes as Ω increases. If Ω/ω = ρ, the equation (4) becomesR = Gωv2{ρ(ρ− 1)2},2gcc(ρ − 1)(4a)which is of the formR = Gω (v2/2g) K,where K depends only on the ratio of the sections of the stream and plane.For example, let cc= 0.85, a value which is probable, if we allow that the sides of the pipe act as internal borders to an orifice. ThenK = ρ(1.176ρ− 1)2.ρ − 1ρ =K =1∞23.6631.7541.2951.1010.94502.001003.50The assumption that the coefficient of contraction ccis constant for different values of ρ is probably only true when ρ is not very large. Further, the increase of K for large values of ρ is contrary to experience, and hence it may be inferred that the assumption that all the filaments have a common velocity v1at the section A1A1and a common velocity v at the section A2A2is not true when the stream is very much larger than the plane. Hence, in the expressionR = KGωv2/ 2g,K must be determined by experiment in each special case. For a cylindrical body putting ω for the section, ccfor the coefficient of contraction, cc(Ω − ω) for the area of the stream at A1A1,v1= vΩ / cc(Ω − ω); v2= vΩ / (Ω −ω);or, putting ρ = Ω/ω,v1= vρ / cc(ρ − 1), v2= vρ / (ρ − 1).ThenR = K1Gωv2/ 2g,whereK1= ρ{ (ρ)2(1− 1)2+(ρ− 1)2}.ρ − 1ccρ − 1Taking cc= 0.85 and ρ = 4, K1= 0.467, a value less than before. Hence there is less pressure on the cylinder than on the thin plane.Fig. 166.§ 165.Distribution of Pressure on a Surface on which a Jet impinges normally.—The principle of momentum gives readily enough the total or resultant pressure of a jet impinging on a plane surface, but in some cases it is useful to know the distribution of the pressure. The problem in the case in which the plane is struck normally, and the jet spreads in all directions, is one of great complexity, but even in that case the maximum intensity of the pressure is easily assigned. Each layer of water flowing from an orifice is gradually deviated (fig. 166) by contact with the surface, and during deviation exercises a centrifugal pressure towards the axis of the jet. The force exerted by each small mass of water is normal to its path and inversely as the radius of curvature of the path. Hence the greatest pressure on the plane must be at the axis of the jet, and the pressure must decrease from the axis outwards, in some such way as is shown by the curve of pressure in fig. 167, the branches of the curve being probably asymptotic to the plane.For simplicity suppose the jet is a vertical one. Let h1(fig. 167) be the depth of the orifice from the free surface, and v1the velocity of discharge. Then, if ω is the area of the orifice, the quantity of water impinging on the plane is obviouslyQ = ωv1= ω √ (2gh1);that is, supposing the orifice rounded, and neglecting the coefficient of discharge.The velocity with which the fluid reaches the plane is, however, greater than this, and may reach the valuev = √ (2gh);where h is the depth of the plane below the free surface. The external layers of fluid subjected throughout, after leaving the orifice, to the atmospheric pressure will attain the velocity v, and will flow away with this velocity unchanged except by friction. The layers towards the interior of the jet, being subjected to a pressure greater than atmospheric pressure, will attain a less velocity, and so much less as they are nearer the centre of the jet. But the pressurecan in no case exceed the pressure v2/2g or h measured in feet of water, or the direction of motion of the water would be reversed, and there would be reflux. Hence the maximum intensity of the pressure of the jet on the plane is h ft. of water. If the pressure curve is drawn with pressures represented by feet of water, it will touch the free water surface at the centre of the jet.Fig. 167.Suppose the pressure curve rotated so as to form a solid of revolution. The weight of water contained in that solid is the total pressure of the jet on the surface, which has already been determined. Let V = volume of this solid, then GV is its weight in pounds. ConsequentlyGV = (G/g) ωv1v;V = 2ω √ (hh1).We have already, therefore, two conditions to be satisfied by the pressure curve.Fig. 168.—Curves of Pressure of Jets impinging normally on a Plane.Some very interesting experiments on the distribution of pressure on a surface struck by a jet have been made by J. S. Beresford (Prof. Papers on Indian Engineering, No. cccxxii.), with a view to afford information as to the forces acting on the aprons of weirs. Cylindrical jets1⁄2in. to 2 in. diameter, issuing from a vessel in which the water level was constant, were allowed to fall vertically on a brass plate 9 in. in diameter. A small hole in the brass plate communicated by a flexible tube with a vertical pressure column. Arrangements were made by which this aperture could be moved1⁄20in. at a time across the area struck by the jet. The height of the pressure column, for each position of the aperture, gave the pressure at that point of the area struck by the jet. When the aperture was exactly in the axis of the jet, the pressure column was very nearly level with the free surface in the reservoir supplying the jet; that is, the pressure was very nearly v2/2g. As the aperture moved away from the axis of the jet, the pressure diminished, and it became insensibly small at a distance from the axis of the jet about equal to the diameter of the jet. Hence, roughly, the pressure due to the jet extends over an area about four times the area of section of the jet.Fig. 168 shows the pressure curves obtained in three experiments with three jets of the sizes shown, and with the free surface level in the reservoir at the heights marked.Height from FreeSurface to BrassPlate in inches.Distance from Axisof Jet in inches.Pressure in inchesof Water.Experiment 1. Jet .475 in. diameter.43040.5”.0539.40”.137.5-39.5”.1535”.233.5-37”.2531”.321-27”.3521”.414”.458”.53.5”.551”.60.5”.650Experiment 2. Jet .988 in. diameter.42.15042”.0541.9”.141.5-41.8”.1541”.240.3”.2539.2”.337.5”.3534.8”.452742.25.523”.5518.5”.613”.658.3”.75”.753”.82.242.15.851.6”.951Experiment 3. Jet 19.5 in. diameter.27.15026.9”.0826.9”.1326.8”.1826.5-26.6”.2326.4-26.5”.2826.3-26.627.3326.2”.3825.9”.4325.5”.4825”.5324.5”.5824”.6323.3”.6822.5”.7321.8”.7821”.8320.3”.8819.3”.9318”.981726.51.1313.5”1.1812.5”1.2310.8”1.289.5”1.338”1.387”1.436.3”1.485”1.534.3”1.583.5”1.92As the general form of the pressure curve has been already indicated, it may be assumed that its equation is of the formy = ab−x2.But it has already been shown that for x = 0, y = h, hence a = h. To determine the remaining constant, the other condition may be used, that the solid formed by rotating the pressure curve represents the total pressure on the plane. The volume of the solid isV =∫∞02πxy dx= 2πh∫∞0b−x2x dx= (πh / logeb)[−b−x2]∞0= πh / logeb.Using the condition already stated,2ω √ (hh1) = πh / logeb,logεb = (π/2ω) √ (h/h1).Putting the value of b in (2) in eq. (1), and also r for the radius of the jet at the orifice, so that ω = πr2, the equation to the pressure curve isy = hε−1/2√(h / h1) (x2/ r2).§ 166.Resistance of a Plane moving through a Fluid, or Pressure of a Current on a Plane.—When a thin plate moves through the air, or through an indefinitely large mass of still water, in a direction normal to its surface, there is an excess of pressure on the anterior face and a diminution of pressure on the posterior face. Let v be the relative velocity of the plate and fluid, Ω the area of the plate, G the density of the fluid, h the height due to the velocity, then the total resistance is expressed by the equationR = fGΩv2/ 2g pounds = fGΩh;where f is a coefficient having about the value 1.3 for a plate moving in still fluid, and 1.8 for a current impinging on a fixed plane, whether the fluid is air or water. The difference in the value of the coefficient in the two cases is perhaps due to errors of experiment. There is a similar resistance to motion in the case of all bodies of “unfair“ form, that is, in which the surfaces over which the water slides are not of gradual and continuous curvature.The stress between the fluid and plate arises chiefly in this way.The streams of fluid deviated in front of the plate, supposed for definiteness to be moving through the fluid, receive from it forward momentum. Portions of this forward moving water are thrown off laterally at the edges of the plate, and diffused through the surrounding fluid, instead of falling to their original position behind the plate. Other portions of comparatively still water are dragged into motion to fill the space left behind the plate; and there is thus a pressure less than hydrostatic pressure at the back of the plate. The whole resistance to the motion of the plate is the sum of the excess of pressure in front and deficiency of pressure behind. This resistance is independent of any friction or viscosity in the fluid, and is due simply to its inertia resisting a sudden change of direction at the edge of the plate.Experiments made by a whirling machine, in which the plate is fixed on a long arm and moved circularly, gave the following values of the coefficientf. The method is not free from objection, as the centrifugal force causes a flow outwards across the plate.ApproximateArea of Platein sq. ft.Values of f.Borda.Hutton.Thibault.0.131.391.24..0.251.491.431.5250.631.64....1.11....1.784There is a steady increase of resistance with the size of the plate, in part or wholly due to centrifugal action.P. L. G. Dubuat (1734-1809) made experiments on a plane 1 ft. square, moved in a straight line in water at 3 to 61⁄2ft. per second. Calling m the coefficient of excess of pressure in front, and n the coefficient of deficiency of pressure behind, so that f = m + n, he found the following values:—m = 1; n = 0.433; f = 1.433.The pressures were measured by pressure columns. Experiments by A. J. Morin (1795-1880), G. Piobert (1793-1871) and I. Didion (1798-1878) on plates of 0.3 to 2.7 sq. ft. area, drawn vertically through water, gave f = 2.18; but the experiments were made in a reservoir of comparatively small depth. For similar plates moved through air they found f = 1.36, a result more in accordance with those which precede.For a fixed plane in a moving current of water E. Mariotte found f = 1.25. Dubuat, in experiments in a current of water like those mentioned above, obtained the values m = 1.186; n = 0.670; f = 1.856. Thibault exposed to wind pressure planes of 1.17 and 2.5 sq. ft. area, and found f to vary from 1.568 to 2.125, the mean value being f = 1.834, a result agreeing well with Dubuat.Fig. 169.§ 167.Stanton’s Experiments on the Pressure of Air on Surfaces.—At the National Physical Laboratory, London, T. E. Stanton carried out a series of experiments on the distribution of pressure on surfaces in a current of air passing through an air trunk. These were on a small scale but with exceptionally accurate means of measurement. These experiments differ from those already given in that the plane is small relatively to the cross section of the current (Proc. Inst. Civ. Eng.clvi., 1904). Fig. 169 shows the distribution of pressure on a square plate. ab is the plate in vertical section. acb the distribution of pressure on the windward and adb that on the leeward side of the central section. Similarly aeb is the distribution of pressure on the windward and afb on the leeward side of a diagonal section. The intensity of pressure at the centre of the plate on the windward side was in all cases p = Gv2/2g ℔ per sq. ft., where G is the weight of a cubic foot of air and v the velocity of the current in ft. per sec. On the leeward side the negative pressure is uniform except near the edges, and its value depends on the form of the plate. For a circular plate the pressure on the leeward side was 0.48 Gv2/2g and for a rectangular plate 0.66 Gv2/2g. For circular or square plates the resultant pressure on the plate was P = 0.00126 v2℔ per sq. ft. where v is the velocity of the current in ft. per sec. On a long narrow rectangular plate the resultant pressure was nearly 60% greater than on a circular plate. In later tests on larger planes in free air, Stanton found resistances 18% greater than those observed with small planes in the air trunk.§ 168.Case when the Direction of Motion is oblique to the Plane.—The determination of the pressure between a fluid and surface in this case is of importance in many practical questions, for instance, in assigning the load due to wind pressure on sloping and curved roofs, and experiments have been made by Hutton, Vince, and Thibault on planes moved circularly through air and water on a whirling machine.Fig. 170.Let AB (fig. 170) be a plane moving in the direction R making an angle φ with the plane. The resultant pressure between the fluid and the plane will be a normal pressure N. The component R of this normal pressure is the resistance to the motion of the plane and the other component L is a lateral force resisted by the guides which support the plane. ObviouslyR = N sin φ;L = N cos φ.In the case of wind pressure on a sloping roof surface, R is the horizontal and L the vertical component of the normal pressure.In experiments with the whirling machine it is the resistance to motion, R, which is directly measured. Let P be the pressure on a plane moved normally through a fluid. Then, for the same plane inclined at an angle φ to its direction of motion, the resistance was found by Hutton to beR = P (sin φ)1.842 cos φ.A simpler and more convenient expression given by Colonel Duchemin isR = 2P sin2φ / (1 + sin2φ).Consequently, the total pressure between the fluid and plane isN = 2P sin φ / (1 + sin2φ) = 2P / (cosec φ + sin φ),and the lateral force isL = 2P sin φ cos φ / (1 + sin2φ).In 1872 some experiments were made for the Aeronautical Society on the pressure of air on oblique planes. These plates, of 1 to 2 ft. square, were balanced by ingenious mechanism designed by F. H. Wenham and Spencer Browning, in such a manner that both the pressure in the direction of the air current and the lateral force were separately measured. These planes were placed opposite a blast from a fan issuing from a wooden pipe 18 in. square. The pressure of the blast varied from6⁄10to 1 in. of water pressure. The following are the results given in pounds per square foot of the plane, and a comparison of the experimental results with the pressures given by Duchemin’s rule. These last values are obtained by taking P = 3.31, the observed pressure on a normal surface:—Angle between Plane and Direction of Blast15°20°60°90°Horizontal pressure R0.40.612.733.31Lateral pressure L1.61.961.26..Normal pressure √ (L2+ R2)1.652.053.013.31Normal pressure by Duchemin’s rule1.6052.0273.2763.31

§ 153. When a stream of fluid in steady motion impinges on a solid surface, it presses on the surface with a force equal and opposite to that by which the velocity and direction of motion of the fluid are changed. Generally, in problems on the impact of fluids, it is necessary to neglect the effect of friction between the fluid and the surface on which it moves.

During Impact the Velocity of the Fluid relatively to the Surface on which it impinges remains unchanged in Magnitude.—Consider a mass of fluid flowing in contact with a solid surface also in motion, the motion of both fluid and solid being estimated relatively to the earth. Then the motion of the fluid may be resolved into two parts, one a motion equal to that of the solid, and in the same direction, the other a motion relatively to the solid. The motion which the fluid has in common with the solid cannot at all be influenced by the contact. The relative component of the motion of the fluid can only be altered in direction, but not in magnitude. The fluid moving in contact with the surface can only have a relative motion parallel to the surface, while the pressure between the fluid and solid, if friction is neglected, is normal to the surface. The pressure therefore can only deviate the fluid, without altering the magnitude of the relative velocity. The unchanged common component and, combined with it, the deviated relative component give the resultant final velocity, which may differ greatly in magnitude and direction from the initial velocity.

From the principle of momentum, the impulse of any mass of fluid reaching the surface in any given time is equal to the change of momentum estimated in the same direction. The pressure between the fluid and surface, in any direction, is equal to the change of momentum in that direction of so much fluid as reaches the surface in one second. If Pais the pressure in any direction, m the mass of fluid impinging per second, vathe change of velocity in the direction of Padue to impact, then

Pa= mva.

If v1(fig. 152) is the velocity and direction of motion before impact, v2that after impact, then v is the total change of motion due to impact. The resultant pressure of the fluid on the surface is in the direction of v, and is equal to v multiplied by the mass impinging per second. That is, putting P for the resultant pressure,

P = mv.

Let P be resolved into two components, N and T, normal and tangential to the direction of motion of the solid on which the fluid impinges. Then N is a lateral force producing a pressure on the supports of the solid, T is an effort which does work on the solid. If u is the velocity of the solid, Tu is the work done per second by the fluid in moving the solid surface.

Let Q be the volume, and GQ the weight of the fluid impinging per second, and let v1be the initial velocity of the fluid before striking the surface. Then GQv12/2g is the original kinetic energy of Q cub. ft. of fluid, and the efficiency of the stream considered as an arrangement for moving the solid surface is

η = Tu / (GQv12/ 2g).

§ 154.Jet deviated entirely in one Direction.—Geometrical Solution(fig. 153).—Suppose a jet of water impinges on a surface ac with a velocity ab, and let it be wholly deviated in planes parallel to the figure. Also let ae be the velocity and direction of motion of the surface. Join eb; then the water moves with respect to the surface in the direction and with the velocity eb. As this relative velocity is unaltered by contact with the surface, take cd = eb, tangent to the surface at c, then cd is the relative motion of the water with respect to the surface at c. Take df equal and parallel to ae. Then fc (obtained by compounding the relative motion of water to surface and common velocity of water and surface) is the absolute velocity and direction of the water leaving the surface. Take ag equal and parallel to fc. Then, since ab is the initial and ag the final velocity and direction of motion, gb is the total change of motion of the water. The resultant pressure on the plane is in the direction gb. Join eg. In the triangle gae, ae is equal and parallel to df, and ag to fc. Hence eg is equal and parallel to cd. But cd = eb = relative motion of water and surface. Hence the change of motion of the water is represented in magnitude and direction by the third side of an isosceles triangle, of which the other sides are equal to the relative velocity of the water and surface, and parallel to the initial and final directions of relative motion.

Special Cases

§ 155. (1)A Jet impinges on a plane surface at rest, in a direction normal to the plane(fig. 154).—Let a jet whose section is ω impinge with a velocity v on a plane surface at rest, in a direction normal to the plane. The particles approach the plane, are gradually deviated, and finally flow away parallel to the plane, having then no velocity in the original direction of the jet. The quantity of water impinging per second is ωv. The pressure on the plane, which is equal to the change of momentum per second, is P = (G/g) ωv2.

(2)If the plane is moving in the direction of the jet with the velocity±u, the quantity impinging per second is ω(v ± u). The momentum of this quantity before impact is (G/g)ω(v ± u)v. After impact, the water still possesses the velocity ±u in the direction of the jet; and the momentum, in that direction, of so much water as impinges in one second, after impact, is ±(G/g) ω (v ± u)u. The pressure on the plane, which is the change of momentum per second, is the difference of these quantities or P = (G/g) ω (v ± u)2. This differs from the expression obtained in the previous case, in that the relative velocity of the water and plane v ± u is substituted for v. The expression may be written P = 2 × G × ω (v ± u)2/2g, where the last two terms are the volume of a prism of water whose section is the area of the jet and whose length is the head due to the relative velocity. The pressure on the plane is twice the weight of that prism of water. The work done when the planeis moving in the same direction as the jet is Pu = (G/g) ω (v − u)2u foot-pounds per second. There issue from the jet ωv cub. ft. per second, and the energy of this quantity before impact is (G/2g) ωv3. The efficiency of the jet is therefore η = 2(v − u)2u/v3. The value of u which makes this a maximum is found by differentiating and equating the differential coefficient to zero:—

dη / du = 2 (v2− 4vu + 3u2) / v3= 0;

∴ u = v or1⁄3v.

The former gives a minimum, the latter a maximum efficiency.

Putting u =1⁄3v in the expression above,

η max. =8⁄27.

(3) If, instead of one plane moving before the jet, a series of planes are introduced at short intervals at the same point, the quantity of water impinging on the series will be ωv instead of ω(v − u), and the whole pressure = (G/g) ωv (v − u). The work done is (G/g)ωvu (v − u). The efficiency η = (G/g) ωvu (v − u) ÷ (G/2g) ωv3= 2u(v-u)/v2. This becomes a maximum for dη/du = 2(v − 2u) = 0, or u =1⁄2v, and the η =1⁄2. This result is often used as an approximate expression for the velocity of greatest efficiency when a jet of water strikes the floats of a water wheel. The work wasted in this case is half the whole energy of the jet when the floats run at the best speed.

§ 156. (4)Case of a Jet impinging on a Concave Cup Vane, velocity of water v, velocity of vane in the same direction u (fig. 155), weight impinging per second = Gw (v − u).

If the cup is hemispherical, the water leaves the cup in a direction parallel to the jet. Its relative velocity is v − u when approaching the cup, and −(v − u) when leaving it. Hence its absolute velocity when leaving the cup is u − (v − u) = 2u − v. The change of momentum per second = (G/g) ω (v − u) {v − (2u − v)} = 2(G/g) ω (v − u)2. Comparing this with case 2, it is seen that the pressure on a hemispherical cup is double that on a flat plane. The work done on the cup = 2(G/g) ω (v − u)2u foot-pounds per second. The efficiency of the jet is greatest when v = 3u; in that case the efficiency =16⁄27.

If a series of cup vanes are introduced in front of the jet, so that the quantity of water acted upon is ωv instead of ω(v − u), then the whole pressure on the chain of cups is (G/g) ωv {v − (2u − v)} = 2(G/g)ωv (v − u). In this case the efficiency is greatest when v = 2u, and the maximum efficiency is unity, or all the energy of the water is expended on the cups.

§ 157. (5)Case of a Flat Vane oblique to the Jet(fig. 156).—This case presents some difficulty. The water spreading on the plane in all directions from the point of impact, different particles leave the plane with different absolute velocities. Let AB = v = velocity of water, AC = u = velocity of plane. Then, completing the parallelogram, AD represents in magnitude and direction the relative velocity of water and plane. Draw AE normal to the plane and DE parallel to the plane. Then the relative velocity AD may be regarded as consisting of two components, one AE normal, the other DE parallel to the plane. On the assumption that friction is insensible, DE is unaffected by impact, but AE is destroyed. Hence AE represents the entire change of velocity due to impact and the direction of that change. The pressure on the plane is in the direction AE, and its amount is = mass of water impinging per second × AE.

Let DAE = θ, and let AD = vr. Then AE = vrcos θ; DE = vrsin θ. If Q is the volume of water impinging on the plane per second, the change of momentum is (G/g) Qvrcos θ. Let AC = u = velocity of the plane, and let AC make the angle CAE = δ with the normal to the plane. The velocity of the plane in the direction AE = u cos δ. The work of the jet on the plane = (G/g) Qvrcos θ u cos δ. The same problem may be thus treated algebraically (fig. 157). Let BAF = α, and CAF = δ. The velocity v of the water may be decomposed into AF = v cos α normal to the plane, and FB = v sin α parallel to the plane. Similarly the velocity of the plane = u = AC = BD can be decomposed into BG = FE = u cos δ normal to the plane, and DG = u sin δ parallel to the plane. As friction is neglected, the velocity of the water parallel to the plane is unaffected by the impact, but its component v cos α normal to the plane becomes after impact the same as that of the plane, that is, u cos δ. Hence the change of velocity during impact = AE = v cos α − u cos δ. The change of momentum per second, and consequently the normal pressure on the plane is N = (G/g) Q(v cos α − u cos δ). The pressure in the direction in which the plane is moving is P = N cos δ = (G/g)Q (v cos α − u cos δ) cos δ, and the work done on the plane is Pu = (G/g)Q(v cos α − u cos δ) u cos δ, which is the same expression as before, since AE = vrcos θ = v cos α − u cos δ.

In one second the plane moves so that the point A (fig. 158) comes to C, or from the position shown in full lines to the position shown in dotted lines. If the plane remained stationary, a length AB = v of the jet would impinge on the plane, but, since the plane moves in the same direction as the jet, only the length HB = AB − AH impinges on the plane.

But AH = AC cos δ / cos α = u cos δ / cos α, and therefore HB = v − u cos δ / cos α. Let ω = sectional area of jet; volume impinging on plane per second = Q = ω(v − u cos δ / cos α) = ω (v cos α − u cos δ) / cos α. Inserting this in the formulae above, we get

(1)

(2)

(3)

Three cases may be distinguished:—

(a) The plane is at rest. Then u = 0, N = (G/g) ωv2cos α; and the work done on the plane and the efficiency of the jet are zero.

(b) The plane moves parallel to the jet. Then δ = α, and Pu = (G/g)ωu cos2α (v − u)2, which is a maximum when u =1⁄3v.

When u =1⁄3v then Pu max. =4⁄27(G/g)ωv3cos2α, and the efficiency = η =4⁄9cos2α.

(c) The plane moves perpendicularly to the jet. Then δ = 90° − α; cos δ = sin α; and Pu = G/g ωu (sin α / cos α) (v cos α − u sin α)2. This is a maximum when u =1⁄3v cos α.

When u =1⁄3v cos α, the maximum work and the efficiency are the same as in the last case.

§ 158.Best Form of Vane to receive Water.—When water impinges normally or obliquely on a plane, it is scattered in all directions after impact, and the work carried away by the water is then generally lost, from the impossibility of dealing afterwards with streams of water deviated in so many directions. By suitably forming the vane, however, the water may be entirely deviated in one direction, and the loss of energy from agitation of the water is entirely avoided.

Let AB (fig. 159) be a vane, on which a jet of water impinges at the point A and in the direction AC. Take AC = v = velocity ofwater, and let AD represent in magnitude and direction the velocity of the vane. Completing the parallelogram, DC or AE represents the direction in which the water is moving relatively to the vane. If the lip of the vane at A is tangential to AE, the water will not have its direction suddenly changed when it impinges on the vane, and will therefore have no tendency to spread laterally. On the contrary it will be so gradually deviated that it will glide up the vane in the direction AB. This is sometimes expressed by saying that the vanereceives the water without shock.

§ 159.Floats of Poncelet Water Wheels.—Let AC (fig. 160) represent the direction of a thin horizontal stream of water having the velocity v. Let AB be a curved float moving horizontally with velocity u. The relative motion of water and float is then initially horizontal, and equal to v − u.

In order that the float may receive the water without shock, it is necessary and sufficient that the lip of the float at A should be tangential to the direction AC of relative motion. At the end of (v − u)/g seconds the float moving with the velocity u comes to the position A1B1, and during this time a particle of water received at A and gliding up the float with the relative velocity v − u, attains a height DE = (v − u)2/2g. At E the water comes to relative rest. It then descends along the float, and when after 2(v − u)/g seconds the float has come to A2B2the water will again have reached the lip at A2and will quit it tangentially, that is, in the direction CA2, with a relative velocity −(v − u) = −√ (2gDE) acquired under the influence of gravity. The absolute velocity of the water leaving the float is therefore u − (v − u) = 2u − v. If u =1⁄2v, the water will drop off the bucket deprived of all energy of motion. The whole of the work of the jet must therefore have been expended in driving the float. The water will have been received without shock and discharged without velocity. This is the principle of the Poncelet wheel, but in that case the floats move over an arc of a large circle; the stream of water has considerable thickness (about 8 in.); in order to get the water into and out of the wheel, it is then necessary that the lip of the float should make a small angle (about 15°) with the direction of its motion. The water quits the wheel with a little of its energy of motion remaining.

§ 160.Pressure on a Curved Surface when the Water is deviated wholly in one Direction.—When a jet of water impinges on a curved surface in such a direction that it is received without shock, the pressure on the surface is due to its gradual deviation from its first direction. On any portion of the area the pressure is equal and opposite to the force required to cause the deviation of so much water as rests on that surface. In common language, it is equal to the centrifugal force of that quantity of water.

Case1.Surface Cylindrical and Stationary.—Let AB (fig. 161) be the surface, having its axis at O and its radius = r. Let the water impinge at A tangentially, and quit the surface tangentially at B. Since the surface is at rest, v is both the absolute velocity of the water and the velocity relatively to the surface, and this remains unchanged during contact with the surface, because the deviating force is at each point perpendicular to the direction of motion. The water is deviated through an angle BCD = AOB = φ. Each particle of water of weight p exerts radially a centrifugal force pv2/rg. Let the thickness of the stream = t ft. Then the weight of water resting on unit of surface = Gt ℔; and the normal pressure per unit of surface = n = Gtv2/gr. The resultant of the radial pressures uniformly distributed from A to B will be a force acting in the direction OC bisecting AOB, and its magnitude will equal that of a force of intensity = n, acting on the projection of AB on a plane perpendicular to the direction OC. The length of the chord AB = 2r sin1⁄2φ; let b = breadth of the surface perpendicular to the plane of the figure. The resultant pressure on surface

which is independent of the radius of curvature. It may be inferred that the resultant pressure is the same for any curved surface of the same projected area, which deviates the water through the same angle.

Case2.Cylindrical Surface moving in the Direction AC with Velocity u.—The relative velocity = v − u. The final velocity BF (fig. 162) is found by combining the relative velocity BD = v − u tangential to the surface with the velocity BE = u of the surface. The intensity of normal pressure, as in the last case, is (G/g) t (v − u)2/r. The resultant normal pressure R = 2(G/g) bt (v − u)2sin1⁄2φ. This resultant pressure may be resolved into two components P and L, one parallel and the other perpendicular to the direction of the vane’s motion. The former is an effort doing work on the vane. The latter is a lateral force which does no work.

P = R sin1⁄2φ = (G/g) bt (v − u)2(1 − cos φ);L = R cos1⁄2φ = (G/g) bt (v − u)2sin φ.

The work done by the jet on the vane is Pu = (G/g) btu (v − u)2(1 − cos φ), which is a maximum when u =1⁄3v. This result can also be obtained by considering that the work done on the plane must be equal to the energy lost by the water, when friction is neglected.

If φ = 180°, cos φ = −1, 1 − cos φ = 2; then P = 2(G/g) bt (v − u)2, the same result as for a concave cup.

§ 161.Position which a Movable Plane takes in Flowing Water.—When a rectangular plane, movable about an axis parallel to one of its sides, is placed in an indefinite current of fluid, it takes a position such that the resultant of the normal pressures on the two sides of the axis passes through the axis. If, therefore, planes pivoted so that the ratio a/b (fig. 163) is varied are placed in water, and the angle they make with the direction of the stream is observed, the position of the resultant of the pressures on the plane is determined for different angular positions. Experiments of this kind have been made by Hagen. Some of his results are given in the following table:—

§ 162.Direct Action distinguished from Reaction(Rankine,Steam Engine, § 147).

The pressure which a jet exerts on a vane can be distinguished into two parts, viz∴—

(1) The pressure arising from changing the direct component of the velocity of the water into the velocity of the vane. In fig. 153, § 154, ab cos bae is the direct component of the water’s velocity, or component in the direction of motion of vane. This is changed into the velocity ae of the vane. The pressure due to direct impulse is then

P1= GQ (ab cos bae − ae) / g.

For a flat vane moving normally, this direct action is the only action producing pressure on the vane.

(2) The term reaction is applied to the additional action due to the direction and velocity with which the water glances off the vane. It is this which is diminished by the friction between the water and the vane. In Case 2, § 160, the direct pressure is

P1= Gbt (v − u)2/ g.

That due to reaction is

P2= −Gbt (v − u)2cos φ / g.

If φ < 90°, the direct component of the water’s motion is not wholly converted into the velocity of the vane, and the wholepressure due to direct impulse is not obtained. If φ > 90°, cos φ is negative and an additional pressure due to reaction is obtained.

§ 163.Jet Propeller.—In the case of vessels propelled by a jet of water (fig. 164), driven sternwards from orifices at the side of the vessel, the water, originally at rest outside the vessel, is drawn into the ship and caused to move with the forward velocity V of the ship. Afterwards it is projected sternwards from the jets with a velocity v relatively to the ship, or v − V relatively to the earth. If Ω is the total sectional area of the jets, Ωv is the quantity of water discharged per second. The momentum generated per second in a sternward direction is (G/g) Ωv (v − V), and this is equal to the forward acting reaction P which propels the ship.

The energy carried away by the water

=1⁄2(G/g) Ωv (v − V)2.

(1)

The useful work done on the ship

PV = (G/g) Ωv (v − V) V.

(2)

Adding (1) and (2), we get the whole work expended on the water, neglecting friction:—

W =1⁄2(G/g) Ωv (v2− V2).

Hence the efficiency of the jet propeller is

PV/W = 2V / (v + V).

(3)

This increases towards unity as v approaches V. In other words, the less the velocity of the jets exceeds that of the ship, and therefore the greater the area of the orifice of discharge, the greater is the efficiency of the propeller.

In the “Waterwitch” v was about twice V. Hence in this case the theoretical efficiency of the propeller, friction neglected, was about2⁄3.

§ 164.Pressure of a Steady Stream in a Uniform Pipe on a Plane normal to the Direction of Motion.—Let CD (fig. 165) be a plane placed normally to the stream which, for simplicity, may be supposed to flow horizontally. The fluid filaments are deviated in front of the plane, form a contraction at A1A1, and converge again, leaving a mass of eddying water behind the plane. Suppose the section A0A0taken at a point where the parallel motion has not begun to be disturbed, and A2A2where the parallel motion is re-established. Then since the same quantity of water with the same velocity passes A0A0, A2A2in any given time, the external forces produce no change of momentum on the mass A0A0A2A2, and must therefore be in equilibrium. If Ω is the section of the stream at A0A0or A2A2, and ω the area of the plate CD, the area of the contracted section of the stream at A1A1will be cc(Ω − ω), where ccis the coefficient of contraction. Hence, if v is the velocity at A0A0or A2A2, and v1the velocity at A1A1,

vΩ = ccv (Ω − ω);

∴ v1= vΩ / cc(Ω − ω).

(1)

Let p0, p1, p2be the pressures at the three sections. Applying Bernoulli’s theorem to the sections A0A0and A1A1,

Also, for the sections A1A1and A2A2, allowing that the head due to the relative velocity v1− v is lost in shock:—

∴ p0− p2= G (v1− v)2/ 2g;

(2)

or, introducing the value in (1),

(3)

Now the external forces in the direction of motion acting on the mass A0A0A2A2are the pressures p0Ω1− p2Ω at the ends, and the reaction −R of the plane on the water, which is equal and opposite to the pressure of the water on the plane. As these are in equilibrium,

(p0− p2) Ω − R = 0;

(4)

an expression like that for the pressure of an isolated jet on an indefinitely extended plane, with the addition of the term in brackets, which depends only on the areas of the stream and the plane. For a given plane the expression in brackets diminishes as Ω increases. If Ω/ω = ρ, the equation (4) becomes

(4a)

which is of the form

R = Gω (v2/2g) K,

where K depends only on the ratio of the sections of the stream and plane.

For example, let cc= 0.85, a value which is probable, if we allow that the sides of the pipe act as internal borders to an orifice. Then

The assumption that the coefficient of contraction ccis constant for different values of ρ is probably only true when ρ is not very large. Further, the increase of K for large values of ρ is contrary to experience, and hence it may be inferred that the assumption that all the filaments have a common velocity v1at the section A1A1and a common velocity v at the section A2A2is not true when the stream is very much larger than the plane. Hence, in the expression

R = KGωv2/ 2g,

K must be determined by experiment in each special case. For a cylindrical body putting ω for the section, ccfor the coefficient of contraction, cc(Ω − ω) for the area of the stream at A1A1,

v1= vΩ / cc(Ω − ω); v2= vΩ / (Ω −ω);

or, putting ρ = Ω/ω,

v1= vρ / cc(ρ − 1), v2= vρ / (ρ − 1).

Then

R = K1Gωv2/ 2g,

where

Taking cc= 0.85 and ρ = 4, K1= 0.467, a value less than before. Hence there is less pressure on the cylinder than on the thin plane.

§ 165.Distribution of Pressure on a Surface on which a Jet impinges normally.—The principle of momentum gives readily enough the total or resultant pressure of a jet impinging on a plane surface, but in some cases it is useful to know the distribution of the pressure. The problem in the case in which the plane is struck normally, and the jet spreads in all directions, is one of great complexity, but even in that case the maximum intensity of the pressure is easily assigned. Each layer of water flowing from an orifice is gradually deviated (fig. 166) by contact with the surface, and during deviation exercises a centrifugal pressure towards the axis of the jet. The force exerted by each small mass of water is normal to its path and inversely as the radius of curvature of the path. Hence the greatest pressure on the plane must be at the axis of the jet, and the pressure must decrease from the axis outwards, in some such way as is shown by the curve of pressure in fig. 167, the branches of the curve being probably asymptotic to the plane.

For simplicity suppose the jet is a vertical one. Let h1(fig. 167) be the depth of the orifice from the free surface, and v1the velocity of discharge. Then, if ω is the area of the orifice, the quantity of water impinging on the plane is obviously

Q = ωv1= ω √ (2gh1);

that is, supposing the orifice rounded, and neglecting the coefficient of discharge.

The velocity with which the fluid reaches the plane is, however, greater than this, and may reach the value

v = √ (2gh);

where h is the depth of the plane below the free surface. The external layers of fluid subjected throughout, after leaving the orifice, to the atmospheric pressure will attain the velocity v, and will flow away with this velocity unchanged except by friction. The layers towards the interior of the jet, being subjected to a pressure greater than atmospheric pressure, will attain a less velocity, and so much less as they are nearer the centre of the jet. But the pressurecan in no case exceed the pressure v2/2g or h measured in feet of water, or the direction of motion of the water would be reversed, and there would be reflux. Hence the maximum intensity of the pressure of the jet on the plane is h ft. of water. If the pressure curve is drawn with pressures represented by feet of water, it will touch the free water surface at the centre of the jet.

Suppose the pressure curve rotated so as to form a solid of revolution. The weight of water contained in that solid is the total pressure of the jet on the surface, which has already been determined. Let V = volume of this solid, then GV is its weight in pounds. Consequently

GV = (G/g) ωv1v;V = 2ω √ (hh1).

We have already, therefore, two conditions to be satisfied by the pressure curve.

Some very interesting experiments on the distribution of pressure on a surface struck by a jet have been made by J. S. Beresford (Prof. Papers on Indian Engineering, No. cccxxii.), with a view to afford information as to the forces acting on the aprons of weirs. Cylindrical jets1⁄2in. to 2 in. diameter, issuing from a vessel in which the water level was constant, were allowed to fall vertically on a brass plate 9 in. in diameter. A small hole in the brass plate communicated by a flexible tube with a vertical pressure column. Arrangements were made by which this aperture could be moved1⁄20in. at a time across the area struck by the jet. The height of the pressure column, for each position of the aperture, gave the pressure at that point of the area struck by the jet. When the aperture was exactly in the axis of the jet, the pressure column was very nearly level with the free surface in the reservoir supplying the jet; that is, the pressure was very nearly v2/2g. As the aperture moved away from the axis of the jet, the pressure diminished, and it became insensibly small at a distance from the axis of the jet about equal to the diameter of the jet. Hence, roughly, the pressure due to the jet extends over an area about four times the area of section of the jet.

Fig. 168 shows the pressure curves obtained in three experiments with three jets of the sizes shown, and with the free surface level in the reservoir at the heights marked.

As the general form of the pressure curve has been already indicated, it may be assumed that its equation is of the form

y = ab−x2.

But it has already been shown that for x = 0, y = h, hence a = h. To determine the remaining constant, the other condition may be used, that the solid formed by rotating the pressure curve represents the total pressure on the plane. The volume of the solid is

V =∫∞02πxy dx

= 2πh∫∞0b−x2x dx

= (πh / logeb)[−b−x2]∞0

= πh / logeb.

Using the condition already stated,

2ω √ (hh1) = πh / logeb,logεb = (π/2ω) √ (h/h1).

Putting the value of b in (2) in eq. (1), and also r for the radius of the jet at the orifice, so that ω = πr2, the equation to the pressure curve is

y = hε−1/2√(h / h1) (x2/ r2).

§ 166.Resistance of a Plane moving through a Fluid, or Pressure of a Current on a Plane.—When a thin plate moves through the air, or through an indefinitely large mass of still water, in a direction normal to its surface, there is an excess of pressure on the anterior face and a diminution of pressure on the posterior face. Let v be the relative velocity of the plate and fluid, Ω the area of the plate, G the density of the fluid, h the height due to the velocity, then the total resistance is expressed by the equation

R = fGΩv2/ 2g pounds = fGΩh;

where f is a coefficient having about the value 1.3 for a plate moving in still fluid, and 1.8 for a current impinging on a fixed plane, whether the fluid is air or water. The difference in the value of the coefficient in the two cases is perhaps due to errors of experiment. There is a similar resistance to motion in the case of all bodies of “unfair“ form, that is, in which the surfaces over which the water slides are not of gradual and continuous curvature.

The stress between the fluid and plate arises chiefly in this way.The streams of fluid deviated in front of the plate, supposed for definiteness to be moving through the fluid, receive from it forward momentum. Portions of this forward moving water are thrown off laterally at the edges of the plate, and diffused through the surrounding fluid, instead of falling to their original position behind the plate. Other portions of comparatively still water are dragged into motion to fill the space left behind the plate; and there is thus a pressure less than hydrostatic pressure at the back of the plate. The whole resistance to the motion of the plate is the sum of the excess of pressure in front and deficiency of pressure behind. This resistance is independent of any friction or viscosity in the fluid, and is due simply to its inertia resisting a sudden change of direction at the edge of the plate.

Experiments made by a whirling machine, in which the plate is fixed on a long arm and moved circularly, gave the following values of the coefficientf. The method is not free from objection, as the centrifugal force causes a flow outwards across the plate.

There is a steady increase of resistance with the size of the plate, in part or wholly due to centrifugal action.

P. L. G. Dubuat (1734-1809) made experiments on a plane 1 ft. square, moved in a straight line in water at 3 to 61⁄2ft. per second. Calling m the coefficient of excess of pressure in front, and n the coefficient of deficiency of pressure behind, so that f = m + n, he found the following values:—

m = 1; n = 0.433; f = 1.433.

The pressures were measured by pressure columns. Experiments by A. J. Morin (1795-1880), G. Piobert (1793-1871) and I. Didion (1798-1878) on plates of 0.3 to 2.7 sq. ft. area, drawn vertically through water, gave f = 2.18; but the experiments were made in a reservoir of comparatively small depth. For similar plates moved through air they found f = 1.36, a result more in accordance with those which precede.

For a fixed plane in a moving current of water E. Mariotte found f = 1.25. Dubuat, in experiments in a current of water like those mentioned above, obtained the values m = 1.186; n = 0.670; f = 1.856. Thibault exposed to wind pressure planes of 1.17 and 2.5 sq. ft. area, and found f to vary from 1.568 to 2.125, the mean value being f = 1.834, a result agreeing well with Dubuat.

§ 167.Stanton’s Experiments on the Pressure of Air on Surfaces.—At the National Physical Laboratory, London, T. E. Stanton carried out a series of experiments on the distribution of pressure on surfaces in a current of air passing through an air trunk. These were on a small scale but with exceptionally accurate means of measurement. These experiments differ from those already given in that the plane is small relatively to the cross section of the current (Proc. Inst. Civ. Eng.clvi., 1904). Fig. 169 shows the distribution of pressure on a square plate. ab is the plate in vertical section. acb the distribution of pressure on the windward and adb that on the leeward side of the central section. Similarly aeb is the distribution of pressure on the windward and afb on the leeward side of a diagonal section. The intensity of pressure at the centre of the plate on the windward side was in all cases p = Gv2/2g ℔ per sq. ft., where G is the weight of a cubic foot of air and v the velocity of the current in ft. per sec. On the leeward side the negative pressure is uniform except near the edges, and its value depends on the form of the plate. For a circular plate the pressure on the leeward side was 0.48 Gv2/2g and for a rectangular plate 0.66 Gv2/2g. For circular or square plates the resultant pressure on the plate was P = 0.00126 v2℔ per sq. ft. where v is the velocity of the current in ft. per sec. On a long narrow rectangular plate the resultant pressure was nearly 60% greater than on a circular plate. In later tests on larger planes in free air, Stanton found resistances 18% greater than those observed with small planes in the air trunk.

§ 168.Case when the Direction of Motion is oblique to the Plane.—The determination of the pressure between a fluid and surface in this case is of importance in many practical questions, for instance, in assigning the load due to wind pressure on sloping and curved roofs, and experiments have been made by Hutton, Vince, and Thibault on planes moved circularly through air and water on a whirling machine.

Let AB (fig. 170) be a plane moving in the direction R making an angle φ with the plane. The resultant pressure between the fluid and the plane will be a normal pressure N. The component R of this normal pressure is the resistance to the motion of the plane and the other component L is a lateral force resisted by the guides which support the plane. Obviously

R = N sin φ;

L = N cos φ.

In the case of wind pressure on a sloping roof surface, R is the horizontal and L the vertical component of the normal pressure.

In experiments with the whirling machine it is the resistance to motion, R, which is directly measured. Let P be the pressure on a plane moved normally through a fluid. Then, for the same plane inclined at an angle φ to its direction of motion, the resistance was found by Hutton to be

R = P (sin φ)1.842 cos φ.

A simpler and more convenient expression given by Colonel Duchemin is

R = 2P sin2φ / (1 + sin2φ).

Consequently, the total pressure between the fluid and plane is

N = 2P sin φ / (1 + sin2φ) = 2P / (cosec φ + sin φ),

and the lateral force is

L = 2P sin φ cos φ / (1 + sin2φ).

In 1872 some experiments were made for the Aeronautical Society on the pressure of air on oblique planes. These plates, of 1 to 2 ft. square, were balanced by ingenious mechanism designed by F. H. Wenham and Spencer Browning, in such a manner that both the pressure in the direction of the air current and the lateral force were separately measured. These planes were placed opposite a blast from a fan issuing from a wooden pipe 18 in. square. The pressure of the blast varied from6⁄10to 1 in. of water pressure. The following are the results given in pounds per square foot of the plane, and a comparison of the experimental results with the pressures given by Duchemin’s rule. These last values are obtained by taking P = 3.31, the observed pressure on a normal surface:—

Water Motors

In every system of machinery deriving energy from a natural waterfall there exist the following parts:—

1. A supply channel or head race, leading the water from the highest accessible level to the site of the machine. This may be an open channel of earth, masonry or wood, laid at as small a slope as is consistent with the delivery of the necessary supply of water, or it may be a closed cast or wrought-iron pipe, laid at the natural slope of the ground, and about 3 ft. below the surface. In some cases part of the head race is an open channel, part a closed pipe. The channel often starts from a small storage reservoir, constructed near the stream supplying the water motor, in which the water accumulates when the motor is not working. There are sluices or penstocks by which the supply can be cut off when necessary.

2. Leading from the motor there is a tail race, culvert, or discharge pipe delivering the water after it has done its work at the lowest convenient level.

3. A waste channel, weir, or bye-wash is placed at the origin of the head race, by which surplus water, in floods, escapes.

4. The motor itself, of one of the kinds to be described presently, which either overcomes a useful resistance directly, as in the case of a ram acting on a lift or crane chain, or indirectly by actuating transmissive machinery, as when a turbine drives the shafting, belting and gearing of a mill. With the motor is usually combined regulating machinery for adjusting the power and speed to the work done. This may be controlled in some cases by automatic governing machinery.

§ 169.Water Motors with Artificial Sources of Energy.—The great convenience and simplicity of water motors has led to their adoption in certain cases, where no natural source of water power is available. In these cases, an artificial source of water power is created by using a steam-engine to pump water to a reservoir at a great elevation, or to pump water into a closed reservoir in which there is great pressure. The water flowing from the reservoir through hydraulic engines gives back the energy expended, less so much as has been wasted by friction. Such arrangements are most useful where a continuously acting steam engine stores up energy by pumping the water, while the work done by the hydraulic engines is done intermittently.

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