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 = vr cos θ; DE = vr sin θ. If Q is the volume of water impinging on the plane per second, the change of momentum is (G/g) Qvr cos 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) Qvr cos 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 = vr cos 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 get
| N = | G | ω | (v cos α − u cos δ)2; | |
| g | cos α |