is 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:—
∴ u = v or 13 v.
The former gives a minimum, the latter a maximum efficiency.
Putting u = 13v in the expression above,
(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 = 12v, and the η = 12. 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).
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| 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 = 1627.
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.
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| 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 = Gg ωcos α(v cos α − u cos δ)2; | (1) |
| P = | G | ω cos δ | (v cos α − u cos δ)2; | |
| g | cos α |
| Pu = | G | ωu | cos δ | (v cos α − u cos δ)2; |
| g | cos α |
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 cos2 α (v − u)2, which is a maximum when u = 13v.
When u = 13v then Pu max. = 427(G/g)ωv3 cos2α, and the efficiency = η = 49cos2α.
(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 = 13v cos α.
When u = 13v 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 of
