476 APPLIED MECHANICS | 3 
less than 2h on account of the loss of energy in the jet from B due to 
friction. = 
As to the distribution of the pressure on a flat plate struck normally 
by a jet, this is shown approximately in Fig. ‘ 
772, where the intensity of the pressure due to 
the impact of the jet is plotted on the back 
of the plate. At the centre the pressure h, in 
feet of water, is slightly less than the head due 
to the velocity v, that is to say, h is slightly less 
than v?/2q. 
Case II. Same as Case I., except that the 
vane is moving in the same direction as the jet 
with a velocity v,—Loss of momentum of 
water impinging on vane per second in direction 
of motion of jet = —(v — v,), therefore 
g 
tea: Ww s. 0). Fia. 772. 
g 
But W=wA(v—v,), therefore P= whe —v,)?. 
g 
Useful work done per second = Py = 20-0) 
- Kinetic energy of jet per secon A 
mn ce ves Oe 20,(v— v,)? 
Efficiency = 
For a given ae of v the efficiency will be a maximum when : 
v,(v —v,)? is a maximum. 
Let y =0,(v—v,)? = 0,0? — Qvv? + v3. 
pa NE 
w=” 4vv, + 3x4. 
y is @ maximum when 8Y-229, i.e. when v¥,=v or ° Obviously 
dv, 3 4 
vy =5 is the result to take. Therefore the velocity of the vane should be © 
one-third of the velocity of the jet for the highest efficiency. 
u(o- 2) 
3\° 3) 8 
Pan & 
Case III. Direction of jet makes an angle 0 
with vane. Vane at rest (Fig. 773).—Loss of 
momentum per second in direction of normal to ae 
Maximum efficiency, = or 29°6 per cent. 
vane = —vsin 6. ; 6 
Therefore P= ue sin 0. Fig. 773. 
