and the efficiency of the pump is, from (7),
[eta] = gH/V_o w_o = gH/{V (V_o - n_o cot [phi])},
= (V_o² - u_o² cosec² [phi])/{2V_o (V_o - u_o cot [phi]) }, (9).
For [phi] = 90°,
[eta] = (V_o² - u_o²)/2V_o²,
which is necessarily less than ½. That is, half the work expended in
driving the pump is wasted. By recurving the vanes, a plan introduced
by Appold, the efficiency is increased, because the velocity v_o of
discharge from the pump is diminished. If [phi] is very small,
cosec [phi] = cot [phi];
and then
[eta] = (V_o, + u_o cosec [phi])/2V_o,
which may approach the value 1, as [phi] tends towards 0. Equation (8)
shows that u_o cosec [phi] cannot be greater than V_o. Putting u_o =
0.25 [root](2gH) we get the following numerical values of the
efficiency and the circumferential velocity of the pump:--
[phi] [eta] V_o
90° 0.47 1.03 [root](2gH)
45° 0.56 1.06 "
30° 0.65 1.12 "
20° 0.73 1.24 "
10° 0.84 1.75 "
[phi] cannot practically be made less than 20°; and, allowing for the
frictional losses neglected, the efficiency of a pump in which [phi] =
20° is found to be about .60.
§ 210. Case 2. _Pump with a Whirlpool Chamber_, as in fig.
210.--Professor James Thomson first suggested that the energy of the
water after leaving the pump disk might be utilized, if a space were
left in which a free vortex could be formed. In such a free vortex the
velocity varies inversely as the radius. The gain of pressure in the
vortex chamber is, putting r_o, r_w for the radii to the outlet
surface of wheel and to outside of free vortex,
v_o² / r_o² \ v_o² / \
---- ( 1 - ---- ) = ---- ( 1 - k² ),
2g \ r_w² / 2g \ /
if
k = r_o/r_w.
The lift is then, adding this to the lift in the last case,
H = {V_o² - u_o² cosec² [phi] + v_o²(1 - k²)}/2g.