In normal working, w_i = 0. Also, multiplying by the angular velocity,
the work done per second is
M[alpha] = (GQ/g)w_o r_o[alpha].
But the useful work done in pumping is GQH. Therefore the efficiency
is
[eta] = GQH/M[alpha] = gH/w_o r_o[alpha] = gH/w_o V_o. (7)
§ 209. Case 1. _Centrifugal Pump with no Whirlpool Chamber._--When no
special provision is made to utilize the energy of motion of the water
leaving the wheel, and the pump discharges directly into a chamber in
which the water is flowing to the discharge pipe, nearly the whole of
the energy of the water leaving the disk is wasted. The water leaves
the disk with the more or less considerable velocity v_o, and impinges
on a mass flowing to the discharge pipe at the much slower velocity
v_s. The radial component of v_o is almost necessarily wasted. From
the tangential component there is a gain of pressure
(w_o² - v_s²)/2g - (w_o - v_s)²/2g
= v_s(w_o - v_s)g,
which will be small, if v_s is small compared with w_o. Its greatest
value, if v_s = ½w_o, is ½w_o²/2g, which will always be a small part
of the whole head. Suppose this neglected. The whole variation of
pressure in the pump disk then balances the lift and the head u_i²/2g
necessary to give the initial velocity of flow in the eye of the
wheel.
u_i²/2g + H = V_o²/2g - u_o² cosec² [phi]/2g + u_i²/2g,