M = (GQ/g) (woro − wiri).
In normal working, wi = 0. Also, multiplying by the angular velocity, the work done per second is
Mα = (GQ/g) woroα.
But the useful work done in pumping is GQH. Therefore the efficiency is
η = GQH / Mα = gH / woroα = gH / woVo.
(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 vo, and impinges on a mass flowing to the discharge pipe at the much slower velocity vs. The radial component of vo is almost necessarily wasted. From the tangential component there is a gain of pressure
(wo2 − vs2) / 2g − (wo − vs)2 / 2g
= vs (wo − vs) / g,
which will be small, if vs is small compared with wo. Its greatest value, if vs = 1⁄2wo, is 1⁄2wo2/2g, 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 ui2/2g necessary to give the initial velocity of flow in the eye of the wheel.
ui2 / 2g + H = Vo2 / 2g − uo2 cosec2 φ / 2g + ui2 / 2g,