Page:EB1911 - Volume 14.djvu/120

the same as for a turbine. If Q is the quantity pumped, and H the lift,

u_i = 0.25 √2gH.
2πr_id_i = Q / u_i.

(1)

Also in practice

Hence,

di = 1.2 ri . . . .	${\bigg \}}$
ri = .2571 √ (Q / √H).

(2) 

Usually

ro = 2ri,

and

do = di or 1/2 di

according as the disk is parallel-sided or coned. The water enters the wheel radially with the velocity u_i, and

u_o = Q / 2πr_od_o.

(3)

Fig. 211.

Fig. 211 shows the notation adopted for the velocities. Suppose the water enters the wheel with the velocity v_i, while the velocity of the wheel is V_i. Completing the parallelogram, v_ri is the relative velocity of the water and wheel, and is the proper direction of the wheel vanes. Also, by resolving, u_i and w_i are the component velocities of flow and velocities of whir of the velocity v_i of the water. At the outlet surface, v_o is the final velocity of discharge, and the rest of the notation is similar to that for the inlet surface.

Usually the water flows equally in all directions in the eye of the wheel, in that case v_i is radial. Then, in normal conditions of working, at the inlet surface,

 v_i = u_i  w_i = 0 tan θ = u_i / V_i  v_ri = u_i cosec θ = √ (u_i² + V_i²).

(4)

If the pump is raising less or more than its proper quantity, θ will not satisfy the last condition, and there is then some loss of head in shock.

At the outer circumference of the wheel or outlet surface,

v_ro = u_o cosec φ w_o = V_o − u_o cot φ v_o = √ {u_o² + (V_o − u_o cot φ)²}

(5)

Variation of Pressure in the Pump Disk.—Precisely as in the case of turbines, it can be shown that the variation of pressure between the inlet and outlet surfaces of the pump is

h_o − h_i = (V_o² − V_i²) / 2g − (v_ro² − v_ri²) / 2g.

Inserting the values of v_ro, v_ri in (4) and (5), we get for normal conditions of working

h_o − h_i = (V_o² − V_i²) / 2g − u_o² cosec² φ / 2g + (u_i² + V_i²) / 2g
= V_o² / 2g − u_o² cosec² φ / 2g + u_i² / 2g.

(6)

Hydraulic Efficiency of the Pump.—Neglecting disk friction, journal friction, and leakage, the efficiency of the pump can be found in the same way as that of turbines (§ 186). Let M be the moment of the couple rotating the pump, and α its angular velocity; w_o, r_o the tangential velocity of the water and radius at the outlet surface; w_i, r_i the same quantities at the inlet surface. Q being the discharge per second, the change of angular momentum per second is

(GQ/g) (w_or_o − w_ir_i).

Hence

M = (GQ/g) (w_or_o − w_ir_i).

In normal working, w_i = 0. Also, multiplying by the angular velocity, the work done per second is

Mα = (GQ/g) w_or_oα.

But the useful work done in pumping is GQH. Therefore the efficiency is

η = GQH / Mα = gH / w_or_oα = gH / w_oV_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 = 1/2w_o, is 1/2w_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 φ / 2g + u_i² / 2g,

H = V_o² / 2g − u_o² cosec² φ / 2g

or

V_o = √ (2gH + u_o² cosec² φ).

(8)

and the efficiency of the pump is, from (7),

η = gH / V_ow_o = gH / {V (V_o − n_o cot φ) },
  = (V_o² − u_o² cosec² φ) / {2V_o (V_o − u_o cot φ) },

(9)

For φ = 90°,

η = (V_o² − u_o²) / 2V_o²,

which is necessarily less than 1/2. 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 φ is very small,

cosec φ = cot φ;

and then

η = (V_o, + u_o cosec φ) / 2V_o,

which may approach the value 1, as φ tends towards 0. Equation (8) shows that u_o cosec φ cannot be greater than V_o. Putting u_o = 0.25 √(2gH) we get the following numerical values of the efficiency and the circumferential velocity of the pump:—