Page:EB1911 - Volume 14.djvu/120

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

ui = 0.25 √ 2gH. 2ridi = Q / ui. (1)

Also in practice

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

uo = Q / 2rodo. (3)

Fig. 211 shows the notation adopted for the velocities. Suppose the water enters the wheel with the velocity vi, while the velocity of the wheel is Vi. Completing the parallelogram, vri is the relative velocity of the water and wheel, and is the proper direction of the wheel vanes. Also, by resolving, ui and wi are the component velocities of flow and velocities of whir of the velocity vi of the water. At the outlet surface, vo 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 vi is radial. Then, in normal conditions of working, at the inlet surface,

(4)

If the pump is raising less or more than its proper quantity, 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,

(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

ho − hi = (Vo2 − Vi2) / 2g − (vro2 − vri2) / 2g.

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

ho − hi = (Vo2 − Vi2) / 2g − uo2 cosec2 / 2g + (ui2 + Vi2) / 2g = Vo2 / 2g − uo2 cosec2 / 2g + ui2 / 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 its angular velocity; wo, ro the tangential velocity of the water and radius at the outlet surface; wi, ri the same quantities at the inlet surface. Q being the discharge per second, the change of angular momentum per second is

(GQ/g) (woro − wiri).

Hence

M = (GQ/g) (woro − wiri).

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

M = (GQ/g) woro.

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

= GQH / M = gH / woro = 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 = wo, is wo2/2g, 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 / 2g + ui2 / 2g,

H = Vo2 / 2g − uo2 cosec2 / 2g

or

Vo = √ (2gH + uo2 cosec2 ). (8)

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

= gH / Vowo = gH / {V (Vo − no cot ) }, &ensp; = (Vo2 − uo2 cosec2 ) / {2Vo (Vo − uo cot ) }, (9)

For = 90°,

= (Vo2 − uo2) / 2Vo2,

which is necessarily less than. 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 vo of discharge from the pump is diminished. If is very small,

cosec = cot ;

and then

= (Vo, + uo cosec ) / 2Vo,

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