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Mechanics of Fluids

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628 Fluid machines<br />

Fig. 13.26<br />

Suffix 1 again refers to the inlet and suffix 2 to the outlet even though<br />

r2 > r1 for a pump. This expression may be transformed, with the aid<br />

<strong>of</strong> the trigonometric relations for the vector triangles, to the equivalent form<br />

(corresponding to eqn 13.10). The relation so obtained for the work done<br />

on the fluid divided by mass is<br />

= 1<br />

�<br />

(v<br />

2<br />

2 2 − v2 1 ) + (u2 2 − u2 1 ) − R2 2 − R2 �<br />

1<br />

(13.18)<br />

A centrifugal pump rarely has any sort <strong>of</strong> guide vanes at inlet. The fluid<br />

therefore approaches the impeller without appreciable whirl and so the inlet<br />

angle <strong>of</strong> the blades is designed to produce a right-angled vector triangle at<br />

inlet (as shown in Fig. 13.26). At conditions other than those for which the<br />

impeller was designed – for example, a smaller flow rate at the same shaft<br />

speed – the direction <strong>of</strong> the relative velocity R1 does not coincide with that<br />

<strong>of</strong> a blade. Consequently the fluid changes direction abruptly on entering the<br />

impeller, eddies are formed and energy is dissipated. In addition, the eddies<br />

give rise to some back flow into the inlet pipe, thus causing the fluid to have<br />

some whirl before entering the impeller. Moreover, particularly if the pump<br />

is dealing with a fluid <strong>of</strong> high viscosity, some pre-whirl may be caused by<br />

viscous drag between the impeller and the incoming fluid.<br />

Whatever the immediate cause <strong>of</strong> such pre-whirl, however, it will have<br />

come from the impeller. The initial angular momentum <strong>of</strong> the fluid may<br />

therefore be taken as zero. (This argument is equivalent to considering the<br />

inlet boundary <strong>of</strong> the control volume across the pump further upstream <strong>of</strong><br />

the impeller. Even in the enlarged control volume the impeller is the only<br />

thing providing torque.) We may therefore set vw1 = 0 in the Euler relation<br />

13.17 to give<br />

Work done on fluid divided by mass = u2vw2<br />

(13.19)<br />

It may be noted that since u1 no longer enters the expression the work done<br />

is independent <strong>of</strong> the inlet radius.<br />

The increase <strong>of</strong> energy received by the fluid in passing through the pump<br />

is most <strong>of</strong>ten expressed in terms <strong>of</strong> head H, that is, energy/weight. It should<br />

not be forgotten that this quantity depends on g; indeed, a pump in an

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