Question 14.3: A centrifugal pump operating at 1100 rpm against a head of 1......
A centrifugal pump operating at 1100\ rpm against a head of 120\ m\ H_{2}O produces a flow of 0.85\,{\mathrm{m}}^{3}/{\mathrm{s}}.
(a) For a geometrically similar pump, operating at the same speed but with an impeller diameter 30% greater than the original, what flow rate will be achieved?
(b) If the new larger pump described in part (a) is also operated at 1300 rpm, what will be the new values of flow rate and total head?
Specifying for pump 1, D=D_{1}, then for the larger pump, {D}_{2}=1.3\,{ D}_{1}; thus, the new flow rate will be, using equation (14-23),
\frac{\dot{ V}_{2}}{\dot{ V}_{1}}=\frac{\omega_{2}}{\omega_{1}}\left(\frac{D_{2}}{D_{1}}\right)^{3} (14-23)
\dot{ V}_{2}=0.85\ m^{3}/s\biggl(\frac{1.3D_{1}}{D_{1}}\biggr)^{3}
=1.867\ {\mathrm{m}}^{3}/{\mathrm{s}}
For the case with {D}_{2}=1.3\,{ D}_{1} and \omega_{2}=1300\,\mathrm{rpm}, we have, from equation (14-23),
\dot{ V}_{2}=0.85\ { m}^{3}/s\Big(\frac{1300\,\mathrm{rpm}}{1100\,\mathrm{rpm}}\Big)\bigg(\frac{1.3\,D_{1}}{D_{1}}\Big)^{3}
=2.207\,{\mathrm{m}}^{3}/{\mathrm{s}}
The new head is determined, using equation (14-22):
{\frac{h_{2}}{h_{1}}}=\left({\frac{\omega_{2}}{\omega_{1}}}\right)^{2}\left({\frac{D_{2}}{D_{1}}}\right)^{2} (14-22)
=\;120\;\mathrm{m\;H}_{2}O\bigg(\frac{1300\;\mathrm{rpm}}{1100\,\mathrm{rpm}}\bigg)^{2}\;\bigg(\frac{1.3\;D_{1}}{D_{1}}\bigg)^{2}
=283\ m\ {\mathrm{H_{2}O}}