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A single stage centrifugal pump with impeller diameter of 30 cm rotates at 2000 rpm and lifts 3 m3 of water per second to a height of 30 m width an efficiency of 75%.

Find the number of stages and diameter of each impeller of a similar multistage pump to lift 5 m3 of water per second to a height of 200 m when rotating at 1500 rpm.

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Given:

$D_1 = 0.3\hspace{0.05cm}m,\hspace{1cm} Q_2 = 5\hspace{0.05cm}m^3/s, \hspace{1cm}N_1 = 2000\hspace{0.05cm}rpm\\ \textit{Total height}(H) = 200\hspace{0.05cm}m,\hspace{1cm}Q_1 = 3\hspace{0.05cm}m^3/s\\ N_2 = 1500\hspace{0.05cm}rpm,\hspace{1cm}H_{m1} = 30\hspace{0.05cm}m$

To Find:

$\eta_{mano} = 0.75\\ \textit{No. of stages} = D_2 = \hspace{0.05cm}?$

Solution:

Specific speed should be same,

$(\frac{N\sqrt{Q}}{H_{m1}^\frac{3}{4}}) = (\frac{N\sqrt{Q}}{H_{m2}^\frac{3}{4}})$

$(\frac{2000\hspace{0.05cm}\times\hspace{0.05cm}\frac{1}{\sqrt{3}}}{30^\frac{3}{4}}) = (\frac{1500\hspace{0.05cm}\times\hspace{0.05cm}\frac{2}{\sqrt{5}}}{H_{m2}^\frac{3}{4}})$

$H_{m2}^\frac{3}{4} = \frac{1500\hspace{0.05cm}\times\hspace{0.05cm}\sqrt{5}\hspace{0.05cm}\times\hspace{0.05cm}30^\frac{3}{4}}{2000\hspace{0.05cm}\times\hspace{0.05cm}\sqrt{3}} = \frac{1500}{2000}\hspace{0.05cm}\times\hspace{0.05cm}\sqrt{\frac{5}{3}}\hspace{0.05cm}\times\hspace{0.05cm}12.818 = 12.411$

$H_{m2} = (12.411)^\frac{4}{3} = 28.71\hspace{0.05cm}m$

No.of stages = $\frac{\textit{Total Head}}{\textit{Head per stage}} = \frac{200}{28.71} = 6.96\approx7$

$\frac{\sqrt{H_{m1}}}{D_1N_1} = \frac{\sqrt{H_{m2}}}{D_2N_2} = \frac{\sqrt{30}}{0.3\hspace{0.05cm}\times\hspace{0.05cm}2000} = \frac{\sqrt{28.71}}{D_2\hspace{0.05cm}\times\hspace{0.05cm}1500}$

$D_2 = \frac{0.3\hspace{0.05cm} \times\hspace{0.05cm}2000 \hspace{0.05cm}\times\hspace{0.05cm}\sqrt{28.71}}{1500\hspace{0.05cm} \times\hspace{0.05cm} \sqrt{30}} = 0.3913\hspace{0.05cm}m = 391.3\hspace{0.05cm}mm$