A single stage centrifugal pump with impeller diameter of 30 cm rotates at 2000 r.p.m and lifts $3 m^3$ of water per second to a height of 30 m with an efficiency of 75% find the number of stages and diameter of each impeller of a similar multistage pump to lift $5 m^3$ of water per second to a height of 200 meters when rotating at 1500 r.p.m.

Solution Given :

Single - stage pump:

1] Diameter of impeller, $D_1$ = 30 cm = 0.30 m

2] Speed $N_1$ = 2000 r.p.m

3] Discharge $Q_1 = 3m^3/s$

4] Height, $Hm_1 = 30 m$

5] Efficiency n man = 75% = 0.75.

Multistage similar pump:

1] Discharge $Q_2 = 5m^3/s$

2] Total Height, = 200 m

3] Speed $N_2$ = 1500 r.p.m

4] Let the height per stage = H_{m2}

5] Diameter of each impeller = $D_2$

Step no (1) specific speed should be same.

$\therefore$ $(\frac{N \sqrt Q}{H_m^{3/4}} = (\frac{N \sqrt Q}{H_m^{3/4}})_2$

$\therefore$ $\frac{N \sqrt Q_1}{H_{m1}^{3/4}} = \frac{N_2 \sqrt Q_2}{H_{m2}^{3/4}}$

$\therefore$ $\frac{2000 \times \sqrt 3}{30^{3/4}} = \frac{1500 \times \sqrt 5}{H_{m2}^{3/4}}$

$\therefore$ $H_{m2}^{3/4} = \frac{1500 \times \sqrt 5 \times 30^{3/4}}{2000 \times \sqrt 3}$

$\therefore$ $H_{m2}^{3/4} = \frac{1500}{2000} \times \sqrt\frac{5}{3} \times 12.818$

$\therefore$ $H_{m2} = (12.411)^{4/3}$

$\therefore$ $[H_{m2} = 28.71m]$

Step No (2) Number of stages.

$= \frac{Total head}{head per stage}$

$= \frac{200}{28.71}$

= 6.96 $\approx$ 7

Using equation,

$\frac{\sqrt {H_{m1}}}{D_1 N_1} = \frac{\sqrt {H_{m2}}}{D_2 N_2}$

$\frac{ \sqrt{30}}{0.30 \times 2000} = \frac{ \sqrt {28.71}}{D_2 \times 1500}$

$\therefore$ $D_2 = \frac{0.30 \times 2000 \times \sqrt 28.71}{1500 \times \sqrt 30}$

$[ D_2 = 0.3913m]$

or

$[ D_2 = 391.3 mm]$