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Two geometrically similar pumps are running at same speed at 1000 rpm. One pump has an impeller diameter of 0.30m and lifts water at the rate of 20 lit/s against a head of 15 m.

Determine the head and impeller diameter of the other pump to deliver half the discharge.

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 08

Years: MAY 2016

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Given :

For pump no. 1

Speed $N_1= 1000rpm$

diameter $D_1 = 0.30 m$

Discharge $Q_2 = 20lits/s=0.02m^3/sec $

Head $Hm_1 = 15m$

For pump no.2

Speed $N_2=1000 r.p.m$

Discharge $Q_2 = \dfrac {Q_1}2 = \dfrac {20}2 = 10 litres/sec = 0.01m^3/s$

Let $D_2$ = Diameter of impeller

$Hm_2 = $ Head developed

$\dfrac {N_1\sqrt{Q_1}}{Hm^{3/4}} = \dfrac {N_2\sqrt{Q_2}}{Hm_2^{3/4}}\\ \therefore \dfrac {1000\times \sqrt{0.02}}{15^{3/4}}= \dfrac {1000\times \sqrt{0.01}}{Hm_2^{3/4}}\\ Hm_2^{3/4}= \dfrac {1000\times \sqrt{0.01}\times 15^{3/4}}{1000\times \sqrt{0.02}}= \sqrt{\dfrac {0.01}{0.02}}\times 7.622=5.389\\ Hm_2 = (5.389)^{4/3}=9.44m \\ (\dfrac {\sqrt {Hm}}{DN})_1=(\dfrac {\sqrt {Hm}}{DN})_2 \space or \space \dfrac {\sqrt {Hm_1}}{D_1N_1}=\dfrac {\sqrt {Hm_2}}{D_2N_2} \\ \dfrac {\sqrt{15}}{0.03\times 1000} = \dfrac {\sqrt{9.44}}{D_2\times 1000}\\ D_2 = \dfrac {\sqrt{9.44}\times 0.3}{\sqrt{15}}= 0.238m=238 mm$

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