Question #113379

A centrifugal water pump operates at 30 m3/hr and at 1440 RPM.The pump operating efficiency is 65% and motor efficiency is 89%.The discharge pressure gauge shows 3.4 kg/cm2.The suction is 3mtr below the pump centreline.If the speed of the pump is reduced by 25%, estimate pump flow, head and motor power.Assume motor and pump efficiency remains same at the reduced speed.

Expert's answer

We know that, The relation between pump efficiency and motor efficiency as

pump efficiency= 65 % and motor efficiency = 89 %, N=1440 rpm, Q=30m3hrQ= 30 \frac{m^3}{hr}


ηp=(p2p1)QP×ηm\eta_p=\frac{(p_2-p_1)Q}{P\times\eta_m}


0.65=(1000×9.8 3)×30P×3600×0.890.65= \frac{ (1000\times 9.8 \ 3) \times 30}{P\times 3600 \times 0.89}


P= 423.5 Pa and as per question , N2=_2= 1440 -25×1440100=1080rpm\frac{25 \times 1440}{100}=1080 rpm

we know that


P1P2=(N1N2)3\frac{P_1}{P_2}= (\frac{N_1}{N_2})^3


P2= 423.5×(10803)14403=178.66423.5 \times \frac{(1080^3)}{1440^3}=178.66


and head is proportional to speed square


H2H1=(N2N1)2\frac {H_2}{H_1}= (\frac{N_2}{N_1})^2


H2=1.6875mH_2= 1.6875 m


Q2Q1=(N2N1)\frac{Q_2}{Q_1}=(\frac{N_2}{N_1})


Q_2= 30×10801440=22.5m3hr30 \times \frac{1080}{1440}= 22.5 \frac{m^3}{hr}




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