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.
1
Expert's answer
2020-05-03T15:27:15-0400

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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