The following design and operating data were taken for a large industrial centri
ID: 2250429 • Letter: T
Question
The following design and operating data were taken for a large industrial centrifugal water pump:
Required water flow rate varies from 500 m3/h to 700 m3/h. For the pump to deliver 550 m3/h, the flowrate has been reduced by partially closing the delivery valve. Motor efficiency is 93%.
a. Calculate the pump operating efficiency (the ratio of pumping power to operating pump input power delivered TO THE MOTOR.
b. Pump operating efficiency is low because of the partial valve closing to control the flowrate. What would be the best way to improve pump operating efficiency?
c. The ratio of two pump volumetric flowrates would be equal to the ratio of two pump shaft speeds to what power? The ratio of two pump operating heads would be equal to the ratio of two pump shaft speeds to what power? The ratio of two pump operating powers would be equal to the ratio of two pump shaft speeds to what power?
d. Suppose the pump energy improvement idea suggested in part b. was used for the pump. From the relations in part c., calculate the pump speed required to give a volumetric flowrate 550 m3/h. That is, (550 m3/h / design flowrate) = (RPM to deliver 550 m3/h / design RPM). From this, calculate the pumping power needed to deliver 550 m3/h using the design pumping power ratio. The difference in the operating values of the pumping power (124 kW and the pumping power with the new energy efficiency feature) is the energy savings.
Parameter Design Operating Flow Q (m^3/h) 800 550 Head H (m) 55 24 (after delivery valve) Power P (kW) 160 124 RPM 1485 1485Explanation / Answer
(a)
power input to pump = power input to motor * efficiency of motor
= 124 * 0.93
= 115.32 kW
Pump operating efficiency = (density * Q * g * H / pump input power ) operating point * 100
= {1000 * (550/3600) * 9.81 * 24 / (115.32*1000)}* 100
= 31.19 %
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(b)
In this case the speed of pump is constant ( design and operational speed are same). The best way to improve the pump efficiency is by changing its roational speed. it can be achived by utilising variable speed pump.
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(c)
(Q / D3N ) pump1 = (Q / D3N ) pump2
therefore volumetric flowrates (Q) is proportional (1st power) to pump shaft speeds (N).
(H / D2N2 ) pump1 = (H / D2N2) pump2
therefore pump operating heads (H) is proportional to square (2nd power) of pump shaft speeds(N).
(P / D5N3 ) pump1 = (P / D5N3) pump2
therefore pump operating powers (p) is proportional to cube (3rd power) of pump shaft speeds(N).
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(d)
Here the pump is same thefore: D = constant
(Q1 / N1 ) = (Q2 / N2 )
800 / 1485 = 550 / N2
N2 = 1021 RPM = new operating speed for variable speed pump
(P1 / N13 ) = (P2 / N23)
160 / 14853 = P2 / 10213
P2 = 52 kW = power consumed in variable speed pump
The difference in the operating values of the pumping power (Energy saving) = 124 - 52 = 72 kW
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