A centrifugal pump performance data is shown in the following table for water at

Question

A centrifugal pump performance data is shown in the following table for water at 77degreeF. If the suction to discharge elevation difference is zero and the suction & discharge pipe diameters are the same, what is the pressure difference in psi between discharge and suction? Calculate the fluid power in hp, pump efficiency and rpm. If this is the best efficiency of the pump, compute the pump specific speed. Specific weight of water = 62.4 lb/ft^3, 1 cubic ft = 7.49 gallons, 1 hp = 550 lb.ft/s.

Explanation / Answer

If a pump suction and discharge elevation is zero, pipe size for suction and discharge being the same what is causing difference in pressures. First we have not been told about the length of the discharge pipe. They can create sizable amount of pressure head requirement. Secondly when we impart energy thru impeller on the water, it creates small amount of pressure difference even if there is head requirement is zero.

1 gpm = 3.78541 lit q 20 gpm 4.542492 m3/h h 6 ft 1.8288 m g 32.2 ft/sec/sec 9.81 m/sec/sec 62.4 lb/cuft 1000 kg/m3 Hydrulic power 0.030345064 hp 0.030345064 hp If impeller is absorbing 0.2 lbft torque the power absrobed by impeller = 2nT/33000
We also observe that standard motor speeds are 1750 rpm and 1150 rpm. By calculation we note that at 1750 rpm, impeller would absrob 0.067hp and at 1150 repm the impeller would absrob 0.044 hp. The pump must be running at 1750 rpm other wise they could have provided smaller than 0.08 hp electric motor to run this pump. The efficency of the pump = real Power for lifting the water / impeller power = 0.030/0.067 = 0.447= 44.7% efficiency. The specific Speed of the pump Ns = n . Q1/2 /h3/4 for q= 4.54m3/h, n=1750, h=1.828
Ns = 2371.7 (SI units) = 0.861 * 2371.7 = 2042 US units.

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