Problem 4.9.1

A centrifugal pump has 10-inch inlet diameter and 5-inche outlet diameter. The measured flow rate is 818 gallons per minute. Inlet pressure is measured as 5 inches of mercury above atmospheric pressure. The discharge pressure, at 4 feet above the pump outlet, is 30.7 psi (absolute). The pump input horsepower is 10 hp. If the pump speed is increased from 1750 rpm to 3500 rpm, what is the (a) pump efficiency, (b) flow rate, (c), net head, and (d) brake horsepower?

Solution:

Pump inlet pressure is 2469 pound-force per square foot. Discharge pressure is 4421 psf. Volumetric flow rate is 818 gallons per minute or 1.823 cubic feet per second. Mass flow rate can be calculated as 113.8 pound per second.

Inlet cross-sectional area is 0.545 square foot. Inlet velocity could be calculated as 3.342 feet per second. Outlet cross-sectional area is 0.136 square foot. Outlet velocity could be calculated as 13.367 feet per second. Head developed by pump is 37.87 foot pound-force per pound-mass. Water horsepower, WHP, is 7.834 hp. Brake horsepower, BHP, is 10 hp. Efficiency can be calculated as 7.834 over 10 or 78.3 percent.

Volumetric flow rate is directly proportional to impeller speed. When impeller speed is doubled, volumetric flow rate is also doubled, or 818 times 2 or 1636 gallons per minute. Head developed is proportional to square of the impeller speed, or 37.87 times 2² or 151.46 foot pound-force per pound-mass. Assuming efficiency is not affected, brake horsepower is proportional to cube of the impeller speed, or 10 times 2³ or 80 hp.