Question

Problem 2 (25 points) A pump operating at 1170 rpm discharges 2000 gpm at a total head of 109 feet. If the pump speed is increased to 1770 rpm, what is the brake horsepower (BHP, in kW) if the pump efficiency is 72% at the new pump speed? Water temperature is 70 'F. Round your answer to the nearest whole kilowatt. (Conversion factor: 1 horsepower = 745.7 W)

Answer 1

Given Q = 2000 gpm N, =1170 rpm H = 109 feet N = 1770rpm Find Q2 N, _ Q N₂ 2 1170 rpm 2000 gpm 1770 rpm Q2 1770 Q2 = 2000 gpmx 1170 Q2 = 3025.64 gpm Find H2 N N, 以日 2 1170 rpm 1770 rpm 109 feet H

1770 H2 =109 ft x1 1170 H2 = 249.460 feet Find Break Horse power by using the below formula BHP 1000, H, 3960n 100x 3025.64 x 249.460 3960 x 72 264.722 hp = Therefore, the break horse power required is 264.722hp

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