A pump operating at 80% efficiency delivers 30 gal/min of water from a reservoir to an open-air
Question:
A pump operating at 80% efficiency delivers 30 gal/min of water from a reservoir to an open-air storage tank at a chemical plant 1 mi away. A 3-in, schedule-40 pipe is used, and the frictional losses are 200 ft lbf/lb. The elevation of the liquid level in the tank is 873 ft above sea level, and the elevation of the liquid level in the reservoir is 928 ft above sea level.
What is the minimum horsepower required for the pump?
The elevation of the reservoir is fixed. What elevation of the liquid level of the tank would make the pump unnecessary?
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Step by Step Answer:
To calculate the minimum horsepower required for the pump and determine the elevation of the liquid level in the tank that would make the pump unnecessary we can use the principles of fluid mechanics ...View the full answer
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Related Book For 
Analysis Synthesis And Design Of Chemical Processes
ISBN: 9780134177403
5th Edition
Authors: Richard Turton, Joseph Shaeiwitz, Debangsu Bhattacharyya, Wallace Whiting
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