TFS · Hydraulic & Fluid Applications · Problem 1PDFSolution in PDF ↓
TFS · Hydraulic & Fluid Applications · Problem 1
Problem & Solution
Video Synthesis
Topic: Pump system — A 10-horse power pump with 90% efficiency transports water to an elevation of 200 feet above the pump.
Approach: the way we want to think about this problem is we have a pump which is using some power to do some work to increase the elevation of some water, which in so doing increases its potential energy by vir
Key values: 62.4 pounds per cubic foot, 200 feet above the pump, , foot pounds is truly a unit of energy or work and that
Reference: reference handbook
✅ Answer: C
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Student questions asked in live office hours about this problem
OH 114
OH 119 · April 28, 2026
Q: When computing potential energy using PE = mgh with mass given in lbm, why doesn't the result carry the correct units, and how does G_c resolve this?
A: Multiplying lbm × (ft/s²) × ft yields lbm·ft²/s², which is not foot-pound-force (the correct US customary energy unit). G_c = 32.2 lbm·ft/(lbf·s²) placed in the denominator cancels lbm and ft/s², leaving lbf·ft. Conveniently, since G_c and g both equal 32.2 numerically at Earth's surface, they cancel each other for potential energy — so you can skip G_c entirely and simply use weight (in lbf) × height (in ft).
Q: I tried solving for volumetric flow rate Q and multiplying by time to get volume, but I can't get the units to work out properly—should I use gravity g or the gravitational constant gc?
A: You can use that approach, but you need to understand when gc comes into play. When you have g (32.2 ft/s²) as acceleration, it's just the unit of acceleration; gc (32.2 lbm·ft/lbf·s²) only appears when you have a mismatch between pound-mass and pound-force in your equation—they always go on opposite sides to cancel properly.