HVAC · Supporting-Topics · Problem 12PDFSolution in PDF ↓
HVAC · Supporting-Topics · Problem 12
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Student questions asked in live office hours about this problem
OH 89: HVAC: Supporting Topics #12
Q: How did you derive the simplified hydraulic horsepower equation WHP = ΔP × Q? We've typically solved pump problems with the 1714 coefficient in the denominator.
A: The 1714 in the denominator is just a unit conversion factor rolled into a rule of thumb — when you carry all the units explicitly, power equals pressure times volumetric flow rate, and the 1714 disappears once everything is in consistent units. The version with 1714 assumes ΔP in psi and Q in GPM, giving horsepower directly; the simplified form ΔP × Q is the foundational equation that works when you're careful about units. It's the same idea, just different packaging — and the foundational form is always more flexible.
OH 96: HVAC: Supporting Topics #12
Q: I solved the problem differently — I converted displacement to GPM, found water horsepower, divided by pump efficiency, and plugged into a power formula. I got a very similar answer but need help visualizing the unit conversion on the test.
A: Your approach is valid — any path that correctly converts to consistent units and applies the efficiency will get you there. For unit conversions on the actual exam, write out all units at every step and watch them cancel; that discipline catches errors faster than anything else. The key chain is: displacement (in³/rev) × speed (rpm) → flow rate (in³/min) → GPM → water HP → brake HP via efficiency.