HVAC · Practice-Exam-2 · Problem 46PDFSolution in PDF ↓
HVAC · Practice-Exam-2 · Problem 46
Problem & Solution
Video Synthesis
Problem: 100 GPM of a liquid with a specific gravity of 0.8 is supplied by a pump operating with a differential pressure of 10 psi.
Approach: So to find the hydraulic horsepower, I really want to drive this point home.
Key step: 100 GPM of a liquid with a specific gravity of 0.8 is supplied by a pump operating with a differential pressure of 10 psi.
Watch out: And if we use this equation and it's not water, and it's got a specific gravity other than 1, we have to include SG.
✅ Answer: D
Office Hours
1
Student questions asked in live office hours about this problem
OH 122 · June 8, 2026
Q: When calculating water horsepower using the delta-H formula (WHP = Q·ΔH·SG / 3960), the student converted 10 psi to head using the water conversion (23.1 ft) and got 0.47 HP — why is this wrong for a fluid with SG = 0.8?
A: The error is using the water-based head conversion (23.1 ft) instead of converting pressure to head using the actual fluid's specific weight. Head must be calculated as H = P / γliquid, where γliquid = SG × 62.4 lb/ft³, giving H = 28.8 ft for SG = 0.8. Alternatively, the delta-P formula (WHP = Q·ΔP / 1714) avoids this issue entirely since PSI already accounts for the fluid's actual pressure regardless of density.