Problem: A pump transfer is 100GPM of water from an open reservoir at atmospheric pressure.
Given: 100GPM of water from an open reservoir at atmospheric pressure; 5 feet; 30 feet and the working pressure at the outle...
Approach: The static suction lift is 5 feet.
Key formula: equation for the hydraulic horsepower
Calc: The pump efficiency is 85% and the motor efficiency is 80%.
Calc: And then we have a pump which is located 5 feet above the top of the reservoir.
Result: So we would round up to the two horsepower motor which is answer choice B.
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Student questions asked in live office hours about this problem
OH 46: HVAC: Fluids-24
Q: I solved problem 24 two ways and got 1.7 HP the first way but a suspicious 1.37 HP the second way—what's wrong with the alternative method?
A: Going straight for P2−P1/γ doesn't represent all the energy the pump adds—it only captures the static pressure difference and misses the elevation and loss terms. The instinct to check with two methods is good, but the 'alternative' method is actually incomplete, not just a rounding variant. Any time there's a pump in the system, the whole Bernoulli equation with the HA term is needed.
OH 96: HVAC: Fluids Module #24
Q: Why don't you factor the suction lift of 5 feet into P1 when calculating the pump head?
A: The suction lift is an elevation difference, not a static pressure—it belongs in the Z term (Z2−Z1), not in the pressure term. For the static pressure at the open reservoir surface, imagine the pump isn't running: the pressure there is just one atmosphere, because no column of water from the pipe is actually sitting on the entire reservoir surface. Keep elevation and static pressure as separate terms in Bernoulli, and this kind of confusion resolves itself.