Problem: A viscous liquid with a specific gravity of 1.1 is pumped with a volume flow rate of 100gpn.
Given: 5 psi vacuum; 3 feet above the pump outlet is 10 psi g; 4 inches on the suction side to 2 inches on the discharge side
Approach: Let's start with a picture to keep everything organized.
Calc: The diameter of the piping reduces from 4 inches on the suction side to 2 inches on the discharge side.
Calc: The pump has a mechanical efficiency of 80%.
Result: We have two situations one where the diameter is 2-inch nominal this is on the discharge side and if we go to the actual table the diameter is 2.07...
Office Hours
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Student questions asked in live office hours about this problem
OH 16: 3. FLUIDS-21
Q: When dealing with non-water fluids in the water horsepower formula, should I calculate delta H as if the fluid is water to avoid double-counting specific gravity at the end?
A: The key decision is to commit to one approach at the start: either do everything in PSI using Q·ΔP/1714 (no SG needed), or work entirely in feet of the actual fluid and include SG once in the Q·ΔH·SG/3960 formula. If static pressure is given in PSI and elevation in feet, you need to convert to a common unit before mixing terms. Double-counting happens when SG sneaks into both the head calculation and the WHP formula—pick one place for it.
OH 32: HVAC_FLUIDS-21
Q: Is double-counting of specific gravity actually occurring in problem 21, or does it cancel out as your colleague suggested?
A: It does correct itself for the exact reason described: including SG in the modified Bernoulli equation puts head in feet of the viscous liquid, and including it in the WHP formula corrects for the water-based 3960 constant. Both uses are valid and serve different purposes, so they don't double-count—they're each doing independent work. The approach works as long as you're consistent about what each SG term is correcting.
OH 77: HVAC: Fluids #21
Q: When removing specific gravity from the delta P term, I get a delta H of 34.6 feet of water and a BHP greater than 1.4 HP—why does the video say the term gets smaller?
A: Everything you said is exactly right—removing SG from the pressure term makes it larger, not smaller, because water is less dense than a fluid with SG = 1.1. There's a clarification posted below the solution video correcting this error. The correct answer is D, not the C circled in the video.
OH 83: HVAC: Fluids Module #21
Q: The problem doesn't state the pipe material—can we assume schedule 40 steel, and could using the nominal diameter instead of actual diameter affect the answer?
A: You're right to question the assumption—the problem should have specified the material, and that's a flaw in the problem statement. For larger pipes (say, 8 inches or more), nominal and actual diameters are very close and the difference is negligible; for smaller pipes, the percentage difference gets squared when converting diameter to area, so it can matter significantly. Hopefully the exam will be clearer about pipe material, but be more careful with small pipes.
OH 96: HVAC: Fluids Module #21
Q: Why don't you apply the specific gravity correction to the 3-foot elevation term as you did in Fluids 14, and why is SG factored into the pressure differential but not the height?
A: This is a correction to the original solution—there was an error where SG was included in the pressure term but not in the velocity or elevation terms, which is inconsistent. The right approach is either include SG in all Bernoulli terms (pressure, velocity, elevation) and not in the final WHP formula, or exclude it from Bernoulli entirely and include it once in the WHP formula. Mixing approaches—counting SG in some terms but not others—is a very common mistake.
OH 111: HVAC: Fluids Module #21
Q: Why isn't it necessary to apply specific gravity to the static pressure difference given in PSI, and does this mean the original video solution was correct after all?
A: The pressure in PSI already stands on its own; when converting it to feet of fluid, divide by (SG × 2.31)—not just 2.31—to get feet of the actual viscous fluid rather than feet of water. Working through it carefully, the correct head added by the pump is ~35.8 feet of the viscous fluid, yielding ~1.24 HP, which is answer choice C—so the original solution was right and a later correction was incorrectly applied. Going forward, chart your course early and be consistent, whether you work in PSI or in feet of fluid.