A: You have to commit to one unit system throughout—either convert everything to PSI or work in feet of that specific fluid, but don't mix approaches mid-calculation. Applying specific gravity at the end is only valid if none of the individual Bernoulli terms already account for it. Inconsistency is what causes the double-counting problem here.

A: When static pressure is given in PSI and elevation is in feet of a non-water fluid, you must choose one consistent unit system before adding Bernoulli terms. If you work in PSI, convert the oil column using the fluid's actual specific weight (SG × 62.4); the 2.31 rule of thumb is only valid for water. Either path gives the right answer as long as you're consistent throughout.

A: Yes—Q·ΔP/1714 works for any fluid as long as Q is in GPM and ΔP is in PSI, since PSI is PSI regardless of fluid type. The formula Q·ΔH/3960 is water-only; the general form for non-water fluids using head is Q·ΔH·SG/3960. Know which version requires specific gravity and which doesn't.

A: Find the minimum conversions needed and stay consistent—if working in PSI, don't add atmospheric pressure since it cancels in the delta P subtraction. For the elevation term with a non-water fluid, multiply by SG/2.31 to convert feet of that fluid to PSI, not just 1/2.31. Avoid mixing unit systems mid-calculation.

A: Adding 14.7 gives absolute pressure, but since you're finding the difference P2−P1, atmosphere cancels out when both pressures are in gauge units. A reading of −2 inches mercury signals gauge pressure, since negative absolute pressures are physically impossible. Stay consistent in your pressure reference and you don't need to convert to absolute at all.

A: Q·ΔP/1714 works for any fluid because delta P is already in PSI—specific gravity is inherently embedded in the pressure value. The Q·ΔH/3960 formula is water-only; for non-water fluids using head in feet, use Q·ΔH·SG/3960. For break horsepower, just divide water horsepower by efficiency—no extra SG is needed once delta P is already in PSI.

A: Both are equivalent—expanding discharge and suction heads into their Bernoulli components (P/γ + V²/2g + Z) and subtracting gives you the same formula. Choose the approach that fits the given information and verify that different methods reconcile. Specific gravity enters through the γ term when the fluid isn't water.

A: Q·ΔP/1714 works for any fluid because you've already distilled the pressure into PSI, which has nothing to do with water specifically. The 'water only' label belongs to Q·ΔH/3960, where the 3960 constant has water's specific weight baked in. With practice, the distinction becomes instinct.

A: The work shows Z1−Z2 on one line (correct and negative), then it's flipped to Z2−Z1 on the next—that changes the signs and is algebraically invalid. For a pump where state 2 is higher than state 1, Z1−Z2 is negative (e.g., −10 ft), and that negative sign matters for the final answer. Commit to one arrangement and carry it through consistently.

A: W = Q·γ·H is valid in principle, but using it with GPM and feet requires additional unit conversions that aren't automatic—this is why Q·ΔP/1714 is preferred, since it works directly in GPM and PSI for any fluid. Both approaches give the same result; pick the one you can execute reliably under exam conditions.

A: The 0.86 factor comes directly from the specific gravity definition: SG = γ_oil/γ_water, so γ_oil = 0.86 × γ_water. A 10-foot column of oil exerts the same pressure as an 8.6-foot column of water—the denser the fluid, the more pressure per foot of height. It's not a memorized conversion; it's the direct application of the specific gravity definition.

A: Q·ΔH·SG/3960 is the general form for any fluid using head in feet—SG is included because a non-water column exerts different pressure than a water column. Once you convert to PSI and use Q·ΔP/1714, SG is already embedded and should never be added again—that would double-count it. Either include SG in the feet-of-head formula or convert to PSI first; never do both.

A: The equation is missing the specific gravity—since the fluid is fuel oil (SG = 0.86), multiplying by 0.86 reduces the answer by ~14% and brings it in line. The preferred approach is to stay in feet of the actual fluid throughout and apply SG once at the end in the water horsepower formula, rather than converting each term to 'feet of water' individually. If you prefer PSI, convert all terms and use Q·ΔP/1714—never mix feet-of-water and feet-of-fluid.

A: The gauge at P2 only measures static pressure at that location; the pump still has to do work to raise the fluid against gravity, which the gauge doesn't capture. In Bernoulli, energy added by the pump must account for static pressure difference plus elevation change plus losses—a pump moving fluid higher does more work than one on a flat system, even with identical gauge readings. Think of it as total energy: the pump supplies the difference in static, kinetic, and potential energy combined.