16.4 Pump Hydraulics & Horsepower

16.4 Pump Hydraulics & Horsepower

Quick Answer: Total dynamic head (TDH) ≈ static head + friction (and minor) losses. Water horsepower (\text{WHP} = (Q_{\text{gpm}} \times H_{\text{ft}}) / 3960). Brake horsepower accounts for pump efficiency: (\text{BHP} = \text{WHP} / \eta_{\text{pump}}). Wire-to-water efficiency folds in motor losses. Measure field gpm from tank drop; velocity (v = Q/A). Florida exams weight math heavily—drill setups until unit conversions are automatic.

Pumps appear on water, wastewater, and distribution outlines: lift stations, high-service pumps, boosters, and backwash pumps. This section closes the operator-math chapter with power and hydraulic rate calculations.

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1. Head Components

[ \text{TDH} \approx H_{\text{static}} + H_{\text{friction}} + H_{\text{minor}} ]

Pressure as head: (H_{\text{ft}} ≈ p_{\text{psi}} × 2.31).

Worked Example 1 — Static + Friction

Lift 45 ft static. Friction and minor losses total 18 ft at the design flow. Discharge pressure requirement already converted into the static/friction breakdown given.

[ \text{TDH} = 45 + 18 = 63\ \text{ft} ]

Worked Example 2 — Pressure to Head

A pump must overcome 30 psi discharge pressure difference (simplified stem) with negligible elevation change stated separately as included.

[ H = 30 × 2.31 = 69.3\ \text{ft} ]

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2. Water Horsepower (WHP)

Water horsepower is the power delivered to the water:

[ \text{WHP} = \frac{Q\ (\text{gpm}) \times H\ (\text{ft})}{3960} ]

Worked Example 3 — WHP

Pump delivers 500 gpm against 80 ft TDH.

[ \text{WHP} = \frac{500 \times 80}{3960} = \frac{40{,}000}{3960} \approx 10.1\ \text{hp} ]

Worked Example 4 — WHP from MGD

Flow 1.44 MGD, head 100 ft.

[ Q_{\text{gpm}} = 1.44 / 0.00144 = 1{,}000\ \text{gpm} ]

[ \text{WHP} = \frac{1000 \times 100}{3960} \approx 25.3\ \text{hp} ]

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3. Brake Horsepower, Motor Load & Efficiency

Pump efficiency (\eta_p) (decimal) relates WHP to shaft power:

[ \text{BHP} = \frac{\text{WHP}}{\eta_p} ]

Motor efficiency (\eta_m):

[ \text{Motor input hp} \approx \frac{\text{BHP}}{\eta_m} ]

Wire-to-water efficiency (overall):

[ \eta_{\text{w2w}} = \frac{\text{WHP}}{\text{Electrical input hp}} = \eta_p \times \eta_m\ \text{(approx., if only those losses)} ]

Or from field data:

[ \eta_{\text{w2w}} = \frac{\text{WHP}}{\text{kW input} \times 1.341} ]

(since 1 kW ≈ 1.341 hp).

Worked Example 5 — BHP

WHP = 10.1 hp (Example 3). Pump efficiency 70%.

[ \text{BHP} = 10.1 / 0.70 \approx 14.4\ \text{hp} ]

Select a motor above BHP (service factor, future head, code practice)—exam may ask calculated BHP, not nameplate choice.

Worked Example 6 — Wire-to-Water

WHP = 25 hp. Measured electrical input = 30 kW.

[ \text{Input hp} = 30 × 1.341 = 40.23\ \text{hp} ]

[ \eta_{\text{w2w}} = 25 / 40.23 ≈ 0.62 = 62% ]

Worked Example 7 — Combined Efficiencies

WHP = 12 hp, (\eta_p = 75%), (\eta_m = 90%).

[ \text{BHP} = 12 / 0.75 = 16\ \text{hp} ]

[ \text{Input hp} = 16 / 0.90 ≈ 17.8\ \text{hp} ]

[ \eta_{\text{w2w}} = 0.75 × 0.90 = 0.675 = 67.5% ]

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4. Flow from Tank Drop (Field gpm)

Same geometry as chemical calibration, applied to clearwells, wet wells, or tanks:

[ Q_{\text{gpm}} = \frac{V_{\text{gal}}}{t_{\text{min}}} ]

with

[ V_{\text{gal}} = 0.785 D^2 H \times 7.48\ \text{(cylinder)} ]

or (L × W × H × 7.48) for rectangles.

Worked Example 8 — Rectangular Clearwell Drop

Basin 40 ft × 25 ft. Level falls 0.40 ft in 10 minutes while high-service pumps run and plant production is isolated (exam idealization).

[ V_{\text{ft}^3} = 40 × 25 × 0.40 = 400\ \text{ft}^3 ]

[ V_{\text{gal}} = 400 × 7.48 = 2{,}992\ \text{gal} ]

[ Q = 2{,}992 / 10 = 299\ \text{gpm} ]

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5. Velocity = Q / A

[ v = \frac{Q}{A} ]

Unit consistency is critical.

Common teaching conversion:

[ Q_{\text{cfs}} ≈ \frac{Q_{\text{gpm}}}{449} ]

(some sheets use 448.8).

Worked Example 9 — Force Main Velocity

Flow 600 gpm in a pipe with internal area 0.349 ft² (approx. 8-in ID order of magnitude).

[ Q_{\text{cfs}} = 600 / 448.8 ≈ 1.34\ \text{cfs} ]

[ v = 1.34 / 0.349 ≈ 3.8\ \text{ft/s} ]

Worked Example 10 — Area from Diameter Then Velocity

Pipe D = 12 in = 1.0 ft, flow 1,000 gpm.

[ A = 0.785 × 1.0^2 = 0.785\ \text{ft}^2 ]

[ Q_{\text{cfs}} = 1000 / 448.8 ≈ 2.23\ \text{cfs} ]

[ v = 2.23 / 0.785 ≈ 2.84\ \text{ft/s} ]

Operator context: Very high velocities increase friction head and water-hammer risk; very low velocities may allow settling in force mains—exam theory plus design standards.

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6. Master Formula Table — Pump & Hydraulics Math

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7. Florida Exam Math Weight — Practice Strategy

FDEP water/wastewater operator exams are multiple-choice but calculation-dense. Math appears inside treatment, distribution, and wastewater process stems—not only in a labeled “math section.”

High-yield practice plan:

1. Memorize the “big five” setups: pounds formula, V/Q detention, % removal, WHP/BHP, tank volume × 7.48.
2. Always rewrite the given flow in the unit the formula needs before multiplying.
3. Work backward from answer choices when stuck: plug each plausible Q or H only if time is short—but prefer forward setup.
4. Use the official formula sheet in timed practice so eye-find speed is real.
5. Log every miss by trap type (unit, %, efficiency as percent vs decimal, diameter).
6. Target accuracy before speed, then add a timer (Class A/B/C: 100 items / 3 hours ≈ 1.8 min each—including reading).
7. Mix water and wastewater stems—same 8.34 and 3960, different stories.
8. Re-work this chapter’s examples with different numbers until the steps feel mechanical.

Worked Example 11 — End-to-End Pump Item

Static lift 50 ft, friction 25 ft, flow 800 gpm, pump efficiency 65%. Find BHP.

[ \text{TDH} = 75\ \text{ft} ]

[ \text{WHP} = (800 × 75) / 3960 = 60{,}000 / 3960 ≈ 15.15\ \text{hp} ]

[ \text{BHP} = 15.15 / 0.65 ≈ 23.3\ \text{hp} ]

Worked Example 12 — Sanity Check

If calculated BHP is 2 hp for a large raw-water pump at 5,000 gpm and 100 ft head, you almost certainly forgot to convert units or misplaced a zero—WHP alone is (5000×100/3960 ≈ 126) hp before efficiency.

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8. Chapter 16 Integration

Operators who treat math as a single toolkit—not isolated tricks—finish Florida exams with time left to recheck the two or three hardest calculations.