Flow Rate, Pump Horsepower & Tank Volume Calculations

13.2 Flow Rate, Pump Horsepower & Tank Volume Calculations

Quick Answer: Flow continuity is Q = V × A. Water horsepower is (gpm × TDH) ÷ 3960; brake horsepower divides by pump efficiency. Tank volume uses geometry (L×W×H or πr²h) then ×7.48 for gallons. Pressure head converts with psi = head_ft ÷ 2.31.

Distribution and plant operators constantly convert between pipe velocity, pump power, and tank capacity. TCEQ math items reward clean unit handling more than exotic algebra. This section works Q = VA, gpm ↔ MGD, rectangular and cylindrical volumes, water and brake horsepower, and force/pressure basics.

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Continuity: Q = V × A

Discharge (flow rate) equals velocity times cross-sectional area:

[ Q = V \times A ]

If velocity is in ft/s and area in ft², then Q is in cfs (ft³/s). Convert to gpm when needed:

[ 1\ \text{cfs} = 448.8\ \text{gpm} ]

(often rounded to 450 gpm on some field sheets; prefer 448.8 when choices are tight).

Area of a circular pipe:

[ A = \pi r^2 = \pi \left(\frac{D}{2}\right)^2 ]

Use diameter and radius in feet when you want ft².

Worked Example 1 — Pipe Flow from Velocity

A 12-inch (1.0 ft) diameter main carries water at 3.5 ft/s. What is the flow in gpm?

Step 1 — radius and area:

[ r = 0.5\ \text{ft},\quad A = \pi (0.5)^2 = 0.7854\ \text{ft}^2 ]

Step 2 — cfs:

[ Q = 3.5 \times 0.7854 = 2.749\ \text{cfs} ]

Step 3 — gpm:

[ 2.749 \times 448.8 \approx 1{,}234\ \text{gpm} ]

Worked Example 2 — Solve for Velocity

Flow in an 8-inch main is 500 gpm. Find velocity in ft/s.

Step 1 — cfs:

[ 500 \div 448.8 \approx 1.114\ \text{cfs} ]

Step 2 — area (D = 8/12 = 0.667 ft, r = 0.333 ft):

[ A = \pi (0.333)^2 \approx 0.349\ \text{ft}^2 ]

Step 3 — velocity:

[ V = \frac{Q}{A} = \frac{1.114}{0.349} \approx 3.19\ \text{ft/s} ]

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gpm and MGD Conversions

There are 1,440 minutes in a day, so:

[ \text{MGD} = \frac{\text{gpm} \times 1{,}440}{1{,}000{,}000} = \text{gpm} \times 0.00144 ]

[ \text{gpm} = \text{MGD} \times \frac{1{,}000{,}000}{1{,}440} = \text{MGD} \times 694.4 ]

Worked Example 3 — Convert Both Ways

A booster station pumps 2,080 gpm. Express as MGD:

[ 2{,}080 \times 0.00144 = 2.995\ \text{MGD} \approx 3.0\ \text{MGD} ]

A plant rated at 0.85 MGD in gpm:

[ 0.85 \times 694.4 = 590.24\ \text{gpm} ]

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Tank Volume — Rectangular and Cylindrical

Rectangular tank (or basin):

[ \text{Volume (ft}^3) = L \times W \times H ]

Cylindrical tank:

[ \text{Volume (ft}^3) = \pi r^2 h ]

Convert to gallons:

[ \text{gallons} = \text{ft}^3 \times 7.48 ]

To MG, divide gallons by 1,000,000. Depth of water (not tank wall height) is the H or h you use for stored volume.

Worked Example 4 — Rectangular Clearwell

A clearwell is 40 ft long × 25 ft wide × 12 ft water depth. How many gallons are stored?

[ V = 40 \times 25 \times 12 = 12{,}000\ \text{ft}^3 ]

[ 12{,}000 \times 7.48 = 89{,}760\ \text{gal} ]

Worked Example 5 — Ground Storage Cylinder

A cylindrical ground storage tank has diameter 30 ft and water depth 18 ft. Find volume in gallons and MG.

[ r = 15\ \text{ft},\quad V = \pi (15)^2 (18) = 12{,}723.5\ \text{ft}^3 ]

[ \text{gal} = 12{,}723.5 \times 7.48 \approx 95{,}172\ \text{gal} ]

[ \text{MG} = 0.0952\ \text{MG} ]

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Water Horsepower and Brake Horsepower

Water horsepower (WHP) is the theoretical power needed to lift water against total dynamic head (TDH):

[ \text{WHP} = \frac{\text{gpm} \times \text{TDH}}{3960} ]

TDH is in feet of head. The constant 3960 comes from unit conversion among gpm, feet, and horsepower for water weighing 8.34 lb/gal.

Pumps are not 100% efficient. Brake horsepower (BHP) is the power that must be delivered to the pump shaft:

[ \text{BHP} = \frac{\text{WHP}}{\text{pump efficiency (decimal)}} ]

Motor efficiency may appear in a second step if the question asks for motor input power:

[ \text{Motor HP input} = \frac{\text{BHP}}{\text{motor efficiency}} ]

Worked Example 6 — Water HP

A pump delivers 450 gpm against 180 ft TDH. Find water horsepower.

[ \text{WHP} = \frac{450 \times 180}{3960} = \frac{81{,}000}{3960} = 20.45\ \text{HP} ]

Worked Example 7 — Brake HP with Efficiency

Same pump is 75% efficient. Brake horsepower:

[ \text{BHP} = \frac{20.45}{0.75} = 27.27\ \text{HP} ]

If the motor is 90% efficient, approximate electrical input horsepower:

[ \frac{27.27}{0.90} = 30.3\ \text{HP} ]

Operators size motors above this calculated load; the exam usually wants the calculated BHP or WHP, not the next catalog motor size unless stated.

Worked Example 8 — Combined Flow and Head

A high-service pump moves 1.2 MGD against 92 psi discharge pressure (assume suction pressure negligible and TDH ≈ pressure head only for this problem). Estimate WHP.

gpm:

[ 1.2 \times 694.4 = 833.3\ \text{gpm} ]

Head from pressure (next subsection):

[ \text{head (ft)} = 92 \times 2.31 = 212.5\ \text{ft} ]

WHP:

[ \frac{833.3 \times 212.5}{3960} \approx 44.7\ \text{HP} ]

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Force, Pressure, and Head

Pressure and elevation head for water are linked by:

[ \text{psi} = \frac{\text{head (ft)}}{2.31} ]

[ \text{head (ft)} = \text{psi} \times 2.31 ]

The factor 2.31 is feet of water per 1 psi (because 1 psi supports about 2.31 ft of water column). Related idea: 0.433 psi per foot of water (since (1 \div 2.31 \approx 0.433)).

Worked Example 9 — Convert Head to psi

A tank water surface is 115 ft above a customer tap (ignore friction for the static case). Static pressure at the tap:

[ \text{psi} = \frac{115}{2.31} \approx 49.8\ \text{psi} ]

Worked Example 10 — Convert psi to Head

A pressure gauge on a hydrant reads 68 psi. Equivalent water head:

[ 68 \times 2.31 = 157.1\ \text{ft} ]

Worked Example 11 — Force on a Valve Gate (Concept Check)

Pressure is force per area: (P = F / A), so (F = P \times A). If a valve disc has area 20 in² and upstream pressure is 60 psi:

[ F = 60\ \text{psi} \times 20\ \text{in}^2 = 1{,}200\ \text{lb} ]

You rarely need this on every TCEQ form, but the relationship explains why high-pressure systems need robust thrust restraint and why psi ↔ feet conversions matter for pump TDH estimates.

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Putting the Pieces Together

A typical multi-step exam item might ask: given tank dimensions, find gallons; convert fill rate from MGD to gpm; then find fill time; or convert discharge pressure to TDH and compute BHP. Work left to right, rewrite every quantity in the unit the formula expects, and only then punch the calculator.

Checklist before you submit an answer:

1. Diameter → radius in feet for pipe or tank area.
2. ft³ → gallons with 7.48; gpm ↔ MGD with 694.4 / 0.00144.
3. WHP uses 3960; BHP divides by efficiency as a decimal.
4. psi and feet use 2.31 (or 0.433 psi/ft)—do not mix the two formulas in one step.

These conversions are the same toolkit used with dosage and detention-time math in the previous section; only the formulas change.