Pump Power Equation

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

5 formulas

10 exam-style examples

~55 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Pump Power Equation within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what pump power equation describes physically and when it applies.
State every one of the 5 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Pump Power Equation is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: pump power equation.
Capstone Studio instructional photograph

Fluid Mechanics — Pump Power Equation: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 5 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

Wo	Quantity produced by "Wo = Qγh/ηt = Qρgh/ηt" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = volumetric flow (m3/s or cfs)" — read its definition and unit from the handbook line directly above the equation.
h	Quantity produced by "h = head (m or ft) the fluid has to be lifted" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Pump brake power

A pump delivers 0.06 m³/s against a total head of 32 m at 74% efficiency. What brake power is required?

Given

Q = 0.06 m³/s
H = 32 m
η = 0.74
γ = 9.81 kN/m³

Find

Brake power (kW)

Start with the thinking

Water power first, then divide by efficiency.
Dividing by efficiency increases the power — a common sign check.

Step-by-step solution

Water power — P_w = γQH = 9.81(0.06)(32)
Evaluate
$P_{w} = 18.8 k W$
Brake power
$P = P_{w} / η = 18.8/0.74$
Result
$P = 25.5 k W$

Answer: P ≈ 25.5 kW

Why the other options are there

13.9 kW (multiplied by efficiency)
18.8 kW (efficiency ignored)

Reference: FE Reference Handbook — Fluid Mechanics — Pump power

Example 2

Pump brake horsepower — Pump Power Equation

A pump delivers 8.5 cfs against 160 ft of head at 64% efficiency. Find the water and brake horsepower.

Given

Q = 8.5 cfs
H = 160 ft
η = 0.64

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (8.5) (160) /550 = 154.3 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 154.3/0.64 = 241.1 h p$

Answer: WHP ≈ 154.3 hp; BHP ≈ 241.1 hp

Why the other options are there

98.8 hp (efficiency multiplied)
2.47 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 3

Water horsepower, shaft power and pump operating cost — Pump Power Equation

A pump delivers 0.400 m³/s against a total dynamic head of 44 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 19 h/day and $0.20/kWh.

Given

Q = 0.400 m³/s
H = 44 m
η = 0.84
19 h/day at $0.20/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.400) (44) = 172.7 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 172.7/0.84 = 205.5 k W$
Annual energy
$205.5 k W \times 19 h / d a y \times 365 = 1, 425, 440 k W h$
Cost — 1,425,440 kWh × $0.20 = $285,088 per year

Answer: P_water = 172.7 kW, P_shaft = 205.5 kW, cost ≈ $285,088/yr

Why the other options are there

145.0 kW (efficiency multiplied)
231.4 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 4

Pump brake horsepower — Pump Power Equation (2)

A pump delivers 2.5 cfs against 151 ft of head at 84% efficiency. Find the water and brake horsepower.

Given

Q = 2.5 cfs
H = 151 ft
η = 0.84

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (2.5) (151) /550 = 42.83 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 42.83/0.84 = 50.99 h p$

Answer: WHP ≈ 42.8 hp; BHP ≈ 51.0 hp

Why the other options are there

36.0 hp (efficiency multiplied)
0.69 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 5

Water horsepower, shaft power and pump operating cost — Pump Power Equation (2)

A pump delivers 0.040 m³/s against a total dynamic head of 31 m at 82% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 15 h/day and $0.14/kWh.

Given

Q = 0.040 m³/s
H = 31 m
η = 0.82
15 h/day at $0.14/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.040) (31) = 12.16 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 12.16/0.82 = 14.83 k W$
Annual energy
$14.83 k W \times 15 h / d a y \times 365 = 81, 220 k W h$
Cost — 81,220 kWh × $0.14 = $11,371 per year

Answer: P_water = 12.2 kW, P_shaft = 14.8 kW, cost ≈ $11,371/yr

Why the other options are there

10.0 kW (efficiency multiplied)
16.3 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 6

Pump brake horsepower — Pump Power Equation (3)

A pump delivers 16.0 cfs against 178 ft of head at 88% efficiency. Find the water and brake horsepower.

Given

Q = 16.0 cfs
H = 178 ft
η = 0.88

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (16.0) (178) /550 = 323.1 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 323.1/0.88 = 367.2 h p$

Answer: WHP ≈ 323.1 hp; BHP ≈ 367.2 hp

Why the other options are there

284.3 hp (efficiency multiplied)
5.18 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 7

Water horsepower, shaft power and pump operating cost — Pump Power Equation (3)

A pump delivers 0.150 m³/s against a total dynamic head of 53 m at 80% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 10 h/day and $0.11/kWh.

Given

Q = 0.150 m³/s
H = 53 m
η = 0.80
10 h/day at $0.11/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.150) (53) = 77.99 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 77.99/0.80 = 97.49 k W$
Annual energy
$97.49 k W \times 10 h / d a y \times 365 = 355, 827 k W h$
Cost — 355,827 kWh × $0.11 = $39,141 per year

Answer: P_water = 78.0 kW, P_shaft = 97.5 kW, cost ≈ $39,141/yr

Why the other options are there

62.4 kW (efficiency multiplied)
104.5 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 8

Pump brake horsepower — Pump Power Equation (4)

A pump delivers 5.0 cfs against 68 ft of head at 84% efficiency. Find the water and brake horsepower.

Given

Q = 5.0 cfs
H = 68 ft
η = 0.84

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (5.0) (68) /550 = 38.57 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 38.57/0.84 = 45.92 h p$

Answer: WHP ≈ 38.6 hp; BHP ≈ 45.9 hp

Why the other options are there

32.4 hp (efficiency multiplied)
0.62 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 9

Water horsepower, shaft power and pump operating cost — Pump Power Equation (4)

A pump delivers 0.130 m³/s against a total dynamic head of 39 m at 74% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 19 h/day and $0.11/kWh.

Given

Q = 0.130 m³/s
H = 39 m
η = 0.74
19 h/day at $0.11/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.130) (39) = 49.74 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 49.74/0.74 = 67.21 k W$
Annual energy
$67.21 k W \times 19 h / d a y \times 365 = 466, 114 k W h$
Cost — 466,114 kWh × $0.11 = $51,272 per year

Answer: P_water = 49.7 kW, P_shaft = 67.2 kW, cost ≈ $51,272/yr

Why the other options are there

36.8 kW (efficiency multiplied)
66.7 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Example 10

Pump brake horsepower — Pump Power Equation (5)

A pump delivers 12.5 cfs against 191 ft of head at 76% efficiency. Find the water and brake horsepower.

Given

Q = 12.5 cfs
H = 191 ft
η = 0.76

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (12.5) (191) /550 = 270.9 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 270.9/0.76 = 356.4 h p$

Answer: WHP ≈ 270.9 hp; BHP ≈ 356.4 hp

Why the other options are there

205.9 hp (efficiency multiplied)
4.34 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Pump Power Equation contains 5 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.