Typical Backward Curved Fans

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

4 formulas

10 exam-style examples

~53 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Typical Backward Curved Fans 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 typical backward curved fans describes physically and when it applies.
State every one of the 4 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. Typical Backward Curved Fans 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.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: typical backward curved fans.
Capstone Studio instructional photograph

Fluid Mechanics — Typical Backward Curved Fans: 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 4 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 = h" — read its definition and unit from the handbook line directly above the equation.
∆P	Quantity produced by "∆P = pressure rise" — read its definition and unit from the handbook line directly above the equation.
ηf	Quantity produced by "ηf = fan efficiency" — 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.

f POWER
where
CONSTANT N, D, ρ
FLOW RATE, Q

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

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans

A pump delivers 0.130 m³/s against a total dynamic head of 53 m at 74% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 4 h/day and $0.09/kWh.

Given

Q = 0.130 m³/s
H = 53 m
η = 0.74
4 h/day at $0.09/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
$P = 1000 (9.81) (0.130) (53) = 67.59 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 67.59/0.74 = 91.34 k W$
Annual energy
$91.34 k W \times 4 h / d a y \times 365 = 133, 355 k W h$
Cost — 133,355 kWh × $0.09 = $12,002 per year

Answer: P_water = 67.6 kW, P_shaft = 91.3 kW, cost ≈ $12,002/yr

Why the other options are there

50.0 kW (efficiency multiplied)
90.6 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 2

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (2)

A pump delivers 0.090 m³/s against a total dynamic head of 37 m at 78% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 14 h/day and $0.12/kWh.

Given

Q = 0.090 m³/s
H = 37 m
η = 0.78
14 h/day at $0.12/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
$P = 1000 (9.81) (0.090) (37) = 32.67 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 32.67/0.78 = 41.88 k W$
Annual energy
$41.88 k W \times 14 h / d a y \times 365 = 214, 013 k W h$
Cost — 214,013 kWh × $0.12 = $25,682 per year

Answer: P_water = 32.7 kW, P_shaft = 41.9 kW, cost ≈ $25,682/yr

Why the other options are there

25.5 kW (efficiency multiplied)
43.8 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 3

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (3)

A pump delivers 0.130 m³/s against a total dynamic head of 46 m at 72% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 12 h/day and $0.10/kWh.

Given

Q = 0.130 m³/s
H = 46 m
η = 0.72
12 h/day at $0.10/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
$P = 1000 (9.81) (0.130) (46) = 58.66 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 58.66/0.72 = 81.48 k W$
Annual energy
$81.48 k W \times 12 h / d a y \times 365 = 356, 871 k W h$
Cost — 356,871 kWh × $0.10 = $35,687 per year

Answer: P_water = 58.7 kW, P_shaft = 81.5 kW, cost ≈ $35,687/yr

Why the other options are there

42.2 kW (efficiency multiplied)
78.6 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 4

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (4)

A pump delivers 0.160 m³/s against a total dynamic head of 9 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 8 h/day and $0.19/kWh.

Given

Q = 0.160 m³/s
H = 9 m
η = 0.84
8 h/day at $0.19/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
$P = 1000 (9.81) (0.160) (9) = 14.13 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 14.13/0.84 = 16.82 k W$
Annual energy
$16.82 k W \times 8 h / d a y \times 365 = 49, 106 k W h$
Cost — 49,106 kWh × $0.19 = $9,330 per year

Answer: P_water = 14.1 kW, P_shaft = 16.8 kW, cost ≈ $9,330/yr

Why the other options are there

11.9 kW (efficiency multiplied)
18.9 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 5

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (5)

A pump delivers 0.310 m³/s against a total dynamic head of 24 m at 72% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 13 h/day and $0.15/kWh.

Given

Q = 0.310 m³/s
H = 24 m
η = 0.72
13 h/day at $0.15/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
$P = 1000 (9.81) (0.310) (24) = 72.99 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 72.99/0.72 = 101.4 k W$
Annual energy
$101.4 k W \times 13 h / d a y \times 365 = 481, 001 k W h$
Cost — 481,001 kWh × $0.15 = $72,150 per year

Answer: P_water = 73.0 kW, P_shaft = 101.4 kW, cost ≈ $72,150/yr

Why the other options are there

52.6 kW (efficiency multiplied)
97.8 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 6

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (6)

A pump delivers 0.300 m³/s against a total dynamic head of 56 m at 88% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 13 h/day and $0.11/kWh.

Given

Q = 0.300 m³/s
H = 56 m
η = 0.88
13 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
$P = 1000 (9.81) (0.300) (56) = 164.8 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 164.8/0.88 = 187.3 k W$
Annual energy
$187.3 k W \times 13 h / d a y \times 365 = 888, 652 k W h$
Cost — 888,652 kWh × $0.11 = $97,752 per year

Answer: P_water = 164.8 kW, P_shaft = 187.3 kW, cost ≈ $97,752/yr

Why the other options are there

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

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 7

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (7)

A pump delivers 0.090 m³/s against a total dynamic head of 50 m at 62% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 8 h/day and $0.09/kWh.

Given

Q = 0.090 m³/s
H = 50 m
η = 0.62
8 h/day at $0.09/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
$P = 1000 (9.81) (0.090) (50) = 44.15 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 44.15/0.62 = 71.20 k W$
Annual energy
$71.20 k W \times 8 h / d a y \times 365 = 207, 909 k W h$
Cost — 207,909 kWh × $0.09 = $18,712 per year

Answer: P_water = 44.1 kW, P_shaft = 71.2 kW, cost ≈ $18,712/yr

Why the other options are there

27.4 kW (efficiency multiplied)
59.2 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 8

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (8)

A pump delivers 0.320 m³/s against a total dynamic head of 15 m at 62% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 9 h/day and $0.11/kWh.

Given

Q = 0.320 m³/s
H = 15 m
η = 0.62
9 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
$P = 1000 (9.81) (0.320) (15) = 47.09 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 47.09/0.62 = 75.95 k W$
Annual energy
$75.95 k W \times 9 h / d a y \times 365 = 249, 490 k W h$
Cost — 249,490 kWh × $0.11 = $27,444 per year

Answer: P_water = 47.1 kW, P_shaft = 75.9 kW, cost ≈ $27,444/yr

Why the other options are there

29.2 kW (efficiency multiplied)
63.1 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 9

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (9)

A pump delivers 0.300 m³/s against a total dynamic head of 15 m at 70% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 10 h/day and $0.13/kWh.

Given

Q = 0.300 m³/s
H = 15 m
η = 0.70
10 h/day at $0.13/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
$P = 1000 (9.81) (0.300) (15) = 44.15 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 44.15/0.70 = 63.06 k W$
Annual energy
$63.06 k W \times 10 h / d a y \times 365 = 230, 185 k W h$
Cost — 230,185 kWh × $0.13 = $29,924 per year

Answer: P_water = 44.1 kW, P_shaft = 63.1 kW, cost ≈ $29,924/yr

Why the other options are there

30.9 kW (efficiency multiplied)
59.2 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

Example 10

Water horsepower, shaft power and pump operating cost — Typical Backward Curved Fans (10)

A pump delivers 0.140 m³/s against a total dynamic head of 54 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 10 h/day and $0.09/kWh.

Given

Q = 0.140 m³/s
H = 54 m
η = 0.84
10 h/day at $0.09/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
$P = 1000 (9.81) (0.140) (54) = 74.16 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 74.16/0.84 = 88.29 k W$
Annual energy
$88.29 k W \times 10 h / d a y \times 365 = 322, 259 k W h$
Cost — 322,259 kWh × $0.09 = $29,003 per year

Answer: P_water = 74.2 kW, P_shaft = 88.3 kW, cost ≈ $29,003/yr

Why the other options are there

62.3 kW (efficiency multiplied)
99.4 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Typical Backward Curved Fans

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

Typical Backward Curved Fans contains 4 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.