Scaling Laws; Affinity Laws

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

13 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Scaling Laws; Affinity Laws 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 scaling laws; affinity laws describes physically and when it applies.
State every one of the 13 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. Scaling Laws; Affinity Laws 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: scaling laws; affinity laws.
Capstone Studio instructional photograph

Fluid Mechanics — Scaling Laws; Affinity Laws: 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 13 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

d n	Quantity produced by "d n =d n" — read its definition and unit from the handbook line directly above the equation.
d n	Quantity produced by "d n =d n" — read its definition and unit from the handbook line directly above the equation.
d n	Quantity produced by "d n = d 2 2n" — read its definition and unit from the handbook line directly above the equation.
d n	Quantity produced by "d n = d 2 2n" — read its definition and unit from the handbook line directly above the equation.
e o	Quantity produced by "e o = e 3 5o" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = volumetric flowrate" — read its definition and unit from the handbook line directly above the equation.
mo	Quantity produced by "mo= mass flowrate" — read its definition and unit from the handbook line directly above the equation.
H	Quantity produced by "H = head" — 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.
Wo	Quantity produced by "Wo= power" — read its definition and unit from the handbook line directly above the equation.
ρ	Quantity produced by "ρ = fluid density" — read its definition and unit from the handbook line directly above the equation.
N	Quantity produced by "N = rotational speed" — 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.

Q Q
ND3 2 ND3 1
mo mo
tND3 2 tND3 1
H H
N 2D 2 2 ND 1
P P
tN 2 D 2 2 tN D 1
Wo Wo
tN3D5 2 tN D 1
where
Subscripts 1 and 2 refer to different but similar machines or to different operating conditions of the same machine.

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 affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws

A pump running at 1764 rpm delivers 0.140 m³/s at 16 m while absorbing 36.5 kW. The same impeller is run at 2961 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1764 rpm, N₂ = 2961 rpm
Q₁ = 0.140 m³/s
H₁ = 16 m
P₁ = 36.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2961/1764 = 1.6786$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.140 (1.6786) = 0.235 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 16 (1.6786)^{2} = 45.1 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 36.5 (1.6786)^{3} = 172.6 k W$

Answer: Q₂ = 0.235 m³/s, H₂ = 45.1 m, P₂ = 172.6 kW

Why the other options are there

H₂ = 26.9 m (linear scaling of head)
P₂ = 102.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 2

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (2)

A pump running at 1759 rpm delivers 0.050 m³/s at 21 m while absorbing 4.0 kW. The same impeller is run at 2446 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1759 rpm, N₂ = 2446 rpm
Q₁ = 0.050 m³/s
H₁ = 21 m
P₁ = 4.0 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2446/1759 = 1.3906$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.050 (1.3906) = 0.070 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 21 (1.3906)^{2} = 40.6 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 4.0 (1.3906)^{3} = 10.8 k W$

Answer: Q₂ = 0.070 m³/s, H₂ = 40.6 m, P₂ = 10.8 kW

Why the other options are there

H₂ = 29.2 m (linear scaling of head)
P₂ = 7.7 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 3

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (3)

A pump running at 1764 rpm delivers 0.030 m³/s at 23 m while absorbing 40.5 kW. The same impeller is run at 2662 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1764 rpm, N₂ = 2662 rpm
Q₁ = 0.030 m³/s
H₁ = 23 m
P₁ = 40.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2662/1764 = 1.5091$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.030 (1.5091) = 0.045 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 23 (1.5091)^{2} = 52.4 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 40.5 (1.5091)^{3} = 139.2 k W$

Answer: Q₂ = 0.045 m³/s, H₂ = 52.4 m, P₂ = 139.2 kW

Why the other options are there

H₂ = 34.7 m (linear scaling of head)
P₂ = 92.2 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 4

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (4)

A pump running at 1153 rpm delivers 0.100 m³/s at 39 m while absorbing 38.0 kW. The same impeller is run at 2094 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1153 rpm, N₂ = 2094 rpm
Q₁ = 0.100 m³/s
H₁ = 39 m
P₁ = 38.0 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2094/1153 = 1.8161$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.100 (1.8161) = 0.182 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 39 (1.8161)^{2} = 128.6 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 38.0 (1.8161)^{3} = 227.6 k W$

Answer: Q₂ = 0.182 m³/s, H₂ = 128.6 m, P₂ = 227.6 kW

Why the other options are there

H₂ = 70.8 m (linear scaling of head)
P₂ = 125.3 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 5

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (5)

A pump running at 1680 rpm delivers 0.160 m³/s at 15 m while absorbing 36.0 kW. The same impeller is run at 2840 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1680 rpm, N₂ = 2840 rpm
Q₁ = 0.160 m³/s
H₁ = 15 m
P₁ = 36.0 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2840/1680 = 1.6905$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.160 (1.6905) = 0.270 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 15 (1.6905)^{2} = 42.9 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 36.0 (1.6905)^{3} = 173.9 k W$

Answer: Q₂ = 0.270 m³/s, H₂ = 42.9 m, P₂ = 173.9 kW

Why the other options are there

H₂ = 25.4 m (linear scaling of head)
P₂ = 102.9 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 6

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (6)

A pump running at 1614 rpm delivers 0.080 m³/s at 23 m while absorbing 55.5 kW. The same impeller is run at 2207 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1614 rpm, N₂ = 2207 rpm
Q₁ = 0.080 m³/s
H₁ = 23 m
P₁ = 55.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2207/1614 = 1.3674$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.080 (1.3674) = 0.109 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 23 (1.3674)^{2} = 43.0 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 55.5 (1.3674)^{3} = 141.9 k W$

Answer: Q₂ = 0.109 m³/s, H₂ = 43.0 m, P₂ = 141.9 kW

Why the other options are there

H₂ = 31.5 m (linear scaling of head)
P₂ = 103.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 7

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (7)

A pump running at 1466 rpm delivers 0.020 m³/s at 13 m while absorbing 6.5 kW. The same impeller is run at 2204 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1466 rpm, N₂ = 2204 rpm
Q₁ = 0.020 m³/s
H₁ = 13 m
P₁ = 6.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2204/1466 = 1.5034$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.020 (1.5034) = 0.030 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 13 (1.5034)^{2} = 29.4 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 6.5 (1.5034)^{3} = 22.1 k W$

Answer: Q₂ = 0.030 m³/s, H₂ = 29.4 m, P₂ = 22.1 kW

Why the other options are there

H₂ = 19.5 m (linear scaling of head)
P₂ = 14.7 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 8

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (8)

A pump running at 1555 rpm delivers 0.030 m³/s at 13 m while absorbing 58.5 kW. The same impeller is run at 1990 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1555 rpm, N₂ = 1990 rpm
Q₁ = 0.030 m³/s
H₁ = 13 m
P₁ = 58.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 1990/1555 = 1.2797$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.030 (1.2797) = 0.038 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 13 (1.2797)^{2} = 21.3 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 58.5 (1.2797)^{3} = 122.6 k W$

Answer: Q₂ = 0.038 m³/s, H₂ = 21.3 m, P₂ = 122.6 kW

Why the other options are there

H₂ = 16.6 m (linear scaling of head)
P₂ = 95.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 9

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (9)

A pump running at 1406 rpm delivers 0.180 m³/s at 13 m while absorbing 21.5 kW. The same impeller is run at 2796 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1406 rpm, N₂ = 2796 rpm
Q₁ = 0.180 m³/s
H₁ = 13 m
P₁ = 21.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2796/1406 = 1.9886$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.180 (1.9886) = 0.358 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 13 (1.9886)^{2} = 51.4 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 21.5 (1.9886)^{3} = 169.1 k W$

Answer: Q₂ = 0.358 m³/s, H₂ = 51.4 m, P₂ = 169.1 kW

Why the other options are there

H₂ = 25.9 m (linear scaling of head)
P₂ = 85.0 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

Example 10

Pump affinity laws for a change in impeller speed — Scaling Laws; Affinity Laws (10)

A pump running at 1291 rpm delivers 0.040 m³/s at 17 m while absorbing 5.5 kW. The same impeller is run at 2631 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1291 rpm, N₂ = 2631 rpm
Q₁ = 0.040 m³/s
H₁ = 17 m
P₁ = 5.5 kW

Find

Q₂, H₂ and P₂

Start with the thinking

Flow scales linearly, head with the square and power with the cube of the speed ratio.
A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

Speed ratio
$N_{2} / N_{1} = 2631/1291 = 2.0380$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.040 (2.0380) = 0.082 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 17 (2.0380)^{2} = 70.6 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 5.5 (2.0380)^{3} = 46.6 k W$

Answer: Q₂ = 0.082 m³/s, H₂ = 70.6 m, P₂ = 46.6 kW

Why the other options are there

H₂ = 34.6 m (linear scaling of head)
P₂ = 22.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

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

Scaling Laws; Affinity Laws contains 13 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.