Similitude

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

33 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 Similitude 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 similitude describes physically and when it applies.
State every one of the 33 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. Similitude 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: similitude.
Capstone Studio instructional photograph

Fluid Mechanics — Similitude: 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 33 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

σ p	Quantity produced by "σ p" — read its definition and unit from the handbook line directly above the equation.
σ m	Quantity produced by "σ m" — read its definition and unit from the handbook line directly above the equation.
FI	Quantity produced by "FI = inertia force" — read its definition and unit from the handbook line directly above the equation.
FP	Quantity produced by "FP = pressure force" — read its definition and unit from the handbook line directly above the equation.
FV	Quantity produced by "FV = viscous force" — read its definition and unit from the handbook line directly above the equation.
FG	Quantity produced by "FG = gravity force" — read its definition and unit from the handbook line directly above the equation.
FE	Quantity produced by "FE = elastic force" — read its definition and unit from the handbook line directly above the equation.
FT	Quantity produced by "FT = surface tension force" — read its definition and unit from the handbook line directly above the equation.
Re	Quantity produced by "Re = Reynolds number" — read its definition and unit from the handbook line directly above the equation.
We	Quantity produced by "We = Weber number" — read its definition and unit from the handbook line directly above the equation.
Ca	Quantity produced by "Ca = Cauchy number" — read its definition and unit from the handbook line directly above the equation.
Fr	Quantity produced by "Fr = Froude number" — 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.

In order to use a model to simulate the conditions of the prototype, the model must be geometrically, kinematically, and
dynamically similar to the prototype system.
To obtain dynamic similarity between two flow pictures, all independent force ratios that can be written must be the same in
both the model and the prototype. Thus, dynamic similarity between two flow pictures (when all possible forces are acting) is
expressed in the five simultaneous equations below.
[] [] [ ] [ ]
ρv 2
P p
ρv 2
[] [] [ ] [ ] [] []
FV p
FV m
vl ρ
vl ρ
[ ] [ ] [ ] [ ] [] []
FG p
FG m
lg p
lg m
[ ] [] [ ] [ ] [] []
FI FI ρv ρv 2
FE p
FE m
Ev p

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

Froude-scaled hydraulic model and prototype conversion — Similitude

A spillway is modelled at a 1:20 scale under Froude similitude. The model shows a velocity of 1.5 m/s and a discharge of 0.040 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:20
V_m = 1.5 m/s
Q_m = 0.040 m³/s

Find

Prototype velocity, discharge and time ratio

Start with the thinking

Free-surface flows are gravity dominated, so Froude number similarity governs, not Reynolds.
Under Froude scaling V_r = √L_r, Q_r = L_r^2.5 and t_r = √L_r.

Step-by-step solution

Formula — V_r = √L_r
Substituting
$V_{p} = 1.5 20 = 6.71 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.040 \times 2 0^{2} .5 = 71.55 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 20 = 4.472 \times m o d e l t im e$

Answer: V_p = 6.71 m/s, Q_p = 71.6 m³/s, time ratio 4.47

Why the other options are there

Q_p = 0.80 m³/s (linear scaling)
V_p = 30.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 2

Pump affinity laws for a change in impeller speed — Similitude

A pump running at 1385 rpm delivers 0.120 m³/s at 21 m while absorbing 60.0 kW. The same impeller is run at 2099 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1385 rpm, N₂ = 2099 rpm
Q₁ = 0.120 m³/s
H₁ = 21 m
P₁ = 60.0 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
$N_{2} / N_{1} = 2099/1385 = 1.5155$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.120 (1.5155) = 0.182 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 21 (1.5155)^{2} = 48.2 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 60.0 (1.5155)^{3} = 208.9 k W$

Answer: Q₂ = 0.182 m³/s, H₂ = 48.2 m, P₂ = 208.9 kW

Why the other options are there

H₂ = 31.8 m (linear scaling of head)
P₂ = 137.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 3

Froude-scaled hydraulic model and prototype conversion — Similitude (2)

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 2.5 m/s and a discharge of 0.120 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:15
V_m = 2.5 m/s
Q_m = 0.120 m³/s

Find

Prototype velocity, discharge and time ratio

Start with the thinking

Free-surface flows are gravity dominated, so Froude number similarity governs, not Reynolds.
Under Froude scaling V_r = √L_r, Q_r = L_r^2.5 and t_r = √L_r.

Step-by-step solution

Formula — V_r = √L_r
Substituting
$V_{p} = 2.5 15 = 9.68 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.120 \times 1 5^{2} .5 = 104.6 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 15 = 3.873 \times m o d e l t im e$

Answer: V_p = 9.68 m/s, Q_p = 104.6 m³/s, time ratio 3.87

Why the other options are there

Q_p = 1.80 m³/s (linear scaling)
V_p = 37.50 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 4

Pump affinity laws for a change in impeller speed — Similitude (2)

A pump running at 1761 rpm delivers 0.180 m³/s at 36 m while absorbing 49.5 kW. The same impeller is run at 2760 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1761 rpm, N₂ = 2760 rpm
Q₁ = 0.180 m³/s
H₁ = 36 m
P₁ = 49.5 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
$N_{2} / N_{1} = 2760/1761 = 1.5673$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.180 (1.5673) = 0.282 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 36 (1.5673)^{2} = 88.4 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 49.5 (1.5673)^{3} = 190.6 k W$

Answer: Q₂ = 0.282 m³/s, H₂ = 88.4 m, P₂ = 190.6 kW

Why the other options are there

H₂ = 56.4 m (linear scaling of head)
P₂ = 121.6 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 5

Froude-scaled hydraulic model and prototype conversion — Similitude (3)

A spillway is modelled at a 1:10 scale under Froude similitude. The model shows a velocity of 2.7 m/s and a discharge of 0.110 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:10
V_m = 2.7 m/s
Q_m = 0.110 m³/s

Find

Prototype velocity, discharge and time ratio

Start with the thinking

Free-surface flows are gravity dominated, so Froude number similarity governs, not Reynolds.
Under Froude scaling V_r = √L_r, Q_r = L_r^2.5 and t_r = √L_r.

Step-by-step solution

Formula — V_r = √L_r
Substituting
$V_{p} = 2.7 10 = 8.54 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.110 \times 1 0^{2} .5 = 34.79 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 10 = 3.162 \times m o d e l t im e$

Answer: V_p = 8.54 m/s, Q_p = 34.8 m³/s, time ratio 3.16

Why the other options are there

Q_p = 1.10 m³/s (linear scaling)
V_p = 27.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 6

Pump affinity laws for a change in impeller speed — Similitude (3)

A pump running at 1223 rpm delivers 0.130 m³/s at 14 m while absorbing 6.5 kW. The same impeller is run at 2932 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1223 rpm, N₂ = 2932 rpm
Q₁ = 0.130 m³/s
H₁ = 14 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
$N_{2} / N_{1} = 2932/1223 = 2.3974$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.130 (2.3974) = 0.312 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 14 (2.3974)^{2} = 80.5 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 6.5 (2.3974)^{3} = 89.6 k W$

Answer: Q₂ = 0.312 m³/s, H₂ = 80.5 m, P₂ = 89.6 kW

Why the other options are there

H₂ = 33.6 m (linear scaling of head)
P₂ = 37.4 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 7

Froude-scaled hydraulic model and prototype conversion — Similitude (4)

A spillway is modelled at a 1:25 scale under Froude similitude. The model shows a velocity of 2.2 m/s and a discharge of 0.100 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:25
V_m = 2.2 m/s
Q_m = 0.100 m³/s

Find

Prototype velocity, discharge and time ratio

Start with the thinking

Free-surface flows are gravity dominated, so Froude number similarity governs, not Reynolds.
Under Froude scaling V_r = √L_r, Q_r = L_r^2.5 and t_r = √L_r.

Step-by-step solution

Formula — V_r = √L_r
Substituting
$V_{p} = 2.2 25 = 11.00 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.100 \times 2 5^{2} .5 = 312.5 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 25 = 5.000 \times m o d e l t im e$

Answer: V_p = 11.00 m/s, Q_p = 312.5 m³/s, time ratio 5.00

Why the other options are there

Q_p = 2.50 m³/s (linear scaling)
V_p = 55.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 8

Pump affinity laws for a change in impeller speed — Similitude (4)

A pump running at 1283 rpm delivers 0.050 m³/s at 34 m while absorbing 45.5 kW. The same impeller is run at 2186 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1283 rpm, N₂ = 2186 rpm
Q₁ = 0.050 m³/s
H₁ = 34 m
P₁ = 45.5 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
$N_{2} / N_{1} = 2186/1283 = 1.7038$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.050 (1.7038) = 0.085 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 34 (1.7038)^{2} = 98.7 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 45.5 (1.7038)^{3} = 225.1 k W$

Answer: Q₂ = 0.085 m³/s, H₂ = 98.7 m, P₂ = 225.1 kW

Why the other options are there

H₂ = 57.9 m (linear scaling of head)
P₂ = 132.1 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 9

Froude-scaled hydraulic model and prototype conversion — Similitude (5)

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 1.6 m/s and a discharge of 0.190 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:15
V_m = 1.6 m/s
Q_m = 0.190 m³/s

Find

Prototype velocity, discharge and time ratio

Start with the thinking

Free-surface flows are gravity dominated, so Froude number similarity governs, not Reynolds.
Under Froude scaling V_r = √L_r, Q_r = L_r^2.5 and t_r = √L_r.

Step-by-step solution

Formula — V_r = √L_r
Substituting
$V_{p} = 1.6 15 = 6.20 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.190 \times 1 5^{2} .5 = 165.6 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 15 = 3.873 \times m o d e l t im e$

Answer: V_p = 6.20 m/s, Q_p = 165.6 m³/s, time ratio 3.87

Why the other options are there

Q_p = 2.85 m³/s (linear scaling)
V_p = 24.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

Example 10

Pump affinity laws for a change in impeller speed — Similitude (5)

A pump running at 1282 rpm delivers 0.150 m³/s at 43 m while absorbing 14.5 kW. The same impeller is run at 2473 rpm. Use the affinity laws to predict discharge, head and power.

Given

N₁ = 1282 rpm, N₂ = 2473 rpm
Q₁ = 0.150 m³/s
H₁ = 43 m
P₁ = 14.5 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
$N_{2} / N_{1} = 2473/1282 = 1.9290$
Formula
$\frac{Q _{2}}{Q _{1}} = \frac{N _{2}}{N _{1}}$
Substituting
$Q_{2} = 0.150 (1.9290) = 0.289 m^{3} / s$
Formula
$\frac{H _{2}}{H _{1}} = (\frac{N _{2}}{N _{1}})^{2}$
Substituting
$H_{2} = 43 (1.9290)^{2} = 160.0 m$
Formula
$\frac{P _{2}}{P _{1}} = (\frac{N _{2}}{N _{1}})^{3}$
Substituting
$P_{2} = 14.5 (1.9290)^{3} = 104.1 k W$

Answer: Q₂ = 0.289 m³/s, H₂ = 160.0 m, P₂ = 104.1 kW

Why the other options are there

H₂ = 82.9 m (linear scaling of head)
P₂ = 54.0 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Similitude

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

Similitude contains 33 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.

Similitude

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

=Needs technical review

=Needs technical review

=Needs technical review

=Needs technical review

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=Rep=Rem

=Needs technical review

=Needs technical review

=Needs technical review

=Frp=Frm

====Cap=Cam

====Wep=Wem

σp

σm

FI=inertiaforce

FP=pressureforce

FV=viscousforce

FG=gravityforce

FE=elasticforce

FT=surfacetensionforce

Re=Reynoldsnumber

We=Webernumber

Ca=Cauchynumber

Fr=Froudenumber

l=characteristiclength

v=velocity

ρ=density

σ=surfacetension

Ev=bulkmodulus

µ=dynamicviscosity

P=pressure

g=accelerationofgravity

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section