Dimensional Analysis

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

0 formulas

10 exam-style examples

~45 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Dimensional Analysis 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 dimensional analysis describes physically and when it applies.
State every one of the 0 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. Dimensional Analysis 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: dimensional analysis.
Capstone Studio instructional photograph

Fluid Mechanics — Dimensional Analysis: 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A dimensionally homogeneous equation has the same dimensions on the left and right sides of the equation. Dimensional
analysis involves the development of equations that relate dimensionless groups of variables to describe physical phemona.
Buckingham Pi Theorem: The number of independent dimensionless groups that may be employed to describe a phenomenon
known to involve n variables is equal to the number (n – rr ), where rr is the number of basic dimensions
(e.g., M, L, T) needed to express the variables dimensionally.

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

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 — Dimensional Analysis

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 2.3 m/s and a discharge of 0.090 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:15
V_m = 2.3 m/s
Q_m = 0.090 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
$V_{p} = 2.3 15 = 8.91 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.090 \times 1 5^{2} .5 = 78.43 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 15 = 3.873 \times m o d e l t im e$

Answer: V_p = 8.91 m/s, Q_p = 78.4 m³/s, time ratio 3.87

Why the other options are there

Q_p = 1.35 m³/s (linear scaling)
V_p = 34.50 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 2

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

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 1.2 m/s and a discharge of 0.070 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:15
V_m = 1.2 m/s
Q_m = 0.070 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
$V_{p} = 1.2 15 = 4.65 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.070 \times 1 5^{2} .5 = 61.00 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 15 = 3.873 \times m o d e l t im e$

Answer: V_p = 4.65 m/s, Q_p = 61.0 m³/s, time ratio 3.87

Why the other options are there

Q_p = 1.05 m³/s (linear scaling)
V_p = 18.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 3

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

A spillway is modelled at a 1:10 scale under Froude similitude. The model shows a velocity of 1.5 m/s and a discharge of 0.010 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:10
V_m = 1.5 m/s
Q_m = 0.010 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
$V_{p} = 1.5 10 = 4.74 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.010 \times 1 0^{2} .5 = 3.16 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 10 = 3.162 \times m o d e l t im e$

Answer: V_p = 4.74 m/s, Q_p = 3.2 m³/s, time ratio 3.16

Why the other options are there

Q_p = 0.10 m³/s (linear scaling)
V_p = 15.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 4

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

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 1.4 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 = 1.4 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
$V_{p} = 1.4 15 = 5.42 m / s$
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 = 5.42 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 = 21.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 5

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

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 0.5 m/s and a discharge of 0.030 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:15
V_m = 0.5 m/s
Q_m = 0.030 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
$V_{p} = 0.5 15 = 1.94 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.030 \times 1 5^{2} .5 = 26.14 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 15 = 3.873 \times m o d e l t im e$

Answer: V_p = 1.94 m/s, Q_p = 26.1 m³/s, time ratio 3.87

Why the other options are there

Q_p = 0.45 m³/s (linear scaling)
V_p = 7.50 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 6

Froude-scaled hydraulic model and prototype conversion — Dimensional Analysis (6)

A spillway is modelled at a 1:10 scale under Froude similitude. The model shows a velocity of 1.9 m/s and a discharge of 0.130 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:10
V_m = 1.9 m/s
Q_m = 0.130 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
$V_{p} = 1.9 10 = 6.01 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.130 \times 1 0^{2} .5 = 41.11 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 10 = 3.162 \times m o d e l t im e$

Answer: V_p = 6.01 m/s, Q_p = 41.1 m³/s, time ratio 3.16

Why the other options are there

Q_p = 1.30 m³/s (linear scaling)
V_p = 19.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 7

Froude-scaled hydraulic model and prototype conversion — Dimensional Analysis (7)

A spillway is modelled at a 1:15 scale under Froude similitude. The model shows a velocity of 1.8 m/s and a discharge of 0.170 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:15
V_m = 1.8 m/s
Q_m = 0.170 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
$V_{p} = 1.8 15 = 6.97 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.170 \times 1 5^{2} .5 = 148.1 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 15 = 3.873 \times m o d e l t im e$

Answer: V_p = 6.97 m/s, Q_p = 148.1 m³/s, time ratio 3.87

Why the other options are there

Q_p = 2.55 m³/s (linear scaling)
V_p = 27.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 8

Froude-scaled hydraulic model and prototype conversion — Dimensional Analysis (8)

A spillway is modelled at a 1:20 scale under Froude similitude. The model shows a velocity of 1.3 m/s and a discharge of 0.050 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:20
V_m = 1.3 m/s
Q_m = 0.050 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
$V_{p} = 1.3 20 = 5.81 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.050 \times 2 0^{2} .5 = 89.44 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 20 = 4.472 \times m o d e l t im e$

Answer: V_p = 5.81 m/s, Q_p = 89.4 m³/s, time ratio 4.47

Why the other options are there

Q_p = 1.00 m³/s (linear scaling)
V_p = 26.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 9

Froude-scaled hydraulic model and prototype conversion — Dimensional Analysis (9)

A spillway is modelled at a 1:10 scale under Froude similitude. The model shows a velocity of 0.9 m/s and a discharge of 0.090 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:10
V_m = 0.9 m/s
Q_m = 0.090 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
$V_{p} = 0.9 10 = 2.85 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.090 \times 1 0^{2} .5 = 28.46 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 10 = 3.162 \times m o d e l t im e$

Answer: V_p = 2.85 m/s, Q_p = 28.5 m³/s, time ratio 3.16

Why the other options are there

Q_p = 0.90 m³/s (linear scaling)
V_p = 9.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

Example 10

Froude-scaled hydraulic model and prototype conversion — Dimensional Analysis (10)

A spillway is modelled at a 1:20 scale under Froude similitude. The model shows a velocity of 1.3 m/s and a discharge of 0.010 m³/s. Find the corresponding prototype velocity, discharge and time scale.

Given

Length scale L_r = 1:20
V_m = 1.3 m/s
Q_m = 0.010 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
$V_{p} = 1.3 20 = 5.81 m / s$
Formula
$Q_{r} = L_{r}^{5/2}$
Substituting
$Q_{p} = 0.010 \times 2 0^{2} .5 = 17.89 m^{3} / s$
Formula — t_r = √L_r
Substituting
$t_{p} = 20 = 4.472 \times m o d e l t im e$

Answer: V_p = 5.81 m/s, Q_p = 17.9 m³/s, time ratio 4.47

Why the other options are there

Q_p = 0.20 m³/s (linear scaling)
V_p = 26.00 m/s (velocity scaled linearly)

Reference: FE Reference Handbook — Fluid Mechanics → Dimensional Analysis

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

Dimensional Analysis contains 0 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.