Dimensional Analysis

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

0 formulas

10 exam-style examples

~45 min

All Fluid Mechanics lectures

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

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

$Length scale L_r = 1:15$
$V_m = 2.3 m/s$
$Q_m = 0.090 m^{3}/s$

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
$V_p = 2.3\sqrt15 = 8.91 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.090 \times 15^2.5 = 78.43 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt15 = 3.873 \times model time$

Answer:

V_p = 8.91 m/s, Q_p = 78.4 m^{3}/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

$Length scale L_r = 1:15$
$V_m = 1.2 m/s$
$Q_m = 0.070 m^{3}/s$

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
$V_p = 1.2\sqrt15 = 4.65 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.070 \times 15^2.5 = 61.00 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt15 = 3.873 \times model time$

Answer:

V_p = 4.65 m/s, Q_p = 61.0 m^{3}/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

$Length scale L_r = 1:10$
$V_m = 1.5 m/s$
$Q_m = 0.010 m^{3}/s$

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
$V_p = 1.5\sqrt10 = 4.74 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.010 \times 10^2.5 = 3.16 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt10 = 3.162 \times model time$

Answer:

V_p = 4.74 m/s, Q_p = 3.2 m^{3}/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

$Length scale L_r = 1:15$
$V_m = 1.4 m/s$
$Q_m = 0.190 m^{3}/s$

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
$V_p = 1.4\sqrt15 = 5.42 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.190 \times 15^2.5 = 165.6 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt15 = 3.873 \times model time$

Answer:

V_p = 5.42 m/s, Q_p = 165.6 m^{3}/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

$Length scale L_r = 1:15$
$V_m = 0.5 m/s$
$Q_m = 0.030 m^{3}/s$

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
$V_p = 0.5\sqrt15 = 1.94 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.030 \times 15^2.5 = 26.14 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt15 = 3.873 \times model time$

Answer:

V_p = 1.94 m/s, Q_p = 26.1 m^{3}/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

$Length scale L_r = 1:10$
$V_m = 1.9 m/s$
$Q_m = 0.130 m^{3}/s$

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
$V_p = 1.9\sqrt10 = 6.01 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.130 \times 10^2.5 = 41.11 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt10 = 3.162 \times model time$

Answer:

V_p = 6.01 m/s, Q_p = 41.1 m^{3}/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

$Length scale L_r = 1:15$
$V_m = 1.8 m/s$
$Q_m = 0.170 m^{3}/s$

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
$V_p = 1.8\sqrt15 = 6.97 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.170 \times 15^2.5 = 148.1 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt15 = 3.873 \times model time$

Answer:

V_p = 6.97 m/s, Q_p = 148.1 m^{3}/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

$Length scale L_r = 1:20$
$V_m = 1.3 m/s$
$Q_m = 0.050 m^{3}/s$

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
$V_p = 1.3\sqrt20 = 5.81 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.050 \times 20^2.5 = 89.44 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt20 = 4.472 \times model time$

Answer:

V_p = 5.81 m/s, Q_p = 89.4 m^{3}/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

$Length scale L_r = 1:10$
$V_m = 0.9 m/s$
$Q_m = 0.090 m^{3}/s$

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
$V_p = 0.9\sqrt10 = 2.85 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.090 \times 10^2.5 = 28.46 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt10 = 3.162 \times model time$

Answer:

V_p = 2.85 m/s, Q_p = 28.5 m^{3}/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

$Length scale L_r = 1:20$
$V_m = 1.3 m/s$
$Q_m = 0.010 m^{3}/s$

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
$V_p = 1.3\sqrt20 = 5.81 m/s$
Formula
$Q_r = L_r^{5/2}$
Substituting
$Q_p = 0.010 \times 20^2.5 = 17.89 m^{3}/s$
Formula — t_r = √L_r
Substituting
$t_p = \sqrt20 = 4.472 \times model time$

Answer:

V_p = 5.81 m/s, Q_p = 17.9 m^{3}/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