11.5 Dimensional Analysis, Similitude & Model Testing

Why Dimensionless Groups Exist

The drag on a submerged body depends on its size, the fluid density and viscosity, and the velocity — five variables. Investigating that experimentally by varying each in turn would need an impractical number of tests. Dimensional analysis reduces the five variables to two dimensionless groups, so a single curve captures the entire relationship.

This is not merely a labour-saving device. It is also the reason a model tested in a laboratory can predict the behaviour of a full-size machine, and the reason correlations such as the Nusselt-Reynolds-Prandtl relations of heat transfer take the form they do.

Fundamental Dimensions

In the MLT system, every mechanical quantity reduces to mass $M$, length $L$ and time $T$.

Note that torque and energy share dimensions but are physically distinct — dimensional identity does not imply physical equivalence.

Buckingham's Pi Theorem

If a physical problem involves $n$ variables and these contain $m$ fundamental dimensions, the relationship can be expressed in terms of $(n - m)$ independent dimensionless $\Pi$ groups.

Procedure

1. List all $n$ variables and write their dimensions.
2. Determine $m$, the number of fundamental dimensions present.
3. Choose $m$ repeating variables. They must together contain all $m$ dimensions, must not themselves form a dimensionless group, and conventionally include one geometric variable (a length), one kinematic (a velocity) and one dynamic (density). The dependent variable must never be a repeating variable.
4. Form $(n-m)$ groups, each consisting of the repeating variables raised to unknown powers multiplied by one non-repeating variable.
5. Solve for the exponents by requiring each group to be dimensionless.

Worked example: drag force

Drag $F$ depends on velocity $v$, length $l$, density $\rho$ and viscosity $\mu$. Here $n = 5$ and $m = 3$, so there are $5 - 3 = 2$ groups.

Take $l$, $v$ and $\rho$ as repeating variables. The two groups turn out to be

$\Pi_1 = \frac{F}{\rho v^2 l^2}, \qquad \Pi_2 = \frac{\mu}{\rho v l} = \frac{1}{Re}$

So the general result is

$\frac{F}{\rho v^2 l^2} = \phi\!\left(Re\right)$

which, written with the conventional factor of one half and area $A$, is exactly the familiar drag equation

$F = C_D\,\tfrac{1}{2}\rho v^2 A, \qquad C_D = f(Re)$

Dimensional analysis has produced the form of the law without any experiment; experiment is needed only to find the function $f$.

The Principal Dimensionless Numbers

Each number is a ratio of inertia force to some other force, and each governs a different class of flow.

In heat transfer the same logic gives Nusselt, Prandtl, Grashof and Stanton numbers, which is why the two subjects share a methodology.

Critical values worth recalling

Similitude

For a model test to predict prototype behaviour, three conditions must hold in sequence.

1. Geometric similarity. The model is a true scale replica: every length ratio is the same, $L_r = L_m/L_p$. This includes surface roughness, which is often the hardest part to satisfy.

2. Kinematic similarity. Velocity and acceleration at corresponding points have the same ratio and the same direction. Streamline patterns are geometrically similar.

3. Dynamic similarity. Force ratios at corresponding points are equal. In practice this means making the governing dimensionless number equal in model and prototype.

All three are required; geometric similarity alone is not enough, and this hierarchy is a standard objective item.

Which number to match

It is generally impossible to satisfy Reynolds and Froude simultaneously with the same fluid. Froude similarity requires $v_r = \sqrt{L_r}$, while Reynolds similarity requires $v_r = 1/L_r$. For a 1:25 ship model, Froude gives $v_r = 0.2$ but Reynolds would demand $v_r = 25$. Naval architects resolve this by scaling on Froude for wave-making resistance and correcting the viscous friction component analytically — a genuine limitation of model testing rather than a mathematical curiosity.

Scale ratios under Froude similarity

With $L_r$ the length scale, matching Froude numbers gives:

Worked example

A spillway model is built to a scale of 1:36. If the model discharge is 0.5 m$^3$/s, find the prototype discharge.

A spillway has a free surface, so Froude similarity applies and $Q_r = L_r^{5/2}$. With $L_p/L_m = 36$:

$Q_p = Q_m \times 36^{5/2} = 0.5 \times 7776 = 3888 \text{ m}^3/\text{s}$

Worked example: pipe flow

A 1:5 scale model of a mine ventilation duct is tested with the same air. Reynolds similarity requires $\rho v L/\mu$ to match, so with the same fluid

$v_m L_m = v_p L_p \;\Rightarrow\; v_m = 5\,v_p$

The model must be run at five times the prototype velocity — often impractical, which is why such models are frequently tested in water or in pressurised air to raise the density and bring the required velocity down.