Dimensional analysis

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Every relation between n physical variables can be written in a an equivalent nondimensional form,

and the number of the dimensionless variables of this new relation is always smaller than the original one.

What's the point?

Reduction of the number of variables

Engineers often have to make experiments, because the governing equations of fluid flow are impossible to solve analytically in many important applications. The outcome of an experiment depends on the physical properties of the fluid, such as viscosity, density ond so on. To fully understand how a variation of these properties will effect the outcome, one might think that many experiments with different values for each of these parameters would have to be made. Let's say, for example, we want to measure the drag on a smooth sphere in an incompressible fluid, e.g. in a large and deep river in which the fluid flows over the sphere with a certain velocity. The drag will depend on the diameter of the sphere, the velocity of the fluid, the density of the fluid and the its viscosity. If we hold three parameters fixed and see how the result changes for 10 different values of the fourth parameter (e.g. velocity) , we would conclude in what way an alteration of velocity affects the drag, given the certain values for the rest of the variables. But what if we changed the diameter of the sphere to a different value now? Would we have to do ten more experiments for finding the drag in function of the velocity with the new sphere? Considering the time we have to spend for conducting a single experiment, and the number of possible combinations of the four parameters (10^4 = 10000), we would maybe regret our choice of becoming experimental physicists,-). The good thing about dimensional analysis is that it shows us that we don't need to try out every possible combination of parameters.

Transforming data obtained from models to the prototype

Dimensional anlysis tells us how to build a model so that we can convert the data from the experiment with the model to the actual prototype.

How do we obtain the dimensionless Parameters?

Every relation between n dimensional variables can be expressed in dimensionless form:

$f\left(x_{1},x_{2}\ldots x_{n}\right)$

can be expressed in dimensionless form:

$g\left(\pi_{1},\pi_{2}\ldots \pi_{k}\right)$

Let's consider the example from above:

We will state that that the drag force is a function of diameter, velocity, density and viscosity:

$F=f\left(D,U_{\infty},\rho, \mu \right)$

We have five variables: $n=5$ .

Diameter of the sphere: $D$
Velocity of the river: $U_{\infty}$
Density of the Medium: $\rho$
Viscosity of the fluid: $\mu$

Now we find the corresponding dimensions to each variable:

$\begin{Bmatrix} F & D & U_{\infty} & \rho& \mu \\ MLT^{-2} & L & LT^{-1} & ML^{-3} & ML^{-1}T^{-1}\end{Bmatrix}$

Where $M$ is mass, $L$ is length, $T$ is time and $\Theta$ is temperature (in this problem, we don't consider temperature.

As mentioned before, the advantage of the dimensionless form is that the number of variables is always less. The Pi-Theorem show us how many variables will be left:

$k=n-rank$

where $rank$ is the rank of the Matrix $c_{ij}$ :

	M	L	T	$\Theta$
$x_{1}$	$c_{11}$	$c_{12}$	$c_{13}$	$c_{14}$
$x_{1}$	$c_{21}$	$c_{22}$	$c_{23}$	$c_{24}$
		...
$x_{1}$	$c_{n1}$	$c_{n2}$	$c_{n3}$	$c_{n4}$

In our case, this is:

	M	L	T	$\Theta$
$F$	$1$	$1$	$-1$	$0$
$D$	$0$	$1$	$0$	$0$
$U_{\infty}$	$0$	$1$	$-1$	$0$
$\rho$	$1$	$-3$	$0$	$0$
$\mu$	$1$	$-1$	$-1$	$0$

The rank of this matrix is three, so $k=2$ .

This means that we will have 2 dimensionless Parameters (also called "pi products") in our dimensionless relation. We now have to choose 5-2=3 variables that will appear (with a certain exponent which can sometimes be 0) in both of the pi products. The important thing is that these variables thrown together contain all of the basic dimensions used (MLT) and that we cannot build a dimesionless parameter out of them. Let's see what choices we can make:

$FLU_{\infty}$

They contain obviously all parameters, but only $F$ contains $M$ , so

we cannot multiply them to get a dimensionless parameter. Thus, we could choose them to appear in all

pi products.

$LU_{\infty}\rho$

That works too.

Other combinations work as well.

The other two variables left will appear each in a different pi product. Since we are interested in expressing the drag force, we choose $F$ to be one of the two variables that will only appear in one of the pi products. In all the books I consulted, the other variable that appears only in one of the pi products is the viscosity, $\mu$ . That means that usually engineers choose $LU\rho$ as the three variables that appear in each pi product. I don't know why this is a good choice, do you? My guess is that if we make this choice, then one of the pi products is (as you will see) $\frac{\rho U_{\infty}L}{\mu}$ , which is called Reynolds number. This dimensionless pi product is very important, and it also appears in the dimensionless form of the Navier-Stokes equation. So we take $LU_{\infty}\rho$ and multiplicate them with $F$ to obtain the first pi product, and then we take $LU_{\infty}\rho$ and build the second pi product together with $\mu$ .

Let's see how the pi products look like: We know that the first one will be made up of $FLU_{\infty}\rho$ and that it needs to be dimensionless, so we have to find the correct exponents for the variables:

$\pi_{1} = FL^{a}U_{\infty}^{b}\rho^{c} =(MLT^{-2})(L)^{a}(LT^{-1})^{b}(ML^{-3})^{c}$

$a+b-3c+1 =$	0
$c+1 =$	0
$-b-2=$	0

Note that if we did everything right, then this system as exactly one solution:

$a=-2 \qquad b=-2 \qquad c=-1$

So

$\pi_{1}=\frac{F}{\rho U_{\infty}^{2}L^{2}}$

If we do the same to get the other pi product:

$\pi_{2}=\frac{\rho U_{\infty} L}{\mu}$

We finally obtained the dimensionless form of the relationship:

$\frac{F}{\rho U_{\infty}^{2}L^{2}}=h\left(\frac{\rho U_{\infty} L}{\mu}\right)$

Imagine we want to make our experiment with water (there are obvious reasons for that: it's cheap and doesn't harm our health), but we are interested in the drag force caused by a much less viscous $\left(\mu_{o}\right)$ fluid with the same density at a certain velocity $u_{o}$ : We could still make the experiment with water, but we would have to increase either the diameter of the sphere or the velocity of the "river" (or the density, but that's complicated or impossible). Doing this, (in this case we chose velocity as an example) we obtain the same Reynolds numbers.