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Fluid Dynamics – Week 6: Dimensional Analysis

Hello everybody and welcome back, once again, to another instalment in my fluid dynamics blog! 🙂 Without further ado, let’s get started!

Dimensional Analysis

Dimensional analysis is a means by which we can understand the properties of physical quantities, but instead of looking at their exact units, we can measure them with regards to their actual dimensions.

Physical quantities are typically composed of mass M, length L, and time T in fluid dynamics, but to varying powers. Other dimensions do exist, but we won’t consider them here.

This leads on nicely to…

Dimensional Homogeneity

This is where we examine an equation composed of physical quantities, and determine whether it is valid or not. The validity of the equation itself is based on whether both sides of the equation equate to having the same dimensions.

For example, from the lecture notes, the following equation was given:

L = C\rho Uw

where L = lift force, C = circulation, \rho = density, U = speed, and w = wing length.

We don’t know the dimensions of circulation off the top of our heads, but we can calculate it based on the various dimensions of the physical quantities that we do know.

Therefore, by rearranging the equation above to make C the subject, we get:

C = \dfrac{L}{\rho Uw}

which we can then write in terms of dimensions as:

C = \dfrac{(MLT^{-2})}{(ML^{-3})(LT^{-1})(L)}

Upon inspection of the dimensions, we can see that the mass factor cancels. After simplification of the powers, we can see that circulation must have dimensions of:

C = L^2 T^{-1}

Based on the units of the other physical quantities in the equation,  it would then be possible to deduce the units of the circulation.

Buckingham’s Pi Groups

In the previous example, it was relatively easy to calculate the dimensions, since we already knew the equation, and thus the proportions of components involved, based on their powers. But what if we knew the physical properties of the components, but not their proportions?

This is where the idea of dimensionless groups and Buckingham’s \pi theorems come into play.

Let us first consider the force F on a propeller, as a function of the following:

F = \phi(d, u, \rho, N, \mu)

where d = diameter, u = forward velocity of the propeller, \rho = density, N = rotation rate (in revolutions per second) and finally \mu = viscosity.

This can subsequently be written as:

\phi_1(F, d, u, \rho, N, \mu) = 0

where \phi, \phi_1 are currently unknown functions. Since we can now combine the components of \phi_1 in such a way that they equal zero, we have created our first dimensionless group.

These dimensionless groups are commonly found using Buckingham’s \pi theorems. The two theorems are stated below, as taken directly from the lecture materials:

Buckingham’s first \pi theorem:

‘A relationship between m physical variables can be expressed as a relationship between m-n non-dimensional groups of variables (called \pi groups), where n is the number of fundamental dimensions (such as mass, length and time) required to express the variables.’

Buckingham’s second \pi theorem:

‘Each dimensionless group is a function of n governing or repeating variables plus one of the remaining variables.’

Oh, and one last thing before we start with the example. As the theorems mentioned, we have n repeating variables. For most fluid dynamics cases, we consider \rho, u \text{ and } d as the repeating variables, although we may sometimes also consider the viscosity over the density in certain situations.

From our equation for \phi_1, we can deduce that n (the number of repeating variables) = 3, and that m = 6 (the total number of physical variables). Therefore, from the first theorem, we have 3\pi groups, such that:

\phi_1(\pi_1, \pi_2, \pi_3) = 0

From this, we can write our 3\pi groups in terms of the repeating variables, and one of the non-repeating variables, like so:

\pi_1 = \rho^A u^B d^C F

\pi_2 = \rho^A u^B d^C N

\pi_3 = \rho^A u^B d^C \mu

where the A’s, B’s and C’s are all different in each case.

As these groups are dimensionless, then they must have a dimension of M^0 L^0 T^0, and thus, using dimensional homogeneity, we can write the groups in terms of the powers of their dimensions:

\text{For } \pi_1 : M^0 L^0 T^0 = (ML^{-3})^A (LT^{-1})^B (L)^C MLT^{-2}

Due to the power equality that is evident on both sides, we get:

A = -1, B = -2, \text{ and } C = -2

and thus we get:

\pi_1 = \dfrac{F}{\rho u^2 d^2}

And for completeness, here are the results of the other \pi groups:

\pi_2 = \dfrac{Nd}{u}

\pi_3 = \dfrac{\mu}{\rho ud}

Example Sheet Questions

Below you will find my scanned answers to question 1 of the Dimensional Analysis Example Sheet.

Dimensional Analysis Example Sheet - Part 1

Dimensional Analysis Example Sheet – Part 1

Dimensional Analysis Example Sheet - Part 2

Dimensional Analysis Example Sheet – Part 2

Reflection

I feel that I understand the initial concepts of dimensional analysis, particularly dimensional homogeneity, that are raised here, mostly because I covered this last year briefly as part of the Vector Calculus module.

Although I must say that I have struggled with the concepts of Buckingham’s \pi groups, due to not thoroughly covering the subject material over the course of the consolidation week. This was mostly due to imminent deadlines that I decided to focus on, rather than the self-taught material. However, this did cause problems during the lectures (particularly the quizzes) when I really had no idea what was going on.

Upon going back over this material though, I feel a lot more confident that I understand the material, and I will continue to work on this topic during the revision that lies ahead. I’ve only shown the first question of the Dimensional Analysis here, but it is evidence that I have shown progress in my understanding of the topic. I hope I will be able to further this understanding when I cover the ideas discussed in the lecture notes on similarity during revision.

I’m fairly pleased overall with this blog post, but with the deadline looming, I’m going to carry on working on drastically simplifying future blog posts, as well as improving my older ones marginally.

Thanks once again everybody! And before I forget, have a fantastic Christmas break! 🙂

References and Sources:

1) The vast majority of the concepts talked about in this blog post are derived from notes taken from Chapter 1 of the Inviscid Flow PowerPoints on dimensional analysis, the supplementary material, and my own notes derived from the lectures.