Dimensional Analysis, Similarity and Scaling Parameters

In today’s lesson, we will be discussing the following topics:

1. Dimensional Homogeneity

2. Definitions (Similarity, Dimensional Variables, Dimensional Constants, Pure Constants)

3. Scaling Parameters

4. Example 1: Using similarity

5. Example 2: Selecting and using scaling parameters

Video coming soon

Dimensional Homogeneity

First, what is Dimensional Analysis?

Dimensional analysis is a way that we can reduce the number of variables we are working with. For example, if you have a situation in which there are 10 variables affecting the outcome, using dimensional analysis can reduce that number to less than 10, making it easier to solve and/or analyze some of the variables in the situation. This is a way to reduce time and money used in figuring out which variables are affecting a specific situation. Dimensional analysis also gives us scaling laws, which can help to convert information about a small model into a more expensive prototype. This definitely helps in knowing how certain variables affect a design (eg. to know whether a million dollar airplane has enough lift force, you can test out the variables on a smaller model and use scaling parameters to convert the information for use with the actual plane).

Using dimensional analysis will reduce the problem to k variables. n is the number of variables you started with. Reducing it to k variables means you have reduced (n-k) variables. (n-k) is the number of different dimensions. These can also be called the basic, primary or fundamental dimensions. See the example below with numbers:

See how dimensional analysis can help reduce the parameters down, like this example. Here, we have reduced the equation down into just g*Re, where Re is the Reynolds Number. This is easier than having to deal with many different variables, like F, p, L, and V in the left most equation:

Now when we say basic, primary or fundamental dimensions, we are specifically referring to four basic dimensions in fluid mechanics. It uses Mass (M), Length (L), Time (T) and Temperature (Θ). The name of this system is:

You may also see FLTΘ, where F stands for Force instead of Mass, but we will use the MLTΘ system for now. The purpose of these systems is to help reduce n number of variables in a problem to group them to a dimensionless form.

Before we learn the steps, take a look at the following example.

This image above shows the units of a Newton – 1 kg*m / s^2. If you convert the kg (mass), meters (length) and seconds (time) into their basic units, it becomes the equation on the right side. We will do examples on how exactly to use dimensional analysis to reduce down the variables in an equation in the following lessons.

Principle of Dimensional Homogeneity

Now, what is Dimensional Homogeneity? First, the Principle of Dimensional Homogeneity is a rule that states the following:

This just means that if you were to add up the different terms in an equation, they would all have the same units so the left and right sides of the equation can equal one another! You’ve probably even used this method on a test before if you weren’t given equations – making sure the units of a term match up to the units of the answer to make sure you have the right equation.

Here is an example of an equation that is dimensionally homogeneous:

All the terms add up to the same basic units, as you will see in the following images. Let’s change these variables into their basic units using the MLTΘ system.

Definitions

Now let’s get onto some definitions you’ll need to know:

Similarity: Dimensional Similarity can come in different forms. Geometric similarity, Kinematic similarity or Dynamic similarity. This is a way to scale up a model into a prototype design, because performing dimensional analysis on a process can show what variables are affecting the system.

Dimensional Variables: These are variables that have dimensions (eg. density, velocity, time) and are varying. They can be nondimensionalized (have no dimensions) using the dimensional analysis technique. For example, in the equation below, S and t are the dimensional variables because they are not constants:

Dimensional Constants: These are variables that are constant. They are normally used to help nondimensionalize the variables in the problem. They also have dimensions. For example, in the equation below, they are So, Vo, and g.

Pure Constants: Pure constants have no dimensions and never did. They can come about through mathematical manipulations, such as an exponent that ends up on a variable after integrating it. Pi, e, and other mathematical constants, as well as the arguments (the values that go into) the functions ln, exp, cos, or other mathematical functions are also pure constants.

So, we didn’t go through the process of how an equation can be nondimensionalized. This process requires the use of scaling parameters. Let’s discuss those first before jumping into an example of nondimensionalizing an equation.

Scaling Parameters

Scaling parameters (also known as repeating variables or scaling variables) help to non-dimensionalize equations (they are used to define dimensionless variables). These are selected based on the dimensions in the original equation. After using scaling parameters, we will only be left with the basic parameters whose effect in the equation is significant.

Advantages to non-dimensionalizing equations include:

(a) Relationships between different parameters in the equation are identified.

(b) The number of parameters are fewer as compared to a dimensionalized equation.

To select scaling parameters, we look at the primary dimensions from the equation we are nondimensionalizing.

There are some common scaling parameters that are used:

• Characteristic Length (L)
• Characteristic Velocity (V)
• Density
• Viscosity
• Gravitational Acceleration
• Reference Pressure Difference (drawn below)

Here are the rules on how to select a scaling parameters/variables:

1) They must not form a dimensionless group among themselves, but adding one more variable will make it dimensionless.

2) Output variables should not be selected. For example, if you have an equation y = x^2, you have isolated y and that is the output, so should not be selected as the scaling parameter.

3) Choose common variables as the scaling parameters so they will appear in all the dimensionless groups. For example, choose density, length or velocity, rather than surface tension, surface roughness or speed of sound.

Example 1: Using similarity

1) We first need to refer to the equation in the question (5.2). This is given below from the textbook. The equation circled in black is equal to a number known as Re, the Reynolds Number. The Force Coefficient, Cf, is also given (as dependent on Re).

We have to assume that the small model and scaled/larger model have achieved similarity. This means that their Reynolds Number and Force Coefficient is the same, and that we can set the Reynolds Number equation for each model equal to one another to solve for the drag force and velocity of the original/not scaled model, which is what we are asked to find.

2) This equation uses p (density), V (velocity), L (length), and u (viscosity). This equation will be used for both the model, and the scaled model (which is 100x larger than the model, as mentioned in the question). The small model is in water, and the larger one is in glycerin. We know the temperature is 20C, so we can find the properties (viscosity and density) of these fluids at 20C. You can use tables given in the textbook appendix or you can find these values online.

The textbook example uses a table given in the book, to get the following properties:

3) Using the knowledge that we can make three equations based on the equation given in Step 1, we will do just that. The original model is labeled with an m, and the scaled model is labeled with an s. Because the Reynolds Numbers of both equations are equal, then the left side of the first equation must also be equal. That’s the part of the equation I am using, but we are also given the Force Coefficient and the Reynolds Number equations here (hence three possible equations):

Now I have substituted in some of the given values. I have also done the proper conversions. Force is in Newtons, which means other units must be in kilograms (mass), meters (length), and seconds (time). The length of the scaled model was not explicitly given as 100mm, but it was told the scale is 100 times larger, which means the length is 100 times larger than the original 1mm.

Rewriting the above equation after substituting in the numbers and substituting in the velocities (and converting to appropriate units) gives the equation below.

4) Let’s set the Reynolds Number equations equal to one another for the original and the scaled models. I have substituted in the values we knew from earlier, including the viscosity we got from the textbook:

5) Solve for the velocity of the original model, Vm. This is one of the final answers.

6) Use this velocity in the first equation to find the force:

7) The final answer for the drag force is Fm =

Example 2: Selecting and using scaling parameters

Suppose we are given the equation below. How many variables and parameters do we have here? S and t are considered the variables (like x and y) and Vo, So, and g are the parameters (they would be given as constants in a question).

To study the effect of a specific variable on the output (S) and non-dimensionalize the equation, we need to see how many dimensions are in the equation overall, using MLTΘ. None of these parameters are using mass, because this is a kinematics equation that does not use mass (you would be told which variables use which units in your specific question though).

Let’s take a look at the base dimensions of each variable here in red. The labels in blue are the general units that are used (for example, Vo is velocity, which is usually meters/second (m/s) and the base unit is Length/Time, L/T):

So we only have dimensions of T and L here. Because we only have two dimensions, we select two of the three parameters in this equation (So, Vo or g) to use as scaling parameters. The third parameter we don’t choose is the one that we will be studying the effect of on the output.

We have three choices:

(a) Study the effect of g. So and Vo are the scaling parameters.

(b) Study the effect of So. Vo and g are the scaling parameters.

(c) Study the effect of Vo. So and g are the scaling parameters.

Let’s say we choose to study the effect of g. Then So and Vo are our scaling parameters. To use these scaling parameters, we will non-dimensionalize the two variables S and t. To nondimensionalize means we need to cancel out the units in each variable, using S and t themselves, as well as including the scaling parameters we chose of So and Vo. We will define the nondimensionalized variables with an asterisk, so we will find S* and t*.

The reason I chose the equations I did for S* and t* are because we need to use S, t, So, and Vo in the equations (the original variables and the scaling parameters we chose). Using all four of these, we need to figure out a way to cancel out the dimensions so that S* and t* become dimensionless. Notice how I have crossed out the dimensions in red, after multiplying and dividing through where it needed to be simplified. All the units of both equations cancel out in the numerator and denominator of each equation, so these are the correct equations for the scaling parameters.

Where can this be used? You can substitute S* and t* back into the original equation, like so. Divide the equation through by So (in blue) to start making all terms dimensionless. Now instead of t in the last term, I will replace it with a t*. Remember S and t are the “x and y” of the equation, so you do have to replace the t in the last term with a t*.

If we rearrange the equation we used earlier for t* in order to get t, we get:

Substituting this into the equation, we get:

Simplifying it, we get the following equation. We want to make sure that the entire equation has no S or t, only S* and t*. Now I have removed t* (because we need to use that variable) and 1/2 from the last term in the equation.

The remaining values that had dimensions were put together. These included g, So and Vo^2. The given equation is in green below. I have grouped them into a bracket, and labeled it as the greek letter alpha in the picture. This dimensionless parameter alpha shows the effect of gravity (g), which is what we wanted.

Substituting alpha in gives the final equation:

You could then also graph S* versus t* on a graph. S* on the y-axis, t* on the x-axis. The graph would look like the following: