In today’s lesson, we will be discussing the following topics:
1. Buckingham Pi Theorem
2. Example 1 – Using Buckingham Pi Theorem
3. Alternate Method by Ipsen
4. Example 2 – Ipsen’s Method
Video coming soon
Buckingham Pi Theorem
There are many different ways to reduce the number of dimensional variables in an equation (nondimensionalize the equations). To do so, we create a smaller number of dimensionless groups. The theorem that allows us to do this is called the Buckingham Pi (π) Theorem. It can also be called the repeating variable method of dimensional analysis.
There are 6 steps to this theorem:
1) List and count the number of variables in the problem (n).
2) List the dimensions of each variable according to the {MLTΘ} or {FLTΘ} system for base units. The MLTΘ system uses Mass (M), Length (L), Time (T) and Temperature (Θ). The FLTΘ system uses Force instead of Mass. A list is given below with different variables and their base unit of either M, L, T, Θ or F, L, T, Θ for both systems.
3) Take an initial guess as to the number of different dimensions present. Look for the number of variables (j) that do not form a pi product. Reduce j by 1 and try again if this number does not work.
4) Select j scaling parameters (aka repeating variables) that do not form a pi product. All the pi groups will contain these j variables in different groups, so pick something like density, velocity or length. Don’t choose something obscure, such as surface tension.
5) Add one variable to your j value. Form a power product with this. Find the number of exponents that make the product dimensionless. Tip: Arrange for your output or dependant variable (eg. force, pressure, drop, torque, power, etc) to appear in the numerator. The graphs look better this way. Do this sequentially, one variable at a time, and you will find all the desired pi products.
6) Write out the final dimensionless function, and check the terms to make sure all pi groups are dimensionless.
Let’s do an example and it’ll make a lot more sense!
Example 1 – Using Buckingham Pi Theorem
1) How many variables is the force F a function of? There are 4: L, V, p, u.
F is another variable, so there are 5. Therefore, n = 5.
2) Write out the basic dimensions of each variable using the MLTΘ system. This means convert each of the dimensions to either a mass (M), length (L), time (T), or theta (Θ).
3) Find j. How many dimensions do we have? We can have up to 4: M, L, T and Θ. We don’t have anything with dimensions of Θ, so when we find j, it will be less than or equal to 3. This is because there are 3 dimensions being used here: M, L, T. Therefore, j = 3.
4) Use the equation n – j = k to find k. n = 5 and j = 3, so 5-3 = 2. This is where the Buckingham Pi Theorem comes into play, because it indicates that there are (n-j) pi terms. A pi term is a set of numbers that cancel out to become dimensionless.
Therefore, k = 2. The k value indicates the number of independent dimensionless groups. Here, there will be 2 groups. A dimensionless group is a simply a group of numbers where all the units cancel out, to make it dimensionless. For example, the Reynolds Number is a well-known dimensionless group.
5) Now since j = 3, we are going to choose a set of 3 repeating variables from the 5 total variables we had in the beginning. A set of repeating variable are those which will always have dimensions. For example, if you have velocity (m/s) and time (s) then the meters (m) will never cancel out when you multiply the two, thus these variables would form a set of repeating variables because they would never be dimensionless. We are using L, U, and p. Here I have multiplied these variables through by their base units of M, L, T to confirm that they do indeed form a set of repeating variables. You can see that the units do not all cancel out.
Note: Do not choose the dependent variable (in this case, F) to be a repeating variable when you do these problems.
6) Combine these with one more variable, to find two pi products, since k = 2. To form a pi term, we multiply one of the non-repeating variables by the product of the repeating variables. The product of the repeating variables was M/LT which is what we ended up in the image above with when we multiplied L, U, and p together.
We have two non-repeating variables left – F and u (u as in viscosity – do not confuse with the large U that we used for velocity in the repeating variables). We will multiply both of these by the non-repeating variables.
7) The variables need to be raised to certain exponents (either positive or negative) to make it all dimensionless to form a pi group/pi product. A pi group, or pi product, is when the repeating variables are multiplied in some way by the non-repeating variables to make it dimensionless.
Because you can change the exponents on these variables, you should be able to do this twice because k = 2, to form two pi products – one with each F and u. Let’s write out the pi group equations. I have made two equations, with L, U, and p in both (since these are the repeating variables). Each non-repeating variable, F and u, are multiplied in each equation. This forms two pi groups/pi products, since k = 2 which we calculated earlier. I have given random variables to the exponents, because we need to figure out what numbers they should be.
8) We now need to solve for the unknown exponent variables. The dependent variable usually gets an exponent of 1. In the original equation, all the variables were a function of force F. So F is the dependent variable, and the rest of the variables are independent.
In the first equation that contains F, it will have an exponent of 1. You will have to take a guess as to what the other exponents will have to be in order to cancel out F. I have rewritten the first pi equation, but using the base units of MLT rather than the variables. Now we can see what a, b and c need to be in order to cancel out the ML/T^2 units that came from F.
We definitely need two T’s in the numerator to cancel out the T^2 in the denominator, so L/T must have an exponent of two, to get that squared onto the T as well. We will also need to flip the fraction, so the T is on the top. This means the exponent will be -2.
For the M in numerator of ML/T^2, we need an M somewhere else in the denominator to cancel it out. This means that the M/L^3 term will have to have an exponent of -1, because there is only one M in the numerator, so we need one in the denominator to cancel out. Rewriting all this (and keeping the exponent on the L as a 1), we get the equation on the right side:
So we still have too many L’s. We have an L^3 left in the numerator.
The only way to cancel this out without affecting any other variables is to change the exponent on the L to a -2. This way, it ends up going to the denominator and canceling out with the other L’s.
So the final equation for π1 will have the following exponents. You can rewrite all the base units using the MLT system out, and double check that they all cancel out to verify this.
You can also do this algebraically, as I will show you how to do for the π2 equation. Because viscosity is not the dependent variable like force was, we can’t confirm that the exponent should be a 1 on it. So it is easier to solve for the exponents algebraically, since we can’t confirm any of the exponents yet.
Select a random exponent for the viscosity (the non-repeating variable). I chose -1. You can only make three equations – one for each length, mass and time. Note that the coefficients (eg. the 3 in the 3g in the length equation) come from the exponents on the MLT. In density, the L has an exponent of 3, so when I used the variable g as the density exponent, the g had a 3 in front of it.
Now we can only have three equations. If we left the viscosity exponent as an unknown, we wouldn’t be able to solve the system of equations with 4 variables and 3 equations. Because the exponent is -1, the fraction flips. L and T will be positive, and M will be negative. I have added the appropriate +1 and -1 to the end of each equation. Set all the equations equal to 0, because you want it to be dimensionless. Solving for all the variables algebraically, and substituting them back into the π2 equation gives:
Now we can equation the Pi1 and Pi2 equations. This gives us the original equation from the beginning of the question. Recall this was the original equation that we needed to get to.
Alternate Method by Ipsen & Example 2
There is another way to do dimensional analysis, so let us cover this method as well. Here, we find all the pi groups at once. Let’s go through an example so we can see how it is done.
To rewrite the function in dimensionless form, we first write out the dependent variable (moment) as a function of its independent variables. We write out the base units using MLT for all of these in red. The variables alpha and k are dimensionless, so there are no units. Alpha is an angle (MLT units don’t apply), and the specific heat ratio is a ratio which divides two specific heats (you can Google this if you’re curious), therefore leaving no units here either. You can label these as dimensions of 1.
In blue, I have added the general units used, for example velocity uses base units of length/time (L/T) in red, but in blue I have put meters/second (m/s) because that’s the length and time that is generally used. This can help you see where I got the MLT units from.
1) Start with mass M and choose a variable that contains M. Let’s choose density. Divide all variables that include dimensions of M, by M.
2) Now we have to eliminate the variable we chose, which was density. We do this by multiplying density (meaning, units of L^3 onto the left side of the equation), since that is the remaining unit for the density after we divided out the M. Now the density disappears on the right side, and the new L^3 appears in the numerator of the units on the left side.
3) Now we will eliminate T. We have to choose a variable that contains T in it first. We will choose a. Bring the a to the left side by dividing out L/T. Dividing out the units of density and a gives the following answer.
But we solely have to use the variable a to get rid of the dimension T on the left side. We still have L^3/T left when we clean up the units, so we need another T multiplied on the left to remove the T from the denominator there.
4) To get another T on the left side of the equation, we can change the exponents on the variables we move over from the right. Let’s put a squared on the a. We already divided by L/T when we divided the a on the left once, so let’s divide by another L/T. This gives the following:
5) We can’t forget to divide a^2 units (L^2/T^2) by any variables on the right side that contain T, since we are trying to eliminate T. When a was present on the right side, we tried to move it to the left by dividing by an L^2/T^2. On the left side as well, we need to divide by an a^2. The one a that was there cancelled out, but there would still be an a left on the right side denominator. This combined with the velocity to give V/a, which makes it dimensionless since both V and a have the same units of L/T.
6) To eliminate L now, we can use the C on the right side. We still have units of L^3 on the left side, so to get rid of that we will use units of L^3 from the right side. This means we need to put an exponent of 3 (cubed) on the C. C^3 will give units of L^3. Then we will again have to divide C^3 by all the L-containing terms on the right side, of which there is only C. This is what we are left with, and all the units on both the left and right are dimensionless:
7) V/a is a quantity known the Mach Number (Ma), and the term on the left is known as the Moment Coefficient (Cm). We can replace these (it isn’t necessary though – your final answer could have been at Step 6). After replacing, this is the final answer with 4 pi groups. Three on the right side, one on the left. Each group (or single) variable is dimensionless. This shows the the Moment Coefficient is a function of the variables on the right side.
