# Determining Pi Terms

There are many different methods that can be used to determine Pi terms, however I am only going to discuss the method of repeating variables.  The method of repeating variables is a several step process that will systematically determine the pi term.

###### Step 1: List out all of the variables need for the problem.

This step is the most important step as well as the most difficult.  During this step it is imperative that all variables are included.  As a result, if a variable is missed the dimensional analysis will not be correct.  To determine these variables you will need to apply your current knowledge of the problem as well as current physical laws. These variables can be both dimensional and non-dimensional constants that describe the phenomenon of interest.  However, to minimize the number of variables that need to be included in the experiment it is important to make sure that they are independent of one another.  For example density and specific weight are not independent of each other.  On the other hand, viscosity and density are.

###### Step 2: Express each variable as basic dimensions.

In order to perform a dimensional analysis you will need to express all variables as basic dimensions.  This can be done two different ways.  First, you can express it as mass, $M$, length, $L$, and time $T$.  On the other hand if you take Newton’s second law of motion where $F=MLT^{-2}$, than you can use force, $F$, length, $L$, and time $T$ as the basic dimensions.   For example if one of the variables that you are interested in is density, than the basic dimensions would be $ρ=FL^{-4}T^2$.

###### Step 3:  How many pi terms are required?

In order to do this you will need to apply the Buckingham pi theorem, which states the number of pi terms equals $k-r$.  The variable $k$ represents the number of variables.  On the other hand $r$ represents the number of reference dimensions.   Reference dimension are related to the number of basic dimensions.  As a result, most problems will have three reference dimensions, but it is possible for a problem to have two.

###### Step 4: Select the number of repeating variables.  This number has to equal the number of reference dimensions.

In this step you will combine variable from the list you created in step 1 to form a pi term.  When you do this all of the required reference dimensions must be included within the group of repeating variables.  In addition, the repeating variable must be dimensionally independent of each other.   As a result, the repeating variables cannot be combined to produce a dimensionless product.  The reason for this is because when performing a dimensional analysis we are interested in seeing how one particular variable could influence other variables.

In addition, when you are choosing a repeating variable you need to be aware of what would be considered a dependent variable.  When you are looking at a problem you will generally be interesting in seeing how a particular variable will influence other variables.  In turn, this variable will be a dependent variable.  Due to this fact you want it to only appear in one of the pi terms.  As a result, it cannot be used as a repeating variable.

###### Step 5: Form the pi term.

In this step you will form the pi term.  This is done by multiplying one of the non-repeating variables by the product of the repeating variables.   The repeating variables will be raised to a certain exponential to make the combination dimensionless.  The following mathematical expression will result, $u_iu_1^au_2^bu_3^c$.  In this expression $u_i$ represents the non-repeating variable.  On the hand, $u_1$, $u_2$, and $u_3$ are the repeating variables.

###### Step 6:  Repeat step 5 for the remaining non-repeating variables.

During this step you should end up having the number of pi terms that was determined in step 3.  If you don’t than you made a mistake somewhere.

###### Step 7: Is everything dimensionless?

Next, using basic dimensions you will want to make sure that all of your resulting pi terms are dimensionless.

###### Step 8:  Create an expression for the relationship between the pi terms.

(Eq 1) $Π_1=Φ(Π_2,~Π_3,…,Π_{k-r}$

The above equation is the typical mathematical form that would be used to express the relationship between pi terms.

### Example

Using the steps above, the pi terms will be determined for an incompressible Newtonian fluid that has steady flow through a long smooth-walled, horizontal circular pipe.  During this analysis the pressure drop per unit length will be of interest.

###### Step 1: List out all of the variables need for the problem.

In order to determine the pressure drop per unit length the following variables will be needed.  These variables are the  pipe diameter, $D$, fluid density, $ρ$, fluid viscosity, $μ$, and fluid velocity $v$.  The variables were determined using current knowledge.  As a result, these variables can be expressed in the following mathematical function.

$Δp_l=f(D,~ρ,~μ,~v)$

###### Step 2: Express each variable as basic dimensions.

$Δp_l=FL^{-3}$

$D=L$

$ρ=FL^{-4}T^2$

$μ=FL^{-2}T$

$v=LT^{-1}$

###### Step 3:  How many pi terms are required?

There are 5 variables $k=5$.  In addition, there are 3 required reference dimensions $r=3$.  As a result, $5-3=2$ so there are 2 pi terms required.

###### Step 4: Select the number of repeating variables.  This number has to equal the number of reference dimensions.

Do to the fact that there are 3 reference dimensions we need to select 3 repeating variable.  Since we don’t want to use the dependent variable $Δp_l$ we will have the following variables to choose from; $D$, $ρ$, $μ$, and $v$.  In addition, the variables chosen must be independent of each other.  For this case I am going to choose $D$, $ρ$, and $v$ since they are the simplest dimensionally and independent of each other.

###### Step 5: Form the pi term.

$Π_1 = Δp_lD^av^bρ^c$

The above equation will rewritten as basic dimensions to solve the exponential that will cause this combination of variables to become dimensionless.

$(FL^{-3})(L)^a(LT^{-1})^b(FL^{-4}T^2)^c=F^0L^0T^0$

This equation can be rewritten into three separate equations to solve for the exponents $a$, $b$, and $c$.

$1+c=0$ (for $F$)

$-3+a+b-4c=0$ (for $L$)

$-b+2c=0$ (for $T$)

In turn, this will result the following values; $a=1$, $b=-2$, and $c=-1$.  As a result, the first pi term will be the following.

$Π_1=\frac{Δp_lD}{ρv^2}$

###### Step 6:  Repeat step 5 for the remaining non-repeating variables.

$Π_2=μD^av^bρ^c$

or

$(FL^{-2}T)(L)^a(LT^{-1})^b(FL^{-4}T^2)^c=F^0L^0T^0$

in turn,

$1+c=0$ (for F)

$-2+a+b-4c=0$ (for L)

$1-b+2c=0$ (for T)

From this the values for the exponents are as follows; $a=-1$, $b=-1$, and $c=-1$.  As a result, the second pi term will be as follows.

$Π_2=\frac{μ}{Dvρ}$

###### Step 7: Is everything dimensionless?

$Π_1=\frac{Δp_lD}{ρv^2}$$=\frac{(FL^{-3}(L)}{(FL^{-4}T^2)(LT^{-1})^2}=F^0L^0T^0 Π_2=\frac{μ}{Dvρ}$$=\frac{(FL^{-2}T)}{(L)(LT^{-1})(FL^{-4}T^2}=F^0L^0T^0$

###### Step 8:  Create an expression for the relationship between the pi terms.

$\frac{Δp_lD}{ρv^2}=Φ\left(\frac{μ}{Dvρ}\right)$

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