We were discussing the basic concept of streamline and equipotential line and dimensional
homogeneity in the subject of fluid mechanics, in our recent
posts.

Now we will go ahead to understand the basic concept of Buckingham π theorem with the help of this post.

Now we will go ahead to understand the basic concept of Buckingham π theorem with the help of this post.

### Buckingham π theorem

In an equation, if the variables are more than the numbers of fundamental dimensions i.e. M, L and T. The Rayleigh’s method of dimensional analysis will be more laborious and this problem was resolved by one theorem or concept and that theorem, as stated below, was termed as Buckingham π theorem.

According to Buckingham π theorem, if there are n variables (Independent and dependent variables) in a physical phenomenon and if these variables contain m fundamental dimensions i.e. M, L and T. Then the variables are arranged in to (n-m) dimensionless terms and each term will be termed as π term.

Let X

_{1}, X

_{2}, X

_{3}….X

_{n}are the variables involved in a physical problem. Let us think that X

_{1}is the dependent variable and X

_{2}, X

_{3}….X

_{n}are the independent variables on which X

_{1}will be dependent.

We can also say that X

X

_{1}will be a function of X_{2}, X_{3}….X_{n}and mathematically we can write as mentioned here.X

_{1}= f(X_{2}, X_{3}….X_{n})### Example

Let us discuss one example here to
understand the concept of Buckingham π theorem.

The power required by an agitator in a tank is a function of following variables as mentioned here.

Diameter of the agitator (D)

Number of the rotations of the impeller per unit time (N)

Viscosity of liquid (µ)

Density of liquid (ρ)

We will secure here one relation between power required by agitator and above mentioned four variables by using the concept of Buckingham π theorem.

There are total five variables here. Power (P) is dependent variable and rest four variables (D, N, µ and ρ) are independent variables. Power (P) will be dependent over the above mentioned four variables.

Number of variables = 5

Number of fundamental dimensions = 3

Number of dimensionless groups = 5-3 = 2

We will select here the variables so as to represent the dimensions, let us select N, D and ρ.

N = [T

^{-1}]
T = [N

^{-1}]
D = [L]

L = [D]

ρ = [ML

^{-3}]
M = ρ [L

For the other variables,

^{3}] = ρ [D^{3}]For the other variables,

###
*Dimension of the power P will be [ML*^{2}T^{-3}]

^{2}T

^{-3}]

Therefore P M

^{-1}L^{-2}T^{3}will be dimensionless
Therefore the П

_{1}term will be given as mentioned here
П

_{1 }= P M^{-1}L^{-2}T^{3}
П

_{1 }= P ρ^{-1}D^{-3}D^{-2}N^{-3}
П

_{1 }= P ρ^{-1}D^{-5}N^{-3}
П

_{1 }= P / (ρ D^{5}N^{3})###
*Dimension of the viscosity μ will be [ML*^{-1}T^{-1}]

^{-1}T

^{-1}]

Therefore μ will be μ [M

^{-1}LT] will be dimensionless term
Therefore the П

_{2}term will be given as mentioned here
П

_{2 }= μ [M^{-1}LT] = μ ρ^{-1}D^{-3}DN^{-1}
П

So, we have determined the relation between the variables with the help of Buckingham π theorem.

_{2 }= μ / (ρ D^{2}N)So, we have determined the relation between the variables with the help of Buckingham π theorem.

We will see another important topic in the field of fluid mechanics i.e. differentiate between model and prototype with the help of our next post.

Do you have any suggestions? Please write in comment box.

### Reference:

Fluid mechanics, By R. K. Bansal