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Saturday, 14 April 2018

STATE BUCKINGHAM PI THEOREM

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. 

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 X1, X2, X3….Xn are the variables involved in a physical problem. Let us think that X1 is the dependent variable and X2, X3….Xn are the independent variables on which X1 will be dependent.

We can also say that X1 will be a function of X2, X3….Xn and mathematically we can write as mentioned here. 

X1 = f(X2, X3….Xn)

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 [L3] = ρ [D3]

For the other variables,

Dimension of the power P will be [ML2T-3]

Therefore P M-1L-2T3 will be dimensionless

Therefore the П1 term will be given as mentioned here

П= P M-1L-2T3
П= P ρ-1 D-3 D-2N-3
П= P ρ-1 D-5 N-3
П= P / (ρ D5 N3)

Dimension of the viscosity μ will be [ML-1T-1]

Therefore μ will be μ [M-1LT] will be dimensionless term

Therefore the П2 term will be given as mentioned here

П= μ [M-1LT] = μ ρ-1 D-3DN-1

П= μ / (ρ D2N) 

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. 

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