Buckingham Pi theorem
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Buckingham Pi-theorem
Buckingham Pi-theorem states that, “If there are n variables (dependent and independent variables) in a dimensionally homogenous equation and if these variable contain m fundamental dimensionless terms are called Π terms.”
Step 1: Mathematically, if any variable X1 depends on independent variables X2,X3,X4,.......,Xn’. Then, the functional relationship between dependent and independent variable is expressed as
X1 = f (X2,X3.X4,.........,Xn)
Equation can be written as,
f1 (X1,X2,X3,X4,..........Xn) ∫ = 0
It is dimensionally homogenous equation and contains n variables.
Step 2: The equation can be written in terms of number of dimensionless Π term which is equal to (n-m).
f (Π1, Π2,......Πn ' m) = 0
Step 3: Each Π-term contains m + 1 variables, where m is the number of fundamental dimensions and it is also called as repeating variable.
Step 4: If X2,X3,X4 are repeating variables, then each Π term is written as:
Π1 = Xa1 . Xb1 . Xc1. X
Π2 = X2a2 . X3b2 . X4c2. X5
Πn-m = X1a n-m X2b n-m X3c n-m . X n-m
Where X4,X5..........Xn-m are non repeating variables.
Step 5: Where equation is solved by principle of dimensional homogeneity and the value of a, b, c are obtained.
Step 6: Substituting the value of a, b, c in their corresponding Π1, Π2, Π3,......etc.
Step 7: The value of Π1, Π2, Π3.........etc. are substitute in Equation (5.4.3)
Step 8: The required expression can obtained by expressing any one of the Π terms as a function of others.
Π1 = Φ [ Π2, Π3,........Πn-m ]
Step 1: Mathematically, if any variable X1 depends on independent variables X2,X3,X4,.......,Xn’. Then, the functional relationship between dependent and independent variable is expressed as
X1 = f (X2,X3.X4,.........,Xn)
Equation can be written as,
f1 (X1,X2,X3,X4,..........Xn) ∫ = 0
It is dimensionally homogenous equation and contains n variables.
Step 2: The equation can be written in terms of number of dimensionless Π term which is equal to (n-m).
f (Π1, Π2,......Πn ' m) = 0
Step 3: Each Π-term contains m + 1 variables, where m is the number of fundamental dimensions and it is also called as repeating variable.
Step 4: If X2,X3,X4 are repeating variables, then each Π term is written as:
Π1 = Xa1 . Xb1 . Xc1. X
Π2 = X2a2 . X3b2 . X4c2. X5
Πn-m = X1a n-m X2b n-m X3c n-m . X n-m
Where X4,X5..........Xn-m are non repeating variables.
Step 5: Where equation is solved by principle of dimensional homogeneity and the value of a, b, c are obtained.
Step 6: Substituting the value of a, b, c in their corresponding Π1, Π2, Π3,......etc.
Step 7: The value of Π1, Π2, Π3.........etc. are substitute in Equation (5.4.3)
Step 8: The required expression can obtained by expressing any one of the Π terms as a function of others.
Π1 = Φ [ Π2, Π3,........Πn-m ]
Selection of Repeating Variables:
The following points should be considered for selecting the repeated variables:
1. The variables should not be dimensionless.
2. No two variables should have same dimensions.
3. As far as possible, the dependant variables should not be selected as repeated variables.
4. The repeated variables is selected in such a way that is contains one from each.
(i) Geometric properties i.e. length, diameter, height etc.
(ii) Flow properties i.e. velocity, acceleration, rotational speed etc.
(iii) Fluid properties i.e. mass density, specific weight, viscosity etc.
The choice of repeating variable for most of fluid mechanics problems may be:
i) d, v, ρ (ii) d, v, μ (iii) d, N, ρ (iv) L, v, μ (v) H, g, ρ
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1. The variables should not be dimensionless.
2. No two variables should have same dimensions.
3. As far as possible, the dependant variables should not be selected as repeated variables.
4. The repeated variables is selected in such a way that is contains one from each.
(i) Geometric properties i.e. length, diameter, height etc.
(ii) Flow properties i.e. velocity, acceleration, rotational speed etc.
(iii) Fluid properties i.e. mass density, specific weight, viscosity etc.
The choice of repeating variable for most of fluid mechanics problems may be:
i) d, v, ρ (ii) d, v, μ (iii) d, N, ρ (iv) L, v, μ (v) H, g, ρ
For more help in Buckingham Pi-theorem click the button below to submit your homework assignment