Buckingham's theorem

“IN THE NAME OF ALLAH , THE MOST
BENIFICIENT , THE MOST MERCIFUL”

Topic:
Buckingham’s ∏
Theorem

History
A more generalized method of dimension
analysis developed by E. Buckingham and
others and is most popular now. This arranges
the variables into a lesser number of
dimensionless groups of variables. Because
Buckingham used ∏ (pi) to represent the
product of variables in each groups, we call
this method Buckingham pi theorem.

Explanation
Statement:
“If ‘n’ is the total number of variables in a
dimensionally homogenous equation containing ‘m’
fundamental dimensions, then they may be grouped
into (n-m) dimensionless terms which are called ∏
terms.
f(X1, X2, ……Xn) = 0
then the functional relationship will be written as
Ф (∏1 , ∏2 ,………….∏n-m) = 0
The final equation obtained is in the form of:
∏1= f (∏2,∏3,………….∏n-m) = 0
Suitable where n ≥ 4
Not applicable if (n-m) = 0

Procedure:
• List all physical variables and note ‘n’ and ‘m’.
n = Total no. of variables
m = No. of fundamental dimensions (That is,
[M], [L], [T])
• Compute number of ∏-terms by (n-m)
• Write the equation in functional form
• Write equation in general form
• Select repeating variables. Must have all of the
‘m’ fundamental dimensions and should not form
a ∏ among themselves
• Solve each ∏-term for the unknown exponents by
dimensional homogeneity.

Example:
Let us apply Buckingham’s ∏ method to an example
problem that of the drag forces FD exerted on a
submerged sphere as it moves through a viscous
fluid.
We need to follow a series of following steps when
applying Buckingham’s ∏ theorem.

Step 1:
Visualize the physical problem, consider the factors that are of
influence and list and count the n variables.
We must first consider which physical factors influence the
drag force. Certainly, the size of the sphere and the velocity of
the sphere must be important. The fluid properties involved
are the density ρ and the viscosity μ. Thus we can write
f (FD, D, V, ρ, μ) = 0
Here we used D, the sphere diameter, to represent sphere
size, and f stands for “some function”. We see that n = 5.

Step 2 & 3:
Step 2: Choose a dimensional system (MLT or FLT) and list the
dimensions of each variables. Find m, the number of
fundamental dimensions involved in all the variables.
Choosing the MLT system, the dimensions are respectively
MLT-2 , L , LT-1 , ML-3 , ML-1T-1
We see that M, L and T are involved in this example. So m = 3.
Step 3: Determine n-m, the number of dimensionless ∏ groups
needed. In our example this is 5 – 3 = 2, so we can write Ф(∏1.
∏2) = 0

Step 4:
Form the ∏ groups by multiplying the product
of the primary (repeating) variables, with
unknown exponents, by each of the remaining
variables, one at a time. We choose ρ, D, and V
as the primary variables. Then the ∏ terms are
∏1 = Da Vb ρc FD
∏2 = Da Vb ρc μ-1

Step 5:
To satisfy dimensional homogeneity, equate the
exponents of each dimension on both sides on each pi
equation and so solve for the exponents
∏1 = Da Vb ρc FD = (L)a (LT-1)b (ML-3)c (MLT-2) = M0L0T0
Equate exponents:
L: a +b -3c +1 = 0
M: c +1 = 0
T: -b -2 = 0
We can solve explicitly for
b = -2, c = -1, a = -2
Therefore
∏1 = D-2 V-2 ρ-1 FD = FD/(ρ V2 D2)

Cont.… :
Finally, add viscosity to D, V, and ρ to find ∏2 . Select any power you
like for viscosity. By hindsight and custom, we select the power -1 to
place it in the denominator
∏1 = Da Vb ρc μ-1 = (L)a (LT-1)b (ML-3)c (ML-1T-1 )-1 = M0L0T0
Equate exponents:
L: a +b -3c +1 = 0
M: c -1 = 0
T: -b +1 = 0
We can solve explicitly for
b = 1, c = 1, a = 1
Therefore,
∏2 = D1 V1 ρ1 μ-1 = (D V ρ)/(μ) =R = Reynolds Number
R = Reynolds Number= Ratio of inertia forces to viscous forces
Check that all ∏s are in fact dimensionless

Cont.… :
Rearrange the pi groups as desired. The pi theorem states that the
∏s are related. In this example hence
FD/(ρ V2 D2) = Ф (R)
So that FD = ρ V2 D2 Ф (R)
We must emphasize that dimensional analysis does not provide a
complete solution to fluid problems. It provides a partial solution
only. The success of dimensional analysis depends entirely on the
ability of the individual using it to define the parameters that are
applicable. If we omit an important variable. The results are
incomplete, and this may lead to incorrect conclusions. Thus, to use
dimensional analysis successfully, one must be familiar with the
fluid phenomena involved.

Buckingham's theorem

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