6.4 Methods

There are a lot of methods can be used to reduce into a smaller number of dimensionless parameters. We will discuss about the two commonly used methods.

6.4.1 Rayleigh’s Method
A basic method to dimensional analysis method and can be simplified to yield dimensionless groups controlling the phenomenon.
The Rayleigh’s method is based on the following steps:
1. First of all, write the functional relationship according to the given data.

2. Now write the equation in terms of a constant with exponents i.e. powers a, b,c,...

3. With the help of the principle of dimensional homogeneity, find out the values of a, b,
c… by obtaining simultaneous equation and simplify it.

4. Now substitute the values of these exponents in the main equation, and simplify it.


Example 1
Express dimensionless equation for the speed v with a wave pressure travels through a fluid. Consider the physical factors probably influence the speeds are compressibility, E, (or K), density,p and kinematics viscosity, v.

Solution: 
Write the fundamental dimension (used MLTθ) for all dimensions given

V= [L/T]          K= [F/L^2 ]= [M/(LT^2 )]    p = [M/L^3 ]    v = [L^2/T]

Let write the equation like this:  V = CK^a p^b v^d

Inserts fundamental dimension into the equation while C is dimensionless constant.
L/T = (M/(LT^2 ))^a (M/L^3 )^b (L^2/T)^d


To satisfy dimensional homogeneity, net power of each dimension must be identical on both sides of this equation. Thus,
For M: 0 = a + b
For L:   1 = -a + (-3b) + 2d
For T:  -1 = -2a – d
a = 1/2     b = -  1/2     d = 0

V = CK^(1/2) p^(-1/2) v^0

Or

V = C √(K/p)

Therefore, wave speed is not affected by the fluid’s kinematics viscosity, v.


6.4.2 The Buckingham pi Theorem

When a large number of variables are involved, Raleigh’s method becomes lengthy. In such circumstances, the Buckingham’s method is useful. This method expressed the variables related to a dimensional homogeneous equation as:

xr: f(xr,xr....x.)

where, the dimension at each section is the same.

Example:
The drag  D, which is a force, on a ship. What exactly is the drag a function of?

Say that we have n number of quantities (e.g. 6 quantities, which are D,l,ρ,μ,V, and g) and m number of dimensions (e.g. 3 dimensions, which are M, L, and T). These quantities can be reduced to (n - m) independent dimensionless groups.

Step 1: Setup π groups
For the MLT System, m = 3, so choose A1, A2, and A3 as the repeating variables.

Using the Buckingham π Theorem on the Drag Equation:
f(D, l, ρ, μ, V, g) = 0
Where m = 3, n = 6, so there will be n - m = 3 π groups.

We will select ρ, V, and l as the repeating variables (RV), leaving the remaining quantities as D, μ, and g. Note that if the analysis does not work out, we could always go back and repeat using new RVs. Thus,

Which are all dimensionless quantities, i.e. having units of

Step 2: Determine π groups

For the first π group,
π1
M^0 T^0 L^0=(〖M/L^3 )〗^(x_1 ) (L/T)^(y_1 ) (L)^(z_1 ) (ML/T^2 )

Expanding and collecting like units, we can solve for the exponents:
For M: 0 = x1 + 1 ⇒ x1 = -1
For T: 0 = -y1 - 2 ⇒ y1 = -2
For L: 0 = -3x1 + y1 + z1 + 1 ⇒ z1 = 3(-1) - (-2) - 1 = -2
Therefore, we find that the exponents x1, y1, and z1 are -1, -2, and -2 respectively.

This means that the first dimensionless π group, π1, is

= D/(p V^2 l^2)

For the second π group,
π2 M^0 L^0 T^0=(〖M/L^3 )〗^(x_2 ) (〖L/T)〗^(y_2 ) (〖L)〗^(z_2 ) (M/LT)

Solving for the exponents,
For M: x2 + 1 = 0 ⇒ x2 = -1
For T: -y2 - 1 = 0 ⇒ y2 = -1
For L: -3x2 + y2 + z2 - 1 = 0 ⇒ z2 = 1 - (-1) + 3(-1) = -1

Thus,
π_2=ρ^(-1) V^(-1) l^(-1)=μ/( ρVl)=v/Vl    
   
However, we will now invert π2 so that,
π_2 = (Vl/v)^-1 = Reynolds Number      
It is permissible to exponentiate any π group, e.g. π-1, π½, π2, etc., to form a new group, as this does not alter the functional form.

For the third π group,
π3 M^0 L^0 T^0=(〖M/L^3 )〗^(x_3 ) (〖L/T)〗^(y_3 ) (〖L)〗^(z_3 ) (L/T^2 )
Solving for the exponents,
For M: x3 = 0 ⇒ x3 = 0
For T: -y3 - 2 = 0 ⇒ y3 = -2
For L: -3x3 + y3 + z3 + 1 = 0 ⇒ z3 = -1 - (-2) = 1

Thus,
π_3=ρ^0 V^(-2) lg=lg/V^2            
Raising it to the power of -½,
√(1/π_3 )= V/√lg =Froude Number    
          
Thus, the three π groups can be written together as
f(D/(ρl^2 V^2 ),Re,Fr)=0

Finally,
D/(ρl^2 V^2 )=f(Re,Fr)              

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