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An example: Taking the example discussed above of force F induced on a propeller blade, we have the equation 0 = φ (K, d, u, ρ , N, μ ) There are m - n = 3 π groups, so φ ( π 1 , π 2 , π 3 ) = 0

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An example

  • An example:

    Taking the example discussed above of force F induced on a propeller blade, we have the equation

    0 = φ (K, d, u, ρ, N, μ)

    There are m - n = 3 π groups, so

    φ( π1 , π2 , π3 ) = 0

  • The choice of ρ, u, d as the repeating variables satisfies the criteria above. They are measurable, good

    design parameters and, in combination, contain all the dimension M,L and T. We can now form the three

    groups according to the 2nd theorem,


An example

As the πgroups are all dimensionless i.e. they have dimensions M0L0T0 we can use the principle of

dimensional homogeneity to equate the dimensions for each πgroup.

For the first π group, in terms of dimensions


An example

M0L0T0 = (ML-3)a1 (LT-1)b1 (L)c1 MLT-2

For each dimension (M, L or T) the powers must be equal on both sides of the equation, so

For M: 0 = a1 + 1

a1 = -1

For L: 0 = -3a1 + b1 + c1 +1

0 = 4 + b1 + c1

For T: 0 = -b1 – 2

b1 = -2


An example

c1 = -4 – b1 = -2

Giving π1 as

For π2 we have,

M0L0T0 = (ML-3)a2 (LT-1)b2 (L)c2 T-1

For M: 0 =a2

a2 = 0


An example

For L: 0 = -3a2 – b2 + c2

0 = - b2 + c2

For T: 0 = -b2 – 1

b2 = -1

c2 = 1

Giving π2 as

For π3 we have


An example

M0L0T0 = (ML-3)a3 (LT-1)b3 (L)c3 ML-1T—1

For M: 0 = a3 + 1

a3 = -1

For L: 0 = -3a3 + b3 + c3 -1

0 = b3 + c3 +2

For T: 0 = -b3 -1

b3 = -1

c3 = -1

Giving π3 as


An example

  • Thus the problem may be described by the following function of the three non-dimensional π groups,


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