The power P required to run a compressor varies with compres

The power, P, required to run a compressor varies with compressor diameter D, angular velocity, volume flowrate, fluid density, and fluid viscosity. Develop a relation between these variables using the Buckingham Pi Method, where fluid viscosity and angular velocity appear in only one dimensionless parameter.

Solution

Dimension of Dia D (meters) is [L].

Dimension of Angular Velocity (rad/s) w = [T^-1]

Dimension of flowrate (m³/s) Q is [L³T^-1]

Dimension of density (kg/m³), rho is [ML^-3]

Dimension of fluid viscosity (Pa-s), u is [ML^-1T^-1]

Dimension of Power (Watt), P is [ML²T^-3]

There are 6 variables and 3 basic dimensions (M,L,T) involved here.

Hence, by Pi theorem we can have 6-3 = 3 dimensionless parameters.

We choose

Pi1 = f(P, Q, rho, D)

Pi2 = f(P, Q, rho, w)

Pi3 = f(P, Q, rho, u)

We write,

Pi1 = [ML²T^-3]^a1 [L³T^-1]^a2 [ML^-3]^a3  [L]

Pi2 = [ML²T^-3]^b1 [L³T^-1]^b2 [ML^-3]^b3  [T^-1]

Pi3 = [ML²T^-3]^c1 [L³T^-1]^c2 [ML^-3]^c3  [ML^-1T^-1]

Now we get,

a1 + a3 = 0

2a1 + 3a2 - 3a3 +1 = 0

-3a1 - a2 = 0

Solving these, a1 = 1/4, a2 = -3/4, a3 = -1/4

Next for Pi2 we get,

b1 + b3 = 0

2b1 + 3b2 - 3b3 = 0

-3b1 - b2 -1 = 0

Solving these we get, b1 = -3/4, b2 = 5/4, b3 = 3/4

Next for Pi3 we get,

c1 + c3 + 1 = 0

2c1 - c2 - 3c3 - 1 = 0

-3c1 - c2 -1 = 0

Solving these we get, c1 = -3/8, c2 = 1/8, c3 = -5/8

Therefore we get

Pi1 = P^1/4 Q^-3/4 rho^-1/4 D

Pi2 = P^-3/4 Q^5/4 rho^3/4 w

Pi3 = P^-3/8 Q^1/8 rho^-5/8 u

Eliminating Q from Pi1 and Pi2 we get,

Pi₁^-4/3 / Pi₂^4/5 = (P^1/4 Q^-3/4 rho^-1/4 D)^-4/3 * (P^-3/4 Q^5/4 rho^3/4 w)^-4/5

= P^7/15 rho^-4/15 D^-4/3 w^-4/5

Raising it to the power 15/4 we get

P^7/4 rho^-1 D^-5 w^-3

or P^7/4 / (rho*D⁵*w³)