Question

In a hole meter a flat disk with a central diameter opening is placed transversely to a pipe of diameter D, and the pressure drop ??? is measured through the opening. It is postulated that ??? is a function of the average velocity of the fluid in pipe V, the density of the flow ?, the viscosity of the fluid ? and the parameters of the pipe and the opening. D and Do, respectively. In this way

?p= ? (V,?,?,D,Do)

Find an acceptable set of dimensionless groups that relate the various factors.

Answer 1

?p= ? (V,?,?,D,Do)

We can write it as

?(?P, D, Do, ?, µ,V)= O

Total number of variables n = 6

Number of fundamental dimensions m = 3

Number of dimensionless groups = n -m

= 6 - 3

= 3

?(?1, ?2, ?3)............eq1

Chosen variables = D, V and ?

Dimension of D = [L]

Dimension of V = [LT-1]

Dimension of ? = [ML-3]

Dimensions of ?P = [ML -1T-2]

Fundamental dimensions in terms of D, V, ?
[L] = D
[M] = ?D3
[T] = DV-1

?1 = ?P (?D3)-1 (D) (DV-1)2 = ?P/?V2
Dimensions of Do = [L]

Dimensionless term of Do = Do[L ]-1

?2 = Do/D

Dimensions of µ = [ML-1T-1]

Dimensionless term of µ = µ[M-1LT]

?3 = µ (?D3)-1 (D) (DV-1)2 = µ/DV?

Now from Eq 1

?(?1, ?2, ?3)

? ( ?P/?v2 , Do /D , µ / V?D)

?P / ?v2 = ? ( Do/D, V?D/µ)