Question

At a sudden contraction in a pipe the diameter changes from D 1 to D 2 . The pressure drop, Δ p , which develops across the contraction is a function of D 1 and D 2 , as well as the velocity, V , in the larger pipe, and the fluid density, rho, and viscosity, μ . Use D 1 , V , and μ as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

Answer 1

Velocity in the smaller pipe should not be included as an additional variable.

Step-by-step explanation:

`$\Delta P = f(D_1, D_2, V, \rho, \mu)$`

The dimensional formula of the variables are

`$\Delta P = FL^(-2) , D_1 = L, D_2=L, V=LT_(-1), \rho =FL^(-4)T^2, \mu = FL^(-2)T$`

Now using Buchingham's Pi Theorem, 6 - 3 = 3 dimensional parameters are required.

Use, D_1, V, \mu as the repeating variables.

Therefore,
`$\pi = \Delta PD_1^aV^b \mu^c$`

`$(FL^(-2))(L)^a(LT^(-1))^b(FL^(-2)T)^c = F^0L^0T^0$`

From this

1+c=0

-2+a+b-2c=0

-b+c=0

c=-1, b = -1, a = 1

Now,
`$\pi_1=(\Delta PD_1)/(V\mu)=((ML^(-1)T^(-2))L)/((LT^(-1))(ML^(-1)T^(-1)))=M^0L^0T^0$`

For
`$\pi_2 = D_2D_1^aV^b\mu^c$`

`$F^0L^0T^0=L(L)^a(LT^(-1))^b(FL^(-2)T)^c$`

c = 0

1 + a + b - 2c = 0

-b + c = 0

Therefore, a = -1, b = 0, c = 0

`$\pi_2 = (D_2)/(D_1)$`

For
`$\pi_3$`

`$\pi_3 = \rho, D_1^a V^b\mu^c$`

`$F^0L^0T^0 = (FL^(-4)T^2)(L)^a(LT^(-1))^b(FL^(-2)T)^c$`

1 + c = 0

-4 + a + b - 2c = 0

2-b+c=0

c=-1, b=1, a = 1

Therefore,
`$\pi_3 = (\rho D_1V)/(\mu)$`

Now checking,

`$\pi_3 = ((ML^(-3))(L)(LT^(-1)))/(ML^(-1)T^(-1)) = M^0L^0T^0$`

Therefore,
`$(\Delta P D_1)/(V \mu) = \phi ((D_2)/(D_1), (\rho D_1 V)/(\mu))$`

From continuity equation

`$V(\pi)/(4)D_1^2 = V_s (\pi)/(4)D_2^2$`

`$V_s = V ((D_1)/(D_2))^2$`

`$V_s$` is not independent of
`$D_1,D_2, V$`

Therefore it should not be included as an additional variable.