Question

Show that the pressure loss in a pipe is a function of length (L) and diameter (D) of the pipe, density of the fluid (rho), and the flow velocity (V). Find a relationship for the pressure loss using the Buckingham pi theorem.

Answer 1

`\Delta P / pv^(2) = f ( L/ d , vpd / \mu)`

Step-by-step explanation:

Buckingham pi theorem is used to find the relationship between variables in a physical phenomenon.

First, we need to identify the variables involved in the physical problem of pressure loss (ΔP). These are: d, L, p, μ and v

hence, we have

f(ΔP, d, L, p, μ,v) = 0

Counting the number of variables n, (ΔP, d, L, p, μ,v) we have them = 6

The number of fundamental dimensions (Mass, Length, Time), m = 3 (That is, [M], [L], [T])

By Buckingham's pi theorem,

Number of dimensionless
`(\Pi)` groups = number of variables - number of fundamental dimensions.

∴No. of dimensionless groups = n -m = 6-3 = 3

To apply the Buckingham Pi theorem, the reoccurring set must contain three variables that cannot themselves be formed into a dimensionless groups.

For this problem we have the following variables chosen as the recurring set : d, v and p.

The others were left out for the following reasons:

• Both L and d cannot be chosen as they can be formed into a dimensionless group, (L /d).
• ΔP, p, and v cannot be used since (
  `\Delta P` /
  `pv^(2)`) is dimensionless

By dimensional analysis,

The dimensions of these variables are :

d = [1]

v = [
`LT^(-1)`]

p = [
`ML^(-3)`]

we can form dimensionless groups from each of the remaining variables ΔP, L and μ in turn

Dimensions of ΔP

Dimensions of
`\Delta P =[ML ^(-1)T^(-2)]`

Hence we have that
`\Delta P M^(-1) L T^(2)` is a dimensionless quantity

Plugging in the dimensions of each of the variables, we obtain the first Pi group as

`\Pi _(1) = \Delta P (pd^(3))^(-1) (d) (dv^(-l))^(2)`

`= \Delta P/pv^(2)`

Dimensions of L

L has the dimensions of [L]

to make it dimensionless, we will have to multiply it by the inverse
`[L^(-1)]`

`L[L ]^(-1)` is therefore dimensionless

The second Pi group we have is

`\Pi_(2) = L/d`

Dimensions of μ

μ has the dimensions of
`[M L^(-1) L^(-1)]`

`\mu [M^(-1)LT]` is therefore dimensionless

substituting in the dimensions of each variable, we have the third dimensionless group as

`\Pi_(3) = \mu(pd^(3))^(-1) (d) (dv^(-1))^(2)`

`\Pi_(3)= \mu /dvp`

Thus we have
`f ( \Delta P/pv^(2) , L/d, \mu/vpd)`

∴ A relationship for the pressure loss is
`\Delta P / pv^(2) = f ( L/ d , vpd / \mu)`