Physics, 23.03.2020 21:28 cynone8254

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.

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Answer from: hlc614

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

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

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explanation:

the masses of the objects and the distance between them

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Show that the pressure loss in a pipe is a function of length (L) and diameter (D) of the pipe, dens...

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