Chapter 5, Problem 5.37P

To determine

The relationship in dimensionless form.

Answer to Problem 5.37P

$\frac{Q}{D^2}\sqrt{\frac{\rho }{\Delta p}}=f\left(\frac{\mu }{D}\sqrt{\frac{1}{\rho \Delta p}},\frac{d}{D}\right)$

Explanation of Solution

Given information:

Volume flow $Q$ depends on,

Pipe diameter $D$, pressure drop $\Delta p$, density $\rho$, viscosity $\mu ,$ and orifice diameter $d$

$D,\rho ,\Delta p$ are repeating variables

To find the relation between these variables,

Basically we have some steps to follow,

1. Find the number of variables $(n)$ given.

2. Find the dimensions of the given variables.

3. Find $j$ which is basically the number of different dimensions of given variables.

4. In other words $j$ is the number of variables which won’t form pi product.

5. Find the number of desired pi products $(k)$ which is given as, $k=n-j$

6. Finally write the dimensionless function obtained.

Calculation:

Write the function,

$Q=f\left(D,\Delta p,\rho ,\mu ,d\right)$

Count the number of variables $n$,

$n=6$

Find dimension of each variable,

$\begin{array}{l}Q\to L^3{T}^{-1}\\ D\to L\\ \Delta p\to M{L}^{-1}{T}^{-2}\\ \rho \to M{L}^{-3}\\ \mu \to M{L}^{-1}{T}^{-1}\\ d\to L\end{array}$

Find $j$, it’s equal to the number of repeating variables, therefore

$j=3$

Find the number of pi products,

$k=n-j=6-3=3$

To obtain the equation,

To find the first pi group combine $D,\rho ,\Delta p$ with volume flow $Q$

${\Pi }_1=D^a\Delta p^b{\rho }^{c}Q={\left(L\right)}^a{\left(M{L}^{-1}{T}^{-2}\right)}^b{\left(M{L}^{-3}\right)}^c\left(L^3{T}^{-1}\right)=M^0{L}^0{T}^0$

Equate exponents,

Length: $a-b-3c+3=0$

Mass: $b+c=0$

Time: $-2b-1=0$

Therefore, we get

$a=-2,b=-\frac{1}{2},c=\frac{1}{2}$

Therefore, the expression will be,

$\begin{array}{l}{\Pi }_1=D^{-2}\Delta p^{-\frac{1}{2}}{\rho }^{\frac{1}{2}}Q\\ {\Pi }_1=\frac{Q}{D^2}\sqrt{\frac{\rho }{\Delta p}}=const\end{array}$

To find the second pi group combine $D,\rho ,\Delta p$ with viscosity $\mu$

${\Pi }_2=D^a\Delta p^b{\rho }^c\mu ={\left(L\right)}^a{\left(M{L}^{-1}{T}^{-2}\right)}^b{\left(M{L}^{-3}\right)}^c\left(M{L}^{-1}{T}^{-1}\right)=M^0{L}^0{T}^0$

Equate exponents,

Length: $a-b-3c-1=0$

Mass: $b+c+1=0$

Time: $-2b-1=0$

Therefore, we get

$a=-1,b=-\frac{1}{2},c=-\frac{1}{2}$

Therefore, the expression will be,

$\begin{array}{l}{\Pi }_2=D^{-1}\Delta p^{-\frac{1}{2}}{\rho }^{-\frac{1}{2}}\mu \\ {\Pi }_2=\frac{\mu }{D}\sqrt{\frac{1}{\rho \Delta p}}=const\end{array}$

To find the third pi group combine $D,\rho ,\Delta p$ with orifice diameter $d$

${\Pi }_3=D^a\Delta p^b{\rho }^{c}d={\left(L\right)}^a{\left(M{L}^{-1}{T}^{-2}\right)}^b{\left(M{L}^{-3}\right)}^c\left(L\right)=M^0{L}^0{T}^0$

Equate exponents,

Length: $a-b-3c+1=0$

Mass: $b+c=0$

Time: $-2b=0$

Therefore, we get

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

Therefore, the expression will be,

$\begin{array}{l}{\Pi }_3=D^{-1}\Delta p^0{\rho }^{0}d\\ {\Pi }_3=\frac{d}{D}=const\end{array}$

The original relation between 6 variables can be reduced in to 3 dimensionless groups as below,

$\frac{Q}{D^2}\sqrt{\frac{\rho }{\Delta p}}=f\left(\frac{\mu }{D}\sqrt{\frac{1}{\rho \Delta p}},\frac{d}{D}\right)$

Conclusion:

The expression between the given variables can be written as,

$\frac{Q}{D^2}\sqrt{\frac{\rho }{\Delta p}}=f\left(\frac{\mu }{D}\sqrt{\frac{1}{\rho \Delta p}},\frac{d}{D}\right)$.