Chapter 5, Problem 5.3CP

To determine

(a)

To rewrite:

In dimensionless parameters, using the pi theorem.

Answer to Problem 5.3CP

The dimensionless function is $\frac{w}{\sqrt{g\delta }}=fcn(\frac{\mu }{\rho \sqrt{g{\delta }^3}},\frac{x}{\delta })$

Explanation of Solution

Given Information:

$w=fcn(x,\rho ,\mu ,g,\delta )$

Reference Prob 4.8:

Oil of viscosity $\mu$ and of density $\rho$, drains steadily down the side of a vertical plate as shown in Figure .Th oil film will be independent of z and of constant thickness $\delta$ after a development region near the top of the plate. It is assumed that the atmosphere offers no shear resistance to the surface of the film and $w=w(x)$.

Concept Used:

The number of pi groups are to be calculated:

$N=k-r$

Where k is the number of variables and r is the number of fundamental references.

On substituting 6 for k and 3 for r ,

$N=3$

Calculation:

Dimensional analysis is applied to find the pi groups.

First pi group:

${\pi }_1={\rho }^a{g}^b{\delta }^{c}w$

Where $\rho$ is the density, acceleration due to gravity is g, film thickness is $\delta$ and the vertical velocity is w.

On substituting $M^0{L}^0{T}^0$ for ${\pi }_1$, $[M{L}^{-3}]$ for $\rho$, $[L{T}^{-2}]$ for g ,[ L ] for $\delta$ and $[L{T}^{-1}]$ for w ,

$M^0{L}^0{T}^0={[M{L}^{-3}]}^a{[L{T}^{-2}]}^b{[L]}^c[L{T}^{-1}]$

$M^0{L}^0{T}^0=[M^a{L}^{-3a+b+c+1}{T}^{-2b-1}]$

On equating M coefficients:

$a=0$

On equating T coefficients:

$\begin{array}{l}-2b-1=0\\ b=-1/2\end{array}$

On equating L coefficients:

$\begin{array}{l}-3a+b+c+1=0\\ -3(0)+(-1/2)+c+1=0\\ c=-1/2\end{array}$

Hence, a = 0, b = -1/2 and c = -1/2

Therefore, the first pi group is as follows:

${\pi }_1={\rho }^0{g}^{-1/2}{\delta }^{-1/2}w$

${\pi }_1=\frac{w}{\sqrt{g\delta }}$

Second pi group:

${\pi }_2={\rho }^a{g}^b{\delta }^c\mu$

Where $\rho$ is the density, acceleration due to gravity is g, film thickness is $\delta$ and the dynamic viscosity is $\mu$.

On substituting $M^0{L}^0{T}^0$ for ${\pi }_2$, $[M{L}^{-3}]$ for $\rho$, $[L{T}^{-2}]$ for g ,[ L ] for $\delta$ and $[M{L}^{-1}{T}^{-1}]$ for $\mu$,

$M^0{L}^0{T}^0={[M{L}^{-3}]}^a{[L{T}^{-2}]}^b{[L]}^c[M{L}^{-1}{T}^{-1}]$

$M^0{L}^0{T}^0=[M^{a+1}{L}^{-3a+b+c-1}{T}^{-2b-1}]$

On equating M coefficients:

$\begin{array}{l}a+1=0\\ a=-1\end{array}$

On equating T coefficients:

$\begin{array}{l}-2b-1=0\\ b=-1/2\end{array}$

On equating L coefficients:

$\begin{array}{l}-3a+b+c-1=0\\ -3(-1)+(-1/2)+c-1=0\\ c=-3/2\end{array}$

Therefore, the second pi group is as follows:

${\pi }_2={\rho }^{-1}{g}^{-1/2}{\delta }^{-3/2}\mu$

${\pi }_2=\frac{\mu }{\rho \sqrt{g{\delta }^3}}$

Third pi group:

${\pi }_3={\rho }^a{g}^b{\delta }^{c}x$

Where $\rho$ is the density, acceleration due to gravity is g, film thickness is $\delta$ and the distance is x.

On substituting $M^0{L}^0{T}^0$ for ${\pi }_3$, $[M{L}^{-3}]$ for $\rho$, $[L{T}^{-2}]$ for g ,[ L ] for $\delta$ and [ L ] for x.

$M^0{L}^0{T}^0={[M{L}^{-3}]}^a{[L{T}^{-2}]}^b{[L]}^c[L]$

$M^0{L}^0{T}^0=[M^a{L}^{-3a+b+c+1}{T}^{-2b}]$

On equating M coefficients:

$a=0$

On equating T coefficients:

$\begin{array}{l}-2b=0\\ b=0\end{array}$

On equating L coefficients:

$\begin{array}{l}-3a+b+c+1=0\\ -3(0)+(0)+c+1=0\\ c=-1\end{array}$

Hence, a = 0, b = 0 and c=- 1

Therefore, the pi group is as follows:

${\pi }_3={\rho }^0{g}^0{\delta }^{-1}x$

${\pi }_3=\frac{x}{\delta }$

Hence, as per the choices:

${\pi }_1=fcn({\pi }_2,{\pi }_3)$

On substituting $\frac{x}{\delta }$ for ${\pi }_3$, $\frac{\mu }{\rho \sqrt{g{\delta }^3}}$ for ${\pi }_2$ and $\frac{w}{\sqrt{g\delta }}$ for ${\pi }_1$,

$\frac{w}{\sqrt{g\delta }}=fcn(\frac{\mu }{\rho \sqrt{g{\delta }^3}},\frac{x}{\delta })$

Hence, the dimensionless function is $\frac{w}{\sqrt{g\delta }}=fcn(\frac{\mu }{\rho \sqrt{g{\delta }^3}},\frac{x}{\delta })$

Conclusion:

The dimensionless function is $\frac{w}{\sqrt{g\delta }}=fcn(\frac{\mu }{\rho \sqrt{g{\delta }^3}},\frac{x}{\delta })$.

To determine

(b)

To verify:

That the exact solution from Prob 4.8 is consistent with the result in part (a).

Answer to Problem 5.3CP

The exact solution from Prob 4.8 is consistent with the result in part (a).

Explanation of Solution

Given Information:

$w=fcn(x,\rho ,\mu ,g,\delta )$

Reference Prob 4.8:

Oil of viscosity $\mu$ and of density $\rho$, drains steadily down the side of a vertical plate as shown in Figure .Th oil film will be independent of z and of constant thickness $\delta$ after a development region near the top of the plate. It is assumed that the atmosphere offers no shear resistance to the surface of the film and $w=w(x)$.

Concept Used:

The exact solution obtained from prob 4.8 is to be considered:

$\frac{w}{\sqrt{g\delta }}=\frac{\rho gx}{2\mu }(x-\delta )$

Calculation:

$\frac{w}{\sqrt{g\delta }}\frac{\mu }{\rho \sqrt{g{\delta }^3}}=\frac{1}{2}\frac{x}{\delta }(\frac{x}{\delta }-2)$

Hence, the exact solution from Prob 4.8 can be expressed in terms of dimensionless function.

Conclusion:

The exact solution from Prob 4.8 is consistent with the result in part (a).