Chapter 9, Problem 98P

To determine

(a)

The velocity profile approach from the outer cylinder wall to the inner cylinder wall.

Answer to Problem 98P

The Velocity profile is $\frac{V\left(y\right)}{h}$.

Explanation of Solution

Given information:

Write the expression for Navier-Stokes equation for cylindrical coordinates along $\theta$ direction.

  $\mu \left[\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r{u}_{\theta }\right)\right)\right]=0$   ....... (I)

Here, the dynamic viscosity is $\mu$, velocity along $\theta$ direction is $u_{\theta }$,the radius is $r$.

Write the expression for gap between the outer and inner radius.

  $h=R_o-R_i$

Here, the gap between the outer and inner radius $h$, the inner radius is $R_i$ and the outer radius is $R_o$.

Write the expression for distance from one end of the wall.

  $y=R_o-h$

Write the expression for speed of the upper plate.

  $V=R_i{\omega }_i$

Here, the angular velocity at inner radius is ${\omega }_i$.

Calculation:

Integrate Equation (I).

  $\begin{array}{l}{\displaystyle \int \mu \left[\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r{u}_{\theta }\right)\right)\right]}={\displaystyle \int 0}\\ \frac{\mu }{r}\frac{\partial }{\partial r}\left(r{u}_{\theta }\right)=C_1\\ \partial \left(r{u}_{\theta }\right)=\frac{C_1}{\mu }r\partial r\end{array}$   ....... (II)

Here, the integration constant is $C_1$.

Integrate Equation (II).

  $\begin{array}{l}{\displaystyle \int \partial \left(r{u}_{\theta }\right)}={\displaystyle \int \frac{C_1}{\mu }r\partial r}\\ r{u}_{\theta }=\frac{C_1}{\mu }\left(\frac{r^2}{2}\right)+C_2\\ u_{\theta }=\frac{C_1}{2\mu }\left(r\right)+\frac{C_2}{r}\end{array}$  ......(III)

Here, the integration constant is $C_2$.

Apply boundary condition in Equation (III).

Substitute $0$ for $u_{\theta }$ and $R_o$ for $r$ in Equation (III).

  $\begin{array}{l}{u}_{\theta }=\frac{C_1}{2\mu }\left(r\right)+\frac{C_2}{r}\\ 0=\frac{C_1}{2\mu }\left(R_o\right)+\frac{C_2}{R_o}\\ -\frac{C_2}{R_o}=\frac{C_1}{2\mu }\left(R_o\right)\\ C_2=-\left(\frac{C_1}{2\mu }\right){\left(R_o\right)}^2\end{array}$   ...... (IV)

Substitute ${\omega }_i{R}_i$ for $u_{\theta }$, $-\left(\frac{C_1}{2\mu }\right){\left(R_o\right)}^2$ for $C_2$ and $R_i$ for $r$ in Equation (III).

  $\begin{array}{l}{u}_{\theta }=\frac{C_1}{2\mu }\left(r\right)+\frac{C_2}{r}\\ {\omega }_i{R}_i=\frac{C_1}{2\mu }\left(R_i\right)+\frac{-\left(\frac{C_1}{2\mu }\right){\left(R_o\right)}^2}{R_i}\\ {\omega }_i{R}_i=\frac{C_1}{2\mu }\left(\frac{R_i^2-R_o^2}{R_i}\right)\\ C_1=\frac{2\mu {\omega }_i{R}_i^2}{R_i^2-R_0^2}\end{array}$

Substitute $\frac{2\mu {\omega }_i{R}_i^2}{R_i^2-R_0^2}$ for $C_1$ in Equation (IV).

  $\begin{array}{c}{C}_2=-\left(\frac{C_1}{2\mu }\right){\left(R_o\right)}^2\\ =-\left(\frac{\frac{2\mu {\omega }_i{R}_i^2}{R_i^2-R_0^2}}{2\mu }\right)R_o^2\\ =-\frac{2R_i^2{R}_o^2}{R_i^2-R_0^2}\end{array}$

Substitute $\frac{2\mu {\omega }_i{R}_i^2}{R_i^2-R_0^2}$ for $C_1$ and $-\frac{2R_i^2{R}_o^2}{R_i^2-R_0^2}$ for $C_2$ in Equation (III).

  $\begin{array}{c}{u}_{\theta }=\frac{\frac{2\mu {\omega }_i{R}_i^2}{R_i^2-R_0^2}}{2\mu }{\left(R_o\right)}^2+\frac{-\frac{2R_i^2{R}_o^2}{R_i^2-R_0^2}}{r}\\ =\frac{{\omega }_i{R}_i^2{R}_o^2}{R_i^2-R_0^2}-\frac{\frac{2R_i^2{R}_o^2}{R_i^2-R_0^2}}{r}\\ =\frac{{\omega }_i{R}_i^2}{R_0^2-R_i^2}\left(\frac{R_o^2}{r}-r\right)\end{array}$

  $u_{\theta }=\frac{{\omega }_i{R}_i^2}{\left(R_o-R_i\right)\left(R_o+R_i\right)}\left(\frac{\left(R_o-r\right)\left(R_o+r\right)}{r}\right)$  ......(V)

Substitute $R_i$ for $r$, $h$ for $\left(R_o-R_i\right)$, $y$ for $\left(R_o-r\right)$ and $R_i$ for $R_o$ in Equation (V).

  $\begin{array}{c}{u}_{\theta }=\frac{{\omega }_i{R}_i^2}{\left(R_o-R_i\right)\left(R_o+R_i\right)}\left(\frac{\left(R_o-r\right)\left(R_o+r\right)}{r}\right)\\ =\frac{{\omega }_i{R}_i^2}{h\left(R_i+R_i\right)}\left(\frac{y\left(R_i+R_i\right)}{R_i}\right)\\ =\frac{{\omega }_i{R}_i^2}{h\left(2R_i\right)}\left(\frac{y\left(2R_i\right)}{R_i}\right)\\ =\frac{{\omega }_i{R}_i\left(y\right)}{h}\end{array}$   ....... (VI)

Substitute $V$ for $R_i{\omega }_i$ in Equation (VI).

  $\begin{array}{c}{u}_{\theta }=\frac{{\omega }_i{R}_i\left(y\right)}{h}\\ =\frac{V\left(y\right)}{h}\end{array}$

Therefore, the final velocity profile is linearly varying from the outer radius.

Conclusion:

The Velocity profile is $\frac{V\left(y\right)}{h}$.

To determine

(b)

The type of flow when the outer wall approaches to infinity and the inner cylinder radius is very small.

Answer to Problem 98P

This type of flow when the outer wall approaches to infinity and the inner cylinder radius is very small is called as linear vortex flow in a cylinder.

Explanation of Solution

Given information:

The outer cylinder wall approaches to infinity and the inner cylinder radius is very small, so $R_i<<R_o$.

Calculation:

Applying boundary condition in Equation (V).

Substitution $R_o$ for $\left(R_o-r\right)$ and $R_o$ for $\left(R_o+r\right)$ Equation (V).

  $\begin{array}{c}{u}_{\theta }=\frac{{\omega }_i{R}_i^2}{\left(R_o-R_i\right)\left(R_o+R_i\right)}\left(\frac{\left(R_o-r\right)\left(R_o+r\right)}{r}\right)\\ =\frac{{\omega }_i{R}_i^2}{\left(R_o\right)\left(R_o\right)}\left(\frac{\left(R_o\right)\left(R_o\right)}{r}\right)\\ =\frac{{\omega }_i{R}_i^2}{r}\end{array}$

Substitute $V$ for $R_i{\omega }_i$ in Equation (VI).

  $\begin{array}{c}{u}_{\theta }=\frac{{\omega }_i{R}_i\left(y\right)}{h}\\ =\frac{Vy}{h}\end{array}$

Therefore, the final velocity profile represents a linear vortex flow in a cylinder.

Conclusion:

This type of flow when the outer wall approaches to infinity and the inner cylinder radius is very small is called as linear vortex flow in a cylinder.