Chapter 9, Problem 94P

To determine

The expression for velocity component $u_{\theta }$.

Answer to Problem 94P

The expression for velocity component $u_{\theta }$ is $\frac{R_o^2{\omega }_o}{R_o^2-R_i^2}\left(r-\frac{R_i^2}{r}\right)$.

Explanation of Solution

Given information:

The flow is steady, laminar, two-dimensional and incompressible. The flow is rotationally symmetric means nothing is function of coordinate $\theta$ and the flow is circular means velocity component $u_r$ is $0$.

The inner cylinder is fixed and angular speed of outer cylinder is ${\omega }_o$, radius of outer cylinder is $R_o$ and radius of inner cylinder is $R_i$.

Write the expression for continuity equation.

  $\frac{1}{r}\frac{\partial \left(r{u}_r\right)}{\partial r}+\frac{1}{r}\frac{\partial \left(u_{\theta }\right)}{\partial \theta }=0$   ....... (I)

Here, velocity components are $u_r$ and $u_{\theta }$, radius is $r$ and angle is $\theta$.

Write the expression for the $\theta$ -component of Navier-Stokes equation.

  $\left[\rho \left(\begin{array}{l}{u}_r\frac{\partial u_{\theta }}{\partial r}+\\ \frac{u_{\theta }}{r}\frac{\partial u_{\theta }}{\partial \theta }+\frac{u_r{u}_{\theta }}{r}\end{array}\right)\right]=\left[-\frac{1}{r}\frac{\partial P}{\partial \theta }+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_{\theta }}{\partial r}\right)-\frac{u_{\theta }}{r^2}\\ +\frac{1}{r^2}\frac{{\partial }^2{u}_{\theta }}{\partial {\theta }^2}-\frac{2}{r^2}\frac{\partial u_r}{\partial \theta }\end{array}\right)+\rho g_{\theta }\right]$  ......(II)

Here, density is $\rho$, time is $t$, dynamic viscosity is $\mu$ and gravitational acceleration in $\theta$ -direction is $g_{\theta }$.

Calculation:

Substitute $0$ for $u_r$ and $0$ for $\frac{\partial \left(u_{\theta }\right)}{\partial \theta }$ in Equation (I).

  $\begin{array}{l}\frac{1}{r}\frac{\partial \left(0\right)}{\partial r}+\frac{1}{r}\left(0\right)=0\\ 0+0=0\\ 0=0\end{array}$  ......(IV)

Since both sides of Equation (IV) are equal, therefore continuity equation is verified.

Substitute $0$ for $u_r$, $0$ for $g_{\theta }$, $0$ for $\frac{\partial P}{\partial \theta }$ and $0$ for $\frac{\partial \left(u_{\theta }\right)}{\partial \theta }$ in Equation (II).

  $\begin{array}{l}\left[\rho \left(\begin{array}{l}\left(0\right)\frac{\partial u_{\theta }}{\partial r}+\\ \frac{u_{\theta }}{r}\left(0\right)+\frac{\left(0\right)u_{\theta }}{r}\end{array}\right)\right]=\left[-\frac{1}{r}\left(0\right)+\mu \left(\begin{array}{l}\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_{\theta }}{\partial r}\right)-\frac{u_{\theta }}{r^2}\\ +\frac{1}{r^2}\left(0\right)-\frac{2}{r^2}\left(0\right)\end{array}\right)+\rho \left(0\right)\right]\\ \mu \left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_{\theta }}{\partial r}\right)-\frac{u_{\theta }}{r^2}\right)=0\\ \left(\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_{\theta }}{\partial r}\right)-\frac{u_{\theta }}{r^2}\right)=0\\ \frac{1}{r}\left(r\frac{{\partial }^2{u}_{\theta }}{\partial r^2}+\frac{\partial u_{\theta }}{\partial r}\right)-\frac{u_{\theta }}{r^2}=0\end{array}$

  $\begin{array}{l}\frac{{\partial }^2{u}_{\theta }}{\partial r^2}+\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}-\frac{u_{\theta }}{r^2}=0\\ \frac{1}{r}\frac{\partial u_{\theta }}{\partial r}+\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}+\frac{{\partial }^2{u}_{\theta }}{\partial r^2}-\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}-\frac{u_{\theta }}{r^2}=0\\ \frac{1}{r}\frac{\partial u_{\theta }}{\partial r}+\frac{1}{r}\left(\frac{\partial u_{\theta }}{\partial r}+r\frac{{\partial }^2{u}_{\theta }}{\partial r^2}\right)-\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}-\frac{u_{\theta }}{r^2}=0\\ \frac{1}{r}\frac{\partial u_{\theta }}{\partial r}+\frac{1}{r}\frac{\partial }{\partial r}\left(r\left(\frac{\partial u_{\theta }}{\partial r}\right)\right)-\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}-\frac{u_{\theta }}{r^2}=0\end{array}$

  $\begin{array}{l}\frac{1}{r}\left(\frac{\partial }{\partial r}\left(u_{\theta }+r\left(\frac{\partial u_{\theta }}{\partial r}\right)\right)\right)-\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}-\frac{u_{\theta }}{r^2}=0\\ \frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(r{u}_{\theta }\right)-\frac{1}{r}\frac{\partial u_{\theta }}{\partial r}-\frac{u_{\theta }}{r^2}=0\\ \frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(r{u}_{\theta }\right)-\frac{1}{r^2}\left(r\frac{\partial u_{\theta }}{\partial r}+u_{\theta }\right)=0\\ \frac{1}{r}\frac{{\partial }^2}{\partial r^2}\left(r{u}_{\theta }\right)-\frac{1}{r^2}\frac{\partial }{\partial r}\left(r{u}_{\theta }\right)=0\end{array}$

  $\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial }{\partial r}\left(r{u}_{\theta }\right)\right)=0$   ....... (V)

Change from partial derivative to total derivative in Equation (V).

  $\frac{d}{dr}\left(\frac{1}{r}\frac{d}{dr}\left(r{u}_{\theta }\right)\right)=0$   ....... (VI)

Integrate Equation (VI) with respect to $r$.

  $\begin{array}{l}\frac{1}{r}\frac{d}{dr}\left(r{u}_{\theta }\right)=C_1\\ \frac{d}{dr}\left(r{u}_{\theta }\right)=C_{1}r\end{array}$   ...... (VII)

Here arbitrary constant is $C_1$.

Again, Integrate Equation (VII) with respect to $r$.

  $\begin{array}{l}r{u}_{\theta }=\frac{C_1{r}^2}{2}+C_2\\ u_{\theta }=\frac{C_{1}r}{2}+\frac{C_2}{r}\end{array}$   ....... (VIII)

Here, arbitrary constant is $C_2$.

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

  $\begin{array}{l}0=\frac{C_1\left(R_i\right)}{2}+\frac{C_2}{\left(R_i\right)}\\ \frac{C_2}{\left(R_i\right)}=\frac{-C_1\left(R_i\right)}{2}\\ C_2=\frac{-C_1{R}_i^2}{2}\end{array}$   ...... (IX)

Substitute $R_o{\omega }_o$ for $u_{\theta }$, $\frac{-C_1{R}_i^2}{2}$ for $C_2$ and $R_o$ for $r$ in Equation (VIII).

  $\begin{array}{l}{R}_o{\omega }_o=\frac{C_1{R}_o}{2}+\frac{\left(\frac{-C_1{R}_i^2}{2}\right)}{R_o}\\ R_o{\omega }_o=\frac{C_1{R}_o}{2}-\frac{C_1{R}_i^2}{2R_o}\\ R_o{\omega }_o=C_1\left(\frac{R_o}{2}-\frac{R_i^2}{2R_o}\right)\\ C_1=\frac{2R_o^2{\omega }_o}{R_o^2-R_i^2}\end{array}$

Substitute $\frac{2R_o^2{\omega }_o}{R_o^2-R_i^2}$ for $C_1$ in Equation (IX).

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

Substitute $\frac{-R_i^2{R}_o^2{\omega }_o}{R_o^2-R_i^2}$ for $C_2$ and $\frac{2R_o^2{\omega }_o}{R_o^2-R_i^2}$ for $C_1$ in Equation (VIII).

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

Conclusion:

The expression for velocity component $u_{\theta }$ is $\frac{R_o^2{\omega }_o}{R_o^2-R_i^2}\left(r-\frac{R_i^2}{r}\right)$.