Chapter 9, Problem 93P

An incompressible Newtonian liquid is confined between two concentric circular cylinders of infinite length-a solid inner cylinder of radius R₁and a hollow, stationary outer cylinder of radius R_v(Fig. 9-93; the z-axis is out of the page). The inner cylinder rotates at angular velocity ${\omega }_l$. The flow is steady, laminar, and two-dimensional in the $r\theta$ -plane. The flow is also ratitionally symmetric, meaning that nothing is a function of coordinate $\theta$and P are functions of radius r only). The flow is also circular, meaning that velocity component ( $u_{\theta }$= 0 everywhere. Generate an exact expression for velocity component $u_{\theta }$ as a function of radius r and the other parameters in the problem. You may ignore gravity. (Hint: The result of Prob. 9-92 is useful.)

FIGURE P9-93

To determine

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

Answer to Problem 93P

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

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 outer cylinder is fixed and angular speed of inner cylinder is ${\omega }_i$, 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 $R_o$ for $r$ in Equation (VIII).

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

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

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

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

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

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

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

Conclusion:

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