Chapter 10, Problem 110P

Calculate the nine components of the viscous stress tensor in cylindrical coordinates (see Chap. 9) for the velocity field of Prob. 10-109. Discuss.

To determine

The nine components of viscous stress tensor in cylindrical coordinate system.

Answer to Problem 110P

The nine components of viscous stress tensor in cylindrical coordinate system are $0$ for ${\tau }_{rr}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{r\theta }$, $0$ for ${\tau }_{rz}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{\theta r}$, $0$ for ${\tau }_{\theta \theta }$, $0$ for ${\tau }_{\theta z}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{z\theta }$.

Explanation of Solution

Given information:

Velocity component for radial direction.

  $u_r=0$

Velocity component for angular direction.

  $u_{\theta }=\frac{\Gamma }{2\pi r}$

Here, radius for angular velocity is $r$, constant is $\Gamma$.

Velocity component in vertical direction.

  $u_z=0$

Write the expression for nine component viscous stress tensor in cylindrical co-ordinates.

  $\tau =\left[\begin{array}{ccc}{\tau }_{rr}& {\tau }_{r\theta }& {\tau }_{rz}\\ {\tau }_{\theta r}& {\tau }_{\theta \theta }& {\tau }_{\theta z}\\ {\tau }_{zr}& {\tau }_{z\theta }& {\tau }_{zz}\end{array}\right]$  ...... (I)

Here, viscous stress tensor in purely radial direction is ${\tau }_{rr}$, viscous stress tensor in angular direction is ${\tau }_{r\theta }$, viscous stress tensor in vertical direction is ${\tau }_{rz}$, viscous stress tensor in angular direction is ${\tau }_{\theta r}$, viscous stress tensor in purely angular direction is ${\tau }_{\theta \theta }$,

viscous stress tensor in radial direction is ${\tau }_{\theta z}$, viscous stress tensor in radial direction ${\tau }_{zr}$, viscous stress tensor in vertical direction is ${\tau }_{z\theta }$, viscous stress tensor in purely vertical direction

Write the expression for viscous stress tensor in radial direction.

  ${\tau }_{rr}=2\mu \left(r\frac{\partial u_r}{\partial r}\right)$  ...... (II)

Here, viscosity is $\mu$.

Write the expression for viscous stress tensor in angular direction.

  ${\tau }_{r\theta }=\mu \left(r\frac{\partial }{\partial r}\left(\frac{u_{\theta }}{r}\right)+\frac{1}{r}\frac{\partial u_r}{\partial \theta }\right)$  ...... (III)

Write the expression for viscous stress tensor in vertical direction.

  ${\tau }_{rz}=\mu \left(\frac{\partial u_r}{\partial z}+\frac{\partial u_z}{\partial r}\right)$  ...... (IV)

Write the expression for viscous stress tensor in angular direction.

  ${\tau }_{r\theta }=\mu \left(\frac{1}{r}\frac{\partial u_{\theta }}{\partial \theta }+\frac{u_r}{r}\right)$  ...... (V)

Write the expression for viscous stress tensor in vertical direction.

  ${\tau }_{\theta z}=\mu \left(\frac{\partial u_{\theta }}{\partial z}+\frac{1}{r}\frac{\partial u_z}{\partial \theta }\right)$  ...... (VI)

Write the expression for viscous stress tensor in purely vertical direction.

  ${\tau }_{zz}=2\mu \left(r\frac{\partial u_z}{\partial z}\right)$  ...... (VII)

Calculation:

Substitute $0$ for $u_r$ in Equation (II).

  $\begin{array}{c}{\tau }_{rr}=2\mu \left(r\frac{\partial \left(0\right)}{\partial r}\right)\\ =2\mu \left(0\right)\\ =0\end{array}$

Substitute $0$ for $u_r$, $\frac{\Gamma }{2\pi r}$ for $u_{\theta }$ in Equation (III).

  $\begin{array}{c}{\tau }_{r\theta }=\mu \left(r\frac{\partial }{\partial r}\left(\frac{\frac{\Gamma }{2\pi r}}{r}\right)+\frac{1}{r}\frac{\partial \left(0\right)}{\partial \theta }\right)\\ =\mu \left(\frac{r\Gamma }{2\pi }\frac{\partial }{\partial r}\left(r^{-2}\right)+0\right)\\ =\frac{-\mu \Gamma }{2\pi r^2}\end{array}$

Substitute $0$ for $u_r$, $0$ for $u_z$.in Equation (IV).

  $\begin{array}{c}{\tau }_{rz}=\mu \left(\frac{\partial \left(0\right)}{\partial z}+\frac{\partial \left(0\right)}{\partial r}\right)\\ =0\end{array}$

Substitute $0$ for $u_r$, $\frac{\Gamma }{2\pi r}$ for $u_{\theta }$ in Equation (V).

  $\begin{array}{c}{\tau }_{r\theta }=\mu \left(\frac{1}{r}\frac{\partial \left(\frac{\Gamma }{2\pi r}\right)}{\partial \theta }+\frac{0}{r}\right)\\ =\mu \left(\frac{1}{r}\frac{\Gamma }{2\pi r}\frac{\partial \left(1\right)}{\partial \theta }+0\right)\\ =0\end{array}$

Substitute $\frac{\Gamma }{2\pi r}$ for $u_{\theta }$, $0$ for $u_z$ in Equation (VI).

  $\begin{array}{c}{\tau }_{\theta z}=\mu \left(\frac{\partial \left(\frac{\Gamma }{2\pi r}\right)}{\partial z}+\frac{1}{r}\frac{\partial \left(0\right)}{\partial \theta }\right)\\ =\mu \left(\frac{\Gamma }{2\pi r}\frac{\partial \left(1\right)}{\partial z}+0\right)\\ =0\end{array}$

Substitute $0$ for $u_z$ in Equation (VII).

  $\begin{array}{c}{\tau }_{zz}=2\mu \left(r\frac{\partial \left(0\right)}{\partial z}\right)\\ =2\mu \left(0\right)\\ =0\end{array}$

Substitute $0$ for ${\tau }_{rr}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{r\theta }$, $0$ for ${\tau }_{rz}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{\theta r}$, $0$ for ${\tau }_{\theta \theta }$, $0$ for ${\tau }_{\theta z}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{z\theta }$, $0$ for ${\tau }_{zz}$ in Equation (I) to find nine components of the viscous stress tensor in cylindrical co-ordinates.

  $\tau =\left[\begin{array}{ccc}0& \frac{-\mu \Gamma }{2\pi r^2}& 0\\ \frac{-\mu \Gamma }{2\pi r^2}& 0& 0\\ 0& 0& 0\end{array}\right]$

Thus, the nine components of viscous stress tensor in cylindrical coordinate system are $0$ for ${\tau }_{rr}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{r\theta }$, $0$ for ${\tau }_{rz}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{\theta r}$, $0$ for ${\tau }_{\theta \theta }$, $0$ for ${\tau }_{\theta z}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{z\theta }$.

Conclusion:

The nine components of viscous stress tensor in cylindrical coordinate system are $0$ for ${\tau }_{rr}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{r\theta }$, $0$ for ${\tau }_{rz}$, $\frac{-\mu \Gamma }{2\pi r^2}$ for ${\tau }_{\theta r}$, $0$ for ${\tau }_{\theta \theta }$, $0$ for ${\tau }_{\theta z}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{zr}$, $0$ for ${\tau }_{z\theta }$.