Chapter 9, Problem 65P

To determine

The expression for the stream function.

Answer to Problem 65P

The stream function is $\frac{r^2}{2}\left(\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+\left(u_{z,entrance}\right)+C\right)$.

Explanation of Solution

Given information:

The velocity component in $\theta$ direction is equal to $0$.

Write the expression for velocity coordinates along $z$ direction.

  $u_z=u_{z,entrance}+\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)z$........... (I)

Write the expression for continuity equation for incompressible flow.

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

Here, the change in distance along $r$ direction is $\partial r$, the change in distance along $\theta$ direction is $\partial \theta$, the change in distance in $z$ direction is $\partial z$, the radial velocity component along $r$ direction is $u_r$, the distance along $r$ direction is $r$, the velocity component along $\theta$ direction is $u_{\theta }$ and the velocity component along $z$ direction is $u_z$.

Write the expression for the radial velocity component along $r$ direction.

  $\begin{array}{l}{u}_r=-\frac{1}{r}\left(\frac{\partial \psi }{\partial z}\right)\\ \frac{\partial \psi }{\partial z}=-r{u}_r\end{array}$............ (III)

Here, the partial derivative of the stream function with respect to the change in distance along $z$ direction is $\frac{\partial \psi }{\partial z}$.

Write the expression for velocity coordinates along $z$ direction.

  $\begin{array}{l}{u}_z=\frac{1}{r}\left(\frac{\partial \psi }{\partial r}\right)\\ \frac{\partial \psi }{\partial r}=r{u}_r\end{array}$............ (IV)

Here, the partial derivative of the stream function with respect to the change in distance along $r$ direction is $\frac{\partial \psi }{\partial z}$.

Calculation:

Substitute $u_{z,entrance}+\left[\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\right]$ for $u_z$ and $0$ for $u_{\theta }$ in Equation (II).

  $\begin{array}{l}\frac{1}{r}\left(\frac{\partial \left(r{u}_r\right)}{\partial r}\right)+\frac{1}{r}\left(\frac{\partial \left(0\right)}{\partial \theta }\right)+\frac{\partial \left(u_{z,entrance}+\left[\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\right]\right)}{\partial z}=0\\ \frac{1}{r}\left(\frac{\partial \left(r{u}_r\right)}{\partial r}\right)+0+\left(0+\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\right)=0\\ \frac{1}{r}\left(\frac{\partial \left(r{u}_r\right)}{\partial r}\right)+\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)=0\\ \left(\frac{\partial \left(r{u}_r\right)}{\partial r}\right)=-r\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\end{array}$........... (V)

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

  $\begin{array}{l}{\displaystyle \int \frac{\partial \left(r{u}_r\right)}{\partial r}}dr={\displaystyle \int -r\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)dr}\\ r{u}_r=-\frac{r^{1+1}}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)+g\left(z\right)\\ r{u}_r=-\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)+g\left(z\right)\end{array}$............. (VI)

Here, the constant is $g\left(z\right)$.

Substitute $0$ for $r$ in Equation (III).

  $\begin{array}{l}\left(0\right)u_r=-\frac{{\left(0\right)}^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)+g\left(z\right)\\ 0=0+g\left(z\right)\\ g\left(z\right)=0\end{array}$

Substitute $0$ for $g\left(z\right)$ in Equation (III).

  $\begin{array}{l}r{u}_r=-\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)+0\\ r{u}_r=-\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\\ u_r=-\frac{r}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\end{array}$............ (VII)

Substitute $-\frac{r}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)$ for $u_r$ in Equation (II).

  $\begin{array}{l}\frac{\partial \psi }{\partial r}=-r\left(-\frac{r}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\right)\\ \frac{\partial \psi }{\partial r}=\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\end{array}$........... (VIII)

Integrate Equation (VII) with respect to $z$.

  $\begin{array}{l}{\displaystyle \int \frac{\partial \psi }{\partial r}}={\displaystyle \int \frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\cdot dz}\\ \psi =\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z^{0+1}+f\left(r\right)\\ \psi =\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+f\left(r\right)\end{array}$............ (IX)

Substitute $\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+f\left(r\right)$ for $\psi$ in Equation (III).

  $\begin{array}{l}\frac{\partial \left(\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+f\left(r\right)\right)}{\partial r}=r{u}_r\\ u_r=\frac{1}{r}\times \frac{\partial \left(\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+f\left(r\right)\right)}{\partial r}\end{array}$........... (X)

Differentiate Equation (X) with respect to $r$.

  $\begin{array}{c}{u}_z=\frac{1}{r}\times \frac{\partial \left(2\times \frac{r^{2-1}}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+f\left(r\right)\right)}{\partial r}\\ =\frac{1}{r}\times \frac{r\times \left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+f\left(r\right)}{\partial r}\\ =\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+\frac{1}{r}\times f\left(r\right)\end{array}$............ (XI)

Here, the constant is $f\left(r\right)$.

Equate Equation (XI) with Equation (V).

  $\begin{array}{l}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+\frac{1}{r}\times f\left(r\right)=u_{z,entrance}+\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)z\\ u_{z,entrance}=\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)z-\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)z+\frac{1}{r}\times f\left(r\right)\\ u_{z,entrance}=\frac{1}{r}\times f\left(r\right)\\ f\left(r\right)=r{u}_{z,entrance}\end{array}$........... (XII)

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

  $\begin{array}{c}{\displaystyle \int f\left(r\right)}={\displaystyle \int r{u}_{z,entrance}}\cdot dr\\ =\frac{r^{1+1}}{1+1}\left(u_{z,entrance}\right)+C\\ =\frac{r^2}{2}\left(u_{z,entrance}\right)+C\end{array}$........... (XIII)

Here, the constant is $C$.

Substitute Equation (XIII) in Equation (IX).

  $\begin{array}{c}\psi =\frac{r^2}{2}\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+\frac{r^2}{2}\left(u_{z,entrance}\right)+C\\ =\frac{r^2}{2}\left(\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+\left(u_{z,entrance}\right)+C\right)\end{array}$

Conclusion:

The stream function is $\frac{r^2}{2}\left(\left(\frac{u_{z,exit}-u_{z,entrance}}{L}\right)\times z+\left(u_{z,entrance}\right)+C\right)$.