Chapter 9, Problem 63EP

To determine

The expression for stream function.

The plot some streamlines of the flow.

Answer to Problem 63EP

The expression for the stream function is $a{x}^{2}y-\frac{bx{y}^2}{2}+C$.

The following figure represents the streamlines of the flow.

  

Explanation of Solution

Given information:

The incompressible flow filed for which the velocity $u$ component is $u=a{x}^2-bxy$, where $a=0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}$, $b=0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}$ and the velocity along the $x$ axis is $0$. The stream function along the $x$ axis is $0$. The range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$.

Write the expression for the velocity along $x$ direction.

  $v=-\frac{\partial \psi }{\partial x}$  ....... (I)

Here, the stream function along $x$ direction is $\partial \psi$, the change in distance along $x$ direction is $\partial x$ and the velocity along $x$ direction is $v$.

Write the expression for the velocity along $x$ direction.

  $u=\frac{\partial \psi }{\partial y}$  ...... (II)

Here, the stream function along $y$ direction is $\partial \psi$, the change in distance along $y$ direction is $\partial y$ and the velocity along $y$ direction is $u$.

Write the expression for quadric stream function.

  $y=\frac{a{x}^2\pm \sqrt{a^2{x}^4-2bx\psi }}{bx}$   ....... (III)

Here, the stream function is $\psi$, the distance along $x$ direction is $x$, the distance along $y$ direction is $y$ and the constants are $a$ and $b$.

Calculation:

Substitute $a{x}^2-bxy$ for $u$ in Equation (II).

  $\begin{array}{l}a{x}^2-bxy=\frac{\partial \psi }{\partial y}\\ \frac{\partial \psi }{\partial y}=a{x}^2-bxy\\ \partial \psi =\left(a{x}^2-bxy\right)\cdot \partial y\end{array}$   ...... (IV)

Integrate Equation (IV) with respect to $y$.

  $\begin{array}{l}\partial \psi =\left(a{x}^2-bxy\right)\cdot \partial y\\ {\displaystyle \int \partial \psi }={\displaystyle \int \left(a{x}^2-bxy\right)\cdot \partial y+f\left(x\right)}\\ \psi =a{x}^{2}y-\frac{bx{y}^{1+1}}{2}+f\left(x\right)\\ \psi =a{x}^{2}y-\frac{bx{y}^2}{2}+f\left(x\right)\end{array}$   ....... (V)

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

Substitute $a{x}^{2}y-\frac{bx{y}^2}{2}+f\left(x\right)$ for $\psi$ and $0$ for $v$ in Equation (I).

  $\begin{array}{l}0=-\frac{\partial \left(a{x}^{2}y-\frac{bx{y}^2}{2}+f\left(x\right)\right)}{\partial x}\\ -\frac{\partial \left(a{x}^{2}y-\frac{bx{y}^2}{2}+f\left(x\right)\right)}{\partial x}=0\end{array}$   ....... (VI)

Differentiate Equation (V) with respect to $x$.

  $\begin{array}{l}-\frac{\partial \left(a{x}^{2}y-\frac{bx{y}^2}{2}+f\left(x\right)\right)}{\partial x}=0\\ -2a{x}^{2-1}y+\frac{b{x}^{1-1}{y}^2}{2}-\frac{\partial f\left(x\right)}{\partial x}=0\\ -2a{x}^{1}y+\frac{b{y}^2}{2}-\frac{\partial f\left(x\right)}{\partial x}=0\end{array}$   ....... (VII)

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

  $\begin{array}{l}-2a{x}^1\left(0\right)+\frac{b{\left(0\right)}^2}{2}-\frac{\partial f\left(x\right)}{\partial x}=0\\ -\frac{\partial f\left(x\right)}{\partial x}=0\\ \frac{\partial f\left(x\right)}{\partial x}=0\end{array}$   ....... (VIII)

Substitute $0$ for $y$ and $C$ for $\psi$ in Equation (V).

  $\begin{array}{l}a{x}^2\left(0\right)-\frac{bx{\left(0\right)}^2}{2}+f\left(x\right)=C\\ 0-0+f\left(x\right)=C\\ f\left(x\right)=C\end{array}$   ....... (IX)

Here, the constant is $C$.

Substitute $C$ for $f\left(x\right)$ in Equation (V).

  $\psi =a{x}^{2}y-\frac{bx{y}^2}{2}+C$   ....... (X)

Substitute $C$ for $f\left(x\right)$ in Equation (X).

  $\begin{array}{l}\psi =a{x}^{2}y-\frac{bx{y}^2}{2}+0\\ \psi =a{x}^{2}y-\frac{bx{y}^2}{2}\end{array}$

Substitute $0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}$ for $a$ and $0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}$ for $b$ in Equation (III).

  $\begin{array}{l}y=\frac{\left(0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)x^2\pm \sqrt{{\left(0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)}^2{x}^4-2\left(0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)x\psi }}{\left(0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)x}\\ y=\frac{0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times x^2\pm \sqrt{0.2025{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times x^4-1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times x\psi }}{\left(0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)x}\end{array}$..... (XI)

Substitute $1$ for $x$ and $1$ for $y$ in Equation (XI).

  $\begin{array}{l}1=\frac{0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times {\left(1\right)}^2\pm \sqrt{0.2025{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times {\left(1\right)}^4-1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi }}{\left(0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)\left(1\right)}\\ \pm \sqrt{0.2025{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times {\left(1\right)}^4-1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi }=0.75{\left(\text{ft}\cdot \text{s}\right)}^{-1}-0.45{\left(\text{ft}\cdot \text{s}\right)}^{-1}\\ 0.2025{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times {\left(1\right)}^4-1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi =\pm {\left(0.35{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)}^2\end{array}$.. (XII)

Solve Equation (XII) by taking positive sign.

  $\begin{array}{l}0.2025{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times {\left(1\right)}^4-1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi =\pm {\left(0.35{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)}^2\\ -1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi =-0.08\\ \psi =0.05333\end{array}$

Solve Equation (XII) by taking negative sign.

  $\begin{array}{l}0.2025{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times {\left(1\right)}^4-1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi =-{\left(0.35{\left(\text{ft}\cdot \text{s}\right)}^{-1}\right)}^2\\ -1.50{\left(\text{ft}\cdot \text{s}\right)}^{-1}\times \left(1\right)\psi =-0.325\\ \psi =0.216666\end{array}$

The following table shows that the value of stream function with respect to value of $x$ and $y$.

$x$	$y$	$-\psi$	$+\psi$
$0$	$0$	$0$	$0$
$1$	$1$	$0.216666$	$0.053333$
$2$	$2$	$0.4352$	$0.1052$

Plot the streamlines on y x plane by using Equation (XII) and the range of $x$ is varies from $0$ to $3$ and the range of $y$ is varies from $0$ to $4$.

  

  Figure- $(1)$

The Figure $(1)$ represents the streamlines of the flow.

Conclusion:

The expression for the stream function is $a{x}^{2}y-\frac{bx{y}^2}{2}+C$.

The following figure represents the streamlines of the flow.