Chapter 4, Problem 4.59P

To determine

(a)

Whether ${\nabla }^2\varphi =0$ exist or not, also discuss its meaning

Answer to Problem 4.59P

Yes, ${\nabla }^2\varphi =0$ exist

It’s an incompressible and steady flow.

Explanation of Solution

Given information:

The two dimensional incompressible velocity potential is given as,

$\varphi =xy+x^2-y^2=0$

According to the given information, the velocity potential is,

$\varphi =xy+x^2-y^2=0$

According to the definition,

$u=\frac{\partial \varphi }{\partial x}$

&

$v=\frac{\partial \varphi }{\partial y}$

$u$ & $v$ are the velocity components and $x$ & $y$ are the directions respectively.

The Laplace equation in three dimensions can be represented as,

${\nabla }^2\varphi =\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}+\frac{{\partial }^2\varphi }{\partial z^2}$

Calculation:

According to the explanation,

${\nabla }^2\varphi =\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}$

According to the given information,

$\varphi =xy+x^2-y^2=0$

By differentiating it with respect to $x$,

$\begin{array}{l}\frac{\partial \varphi }{\partial x}=\frac{\partial \left(xy+x^2-y^2\right)}{\partial x}\\ =y+2x\\ \frac{\partial \varphi }{\partial x}=u\end{array}$

To find $\frac{{\partial }^2\varphi }{\partial x^2}$, again differentiate above result with respect to $x$

$\begin{array}{l}\frac{{\partial }^2\varphi }{\partial x^2}=\frac{\partial \left(\frac{\partial \varphi }{\partial x}\right)}{\partial x}\\ =\frac{\partial \left(y+2x\right)}{\partial x}\\ \frac{{\partial }^2\varphi }{\partial x^2}=2\to \left(1\right)\end{array}$

$\frac{{\partial }^2\varphi }{\partial x^2}=\frac{\partial u}{\partial x}\to \left(3\right)$

Similarly,

We can say,

$\begin{array}{l}\frac{\partial \varphi }{\partial y}=\frac{\partial \left(xy+x^2-y^2\right)}{\partial y}\\ =x-2y\end{array}$

$\frac{\partial \varphi }{\partial y}=v$

To find $\frac{{\partial }^2\varphi }{\partial y^2}$, again differentiate above result with respect to $y$

$\begin{array}{l}\frac{{\partial }^2\varphi }{\partial y^2}=\frac{\partial \left(\frac{\partial \varphi }{\partial y}\right)}{\partial y}\\ =\frac{\partial \left(x-2y\right)}{\partial y}\\ \frac{{\partial }^2\varphi }{\partial y^2}=-2\to \left(2\right)\end{array}$

$\frac{{\partial }^2\varphi }{\partial y^2}=\frac{\partial v}{\partial y}\to \left(4\right)$

Therefore, according to the equations $\left(1\right)$ & $\left(2\right)$

$\begin{array}{l}{\nabla }^2\varphi =\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}\\ =2+\left(-2\right)\\ {\nabla }^2\varphi =0\end{array}$

According to the equations, $\left(3\right)$ & $\left(4\right)$

${\nabla }^2\varphi =\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}$

Since, ${\nabla }^2\varphi =0$

We can say that,

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

Therefore, according to the above result,

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ is a continuity equation of two dimensional incompressible steady flow

As the velocity potential satisfies this equation, we can say that this is an incompressible and steady flow.

Conclusion:

Yes, ${\nabla }^2\varphi =0$

As the velocity potential satisfies this equation, we can say that this is an incompressible and steady flow.

To determine

(b)

The stream function.

Answer to Problem 4.59P

Explanation of Solution

Given information:

The two dimensional incompressible velocity potential is given as,

$\varphi =xy+x^2-y^2=0$

The incompressible flow in $xy$ plane can be defined as,

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

For a function $\psi (x,y)$, the above equation can be rewritten as,

$\frac{\partial }{\partial x}\left(\frac{\partial \psi }{\partial y}\right)+\frac{\partial }{\partial y}\left(-\frac{\partial \psi }{\partial x}\right)=0$

According to the above equations, the new function $\psi$ can be defined as,

$u=\frac{\partial \psi }{\partial y}v=-\frac{\partial \psi }{\partial x}$

Calculation:

According to the subpart a,

$u=y+2x$

According to the explanation given,

$u=\frac{\partial \psi }{\partial y}=y+2x$

To find $\psi$, the above equation needs to be integrated,

Therefore,

$\psi =\frac{y^2}{2}+2xy+f\left(x\right)$

To find $f\left(x\right)$, differentiate the above equation with respect to $x$

$\frac{\partial \psi }{\partial x}=2y+f^1\left(x\right)\to \left(1\right)$

Similarly,

$v=x-2y$

$v=-\frac{\partial \psi }{\partial x}=x-2y\to \left(2\right)$

According to $\left(1\right)$ & $\left(2\right)$

We can say,

$\begin{array}{l}2y+f^1\left(x\right)=2y-x\\ f^1\left(x\right)=-x\end{array}$

Therefore,

$f\left(x\right)=-\frac{x^2}{2}$

So, the stream function will be,

$\psi =\left(\frac{y^2-x^2}{2}\right)+2xy$

Conclusion:

The stream function is equal to $\psi =\left(\frac{y^2-x^2}{2}\right)+2xy$.

To determine

(c)

The equation of the streamline that passes through $\left(x,y\right)=\left(2,1\right)$ ?

Answer to Problem 4.59P

$\psi =2.5$

Explanation of Solution

Given information:

The two dimensional incompressible velocity potential is given as,

$\varphi =xy+x^2-y^2=0$

According to the subpart b, the stream function is equal to

$\psi =\left(\frac{y^2-x^2}{2}\right)+2xy$

Therefore, by substituting $x$ & $y$ by relevant values, we can find the equation for the stream line.

Calculation:

By substituting $\left(x,y\right)=\left(2,1\right)$ to the stream function found,

$\begin{array}{l}\psi =\left(\frac{y^2-x^2}{2}\right)+2xy\\ =\left(\frac{{\left(1\right)}^2-{\left(2\right)}^2}{2}\right)+2\left(2\right)\left(1\right)\\ =-1.5+4\\ \psi =2.5\end{array}$

Conclusion:

The stream line passes through point $\left(x,y\right)=\left(2,1\right)$ is equal to, $\psi =2.5$.