## Q. 5.40

If stream function for steady flow is given by ψ= y² — x², determine whether the flow is rotational or irrotational. Find the potential function, if the flow is irrotational and vorticity, if it is rotational.

## Verified Solution

Given data:

Stream function ψ=y²-x²

From the definition of stream function ψ, we get

$u=\frac{\partial \psi}{\partial y}$

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

Thus, the velocity components become

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

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

The velocity is then $\vec{V}=u \hat{i}+v \hat{j}=2 y \hat{i}+2 x \hat{j}$

Hence,                      $\frac{\partial v}{\partial x}=2$

$\frac{\partial u}{\partial y}=2$

The rotation is given by

$\omega_z=\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)=\frac{1}{2}(2-2)=0$

Since the rotation is zero, the flow is irrotational and hence the velocity potential exists.

For irrotational flow, the velocity potential $(\phi)$ is defined as

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

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

Thus,                          $u=\frac{\partial \phi}{\partial x}=2 y$

or                        $\phi=2 x y+f_1(y)+C_1$            (5.73)

And                  $v=\frac{\partial \phi}{\partial y}=2 x$

or                          $\phi=2 x y+f_2(x)+C_2$              (5.74)

Comparing Eqs. (5.73) and (5.74), we have

$f_1(y)=0, f_2(x)=0$

Hence, the velocity potential for the flow is

$\phi=2 x y+C$

where C is a constant.