Question 5.42: Velocity potential of a certain flow field is given as: Φ = ......

Given that:

Velocity potential as $\phi=4 x y$
For irrotational flow, the velocity potential $(\phi)$ is defined as

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

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

Thus, the velocity components becomes

u = 4y
v = 4x

Hence,                                                $\frac{\partial u}{\partial x}=0$

$\frac{\partial v}{\partial y}=0$

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

The above velocity field satisfies the continuity equation for incompressible flow and hence the stream function exists.

From the definition of stream function ψ, we get

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

or                              $\psi=\int u d y=\int 4 y d y$

or                              $\psi=2 y^2+f(x)$                (5.8o)

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

 $\psi=-\int v d x=-\int 4 x d x$

or                            $\psi=-2 x^2+g(y)$          (5.81)

Comparing Eqs. (5.80) and (5.81), we have

$\psi=2 y^2-2 x^2$

Hence, the stream function for the flow is

$\psi=2 y^2-2 x^2$

The streamlines are lines of constant tyand for constant $\psi, 2 y^2-2 x^2$= constant represents a hyperbola.
For different values of $\psi$, streamlines are plotted in Fig. 5.28.