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.