Question 5.39: Find out the velocity potential of a uniform stream with vel......

The velocity components becomes

u=V cosθ and v=V sinθ

The flow field is irrotational one and 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}

Hence,                u=\frac{\partial \phi}{\partial x}=V \cos \theta

or                            \phi=(V \cos \theta) x+f_1(y)+C_1                 (5.71)

v=\frac{\partial \phi}{\partial y}=V \sin \theta

or                \phi=(V \sin \theta) y+f_2(x)+C_2           (5.72)

Comparing Eqs. (5.71) and (5.72), we have

f_1(y)=(V \sin \theta) y, f_2(x)=(V \cos \theta) x

Hence, the velocity potential for the flow is

\phi=(V \cos \theta) x+(V \sin \theta) y+C

where C is a constant.