Question 5.37: The velocity component in a two-dimensional flow field for a......

Given data:

Velocity components          u=\frac{y^3}{3}+2 x-x^2 y, \text { and } v=x y^2-2 y-\frac{x^3}{3}

Hence,                    \frac{\partial u}{\partial y}=y^2-x^2 \text {, and }

\frac{\partial v}{\partial x}=y^2-x^2

For a two-dimensional, incompressible flow the condition of irrotationality is

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

Substituting the values of \frac{\partial u}{\partial x} \text { and } \frac{\partial v}{\partial y} in the expression of w_z , we get

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

Hence, the flow is irrotational and the velocity potential exists.

From the definition of velocity potential, we have

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

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

Thus,                      u=\frac{\partial \phi}{\partial x}=\frac{y^3}{3}+2 x-x^2 y

or                          \phi=\frac{x y^3}{3}+x^2-\frac{x^3 y}{3}+f_1(y)+C_3                 (5.69)

And                      v=\frac{\partial \phi}{\partial y}=x y^2-2 y-\frac{x^3}{3}

or                        \phi=\frac{x y^3}{3}-y^2-\frac{x^3 y}{3}+f_2(x)+C_4           (5.70)

Comparing Eqs. (5.69) and (5.70), we have

f_1(y)=-y^2, f_2(x)=x^2

Hence, the velocity potential for the flow is

\phi=\frac{x y^3}{3}+x^2-y^2-\frac{x^3 y}{3}+C

where C is a constant.