Question 5.40: If stream function for steady flow is given by ψ= y² — x², d......

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.