## Q. 5.38

For a two-dimensional irrotational flow, the velocity potential is given by $\phi=\frac{1}{2} \ln \left(x^2+y^2\right)$ . Determine 2 the velocity at point (1, 1).

## Verified Solution

Given that:

Velocity potential                                $\phi=\frac{1}{2} \ln \left(x^2+y^2\right)$

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

$u=\frac{\partial \phi}{\partial x}=\frac{1}{2} \frac{2 x}{x^2+y^2}=\frac{x}{x^2+y^2}$

$v=\frac{\partial \phi}{\partial y}=\frac{1}{2} \frac{2 y}{x^2+y^2}=\frac{y}{x^2+y^2}$

Velocity is given by

$\vec{V}=\frac{x}{x^2+y^2} \hat{i}+\frac{x}{x^2+y^2} \hat{j}$

Velocity at point (1, 1) is then

$\left.\vec{V}\right|_{(1,1)}=\frac{1}{1^2+1^2} \hat{i}+\frac{1}{1^2+1^2} \hat{j}=\frac{1}{2} \hat{i}+\frac{1}{2} \hat{j}$