Chapter 5
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).
Step-by-Step
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}