Holooly Plus Logo

Question 5.38: For a two-dimensional irrotational flow, the velocity potent......

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
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

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}

Related Answered Questions

Question: 5.39

Verified Answer:

The velocity components becomes u=V cosθ and v=V s...
Question: 5.17

Verified Answer:

Given that:            u = 2xy Hence,             ...