Question

In a potential flow field, if the stream function \( \psi = xy^2 \), then the velocity potential \( \phi \) is:

• A. \( \frac{x^2 - y^2}{2} \)
• B. \( \frac{x^2 + y^2}{2} \)
• C. \( y \left(x^2 + \frac{y^2}{3} \right) \)
• D. \( y \left(x^2 - \frac{y^2}{3} \right) \)

Hint:

In potential flow, the velocity potential \( \phi \) can be determined by integrating the stream function \( \psi \) with respect to the appropriate variables and using the boundary conditions.

Answer 1

The Correct Option is D

For potential flow, the stream function \( \psi \) and velocity potential \( \phi \) are related through the equations: \[ u = \frac{\partial \phi}{\partial x}, \quad v = \frac{\partial \phi}{\partial y} \] and the stream function \( \psi \) is related to the velocity components by: \[ u = \frac{\partial \psi}{\partial y}, \quad v = -\frac{\partial \psi}{\partial x} \] The stream function is given as \( \psi = xy^2 \). Now, differentiate \( \psi \) with respect to \( y \) and \( x \): \[ u = \frac{\partial \psi}{\partial y} = 2xy \] \[ v = -\frac{\partial \psi}{\partial x} = -y^2 \] Now, we integrate \( u = 2xy \) with respect to \( x \) to find \( \phi \): \[ \phi = \int 2xy \, dx = x^2y + f(y) \] Next, differentiate \( \phi = x^2y + f(y) \) with respect to \( y \) and equate it to \( v = -y^2 \): \[ \frac{\partial \phi}{\partial y} = x^2 - f'(y) = -y^2 \] Solving for \( f'(y) \), we get: \[ f'(y) = x^2 + y^2 \] Thus, the velocity potential \( \phi \) is: \[ \phi = y \left( x^2 - \frac{y^2}{3} \right) \] Thus, the correct answer is (D).