Topic: Fluid Mechanics
For a two-dimensional, steady, incompressible and irrotational flow, a stream function is given by $(x^2-y^2)$. The potential function of the flow will be :
- A. $2xy+f(x)$
- B. 2xy + constant
- C. $Z(x^2-y^2)$
- D. $2xy+f(y)$
Answer: B. 2xy + constant
Explanation: Given stream function $\psi = x^2 - y^2$.
Velocity components: $u = \frac{\partial\psi}{\partial y} = -2y$ and $v = -\frac{\partial\psi}{\partial x} = -2x$.
Relating velocity components to the potential function $\phi$ using $u = -\frac{\partial\phi}{\partial x}$ and $v = -\frac{\partial\phi}{\partial y}$:
$-\frac{\partial\phi}{\partial x} = -2y \Rightarrow \frac{\partial\phi}{\partial x} = 2y$. Integrating w.r.t $x$: $\phi = 2xy + f(y)$.
$-\frac{\partial\phi}{\partial y} = -2x \Rightarrow \frac{\partial (2xy + f(y))}{\partial y} = 2x \Rightarrow 2x + f'(y) = 2x \Rightarrow f'(y) = 0$.
Therefore, $f(y) = \text{constant}$. The potential function is $\phi = 2xy + \text{constant}$.