← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

Potential flow · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

For a two-dimensional potential flow, the velocity potential is given by $\phi=x^2y-xy^2+\tfrac13(x^3-y^3)$. Determine the velocity components along the directions $x$ and $y$. Also, determine the stream function $\psi$ and check whether $\phi$ represents a possible case of flow or not.

Technique

$u=\phi_x,\,v=\phi_y$; possibility test $\nabla^2\phi=0$; stream function by integrating the Cauchy–Riemann relations $u=\psi_y,\ v=-\psi_x$.

Solution

Setup. With the convention $\vec q=\nabla\phi$, the velocity components are $u=\dfrac{\partial\phi}{\partial x},\ v=\dfrac{\partial\phi}{\partial y}$. The stream function $\psi$ satisfies the Cauchy–Riemann relations $u=\dfrac{\partial\psi}{\partial y},\ v=-\dfrac{\partial\psi}{\partial x}$. The flow is “possible” (incompressible) iff $\phi$ is harmonic, $\nabla^2\phi=0$.

Step 1 — Velocity components

Step 2 — Check whether the flow is possible (harmonic $\phi$)

Since $\nabla^2\phi=0$, the continuity equation $\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}=0$ is satisfied:

Step 3 — Stream function

Integrate $\dfrac{\partial\psi}{\partial y}=u=x^2+2xy-y^2$ with respect to $y$:

Differentiate and match $\dfrac{\partial\psi}{\partial x}=-v$:

Comparing: $f'(x)=-x^2$, so $f(x)=-\dfrac{x^3}{3}$ (constant absorbed). Hence

Answer