Question

The steady velocity field in an inviscid fluid of density 1.5 is given to be $\vec{V} = (y^{2} - x^{2})\hat{i} + (2xy)\hat{j}.$ Neglecting body forces, the pressure gradient at $(x = 1, y = 1)$ is ________________.

• A. $10\hat{j}$
• B. $20\hat{i}$
• C. $-6\hat{i} - 6\hat{j}$
• D. $-4\hat{i} - 4\hat{j}$

Hint:

In inviscid flow without body forces, pressure gradients arise solely from convective acceleration: $\nabla p = -\rho(\vec{V}\cdot\nabla)\vec{V}$.

Answer 1

The Correct Option is C

Using Euler’s equation for an inviscid fluid (no body forces): \[ \nabla p = -\rho\,(\vec{V}\cdot\nabla)\vec{V}. \]
Step 1: Compute directional derivatives.
Given \[ u = y^{2} - x^{2}, \qquad v = 2xy. \] Compute: \[ \frac{\partial u}{\partial x} = -2x, \quad \frac{\partial u}{\partial y} = 2y, \] \[ \frac{\partial v}{\partial x} = 2y, \quad \frac{\partial v}{\partial y} = 2x. \]
Step 2: Compute convective acceleration $(\vec{V}\cdot\nabla)\vec{V$.} At $(x,y)=(1,1)$: \[ u = 0,\ v = 2. \] Thus, \[ a_x = u(-2) + v(2) = 0 + 4 = 4, \] \[ a_y = u(2) + v(2) = 0 + 4 = 4. \] So, \[ (\vec{V}\cdot\nabla)\vec{V} = 4\hat{i} + 4\hat{j}. \]
Step 3: Apply Euler's equation. Density $\rho = 1.5$: \[ \nabla p = -1.5(4\hat{i} + 4\hat{j}) = -6\hat{i} - 6\hat{j}. \]
Final Answer: $-6\hat{i} - 6\hat{j}$