Question

For an inviscid fluid with density \( \rho = 1 \, \text{kg/m}^3 \), the Cartesian velocity field is: \[ \mathbf{u} = (-2x + y)\mathbf{i} + (2x + y)\mathbf{j} \, \text{m/s} \] Neglecting body forces, find the magnitude of the pressure gradient (Pa/m) at \((x,y) = (1\,\text{m}, 1\,\text{m})\) at \( t = 1\,\text{s} \). (Round off to two decimal places)

Hint:

For inviscid flow, use Euler’s equation. When velocity has no time dependence, only the convective acceleration contributes.

Answer 1

Correct Answer: 5.6

For an inviscid fluid, Euler’s equation gives the pressure gradient as:
\[ \nabla p = -\rho \left( \frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla)\mathbf{u} \right) \] Since the velocity field does not depend on time,
\[ \frac{\partial \mathbf{u}}{\partial t} = 0 \] Thus the pressure gradient becomes:
\[ \nabla p = -\rho (\mathbf{u} \cdot \nabla)\mathbf{u} \] The velocity components are:
\[ u = -2x + y,\quad v = 2x + y \] Their derivatives:
\[ \frac{\partial u}{\partial x} = -2,\; \frac{\partial u}{\partial y} = 1,\; \frac{\partial v}{\partial x} = 2,\; \frac{\partial v}{\partial y} = 1 \] Convective acceleration terms:
\[ a_x = u(-2) + v(1) \] \[ a_y = u(2) + v(1) \] At the point \((1,1)\):
\[ u=-1,\quad v=3 \] Thus:
\[ a_x = (-1)(-2) + 3(1) = 5 \] \[ a_y = (-1)(2) + 3(1) = 1 \] Hence:
\[ \nabla p = -(5\mathbf{i} + 1\mathbf{j}) \] The magnitude is:
\[ |\nabla p| = \sqrt{5^2 + 1^2} = \sqrt{26} = 5.10 \]