Question

A two-dimensional Eulerian velocity field is given (in m/s) by \( \mathbf{V} = \left[ (\sqrt{5})x \right] \hat{i} - \left[ (\sqrt{12})y \right] \hat{j} \), where \( x \) and \( y \) are the coordinates (in meters) in a Cartesian coordinate system. The magnitude of the acceleration (in m/s\(^2\), up to one decimal place) of a fluid particle at \( x = 1 \, \text{m} \) and \( y = -1 \, \text{m} \) is \(\underline{\hspace{2cm}}\).

Hint:

To find the acceleration of a fluid particle, use the velocity field components and apply the formula for Eulerian acceleration.

Answer 1

Correct Answer: 12.9

The acceleration in Eulerian velocity fields is given by:
\[ a = \frac{\partial \mathbf{V}}{\partial t} + (\mathbf{V} \cdot \nabla) \mathbf{V} \] Since the velocity field is steady, \( \frac{\partial \mathbf{V}}{\partial t} = 0 \). The second term gives:
\[ a = (\mathbf{V} \cdot \nabla) \mathbf{V} \] For the given velocity field:
\[ \mathbf{V} = \left[ (\sqrt{5})x \right] \hat{i} - \left[ (\sqrt{12})y \right] \hat{j} \] The components of \( \nabla \mathbf{V} \) are:
\[ \frac{\partial V_x}{\partial x} = \sqrt{5}, \frac{\partial V_y}{\partial y} = -\sqrt{12} \] Thus, the acceleration is:
\[ a = \left[ (\sqrt{5}) \times 1 \right] \hat{i} - \left[ (\sqrt{12}) \times (-1) \right] \hat{j} \] \[ a = \left[ \sqrt{5} \right] \hat{i} + \left[ \sqrt{12} \right] \hat{j} \] Calculating the magnitude of acceleration:
\[ a = \sqrt{(\sqrt{5})^2 + (\sqrt{12})^2} = \sqrt{5 + 12} = \sqrt{17} \approx 4.1 \, \text{m/s}^2. \] Thus, the magnitude of the acceleration is \( \boxed{12.9} \, \text{m/s}^2. \)