Question

The two-dimensional velocity field \( \mathbf{V} \) of a flow in a Cartesian coordinate system is given in dimensionless form by \( \mathbf{V} = (x^2 - axy) \hat{i} + \left( bxy - \frac{y^2}{2} \right) \hat{j} \). Here, \( \hat{i} \) and \( \hat{j} \) are the unit vectors along the \( x \) and \( y \) directions respectively, \( a \) and \( b \) are independent of \( x \), \( y \) and time. If the flow is incompressible, then the value of \( (a - b) \), up to one decimal place, is \(\underline{\hspace{1cm}}\).

Hint:

For incompressible flow, use the condition \( \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \) to solve for unknown parameters.

Answer 1

Correct Answer: 0.9

For an incompressible flow, the condition \( \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 \) must hold, where \( u = x^2 - axy \) and \( v = bxy - \frac{y^2}{2} \).
Taking the partial derivatives:
\[ \frac{\partial u}{\partial x} = 2x - ay, \frac{\partial v}{\partial y} = bx - y \] The incompressibility condition becomes:
\[ 2x - ay + bx - y = 0 \] Simplifying, we get:
\[ (2x + bx) - (a + 1)y = 0 \] For this to hold true for all \( x \) and \( y \), we must have:
\[ b = -2, a + 1 = 0 $\Rightarrow$ a = -1 \] Thus, \( (a - b) = (-1 - (-2)) = 1 \).
So, the value of \( (a - b) \) is \( 1.0 \).