Question

An incompressible fluid is flowing between two infinitely large parallel plates separated by 5 mm distance. The bottom plate is stationary and the top plate is moving at a constant velocity of 5 mm/s in the direction parallel to the bottom plate. The flow of the fluid between the plates is steady, two-dimensional, laminar, and the variation of fluid velocity is linear between the plates. A square fluid element of 1 mm side is considered at equal distance from both the plates in the flow field such that one of its sides is parallel to the plates. The magnitude of circulation in mm$^2$/s (in integer) along the edges of the square fluid element is .................

Hint:

For linear Couette flow, vorticity is constant and equal to velocity gradient. Circulation around a small element is simply vorticity times area.

Answer 1

Correct Answer: 1

Step 1: Velocity distribution (Couette flow).
For simple Couette flow: \[ u(y) = \frac{U}{h} y, \] where $U$ = velocity of top plate, $h$ = gap between plates. Here, $U = 5 \,\text{mm/s}, \; h = 5 \,\text{mm}$. So, \[ u(y) = \frac{5}{5} y = y \text{(mm/s)}. \]

Step 2: Vorticity and circulation relation.
Circulation $\Gamma$ around a closed contour is related to vorticity $\zeta$: \[ \Gamma = \zeta \cdot A, \] where $A$ = area of contour.

Step 3: Vorticity for Couette flow.
For 2D flow $u=u(y), v=0$: \[ \zeta = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = - \frac{du}{dy}. \] Here, \[ \frac{du}{dy} = 1 \,\text{s}^{-1}. \] So, \[ \zeta = -1 \,\text{s}^{-1}. \]

Step 4: Compute circulation.
Area of square = $1 \,\text{mm} \times 1 \,\text{mm} = 1 \,\text{mm}^2$. \[ \Gamma = \zeta \cdot A = (-1)(1) = -1 \,\text{mm}^2/\text{s}. \] Since question asks for magnitude, we take: \[ |\Gamma| = 1 \,\text{mm}^2/\text{s}. \] Correction: Rechecking — velocity gradient $\frac{U}{h} = \frac{5}{5} = 1 \,\text{s}^{-1}$. So vorticity = $-1 \,\text{s}^{-1}$. Area = $(1\,\text{mm})^2 = 1 \,\text{mm}^2$. Thus circulation magnitude = 1. Final Answer:
\[ \boxed{1} \]