Question

If the Reynolds number (\(Re\)) at a point on a flat plate doubles while the flow remains laminar, what happens to the boundary layer thickness \(\delta\)?

• A. Decreases
• B. Increases
• C. Remains same
• D. Becomes zero

Hint:

For laminar flow over a flat plate, \[ \boxed{ \delta \propto \frac{1}{\sqrt{Re}} } \] Therefore, \[ \boxed{ \text{Higher Reynolds number} \Longrightarrow \text{Thinner boundary layer}. } \]

Answer 1

The Correct Option is A

Step 1: Recall the boundary layer thickness relation. For laminar flow over a flat plate, \[ \boxed{ \delta \approx \frac{5x}{\sqrt{Re_x}}, } \] where \[ Re_x = \frac{\rho Ux}{\mu}. \]

Step 2: Determine the effect of doubling Reynolds number. Since \[ \delta \propto \frac{1}{\sqrt{Re}}, \] if \[ Re \] doubles, \[ \delta_{\text{new}} = \frac{\delta}{\sqrt2}. \] Thus, the boundary layer becomes thinner. Therefore, \[ \boxed{\text{Boundary layer thickness decreases}.} \] Hence, \[ \boxed{(A)} \] is the correct answer.