Question

Air flows with a velocity of \( 2 \, {m/s} \) over a flat stationary surface parallel to its length of \( 0.5 \, {m} \). Kinematic viscosity of air \( \nu \) is \( 1.5 \times 10^{-5} \, {m}^2/{s} \). Using Blasius solution, the boundary layer thickness at the trailing edge of the surface is ________ mm (rounded off to two decimal places).

Hint:

For laminar flow over a flat plate, Blasius boundary layer thickness at distance \( x \) is approximated by \( \delta(x) \approx \frac{5x}{\sqrt{Re_x}} \). Always compute Reynolds number \( {Re}_x = \frac{Ux}{\nu} \) before substitution.

Answer 1

Use Blasius boundary layer thickness formula at the trailing edge. 
The Blasius solution for boundary layer thickness at a distance \( x \) from the leading edge is: \[ \delta(x) = \frac{5.0x}{\sqrt{{Re}_x}}, \quad {where } {Re}_x = \frac{Ux}{\nu} \] Given: \( U = 2 \, {m/s}, \quad x = 0.5 \, {m}, \quad \nu = 1.5 \times 10^{-5} \, {m}^2/{s} \) \[ {Re}_x = \frac{2 \times 0.5}{1.5 \times 10^{-5}} = \frac{1}{1.5 \times 10^{-5}} = 66666.67 \] \[ \delta = \frac{5 \times 0.5}{\sqrt{66666.67}} = \frac{2.5}{258.2} \approx 0.0094 \, {m} \] \[ \delta = 9.40 \, {mm} \]