A fluid is flowing over a flat plate. At a distance of 8 cm from the leading edge, the Reynolds number is found to be 25600. The thickness of the boundary layer at this point is

Concept:

The thickness of the boundary layer is given by \(\delta = \frac{{5x}}{{\sqrt {R{e_x}} }}\)

For turbulent flow \(delta = \frac{{0.379\;x}}{{R{e^{\frac{1}{5}}}}}\)

Where δ = Boundary layer thickness, x = Distance of boundary layer from the leading edge, Re_x = Reynold’s number at the distance x from the leading edge

Calculation:

Given:

Since in this given case, Reynold’s number is 25600 which is much lesser than the limit for laminar flow (2 × 10⁵), so the flow is laminar.

 x = 8 cm = 0.08 m

∴\( \delta = \frac{{5x}}{{\sqrt {R{e_x}} }} = \frac{{5{\rm{\;}} × {\rm{\;}}0.08}}{{\sqrt {25600} }} = \;\frac{{0.4}}{{160}} = \frac{1}{{400}} = \;0.0025{\rm{\;m}} = 2.5{\rm{\;mm}}\;\)