Question:

The laminar and turbulent boundary layer thickness of a flat plate are: \[ \delta_l = \frac{5x}{Re_x^{1/2}}, \qquad \delta_t = \frac{0.37\,x}{Re_x^{1/5}} \] The fluid has kinematic viscosity \( \nu = 10^{-6}\ \text{m}^2/\text{s} \). A 100 m long plate moves at 10 m/s. The boundary layer thickness at the rear end of the plate is ____________ m (rounded to two decimals).

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Check Reynolds number first. If it is very high, use the turbulent boundary layer expression even if laminar formulas are provided.
Updated On: Dec 2, 2025
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Correct Answer: 0.58

Solution and Explanation

Velocity:
\[ U = 10\ \text{m/s} \]
Length:
\[ x = 100\ \text{m} \]
Kinematic viscosity:
\[ \nu = 10^{-6}\ \text{m}^2/s \]
Compute Reynolds number:
\[ Re_x = \frac{Ux}{\nu} = \frac{10 \times 100}{10^{-6}} = 1 \times 10^{9} \]
Since \(Re_x = 10^9>5\times10^5\), the flow is fully turbulent.
Thus use turbulent boundary layer formula:
\[ \delta = \frac{0.37 x}{Re_x^{1/5}} \]
Compute \(Re_x^{1/5}\):
\[ (10^9)^{1/5} = 10^{9/5} = 10^{1.8} \approx 63.1 \]
Now compute boundary layer thickness:
\[ \delta = \frac{0.37 \times 100}{63.1} \]
\[ \delta = \frac{37}{63.1} \approx 0.586\ \text{m} \]

Final Answer: 0.59\ \text{m}
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