A smooth flat plate of length $L = 1.0\text{ m}$ is held parallel to an incompressible air stream at velocity $V_\infty = 15\text{ m/s}$. The kinematic viscosity of air is $\nu = 1.5 \times 10^{-5}\text{ m}^2/\text{s}$. Using the Blasius exact solution where boundary layer thickness $\delta = \frac{5.0 x}{\sqrt{Re_x}}$, what is the boundary layer thickness $\delta$ (in mm) at the trailing edge $x = 1.0\text{ m}$?

Answer Key:

5.0

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Local Reynolds Number:** * $Re_L = \frac{V_\infty L}{\nu} = \frac{15 \times 1.0}{1.5 \times 10^{-5}} = 10^6$. 2. **Stage 2: Flow Regime Check:** $Re_L = 10^6$ is within laminar transition limits for quiet flow conditions. 3. **Stage 3: Blasius Thickness Formula:** * $\delta = \frac{5.0 L}{\sqrt{Re_L}} = \frac{5.0 \times 1.0}{\sqrt{10^6}} = \frac{5.0}{1000} = 0.005\text{ m}$. * In millimeters: $0.005 \times 1000 = 5.0\text{ mm}$. 4. **Stage 4: Common Trap:** Forgetting to convert meters to millimeters or using momentum thickness $\theta$ ($0.664$) instead of $99\%$ thickness $\delta$. 5. **Stage 5: Official Reference:** Hermann Schlichting, *Boundary-Layer Theory*, Chapter 7.

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