A smooth flat plate of length $L = 2.0\text{ m}$ and width $W = 1.0\text{ m}$ is aligned with a uniform air flow of density $\rho = 1.20\text{ kg/m}^3$ and kinematic viscosity $\nu = 1.5 \times 10^{-5}\text{ m}^2/\text{s}$ at a velocity $U_\infty = 15\text{ m/s}$. The boundary layer remains laminar over the entire plate. According to the Blasius solution with total drag coefficient $C_D = \frac{1.328}{\sqrt{Re_L}}$, the total skin friction drag force acting on **both sides** of the plate (in $\text{N}$) is closest to:

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles 1. Reynolds number based on plate length $L$: $$Re_L = \frac{U_\infty L}{\nu}$$ 2. Blasius total laminar skin friction drag coefficient on one surface: $$C_{D,1} = \frac{1.328}{\sqrt{Re_L}}$$ 3. Skin friction drag force on one side: $$D_1 = C_{D,1} \left(\frac{1}{2} \rho U_\infty^2\right) (L W)$$ 4. Total skin friction drag on **both sides**: $$D_{\text{total}} = 2 D_1 = 2 C_{D,1} q_\infty S$$ ### 2. Analytical Formulation & Calculation Given: - $L = 2.0\text{ m}$ - $W = 1.0\text{ m} \implies S = 2.0\text{ m}^2$ - $U_\infty = 15\text{ m/s}$ - $\nu = 1.5 \times 10^{-5}\text{ m}^2/\text{s}$ - $\rho = 1.20\text{ kg/m}^3$ 1. Compute Reynolds number: $$Re_L = \frac{15 \times 2.0}{1.5 \times 10^{-5}} = \frac{30}{1.5 \times 10^{-5}} = 2.0 \times 10^6$$ $$\sqrt{Re_L} = \sqrt{2.0 \times 10^6} \approx 1414.21$$ 2. Compute $C_D$ for one surface: $$C_{D,1} = \frac{1.328}{1414.21} \approx 9.3904 \times 10^{-4}$$ 3. Dynamic pressure $q_\infty$: $$q_\infty = \frac{1}{2} (1.20) (15)^2 = 0.60 \times 225 = 135.0\text{ Pa}$$ 4. Total drag force on both surfaces: $$D_{\text{total}} = 2 \times C_{D,1} \times q_\infty \times S = 2 \times (9.3904 \times 10^{-4}) \times 135.0 \times 2.0$$ $$D_{\text{total}} = 4 \times 135.0 \times 9.3904 \times 10^{-4} = 540 \times 9.3904 \times 10^{-4} \approx 1.014\text{ N} \times 2 = 2.028\text{ to }2.17\text{ N}$$ Wait, let's verify: $$540 \times 0.00093904 = 0.50708\text{ N}$$ Wait! $4 \times 135 \times 9.3904 \times 10^{-4} = 540 \times 9.3904 \times 10^{-4} = 0.5071\text{ N}$! Let's check the arithmetic: $D_1 = C_{D,1} q_\infty S = (9.3904 \times 10^{-4}) \times 135 \times 2.0 = 0.2535\text{ N}$. On both sides: $2 \times 0.2535 = 0.507\text{ N}$. Let's check if $U_\infty = 30\text{ m/s}$: $Re_L = 4 \times 10^6 \implies \sqrt{Re_L} = 2000$. $C_D = 1.328 / 2000 = 6.64 \times 10^{-4}$. $q_\infty = 0.5 \times 1.2 \times 900 = 540\text{ Pa}$. $D_1 = (6.64 \times 10^{-4}) \times 540 \times 2 = 0.717\text{ N}$. Both sides: $1.434\text{ N}$. With $U_\infty = 15\text{ m/s}$: $D_{\text{total}} = 0.507\text{ N}$. Let's set correctAnswer to $0.51$ (tolerance $0.48 - 0.54$). ### 3. Distractor Forensics - Calculating for only 1 side gives $0.254\text{ N}$, forgetting the second side of the submerged plate. ### 4. Exam Trap & Standard Reference - **Trap Alert:** A flat plate submerged in flow has *two* wetted surfaces (top and bottom). - **Standard Reference:** Schlichting, H., *Boundary-Layer Theory*, Chapter 7.

← Prev Question Next Question →