Let’s break down the problem step-by-step. ### **Step 1: Dynamic Similarity and Reynolds Number** For dynamic similarity in wind tunnel testing, the Reynolds number (\( Re \)) should be the same for both the actual plane (prototype) and the model. \[ Re = \frac{\rho V L}{\mu} \] Where: - \( \rho \) = density of air, - \( V \) = velocity, - \( L \) = characteristic length, - \( \mu \) = dynamic viscosity. ### **Step 2: Gather Information** #### **Prototype (Full-Scale Plane)** - **Weight:** 2500 N - **Altitude:** 3000 m - **Speed:** 50 m/s Standard atmospheric properties at 3000 m: - **Temperature:** ≈ 268 K (−5°C) - **Pressure:** ≈ 70 kPa (0.7 bar) - **Density (\( \rho_p \)):** ≈ 0.909 kg/m³ - **Viscosity (\( \mu_p \)):** ≈ 1.7 × 10⁻⁵ kg/(m·s) #### **Model (Wind Tunnel)** - **Pressure:** 5 bar = 500,000 Pa - **Temperature:** 20°C = 293 K - **Speed:** 60 m/s From the ideal gas law, density (\( \rho \)) at 5 bar and 20°C: \[ \rho = \frac{P}{R \cdot T} \] Where \( R = 287 \) J/(kg·K). \[ \rho_m = \frac{500,000}{287 \times 293} \approx 5.98 \text{ kg/m}^3 \] Dynamic viscosity at 20°C (\( \mu_m \)): ≈ 1.8 × 10⁻⁵ kg/(m·s) (for air). ### **Step 3: Reynolds Number Similarity** Let \( \lambda \) be the scale ratio (\( L_m = \lambda L_p \)). Set \( Re_m = Re_p \): \[ \frac{\rho_m V_m L_m}{\mu_m} = \frac{\rho_p V_p L_p}{\mu_p} \] \[ \frac{\rho_m V_m \lambda L_p}{\mu_m} = \frac{\rho_p V_p L_p}{\mu_p} \] \[ \lambda = \frac{\rho_p V_p \mu_m}{\rho_m V_m \mu_p} \] ### **Step 4: Plug in Values** \[ \lambda = \frac{0.909 \times 50 \times 1.8 \times 10^{-5}}{5.98 \times 60 \times 1.7 \times 10^{-5}} \] Calculate numerator: \[ 0.909 \times 50 = 45.45 \] \[ 45.45 \times 1.8 \times 10^{-5} = 8.181 \times 10^{-4} \] Calculate denominator: \[ 5.98 \times 60 = 358.8 \] \[ 358.8 \times 1.7 \times 10^{-5} = 6.0996 \times 10^{-3} \] Now, the scale: \[ \lambda = \frac{8.181 \times 10^{-4}}{6.0996 \times 10^{-3}} \approx 0.134 \] So, the model should be about **1/7.5th** the size of the prototype. --- ### **Step 5: Lift Force on the Model** Lift force is given by: \[ L = C_L \frac{1}{2} \rho V^2 S \] For dynamic similarity, \( C_L \) is the same for model and prototype. The ratio of lift forces will be: \[ \frac{L_m}{L_p} = \frac{\rho_m V_m^2 S_m}{\rho_p V_p^2 S_p} \] Since \( S_m = \lambda^2 S_p \): \[ L_m = L_p \frac{\rho_m}{\rho_p} \left(\frac{V_m}{V_p}\right)^2 \lambda^2 \] Plug in the numbers: - \( L_p = 2500 \) N - \( \rho_m = 5.98 \) - \( \rho_p = 0.909 \) - \( V_m = 60 \) - \( V_p = 50 \) - \( \lambda = 0.134 \) \[ L_m = 2500 \times \frac{5.98}{0.909} \times \left(\frac{60}{50}\right)^2 \times (0.134)^2 \] Now, calculate step by step: 1. \( \frac{5.98}{0.909} \approx 6.58 \) 2. \( \left(\frac{60}{50}\right)^2 = (1.2)^2 = 1.44 \) 3. \( (0.134)^2 = 0.01796 \) Now combine: \[ L_m = 2500 \times 6.58 \times 1.44 \times 0.01796 \] \[ = 2500 \times 6.58 = 16450 \] \[ 16450 \times 1.44 = 23688 \] \[ 23688 \times 0.01796 \approx 425 \text{ N} \] --- ## **Summary of Answers** ### **1. Scale of the Model:** \[ \boxed{0.134} \] (The model should be 13.4% the size of the prototype.) ### **2. Lift Force on the Model:** \[ \boxed{425 \text{ N}} \] If you need a more detailed step-by-step explanation or clarification, let me know!