Let's break down and solve the problem step by step: ## **Given Data** - **Airfoil:** NACA 23012 - **Chord length, \( c \):** 0.19 m - **Wing span, \( b \):** 2 m - **Lift measured, \( L \):** 20.0 N - **Air pressure, \( p \):** 1 atm = 101,325 Pa - **Temperature, \( T \):** 303 K - **Velocity, \( V \):** 42 m/s ## **Step 1: Calculate Air Density (\( \rho \))** Using the ideal gas law: \[ \rho = \frac{p}{R T} \] Where \( R = 287 \, \text{J/(kg·K)} \) for air. \[ \rho = \frac{101325}{287 \times 303} = \frac{101325}{87081} \approx 1.163 \,\text{kg/m}^3 \] --- ## **Step 2: Calculate Wing Area (\( S \))** \[ S = b \times c = 2 \times 0.19 = 0.38 \, \text{m}^2 \] --- ## **Step 3: Calculate Lift Coefficient (\( C_L \))** The lift equation: \[ L = \frac{1}{2} \rho V^2 S C_L \] Solving for \( C_L \): \[ C_L = \frac{2L}{\rho V^2 S} \] Plug in the numbers: - \( L = 20 \) N - \( \rho = 1.163 \) kg/m³ - \( V = 42 \) m/s - \( S = 0.38 \) m² \[ C_L = \frac{2 \times 20}{1.163 \times 42^2 \times 0.38} \] \[ = \frac{40}{1.163 \times 1764 \times 0.38} \] \[ = \frac{40}{1.163 \times 670.32} \] \[ = \frac{40}{780.482} \] \[ \approx 0.0513 \] --- ## **Step 4: Find the Angle of Attack (\( \alpha \)) Using the Graph** From the **first graph** (Lift coefficient \( C_L \) vs. Angle of Attack for NACA 23012): - Locate \( C_L = 0.051 \) on the vertical axis. - Trace horizontally to the curve, then down to the horizontal axis (angle of attack). **From the graph:** - The curve is nearly linear for small angles. - \( C_L = 0 \) at about \(-2^\circ\). - \( C_L \approx 1.0 \) at about \(10^\circ\). So, the slope (lift curve slope) is approximately: \[ \frac{\Delta C_L}{\Delta \alpha} \approx \frac{1.0}{12} \approx 0.083 \text{ per degree} \] For \( C_L = 0.051 \): \[ \alpha = \frac{C_L}{0.083} \approx \frac{0.051}{0.083} \approx 0.61^\circ \] --- ## **Step 5: Final Answer** **The angle of attack required is approximately:** \[ \boxed{0.6^\circ} \] --- ### **Summary of Steps:** 1. **Calculated air density \(\rho = 1.163 \, \text{kg/m}^3\).** 2. **Calculated wing area \(S = 0.38 \, \text{m}^2\).** 3. **Calculated lift coefficient \(C_L \approx 0.051\).** 4. **Used the graph to find the corresponding angle of attack: \(\alpha \approx 0.6^\circ\).** If you need to show the process using the graph visually, let me know!

Let's carefully solve the problem step by step with detailed calculations: --- ### **Given Data:** - **Airfoil:** NACA 23012 - **Chord length, \( c \):** 0.19 m - **Wing span, \( b \):** 2.0 m - **Lift force, \( L \):** 20.0 N - **Air pressure, \( p \):** 1 atm = 101,325 Pa - **Temperature, \( T \):** 303 K - **Flow velocity, \( V \):** 42 m/s --- ### **Step 1: Calculate Air Density (\( \rho \))** Using the ideal gas law: \[ \rho = \frac{p}{R T} \] where: - \( R = 287 \, \text{J/(kg·K)} \) Calculate: \[ \rho = \frac{101,325}{287 \times 303} = \frac{101,325}{86,961} \approx 1.164 \, \text{kg/m}^3 \] --- ### **Step 2: Calculate Wing Area (\( S \))** \[ S = b \times c = 2.0 \times 0.19 = 0.38 \, \text{m}^2 \] --- ### **Step 3: Calculate Lift Coefficient (\( C_L \))** The lift equation: \[ L = \frac{1}{2} \rho V^2 S C_L \] Rearranged for \( C_L \): \[ C_L = \frac{2L}{\rho V^2 S} \] Plugging in the numbers: \[ C_L = \frac{2 \times 20}{1.164 \times (42)^2 \times 0.38} \] Calculate denominator: \[ (42)^2 = 1764 \] \[ 1.164 \times 1764 = 2053.7 \] \[ 2053.7 \times 0.38 = 780.7 \] Now: \[ C_L = \frac{40}{780.7} \approx 0.0513 \] --- ### **Step 4: Determine the Angle of Attack (\( \alpha \))** Using the **graph for NACA 23012** (from the provided appendix): - At small angles, \( C_L \) varies approximately linearly with \( \alpha \). - From the graph, the lift curve slope (\( \frac{dC_L}{d\alpha} \)) is roughly: \[ \text{At } \alpha = 0^\circ, C_L \approx 0 \] \[ \text{At } \alpha \approx 10^\circ, C_L \approx 1.0 \] Thus: \[ \frac{\Delta C_L}{\Delta \alpha} \approx \frac{1.0 - 0}{10^\circ - 0} = 0.1 \text{ per degree} \] Alternatively, from the graph's more precise data, the slope is about **0.083 per degree**. Using this slope: \[ \alpha = \frac{C_L}{\text{slope}} \approx \frac{0.0513}{0.083} \approx 0.618^\circ \] --- ### **### **Final Answer:** \[ \boxed{ \text{The required angle of attack, } \alpha \approx \mathbf{0.62^\circ} } \] --- ### **Summary:** - **Air density:** 1.164 kg/m³ - **Wing area:** 0.38 m² - **Lift coefficient:** approximately 0.0513 - **Corresponding angle of attack:** approximately **0.62 degrees** This small angle of attack indicates the wing is nearly level at the given lift force.