Let's break down the problem and solve step by step. ### **Given Data** - **Wingspan (\(b\))** = 14.85 m - **Wing area (\(S\))** = 11.45 m² - **Maximum weight (\(W_{max}\))** = 1350 kgf - **Fuel weight (\(W_f\))** = 295 kgf - **Power (P)** = 85 hp - **Oswald efficiency factor (\(e\))** = 0.7 - **Zero-lift drag coefficient (\(C_{D0}\))** = 0.03 - **Propeller efficiency (\(\eta_p\))** = 0.9 - **Specific fuel consumption (\(c\))** = 0.2 kg/hp/hr - **Gravity (\(g\))** ≈ 9.81 m/s² --- ## **Step 1: Convert Units** 1. **Weight to Newtons** \( W = 1350 \text{ kgf} \times 9.81 = 13,243.5 \text{ N} \) \( W_f = 295 \text{ kgf} \times 9.81 = 2,894.0 \text{ N} \) 2. **Power (hp to Watts)** \( 1 \text{ hp} = 745.7 \text{ W} \) \( 85 \text{ hp} = 85 \times 745.7 = 63,384.5 \text{ W} \) --- ## **Step 2: Maximum Endurance Formula for Propeller Aircraft** For **maximum endurance** (E) for a propeller-driven aircraft: \[ E = \frac{\eta_p}{c} \int_{W_{final}}^{W_{initial}} \frac{dW}{P_{req}} \] For small fuel weight compared to the aircraft weight, we approximate: \[ E_{max} \approx \frac{\eta_p}{c} \frac{W_{f}}{P_{req}} \] Where: - \(\eta_p\) = propeller efficiency - \(c\) = specific fuel consumption (kg/hp/hr) - \(W_{f}\) = fuel weight (N) - \(P_{req}\) = power required for maximum endurance (minimum power required for level flight) --- ## **Step 3: Calculate Power Required at Maximum Endurance** Maximum endurance occurs at **minimum power required**. For a propeller-driven aircraft, this is at a speed where: \[ P_{req} = \frac{D \cdot V}{\eta_p} \] But for endurance calculations, we use the minimum power required, which occurs at a specific CL: For minimum power: \[ C_L^{3/2}/C_D = \text{max} \] But let's use a simplified power required formula at steady, level flight: \[ P_{req} = \frac{2 W^{3/2}}{\sqrt{\rho S} \sqrt{\pi e AR} C_{D0}^{1/2}} \] But since that's complicated, let's use the standard endurance formula: \[ E = \frac{\eta_p}{c} \frac{W_f}{P_{req}} \] We need to find \(P_{req}\) at the condition for minimum power required. --- ## **Step 4: Calculate Power Required for Level Flight** The minimum power required for level flight (for endurance) occurs at: \[ C_L = \sqrt{\frac{3 C_{D0} \pi e AR}{}} \] But, to simplify, let's use: \[ P_{req} = \frac{2 W^{3/2}}{\rho^{1/2} S^{1/2} (C_{L}^{1/2})} \] But we need density at sea level: - \(\rho = 1.225 \text{ kg/m}^3\) (sea level) --- #### **Aspect Ratio (AR):** \[ AR = \frac{b^2}{S} = \frac{(14.85)^2}{11.45} = \frac{220.5225}{11.45} = 19.26 \] --- #### **To Find \(C_L\) and \(V\) at minimum power required:** For minimum power required: \[ C_{L_{minP}} = \sqrt{\frac{3 C_{D0} \pi e AR}{}} \] Plug in values: \[ C_{L_{minP}} = \sqrt{3 \times 0.03 \times \pi \times 0.7 \times 19.26} \] \[ = \sqrt{3 \times 0.03 \times 3.1416 \times 0.7 \times 19.26} \] \[ = \sqrt{3 \times 0.03 \times 3.1416 \times 13.482} \] \[ = \sqrt{3 \times 0.03 \times 42.395} \] \[ = \sqrt{3 \times 1.27185} \] \[ = \sqrt{3.8156} \] \[ = 1.955 \] --- #### **Find Velocity at this \(C_L\):** \[ C_L = \frac{W}{0.5 \rho V^2 S} \] \[ V = \sqrt{ \frac{2W}{\rho S C_L} } \] Plug in values: - \(W = 13,243.5\) N - \(\rho = 1.225\) kg/m³ - \(S = 11.45\) m² - \(C_L = 1.955\) \[ V = \sqrt{ \frac{2 \times 13,243.5}{1.225 \times 11.45 \times 1.955} } \] \[ = \sqrt{ \frac{26,487}{27.450} } \] \[ = \sqrt{ 965.11 } \] \[ = 31.07 \text{ m/s} \] --- #### **Calculate Drag at this condition:** \[ C_D = C_{D0} + \frac{C_L^2}{\pi e AR} \] \[ = 0.03 + \frac{(1.955)^2}{\pi \times 0.7 \times 19.26} \] \[ = 0.03 + \frac{3.824}{42.395} \] \[ = 0.03 + 0.0902 \] \[ = 0.1202 \] --- \[ D = 0.5 \rho V^2 S C_D \] \[ = 0.5 \times 1.225 \times (31.07)^2 \times 11.45 \times 0.1202 \] \[ = 0.5 \times 1.225 \times 966.34 \times 11.45 \times 0.1202 \] \[ = 0.5 \times 1.225 \times 966.34 = 0.5 \times 1,183.26 = 591.63 \] \[ 591.63 \times 11.45 = 6,774.15 \] \[ 6,774.15 \times 0.1202 = 814.31 \text{ N} \] --- #### **Power Required:** \[ P_{req} = D \times V = 814.31 \times 31.07 = 25,316.7 \text{ W} \] \[ = \frac{25,316.7}{745.7} = 33.97 \text{ hp} \] --- ## **Step 5: Maximum Endurance Calculation** Recall: \[ E_{max} = \frac{\eta_p}{c} \cdot \frac{W_f}{P_{req}} \] - \(\eta_p = 0.9\) - \(c = 0.2 \text{ kg/hp/hr}\) - \(W_f = 295 \text{ kgf}\), but since the specific fuel consumption is per hour, and the fuel is in kgf, let's use mass: 295 kg - \(P_{req} = 33.97 \text{ hp}\) \[ \text{Fuel consumption per hour at } P_{req} = 33.97 \times 0.2 = 6.794 \text{ kg/hr} \] \[ \text{Total endurance} = \frac{295}{6.794} = 43.41 \text{ hr} \] --- ## **Final Answer** \[ \boxed{43.4 \text{ hr}} \] --- ### **Summary Table** | Quantity | Value | |-------------------------|-----------| | Power required (hp) | 33.97 | | Fuel consumption (kg/hr)| 6.794 | | Fuel weight (kg) | 295 | | Endurance (hr) | 43.4 | --- **The maximum endurance at sea level is \(\boxed{43.4 \text{ hr}}\).**