Let's break down the steps for **Part B** (engine efficiency): ## **Given Data** - Jet speed, \( v = 400 \ \text{m/s} \) - Fuel consumption rate, \( \dot{m}_f = 1.6 \ \text{kg/s} \) - Exhaust speed relative to plane, \( u = 1200 \ \text{m/s} \) - Engine consumes 1 kg of fuel per 50 kg of air. - Thrust \( T = 65.9 \ \text{kN} = 65900 \ \text{N} \) --- ## **Step 1: Find Air Intake Rate** Given 1 kg fuel for every 50 kg air, \[ \dot{m}_a = 50 \times \dot{m}_f = 50 \times 1.6 = 80\ \text{kg/s} \] **Total exhaust mass flow:** \[ \dot{m}_e = \dot{m}_a + \dot{m}_f = 80 + 1.6 = 81.6\ \text{kg/s} \] --- ## **Step 2: Engine Efficiency Formula** Propulsive/engine efficiency (\( e \)): \[ e = \frac{\text{Useful power output}}{\text{Total power output}} \] - **Useful power output:** the power used to move the jet forward: \[ \text{Useful power} = T \cdot v \] - **Total power output:** the rate at which the engine imparts kinetic energy to the exhaust gases: \[ \text{Total power} = \frac{1}{2} \dot{m}_e (u^2 - v^2) \] --- ## **Step 3: Calculate Each Term** ### **Useful Power Output** \[ \text{Useful power} = T \cdot v = 65900\ \text{N} \times 400\ \text{m/s} = 26,\!360,\!000\ \text{W} \] ### **Total Power Output** \[ u^2 - v^2 = (1200)^2 - (400)^2 = 1,\!440,\!000 - 160,\!000 = 1,\!280,\!000 \] \[ \text{Total power} = \frac{1}{2} \cdot 81.6\ \text{kg/s} \cdot 1,\!280,\!000\ \text{(m}^2/\text{s}^2) \] \[ = 0.5 \times 81.6 \times 1,280,000 \] \[ = 40.8 \times 1,280,000 \] \[ = 52,224,000\ \text{W} \] --- ## **Step 4: Calculate Efficiency** \[ e = \frac{26,360,000}{52,224,000} = 0.505 \] (Three significant figures) --- ## **Final Answer** \[ \boxed{0.505} \] or \[ \boxed{50.5\%} \] --- ### **Summary Table** | Quantity | Value | |-----------------|---------------| | Useful Power | 26,360,000 W | | Total Power | 52,224,000 W | | Efficiency (\(e\)) | 0.505 (50.5%) | **Let me know if you need more explanation on any step!**