Let's break down the given problem and solve for **thrust** and **TSFC (Thrust Specific Fuel Consumption)** step by step. --- ## **Given Data** - **Mach number (\( M_0 \))**: 0.75 - **Ambient conditions (sea level):** - \( T_0 = 288.15 \) K - \( P_0 = 101325 \) Pa - **Core air mass flow rate (\( \dot{m}_{core} \)):** 74.83 kg/s - **Bypass ratio (\( \alpha \))**: 1.20 - **Compressor pressure ratio (\( r_c \))**: 15 - **Fuel heating value (\( h_{PR} \))**: 41,400 kJ/kg = 41,400,000 J/kg - **Burner exit total temperature (\( T_{t4} \))**: 1389 K - **Afterburner exit total temperature (\( T_{t7} \))**: 1779 K --- ## **Step 1: Find Total Mass Flow Rate** \[ \dot{m}_{total} = \dot{m}_{core} + \dot{m}_{bypass} = \dot{m}_{core} \times (1 + \alpha) \] \[ \dot{m}_{total} = 74.83 \times (1 + 1.2) = 74.83 \times 2.2 = \boxed{164.63 \text{ kg/s}} \] \[ \dot{m}_{bypass} = \alpha \times \dot{m}_{core} = 1.2 \times 74.83 = \boxed{89.80 \text{ kg/s}} \] --- ## **Step 2: Find Freestream Velocity (\( V_0 \))** For air, \( \gamma = 1.4 \), \( R = 287 \) J/kg·K: \[ a_0 = \sqrt{\gamma R T_0} = \sqrt{1.4 \times 287 \times 288.15} = 340.1 \text{ m/s} \] \[ V_0 = M_0 \times a_0 = 0.75 \times 340.1 = \boxed{255.1 \text{ m/s}} \] --- ## **Step 3: Find Fuel Mass Flow Rate (\( \dot{m}_f \))** Apply energy balance at the **combustor**: \[ \dot{m}_f = \dot{m}_{core} \frac{c_p (T_{t4} - T_{t3})}{\eta_b h_{PR}} \] But \( T_{t3} \) (compressor exit temperature) is not given directly. Assume \( c_p \approx 1005 \) J/kg·K for air, \( \gamma = 1.4 \). ### Find \( T_{t2} \) and \( T_{t3} \): \[ T_{t2} = T_0 \left[1 + \frac{\gamma - 1}{2} M_0^2\right] = 288.15 \left[1 + 0.2 \times 0.75^2\right] = 288.15 \left[1 + 0.1125\right] = 320.7 \text{ K} \] \[ T_{t3} = T_{t2} \times r_c^{(\gamma-1)/\gamma} \] \[ (\gamma-1)/\gamma = 0.286 \] \[ T_{t3} = 320.7 \times 15^{0.286} = 320.7 \times 2.030 = 651 \text{ K} \] --- ### Calculate \( \dot{m}_f \): \[ \dot{m}_f = 74.83 \times \frac{1005 \times (1389 - 651)}{41,400,000} \] \[ = 74.83 \times \frac{1005 \times 738}{41,400,000} \] \[ = 74.83 \times \frac{741690}{41,400,000} \] \[ = 74.83 \times 0.01791 = \boxed{1.34 \text{ kg/s}} \] --- ## **Step 4: Find Jet Exit Velocity (\( V_9 \))** For an ideal engine (perfect expansion), \[ V_9 = \sqrt{2c_p (T_{t7} - T_0)} \] \[ V_9 = \sqrt{2 \times 1005 \times (1779 - 288.15)} = \sqrt{2 \times 1005 \times 1490.85} \] \[ = \sqrt{2,997,299.25} = \boxed{1,731 \text{ m/s}} \] --- ## **Step 5: Find Thrust (\( F \))** \[ F = (\dot{m}_{total} + \dot{m}_f) V_9 - \dot{m}_{total} V_0 \] \[ F = (164.63 + 1.34) \times 1731 - 164.63 \times 255.1 \] \[ F = 165.97 \times 1731 - 41,999 \] \[ = 287,386 - 41,999 = \boxed{245,387 \text{ N}} \] --- ## **Step 6: Find TSFC** \[ TSFC = \frac{\dot{m}_f}{F} \] \[ TSFC = \frac{1.34}{245,387} = 5.46 \times 10^{-6} \text{ s}^{-1} \] Or as is common (per hour): \[ TSFC = \frac{1.34 \times 3600}{245,387} = 0.0197 \text{ kg/(N·hr)} \] --- ## **Summary Table** | Quantity | Value | |----------------|--------------| | Thrust (F) | **245,387 N** | | TSFC | **0.0197 kg/(N·hr)** | --- ## **Summary** - **Thrust:** \( \boxed{245,400 \ \text{N}} \) - **TSFC:** \( \boxed{0.0197 \ \text{kg/(N·hr)}} \) If you need further breakdown or have a specific step you want more details on, just ask!