McDonnell Aircraft's F2H Banshee fighter first flew in 1947 and some served into the 1960s. The aircraft used 2 Westinghouse J34 -WE-30 turbojets delivering 14,000 N of thrust each.giving a maximum speed of 945km/h and a range of almost 2090 km . The F2H could climb at about 2250m/min and reach a service ceiling of 14,800 m. If the airplane is operating at an altitude of 6100 m with a choked nozzle exit (Me=1, Ye=1.33) and the nozzle exit area A_(e)=0.116m^2 and the exit pressure is p_(e)=62 kPa find the following; (a) gross thrust produced, (b) percentage of gross thrust caused by the pressure imbalance at the nozzle exit, (c) exit mass flow if T_(e)=537.8^circ C (d) ram drag at a flight velocity of 167.6m/s f/a=0.02 (e) net thrust, (f) thrust power produced, (g) exit velocity V_(e) (h) effective exhaust velocity V_(eff) (i) propulsive efficiency n_(p) (i) nth and n_(0) for Q_(f)=43.96MJ/kg) (k) the specific fuel consumption, cj

To solve the given problem, we need to use various principles of fluid dynamics and thermodynamics. Let's go through each part step-by-step.<br /><br />### Given Data:<br />- Altitude: \( h = 6100 \) m<br />- Choked nozzle exit: \( M_e = 1 \)<br />- Specific heat ratio: \( \gamma_e = 1.33 \)<br />- Nozzle exit area: \( A_e = 0.116 \) m²<br />- Exit pressure: \( p_e = 62 \) kPa<br />- Exit temperature: \( T_e = 537.8^\circ C = 810.95 \) K (converted to Kelvin)<br />- Flight velocity: \( V_0 = 167.6 \) m/s<br />- Fuel-to-air ratio: \( f/a = 0.02 \)<br />- Heat of combustion: \( Q_f = 43.96 \) MJ/kg<br /><br />### Constants:<br />- Universal gas constant: \( R = 287 \) J/(kg·K)<br /><br />### Part (a): Gross Thrust Produced<br />Gross thrust (\( F_g \)) is given by:<br />\[ F_g = \dot{m} V_e + (p_e - p_0) A_e \]<br /><br />Since the nozzle is choked, the exit Mach number \( M_e = 1 \), and we can find the exit velocity \( V_e \) using:<br />\[ V_e = M_e \sqrt{\gamma_e R T_e} \]<br />\[ V_e = 1 \times \sqrt{1.33 \times 287 \times 810.95} \]<br />\[ V_e \approx 574.2 \text{ m/s} \]<br /><br />Next, we need the mass flow rate (\( \dot{m} \)):<br />\[ \dot{m} = \rho_e A_e V_e \]<br />Where \( \rho_e \) is the density at the exit, which can be found using the ideal gas law:<br />\[ \rho_e = \frac{p_e}{R T_e} \]<br />\[ \rho_e = \frac{62000}{287 \times 810.95} \]<br />\[ \rho_e \approx 0.267 \text{ kg/m}^3 \]<br /><br />Now, calculate the mass flow rate:<br />\[ \dot{m} = 0.267 \times 0.116 \times 574.2 \]<br />\[ \dot{m} \approx 17.75 \text{ kg/s} \]<br /><br />Finally, calculate the gross thrust:<br />\[ F_g = \dot{m} V_e + (p_e - p_0) A_e \]<br />Assuming standard atmospheric pressure at 6100 m altitude is approximately 47.35 kPa:<br />\[ F_g = 17.75 \times 574.2 + (62000 - 47350) \times 0.116 \]<br />\[ F_g \approx 10191.15 + 1707.4 \]<br />\[ F_g \approx 11900 \text{ N} \]<br /><br />### Part (b): Percentage of Gross Thrust Caused by Pressure Imbalance<br />\[ \text{Pressure thrust} = (p_e - p_0) A_e \]<br />\[ \text{Pressure thrust} = 1707.4 \text{ N} \]<br /><br />Percentage of gross thrust due to pressure imbalance:<br />\[ \text{Percentage} = \left( \frac{1707.4}{11900} \right) \times 100 \]<br />\[ \text{Percentage} \approx 14.35\% \]<br /><br />### Part (c): Exit Mass Flow if \( T_e = 537.8^\circ C \)<br />Already calculated in part (a):<br />\[ \dot{m} \approx 17.75 \text{ kg/s} \]<br /><br />### Part (d): Ram Drag at a Flight Velocity of 167.6 m/s<br />Ram drag (\( D_r \)) is given by:<br />\[ D_r = \dot{m} V_0 \]<br />\[ D_r = 17.75 \times 167.6 \]<br />\[ D_r \approx 2975.4 \text{ N} \]<br /><br />### Part (e): Net Thrust<br />Net thrust (\( F_n \)) is given by:<br />\[ F_n = F_g - D_r \]<br />\[ F_n = 11900 - 2975.4 \]<br />\[ F_n \approx 8924.6 \text{ N} \]<br /><br />### Part (f): Thrust Power Produced<br />Thrust power (\( P_t \)) is given by:<br />\[ P_t = F_n \times V_0 \]<br />\[ P_t = 8924.6 \times 167.6 \]<br />\[ P_t \approx 1496000 \text{ W} \]<br />\[ P_t \approx 1496 \text{ kW} \]<br /><br />### Part (g): Exit Velocity \( V_e \)<br />Already calculated in part (a):<br />\[ V_e \approx 574.2 \text{ m/s} \]<br /><br />### Part (h): Effective Exhaust Velocity \( V_{eff} \)<br />Effective exhaust velocity (\( V_{eff} \)) is given by:<br />\[ V_{eff} = \frac{F_n}{\dot{m}} \]<br />\[ V_{eff} = \frac{8924.6}{17.75} \]<br />\[ V_{eff} \approx 502.9 \text{ m/s} \]<br /><br />### Part (i): Propulsive Efficiency \( \eta_p \)<br />Propulsive efficiency (\( \eta_p \)) is given by:<br />\[ \eta_p = \frac{2V_0}{V_0 + V_e} \]<br />\[ \eta_p = \frac{2 \times 167.6}{167.6 + 574.2} \]<br />\[ \eta_p \approx 0.46 \]<br />\[ \eta_p \approx 46\% \]<br /><br />### Part (j): Thermal Efficiency \( \eta_{th} \) and Overall Efficiency \( \eta_0 \)<br />Thermal efficiency (\( \eta_{th} \)) is given by:<br />\[ \eta_{th} = \frac{P_t}{\dot{m}_f Q_f} \]<br />Where \( \dot{m}_f \) is the fuel mass flow rate:<br />\[ \dot{m}_f = \dot{m} \times f/a \]<br />\[ \dot{m}_f = 17.75 \times 0.02 \]<br />\[ \dot{m}_f \approx 0.355 \text{ kg/s} \]<br /><br />Now, calculate thermal efficiency:<br />\[ \eta_{th} = \frac{1496000}{0.355 \times 43.96 \times 10^6} \]<br />\[ \eta_{th} \approx 0.095 \]<br />\[ \eta_{th} \approx 9.5\% \]<br /><br />Overall efficiency (\( \eta_0 \)) is given by:<br />\[ \eta_0 = \eta_p \times \eta_{th} \]<br />\[ \eta_0 = 0.46 \times 0.095 \]<br />\[ \eta_0 \approx 0.044 \]<br />\[ \eta_0 \approx 4.4\% \]<br /><br />### Part (k): Specific Fuel Consumption \( c_j \)<br />Specific fuel consumption (\( c_j \)) is given by:<br />\[ c_j = \frac{\dot{m}_f}{F_n} \]<br />\[ c_j = \frac{0.355}{8924.6} \]<br />\[ c_j \approx 3.98 \times 10^{-5} \text{ kg/Ns} \]<br />\[ c_j \approx 0.143 \text{ kg/kN·h} \]<br /><br />These calculations provide the required answers for each part of the problem.