QUESTION-2 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) (j) nth and n_(0) for Q_(f)=43.96MJ/kg) (k) the specific fuel consumption, cj.

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QUESTION-2 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) (j) nth and n_(0) for Q_(f)=43.96MJ/kg) (k) the specific fuel consumption, cj.

Answer 1

To solve the given problem, we need to use various principles of jet propulsion and thermodynamics. Let's go through each part step by step. ### Given Data: - Altitude: \( h = 6100 \) m - Choked nozzle exit: \( M_e = 1 \) - Specific heat ratio: \( \gamma_e = 1.33 \) - Nozzle exit area: \( A_e = 0.116 \) m² - Exit pressure: \( p_e = 62 \) kPa - Exit temperature: \( T_e = 537.8^\circ C = 810.95 \) K (converted to Kelvin) - Flight velocity: \( V_0 = 167.6 \) m/s - Fuel-to-air ratio: \( f/a = 0.02 \) - Heating value of fuel: \( Q_f = 43.96 \) MJ/kg ### Constants: - Universal gas constant: \( R = 287 \) J/(kg·K) ### Part (a): Gross Thrust Produced Gross thrust (\( F_g \)) is given by: \[ F_g = \dot{m} V_e + (p_e - p_0) A_e \] Since the nozzle is choked (\( M_e = 1 \)), the exit velocity \( V_e \) can be calculated using: \[ V_e = \sqrt{\gamma_e R T_e} \] \[ V_e = \sqrt{1.33 \times 287 \times 810.95} \] \[ V_e \approx 678.5 \text{ m/s} \] The mass flow rate (\( \dot{m} \)) can be found using the continuity equation for choked flow: \[ \dot{m} = \rho_e A_e V_e \] Where \( \rho_e \) is the density at the exit, which can be found using the ideal gas law: \[ \rho_e = \frac{p_e}{R T_e} \] \[ \rho_e = \frac{62000}{287 \times 810.95} \] \[ \rho_e \approx 0.267 \text{ kg/m}^3 \] Now, calculate the mass flow rate: \[ \dot{m} = 0.267 \times 0.116 \times 678.5 \] \[ \dot{m} \approx 21.0 \text{ kg/s} \] Finally, calculate the gross thrust: \[ F_g = \dot{m} V_e + (p_e - p_0) A_e \] Assuming standard atmospheric pressure at 6100 m altitude (\( p_0 \approx 47.35 \) kPa): \[ F_g = 21.0 \times 678.5 + (62000 - 47350) \times 0.116 \] \[ F_g \approx 14248.5 + 1707.4 \] \[ F_g \approx 15955.9 \text{ N} \] ### Part (b): Percentage of Gross Thrust Caused by Pressure Imbalance \[ \text{Pressure thrust} = (p_e - p_0) A_e \] \[ \text{Pressure thrust} = 1707.4 \text{ N} \] Percentage of gross thrust due to pressure imbalance: \[ \text{Percentage} = \left( \frac{1707.4}{15955.9} \right) \times 100 \] \[ \text{Percentage} \approx 10.7\% \] ### Part (c): Exit Mass Flow if \( T_e = 537.8^\circ C \) Already calculated in part (a): \[ \dot{m} \approx 21.0 \text{ kg/s} \] ### Part (d): Ram Drag at a Flight Velocity of \( 167.6 \) m/s Ram drag (\( F_r \)) is given by: \[ F_r = \dot{m} V_0 \] \[ F_r = 21.0 \times 167.6 \] \[ F_r \approx 3520 \text{ N} \] ### Part (e): Net Thrust Net thrust (\( F_n \)) is given by: \[ F_n = F_g - F_r \] \[ F_n = 15955.9 - 3520 \] \[ F_n \approx 12435.9 \text{ N} \] ### Part (f): Thrust Power Produced Thrust power (\( P_t \)) is given by: \[ P_t = F_n \times V_0 \] \[ P_t = 12435.9 \times 167.6 \] \[ P_t \approx 2085.2 \text{ kW} \] ### Part (g): Exit Velocity \( V_e \) Already calculated in part (a): \[ V_e \approx 678.5 \text{ m/s} \] ### Part (h): Effective Exhaust Velocity \( V_{eff} \) Effective exhaust velocity (\( V_{eff} \)) is given by: \[ V_{eff} = \frac{F_n}{\dot{m}} \] \[ V_{eff} = \frac{12435.9}{21.0} \] \[ V_{eff} \approx 592.2 \text{ m/s} \] ### Part (i): Propulsive Efficiency \( \eta_p \) Propulsive efficiency (\( \eta_p \)) is given by: \[ \eta_p = \frac{2V_0}{V_0 + V_e} \] \[ \eta_p = \frac{2 \times 167.6}{167.6 + 678.5} \] \[ \eta_p \approx 0.396 \text{ or } 39.6\% \] ### Part (j): Thermal Efficiency \( \eta_{th} \) and Overall Efficiency \( \eta_0 \) Thermal efficiency (\( \eta_{th} \)) is given by: \[ \eta_{th} = \frac{P_t}{\dot{m}_f Q_f} \] Where \( \dot{m}_f \) is the fuel mass flow rate: \[ \dot{m}_f = \dot{m} \times f/a \] \[ \dot{m}_f = 21.0 \times 0.02 \] \[ \dot{m}_f = 0.42 \text{ kg/s} \] Now, calculate thermal efficiency: \[ \eta_{th} = \frac{2085.2 \times 10^3}{0.42 \times 43.96 \times 10^6} \] \[ \eta_{th} \approx 0.113 \text{ or } 11.3\% \] Overall efficiency (\( \eta_0 \)) is given by: \[ \eta_0 = \eta_p \times \eta_{th} \] \[ \eta_0 = 0.396 \times 0.113 \] \[ \eta_0 \approx 0.045 \text{ or } 4.5\% \] ### Part (k): Specific Fuel Consumption \( c_j \) Specific fuel consumption (\( c_j \)) is given by: \[ c_j = \frac{\dot{m}_f}{F_n} \] \[ c_j = \frac{0.42}{12435.9} \] \[ c_j \approx 3.38 \times 10^{-5} \text{ kg/N·s} \] This concludes the solution to the problem.

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