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.### Given Data:- Altitude: m- Choked nozzle exit: - Specific heat ratio: - Nozzle exit area: m²- Exit pressure: kPa- Exit temperature: K (converted to Kelvin)- Flight velocity: m/s- Fuel-to-air ratio: - Heat of combustion: MJ/kg### Constants:- Universal gas constant: J/(kg·K)### Part (a): Gross Thrust ProducedGross thrust ( ) is given by: Since the nozzle is choked, the exit Mach number , and we can find the exit velocity using: Next, we need the mass flow rate ( ): Where is the density at the exit, which can be found using the ideal gas law: Now, calculate the mass flow rate: Finally, calculate the gross thrust: Assuming standard atmospheric pressure at 6100 m altitude is approximately 47.35 kPa: ### Part (b): Percentage of Gross Thrust Caused by Pressure Imbalance Percentage of gross thrust due to pressure imbalance: ### Part (c): Exit Mass Flow if Already calculated in part (a): ### Part (d): Ram Drag at a Flight Velocity of 167.6 m/sRam drag ( ) is given by: ### Part (e): Net ThrustNet thrust ( ) is given by: ### Part (f): Thrust Power ProducedThrust power ( ) is given by: ### Part (g): Exit Velocity Already calculated in part (a): ### Part (h): Effective Exhaust Velocity Effective exhaust velocity ( ) is given by: ### Part (i): Propulsive Efficiency Propulsive efficiency ( ) is given by: ### Part (j): Thermal Efficiency and Overall Efficiency Thermal efficiency ( ) is given by: Where is the fuel mass flow rate: Now, calculate thermal efficiency: Overall efficiency ( ) is given by: ### Part (k): Specific Fuel Consumption Specific fuel consumption ( ) is given by: These calculations provide the required answers for each part of the problem.