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/m in 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,Y_(e)=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 this problem, we need to apply principles from fluid dynamics and propulsion. Let's go through each part step by step.<br /><br />### Given Data:<br />- Thrust per engine: \( F = 14,000 \, \text{N} \)<br />- Altitude: \( h = 6100 \, \text{m} \)<br />- Nozzle exit Mach number: \( M_e = 1 \)<br />- Specific heat ratio: \( \gamma = 1.33 \)<br />- Nozzle exit area: \( A_e = 0.116 \, \text{m}^2 \)<br />- Exit pressure: \( p_e = 62 \, \text{kPa} \)<br />- Exit temperature: \( T_e = 537.8^\circ C = 810.95 \, \text{K} \) (converted to Kelvin)<br />- Flight velocity: \( V_0 = 167.6 \, \text{m/s} \)<br />- Fuel-to-air ratio: \( f/a = 0.02 \)<br />- Heat of combustion: \( Q_f = 43.96 \, \text{MJ/kg} = 43,960,000 \, \text{J/kg} \)<br /><br />### Assumptions:<br />- The nozzle is choked, so the exit Mach number is 1.<br />- The ambient pressure at 6100 m altitude can be approximated using standard atmospheric tables or equations.<br /><br />### Calculations:<br /><br />#### (a) Gross Thrust Produced<br />The gross thrust (\( F_g \)) is given by:<br />\[ F_g = \dot{m} V_e + (p_e - p_a) A_e \]<br />where \( \dot{m} \) is the mass flow rate, \( V_e \) is the exit velocity, and \( p_a \) is the ambient pressure.<br /><br />First, calculate the exit velocity \( V_e \) using the isentropic flow relations for a choked nozzle:<br />\[ V_e = \sqrt{\gamma R T_e} \]<br />where \( R = \frac{R_{\text{univ}}}{M} \) is the specific gas constant for air, with \( R_{\text{univ}} = 8.314 \, \text{J/(mol K)} \) and \( M \approx 28.97 \, \text{g/mol} \).<br /><br />Calculate \( V_e \):<br />\[ V_e = \sqrt{1.33 \times \frac{8.314}{0.02897} \times 810.95} \]<br /><br />Next, find the ambient pressure \( p_a \) at 6100 m using the barometric formula or standard atmosphere tables.<br /><br />Finally, calculate \( F_g \) using the above formula.<br /><br />#### (b) Percentage of Gross Thrust Caused by Pressure Imbalance<br />Calculate the pressure thrust component:<br />\[ F_p = (p_e - p_a) A_e \]<br /><br />Percentage of gross thrust due to pressure imbalance:<br />\[ \text{Percentage} = \left(\frac{F_p}{F_g}\right) \times 100\% \]<br /><br />#### (c) Exit Mass Flow<br />Use the continuity equation for choked flow:<br />\[ \dot{m} = \frac{p_e A_e}{\sqrt{T_e}} \sqrt{\frac{\gamma}{R}} \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}} \]<br /><br />#### (d) Ram Drag<br />Ram drag (\( D_r \)) is given by:<br />\[ D_r = \dot{m} V_0 \]<br /><br />#### (e) Net Thrust<br />Net thrust (\( F_n \)) is the difference between gross thrust and ram drag:<br />\[ F_n = F_g - D_r \]<br /><br />#### (f) Thrust Power Produced<br />Thrust power (\( P_t \)) is given by:<br />\[ P_t = F_n \times V_0 \]<br /><br />#### (g) Exit Velocity \( V_e \)<br />Already calculated in part (a).<br /><br />#### (h) Effective Exhaust Velocity \( V_{eff} \)<br />Effective exhaust velocity is given by:<br />\[ V_{eff} = \frac{F_n}{\dot{m}} \]<br /><br />#### (i) Propulsive Efficiency \( n_p \)<br />Propulsive efficiency is given by:<br />\[ n_p = \frac{2}{1 + \frac{V_e}{V_0}} \]<br /><br />#### (j) Thermal Efficiency \( \eta_{th} \) and Overall Efficiency \( \eta_0 \)<br />Thermal efficiency:<br />\[ \eta_{th} = \frac{P_t}{\dot{m} f/a \cdot Q_f} \]<br /><br />Overall efficiency:<br />\[ \eta_0 = \eta_{th} \times n_p \]<br /><br />#### (k) Specific Fuel Consumption \( c_j \)<br />Specific fuel consumption is given by:<br />\[ c_j = \frac{\dot{m} f/a}{F_n} \]<br /><br />These calculations require numerical values for some constants and assumptions about the ambient conditions. You may need to refer to standard atmospheric tables for precise values of ambient pressure and temperature at 6100 m.