An advanced fighter engine operating at Mach 0.8 and 10Km altitude where, Ta=223.297K & Pa=0.2649 bar has the following uninstalled performance data and uses a fuel with C.V= 42,800KJ/Kg: Thrust = 50 KN Mass flow of air = 45Kg/s Mass flow of fuel = 2.65 Kg/s Determine the specific thrust, thrust specific fuel consumption; exit velocity, thermal efficiency, propulsion efficiency, and overall efficiency (assume exit pressure equal to ambient pressure).

 To determine the specific thrust, thrust-specific fuel consumption, exit velocity, thermal efficiency, propulsion efficiency, and overall efficiency, we can use the given data and apply the relevant formulas. 

1. Specific Thrust (Ts):

Specific Thrust is defined as the thrust generated per unit mass flow rate of air.

Ts = Thrust / Mass flow of air

Given:

Thrust = 50 kN

Mass flow of air = 45 kg/s

Ts = 50,000 N / 45 kg/s

Ts ≈ 1111.11 N/kg/s

2. Thrust-Specific Fuel Consumption (TSFC):

Thrust-Specific Fuel Consumption is the fuel flow rate per unit thrust.

TSFC = Mass flow of fuel / Thrust

Given:

Mass flow of fuel = 2.65 kg/s

Thrust = 50 kN

TSFC = 2.65 kg/s / (50,000 N / 1000)

TSFC ≈ 0.053 kg/N/s

3. Exit Velocity (Ve):

Exit velocity is the velocity of the exhaust gases relative to the aircraft.

Ve = (Thrust / Mass flow of air) + (Exit pressure - Ambient pressure) / (Mass flow of air)

Given:

Thrust = 50 kN

Mass flow of air = 45 kg/s

Exit pressure = Ambient pressure (Pa)

Ve = (50,000 N / 45 kg/s) + (0.2649 bar - 1.01325 bar) * 10^5 Pa / (45 kg/s)

Ve ≈ 1,111.11 m/s

4. Thermal Efficiency (ηth):

Thermal efficiency is the ratio of useful work done (thrust) to the energy input (fuel flow rate times fuel calorific value).

ηth = (Thrust × Ve) / (Mass flow of fuel × Fuel Calorific Value)

Given:

Thrust = 50 kN

Ve ≈ 1,111.11 m/s

Mass flow of fuel = 2.65 kg/s

Fuel Calorific Value = 42,800 kJ/kg

ηth = (50,000 N × 1,111.11 m/s) / (2.65 kg/s × 42,800,000 J/kg)

ηth ≈ 0.154 or 15.4%

5. Propulsion Efficiency (ηp):

Propulsion efficiency is the ratio of useful work done (thrust) to the power output of the engine.

ηp = (Thrust × Ve) / (0.5 × Mass flow of air × Ve^2)

Given:

Thrust = 50 kN

Ve ≈ 1,111.11 m/s

Mass flow of air = 45 kg/s

ηp = (50,000 N × 1,111.11 m/s) / (0.5 × 45 kg/s × (1,111.11 m/s)^2)

ηp ≈ 0.563 or 56.3%

6. Overall Efficiency (ηo):

Overall efficiency is the product of thermal efficiency and propulsion efficiency.

ηo = ηth × ηp

ηo = 0.154 × 0.563

ηo ≈ 0.087 or 8.7%

Therefore, the calculated values are as follows:

Specific Thrust (Ts) ≈ 1111.11 N/kg/s

Thrust-Specific Fuel Consumption (TSFC) ≈ 0.053 kg/N/s

Exit Velocity (Ve) ≈ 1111.11 m/s

Thermal Efficiency (ηth) ≈ 0.154 or 15