A turbo-jet engine consumes air at the rate of 60.2 kg/s when flying at a speed of 1000 km/h. Calculate : (i) Exit velocity of the jet when the enthalpy change for the nozzle is 230 kJ/kg and velocity co-efficient is 0.96. (ii) Fuel flow rate in kg/s when air-fuel ratio is 70 : 1 (iii) Thrust

Question

A turbo-jet engine consumes air at the rate of 60.2 kg/s when flying at a speed of 1000 km/h. Calculate :
(i) Exit velocity of the jet when the enthalpy change for the nozzle is 230 kJ/kg and velocity co-efficient is 0.96.
(ii) Fuel flow rate in kg/s when air-fuel ratio is 70 : 1
(iii) Thrust specific fuel consumption
(iv) Thermal efficiency of the plant when the combustion efficiency is 92% and calorific value of the fuel used is 42000 kJ/kg.
(v) Propulsive power (vi) Propulsive efficiency
(vii) Overall efficiency.

Accepted Answer

Rate of air consumption, [latex] \dot{m}_{a} [/latex] = 60.2 kg/s
Enthalpy change for nozzle, ∆h = 230 kJ/kg
Velocity coefficient, z = 0.96
Air-fuel ratio = 70 : 1
Combustion efficiency, [latex] η _{combustion} [/latex] = 92%
Calorific value of fuel, C.V. = 42000 kJ/kg

Aircraft velocity, [latex]C_{a} = \frac{1000 × 1000}{60 × 60} [/latex] = 277.8 m/s

(i) Exit velocity of jet, [latex]C_{j} [/latex] :

[latex]C_{j} = z \sqrt{2 ∆h × 1000}[/latex] , where ∆h is in kJ

= 0.96 [latex] \sqrt{2 × 230 × 1000}[/latex] = 651 m/s.

i.e., Exit velocity of jet = 651 m/s.

(ii) Fuel flow rate :

Rate of fuel consumption, [latex] \dot{m}_{f} = \frac{Rate of air consumption }{Air-fuel ratio} [/latex]

= [latex] \frac{60.2}{70} [/latex] = 0.86 kg/s.

(iii) Thrust specific fuel consumption :
Thrust is the force produced due to change of momentum.
Thrust produced =[latex] \dot{m}_{a} (C_{j} - C_{a} ) [/latex] , neglecting mass of fuel.

= 60.2 (651 – 277.8) = 22466.6 N.
∴ Thrust specific fuel consumption

= [latex] \frac{Fuel consumption }{Thrust} = \frac{0.86}{22466.6} [/latex]

= 3.828 × [latex] 10^{–5}[/latex] kg/N of thrust/s.

(iv) Thermal efficiency, [latex] η_{thermal} [/latex]:

[latex] η_{thermal} = \frac{Work output }{Heat supplied} [/latex]

= [latex] \frac{Gain in kinetic energy per kg of air}{Heat supplied by fuel per kg of air } [/latex]

= [latex] \frac{(C_{j}² - C_{a}² ) }{( \frac{m_{f}}{m_{a}}) × C.V. × η_{combustion } × 1000} [/latex]

= [latex] \frac{(651² - 277.8² )}{2 × \frac{1}{70} × 42000 × 0.92 × 1000} [/latex] = 0.3139 or 31.39%

i.e., Thermal efficiency = 31.39%.

(v) Propulsive power :

Propulsive power = [latex] \dot{m}_{a} × (\frac{(C_{j}² - C_{a}² ) }{2}) = \frac{60.2}{1000} × ( \frac{(651² - 277.8² )}{2})[/latex] kW

= 10433.5 kW.

(vi) Propulsive efficiency, [latex] η_{prop.} [/latex] :

[latex] η_{prop.} = \frac{Thrust power }{Propulsive power} = \frac{2 C_{a}}{C_{j} + C_{a}} [/latex] ...[Eq.(18)]

= [latex] \frac{2 × 277.8}{651 + 277.8}[/latex] = 0.598 or 59.8%.

(vii) Overall efficiency, [latex] η_{0} [/latex] :

[latex] η_{0} = \frac{Thrust work }{Heat supplied by fuel } = \frac{(C_{j} - C_{a} ) C_{a}}{( \frac{m_{f}}{m_{a}}) × C.V. × η_{combustion }}[/latex] ...(22)

= [latex] \frac{(651 - 277.8 ) × 277.8}{\frac{1}{70} × 42000 × 0.92 × 1000} [/latex] = 0.1878 or 18.78%.

Question 11.19: A turbo-jet engine consumes air at the rate of 60.2 kg/s whe...

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