A turbo-jet engine travels at 216 m/s in air at 0.78 bar and – 7.2°C. Air first enters diffuser in which it is brought to rest relative to the unit and it is then compressed in a compressor through a pressure ratio of 5.8 and fed to a turbine at 1110°C. The gases expand through the turbine and

Question

A turbo-jet engine travels at 216 m/s in air at 0.78 bar and – 7.2°C. Air first enters diffuser in which it is brought to rest relative to the unit and it is then compressed in a compressor through a pressure ratio of 5.8 and fed to a turbine at 1110°C. The gases expand through the turbine and then through the nozzle to atmospheric pressure (i.e., 0.78 bar). The efficiencies of diffuser, nozzle and compressor are each 90%. The efficiency of turbine is 80%. Pressure drop in the combustion chamber is 0.168 bar. Determine :
(i) Air-fuel ratio ;
(ii) Specific thrust of the unit ;
(iii) Total thrust, if the inlet cross-section of diffuser is 0.12 m².
Assume calorific value of fuel as 44150 kJ/kg of fuel.

Accepted Answer

Refer Fig. 41.
Speed of the aircraft, [latex] C_{a} [/latex] = 216 m/s
Intake air temperature, [latex] T_{1} [/latex]= – 7.2 + 273 = 265.8 K
Intake air pressure, [latex]p_{1}[/latex] = 0.78 bar
Pressure ratio in the compressor, [latex]r_{p} [/latex]= 5.8
Temperature of gases entering the gas turbine, [latex]T_{4} [/latex]= 1110 + 273 = 1383 K
Pressure drop in combustion chamber = 0.168 bar
[latex]η_{d} = η_{n} ; η_{c} [/latex]= 90% ; [latex] η_{t} [/latex]= 80%.
Calorific value of fuel, C.V. = 44150 kJ/kg of coal
(i) Air-fuel ratio :

For ideal diffuser (i.e., process 1-2) the energy equation is given by :

[latex]h_{2} = h_{1} + \frac{C_{a} ²}{2} or h_{2} - h_{1} = \frac{C_{a} ²}{2} or T_{2} - T_{1} =\frac{C_{a} ²}{2 c_{p}} [/latex]

or [latex] T_{2} = T_{1} + \frac{C_{a} ²}{2 c_{p}} = 265.8 + \frac{216²}{2 × 1.005 × 1000} [/latex] = 289 K

For actual diffuser (i.e., process 1-2′),

[latex]η_{d} = ( \frac{h_{2} - h_{1}}{h_{2} ^{′} - h_{1}}) or h_{2} ^{′} - h_{1} = \frac{h_{2} - h_{1}}{η_{d}} [/latex]

or [latex]h_{2} ^{′} = h_{1} + \frac{h_{2} - h_{1}}{η_{d}} = h_{1} + \frac{C_{a} ²}{2 η_{d}}[/latex]

or [latex] T_{2} ^{′} = T_{1} + \frac{C_{a} ²}{2 c_{p} η_{d}} = 265.8 + \frac{216²}{2 × 1.005 × 1000 × 0.9} [/latex] = 291.6 K

Now, [latex]\frac{ T_{2} }{ T_{1} } = (\frac{p_{2}}{p_{1}})^{\frac{γ - 1}{γ }} or \frac{289 }{ 265.8 } = (\frac{p_{2}}{0.78})^{\frac{1.4 - 1}{1.4}} or (1.087)^{3.5} = (\frac{p_{2}}{0.78})[/latex]

or [latex] p_{2} = 0.78 × (1.087)^{3.5} [/latex] = 1.044 bar

Again, [latex]\frac{ T_{3} }{ T_{2} ^{′} } = (r_{p})^{\frac{γ - 1}{γ }} = (5.8)^{\frac{1.4 - 1}{1.4}} = 1.652 or T_{3} [/latex] = 291.6 × 1.652 = 481.7 K

Also, [latex] η_{c} = \frac{ T_{3} - T_{2} ^{′} }{ T_{3} ^{′} - T_{2} ^{′} } or T_{3} ^{′} = T_{2} ^{′} + \frac{ T_{3} - T_{2} ^{′} }{ η_{c}} = 291.6 + \frac{481.7 - 291.6 }{0.9} [/latex] = 502.8 K

Assume [latex] c_{pg} = c_{pa} = c_{p}[/latex]
Heat supplied [latex] = (m_{a} + m_{f}) c_{p} T_{4} – m_{a} c_{p} T_{3} ^{′} = m_{f} × C[/latex]
or [latex] m_{a} c_{p} T_{4} + m_{f} c_{p} T_{4} – m_{a} c_{p} T_{3} ^{′} = m_{f} × C [/latex]
or [latex]m_{a} c_{p} (T_{4} – T_{3} ^{′} ) = m_{f} (C – c_{p} T_{4})[/latex]

or [latex] \frac{m_{a}}{m_{f}} = \frac{C – c_{p} T_{4}}{ c_{p} (T_{4} – T_{3} ^{′} )} = \frac{44150 - 1.005 × 1383}{1.005 (1383 - 502.8)} [/latex] = 48.34

∴ Air-fuel ratio = 48.34.
(ii) Specific thrust of the unit :
[latex]p_{4} = p_{3} [/latex] – 0.168 = 5.8 × 1.044 – 0.168 = 5.88 bar
Assume that the turbine drives compressor only (and not accessories also as is the usual case)
∴ [latex] c_{p} (T_{3} ^{′} – T_{2} ^{′} ) = c_{p} (T_{4} – T_{5} ^{′} )[/latex]
or [latex]T_{3} ^{′} – T_{2} ^{′} = T_{4} – T_{5} ^{′} or T_{5} ^{′} = T_{4} – (T_{3} ^{′} – T_{2} ^{′} )[/latex]
= 1383 – (502.8 – 291.6) = 1171.8 K

Also, [latex] η_{t} = \frac{ T_{4} - T_{5} ^{′} }{ T_{4} - T_{5} } [/latex]

or [latex]T_{5} = T_{4} - \frac{T_{4} - T_{5} ^{′} }{η_{t}} = 1383 – \frac{1383 - 1171.8}{0.8} [/latex] = 1119 K

Now, [latex]\frac{ T_{4} }{ T_{5}} = ( \frac{ p_{4}}{p_{5}})^{\frac{γ - 1}{γ }} = ( \frac{ 5.88}{p_{5}})^{\frac{1.4 - 1}{1.4}} [/latex]

or [latex]( \frac{1383 }{1119})^{3.5} = \frac{ 5.88}{p_{5}} or p_{5} [/latex] = 2.8 bar

Again, [latex]\frac{ T_{5} ^{′} }{ T_{6}} = ( \frac{ p_{5}}{p_{6}})^{\frac{γ - 1}{γ }} = (\frac{2.8 }{0.78})^{\frac{1.4 - 1}{1.4}} [/latex] = 1.44

or [latex] T_{6} = \frac{ T_{5} ^{′} }{1.44} = \frac{1171.8 }{1.44} [/latex] = 813.75 K

and [latex] η_{n} = \frac{T_{5} ^{′} - T_{6} ^{′} }{ T_{5} ^{′} - T_{6}} or T_{6} ^{′} = T_{5} ^{′} - η_{n} (T_{5} ^{′} - T_{6})[/latex]

= 1171.8 – 0.9 (1171.8 – 813.75) = 849.5 K
Velocity at the exit of the nozzle,

[latex] C_{j} = 44.72 \sqrt{h_{5} ^{′} - h_{6} ^{′} } = 44.72 \sqrt{c_{p} (T_{5} ^{′} - T_{6} ^{′} )}[/latex]

[latex] = 44.72 \sqrt{1.005 (1171.8 - 849.5)} [/latex] = 804.8 m/s

Specific thrust = (1 + [latex]m_{f}) × C_{j} = (1 + \frac{1}{48.34} ) [/latex]× 804.8

= 821.45 N/kg of air/s.

(iii) Total Thrust :
Volume of flowing air, [latex]V_{1} [/latex] = 0.12 × 216 = 92 m³/s

Mass flow, [latex]m_{a} = \frac{p_{1} V_{1}}{R T_{1}} = \frac{0.78 × 10^{5} × 25.92}{(0.287 × 1000) × 265.8} [/latex] = 26.5 kg/s

∴ Total thrust = 26.5 × 821.45 = 21768.4 N.

Question 11.22: A turbo-jet engine travels at 216 m/s in air at 0.78 bar and...

Related Answered Questions

Calculate the thermal efficiency and work ratio of the plant is example 25.4, assuming that cp for the combustion process is 1.11 kJ/kg K. ...

Verified Answer:

The following data pertain to a jet engine flying at an altitude of 9000 metres with a speed of 215 m/s. Thrust power developed 750 kW Inlet pressure and temperature 0.32 bar, – 42°C Temperature of gases leaving the combustion chamber 690°C Pressure ratio 5.2 Calorific value of fuel 42500 ...

Verified Answer:

Verified Answer:

Verified Answer:

A gas turbine employs a heat exchanger with a thermal ratio of 72%. The turbine operates between the pressures of 1.01 bar and 4.04 bar and ambient temperature is 20°C. Isentropic efficiencies of compressor and turbine are 80% and 85% respectively. The pressure drop on each side of the heat ...

Verified Answer:

In a gas turbine the compressor takes in air at a temperature of 15°C and compresses it to four times the initial pressure with an isentropic efficiency of 82%. The air is then passed through a heat exchanger heated by the turbine exhaust before reaching the combustion chamber. In the heat ...

Verified Answer:

(a) Why are the back work ratios relatively high in gas turbine plants compared to those of steam power plants ? (b) In a gas turbine plant compression is carried out in two stages with perfect intercooling and expansion in one stage turbine. If the maximum temperature (Tmax K) and minimum ...

Verified Answer:

In a jet propulsion unit air is drawn into the rotary compressor at 15°C and 1.01 bar and delivered at 4.04 bar. The isentropic efficiency of compression is 82% and the compression is uncooled. After delivery the air is heated at constant pressure until the temperature reaches 750°C. ...

Verified Answer:

The following data pertain to a turbo-jet flying at an altitude of 9500 m : Speed of the turbo-jet = 800 km/h Propulsive efficiency = 55% Overall efficiency of the turbine plant = 17% Density of air at 9500 m altitude = 0.17 kg/m³ Drag on the plane = 6100 N Assuming calorific value of the ...

Verified Answer:

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 ...

Verified Answer: