Question 11.22: A turbo-jet engine travels at 216 m/s in air at 0.78 bar and...
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.
Refer Fig. 41.
Speed of the aircraft, C_{a} = 216 m/s
Intake air temperature, T_{1} = – 7.2 + 273 = 265.8 K
Intake air pressure, p_{1} = 0.78 bar
Pressure ratio in the compressor, r_{p} = 5.8
Temperature of gases entering the gas turbine, T_{4} = 1110 + 273 = 1383 K
Pressure drop in combustion chamber = 0.168 bar
η_{d} = η_{n} ; η_{c} = 90% ; η_{t} = 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 :
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}}or T_{2} = T_{1} + \frac{C_{a} ²}{2 c_{p}} = 265.8 + \frac{216²}{2 × 1.005 × 1000} = 289 K
For actual diffuser (i.e., process 1-2′),
η_{d} = ( \frac{h_{2} – h_{1}}{h_{2} ^{′} – h_{1}}) or h_{2} ^{′} – h_{1} = \frac{h_{2} – h_{1}}{η_{d}}or h_{2} ^{′} = h_{1} + \frac{h_{2} – h_{1}}{η_{d}} = h_{1} + \frac{C_{a} ²}{2 η_{d}}
or T_{2} ^{′} = T_{1} + \frac{C_{a} ²}{2 c_{p} η_{d}} = 265.8 + \frac{216²}{2 × 1.005 × 1000 × 0.9} = 291.6 K
Now, \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})
or p_{2} = 0.78 × (1.087)^{3.5} = 1.044 bar
Again, \frac{ T_{3} }{ T_{2} ^{′} } = (r_{p})^{\frac{γ – 1}{γ }} = (5.8)^{\frac{1.4 – 1}{1.4}} = 1.652 or T_{3} = 291.6 × 1.652 = 481.7 K
Also, 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} = 502.8 K
Assume c_{pg} = c_{pa} = c_{p}
Heat supplied = (m_{a} + m_{f}) c_{p} T_{4} – m_{a} c_{p} T_{3} ^{′} = m_{f} × C
or m_{a} c_{p} T_{4} + m_{f} c_{p} T_{4} – m_{a} c_{p} T_{3} ^{′} = m_{f} × C
or m_{a} c_{p} (T_{4} – T_{3} ^{′} ) = m_{f} (C – c_{p} T_{4})
or \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)} = 48.34
∴ Air-fuel ratio = 48.34.
(ii) Specific thrust of the unit :
p_{4} = p_{3} – 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)
∴ c_{p} (T_{3} ^{′} – T_{2} ^{′} ) = c_{p} (T_{4} – T_{5} ^{′} )
or T_{3} ^{′} – T_{2} ^{′} = T_{4} – T_{5} ^{′} or T_{5} ^{′} = T_{4} – (T_{3} ^{′} – T_{2} ^{′} )
= 1383 – (502.8 – 291.6) = 1171.8 K
Also, η_{t} = \frac{ T_{4} – T_{5} ^{′} }{ T_{4} – T_{5} }
or T_{5} = T_{4} – \frac{T_{4} – T_{5} ^{′} }{η_{t}} = 1383 – \frac{1383 – 1171.8}{0.8} = 1119 K
Now, \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}}
or ( \frac{1383 }{1119})^{3.5} = \frac{ 5.88}{p_{5}} or p_{5} = 2.8 bar
Again, \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}} = 1.44
or T_{6} = \frac{ T_{5} ^{′} }{1.44} = \frac{1171.8 }{1.44} = 813.75 K
and η_{n} = \frac{T_{5} ^{′} – T_{6} ^{′} }{ T_{5} ^{′} – T_{6}} or T_{6} ^{′} = T_{5} ^{′} – η_{n} (T_{5} ^{′} – T_{6})
= 1171.8 – 0.9 (1171.8 – 813.75) = 849.5 K
Velocity at the exit of the nozzle,
= 44.72 \sqrt{1.005 (1171.8 – 849.5)} = 804.8 m/s
Specific thrust = (1 + m_{f}) × C_{j} = (1 + \frac{1}{48.34} ) × 804.8
= 821.45 N/kg of air/s.
(iii) Total Thrust :
Volume of flowing air, V_{1} = 0.12 × 216 = 92 m³/s
Mass flow, m_{a} = \frac{p_{1} V_{1}}{R T_{1}} = \frac{0.78 × 10^{5} × 25.92}{(0.287 × 1000) × 265.8} = 26.5 kg/s
∴ Total thrust = 26.5 × 821.45 = 21768.4 N.
