Question

2. An aircraft with a single turbojet engine (with an inlet area of 1 m2) is flying at cruising condition with a flight Mach number of 0.7. The ambient temperature and pressure are 250 K and 100 kPa, respectively. The engine compressor pressure ratio is 12, and the turbine inlet temperature is 1200 K. Assume all mechanical components are operating at isentropic condition and the specific heat can be considered a constant (throughout the entire engine) of 1 kJ/(kg K). Also, the heat input due to fuel combustion (lower heating value) is 40,000 kJ per kg of fuel burned. Assume the exhaust nozzle is a converging-diverging (C-D) nozzle and the exhaust gas expands isentropically so that the exit pressure is equal to the ambient pressure. (R-287 J/(kg K), y-1.4). Determine the following (a) Air mass flow rate (b) Temperature at the outlet of the compressor (T2) (c) Fuel-air ratio (100% combustion) (d) Temperature at the outlet of the turbine (T4) (e) Exhaust gas temperature at the exit (Ts) (f) Exit area of the C-D nozzle (As) (g) Exit velocity of the nozzle (h) Thrust of the engine Exhaust compressor turbine nozzle diffuser ambient Combustion chamber 0 4 5

Answer 1

Find the velocity by using the Mach number kRT 0.7 = 1.4x 287x 250 221.85m/s Apply energy equation to the process 0-1 T = 250+221.852 2x1000 T = 274, 608K Apply isentropic relation to the process 0-1 Pi To Po 1.4-1 P 274.608 100 250 P=138.9kPa Find the air mass flow rate RT 287x 250 467-x1x 221.85 100x103 m 159. 177kg/s

Find the temperature at the outlet of the comp ressor, 1.4-1 T. 1214 274.068 7, = 557.437 K Find the pressure at state 2 =12x138. 901 p Ps = 1666 814kPa Apply energy equation to the process 2-3 m x1x557.437+mx 40000x1200 ma 1x1200 2x 1x 557.437 40000 ×40000-1+-|1200 557.4374 (40000-1200) = 1200-557.437 = 0.0165

Find the temperature at state 4 by equating the work output from turbine is equal to the work nput to the compressor n, (557437-274.608)-(m, +0.0165%)(1200-7,) 282829 = 1.0165 (T,-1200) T4 -921.7619K Apply isentropic relation to the process 3-4 T. T. Pa 1.4-1 14 921.7619 P4 1666. 814 1200 p4 662 092kPa Apply isentropic relation to the process 4-5 T. Ps 662.092 921.7619 100 7,=537. 119K

Apply energy equation to the nozzle and find the exit velocity of the nozzle. 1000(921.7619- 537.119)-5 V877.089 m/s Find the exit area of the nozzle by using mass flow rate AVs Ps /RT 1.0165m-x877.089 4 100/0.287 x537.119 A x877.089 1.0165x159.177- 0.648 A0.119m2 Find the thrust of the engine. = 1.0165×159.177×877.089-159, 177x 221.85 F 106602.582N F 106.6021A