Thermodynamics

emaining part of the energy released shows up as the kinetic energy of theexhaust gases relative to a fixed point on the ground and as an increase inthe enthalpy of the gases leaving the engine.Chapter 9 | 523EXAMPLE 9–9The Ideal Jet-Propulsion CycleA turbojet aircraft flies with a velocity of 850 ft/s at an altitude where the air isat 5 psia and 40°F. The compressor has a pressure ratio of 10, and the temperatureof the gases at the turbine inlet is 2000°F. Air enters the compressorat a rate of 100 lbm/s. Utilizing the cold-air-standard assumptions, determine(a) the temperature and pressure of the gases at the turbine exit, (b) the velocityof the gases at the nozzle exit, and (c) the propulsive efficiency of the cycle.Solution The operating conditions of a turbojet aircraft are specified. Thetemperature and pressure at the turbine exit, the velocity of gases at thenozzle exit, and the propulsive efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The cold-air-standardassumptions are applicable and thus air can be assumed to have constantspecific heats at room temperature (c p 0.240 Btu/lbm · °F and k 1.4).3 Kinetic and potential energies are negligible, except at the diffuser inletand the nozzle exit. 4 The turbine work output is equal to the compressorwork input.Analysis The T-s diagram of the ideal jet propulsion cycle described is shownin Fig. 9–50. We note that the components involved in the jet-propulsioncycle are steady-flow devices.(a) Before we can determine the temperature and pressure at the turbineexit, we need to find the temperatures and pressures at other states:Process 1-2 (isentropic compression of an ideal gas in a diffuser): For convenience,we can assume that the aircraft is stationary and the air is movingtoward the aircraft at a velocity of V 1 850 ft/s. Ideally, the air exits thediffuser with a negligible velocity (V 2 0):h 2 V ¡0222 h 1 V 1220 c p 1T 2 T 1 2 V 2 12T 2 T 1 V 2 12c p 420 R 480 R1850 ft>s2 22 10.240 Btu>lbm #a 1 Btu>lbmR2 25,037 ft 2 >s b 2P 2 P 1 a T k>1k122b 15 psia2 a 480 R 1.4>11.412T 1 420 R b 8.0 psiaProcess 2-3 (isentropic compression of an ideal gas in a compressor):P 3 1r p 21P 2 2 110218.0 psia2 80 psia 1 P 4 2T 3 T 2 a P 1k12>k3b 1480 R2 1102 11.412>1.4 927 RP 2T, °F2000–40321P = const.q in 56q outP = const.FIGURE 9–50T-s diagram for the turbojet cycledescribed in Example 9–9.4s