 Ch 9. Brayton Cycle Multimedia Engineering Thermodynamics BraytonCycle Intercoolingand Reheating
 Chapter 1. Basics 2. Pure Substances 3. First Law 4. Energy Analysis 5. Second Law 6. Entropy 7. Exergy Analysis 8. Gas Power Cyc 9. Brayton Cycle 10. Rankine Cycle Appendix Basic Math Units Thermo Tables Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Meirong Huang Kurt Gramoll ©Kurt Gramoll THERMODYNAMICS - CASE STUDY SOLUTION A turbojet aircraft is held stationary by the brakes to load seriously injured people. The force applied to the brakes needs to be determined. Assumptions: Model the cycle in the aircraft engine as an ideal jet-propulsion cycle. Cold-air-standard assumption is valid for this analysis, so the constant pressure specific heat cP = 1.005 kJ/(kg-K) and the specific heat ratio k = 1.4. The engine operates in steady condition. Kinetic and potential energies are negligible, except at the nozzle exit. The turbine work output equals the work consumed by the compressor. Neglect the effect of the diffuser. Neglect the slight increase in mass due to the addition of fuel. T-s Diagram of the Ideal Jet-propulsion Cycle Model the cycle in the aircraft engine as an ideal jet-propulsion cycle. The T-s diagram of the ideal jet-propulsion cycle described is shown on the left. The braking force equals the thrust developed by the engine when the airplane is stationary. The thrust can be determined by where = mass flow rate of air, given as 8 kg/s       vexit = velocity of air at the exit of the nozzle       vinlet = velocity of air at the inlet of the engine The airplane is stationary on the ground and the air velocity at the inlet of the engine is negligible. Also, the effect of diffuser is negligible. Thus, the thrust can be simplified as The exit velocity of the nozzle can determined by the energy balance of the nozzle. In steady-flow condition, it is The above analysis shows that in order to calculate the exit velocity, temperatures at state 4 and state 5 need to be determined first. State 1:       T1 = 300 K (given)       P1 = 100 kPa (given) State 2:       P2/P1 = rP = 10 (given)       P2 = P1rP = 100(10) = 1,000 kPa Process 1-2: isentropic compression. State 3: Process 2-3: constant pressure heat addition       P3 = P2 = 1,000 kPa Heat transfer into the cycle equals, The temperature at state 3 is State 4: The work produced by the turbine equals the work consumed by the compressor. Thus,       h3 - h4 = h2 - h1       T4 = T3 - (T2 - T1 )           = 1,101.6 - (579.2 -300) = 822.4 K Process 3-4: isentropic expansion in turbine State 5: Process 5-1 is a constant-pressure heat rejection process, thus,       P5 = P1 = 100 kPa Process 4-5 is an isentropic expansion process in nozzle. After determining state 4 and state 5, the exit velocity at the nozzle can be determined as Substitute v5 and mass flow rate of air into the expression of thrust yields,       F = 8(711.5) = 5,692 N To hold the airplane stationary, 5,692 N force needs to be applied to the brakes to balance the thrust developed by the engine.