Thermodynamics

516 | Thermodynamicsη th,Brayton0.70.60.50.40.30.20.1With regenerationWithout regenerationT 1 /T 3 = 0.33T 1 /T 3 = 0.2T 1 /T 3 = 0.255 10 15 20 25Pressure ratio, r pFIGURE 9–40Thermal efficiency of the idealBrayton cycle with and withoutregeneration.When the cold-air-standard assumptions are utilized, it reduces toP T 5 T 2T 4 T 2(9–24)A regenerator with a higher effectiveness obviously saves a greateramount of fuel since it preheats the air to a higher temperature prior to combustion.However, achieving a higher effectiveness requires the use of alarger regenerator, which carries a higher price tag and causes a larger pressuredrop. Therefore, the use of a regenerator with a very high effectivenesscannot be justified economically unless the savings from the fuel costsexceed the additional expenses involved. The effectiveness of most regeneratorsused in practice is below 0.85.Under the cold-air-standard assumptions, the thermal efficiency of anideal Brayton cycle with regeneration ish th,regen 1 a T 1T 3b1r p 2 1k12>k(9–25)Therefore, the thermal efficiency of an ideal Brayton cycle with regenerationdepends on the ratio of the minimum to maximum temperatures as wellas the pressure ratio. The thermal efficiency is plotted in Fig. 9–40 for variouspressure ratios and minimum-to-maximum temperature ratios. This figureshows that regeneration is most effective at lower pressure ratios andlow minimum-to-maximum temperature ratios.EXAMPLE 9–7Actual Gas-Turbine Cycle with RegenerationT, K130030012a5q in34aq regen = q savedFIGURE 9–41T-s diagram of the regenerativeBrayton cycle described inExample 9–7.sDetermine the thermal efficiency of the gas-turbine described in Example9–6 if a regenerator having an effectiveness of 80 percent is installed.Solution The gas-turbine discussed in Example 9–6 is equipped with aregenerator. For a specified effectiveness, the thermal efficiency is to bedetermined.Analysis The T-s diagram of the cycle is shown in Fig. 9–41. We first determinethe enthalpy of the air at the exit of the regenerator, using the definitionof effectiveness:Thus,P h 5 h 2ah 4a h 2a0.80 1h 5 605.392 kJ>kg1880.36 605.392 kJ>kg S h 5 825.37 kJ>kgq in h 3 h 5 11395.97 825.372 kJ>kg 570.60 kJ>kgThis represents a savings of 220.0 kJ/kg from the heat input requirements.The addition of a regenerator (assumed to be frictionless) does not affect thenet work output. Thus,h th w netq in210.41 kJ>kg 0.369 or 36.9%570.60 kJ>kg