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ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma.

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Presentation on theme: "ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma."— Presentation transcript:

1 ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

2 Brayton Cycles George Brayton (1870) Applications of Gas-Turbine Engines Propulsion Power Generation - The first gas turbine for electric power generation was installed in 1949 in Oklahoma. - Before 1980s, gas power plants were mainly used for peak-load power production. - It is forecast that more than half of all power plants to be installed in the future are gas power plants.

3 Brayton Cycles George Brayton (1870) Open cycle Closed Cycle

4 Brayton Cycles

5 Staedy-flow q – w = h e – h i q in = h 3 – h 2 q out = h 4 – h 1 1 2 3 4 T S Q in Q out = c p (T 3 – T 2 ) = c p (T 4 – T 1 )

6 Brayton Cycles The thermal efficiency increases with the pressure ratio. The highest temperature in the cycle is limited by the maximum temperature that the turbine blades can withstand.

7 Example 1 A gas power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K At the compression inlet and 1300 K at the turbine inlet. Utilizing the air-standard assumptions, determine (a) the gas temperature at the exits of the compressor and the turbine, (b) the back work ratio, (c) the thermal efficiency of this cycle.

8 Example 1 (continued) Process 1-2: Isentropic compression Table A-17 h 1 = 300.19 kJ/kg p r1 = 1.386 State 1: air at the inlet of compressor, T 1 = 300 K Table A-17 T 2 = 540 K h 2 = 544.35 kJ/kg (a)

9 Example 1 (continued) Process 3-4: Isentropic expanion Table A-17 T 4 = 770 K h 4 = 789.11 kJ/kg w comp = h 2 – h 1 = 544.35 – 300.19 = 244.16 kJ/kg State 3: air at the inlet of turbine, T 3 = 1300 K Table A-17 h 3 = 1395.97 kJ/kg p r3 = 330.9

10 Example 1 (continued) w turb = h 3 – h 4 = 1395.97 – 789.11 = 606.86 kJ/kg (c) (b) q in = h 3 – h 2 = 1395.97 – 544.35 = 851.62 kJ/kg w net = w turb – w comp = 606.86 – 244.16 = 362.7 kJ/kg = 1 – r p (1-k)/k = 1 – (8) -(0.4/1.4) = 0.448

11 Brayton Cycles with Regeneration

12 Brayton Cycles with Intercooling, Reheating, and Regeneration


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