FTFS Chap08 P148 - Chapter 8 Power and Refrigeration Cycles...

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Unformatted text preview: Chapter 8 Power and Refrigeration Cycles Review Problems 8-148 A turbocharged four-stroke V-16 diesel engine produces 4000 hp at 1050 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined. Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles, we have ( a ) cycle) mech kJ/cyl 10.7 (= hp 1 Btu/min 42.41 rev/min) (1050 cylinders) (16 hp 4000 cycles) mechanical of (No. cylinders) of (No. produced power Total mechanical = = = cycle mech Btu/cyl 10.1 w ( b ) cycle) therm kJ/cyl 21.3 (= hp 1 Btu/min 42.41 rev/min) (1050/2 cylinders) (16 hp 4000 cycles) amic thermodyn of (No. cylinders) of (No. produced power Total mic thermodyna = = = cycle therm Btu/cyl 20.2 w 8-149 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k =1.4 (Table A-2). Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as ( 29 ( 29 ( 29 ( 29 ( 29 k k p k k k k p k k r T P P T T r T P P T T / 1 3 / 1 3 4 3 4 / 1 1 / 1 1 2 1 2 1---- = = = = Setting T 2 = T 4 and solving for r p gives ( 29 23.0 = = =- 1.4/0.8 1 2 / 1 3 K 300 K 1800 k k p T T r Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 23. 8-119 s T 1 2 4 3 q in q out T 3 T 1 Chapter 8 Power and Refrigeration Cycles 8-150 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-21. Analysis (b) We treat air as an ideal gas with variable specific heats, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 kJ/kg 187.28 =- =- = =- =- = =- +- =- +- = + = = = = = = = = = = = = = = = = = = 27 . 736 55 . 923 kJ/kg 736.27 19 . 300 46 . 1036 kJ/kg 923.55 93 . 932 97 . 1395 07 . 214 58 . 674 kJ/kg 1036.46 3 . 110 330.9 kPa 300 kPa 100 9 . 330 kJ/kg, 1395.97 kJ/kg 1022.82 1300 kJ/kg 932.93 kJ/kg 674.58674....
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FTFS Chap08 P148 - Chapter 8 Power and Refrigeration Cycles...

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