4_8esm_09_043.pdf

4_8esm_09_043.pdf - 9.43 An ideal air standard Brayton...

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Unformatted text preview: 9.43 An ideal air standard Brayton cycle operates at steady state with compressor inlet conditions of 300 K and 100 kPa and a fixed turbine inlet temperature of 1700 K. For the cycle, (a) determine the net mm per unit mass flowing, in kJ/kg, and the thermal efficiency for a compressor pressure ratio of 8. (b) plot the net mm per unit mass flowing, in kJ/kg, and the thermal efficiency, each versus com ssor ssure ratio ran 'n from 2 to 50. KNOWN: An ideal air standard Brayton cycle operates with fixed compressor inlet conditions of 300 K and 100 kPa and fixed turbine inlet temperature of 1700. FIND: (a) the net power per unit mass flowing and the thermal efficiency for a compressor pressure ratio of 8 and (b) plot the net power per unit mass flowing and the thermal efficiency, each versus compressor pressure ratio ranging from 2 to 50. SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: 1. Each component is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines. 2. Air, modeled as an ideal gas, is the working fluid. 3. All processes of the working fluid are internally reversible. 4. The compressor and turbine operate adiabatically. 5. Kinetic and potential energy effects are negligible. ANALYSIS: (a) The net work of the cycle per unit of mass flow using an air standard analysis 3’8 Problem 9.43 (Continued) — Page 2 i{jg-““013410-01241) State 1: T1 = 300 K —v From Table A-22: h1= 300.19 kJ/kg and p” = 1.3860. Process 1-2 is an isenu'opic process. Thus £1: M 3 pf; = pnfl =1.3sso(s)=11.oss P] pr] pl From Table A-22 (interpolated): h; = 544.18 kJ/kg. State 3: T3 = 1700 K —» From Table A-22: h; = 1880.1 kJ/kg and pg = 1025. Process 3-4 is an isentropic process. Thus 1?; = 2.1.: 1m 2 p“ = p” fl = 1025(1): 128.125 P3 P2 Pr: P2 8 From Table A-22 (interpolated): h4 = 1079.85 kJ/kg. Solving for net work of the cycle per unit of mass flow 171 kg kg kg kg Thermal efficiency is k] - . - 556.3— W / W / ' r7= ———?’°‘° .m =————"’“° m =ka kg H = .4164 (41.64%) Qin ’m hs ‘hz 1880.1——544.18— kg kg IT Code I‘ANALYSIS: Air Standard Analysis'! FP = 8 mdot = 1 ll kgls I' State 1 "I p1 = 100 II kPa T1 = 300 ll K 51 = s_Tp("Air".T1,p1)ll le(kg-K) m = h_T("Air",T1) II kJIkg 2 W “‘th =[IBSOJE—1079.85E)—[544.18E—300.19—kl]= 556.3 kJ/_kg ‘17 Problem 9.43 (Continued) — Page 3 I“ State 2 ‘I 32 = 51 ll kJI(kg-K) p2 = rp ‘ p1 II kPa s_Tp("Aii".T1,p1) = s__Tp("AiI".T2,p2) II Retums T2 in K h2 = h_T("Ail".T2) II lekg I‘ State 3 '/ T3 = 1700 II K p3 = p2 II kPa 53 = s_Tp("Aif',T3,p3) II kJI(kg—K) h3 = h__T("Air'.T3) II kJIkg I' State 4 ‘I p4 = p1 ll kPa s_Tp("Air".T3.p3) = s_Tp("Air".T4,p4) ll Returns T4 in K M = h__T("Air“.T4) u kJIkg s4 = $3 ll kJI(kg-K) I” Energy Transfers and Cycle Perfonnanee 'l Wdotcydepenndot = 013 - M) - (h2 - M) I/ kJIkg Qdoiinperrndot = h3 - h2 II kJIkg eta = Wdetcydepermdotl Qdotinpermdot 1T Results for pressure gtio of 8 (Note: The values in this table compare favorably to those calculawd above using Air Table data The specific heat functions used in [Tare not exactly the ones used to generate the Air Tables. That and round-off account for the slight differences.) Problem 9.45 (Continued) - Page 4 -25 l for . m 4.. 7 6w 1 mm “6.6.6. M] 72 4|..l. 277m ”4.8.1m m1mm-7. 72211.1.2. 77“ 5.6.9. .7m 622 77W $51271 m 7.7% 7 4|.4I.2277m 7 $774.1 a 2% 7.7.6.2 71.1.2 1117 Plots Power Output Per Unit Mass Flow IIIIIIIIIIIIIII Pusan Ratio vs Not PoworOutput Per Unlt Mas Flow 8&2. 3E 3m: =5 an. arm H38 82 2 4 6 8 10121416182022243233032343638404244464850E Pressure Ratio ; 3! Problem 9.43 (Continued) — Page 5 Thermal Efficiency Pro-um Ratio vs cycle Thom! Elncloncy 2 4 6 31012141818m222423283032343638404244464350I PM“!!! Ratio ’ Discussion Maximum net work output per unit mass flow (640.9 kJ/kg) occurs at a pressure ratio of 27 using air standard analysis. Thermal efficiency continues to increase with increasing pressure ratio. 81 ...
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