ME
4_8esm_09_043.pdf

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

• Homework Help
• 5

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

• Fall '08
• FEVE
• Thermodynamics, pressure ratio, compressor inlet, unit mass, air standard analysis

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern