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