tgas 2
7
9-13 9-34 An ide a l Otto cycle with air as th e working fluid has a com pres sion ratio of 8. The pre ssu re a nd tem perature a t the end of the he at a ddition process, the net work output, the therm al effici ency, and the m ea n effective pressure f or the cycl e a re to be dete rmi ned. Assum pt ions 1 The a ir-sta nda rd assum ptions a re app licab le. 2 K ine tic and pote nti al en ergy cha nge s are negligible. 3 Air is a n ide al gas with va riab le s pe cific he a ts. Properties The ga s const an t of air is R =0.287 kJ /kg.K. T he prop e rties o f a ir are g ive n in Ta ble A-17. Analysis ( a ) Process 1 -2: ise ntropic com pres sion. 621.2 kJ /k g 214.07 K 3 0 0 1 1 1 r u T P 75 0 kJ/kg 2 3 1 k Pa 1705 kP a 9 5 K 300 K 673.1 8 kJ/kg 491.2 K 673.1 65 . 77 2 . 621 8 1 1 1 1 2 2 1 2 1 1 1 2 2 2 2 2 1 2 1 1 2 P T T P T P T P u T r r r r Process 2-3: =constan t hea t add ition. 588 . 6 kJ /k g 241.2 1 750 2 . 491 3 3 in 2 3 , 2 3 2 3 in 2 3 , r T q u u u u q K 1539 kPa 3898 kPa 1705 K 673.1 K 1539 2 2 3 3 2 2 2 3 3 3 P T T P T P T P ( b ) Proces s 3 -4: isen tropic exp an sion. kJ /k g 571.69 K 774.5 70 . 52 588 . 6 8 4 4 2 1 3 3 4 u T r r r r Process 4-1: =constan t hea t rejection. q u u ou t 571.69 214.07 357.62 k J /kg 4 1 kJ / kg 392.4 6 2 . 357 750 ou t in ou t net, q q w ( c ) 52.3% kJ /kg 750 kJ/kg 392.4 in ou t net, th q w ( d ) kPa 495.0 k J m k Pa 1/8 1 /k g m 0.906 kJ/kg 392.4 ) / 1 1 ( MEP /k g m 0.906 k Pa 9 5 K 3 0 0 K /k g m k Pa 0.287 3 3 1 o ut net, 2 1 out net, max 2 m in max 3 3 1 1 1 r w w r P R T
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Transcript of tgas 2
7/30/2019 tgas 2
http://slidepdf.com/reader/full/tgas-2 1/6
9-13
9-34 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and
temperature at the end of the heat addition process, the net work output, the thermal efficiency, and themean effective pressure for the cycle are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes arenegligible.3Air is an ideal gas with variable specific heats.
PropertiesThe gas constant of air isR =0.287 kJ/kg.K. The properties of air are given in Table A-17.
Analysis(a) Process 1-2: isentropic compression.
621.2
kJ/kg214.07K 300
1
1
1r
u T
P
750 kJ/kg2
3
1
kPa1705kPa95K 300
K 673.18
kJ/kg491.2
K 673.1 65.772.621
8
11
11
2
2
12
1
11
2
22
2
2
1
2
112
P T
TP
T
P
T
P
u
T
rrrr
Process 2-3: =constant heat addition.
588.6
kJ/kg241.217502.491
3
3
in23,2323in23,r
Tquuuuq
K 1539
kPa3898 kPa1705K 673.1
K 15392
2
33
2
22
3
33 P T
TP
T
P
T
P
(b) Process 3-4: isentropic expansion.
kJ/kg571.69
K 774.570.52588.68
4
4
2
1334 u
Tr rrr
Process 4-1: =constant heat rejection.
q u uout 571.69 214.07 357.62kJ /kg4 1
kJ /kg392.462.357750outinoutnet, qqw
(c) 52.3%kJ/kg750
kJ/kg392.4
in
outnet,th
q
w
(d)
kPa495.0 kJ
mkPa
1/81/kgm0.906
kJ/kg392.4
)/11(MEP
/kgm0.906kPa95
K 300K /kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
r
ww
r
P
RT
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9-14
9-35 EES Problem 9-34 is reconsidered. The effect of the compression ratio on the net work output and
thermal efficiency is to be investigated. Also, T-sandP- diagrams for the cycle are to be plotted.
AnalysisUsing EES, the problem is solved as follows:
"Input Data" T[1]=300 [K]P[1]=95 [kPa]q_23 =750 [kJ /kg]{r_comp =8} "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1])s[2]=s[1]
T[2]=temperature(air, s=s[2], P=P[2])P[2]*v[2]/T[2]=P[1]*v[1]/T[1]P[1]*v[1]=R*T[1]R=0.287 [kJ /kg-K]V[2] =V[1]/ r_comp"Conservation of energy for process 1 to 2" q_12 - w_12 =DELTAu_12q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1])"Process 2-3 is constant volume heat addition" v[3]=v[2]s[3]=entropy(air, T=T[3], P=P[3])P[3]*v[3]=R*T[3]"Conservation of energy for process 2 to 3" q_23 - w_23 =DELTAu_23w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2])"Process 3-4 is isentropic expansion" s[4]=s[3]s[4]=entropy(air,T=T[4],P=P[4])P[4]*v[4]=R*T[4]"Conservation of energy for process 3 to 4" q_34 -w_34 =DELTAu_34q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3])"Process 4-1 is constant volume heat rejection" V[4] =V[1]
"Conservation of energy for process 4 to 1" q_41 - w_41 =DELTAu_41w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4])q_in_total=q_23q_out_total =-q_41w_net =w_12+w_23+w_34+w_41Eta_th=w_net/q_in_total*100"Thermal efficiency, in percent" "The mean effective pressure is:" MEP =w_net/(V[1]-V[2])"[kPa]"
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9-15
rcomp th MEP [kPa] wnet [kJ /kg]5 43.78 452.9 328.46 47.29 469.6 354.77 50.08 483.5 375.68 52.36 495.2 392.79 54.28 505.3 407.110 55.93 514.2 419.5
4.5 5.0 5.5 6.0 6.5 7.0 7.5
200
400
600
800
1000
1200
1400
1600
s [kJ /kg-K]
Air
1
2
3
4
10-2 10-1 100 101 102
101
102
103
104
v [m3/kg]
300 K
1500 K
Air
1
2
3
4
s2
=s1
=5.716 kJ /kg-K
s4
=33
=6.424 kJ /kg-K
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9-16
5 6 7 8 9 10
320
340
360
380
400
420
rcomp
net
]
5 6 7 8 9 10450
460
470
480
490
500
510
520
rcomp
5 6 7 8 9 1042
44
46
48
50
52
54
56
rcomp
th
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9-17
9-36 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and
temperature at the end of the heat addition process, the net work output, the thermal efficiency, and themean effective pressure for the cycle are to be determined.
Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes arenegligible.3Air is an ideal gas with constant specific heats.
Properties The properties of air at room temperature are cp =1.005 kJ/kg·K, c =0.718 kJ/kg·K, R =0.287 kJ/kg·K, and k=1.4 (Table A-2).
Analysis(a) Process 1-2: isentropic compression.
kPa1745kPa95K 300
K 6898
K 6898K 300
11
2
2
1
21
11
2
22
0.4
1
2
112
P T
TP
T
P
T
P
T T
k
P
750 kJ/kg
2
3
1
Process 2-3: =constant heat addition.
K 1734
K 689K kJ/kg0.718kJ/kg750
3
3
2323in23,
T
T
T Tcuuq v
kPa4392 kPa1745K 689
K 17342
2
33
2
22
3
33 P T
TP
T
P
T
P
(b) Process 3-4: isentropic expansion.
K 7558
1
K 1734
0.41
4
3
34
k
T T
Process 4-1: =constant heat rejection.
kJ/kg327K 300755K kJ/kg0.7181414out T Tcuuq
kJ /kg423 327750outinoutnet, qqw
(c) 56.4%kJ/kg750
kJ/kg423
in
outnet,th
q
w
(d)
kPa534 kJ
mkPa
1/81/kgm0.906
kJ/kg423
)/11(MEP
/kgm0.906kPa95
K 300K /kgmkPa0.287
3
31
outnet,
21
outnet,
max2min
max3
3
1
11
r
ww
r
P
RT
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