Brayton Cycle with Intercooling, Reheating, and …€¦ · Brayton Cycle with Intercooling,...

9-70

Brayton Cycle with Intercooling, Reheating, and Regeneration 9-101C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle. 9-102C (a) decrease, (b) decrease, and (c) decrease. 9-103C (a) increase, (b) decrease, and (c) decrease. 9-104C (a) increase, (b) decrease, (c) decrease, and (d) increase. 9-105C (a) increase, (b) decrease, (c) increase, and (d) decrease. 9-106C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output. 9-107C (c) The Carnot (or Ericsson) cycle efficiency.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

9-71

9-108 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then,

( )( )

( )

( ) ( )( ) ( ) kJ/kg 62.86636.94679.127722

kJ/kg 2.142219.30026.41122

kJ/kg 946.3633.7923831

238kJ/kg 77.7912

K 2001

kJ/kg 411.26158.4386.13

386.1kJ/kg 300.19

K 300

65outT,

12inC,

865

6

755

421

2

11

56

5

12

1

=−=−=

=−=−=

==→=

==

===

→=

==→===

==

→=

hhw

hhw

hhPPP

P

Phh

T

hhPPP

P

Ph

T

rr

r

rr

r

2

9

10

7

6 8

300 K

1200 K qin 5

1

4

3

T

s

Thus, 33.5%===kJ/kg 662.86kJ/kg 222.14

outT,

inC,bw w

wr

( ) ( ) ( ) ( )

36.8%===

=−=−=

=−+−=−+−=

kJ/kg 1197.96kJ/kg 440.72

kJ/kg 440.72222.1486.662

kJ/kg 1197.9636.94679.127726.41179.1277

in

netth

inC,outT,net

6745in

qw

www

hhhhq

η

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes

( ) ( )( )

55.3%===

=−=−=

=−=−=

kJ/kg 796.63kJ/kg 440.72

kJ/kg 796.6333.40196.1197

kJ/kg 33.40126.41136.94675.0

in

netth

regenoldin,in

48regen

qw

qqq

hhq

η

ε


9-72

9-109 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then,

( )( )

( )( ) ( )

( )

( )( )(

( ) ( )( ) ( ) kJ/kg 63.44507.99679.127722

kJ/kg 77.68219.30003.43922

kJ/kg 96.07936.94679.127785.079.1277

kJ/kg 946.3633.7923831

238kJ/kg 1277.79K 1200

kJ/kg 9.034380.0/19.30026.41119.300

/

kJ/kg 411.26158.4386.13

386.1kJ/kg 300.19K 300

65outT,

12inC,

6558665

65

865

6

755

1214212

12

421

2

11

56

5

12

1

=−=−=

=−=−=

=−−=

−−==→−−

=

==→=

==

===→=

=−+=

−+==→−−

=

==→===

==→=

hhw

hhw

hhhhhhhhh

hhPPP

P

PhhT

hhhhhhhhh

hhPPP

P

PhT

sTs

T

rr

r

Css

C

ssrr

r

ηη

ηη

)

T

4

8

22s

9

10

7

66s

8qin

5

1

4

3s

Thus, 49.3%===kJ/kg 563.44kJ/kg 277.68

outT,

inC,bw w

wr

( ) ( ) ( ) ( )

25.5%===

=−=−=

=−+−=−+−=

kJ/kg 1120.48kJ/kg 285.76

kJ/kg 5.762868.27744.563

kJ/kg 1120.48996.071277.79439.031277.79

in

netth

inC,outT,net

6745in

qw

www

hhhhq

η

(b) When a regenerator is used, rbw remains the same. The thermal efficiency in this case becomes

( ) ( )( )

40.7%===

=−=−=

=−=−=

kJ/kg 702.70kJ/kg 285.76

kJ/kg .7070278.41748.1120

kJ/kg 17.78403.43907.99675.0

in

netth

regenoldin,in

48regen

qw

qqq

hhq

η

ε


9-73

9-110 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.

( )( )

( )

( ) ( )( ) ( )

kg/s 249.6===

=−=−=

=−=−=

=−=−=

==→=

==

===→=

==→===

==→=

kJ/kg 440.72kJ/s 110,000

kJ/kg 440.72222.1468.662

kJ/kg 662.86946.361277.7922

kJ/kg 222.1419.30026.41122

kJ/kg 6.369433.7923831

238 kJ/kg, 1277.79K 0120

kJ/kg 11.264158.4386.13

386.1 kJ/kg, 00.193K 003

net

net

inC,outT,net

65outT,

12inC,

865

6

755

421

2

11

56

5

12

1

wW

m

www

hhw

hhw

hhPPP

P

PhhT

hhPPP

P

PhT

rr

r

rr

r

&&

2

7

6 8

300 K

1200 K 5

1

4

3

T

s

9-111 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 Argon is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are negligible. Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K and k = 1.667 (Table A-2a). Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp = 9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.

( )( )( )

( )( )

( ) ( ) ( )( )

( ) ( ) ( )( )

kg/s 404.7===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=

=

=

==

=

−

−

kJ/kg 271.8kJ/s 110,000

kJ/kg 271.83.1721.444

kJ/kg 444.1K773.21200KkJ/kg 0.5203222

kJ/kg 172.3K300465.6KkJ/kg 0.5203222

K 773.231K 1200

K 465.63K 300

net

net

inC,outT,net

6565outT,

1212inC,

70.667/1.66/1

5

656

70.667/1.66/1

1

212

wW

m

www

TTchhw

TTchhw

PP

TT

PP

TT

p

p

kk

kk

&&

2

7

6 8

300 K

1200 K 5

1

4

3

T

s


9-74

Jet-Propulsion Cycles 9-112C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity. 9-113C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio. 9-114C It reduces the exit velocity, and thus the thrust. 9-115E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 0.24 Btu/lbm.R and k = 1.4 (Table A-2Ea). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V1 = 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V2 ≅ 0). Diffuser:

( )

( )( )( )

( )( ) psia 11.19

R 470R 537.3

psia 7

R 537.4/sft 25,037

Btu/lbm 1RBtu/lbm 0.242

ft/s 900470

2

2/02

0

2/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

12

222

211

outin

(steady) 0systemoutin

=

=

=

=

⋅+=+=

−−=

−+−=

+=+

=

∆=−

−kk

p

p

TT

PP

cV

TT

VTTc

VVhh

VhVh

EE

EEE&&

&&&

T

6

qout

5

qin

3

4

2

1s

Compressor:

( )( ) ( )( )( )

( )( ) R 1118.313R 537.4

psia 145.5psia 11.1913

0.4/1.4/1

2

323

243

==

=

====

− kk

p

PP

TT

PrPP


9-75

Turbine:

( ) ( )54235423outturb,incomp, TTcTTchhhhww pp −=−→−=−→=

or,

( )( ) psia 55.2=

=

=

=+−=+−=

− 1.4/0.41/

4

545

2345

R 2400R 1819.1

psia 145.5

R 1819.14.5371118.34002kk

TT

PP

TTTT

(b) Nozzle:

( )( )

( ) 2/02

0

2/2/

R 1008.6psia 55.2

psia 7R 1819.1

2656

025

26

56

266

255

outin


0.4/1.4/1

5

656

VTTc

VVhh

VhVh

EE

EEE

PP

TT

p

kk

+−=

−+−=

+=+

=

∆=−

=

=

=

−

&&

&&&

or,

( )( )( ) s/ft 3121=

−⋅=

Btu/lbm 1/sft 25,037

R1008.61819.1RBtu/lbm 0.240222

6V

(c) The propulsive efficiency is the ratio of the propulsive work to the heat input,

( )

( )[ ]( )

( ) ( )( )

%25.9===

=−⋅=−=−=

=

−=

−=

Btu/lbm 307.6Btu/lbm 79.8

Btu/lbm 307.6R1118.32400RBtu/lbm 0.24

Btu/lbm 79.8/sft 25,037

Btu/lbm 1ft/s 900ft/s 9003121

in

3434in

22

aircraftinletexit

qw

TTchhq

VVVw

pp

p

p

η


9-76

9-116E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air are given in Table A-17E. Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V1= 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).

Diffuser: 8548.0

Btu/lbm 112.20R 470

1

11

==→=

rPhT

T

( )

( ) psia 11.220.85481.3698psia 7

3698.1Btu/lbm 128.48/sft 25,037

Btu/lbm 12ft/s 900

20.1122

20

2/2/

1

2

2

12

r22

221

12

21

022

12

222

211

outin


=

=

=

=→=

+=+=

−+−=

+=+

=

∆=−

r

r

P

PPP

PV

hh

VVhh

VhVh

EE

EEE&&

&&&

6

qout

5

qin

3

4

2

1s

Compressor:

( )( ) ( )( )

( ) Btu/lbm 267.5680.171.36811.22145.8

psia 145.8psia 11.2213

32

3

243

23=→=

=

=

====

hPPP

P

PrPP

rr

p

Turbine: T 6.367Btu/lbm 22.617

R 24004

44 =

=→=

rPh

5423

outturb,incomp,

hhhh

ww

−=−

=

or,

( ) psia 56.6=

=

=

=→=+−=+−=

367.6142.7psia 145.8

7.142 Btu/lbm14.47848.12856.26722.617

4

5

5

45

2345

r

r

r

P

PPP

Phhhh


9-77

(b) Nozzle:

20

2/2/

Btu/lbm 266.9317.66psia 56.6

psia 7)7.142(

025

26

56

266

255

outin


65

656

VVhh

VhVh

EE

EEE

hPP

PP rr

−+−=

+=+

=

∆=−

=→=

=

=

&&

&&&

or,

( ) ( ) ft/s 3252=

−=−=

Btu/lbm 1/sft 25,037

Btu/lbm)93.26614.478(2222

656 hhV

(c) The propulsive efficiency is the ratio of the propulsive work to the heat input,

( )

[ ]

24.2%===

=−=−=

=

−=

−=

Btu/lbm 349.66Btu/lbm 84.55

Btu/lbm 349.6656.26722.617

Btu/lbm 84.55/sft 25,037

Btu/lbm 1ft/s) (900ft/s) 900)(3252

in

34

22

aircraftinletexit

qw

hhq

VVVw

pp

in

p

η


9-78

9-117 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser:

( )( )

( )( )( )

( ) kPa 62.6K 241K 291.9

kPa 32

K 291.9/sm 1000

kJ/kg 1KkJ/kg 1.0052

m/s 320K 241

2

2/02

02/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

122

222

11

outin(steady) 0

systemoutin

=

=

=

=

⋅+=+=

−−=

−+−=→+=+

=→∆=−

−kk

p

p

TT

PP

cV

TT

VTTc

VVhhVhVh

EEEEE &&&&&

T

6

5

Qi·

3

4

2

1s

Compressor:

( )( ) ( )( )( )

( )( ) K 593.712K 291.9

kPa 751.2kPa 62.612

0.4/1.4/1

2

323

243

==

=

====

− kk

p

PP

TT

PrPP

Turbine:

or, ( ) ( )

K1098.2291.9593.714002345

54235423outturb,incomp,

=+−=+−=

−=−→−=−→=

TTTT

TTcTTchhhhww pp

Nozzle:

( )( )

( ) 2/02

0

2/2/

K 568.2kPa 751.2

kPa 32K 1400

2656

025

26

56

266

255

outin(steady) 0

systemoutin

0.4/1.4/1

4

646

VTTcVV

hh

VhVh

EEEEE

PP

TT

p

kk

+−=→−

+−=

+=+

=→∆=−

=

=

=

−

&&&&&

or, ( )( )( ) m/s1032 kJ/kg 1

/sm 1000K568.21098.2KkJ/kg 1.0052

22

6 =

−⋅=V

(b) ( ) ( )( ) ( ) kW 13,670=

−=−=

22aircraftinletexit/sm 1000

kJ/kg 1m/s 320m/s3201032kg/s 60VVVmW p &&

(c) ( ) ( ) ( )( )( )

kg/s 1.14===

=−⋅=−=−=

kJ/kg 42,700kJ/s 48,620

HV

kJ/s 48,620K593.71400KkJ/kg 1.005kg/s 60

infuel

3434in

Qm

TTcmhhmQ p

&&

&&&


9-79

9-118 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0). Diffuser:

( )

( )( )( )

( )( ) kPa 62.6

K 241K 291.9

kPa 32

K 291.9/sm 1000

kJ/kg 1KkJ/kg 1.0052

m/s 320K 241

2

2/02

0

2/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

12

222

211

outin


=

=

=

=

⋅+=+=

−−=

−+−=

+=+

=

∆=−

−kk

p

p

TT

PP

cV

TT

VTTc

VVhh

VhVh

EE

EEE&&

&&&

T

5

6

5s

Qi·

3

4

2

1s

Compressor:

( )( ) ( )( )( )

( )( ) K 593.712K 291.9

kPa 751.2kPa 62.612

0.4/1.4/1

2

323

243

==

=

====

− kk

s

p

PP

TT

PrPP

( )( )

( ) ( ) ( ) K 669.20.80/291.9593.71.929/2323

23

23

23

23

=−+=−+=

−

−=

−−

=

Cs

p

spsC

TTTT

TTcTTc

hhhh

η

η

Turbine:

or, ( ) ( )

K 1022.7291.9669.214002345

54235423outturb,incomp,

=+−=+−=

−=−→−=−→=

TTTT

TTcTTchhhhww pp

( )( )

( ) ( )( )

( ) kPa 197.7K 1400K 956.1

kPa 751.2

K 956.185.0/1022.714001400/1.4/0.41/

4

545

5445

54

54

54

54

=

=

=

=−−=−−=

−

−=

−−

=

−kks

Ts

sp

p

sT

TT

PP

TTTT

TTcTTc

hhhh

η

η


9-80

Nozzle:

( )( )

( ) 2/02

0

2/2/

K 607.8kPa 197.7

kPa 32K 1022.7

2656

025

26

56

266

255

outin


0.4/1.4/1

5

656

VTTc

VVhh

VhVh

EE

EEE

PP

TT

p

kk

+−=

−+−=

+=+

=

∆=−

=

=

=

−

&&

&&&

or,

( )( )( ) m/s 913.2=

−⋅=

kJ/kg 1/sm 1000

K607.81022.7KkJ/kg 1.005222

6V

(b) ( )( )( ) ( )

kW 11,390=

−=

−=

22

aircraftinletexit

/sm 1000kJ/kg 1

m/s 320m/s320913.2kg/s 60

VVVmW p &&

(c) ( ) ( ) ( )( )( )

kg/s 1.03===

=−⋅=−=−=

kJ/kg 42,700kJ/s 44,067

HV

kJ/s 44,067K669.21400KkJ/kg 1.005kg/s 60

infuel

3434in

Qm

TTcmhhmQ p

&&

&&&


9-81

9-119 A turbojet aircraft that has a pressure rate of 12 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A17.

TAnalysis (a) Using variable specific heats for air,

5

4

qin

2

3

1

Compressor: T 386.1

kJ/kg 300.19K 300

1

11

==→=

rPh

( )( )

( )( )

27.396 kJ/kg 1464.65854610.65

kJ/kg 854kg/s 10

kJ/s 8540

kJ/s 8540kJ/kg 42,700kg/s 0.2HV

kJ/kg 610.6563.16386.112

3

12

2323in

inin

fuelin

21

2

=→

=+=+=→−=

===

==×=

=→===

r

in

rr

Pqhhhhq

mQ

q

mQ

hPPP

P

&

&

&&

s

Turbine:

or, kJ/kg 741.17300.19610.65464.6511234

4312outturb,incomp,

=+−=+−=

−=−→=

hhhh

hhhhww

Nozzle:

( )

20

2/2/

kJ/kg 41.79702.3312127.396

024

25

45

255

244

outin


53

535

VVhh

VhVh

EE

EEE

hPP

PP rr

−+−=

+=+

=

∆=−

=→=

=

=

&&

&&&

or,

( ) ( )( ) m/s 908.9kJ/kg 1

/sm 1000kJ/kg741.171154.1922

22

545 =

−=−= hhV

Brake force = Thrust = ( ) ( )( ) N 9089=

⋅−=−

2inletexitm/skg 1N 1

m/s0908.9kg/s 10VVm&


9-82

9-120 EES Problem 9-119 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" W_dot_net = 0 [kW]


9-83

"Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) "[N]" BrakeForce = Thrust "[N]" "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2])

BrakeForce

[N]

m [kg/s]

T3 [K]

T1 [C]

9971 11.86 1164 -20 9764 11.41 1206 -10 9568 10.99 1247 0 9383 10.6 1289 10 9207 10.24 1330 20 9040 9.9 1371 30

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]

T [K

]

95

kPa

114

0 kP

a

Air

1

2s

3

4s

5s

-20 -10 0 10 20 309000

9200

9400

9600

9800

10000

T1 [C]

Bra

keFo

rce

[N]


9-84

9-121 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 300 m/s. Taking the entire engine as our control volume and writing the steady-flow energy balance yield

T h

T h1 1

2 2

280 28013

700 713 27

= → =

= → =

K k

K k

.

.

J / kg

J / kg 15,000 kJ/s

( ) ( )

−+−=

−+−=

+=++

=

∆=−

22

222

21

22

12in

222

211in

outin


/sm 1000kJ/kg 1

2m/s 300

13.28027.713kg/s 16kJ/s 000,15

2

)2/()2/(

V

VVhhmQ

VhmVhmQ

EE

EEE

&&

&&&

&&

&&&

7°C 300 m/s16 kg/s

427°C

1 2

It gives V 2 = 1048 m/s Thus, ( ) ( )( ) N 11,968=−=−= m/s3001048kg/s 1612 VVmFp &


9-85

Second-Law Analysis of Gas Power Cycles 9-122 The total exergy destruction associated with the Otto cycle described in Prob. 9-34 and the exergy at the end of the power stroke are to be determined. Analysis From Prob. 9-34, qin = 750, qout = 357.62 kJ/kg, T1 = 300 K, and T4 = 774.5 K. The total exergy destruction associated with this Otto cycle is determined from

( ) kJ/kg 245.12=

−=

−=

K 2000kJ/kg 750

K 300kJ/kg 357.62

K 300inout0destroyed

HL Tq

Tq

Tx

Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from ( ) ( ) ( )040040044 vv −+−−−= PssTuuφ

where

( ) KkJ/kg 0.7081K 300K 774.5ln KkJ/kg 0.2871.70203.68232

lnlnln

0kJ/kg357.62

1

414

41

1414

1

4141404

1404

out1404

⋅=⋅−−=

−−=−−=−−=−=−

=−=−==−=−

TT

RssTT

RssPP

Rssssss

quuuu

oooooo

v

v

vvvv

Thus, ( ) ( )( ) kJ/kg 145.2=+⋅−= 0KkJ/kg 0.7081K 300kJ/kg 357.624φ

9-123 The total exergy destruction associated with the Diesel cycle described in Prob. 9-47 and the exergy at the end of the compression stroke are to be determined. Analysis From Prob. 9-47, qin = 1019.7, qout = 445.63 kJ/kg, T1 = 300 K, v1 = 0.906 m3/kg, and v 2 = v 1 / r = 0.906 / 12 = 0.0566 m3/kg. The total exergy destruction associated with this Otto cycle is determined from

( ) kJ/kg 292.7=

−=

−=

K 2000kJ/kg 1019.7

K 300kJ/kg 445.63


HL Tq

Tq

Tx

Noting that state 1 is identical to the state of the surroundings, the exergy at the end of the compression stroke (state 2) is determined from

( ) ( ) ( )( ) ( ) ( )( ) ( )( )

kJ/kg 348.6=

⋅−+−−=

−+−−−=−+−−−=

33

12012012

020020022

mkPa 1kJ 1

/kgm0.9060.0566kPa 950214.07643.3

vv

vv

PssTuuPssTuuφ


9-86

9-124E The exergy destruction associated with the heat rejection process of the Diesel cycle described in Prob. 9-49E and the exergy at the end of the expansion stroke are to be determined. Analysis From Prob. 9-49E, qout = 158.9 Btu/lbm, T1 = 540 R, T4 = 1420.6 R, and v 4 = v 1. At Tavg = (T4 + T1)/2 = (1420.6 + 540)/2 = 980.3 R, we have cv,avg = 0.180 Btu/lbm·R. The entropy change during process 4-1 is

( ) RBtu/lbm 0.1741R 1420.6

R 540lnRBtu/lbm 0.180lnln0

4

1

4

141 ⋅−=⋅=+=−

vvR

TTcss v

Thus,

( ) Btu/lbm 64.9=

+⋅−=

+−=

R 540Btu/lbm 158.9RBtu/lbm 0.1741R54041,

41041 destroyed,R

R

Tq

ssTx

Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from ( ) ( ) ( )040040044 vv −+−−−= PssTuuφ

where

RBtu/lbm 0.1741

0RBtu/lbm 9.581

1404

1404

out1404

⋅=−=−=−=−

⋅==−=−

ssss

quuuuvvvv

Thus, ( ) ( )( ) Btu/lbm 64.9=+⋅−= 0RBtu/lbm 0.1741R540Btu/lbm 158.94φ

Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.


9-87

9-125 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-73 is to be determined. Analysis From Prob. 9-73, qin = 584.62 kJ/kg, qout = 478.92 kJ/kg, and

KkJ/kg2.67602kJ/kg789.16

KkJ/kg3.13916K1160

KkJ/kg2.47256kJ/kg6.364

KkJ/kg.734981K310

44

33

22

11

⋅=→=

⋅=→=

⋅=→=

⋅=→=

o

o

o

o

sh

sT

sh

sT

Thus,

( )

( ) ( ) ( )( )

( )

( )

( ) ( ) ( )( )

( ) kJ/kg 206.0

kJ/kg 38.76

kJ/kg 87.35

kJ/kg 40.83

=

+−=

+−−=

+−==

=⋅−−=

=

−−=−==

=

−−=

−+−−=

+−==

=⋅−−=

=

−−=−==

K 310kJ/kg 478.92

2.676021.73498K 290

ln

1/8lnKkJ/kg 0.2873.139162.67602K 290

ln

K 1600kJ/kg 584.62

2.472563.13916K 290

ln

8lnKkJ/kg 0.2871.734982.47256K 290

ln

out0

4

1410

41,41041gen,041 destroyed,

3

434034034gen,034 destroyed,

in0

2

3230


1


LR

R

HR

R

Tq

PP

RssTT

qssTsTx

PP

RssTssTsTx

Tq

PP

RssTT

qssTsTx

PP

RssTssTsTx

oo

oo

oo

oo

9-126 The total exergy destruction associated with the Brayton cycle described in Prob. 9-93 and the exergy at the exhaust gases at the turbine exit are to be determined. Analysis From Prob. 9-93, qin = 601.94, qout = 279.68 kJ/kg, and h6 = 579.87 kJ/kg. The total exergy destruction associated with this Otto cycle is determined from

( ) kJ/kg 179.4=

−=

−=

K 1800kJ/kg 601.94

K 300kJ/kg 279.68


HL Tq

Tq

Tx

Noting that h0 = h@ 300 K = 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from

( ) ( ) 06

026

060066 2gz

VssThh ++−−−=φ

where s s s s s s RPP6 0 6 1 6 1

6

1

0

− = − = − − = − = ⋅o o ln 2.36275 1.70203 0.66072 kJ / kg K

Thus, ( )( ) kJ/kg 81.5=⋅−−= KkJ/kg 0.66072K 300300.1979.8756φ


9-88

9-127 EES Problem 9-126 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: "Input data" T_o = 300 [K] T_L = 300 [K] T_H = 1400 [K] T[3] = 1200 [K] {Pratio = 10} T[1] = 300 [K] C_P=1.005 [kJ/kg-K] P[1]= 100 [kPa] P_o=P[1] Eta_reg = 1.0 Eta_c =1.0"Compressor isentorpic efficiency" Eta_t =1.0"Turbien isentropic efficiency" MM=MOLARMASS(Air) R=R_u/MM R_u=8.314 [kJ/kmol-K] C_V=C_P - R k=C_P/C_V "Isentropic Compressor anaysis" "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit" T_s[2]=T[1]*(Pratio)^((k-1)/k) Eta_c = w_compisen/w_comp "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:" w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow" q_in_noreg = C_P*(T[3]-T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "Turbine analysis" "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio T_s[4]=T[3]*(1/Pratio)^((k-1)/k) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow" w_turbisen=C_P*(T[3] - T_s[4]) "Actual Turbine analysis:" w_turb= C_P*(T[3]-T[4]) "Cycle analysis" w_net=w_turb-w_comp "[kJ/kg]" Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency" Bwr=w_comp/w_turb "Back work ratio"


9-89

"With the regenerator the heat added in the external heat exchanger is" q_in_withreg = C_P*(T[3]-T[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" "h[2] + h[4]=h[5] + h[6]" T[2] + T[4]=T[5] + T[6] P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]" "Irreversibility associated with the Brayton cycle is determined from:" q_out_withreg = q_in_withreg - w_net i_withreg = T_o*(q_out_withreg/T_L - q_in_withreg/T_H) q_out_noreg = q_in_noreg - w_net i_noreg = T_o*(q_out_noreg/T_L - q_in_noreg/T_H) "Neglecting the ke and pe of the exhaust gases, the exergy of the exhaust gases at the exit of the regenerator is:" "Psi_6 = (h[6] - h_o) - T_o(s[6] - s_o)" Psi_exit_withreg = C_P*(T[6] - T_o) - T_o*(C_P*ln(T[6]/T_o)-R*ln(P[6]/P_o)) Psi_exit_noreg = C_P*(T[4] - T_o) - T_o*(C_P*ln(T[4]/T_o)-R*ln(P[4]/P_o))

inoreg iwithreg Pratio Ψexit,noreg [kJ/kg]

Ψexit,withreg [kJ/kg]

ηth,noreg [%]

ηth,withreg [%]

270.8 97.94 6 157.8 47.16 40.05 58.3 244.5 113.9 7 139.9 56.53 42.63 56.42 223 128.8 8 125.5 65.46 44.78 54.73 205 142.7 9 113.6 74 46.61 53.18

189.6 155.9 10 103.6 82.17 48.19 51.75 176.3 168.4 11 95.05 90.02 49.58 50.41 164.6 180.2 12 87.62 97.58 50.82 49.17 154.2 191.5 13 81.11 104.9 51.93 47.99 144.9 202.3 14 75.35 111.9 52.94 46.88

4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T [K

]

100 kPa

1000 kPa

Air

2s

1

2 5

3

44s

6


9-90

6 7 8 9 10 11 12 13 1475

115

155

195

235

275

Pratio

i [kJ

/kg]

No regeneration

With regeneration

6 7 8 9 10 11 12 13 1440

60

80

100

120

140

160

Pratio

Psi

exit

[kJ/

kg]

No regeneration

With regeneration

6 7 8 9 10 11 12 13 1440

44

48

52

56

60

Pratio

η th

[%

]

With regeneration

No regeneration


9-91

9-128 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-98 and the exergy at the end of the exhaust gases at the exit of the regenerator are to be determined. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis From Prob. 9-98, qin = 539.23 kJ/kg, qout = 345.17 kJ/kg, and

KkJ/kg 08336.2kJ/kg 738.56

KkJ/kg 68737.2kJ/kg 88.797

KkJ/kg 17888.3K 1200

KkJ/kg .373482kJ/kg 04.586

KkJ/kg 1.70203K 300

55

44

33

22

11

⋅=→=

⋅=→=

⋅=→=

⋅=→=

⋅=→=

o

o

o

o

o

sh

sh

sT

sh

sT

and, from an energy balance on the heat exchanger,

KkJ/kg 2.47108

kJ/kg 645.36.04586738.5688.797

6

66425

⋅=→

=+−=→−=−os

hhhhh

Thus,

( )

( ) ( ) ( )( )

( )

( ) ( ) ( )( )( ) ( )[ ] ( ) ( )[ ]

( )( )

( )

( ) kJ/kg 114.5

kJ/kg 42.78

kJ/kg 5.57

kJ/kg 31.59

kJ/kg 22.40

=

+−=

+−−=

+−==

=

−−=

−−−=

−−==

=−+−=

−+−=−+−==

=⋅−−=

−−=−==

=⋅−−=

−−=−==

K 300kJ/kg 345.17

2.471081.70203K 300

ln

K 1260kJ/kg 539.23

2.608333.17888K 300

ln

2.687372.471082.373482.60833K 300

1/8lnKkJ/kg0.2873.178882.68737K 300

ln

8lnKkJ/kg 0.2871.702032.37348K 300

ln

out0

6

1610


0

5

3530


4625046250regengen,0regen destroyed,

3


1


LR

R

H

in

R

R

Tq

PP

RssTT

qssTsTx

Tq

PP

RssTT

qssTsTx

ssssTssssTsTx

PP

RssTssTsTx

PP

RssTssTsTx

oo

oo

oooo

oo

oo

Noting that h0 = h@ 300 K = 300.19 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from

( ) ( ) 06

026

060066 2gzVssThh ++−−−=φ

where KkJ/kg 0.769051.702032.47108ln0

1

6161606 ⋅=−=−−=−=−

PP

Rssssss oo

Thus, ( )( ) kJ/kg 114.5=⋅−−= KkJ/kg 0.76905K 300300.1936.6456φ


9-92

9-129 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law 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 constant specific heats.

700 kPa

Diesel fuel

100 kPa

Compres

4

3 2

Turbine

Combustion chamber

1

Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first calculate the exit temperature for the isentropic case

( ) K 6.505kPa 100kPa 700K 303

1)/1.357-(1.357/)1(

1

212 =

=

=

− kk

s PP

TT

0.881=−−

=−−

=K)303533(K)3036.505(

12

12

TTTT s

Cη

(b) The total mass flowing through the turbine and the rate of heat input are

kg/s 81.12kg/s 21.0kg/s 6.1260

kg/s 6.12kg/s 6.12AF

=+=+=+=+= aafat

mmmmm

&&&&&

kW 85557)kJ/kg)(0.9 00kg/s)(42,0 21.0(HVin === cf qmQ η&& The temperature at the exit of combustion chamber is K 1144)K533kJ/kg.K)( 3kg/s)(1.09 81.12(kJ/s 8555)( 3323in =→−=→−= TTTTcmQ p&&

The temperature at the turbine exit is determined using isentropic efficiency relation

( ) K 7.685kPa 700kPa 100K 1144

1)/1.357-(1.357/)1(

3

434 =

=

=

− kk

s PP

TT

K 4.754K)7.6851144(

K)1144(85.0 4

4

43

43 =→−−

=→−−

= TT

TTTT

sTη

The net power and the back work ratio are kW 3168)K30333kJ/kg.K)(5 3kg/s)(1.09 6.12()( 12inC, =−=−= TTcmW pa&

&

kW 5455)K4.754144kJ/kg.K)(1 3kg/s)(1.09 81.12()( 43outT, =−=−= TTcmW p&&

kW 2287=−=−= 31685455inC,outT,net WWW &&&

0.581===kW 5455kW 3168

outT,

inC,bw W

Wr &

&

(c) The thermal efficiency is 0.267===kW 8555kW 2287

in

netth Q

W&

&η

The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,

735.0K 1144K 30311

3

1max =−=−=

TT

η

and 0.364===735.0267.0

max

thII η

ηη


9-93

9-130 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases 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 constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.349 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are

xcc

c

c

dcr VVVV

VV

VV===→

+=→

+= 2

33

m 0002154.0m 0028.014

43

1 m 003015.00028.00002154.0 VVVV ==+=+= dc

Process 1-2: Isentropic compression

( )( )

( )( ) kPa 334114kPa 95

K 9.82314K 328

1.349

2

112

1-1.3491

2

112

==

=

==

=

−

k

k

PP

TT

v

v

v

v

Process 2-x and x-3: Constant-volume and constant pressure heat addition processes:

K 2220kPa 3341kPa 9000K) 9.823(

22 ===

PP

TT xx

kJ/kg 1149K)9.8232220(kJ/kg.K) (0.823)( 2-2 =−=−= TTcq xx v

x

Qout

V

P

4

3

2Qin

1

K 3254=→−=→−==− 3333-2 K)2220(kJ/kg.K) (0.823kJ/kg 1149)( TTTTcqq xpxx

(b) kJ/kg 2298114911493-2in =+=+= − xx qqq

3333 m 0003158.0

K 2220K 3254)m 0002154.0( ===

xx T

TVV

Process 3-4: isentropic expansion.

( )

( ) kPa 9.428m 0.003015m 0.0003158kPa 9000

K 1481m 0.003015m 0.0003158K 3254

1.349

3

3

4

334

1-1.349

3

31

4

334

=

=

=

=

=

=

−

k

k

PP

TT

V

V

V

V

Process 4-1: constant voume heat rejection. ( ) ( )( ) kJ/kg 7.948K3281481KkJ/kg 0.82314out =−⋅=−= TTcq v

The net work output and the thermal efficiency are kJ/kg 1349=−=−= 7.9482298outinoutnet, qqw

0.587===kJ/kg 2298kJ/kg 1349

in

outnet,th q

wη


9-94

(c) The mean effective pressure is determined to be

( )( )kg 0.003043

K 328K/kgmkPa 0.287)m 015kPa)(0.003 95(

3

3

1

11 =⋅⋅

==RTP

mV

kPa 1466=

⋅

−=

−=

kJmkPa

m)0002154.0003015.0(kJ/kg) kg)(1349 (0.003043

MEP3

321

outnet,

VV

mw

(d) The power for engine speed of 3500 rpm is

kW 120=

==

s 60min 1

rev/cycle) 2((rev/min) 3500kJ/kg) kg)(1349 003043.0(

2netnetnmwW&&

Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.

908.0K 3254

K )27325(113

0max =

+−=−=

TT

η

and 0.646===908.0587.0

max

thII η

ηη

The rate of exergy of the exhaust gases is determined as follows

( )

( )( ) kJ/kg 6.567100

9.428ln287.0298

1481kJ/kg.Kln 110.1()298(29814810.823

lnln)(0

4

0

4004040044

=

−−−=

−−−=−−−=

PP

RTT

cTTTcssTuux pv

kW 50.4=

==

s 60min 1

rev/cycle) 2((rev/min) 3500kJ/kg) kg)(567.6 003043.0(

244nmxX&&


9-95

9-131 A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion chamber, the net power output, the back work ratio, the thermal efficiency, the second law efficiency, the exergy efficiencies of the compressor, the turbine, and the regenerator, and the rate of the exergy of the combustion gases at the regenerator exit 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 constant specific heats. Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b). Analysis (a) For the compressor and the turbine:

( ) K 6.505kPa 100kPa 700K 303

1.3571-1.3571

1

212 =

=

=

−k

k

s PPTT

65

400°C 700 kPa 260°C

Regenerator

871°C

Compress.

432

Turbine

Combustion chamber

1100 kPa 30°C

0.881=−−

=−−

=K)303533(K)3036.505(

12

12

TTTT s

Cη

( ) K 6.685kPa 700kPa 100K 1144

1)/1.357-(1.357/)1(

3

434 =

=

=

− kk

s PP

TT

K 4.754K)6.6851144(

K)1144(85.0 4

4

43

43 =→−−

=→−−

= TT

TTTT

sTη

(b) The effectiveness of the regenerator is

0.632=−−

=−−

=K)5334.754(

K)533673(

24

25regen TT

TTε

(c) The fuel rate and air-fuel ratio are

kg/s 1613.0673)K144kJ/kg.K)(1 093.1)(6.12(7)kJ/kg)(0.9 000,42(

)()( 53HVin

=→−+=

−+==

fff

pafcf

mmm

TTcmmqmQ

&&&

&&&& η

78.14===1613.0

6.12AFf

a

mm&

&

Also, kg/s 76.121613.06.12 =+=+= fa mmm &&&

kW 65707)kJ/kg)(0.9 00kg/s)(42,0 1613.0(HVin === cf qmQ η&& (d) The net power and the back work ratio are

kW 3168)K30333kJ/kg.K)(5 3kg/s)(1.09 6.12()( 12inC, =−=−= TTcmW pa&&

kW 5434)K4.754144kJ/kg.K)(1 3kg/s)(1.09 76.12()( 43outT, =−=−= TTcmW p&&

kW 2267=−=−= 31685434inC,outT,net WWW &&&

0.583===kW 5434kW 3168

outT,

inC,bw W

Wr

&

&

(e) The thermal efficiency is

0.345===kW 6570kW 2267

in

net

QW

th &

&η

(f) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,


9-96

735.0K 1144K 30311

3

1max =−=−=

TT

η

and 0.469===735.0345.0

max

thII η

ηη

(g) The exergy efficiency for the compressor is defined as the ratio of stream exergy difference between the inlet and exit of the compressor to the actual power input:

[ ] ( )

kW 2943100700ln287.0

303533ln)093.1()303()303533)(093.1()6.12(

lnln)(1

2

1

201212012C

=

−

−−=

−−−=−−−=∆

PP

RTT

cTTTcmssThhmX ppaa &&&

0.929==∆

=kW 3168kW 2943

inC,

CCII, W

X&

&η

The exergy efficiency for the turbine is defined as the ratio of actual turbine power to the stream exergy difference between the inlet and exit of the turbine:

( )

kW 5834100700ln287.0

754.41144ln)093.1()303()4.7541144)(093.1()76.12(

lnln4

3

4

3043T

=

−

−−=

−−−=∆

PP

RTT

cTTTcmX pp&&

0.932==∆

=kW 5834kW 5434

T

inT,, X

WTII &

&η

An energy balance on the regenerator gives

K 2.616)4.754)(093.1)(76.12()53373)(1.093)(66.12(

)()(

66

6425

=→−=−

−=−

TT

TTcmTTcm ppa &&

The exergy efficiency for the regenerator is defined as the ratio of the exergy increase of the cold fluid to the exergy decrease of the hot fluid:

( )

kW 10730616.2754.4ln)093.1()303()2.6164.754)(093.1()76.12(

0ln6

4064hotregen,

=

−

−−=

−−−=∆

TT

cTTTcmX pp&&

( )

kW 8.9540533673ln)093.1()303()533673)(093.1()76.12(

0ln2

5025coldregen,

=

−

−−=

−−−=∆

TT

cTTTcmX pp&&

0.890==∆

∆=

kW 1073kW 8.954

hotregen,

coldregen,, X

XTII &

&η

The exergy of the combustion gases at the regenerator exit:

( )

kW 1351=

−

−−=

−−−=

0303

616.2ln)093.1()303()3032.616)(093.1()76.12(

0ln0

60066 T

TcTTTcmX pp&&


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