SM_Chap09

9-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition in SI Units

Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 9 GAS POWER CYCLES

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PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

9-2

Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions, Reciprocating Engines

9-1C It represents the net work on both diagrams.

9-2C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) all the processes are internally reversible, (3) the combustion process is replaced by the heat addition process, and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state.

9-4C The cold air standard assumptions involves the additional assumption that air can be treated as an ideal gas with constant specific heats at room temperature.

9-5C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center.

9-6C It is the ratio of the maximum to minimum volumes in the cylinder.

9-6C The MEP is the fictitious pressure which, if acted on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle.

9-7C Yes.

9-8C Assuming no accumulation of carbon deposits on the piston face, the compression ratio will remain the same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a car gets older as a result of wear and tear.

9-9C The SI and CI engines differ from each other in the way combustion is initiated; by a spark in SI engines, and by compressing the air above the self-ignition temperature of the fuel in CI engines.

9-10C Stroke is the distance between the TDC and the BDC, bore is the diameter of the cylinder, TDC is the position of the piston when it forms the smallest volume in the cylinder, and clearance volume is the minimum volume formed in the cylinder.



9-3

9-11 The maximum possible thermal efficiency of a gas power cycle with specified reservoirs is to be determined.

Analysis The maximum efficiency this cycle can have is

64.0%==+

+−=−= 1

T0.640

K )273(500K )273(51Carnotth,

H

L

Tη

specified processes. The cycle is to be

2 Kinetic and potential energy changes are negligible. 3 Air is

erties The properties of air are given as R = 0.287 kPa·m3/kg·K, c = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k =

the cycle are shown in the figures.

(b) Process 1-2: Isentropic com

TTmcw in −= v

9-12 An air-standard cycle executed in a piston-cylinder system is composed of three sketcehed on the P-v and T-s diagrams and the back work ratio are to be determined.

Assumptions 1 The air-standard assumptions are applicable. an ideal gas with constant specific heats.

Prop p1.4.

Analysis (a) The P-v and T-s diagrams of

pression

)( 12,21−

11

s

T

3 2

1

v

P

32

1

1−

−⎞⎛

kv

2

112 =⎟⎟

⎠⎜⎜⎝

= krTTTv

Process 2-3: Constant pressure heat addition

2,3 TTmRPPdw out −=−== ∫− VVv

rk ratio is

3

)()( 23232 2

The back wo

)()(

23

12

,32

,21

TTmRTTmc

ww

rout

inbw −

−==

−

− v

Noting that

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

1

thus,and and =−=c

kccR pp v −

=k

Rcc

vv

From ideal gas relation,

rTT

===2

1

2

3

2

3

vv

vv

Substituting these into back work relation,

( )

( )0.256=

−−

−=

−−

−=

−

⎟⎠⎞

⎜⎝⎛ −

−=

−−

−=

−

−−

1661

14.11

11

11

1

11

11

)1/()/1(1

1

4.0

11

23

21

2

2

rr

krr

k

TTTT

TT

RkRr

kk

bw

preparation. If you are a student using this Manual, you are using it without permission.

9-4

9-13 The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the back work ratio and the thermal 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.

Properties The properties of air are given as R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and

k = 1.4.

Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures.

(b) During process 1-3, we have


=−⋅−=

−−=− TTRP VV

uring process 2-3, we have

,13 −=−= ∫− Pdw in v

kJ/kg 516.600)K21K)(300kJ/kg 287.0(

)()( 31311

1

3

s

T

32

1

v

P

3

2

1

D

kJ/kg 8.1172n7K)(2100)KlkJ/kg 287.0(

7ln7

lnln2

2

2

33

2

3

2,32 == ∫∫− Pdw out v

=⋅=

=== RTRTRTdRTV

V

V

Vv

V

The back work ratio is then

0.440===−

−

kJ/kg 8.1172kJ/kg 6.516

,32

,13

out

inbw w

wr

Hprocess 1-3,

eat input is determined from an energy balance on the cycle during

kJ/kg 1172.8300)K)(2100kJ/kg 718.0()( ,3213

,3231

31,32,31

+−⋅=

+−=

+∆=−

∆=−

−

−−

−−−

outv

out

outin

wTTcwu

uwq

he net work output is

,31− inq

k 2465= J/kg

T

kJ/kg 2.6566.5168.1172,13,32 =−=−= −− inoutnet www

(c) The thermal efficiency is then

26.6%==== 266.0kJ 2465kJ 656.2

in

netth q

wη


9-5

9-14 The three processes of an ideal gas power cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the maximum temperature, expansion and compression works, and thermal efficiency are to be determined.

Assumptions 1 Kinetic and potential energy changes are negligible. 2 The ideal gas has constant specific heats.

Properties The properties of ideal gas are given as R = 0.3 kJ/kg.K, cp = 0.9 kJ/kg.K, cv = 0.6 kJ/kg·K, and

k = 1.5.


(b) The maximum temperature is determined from


K 734.8=+==⎟⎟⎠

⎜⎜⎝

== 12

112max 27(rTTTT

V

⎞ −−−

11.511

K)(6) 273kk

V

) An energy balance during process 2-3 gives

voutin TTTTcuwq 232332,32,32 since 0)(−−− ==−=∆=−

Then, the work of compression is

s

T

32

1

v

P

3

2

1

⎛

(c

ouin wq ,32,32 −− = t

kJ/kg 395.0=⋅=

=== ∫K)ln6 K)(734.8kJ/kg 3.0(

lnln 22

32

22

rRTRTdRTPdV

Vv

Vv

(d) The work during isentropic compression is determined from an energy

−=∆= −

300))( 1221,2 TTcuw vin

) Net work output is

== ∫−−

33

,32,32 wq outin

balance during process 1-2:

−1

kJ/kg 260.9=

⋅= K)(734.8kJ/kg 6.0( −

(e

kJ/kg 1.1349.2600.395,21,32 =−=−= −− inoutnet www

The thermal efficiency is then

33.9%==== 339.0kJ 395.0kJ 134.1

in

netth q

wη


9-6

9-15 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal 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 variable specific heats.

Properties The properties of air are given in Table A-17.

Analysis (b) The properties of air at various states are

v

P

1

2

4

3qin

qout

s

T

1

24

3qin

qout

( )

( )

( ) kJ/kg 71.73979.32601.9kPa 1835.6

kPa 100

kPa 6.1835kPa 600K 490.3K 1500

9.601kJ/kg 41.1205

K 1500

K 3.490kJ/kg 352.29

841.71.3068kPa 100kPa 600

3068.1kJ/kg .17295

K 2951 ⎯→⎯=T

43

4

22

323

33

2

2

1

2

1

34

3

12

1

=⎯→⎯===

=

==

⎯→⎯=

==

⎯→⎯===

==

hPPP

P

TT

Pu

T

Tu

PPP

P

Ph

rr

r

rr

r

From energy balances,

(c) Then the thermal efficiency becomes

32233 ==⎯→⎯= PT

PPP vv

T

kJ/kg 408.6=−=−=

=−=−=

=−=−=

5.4441.853

kJ/kg .544417.29571.739

kJ/kg 1.85329.35241.1205

outinoutnet,

14out

23in

qqw

hhq

uuq

47.9%==== 479.0kJ/kg 853.1kJ/kg 408.6

in

outnet,th q

wη



9-7

9-16 Problem 9-15 is reconsidered. The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted.

Analysis Using EES, the problem is solved as follows:

"Input Data" T[1]=295 [K] P[1]=100 [kPa] P[2] = 600 [kPa] T[3]=1500 [K] P[4] = 100 [kPa] "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] "Conservation of energy for process 1 to 2" q_12 -w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3" q_23 -w_23 = DELTAu_23 w_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]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=0.287*T[4]} "Conservation of energy for process 3 to 4" q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant pressure heat rejection" {P[4]*v[4]/T[4]=P[1]*v[1]/T[1]} "Conservation of energy for process 4 to 1" q_41 -w_41 = DELTAu_41 w_41 =P[1]*(v[1]-v[4]) "constant pressure process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent"



9-8

T3 [K]

ηth qin,total [kJ/kg]

Wnet [kJ/kg]

1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500

47.91 48.31 48.68 49.03 49.35 49.66 49.95 50.22 50.48 50.72 50.95

852.9 945.7 1040 1134 1229 1325 1422 1519 1617 1715 1813

408.6 456.9 506.1 556

606.7 658.1 710.5 763

816.1 869.8 924

1500 1700 1900 2100 2300 250047.5

48

48.5

49

49.5

50

50.5

51

T[3] [K]

ηth

1500 1700 1900 2100 2300 2500800

1020

1240

1460

1680

1900

T[3

g]

] [K]

q in,

tota

l [k

J/k



9-9

1500 1700 1900 2100 2300 2500400

500

600

700

800

900

1000

T[3] [K]

wne

t [k

J/kg

]

5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50

500

1000

1500

2000

s [kJ/kg-K]

T [K

]

100 kPa

600 kPa

Air

1

2

3

4

10-2 10-1 100 101 102101

102

103

104

v [m3/kg]

P [k

Pa]

295 K 1500 K

Air

1

2

3

4



9-10

9-17 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the heat rejected and the thermal efficiency are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

Analysis (b) The temperature at state 2 and the heat input are


( )( ) K 579.2

kPa 100kPa 1000

K 3000.4/1.4/1

1

212 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

( ) ( )( )( )( )2.579KkJ/kg 1.005kg 0.004kJ 2.76 3 ⎯→⎯−⋅= T K 12663

2323in

=

−=−=

T

TTmchhmQ p

Process 3-1 is a straight line on the P-v diagram, thus the w31 is simply the area under the process curve,

( )

( )

kJ/kg 7.273=

KkJ/kg 0.287kPa 1000K 1266

kPa 100K 300

2kPa 1001000

22area

3

3

1

11331

1331

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠

⎞⎜⎝

⎛ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=−

+==

PRT

PRTPPPP

w vv

Energy bal nce for process 3-1 gives

−−−=

−=−−→

K1266-300KkJ/kg 0.718273.7kg 0.004)(

)(

31out31,31out31,out31,

31out31,out31,systemoutin

TTcwmTTmcmwQ

uumWQ

vv

) The thermal efficiency is then

a

( )[ ]( ) ( )( )[ ] kJ 1.679=⋅+−=

−+−

v

P

32

1

qin

qout

s

T

32

1 qout

qin

⎯⎯∆=− EEE

=

(c

39.2%=−=−=kJ 2.76kJ 1.679

11in

outth Q

Qη


9-11

9-18 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is to be determined.

Assumptions Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg.K, and k = 1.4 (Table A-2).

Analysis The minimum pressure in the cycle is P3 and the maximum pressure is P1. Then,


or

( )

( )( ) kPa 1888

K 300K 1100

kPa 201.4/0.41/

3

232

/1

3

2

3

2⎟⎟⎠

⎜⎜⎝

=PT

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎞⎛

−

−

kk

kk

TT

PP

PT

t input is determined from

hen,

s

T

3

2qin

qout4

11100

300 The hea

( )

( ) ( )( )( )

( )( ) kJ 63.8===

==−=−=

=⋅=−=

⋅=⋅−=−=−

kJ 87.730.7273

.7%727273.0K 1100K 300

11

kJ 73.87KkJ/kg 0.1329K 1100kg 0.6

KkJ/kg 0.1329kPa 3000kPa 1888lnKkJ/kg 0.287lnln

inthoutnet,

th

12in

1

20

1

212

QW

TT

ssmTQ

PP

RTT

css

H

L

H

p

η

η

T


9-12

9-19 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimum pressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the second-law efficiency of an actual cycle operating between the same temperature limits are to be determined.


Properties The properties of air at room temperatures are R = 0.287 kJ/kg.K and k = 1.4 (Table A-2).

Analysis (a) The minimum temperature is determined from

( )( ) ( )( ) K 350K750KkJ/kg 0.25kJ/kg 10012net =⎯→⎯−⋅=⎯→⎯−−= LLLH TTTTssw

The pressure at state 4 is determined from

or

( )

( )

kPa 1.110K 350K 750

kPa 800 4

1.4/0.4

4

1/

4

141

/1

4

1

4

1

=⎯→⎯⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

PP

TT

PP

PP

TT

kk

kk

s

T

3

2qin

4

1750 K

wnet=100 kJ/kg

qout

The minimum pressure in the cycle is determined from

( ) kPa 46.1=⎯→⎯⋅−=⋅−

−=∆−=∆ 3412 lncss p

33

3

40

3

4

kPa 110.1lnKkJ/kg 0.287KkJ/kg 25.0

ln

PP

PP

RTT

(b) The heat rejection from the cycle is

kgkJ/ 87.5==∆= kJ/kg.K K)(0.25 350(12out sTq L )

) The thermal efficiency is determined from

(c


0.533=−=−=K 350

11 LTη

K 750HT

(d) The power output for the Carnot cycle is

Then, the second-law efficiency of the actual cycle becomes

th

kW 9000kJ/kg) kg/s)(100 90(netCarnot === wmW &&

0.578===kW 9000kW 5200

Carnot

actualII W

W&

&η


9-13

9-20 An ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature of the low-temperature energy reservoir, the cycle's thermal efficiency, and the amount of heat that must be supplied per cycle are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis The temperature of the low-temperature reservoir can be found by applying the isentropic expansion process relation

K 481.1=⎟⎠⎞

⎜⎝⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−− 14.11

1

221 12

1K) 2731027(k

TTv

v

s

T

3

2qin

qout4

11300 K Since the Carnot engine is completely reversible, its efficiency is

0.630=+

−=−=K 273)(1027

K 481.111Carnotth,HTLT

η

The work output per cycle is

kJ/cycle 20min 1cycle/min 1500net

⎠⎝n&s 60kJ/s 500net =⎟

⎞⎜⎛==

WW

&

ccording the definition of the cycle efficiency,

A to

kJ/cycle 31.75===⎯→⎯=0.63kJ/cycle 20

Carnotth,

netin

in

netCarnotth, η

ηW

QQW



9-14

9-21 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to be determined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained.


Properties The properties of air are given as R = 0.3 kJ/kg·K and cv = 0.3 kJ/kg·K.


(b) Noting that

1.429

7.00.1

K kJ/kg 0.13.07.0

===

⋅=+=+=

v

v

cc

k

Rcc

p

p

s

T

32

1

v

P

32

1Process 1-2: Isentropic compression

K 4.584)5)(K 293( 429.011

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

−k

k

rTTTvv

kJ/kg 204.0=−⋅=−=−1 K )2934.584)(KkJ/kg 7.0()( 12in2, TTcw v

rom ideal gas relation,

0=q −21

F

2922)5)(4.584(3133 =⎯→⎯=== Tr

T v

v

v

v

22=

T

Constant pressure heat addition

−=−== ∫−

K )4.

)()( 23232

3

out3,2 TTmRPPdw VVv

huq

p

out

ant volume heat rejection

2

Process 2-3:

kJ/kg 701.3=−⋅= 5842922)(KkJ/kg 3.0(2

kJ/kg 2338=−⋅=−=

∆=∆+= −−−−

K )4.5842922)(KkJ/kg 1()( 23

3232,32in3,2

TTcw

Process 3-1: Const

=∆= −− (31out1,3 cuq v kJ/kg 1840.3=⋅=− K 293)-K)(2922kJ/kg 7.0()13 TT

(c) Net work is

0=−13w

KkJ/kg 3.4970.2043.701in2,1out3,2net ⋅=−=−= −− www


21.3%==== 213.0kJ 2338kJ 497.3

in

netth q

wη



9-15

(d) The expression for the cycle thermal efficiency is obtained as follows:


⎟⎠

⎞⎜⎝

⎛ −−

−⎟⎠⎞

⎜⎝⎛ −=

⎟⎠

⎞⎜⎝

⎛ −−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−

−−=

−−−−

=

−==

−

−

−

−

−−

−−

−

−−

1

1

11

1

11

11

111

11

11

11

1

23

1223

in

in2,1out3,2

in

netth

11)1(

111

11)1(

1

1)1(

1

)1(

1

)()(

)()()(

k

kp

kp

kp

kk

v

p

kkp

kv

p

p

v

rrkk

rrkcR

rTT

rkcR

rrTc

rTT

rTc

cR

rTrrTcTrTc

cR

TTcTTcTTR

qww

qw

η

since

111kc

cc

cccR

p

v

p

vp

p−=−=

−=


9-16

Otto Cycle

9-22C For actual four-stroke engines, the rpm is twice the number of thermodynamic cycles; for two-stroke engines, it is equal to the number of thermodynamic cycles.

9-23C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.

9-24C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heat addition, (3) isentropic expansion, and (4) v = constant heat rejection.

9-25C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.

9-26C It increases with both of them.

9-27C Because high compression ratios cause engine knock.

9-28C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k = 1.667.

9-29C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline engines.



9-17

9-30 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis The definition of cycle thermal efficiency reduces to

v

P

4

1

3

2

qin

qout

61.0%==−=−=−− 10.51th kr

6096.01111 11.4η

The rate of heat addition is then

kW 148===0.6096

kW 90

th

netin η

WQ

&&

9-31 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.

Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is

erties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-

nalysis The definition of cycle thermal efficiency reduces to

an ideal gas with constant specific heats.

Prop2a).

A

v

P

4

1

3

2

qin

qout

57.5%==−=−=1111thη

−−5752.0

8.5 11.41kr

he rate of heat addition is then

T

kW 157===0.5752

kW 90

th

netin η

WQ

&&



9-18

9-32 The two isentropic processes in an Otto cycle are replaced with polytropic processes. The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined.


Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg·K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis The temperature at the end of the compression is

v

P

4

1

3

2


K 4.537K)(8) 288(12

12 ==⎟⎟⎠

⎜⎜⎝

= rTTTv

13.111

1 =⎞⎛ −−

−n

nv

And the temperature at the end of the expansion is

K 4.7898

K) 1473(34

34 =⎟⎠

⎜⎝

=⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=r

TTTv

11 13.1113 ⎞⎛⎞⎛⎞⎛ −−− nn

v

he integr of the work expression for the polytropic compression gives

T al

kJ/kg 6.238)18(13.1

K) K)(288kJ/kg 287.0(11

13.11 ⎤⎡ −n

2

1121 =−

−⋅

=⎥⎥

⎦⎢⎢

⎣−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

= −− n

RTw

v

v

Similarly, the work produced during the expansion is

kJ/kg 0.654181

13.1K) K)(1473kJ/kg 287.0(

113

43 ⎢⎢⎜⎜⎛

−−=− n

RT 13.1

4

3 =⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

−⋅

−=⎥⎥

⎦⎣−⎟⎟

⎠

⎞

⎝

−

wv

v

pplicatio of the first law to each of the four processes gives

84.537)(KkJ/kg 718.0(kJ/kg 6.238)( 122121

1 ⎤⎡ −n

A n

28 kJ/kg 53.59K) =−⋅−=−−= −− TTcwq v

1473)(KkJ/kg 718.0()( 2332 ⋅=−=− TTcq v kJ/kg 8.671K)4.537 =−

kJ/kg 2.163K)4.7891473)(KkJ/kg 718.0(kJ/kg 0.654)( 43434 3 =−⋅−=−−= − TTcwq v −

kJ/kg 718.0()( 1414 ⋅=−=− TTcq v kJ/kg 0.360K)2884.789)(K =−

he head added and rejected from the cycle are

The thermal efficiency of this cycle is then

T

kJ/kg 419.5kJ/kg 835.0

=+=+==+=+=

−−

−−

0.36053.592.1638.671

1421out

4332in

qqqqqq

0.498=−=−=0.8355.41911

in

outth q

qη


9-19

9-33 An ideal Otto cycle is considered. The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined.



Analysis The temperature at the end of the compression is

v

P

4

1

3

2


K 7.661K)(8) 288(12

12 ==⎟⎟⎠

⎜⎜⎝

= rTTTv

14.111

1 =⎞⎛ −−

−k

kv

and the temperature at the end of the expansion is

K 2.648

K) 1473(34

34 =⎟⎠

⎜⎝

=⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=r

TTTv

111 14.1113 ⎞⎛⎞⎛⎞⎛ −−− kk

v

pplicatio of the first law to the heat addition process gives A n

=−= ()( TTcq kJ/kg 582.5=−⋅ K)7.6611473)(KkJ/kg 718.023in v

ilarly, the heat rejected is

Sim

⋅=−= KkJ/kg 718.0()( TTcq kJ/kg 253.6=− K)2882.641)(14v

a

out

l efficiency of this cycle is then The therm

0.565=−=−=5.5826.25311

in

outth q

qη


9-20

9-34 A six-cylinder, four-stroke, spark-ignition engine operating on the ideal Otto cycle is considered. The power produced by the engine is to be determined.


Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K (Table A-1), cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis From the data specified in the problem statement,

v

P

4

1

3

2

143.7140 1

1

2

1 ===v

v

v

v

.r

Since the compression and expansion processes are isentropic,

( ) K 7.636143.7K) 290( 14.111

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

−k

k

rTTTvv

K 6.5207.143

K) 1143(34

34 =⎟⎠

⎜⎝

=⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=r

TTTv

11 14.1113 ⎞⎛⎞⎛⎞⎛ −−− kk

v

pplicatio of the first law to the compression and expansion processes gives

=−−−=

A n

)()( 1243net −−−= TTcTTcw vv

kJ/kg 0.198K)2907.636)(kJ/kg·K 718.0(K)6.5201143)(kJ/kg·K 718.0(

When each cylinder is charged with the air-fuel mixture,


/kgm 8761.0kPa 951

1 ==P

vK) 290)(K/kgmkPa 287.0( 3

31 =

⋅⋅RT

he total air mass taken by all 6 cylinders when they are charged is T

kg 004218.0/kgm 8761.0

m)/4 099.0(m) 089.0()6(4/2∆ πV SB3

2

1cyl ==

πv

Nm 1

cyl ==v

N

The net work produced per cycle is

kJ/cycle 8352.0kJ/kg) kg(198.0 (0.004218netnet === mwW

The power produced is determined from

kW 17.4===rev/cycle 2

rev/s) 2500/60kJ/cycle)( (0.8352

rev

netnet N

nWW

&&

since there are two revolutions per cycle in a four-stroke engine.


9-21

9-35 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 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.

v

P

4

1

3

2Qin Qout

( )( )

( ) ( ) kPa 2338kPa 100K 308K 757.9

9.5

K 757.99.5K 308

11

2

2

12

1

11

2

22

0.41

2

112

=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

PTT

PT

PT

P

TTk

v

vvv

v

v

Process 3-4: isentropic expansion.


( )( )=⎟⎟⎠

⎜⎜⎝

=3

443 9.5K 800TT

vK 1969=

⎞⎛−

0.41k

v

rocess 2-3: v = constant heat addition. P

( ) kPa 6072=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯= kPa 2338

K 757.9K 1969

22

33

2

22

3

33

TP

TT

PT

PP vv

( )( )(b) ( )( )

kg10788.6K 308

m 0.0006kPa 100 43

11 −×=PV

K/kgmkPa 0.287 3

1 ⋅⋅==

RTm

( ) ( ) ( )( )( ) kJ 0.590=−⋅×=−=−= − K757.91969KkJ/kg 0.718kg106.788 42323in TTmcuumQ v

(c) Process 4-1: v = constant heat rejection.

( ) ( )( )( ) kJ0.2 40K308800KkJ/kg 0.718kg106.788)( 41414out =−⋅×−=−=−= −TTmcuumQ v

kJ 0.350240.0590.0outinnet =−=−= QQW

59.4%===kJ 0.590kJ 0.350

in

outnet,th Q

Wη

( )( )kPa 652=⎟

⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

kJmkPa

1/9.51m 0.0006kJ 0.350

)/11(MEP

3

31

outnet,

21

outnet,

max2min

rWW

r

VVV

VVV (d)


9-22

9-36 An Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure 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 room temperature are cp = 1.005 kJ/kg·K, cv = 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.

( )( )

( ) ( ) 2338kPa 100K 308K 757.9

9.5

K 757.99.5

11

2

2

12

1

11

2

22

0.41

=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯=

=⎞⎛

−

PTT

PT

PT

P

k

v

vvv

v

rocess 3- polytropic expansion.

kPa v

P

4

1

3

2

Qin

Qout

Polytropic

800 K

308 KK 3082

112 =⎟⎟

⎠⎜⎜⎝

= TTv

P 4:

( )( )( )( )

kg10788.6K 308K/kgmkPa 0.287

m 43

3

1

11 −×=⋅⋅

==RT

m

0.0006kPa 100PV

( )( )

( ) ( )( )( )kJ 0.5338

1.351K1759800KkJ/kg 0.287106.788

1

434

34 =−

−⋅×=

−−

=−

nTTmR

W

9.5K 800 35.01

3

443 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

TTn

K 1759v

v

ring the expansion process (This is not realistic, and probably is due to ssuming constant specific heats at room temperature). ) Process 2-3: v = constant heat addition.

Then energy balance for process 3-4 gives

( )( ) ( )

( )( )( ) kJ 0.0664kJ 0.5338K1759800KkJ/kg 0.718kg106.788 4in34,

out34,34out34,34in34,

34out34,in34,

outin

=+−⋅×=

+−=+−=

−=−

∆=−

−QWTTmcWuumQ

uumWQ

EEE system

v

That is, 0.066 kJ of heat is added to the air dua(b

( ) kPa 5426=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯= kPa 2338

K

757.9K 1759

22

33

2

22

3

33 PTT

PT

PT

P vv

3,

=−⋅×=

−=−=−Q

TTmcuumQ v

Therefore,

( ) ( )( )( )( ) kJ 0.4879K757.91759KkJ/kg 0.718kg106.788 4

in23,

2323in2

kJ 0.5543=+=+= 0664.04879.0in34,in23,in QQQ

(c) Process 4-1: v = constant heat rejection.

( ) ( ) ( )( )( ) kJ 0.2398K308800KkJ/kg 0.718kg 106.788 41414out =−⋅×=−=−= −TTmcuumQ v

kJ 0.31452398.05543.0outinoutnet, =−=−= QQW

56.7%===kJ 0.5543kJ 0.3145

in

outnet,th Q

Wη

( )( )kPa 586=⎟

⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

kJmkPa

1/9.51m 0.0006kJ 0.3145

)/11(MEP

3

31

outnet,

21

outnet,

max2min

rWW

r

VVV

VVV (d)


9-23

9-37 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined.




v

P

4

1

3

2

qin

qout

1340 K

300K

2.621kJ/kg 07.214

K 3001

11 =

=⎯→⎯=

r

uT

v

( ) kJ/kg 22.49165.772.62181

222 r rrr v11

22 =⎯→⎯==== uvv

vv

Process 2-3: v = constant heat addition.

==

⎯→⎯=

22.

247.10kJ/kg 94.1058

K 1340 33

3

uT

rv

) Proces -4: isentropic expansion.

−=−= 49194.105823 uuqin kJ/kg 567.72=

(b s 3


( )( 247.1084 === rvvv

v ) kJ/kg 82.480976.81 43

334=⎯→⎯= urrr v

rocess 4- v = constant heat rejection.

P 1:

kJ/kg 75.26607.21482.48014out =−=−= uuq

53.0%==−=−= 530.0kJ/kg 567.72kJ/kg 266.75

11in

outth q

qη

77.6%==−=−= 776.0K 1340K 300

11Cth,H

L

TT

η (c)


9-24

9-38 An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount of heat transferred to the argon during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined.

Assumptions 1 The air-standard assumptions are applicable with argon as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats.

Properties The properties of argon are cp = 0.5203 kJ/kg.K, cv = 0.3122 kJ/kg.K, and k = 1.667 (Table A-2).


v

P

4

1

3

2

qinqout

( )( )8K 3002

12 =⎟⎟⎠

⎜⎜⎝

= TTv

K 12010.6671

1 =⎞⎛

−kv


( ) ( )( ) kJ/kg 43.45=−=−=−= k 0.31222323in TTcuuq v K12011340J/kg.K

) Process 3-4: isentropic expansion. (b

( ) K 8.33481K 1340

0.6671−k

4

334 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= TT

v

v

rocess 4- v = constant heat rejection.

P 1:

)( ) ( ( ) kJ/kg .8510K300334.8kJ/kg.K 0.31221414out =−=−=−= TTcuuq v

75.0%==−=−= 750.0kJ/kg 43.45kJ/kg 10.85

11in

outth q

qη

77.6%==−=−= 776.0K 1340K 300

11Cth,H

L

TT

η (c)



9-25

9-39 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeled as polytropic. The temperature at the end of expansion process, the net work output, the thermal efficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuel consumption are to be determined.


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) Process 1-2: polytropic compression

1

Qin

2

3

4

P

V

Qout

( )( )

( )( )11kPa 100 1.3

2

112 ==⎟⎟

⎠⎜⎜⎝

= PPv

kPa 2258

K 5.63611K 310 1-1.31

2

112

⎞⎛

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

n

n

TT

v

v

v

kJ/kg 3.3121.31

310)KK)(636.5kJ/kg (0.2871

)( 1212 −=

−−⋅

=−

−=

nTTR

w

Process 2-3: constant volume heat addition

( ) K 2255kPa 2258kPa 8000K 636.5

2

32


3 =⎟⎠⎞

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=P

TT

Process 3-4: polytropic expansion.

⎛⎞⎛ P

( )( )( ) kJ/kg 1332K636.52255KkJ/kg 0.823

2323in

=−⋅=−=−= TTcuuq v

( ) K 1098=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠⎜⎜⎝

=4

334 11

1K 2255TTv

⎞⎛− 1-1.31n

v

( ) kPa 2.354111kPa 8000

1.3

1

234 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎛

=n

PPvv

⎠⎝

kJ/kg 11061.31

2255)KK)(1098kJ/kg (0.2871

)( 3434 =

−−⋅

=−−

=nTTR

w

Process 4-1: constant volume heat rejection.

(b) The net work output and the thermal efficiency are

kJ/kg 794=−=−= 3.31211061234outnet, www

59.6%==== 596.0kJ/kg 1332kJ/kg 794

in

outnet,th q

wη

(c) The mean effective pressure is determined as follows

( )( )

( )( )kPa 982=⎟

⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

==⋅⋅

==

kJmkPa

1/111/kgm 0.8897kJ/kg 794

)/11(MEP

/kgm 0.8897kPa 100

K 310K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv


9-26

(d) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are

33

m 00016.0m 0016.0

11 =⎯→⎯+

=⎯→⎯+

= cc

c

c

dcr VV

V

V

VV

31 0016.000016.0 =+=+= dc VVV m 00176.0

The total m ss contained in the cylinder is a

( )( )kg 0.001978

K 310K/kgmkPa 0.287)m 76kPa)/0.001 100( 3PV

31

11 =⋅⋅

==RT

mt

The engine speed for a net power output of 50 kW is

rev/min 3820=⎟⎠⎞

⎜⎝⎛ 60kJ/s 50W&

⋅ min 1s

cycle)kJ/kg

ote that t wo revolutions in one cycle in four-stroke engines.

cle is

==kg)(794 001978.0(

rev/cycle) 2(2net

net

wmn

t&

N here are t

(e) The mass of fuel burned during one cy

kg 0001164.0kg) 001978.0(

16AF =⎯→⎯−

=⎯→⎯−

== ff

f

f

ft

f

a mm

mm

mmmm

Finally, the specific fuel consumption is

g/kWh 267=⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛==

kWh 1kJ 3600

kg 1g 1000

kJ/kg) kg)(794 001978.0(kg 0001164.0sfc

netwmm

t

f



9-27

9-40 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled.


Analysis The temperature at the end of the compression varies with the compression ratio as

11

212 =⎟⎟

⎠⎜⎜⎝

= rTTTv

1

1 −−

⎞⎛ kk

v

T1 is fixed. The temperature rise during the combustion remains constant since the mount of heat addition is fixed. Then, the maximum cycle temperature is given by

he smallest gas specific volume during the cycle is

v

P

4

1

3

2 qout

qinsince a

11in2in3 // −+=+= krTcqTcqT vv

T

r1

3v

v =

When this is combined with the maximum temperature, the maximum pressure is given by

( )11in

13

33 / −+== krTcqRrRTP vvv

9-41 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent.

Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is n ideal ga with constant specific heats.

rocess,

=kPv

heat is lost during the expansion of the gas,

where T4s is the temperature that would occur if the expansion were reversible and adiabatic (n=k). This can only occur when

a s

Analysis During a polytropic p


constant

constant/)1( =

=− nn

n

TP

Pv

v

P

4

1

3

2 qout

qinand for an isentropic process,

constant/)1( =− kkTP

constant

If

sTT 44 >

kn ≤


9-28

Diesel Cycle

9-42C A diesel engine differs from the gasoline engine in the way combustion is initiated. In diesel engines combustion is initiated by compressing the air above the self-ignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine.

9-43C The Diesel cycle differs from the Otto cycle in the heat addition process only; it takes place at constant volume in the Otto cycle, but at constant pressure in the Diesel cycle.

9-44C The gasoline engine.

9-45C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem.

9-46C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process. As the cutoff ratio decreases, the efficiency of the diesel cycle increases.

9-47 An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 1.3. The maximum temperature of the air and the rate of heat addition are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis We begin by using the process types to fix the temperatures of the states.


( ) K 6.95420K) 288( 14.111

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

−k

k

rTTTv

v

K 1241===⎟⎟⎠

⎜⎜⎝

= K)(1.3) 6.954(22

23 crTTTv

⎞⎛ 3v

ombining e first law as applied to the various processes with the process equations gi

C th ves v

P

4

1

2 3qin

qout

6812.0)13.1(4.1

13.1111 4.1 −−kr20

1)1(

11.41th =

−−=

−−=

−c

ck rkr

η

According to the definition of the thermal efficiency,

1−

kW 367===0.6812

kW 250

th

netin η

WQ

&&


9-29

9-48 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined.



Analysis The specific volume of the air at the start of the compression is

v

P

4

1

2

3

qout

x

qin/kgm 051.1

kPa 8011 ===

Pv

K) 293)(K/kgmkPa 287.0( 33

1 ⋅⋅RT

olume at the end of the compression is and the specific v

/kgm 07508.014

12 ===

/kgm 051.1 33

rv

he pressu at the end of the compression is

v

T re

( )


kPa 321914kPa) 80( 4.11

2

112 ===⎟⎟

⎠⎜⎜⎝

= krPPPv

temperature at the end of the compression is

⎞⎛k

v

and the maximum pressure is

x kPa 4829==== kPa) 3219)(5.1(23 PrPP p

The

( ) K 0.84214K) 293( 14.111

112 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −krTTT

v

v1

2== −

−k

and K 1263kPa 3219k 4829K) 0.842(3

2 ⎜⎛=⎟⎟

⎞⎜⎜⎛

=P

TTxPa

2=⎟

⎠⎞

⎝⎠⎝ P

rom the definition of cutoff ratio

==== vvv cxc rr

F

/kgm 09010.0)/kgm 07508.0)(2.1( 332 3

The remaining state temperatures are then

K 1516=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0.075080.09010K) 1263(3

3x

xTTv

v

K 5.5671.051

0.09010K) 1516(14.11

434 ⎜⎜⎝

= TTv

3 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎛ −−kv

pplying t e first law and work expression to the heat addition processes gives

The heat rejected is

A h

kJ/kg 556.5=−⋅+−⋅=

−+−=

K)12631516)(KkJ/kg 005.1(K)0.8421263)(KkJ/kg 718.0(

)()( 32in xpx TTcTTcq v

kJ/kg 1.197K)2935.567)(KkJ/kg 718.0()( 14out =−⋅=−= TTcq v

0.646=−=−=kJ/kg 556.5kJ/kg 197.1

11in

outth q

qη Then,


9-30

9-49 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined.



Analysis The specific volume of the air at the start of the compression is

/kgm 9076.0kPa 801

1 ===P

vK) 253)(K/kgmkPa 287.0( 3

31 ⋅⋅RT

olume at the end of the compression is

v

P

4

1

2

3

qout

x

qinand the specific v

/kgm 06483.014

12 ===

/kgm 9076.0 33

rv

he pressu at the end of the compression is

v

T re

( )


kPa 321914kPa) 80( 4.11

2

112 ===⎟⎟

⎠⎜⎜⎝

= krPPPv

temperature at the end of the compression is

⎞⎛k

v

and the maximum pressure is

x kPa 4829==== kPa) 3219)(5.1(23 PrPP p

The

( ) K 1.72714K) 253( 14.111

112 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −krTTT

v

v1

2== −

−k

and K 1091kPa 3219kP 4829K) 1.727(3

2 ⎜⎛=⎟⎟

⎞⎜⎜⎛

=P

TTxa

2=⎟

⎠⎞

⎝⎠⎝ P

rom the definition of cutoff ratio

==== vvv cxc rr

F

/kgm 07780.0)/kgm 06483.0)(2.1( 332 3

The remaining state temperatures are then

K 1309=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0.064830.07780K) 1091(3

3x

xTTv

v

K 0.4900.9076

0.07780K) 1309(14.11

434 ⎜⎜⎝

= TTv

3 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎛ −−kv

pplying t e first law and work expression to the heat addition processes gives


A h

kJ/kg 480.4=−⋅+−⋅=

−+−=

K)10911309)(KkJ/kg 005.1(K)1.7271091)(KkJ/kg 718.0(

)()( 32in xpx TTcTTcq v


0.646=−=−=kJ/kg 480.4kJ/kg 170.2

11in

outth q

qη Then,


9-31

9-50 An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined.




v

P

4

1

2 3 qin

qout

1700 KAnalysis (a) Process 1-2: isentropic compression.

2.621kJ/kg 07.214

K 3001

11 =

=⎯→⎯=

r

uT

v

( )kJ/kg 83.934

K 7.90113.34 2.621

2.181112

====r rrr vv

vv

11

2

22

==

⎯→⎯hTv

rocess 2- P = constant heat addition. P 3:

1.885===⎯→⎯=K 901.7K 1700

2

3

2

3

2

22

3

33

TT

TP

TP

v

vvv

(b) K 001733

=⎯→⎯=T

rv

kJ/kg 27.94583.9341.1880

761.4kJ/kg 1.1880

23

3

=−=−=

=

hhq

h

in


( ) kJ/kg 91.60297.45761.4885.1

2.18885.1885.1 4

2

4

3

43334

=⎯→⎯===== urrrrr vv

v

vv

v

vv

Process 4-1: v = constant heat rejection.

(c) 58.9%

kJ/kg 388.84

==−=−=

=−=−=

589.0kJ/kg 945.27kJ/kg 388.84

11

07.21491.602

in

outth

14out

qq

uuq

η


9-32

9-51 An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K, and k = 1.4 (Table A-2).



( )( ) K 5.95718.2K 3002

12 =⎟⎟⎠

⎜⎜⎝

= TTv

0.41

1 =⎞⎛

−kv

rocess 2-3: P = constant heat addition. P

1.775===⎯→⎯=K 957.5K 1700

2

3

2

3

2

22

3

33

T TT

TPP

v

vvv

(b) ( ) ( )( ) kJ/kg 2.746K5.9571700kJ/kg.K 1.0052323in =−=−=−= TTchhq p

rocess 3-4: isentropic expansion.

v

P

4

1

2 3qin

P

( ) K 1.67018.21.775K 1700

775.1 23

334 ⎟⎟⎜⎜

⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

vTTT

v

v

v

0.41

4

1

4=⎟

⎠⎞

⎜⎝⎛=

⎠

⎞−− kk

v = constant heat rejection.

(c)

Process 4-1:

( )( )( )

64.4%

kJ/kg 265.7

==−=−=

=−=−=−=

644.0kJ/kg 746.2kJ/kg 265.7

11

K300670.1kJ/kg.K 0.718

in

outth

1414out

qq

TTcuuq

η

v


9-33

9-52 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined.




v

P

4

1

2 3 qin

qout

( )( ) K 971.120K 293 0.41

2

112 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−k

TTV

V

Process 2-3: P = constant heat addition.

2.265K971.1K2200

2

3

223 TT V3223 ===⎯→⎯=

TTPP VVV


3

( )

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

63.5%===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−−

kJ/kg 1235kJ/kg 784.4

kJ/kg 784.46.4501235

kJ/kg 450.6K293920.6KkJ/kg 0.718

kJ/kg 1235K971.12200KkJ/kg 1.005

K 920.620

2.265K 2200265.2265.2

in

outnet,th

outinoutnet,

1414out

2323in

0.41

3

1

4

23

1

4

334

qw

qqw

TTcuuq

TTchhq

rTTTT

p

kkk

η

v

V

V

V

V

( )( )

( ) ( )( )kPa933

kJmkPa

1/201/kgm 0.885kJ/kg 784.4

/11MEP

/kgm 0.885kPa 95

K 293K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

==⋅⋅

==

rww

r

PRT

vvv

vvv

vv (b)



9-34

9-53 A diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure 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 room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression.

v

P

4

1

2 3 qin

qout

Polytropic ( )( ) K 971.120K 293 0.41

2

112 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−k

TTV

V


2.265K 971.1K 2200

2

3

223=⎯→⎯=

TT V


3223 ==TTPP VVV

Process 3-4: polytropic expansion.

3

( )

( ) ( )( )( ) ( )( ) kJ/kg 526.3K 2931026KkJ/kg 0.7181414out =−⋅=−=−= TTcuuq v

te that q

kJ/kg 1235K 971.12200KkJ/kg 1.005

K 102620

2.265K 2200r

2.2652.265

2323in

0.351n

3

1n

4

23

1

4

334

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−−

TTchhq

TTTT

p

n

V

V

V

V

t some heat is also rejected during the polytropic rocess, w ch is determined from an energy balance on process 3-4:

No oup hi

in this case does not represent the entire heat rejected since

( ) ( )( )

( )( )( )

kJ/kg 120.1K 22001026KkJ/kg 0.718kJ/kg

34out34,in34,34out34,in34,

system

=−⋅+

−+=⎯→⎯−=− TTcwquuwq v

h means that 120.1 kJ/kg of heat is transferred to the combustion gases during the expansion process. This is unrealistic since the gas is at a much higher temperature than the surroundings, and a hot gas loses heat during polytropic

pansion. The cause of this unrealistic result is the constant specific heat assumption. If we were to use u data from the ir table, we would obtain

kJ/kg 9631.351

K 22001026KkJ/kg 0.2871

outin

34out34,

∆=−

=−

−⋅=

−−

=

EEEnTTR

w

963=

whic

exa

( ) kJ/kg 1.128)4.18723.781(96334out34,in34, −=−+=−+= uuwq

whic hh is a eat loss as expected. Then qout becomes kJ/kg 654.43.5261.128out41,out34,out =+=+= qqq

and

47.0%===

=−=−=

kJ/kg 1235kJ/kg 580.6

kJ/kg 580.64.6541235

in

outnet,th

outinoutnet,

qw

qqw

η

( )( )

( ) ( )( )kPa 691=⎟

⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

==⋅⋅

==

kJmkPa 1

1/201/kgm 0.885kJ/kg 580.6

/11MEP

/kgm 0.885kPa 95

K 293K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv (b)


9-35

9-54 Problem 9-53 is reconsidered. The effect of the compression ratio on the net work output, mean effective pressure, and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted.


Procedure QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) q_in_total = 0 q_out_total = 0 IF (q_12 > 0) THEN q_in_total = q_12 ELSE q_out_total = - q_12 If q_23 > 0 then q_in_total = q_in_total + q_23 else q_out_total = q_out_total - q_23 If q_34 > 0 then q_in_total = q_in_total + q_34 else q_out_total = q_out_total - q_34 If q_41 > 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41 END "Input Data" T[1]=293 [K] P[1]=95 [kPa] T[3] = 2200 [K] n=1.35 {r_comp = 20} "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_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant pressure heat addition" P[3]=P[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_23 w_23 =P[2]*(V[3] - V[2])"constant pressure process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is polytropic expansion" P[3]/P[4] =(V[4]/V[3])^n 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_34 "q_34 is not 0 for the ploytropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) P[3]*V[3]^n = Const w_34=(P[4]*V[4]-P[3]*V[3])/(1-n) "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_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) w_net = w_12+w_23+w_34+w_41



9-36

Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent" "The mean effective pressure is:" MEP = w_net/(V[1]-V[2])

rcomp ηth MEP [kPa]

wnet [kJ/kg]

14 47.69 970.8 797.9 16 50.14 985 817.4 18 52.16 992.6 829.8 20 53.85 995.4 837.0 22 55.29 994.9 840.6 24 56.54 992 841.5

4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5200400600800

10001200140016001800200022002400

s [kJ/kg-K]

T [K

]

95

kPa

340

.1 kP

a

592

0 kP

a

0.0

44

0.1

0.8

8 m

3/kg

Air

2

1

3

4

10-2 10-1 100 101 102101

102

103

104

101

102

103

104

v [m3/kg]

P [k

Pa] 293 K

1049 K

2200 K

5.69 6.74 kJ/kg-K

Air



9-37

14 16 18 20 22 24790

800

810

820

830

840

850

rcomp

wne

t [k

J/kg

]

14 16 18 20 22 2447

49

51

53

55

57

rcomp

ηth

14 16 18 20 22 24970

975

980

985

990

995

1000

rcomp

MEP

[kP

a]



9-38

9-55 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.

Assumptions 1 The cold air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.


Analysis Process 1-2: isentropic compression.

v

P

4

1

2 3 Qin

Qout

( )( ) K 118122K 343 0.41

2

112 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−k

TTV

V


( )( ) K 2126K 11811.88.1 222

33

23 TT223 ====⎯→⎯= TTT

PPv

vvv


3

( )

( ) ( )

( ) ( )( )( )( )

( )( ) kW 48.0=== kJ/rev 1.251rev/s 2300/60outnet,outnet, WnW &&

Discussion No

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⋅⋅

==

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−−

kJ/rev 251.16198.0871.1

kJ 6198.0K343781KkJ/kg 0.718kg 001971.0

kJ .8711K)11812216)(KkJ/kg 1.005)(kg 001971.0(

kg 001971.0)K 343)(K/kgmkPa 0.287(

)m 0.0020)(kPa 97(

K 78122

8.1K 22162.22.2

outinoutnet,

1414out

2323in

3

3

1

11

4.01

3

1

4

23

1

4

334

QQW

TTmcuumQ

TTmchhmQ

RTP

m

rTTTT

v

p

kkk

V

V

V

V

V

te that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).



9-39

9-56 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.

Assumptions 1 The air-standard assumptions are applicable with nitrogen as the working fluid. 2 Kinetic and potential energy changes are negligible. 3 Nitrogen is an ideal gas with constant specific heats.

Properties The properties of nitrogen at room temperature are cp = 1.039 kJ/kg·K, cv = 0.743 kJ/kg·K, R = 0.2968 kJ/kg·K, and k = 1.4 (Table A-2).

Analysis Process 1-2: isentropic compression.

v

P

4

1

2 3 Qin

Qout


( )( )22K 3432

12 =⎟⎟⎠

⎜⎜⎝

= TTV

K 11810.41

1 =⎞⎛

−kV


( )( ) K 2126K 11811.88.1 222

33

2

22

3

33 =⎯→⎯= TTT

=== TTPP

v

vvv

rocess 3- isentropic expansion.

P 4:

( )

( ) ( )

( ) ( )( )( )( )

( )( ) kW 47.9===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⋅⋅

==

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−− 4.0111

Tkkk

kJ/rev 1.251rev/s 2300/60

kJ/rev 251.16202.0871.1

kJ 6202.0K343781KkJ/kg 0.743kg 001906.0

kJ .8711K)11812216)(KkJ/kg 1.039)(kg 001906.0(

kg 001906.0)K 343)(K/kgmkPa 0.2968(

)m 0.0020)(kPa 97(

K 78122

8.1K 22162.22.2

outnet,outnet,

outinoutnet,

1414out

2323in

3

3

1

11

34

23

4

334

WnW

QQW

TTmcuumQ

TTmchhmQ

RTP

m

rTTT

v

p

&&

V

V

V

V

V

Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).


9-40

9-57 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 1.1. The power produced by the cycle is to be determined.



Analysis We begin by fixing the temperatures at all states.

( ) K 7.92418K) 291( 14.111

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

−k

k

rTTTv

v

v

P

4

1

2

3

qout

x

qin

( ) kPa 514818kPa) 90( 4.11

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= k

k

rPPPv

v

x kPa 5663kPa) 5148)(1.1(23 ==== PrPP p

K 1017kPa 5148kPa 5663K) 7.924(

22 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PP

TT x

xc

x

K 1119K) 1017)(1.1( === TrT 3

K 8.365181.1K) 1119(

14.11

3

1

4

33 4 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−− kc

k

rr

TTTv

v

Applying the first law to each of the processes gives

kJ/kg 0.455K)2917.924)(KkJ/kg 718.0()(


1221 =−⋅=−= TTcw − v

kJ/kg 5.102K)10171119)(KkJ/kg 005.1()( 33 =−⋅=−=− xpx TTcq

)( 333 −−= −− xxx TTcqw v kJ/kg 26.29K)10171119)(KkJ/kg 718.0(5.102 =−⋅−=

kJ/kg 8.540K)8.3651119)(KkJ/kg 718.0()( 434 3 =−⋅=−=− TTcw v

The net work of the cycle is

kJ/kg 1.1150.45526.298.54021343net =−+=−+= −−− wwww x

The mass in the device is given by

kg 003233.0K) 291)(K/kgmkPa 287.0(

)m kPa)(0.003 90(3

3

1

11 =⋅⋅

==RTP

mV

The net power produced by this engine is then

kW 24.8=== cycle/s) 0/60kJ/kg)(400 115.1kg/cycle)( 003233.0(netnet nmwW &&


9-41

9-58 A dual cycle with non-isentropic compression and expansion processes is considered. The power produced by the cycle is to be determined.



Analysis We begin by fixing the temperatures at all states.


( ) K 7.92418K) 291( 14.111

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

−k

k

s rTTTv

v

K 10370.85

K )291(924.7K) 291(1212

12

12 =−

+=−

+=⎯→⎯−−

=η

ηTT

TTTTTT ss

v

P

4

1

2

3

qout

x

qin

( ) kPa 514818kPa) 90( 4.11

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= k

k

rPPPv

v

kPa) 5148)(1.1(23 ==== PrPP px 5663 kPa

K 1141kPa 5148kPa 5663K) 1037(

22 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PP

TT xx

K 1255K) 1141)(1.1(3 === xcTrT

K 3.410181.1K) 1255(

14.11

3

1

4

334 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−− kc

k

s rr

TTTv

v

K 8.494K )3.410(1255)90.0(K) 1255()( 433443

43 =−−=−−=⎯→⎯−−

= ss

TTTTTTTT

ηη


kJ/kg 6.535K)2911037)(KkJ/kg 718.0()( 122 1 =−⋅=−= TTcw v

5 =−⋅

−

00.1()( 33 =−=− xpx TTcq kJ/kg 6.114K)11411255)(KkJ/kg

kJ/kg 75.32K)11411255)(KkJ/kg 718.0(6.114)( 333 =− ⋅−=−−= −− xxx TTcqw v

kJ/kg 8.545K)8.4941255)(KkJ/kg 718.0()( 4343 =−⋅=−=− TTcw v


kJ/kg 95.426.53575.328.54521343 net =−+=−+= −−− wwww x

The mass in the device is given by

kg 003233.0K) 291)(K/kgmkPa 287.0(

)m kPa)(0.003 90(3

3

1

11 =⋅⋅

==RTP

mV

The net power produced by this engine is then

kW 9.26=== cycle/s) 0/60kJ/kg)(400 42.95kg/cycle)( 003233.0(netnet nmwW &&


9-42

9-59 An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, and the thermal efficiency are to be determined.


Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K (Table A-1), cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis Working around the cycle, the germane properties at the various states are

( ) K 4.87715K) 297( 14.111

1

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

−k

k

rTTTv

v

v

P

4

1

2

3

qout

x

qin( ) kPa 434315kPa) 98( 4.11

2

112 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= k

k

rPPPv

v

23 === PrP px kPa 4777kPa) 4343)(1.1( =P

K 1.965kPa 4343kPa 4777K) 4.877(

22 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PP

TT xx

K 1351K)(1.4) 1.965(33 ===⎟⎟

⎠⎜⎜⎝

= cxx

x rTTTv

⎞⎛ v

K 2.523151.4K) 1351(

14.11

3

1

4

33 4 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−− kc

k

rr

TTTv

v


kJ/kg 7.416K)2974.877)(KkJ/kg 718.0()( 1221 =−⋅=−=− TTcw v

kJ/kg 0.63K)4.8771.965)(KkJ/kg 718.0()(


22 =−⋅=−= TTcq

33

− xx v

)( kJ/kg 8.387K)1.9651351)(KkJ/kg 005.1( =−⋅=xp −= TTcq −x

( 333 kJ/kg 8.110K)1.9651351)(KkJ/kg 718.0(kJ/kg 8.387) =−⋅−=

31351)(KkJ/kg 718.0()( 4343

−−= −− xxx TTcqw v

52 kJ/kg 4.594K)2. =−⋅=−=− TTcw v


kJ/kg 289=−+=−+= −−− 7.4168.1104.59421343 net wwww x

and the net heat addition is

Hence, the thermal efficiency is

kJ/kg 451=+=+= −− 8.3870.6332in xx qqq

0.641===kJ/kg 451kJ/kg 289

in

netth q

wη


9-43

9-60 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined.



Analysis (a) Process 1-2: Isentropic compression

1

Qin

2 3

4

Qout

( )( )

( )( ) kPa 504419kPa 95

K 1.95019K 340

1.349

2

112

1-1.3491

2

112

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

k

k

PP

TT

v

v

v

v

The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are

3

3

m 0001778.0=cV

m 0045.019

+=⎯→⎯

+=

c

c

c

dcrV

V

V

VV

m 003378.00032.00001778.0 =+=+= dc VVV 31

The total mass contained in the cylinder is

( )( )kg 0.003288

K 340K/kgmkPa 0.287)m 378kPa)(0.003 95(

3

3

1

11 =⋅⋅

==RTP

mV

The mass of fuel burned during one cycle is

kg 0001134.0kg) 003288.0(

28 =⎯→⎯−

=⎯→⎯−

== ff

f

f

f

f

a mm

mm

mmmm

AF

Process 2-3: constant pressure heat addition


kJ 723.48)kJ/kg)(0.9 kg)(42,500 0001134.0(HVin ==Q = cf qm η

003288.0(kJ 723.4)( TTTTmcv

The cutoff ratio is

kg)(0.823 Q K 2244=⎯→⎯−=⎯→⎯−= 3323in K)1.950(kJ/kg.K)

2.362===K 950.1K 2244

2

3

TT

β

(b) 33

12 m 0001778.0

19m 0.003378

===r

VV

23

14

3323 m 0004199.0)m 0001778.0)(362.2(

PP ==

===VV

VV β


9-44


( )

( ) kPa 9.302m 0.003378m 0.0004199kPa 5044

K 1084m 0.003378m 0.0004199K 2244

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 013.2K3401084KkJ/kg 0.823kg) 003288.0(14out =−⋅=−= TTmcQ v

The net work output and the thermal efficiency are

kJ 2.710=−=−= 013.2723.4outinoutnet, QQW

57.4%==== 5737.0kJ 4.723in

th Qη kJ 2.710outnet,W

(c) The mean effective pressure is determined to be

kPa 847=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

00017.0003378.0(MEP

21 VV kJmkPa

m)78kJ 2.710 3

3outnet,W

) The power for engine speed of 1750 rpm is (d

kW 39.5=⎟⎠⎞⎛ min 1(rev/min) 1750n&


⎜⎝ s 60rev/cycle)

ote that there are two revolutions in one cycle in four-stroke engines

(e) Finally, the specific fuel consumption is

==2(

kJ/cycle) (2.7102netnet WW&

N .

g/kWh 151=⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛==

kWh 1kJ 3600

kg 1g 1000

kJ/kg 2.710kg 0001134.0sfc

netWm f


9-45

9-61 An expression for cutoff ratio of an ideal diesel cycle is to be developed.


Analysis Employing the isentropic process equations,


−= krTT

1 TrTrrcTTcq kk −− −=−=

When this is solved for cutoff ratio, the result is

v

P

4

1

2 3 qin

qout

21

1

while the ideal gas law gives

11

23 TrrrTT kcc

−==

When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is

)()( 11

123in cpp

11

in1Trc

qr kp

c −+=


9-46

9-62 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4 (Table A-2)

Analysis The thermal efficiency of a dual cycle may be expressed as

)()( 32in

thxpx TTcTTcq −+−v

By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as wellas the ideal gas equation of state, the tem

)(11 14out TTcq −

−=−= vη

peratures may be eliminated from the thermal efficiency expression. This yields

the result

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+−

−−=

− 1)1(111

1thpcp

kcp

k rrkrrr

rη

v

P

4

1

2

3

qout

x

qinwhere


2PpP

r x= and x

cr v

v 3=

hen rc = , we obtain

W rp

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛1 r

−+−

−−=

+

− 1)(

11

2

1

1thppp

kp

k rrrkrη

For the case r = 20 and rp = 2,

0.660=⎟⎟⎠

⎞⎜⎜⎝

⎛

−+−

−−=

+

− 12)22(4.112

2011

2

14.1

14.1thη


9-47

9-63 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained.


Analysis The thermal efficiency of a dual cycle may be expressed as

)()( 32in

thxpx TTcTTcq −+−v

By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as wellas the ideal gas equation of state, the tem

)(11 14out TTcq −

−=−= vη

peratures may be eliminated from the thermal efficiency expression. This yields

the result

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−+−

−−=

− 1)1(111

1thpcp

kcp

k rrkrrr

rη

v

P

4

1

2

3

qout

x

qinwhere


2P

rp and Px=

xcr v

v 3=

hen rc = , we obtain

W rp

⎟⎟

⎠

⎞

⎜⎜

⎝ −+−

−−=

− 1)(

11

21thppp

pk rrrkr

η

Rearrangement of this result gives

⎛ +1 1kr

1th2

1

)1(1)(

1 −+

−=−+−

− k

ppp

kp r

rrrk

rη


9-48

9-64 The five processes of the dual cycle is described. The P-v and T-s diagrams for this cycle is to be sketched. An expression for the cycle thermal efficiency is to be obtained and the limit of the efficiency is to be evaluated for certain cases.


Analysis (a) The P-v and T-s diagrams for this cycle are as shown.

s

T

52

1

4

3

v

P

5

1

2

4

qout

3

qin

(b) Apply first law to the closed system for processes 2-3, 3-4, and 5-1 to show:


3 2 4 3

5 1

v p

out v

q C T T C T T= − + −

The cycle thermal efficiency is given by

( ) ( )in

( )q C T T= −

( )( ) ( )

( )( ) ( )

( )( ) ( )

5 1 1 5 1

3 2 4 3 2 3 2 3 4 3

5 1

34 3

1 1

/ 11 1 1

/ 1 / 1

/ 11

/ 1

voutth

in v p

th

C T T T T Tqq C T T C T T T T T kT T T

T TTT k T TT

η

η

− −= − = − = −

− + − − + −

−= −

−

23 2/ 1T T

T− +

Process 1-2 is isentropic; therefore, 1

12 1

1 2

kkT V r

T V

−

−⎛ ⎞= =⎜ ⎟

⎝ ⎠

Process 2-3 is constant volume; therefore,

3 3 3 3T PV P r= = =

Process 3-4 is constant pressure; therefore,

2 2 2 2pT PV P

3 34 4 4 4

4 3 3 3T T T V cPVPV T V r= ⇒ = =

Process 4-5 is isentropic; therefore, 1 1

5 34 4

k k

c cT r VV V− − 1 1 1

2

4 5 1 1 1

k k kcr V r

T V V V V r

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

Process 5-1 is constant volume; however T5/T1 is found from the following. 1

5 5 34 2 1c

kc k k

c p pr

T T T T T rr r r r r

−−⎛ ⎞= = =⎜ ⎟

T T TT T

1 4 3 2 1 ⎝ ⎠


9-49


The ratio T3/T1 is found from the following.

3 3 2

1 2 1

1kp

T T TT T T

r r −= =

The efficiency becomes

( ) ( )1 11 1th kp c

kpr r k rr r− −

11

kc pr r

η−

= −− + −

(c) In the limit as rp approaches unity, the cycle thermal efficiency becomes

( ) ( )

( )

11 1

1

1

1

1lim 1 lim

1 1

11lim 11

p p

p

kc p

th kr rp c

kc

th th Dieselrc

kp

k

r rr r k r

rk r

r r

r

η

η η

−→ →

→

−

−

⎧ ⎫−⎪ ⎪= − ⎨ ⎬− + −⎪ ⎪⎩ ⎭

⎡ ⎤−⎢ ⎥= − =−⎢ ⎥⎣ ⎦

(d) In the limit as rc approaches unity, the cycle thermal efficiency becomes

( ) ( ) ( )1 11 1

1

1

1

1 1lim 1 lim 1

1 1

1lim 1

c p

c

kc p p

th k kr rp c

th th Ottor

kp

k

r r rr r k r r rr r

r

η

η η

− −→ →

→

−

−

⎧ ⎫− −⎪ ⎪= − = −⎨ ⎬− + − −⎪ ⎪⎩ ⎭

= − =

1p

⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭


9-50

Stirling and Ericsson Cycles

9-65C The Stirling cycle.

9-66C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.

9-67C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.

9-68C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.

9-69 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximum pressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined.

Assumptions Air is an ideal gas.

Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).

Analysis (a) The entropy change during process 3-4 is

s

T

3

2qin

qout

4

1 1200 K

300 K

KkJ/kg 0.5K 300

kJ/kg 150

0

out34,34 ⋅−=−=−=−

Tq

ss

and


( ) KkJ/kg 0.5kPa 120

PlnKkJ/kg

ln

4

3

40

4

⋅−=⋅

−PPRT

It yields P4 = 685.2 kPa

sible cycles,

ln3

34 =−T

css p

0.287−=

(b) For rever

( ) kJ/kg 600kJ/kg 150K 300in TTq LH

K 1200outin

t ===⎯→⎯= qT

qTq HL

Thus,

(c) The thermal efficiency of this totally reversible cycle is determined from

ou

kJ/kg 450=−=−= 150600outinoutnet, qqw

75.0%=−=−=K1200K300

11thH

L

TT

η


9-51

9-70 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The net work produced per cycle and the thermal efficiency of the cycle are to be determined.



Analysis Since the specific volume is constant during process 2-3,

s

T

3

2qin

qout

4

1 800 K

300 K

kPa 7.266K 3003

32⎠⎝T

K 800kPa) 100(2 =⎟⎞

⎜⎛==

TPP

Heat is only added to the system during reversible process 1-2. Then,

kJ/kg .6462)KkJ/kg 0.5782)( K800()( 121in =⋅=−= ssTq

KkJ/kg 5782.0kPa 2000kPa 266.7ln)KkJ/kg 0.287(0

lnln1

20

1

212

⋅=

⎟⎠⎞

⎜⎝⎛⋅−=

−=−PP

RTT

css p

he therm reversible cycle is determined from T al efficiency of this totally

0.625=−=−=K 800K 300

11thH

L

TT

η

Then,

kJ 289.1=== kJ/kg) kg)(462.6 (0.625)(1inthnet mqW η



9-52

9-71 An ideal Stirling engine with air as the working fluid operates between the specified temperature and pressure limits. The power produced and the rate of heat input are to be determined.



Analysis Since the specific volume is constant during process 2-3,

s

T

3

2qin

qout

4

1 800 K

300 K

kPa 7.266K 3003

32⎠⎝T

K 800kPa) 100(2 =⎟⎞

⎜⎛==

TPP

Heat is only added to the system during reversible process 1-2. Then,

kJ/kg .6462)KkJ/kg 0.5782)( K800()( 121in =⋅=−= ssTq

KkJ/kg 5782.0kPa 2000kPa 266.7ln)KkJ/kg 0.287(0

lnln1

20

1

212

⋅=

⎟⎠⎞

⎜⎝⎛⋅−=

−=−PP

RTT

css p

he therm reversible cycle is determined from T al efficiency of this totally

0.625K 800K 300

11th =−=−=H

L

TT

η

Then,

kJ 289.1kJ/kg) kg)(462.6 (0.625)(1inth net === mqW η

e is

hile the power produced by the engine is

The rate at which heat is added to this engin

kW 10,020=== cycle/s) 0/60kJ/kg)(130 462.6kg/cycle)( 1(inin nmqQ &&

w

kW 6264=== cycle/s) 1300/60kJ/cycle)( 1.289()netnet nWW &&



9-53

9-72 An ideal Ericsson cycle operates between the specified temperature limits. The rate of heat addition is to be determined.

Analysis The thermal efficiency of this totally reversible cycle is determined from

s

T

3

2qin

qout

4

1 900 K

280 K

0.6889K 900th

HT

According t

K 28011 =−=−= LT

η

o the general definition of the thermal efficiency, the rate of heat

addition is

kW 726===0.6889

kW 500

th

netin η

WQ

&&

9-73 An ideal Ericsson cycle operates between the specified temperature limits. The power produced by the cycle is to be

hen the cycle is repeated 2000 times er minute. Then the work per cycle is

determined.

Analysis The power output is 500 kW w

s

T

3

2qin

qout

4

1 900 K

280 K

p

kJ/cycle 15cycle/s (2000/60)

kJ/s 500netW&net ===

nW

&

hen the cycle is repeated 3000 times per minute, the power output will be

W

kW 750=== kJ/cycle) 5cycle/s)(1 000/603(netnet WnW &&



9-54

9-74 An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat added to the cycle and the net work produced by the cycle are to be determined.


Properties The properties of air at room temperature are R = 0.287 kPa·m3/kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2a).

Analysis Applying the ideal gas equation to the isothermal process 3-4 gives

s

T

3

2qin

qout

4

1

298 K

kPa 600kPa)(12) 50(4

34 v3 ===

vPP

Since process 4-1 is one of constant volume,

K 1788kPa 6004

41⎠⎝⎟

⎠⎜⎝ P

kPa 3600K) 298(1 =⎟⎞

⎜⎛=⎟

⎞⎜⎛

=P

TT

dapting t work integral to the heat addition process gives A he first law and

kJ/kg 1275=⋅=== − K)ln(12) 1788)(KkJ/kg 0.287(ln1

2121in v

vRTwq

Similarly,

kJ/kg 212.5=⎟⎠⎞

⎜⎝⎛⋅== = wq − 12

1K)ln 298)(KkJ/kg 0.287(ln3

4343t v

vRT

he net work is then

Properties The properties of air at room temperature are R = 0.287 kPa·m /kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, nd k = 1.4 (Table A-2a).

Analysis Applying the ideal gas equation to the isothermal process 3-4 gives

ou

T

kJ/kg 1063=−=−= 5.2121275outinnet qqw

9-75 An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat transfer in the regenerator is to be determined.

Assumptions Air is an ideal gas with constant specific heats. 3

a

kPa 600kPa)(12) 50(4

s

T

3

2qin

qout

4

1

298 K

3v3 ===

vPP 4

Since process 4-1 is one of constant volume,

K 1788kPa 600kPa 3600K) 298(

4

141 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PP

TT

Application of the first law to process 4-1 gives

kJ/kg 1070=−⋅=−= K)2981788)(KkJ/kg 718.0()( 41regen TTcq v



9-55

Ideal and Actual Gas-Turbine (Brayton) Cycles

9-76C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropic expansion (in a turbine), and (4) P = constant heat rejection.

9-77C For fixed maximum and minimum temperatures, (a) the thermal efficiency increases with pressure ratio, (b) the net work first increases with pressure ratio, reaches a maximum, and then decreases.

9-78C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It is usually between 0.40 and 0.6 for gas turbine engines.

9-79C In gas turbine engines a gas is compressed, and thus the compression work requirements are very large since the steady-flow work is proportional to the specific volume.

9-80C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.



9-56

9-81 A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.


Analysis (a) Noting that process 1-2 is isentropic,


s

T

1

2 4

3qin

qout

1100 K

290 K

2311.1kJ/kg 16.290

K 0291

11 =

=⎯→⎯=

rPh

T

( )( )kJ/kg 06.561

311.122311.1102

2

112

=== PP

P rr2

==

⎯→⎯hTP K 555.6

roces 3-4 is isentropic, and thus

(b) P s

( )

kJ/kg 62.54945.61107.1161

kJ/kg 9.27016.29006.561

kJ/kg 45.61171.161.167101

1.167kJ/kg 07.1161

K 0011

43outT,

12inC,

43

4

3 =h3

34

3

=−=−=

=−=−=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

=⎯→⎯=

hhw

hhw

hPPP

P

PT

rr

r

Then the back-work ratio becomes

49.3%==== 493.0kJ/kg 549.62kJ/kg 270.9

outT,

inC,bw w

wr

(c)

46.5%====

=−=−=

=−=−=

465.0kJ/kg 600.01kJ/kg 278.72

kJ/kg 72.2789.27062.549

kJ/kg 01.60006.56107.1161

in

outnet,th

inC,outT,outnet,

23in

qw

www

hhq

η


9-57

9-82 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.

s

T

1

2s

4s

3qin

qout

1240 K

295 K

2

4



Analysis (a) Noting that process 1-2s is isentropic,

3068.1kJ/kg .17295

K 2951

11 =

=⎯→⎯=

rPh

T

( )( )

( )

( )( )( )

kJ/kg 83.04707.70293.132487.093.1324

K 6.689 and kJ/kg .0770223.273.272101

3.272kJ/kg 1324.93

K 1240

kJ/kg .6062683.0

17.29526.57017.295

K 564.9 and kJ/kg 570.2607.133068.110

433443

43

443

4

33

1212

12

12

221

2

34

3

12

=−−=

−−=⎯→⎯−−

=

==⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==

⎯→⎯=

=−

+=

−+=⎯→⎯

−−

=

==⎯→⎯===

sTs

T

ssrr

r

C

ssC

ssrr

hhhhhhhh

ThPPP

P

Ph

T

hhhh

hhhh

ThPPP

P

ηη

ηη

Thus,

T = 764.4 K

(c)

4

(b)

kJ/kg 210.4=−=−=

=−=−=

=−=−=

9.4873.698

kJ/kg 487.917.29504.783

kJ/kg .369860.62693.1324

outinoutnet,

14out

23in

qqw

hhq

hhq

30.1%==== 3013.0kJ/kg 698.3kJ/kg 210.4

in

outnet,th q

wη



9-58

9-83 Problem 9-82 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, and the isentropic efficiencies of the turbine and compressor are to be varied and a general solution for the problem by taking advantage of the

be developed.

nalysis Using EES, the problem is solved as follows:

m diagram window"

]

P[1])

compressor, assuming: adiabatic, ke=pe=0"

_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0"

ot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"

ycle work, kW" fficiency"

determined only to produce a T-s plot"

[4]=entropy(air,T=T[4],P=P[4])

diagram window method for supplying data to EES is to

A

"Input data - fro{P_ratio = 10} {T[1] = 295 [K] P[1]= 100 [kPa] T[3] = 1240 [K] m_dot = 20 [kg/s

ta_c = 83/100 EEta_t = 87/100} "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P= "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"

sure ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of presT_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])

mpressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Com_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual

lysis" "External heat exchanger anaP[3]=P[2]"process 2-3 is SSSF constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot "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,s=s_s[4],P=P[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_d "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition oEta=W_dot_net/Q_dot_in"Cycle thermal e

f the net c

Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points areT[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s



9-59

Bwr η Pratio Wc

[kW] Wnet [kW]

Wt [kW]

Qin [kW]

0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700

0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241

2 4 6 8 10 12 14 16 18 200.12

0.16

0.2

0.24

0.28

0.32

0.36

2250

2700

3150

3600

4050

4500

Pratio

η

Wne

t [k

W]

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50

500

1000

1500

s [kJ/kg-K]

T [K

]

100 kPa

1000 kPa

Air

1

2

3

42s 4s



9-60

9-84 A simple Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

Analysis (a) Using the compressor and turbine efficiency relations, ( )

( )( )

( )( )

( )( )

( )( ) ( )

( )( )K 720=

−−=−−=⎯→⎯

−

−=

−−

=

=−

+=

−+=⎯→⎯

−

−=

−−

=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

3.642124087.01240

K 625.883.0

2956.569295

K 642.3101K 1240

K 6.56910K 295

433443

43

43

43

1212

12

12

12

12

0.4/1.4/1

3

434

0.4/1.4/1

1

212

sTsp

p

sT

C

s

p

spsC

kk

s

kk

s

TTTTTTcTTc

hhhh

TTTT

TTcTTc

hhhh

PP

TT

PP

TT

ηη

ηη

s

T

1

2s

4s

3qin

qout

1240 K

295 K

2

4

(b) ( ) ( )( )( ) ( )( )

kJ/kg 190.2=−=−=

=−⋅=−=−=

=−⋅=−=−=

1.4273.617

kJ/kg 427.1K295


720KkJ/kg 1.005

kJ/kg 17.36K625.81240KkJ/kg 1.005

outinoutnet,

1414out

2323

qqw

TTchhq

TTchhq

p

p

(c)

in

30.8%==== 3081.0kJ/kg 617.3kJ/kg 190.2

in

outnet,th q

wη


9-61

9-85 A simple ideal Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats.

Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2a).

Analysis Using the isentropic relations for an ideal gas,

s

T

1

2

4

3 qin

qout

973 K

288 K

K 9.827K)(14) 288( 70.667/1.66/)1(1

/)1(

1

212T ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

−kk

p

kk

rTPP

T

Similarly,

K 5.33814

K) 973(33

34 =⎟⎠

⎜⎝

=⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=pr

TP

TT 11 70.667/1.66/)1(/)1(4 ⎞⎛⎟

⎞⎜⎛⎞⎛

−− kkkkP

t law to the constant-pressure heat addition process 2-3 produces

p

Similarly,

14

Applying he first

q kJ/kg 4.753K)9.827973)(kJ/kg·K .19265()( 23in =−=−= TTc

1926.5()(


kJ/kg 2.262K)2885.338)(kJ/kg·K out =−= TTcq =−p

net work production is then

nd

The

kJ/kg 2.4912.2624.753outinnet =−=−= qqw

a

kW 409=⎟⎠⎞

⎜⎝⎛==

s 60min 1kJ/kg) 2.491(kg/min) 50(netnet wmW &&


9-62

9-86 A simple Brayton cycle with helium has a pressure ratio of 14. The power output is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Helium is an ideal gas with constant specific heats.

Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2a).

Analysis For the compression process,

K 9.827K)(14) 288( 70.667/1.66/)1(1

/)1(

1

212 ===⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

−kk

p

kk

s rTPP

TT

4

s

T

1

2s

3 qin

qout

973 K

288 K

2

K 3.85695.0

2889.827288

)()( 12

1212

12

12

12

=−

+=

−+=⎯→⎯

−

−=

−−

=C

s

p

spsC

TTTT

TTcTTc

hhhh

ηη

For the isentropic expansion process,


K 53⎛

=T .338141K) 973(1

33

44 =⎟

⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝ pr

TPP

T

ssure heat addition process 2-3 produces

=−

imilarly,

70.667/1.66/)1(/)1( −− kkkk

Applying the first law to the constant-pre

.19265()( 23in =−= TTcq p kJ/kg 0.606K)3.856973)(kJ/kg·K

S

kJ/kg 2.262K)2885.338)(kJ/kg·K 1926.5()( 14out =−=−= TTcq p

T orhe net w k production is then

and

kJ/kg 8.3432.2620.606outinnet =−=−= qqw

kW 287=⎟⎠⎞

⎜⎝⎛==

s 60min 1kJ/kg) 8.343(kg/min) 50(netnet wmW &&


9-63

9-87 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The effects of non-isentropic compressor and turbine on the back-work ratio is to be compared.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

Analysis For the compression process,

4s

s

T

1

2s

4

3 qin

qout

873 K

288 K

2

K 8.585K)(12) 288( 0.4/1.4/)1(

1

212 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

K 9.61890.0

2888.585288

)()( 12

1212

12

12=

−=C chh

η 12

=−

+=

−+=⎯→⎯

−

−−

C

s

p

sps TTTT

TTTTchh

η

For the expansion process,


K 2.429121K) 873(4

34 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎞⎜⎜⎛

=s PP

TT 0.4/1.4/)1(

3 ⎠⎝

− kk

K 6.473)2.429873)(90.0(873

)()()(

433443

43

43

43

=−−=

−−=⎯→⎯−

−=

−−

= sTsp

p

s T TTTT

TTcTTc

hhhh

ηη

of compressor and turbine are

The isentropic and actual work

kJ/kg 3.299K)2888.585)(KkJ/kg 1.005()( 12Comp, =−⋅=−= TTcW sps

8KkJ/kg 1.005()( 12Comp =⋅=−= TTcW p

4

289.618)( − kJ/kg 6.332K)

kJ/kg 0.446K)2.429873)(KkJ/kg 1.005()( 43Turb, =−⋅=−= sps TTcW

( 3Turb kJ/kg 4.401K)6.473873)(KkJ/kg 1.005() =−⋅=sTW

compressor and isentropic turbine case is

−= p Tc

The back work ratio for 90% efficient

0.7457===kJ/kg 446.0kJ/kg 332.6

Turb,

Compbw

sWW

r

The back work ratio for 90% efficient turbine and isentropic compressor case is

0.7456===kJ/kg 401.4kJ/kg 299.3

Turb

Comp,bw W

Wr s

The two results are almost identical.


9-64

9-88 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases.




Analysis (a) Using the isentropic relations,

( )( )( )

( )( )

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

kg/s 352===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

kJ/kg 199.1kJ/s 70,000

kJ/kg 99.1175.31184.510

kJ/kg 510.84K491.71000KkJ/kg 1.005

kJ/kg 311.75K300610.2KkJ/kg 1.005

K 491.7121K 1000

K 610.212K 300

outnet,s,

outnet,

inC,s,outT,s,outnet,s,

4343outT,s,

1212inC,s,

0.4/1.4/1

3

434

0.4/1.4/1

1

212

wW

m

www

TTchhw

TTchhw

PP

TT

PP

TT

s

sps

sps

kk

s

kk

s

&&

s

T

1

2s

4s

31000 K

300 K

2

4

(b) The net work output is determined to be

( )( )

kg/s 1037===

=−=

−=−=

kJ/kg 67.5kJ/s 70,000

kJ/kg 67.50.85311.7584.51085.0

/

outnet,a,

outnet,

inC,s,outT,s,inC,a,outT,a,outnet,a,

wW

m

wwwww

a

CT

&&

ηη


9-65

9-89 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined.



Analysis (a) Using the isentropic relations,


T h

T h

1 1

2 2

= ⎯ →⎯ =

= ⎯ →⎯ =

300 K 300.19 kJ / kg

580 K 586.04 kJ / kg

( )

( ) ( )( ) kJ/kg 542.0905.831536.0486.0

kJ/kg 285.8519.30004.586

kJ/kg 05.83973.6711.47471

11.474

kJ/kg1536.0404.586950

7100700

43outT,

12inC,

43

4

323in

1

2

34

3

=−=−=

=−=−=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

=→

=+=⎯→⎯−=

===

sT

srr

r

p

hhw

hhw

hPPP

P

P

hhhq

PP

r

η

s

T

1

2s

4s

3950 kJ/kg

580 K

300 K

2

4

Thus,

52.7%===kJ/kg 542.0kJ/kg 285.85

outT,

inC,bw w

wr

27.0%===

=−=−=

kJ/kg 950kJ/kg 256.15

kJ/kg 256.15285.85.0542

in

outnet,th

inC,outT,net.out

qw

www

η

(b)


9-66

9-90 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined.




Analysis (a) Using constant specific heats,

( )( ) ( )

( )( )

( ) ( )( )( ) ( ) ( )( )( ) kJ/kg 562.2K874.81525.3KkJ/kg 1.00586.0

kJ/kg 281.4K300580KkJ/kg1.005

K 874.871K 1525.3

K 1525.3KkJ/kg 1.005/kJ/kg 950K 580

/

7100700

4343outT,

1212inC,

0.4/1.4/k1k

3

434s

in232323in

1

2

=−⋅=−=−=

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=⋅+=

+=⎯→⎯−=−=

===

−

spTsT

p

pp

p

TTchhw

TTchhw

PP

TT

cqTTTTchhq

PP

r

ηη

s

T

1

2s

4s

3950 kJ/kg

580 K

300 K

2

4

Thus,

50.1%===kJ/kg 562.2kJ/kg 281.4

outT,

inC,bw w

wr

29.6%===

=−=−=

kJ/kg 950kJ/kg 280.8

kJ/kg 280.8281.42.562

in

outnet,th

inC,outT,outnet,

qw

www

η

(b)


9-67

9-91 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.



Analysis For the isentropic compression process,


=== − kkprTT 2 K .1527K)(10) 273( 0.4/1.4/)1(

1

s

T

1

2

4

3 qin

qout273 K

The heat addition is

kJ/kg 500kg/s 1

kW 500inin ===

mQ

q&

&

Applying the first law to the heat addition process,

K 1025

KkJ/kg 1.005kJ/kg 500

)( 23in p

qK 1.527in

23 =⋅

+=+=

−

pcTT

TTq

The temperature at the exit of the turbine is

= c

K 9.5301011 0.4/1.4/)1(

⎞⎛⎞⎛− kk

K) 1025(3 =⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=pr

TT

tic turbine and the compressor produce

4

Applying the first law to the adiaba

1025)(KkJ/kg 1.005()( 43T ⋅=−= TTcw p kJ/kg 6.496K)9.530 =−

=−⋅

The net power produced by the engine is then

inally the thermal efficiency is

1.005()( 12C =−= TTcw p kJ/kg 4.255K)2731.527)(KkJ/kg

kW 241.2=−=−= kJ/kg)4.2556kg/s)(496. 1()( CTnet wwmW &&

F

0.482===kW 500kW 241.2

in

netth Q

W&

&η


9-68

9-92 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined.



Analysis For the isentropic compression process,


=== − kkprTT 2 K .8591K)(15) 273( 0.4/1.4/)1(

1

s

T

1

2

4

3 qin

qout273 K

The heat addition is

kJ/kg 500kg/s 1

kW 500inin ===

mQ

q&

&

Applying the first law to the heat addition process,

K 1089


)( 23in p

qK 8.591in

23 =⋅

+=+=

−

pcTT

TTq

The temperature at the exit of the turbine is

= c

K 3.5021511 0.4/1.4/)1(

⎞⎛⎞⎛− kk

K) 1089(3 =⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=pr

TT

tic turbine and the compressor produce

4

Applying the first law to the adiaba

1089)(KkJ/kg 1.005()( 43T ⋅=−= TTcw p kJ/kg 6.589K)3.502 =−

=−⋅

The net power produced by the engine is then

Finally the thermal efficiency is

1.005()( 12C =−= TTcw p kJ/kg 4.320K)2738.591)(KkJ/kg

kW 269.2=−=−= kJ/kg)4.3206kg/s)(589. 1()( CTnet wwmW &&

0.538===kW 500kW 269.2

in

netth Q

W&

&η


9-69

9-93 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal 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 variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.


⋅=⎭⎬⎫

=°=

=⎯→⎯°=

sPT

hT

kPa 20002 =

⎭⎬⎫=

shP

Process 1-2: Compression

KkJ/kg 749.5kPa 100

C40kJ/kg 6.313C40

11

1

11

kJ/kg 7.736kJ/kg.K 749.512 == ss

2

kJ/kg 4.8116.313

6.

1

Combustion chamber

Turbine

23

4

Compress.

100 kPa 40°C

650°C

2 MPa

3137.73685.0 2212

12 =⎯→⎯−

−=⎯→⎯

−−

= hhhh

hh sη

0 4 =⎯→⎯° h

C

Process 3-4: Expansion

654 =T kJ/kg 2.959C

ss hhh 3

3

43T 88.0

−=⎯→⎯

−=η

hh

4

2.959−

. However, using the following lines in EES together with the isentropic J/kg, T3 = 1421ºC, s3 = 6.736 kJ/kg.K. The solution by hand would require a trial-

error approach. h_3=enthalpy(Air, T=T_3)

=T_3, P=P_2) P_1, s=s_3)

The mass flow rate is determined from

hh 43 −

We cannot find the enthalpy at state 3 directlyefficiency relation, we find h3 = 1873 k

s_3=entropy(Air, Th_4s=enthalpy(Air, P=

( )( )kg/s 99.12

K 27340K/kgmkPa 0.287)s/m 0kPa)(700/6 100(

3

3

1

11 =+⋅⋅

==RTP

mV&

&

=−=−=&

WWW &&

(b) The back work ratio is

The net power output is

kW 6464)kJ/kg6.3134kg/s)(811. 99.12()( 12inC, =−=−= hhmW &&

2.959kg/s)(1873 99.12()( 43outT, hhmW & kW 868,11)kJ/kg

net& kW 5404=−=−= 6464868,11inC,outT,

0.545===kW 868,11

kW 6464

outT,

inC,bw W

Wr

&

&

(c) The rate of heat input and the thermal efficiency are

kW 788,13)kJ/kg4.811kg/s)(1873 99.12()( 23in =−=−= hhmQ &&

39.2%==== 392.0kW 788,13

kW 5404

in

net

QW

th &

&η


9-70

9-94 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the isentropic efficiency of the turbine and the cycle thermal efficiency are to be determined.


Properties The properties of air are given as cv = 0.718 kJ/kg·K, cp = 1.005 kJ/kg·K, R = 0.287 kJ/kg·K, k = 1.4.

Analysis (b) For the compression process,


)( 12

=−

−= TTcmW p&

For the turbine during the isentropic process, 4s

s

T

1

2s

4

3 873 K

303 K

2

Comp&

kW 300,60K)30330)(KkJ/kg 1.005(kg/s) 200( ⋅=

K 9.772kPa 803

34⎝⎟

⎠⎜⎝

s P 0kPa 100K) 1400(

0.4/1.4/)1(4 =⎟

⎠⎞

⎜⎛=⎟

⎞⎜⎛

=− kk

PTT

The actual power output from the turbine is

=+=+= WWW &&

the turbine is then

kW 050,126K)9.7721400)(KkJ/kg 1.005(kg/s) 200()( 43sTurb, =−⋅=−= sp TTcmW &&

CompTurbnet −= WWW &&&

Turb& kW 300,120300,60000,60Turbnet

The isentropic efficiency of

95.4%====,126sTurb,

Turb W&η 954.0

kW 050kW 300,120TurbW&

) The rat input is


(c e of heat

kW 200,160K)]273330(1400)[(KkJ/kg 1.005(kg/s) 200()( 23in =+−⋅=−= TTcmQ p&&

37.5%==== 375.0kW 200,160kW 000,60

in

netth Q

W&

&η


9-71

9-95 A modified Brayton cycle with air as the working fluid operates at a specified pressure ratio. The T-s diagram is to be sketched and the temperature and pressure at the exit of the high-pressure turbine and the mass flow rate of air are to be determined.




K 5.494K)(8) 273( 0.4/1.4/)1(

1

212 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

s

T

1

4

5

3

P4

1500 K

273 K

2

P5

P3 = P2

The power input to the compressor is equal to the power output from the high-pressure turbine. Then,

K 1278.5=−+=−+=−=−

−=− )( 12 cmTTcm p &&

=

5.4942731500

)(

2134

4312

43

outTurb, HPinComp,

TTTTTTTT

TT

WW

p

&&

The pressure at this state is

kPa 457.3=⎟⎠⎝⎟

⎠⎜⎝

⎟⎠

⎜⎝ 3

1433 K 1500TTP

(c) The temperature at state 5 is determined f

⎞⎜⎛=⎟

⎞⎜⎛

=⎯→⎯⎟⎞

⎜⎛

=−− 4.0/4.1)1/(

4)1/(

4 K 5.1278kPa) 100(8kkkk

TrPP

TP

rom

4

K 1.828kPa 3.457

kPa 100K) 5.1278(0.4/1.4/)1(

4

545 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

PP

TT

The net power is that generated by the low-pressure turbine since the power output from the high-pressure turbine is equal the power input to the compressor. Then,

to

kg/s 441.8=−⋅

=−

=

−=

K)1.8285.1278)(KkJ/kg 1.005(kW 000,200

)(

)(

54

Turb LP

54Turb LP

TTcW

m

TTcmW

p

p

&&

&&



9-72

9-96 A simple Brayton cycle with air as the working fluid operates at a specified pressure ratio and between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the volume flow rate of the air into the compressor is to be determined. Also, the effect of compressor inlet temperature on the mass flow rate and the net power output are to be investigated.





( ) K 0.4767K) 273( 0.4/1.4/)1(

11

212 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

4s

s

T

1

2s

4

3 1500 K

273 K

2

K 8.526

27380.0

2 −=

T2730.476

)()(

2

12

12

12

12

Comp

sComp,Comp

=⎯→⎯−

−−

=−

−==

T

TTTT

TTcmTTcm

W

W s

p

sp

&

&

&

&η

or the expansion process, F

K 3.86071K) 1500(

0.4/1.4/)1(

3

434 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

K 3.924

3.8601500 −1500

90

)(

44

4343Turb

=⎯→⎯−

=

−=

−==

TT

TTTTcmW p&&η

Given the net power, the mass flow rate is determined from

)( 4343sTurb,Turb −− TTTTcmW ssp&&

.0

[ ]

[ ]

[ ]kg/s 7.63

)2738.526()3.9241500()KkJ/kg 1.005(

)( 123

net

=−−−⋅

=

−=

TTTcW

mp

&&

4

kW 000,150

()

)()(

()(

4

1243

1243CompTurbnet

−−

−−−=

−−−=−=

T

TTTTcmW

TTcmTTcmWWW

p

pp

&

&&&&&

The specific volume and the volume flow rate at the inlet of the compressor are

)

net&

/kgm 7835.0kPa 100

K) 273)(KkJ/kg 287.0( 3

1

11 =

⋅==

PRT

v

(c) For a fixed compressor inlet velocity and flow area, when the compressor inlet temperature increases, the specific

volume increases since

3m 342.2=== )/kgm 7835.0)(kg/s 7.436( 311 vV m&&

PRT

=v . When specific volume increases, the mass flow rate decreases since vV&

& =m . Note that

volume flow rate is the same since inlet velocity and flow area are fixed ( ). When mass flow rate decreases, the net power decreases since . Therefore, when the inlet temperature increases, both mass flow rate and

the net power decrease.

AV=V&

)( CompTurbnet wwmW −= &&


9-73

Brayton Cycle with Regeneration

9-97C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber.

9-98C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease.

9-99C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, and is defined as ε = qregen, act /qregen, max.

9-100C (b) turbine exit.

9-101C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since in/ QW=η and W increases while Qin remains constant. Steam can be obtained by utilizing the hot exhaust gases.



9-74

9-102 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined.

Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).

Analysis According to the isentropic process expressions for an ideal gas,


=== − kkprTT 2 K 8.530K)(8) 293( 0.4/1.4/)1(

1

s

T

1

2 5

4 qin1073 K

293 K

3

6

qout

K 3.59281K) 1073(1 0.4/1.4/)1(

45 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

When the first law is applied to the heat exchanger, the result is

6523 TTTT −=−

while the regenerator temperature specification gives

=−=−= TT

mult neous solution of these two results gives

K 3.582103.5921053

The si a

K 8.540)8.5303.582(3.592)( 2356 =−−=−−= TTTT

Application of the first law to the turbine and compressor gives

hen,

kJ/kg 244.1K )2938.530)(KkJ/kg 005.1(K )3.5921073)(KkJ/kg 005.1(

)()( 1254net

=−⋅−−⋅=

−−−= TTcTTcw pp

T

kg/s 6145.0kW 150net ===W

m&

& kJ/kg 244.1

he first law to the combustion chamber produces

TTcmQ p&

imilarly,

netw

Applying t

in& kW 303.0=−⋅=−= K)3.5821073)(KkJ/kg 005.1(kg/s) 6145.0()( 34

S

kW 153.0=−⋅=−= K)2938.540)(KkJ/kg 005.1(kg/s) 6145.0()( 16out TTcmQ p&&


9-75

9-103 A Brayton cycle with regeneration produces 150 kW power. The rates of heat addition and rejection are to be determined.



Analysis For the compression and expansion processes we have

K 8.530K)(8) 293( 0.4/1.4/)1(12 === − kk

ps rTT

s

T

1

2 5s

4qin1073 K

293 K

3 6

qout

52s K 3.566

87.02938.530293

)()( 12

1212

12

=−

+=

−+=⎯→⎯

−

−=

C

s

p

spC

TTTT

TTcTTc

ηη

K 3.59281K) 1073(1

45 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎜⎜=s r

TT 0.4/1.4/)1(

⎠

⎞

⎝

⎛− kk

p


K 9.625=)3.5921073)(93.0(1073

)554 −−=

ssp

hen the first law is applied to the heat exchanger, the result is

hile the regenerator temperature specification gives

multaneous solution of these two results gives

()()(

44554 −−=⎯→⎯

−

−= T

pT TTTT

TTcTTc

ηη

W

6523 TTTT −=−

w

K 9.615109.6251053 =−=−= TT

The si

K 3.576)3.5669.615(9.625)( 2356 =−−=−−= TTTT

Application of the first law to the turbine and compressor gives

=−⋅−−⋅=

−−− TTcTT p

Then,

kJ/kg 7.174K )2933.566)(KkJ/kg 005.1(K )9.6251073)(KkJ/kg 005.1(

)()( 1254net = cw p

kg/s 8586.0kJ/kg 174.7kW 150

net

net ===wW

m&

&

Applying the first law to the combustion chamber produces

Similarly,

kW 394.4=−⋅=−= K)9.6151073)(KkJ/kg 005.1(kg/s) 8586.0()( 34in TTcmQ p&&

kW 244.5=−⋅=−= K)2933.576)(KkJ/kg 005.1(kg/s) 8586.0()( 16out TTcmQ p&&


9-76

9-104 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallel-flow and counter-flow arrangements of the regenerator are to be compared.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).

Analysis According to the isentropic process expressions for an ideal gas,


=== − kkprTT 2 K 9.510K)(7) 293( 0.4/1.4/)1(

1

s

T

1

2 5

4qin1000 K

293 K

3

6

qout

K 5.57371K) 1000(1 0.4/1.4/)1(

45 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

When the first law is applied to the heat exchanger as originally arranged, the result is

6523 TTTT −=−

while the regenerator temperature specification gives

=−=−= TT

he simul ution of these two results gives

K 5.56765.573653

T taneous sol

5.5675.5732356 +−=+−= TTTT K 9.5169.510 =

The thermal efficiency of the cycle is then

0.482=−

−−=

− 34in TTq−

−=−=5.5671000

293516.9111 16outth

TTqη

or the rearranged version of this cycle,

n energy balance on the eat exchanger gives

he soluti hese two equations is

6 =T


F

663 −= TT

s

T

1

2 5

4qin1000 K

293 K

3 6

qout

A h

6523 TTTT −=−

T on of t

K 2.5393 =T

K 2.545

0.453=−

−−=

−−

−=−=2.5391000

293545.211134

16

in

outth TT

TTqq

η


9-77

9-105 A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 70 kW are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 The ambient air is 300 K and 100 kPa. 4 The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5 The combustion gases can be treated as air.

Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17.

Analysis The gas turbine cycle with regeneration can be analyzed as follows:


( )( )

( ) kJ/k 76.87350.590.23814

434

=⎯→⎯=⎟⎞

⎜⎛== srr hP

PP g

4

0.238kJ/kg 79.1277

K 1200

kJ/kg 49.446544.5386.14

386.1kJ/kg 19.300

K 300

3

33

21

2

11

3

12

1

⎠⎝

==

⎯→⎯=

=⎯→⎯===

==

⎯→⎯=

r

srr

r

Ph

T

hPPP

P

Ph

and

T

s

T

1

2s4s

3 qin1200 K

300 K

5

42

P

kJ/kg 57.954 76.87379.1277

79.1277 443

212

−− hhh0.80

kJ/kg 07.483 19.300

19.30049.4460.80

443

turb

212

comp

=→−

=→−

=

=→−

−=→

−−

=

hhh

hhhh

hh

s

s

η

η

Then the thermal efficiency of the gas turbine cycle becomes

kJ/kg 35.424)07.48357.954(9.0)( 24regen =−=−= hhq ε

kJ/kg 34.140)19.30007.483()57.95479.1277()()(

kJ/kg 370.37=35.424)07.48379.1277()(

1243inC,outT,outnet,

regen23in

=−−−=−−−=−=

−−=−−=

hhhhwww

qhhq

0.379kJ/kg 370.37kJ/kg 140.34

in

outnet,th %37.9====

qw

η

Finally, the mass flow rate of air through the turbine becomes

kg/s 0.499===kJ/kg 140.34

kW 70

net

netair w

Wm

&&


9-78

9-106 The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the

1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The same, and the properties of combustion gases are the same as

e A-17.

Analysis The properties at various states are

turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined.

Assumptionsmass flow rates of air and of the combustion gases are the those of air.

Properties The properties of air are given in Tabl


( )( )

( ) 47.485.7127.14

14

434

=⎯→⎯=⎟⎠⎞

⎜⎝⎛== srr hP

PP kJ/kg 23.825

5.712kJ/kg 1710.0

K 1561C1288

kJ/kg 25.65310.214356.17.14

4356.1kJ/kg .21303

K 303=C30

3

33

21

2

11

3

12

1

==

⎯→⎯=°=

=⎯→⎯===

==

⎯→⎯°=

r

srr

r

P

Ph

T

hPPP

P

Ph

T

s

T

1

2s 4s

3qin1561 K

303 K

5

4

2

The net work output and the heat input per unit mass are

C7.658K 7.931 kJ/kg 55.96821.30334.665 kJ/kg 34.66566.3720.1038netinout =−=−= wqq

kJ/kg 0.67210381710

kJ/kg 0.10380.359

kJ/kg 372.66

kJ/kg 66.372h 1

s 3600kg/h 1,536,000kW 159,000

41out414t

3223in

th

netin

netnet

°==→=+=+=→−=

=−=−=→−=

===

=⎟⎠⎞

⎜⎝⎛==

Thqhhhq

qhhhhq

wq

mW

w

in

η

&

&

ou

Then the compressor and turbine efficiencies become

94.9%

83.8%

==−

−=

−−

=

==−−

=−−

=

949.021.303672

21.30325.653

838.023.825171055.9681710

12

12

43

43

hhhh

hhhh

sC

sT

η

η

When a regenerator is added, the new heat input and the thermal efficiency become

44.1%====

−=−=

−=

441.0kJ/kg 845.2kJ/kg 372.66

kJ/kg 845.2=8.1921038

kJ/kg 192.8=672.0)-.55(0.65)(968=)(

newin,

netnewth,

regeninnewin,

24regen

qw

qqq

hhq

η

ε

Discussion Note a 65% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 44.1%.


9-79

9-107 Problem 9-106 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted.


"Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 30 [C] P[1]= 100 [kPa] {T[4]=659 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.65 Eta_c = 0.84 "Compressor isentropic efficiency" Eta_t = 0.95 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_compisen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2]"process 2-3 is SSSF constant pressure" "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[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turbisen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])



9-80

"Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and 1 MW = 1000 kW:" W_dot_net*1000=W_dot_turb-W_dot_comp "kJ/s" Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]



9-81

ηt ηc ηth,noreg ηth,withreg Qinnoreg

[kW] Qinwithreg

[kW] Wnet [kW]

0.7 0.75 0.8

0.85 0.9

0.95 1

0.84 0.84 0.84 0.84 0.84 0.84 0.84

0.2044 0.2491 0.2939 0.3386 0.3833 0.4281 0.4728

0.27 0.3169 0.3605 0.4011 0.4389 0.4744 0.5076

422152 422152 422152 422152 422152 422152 422152

319582 331856 344129 356403 368676 380950 393223

86.3 105.2 124.1 142.9 161.8 180.7 199.6

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5-100

100

300

500

700

900

1100

1300

1500

1700

s [kJ/kg-K]

T [C

] 100 kPa

1470 kPa

T-s Diagram for Gas Turbine with Regeneration

1

2s2

5

3

4s

4

6

0.7 0.75 0.8 0.85 0.9 0.95 180

100

120

140

160

180

200

η t

Wne

t [k

W]

0.7 0.75 0.8 0.85 0.9 0.95 1310000

330000

350000

370000

390000

410000

430000

η t

Qin

[kW

]

no regeneration

with regeneration

0.7 0.75 0.8 0.85 0.9 0.95 10.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

ηt

ηth

no regeneration

with regeneration



9-82

9-108 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are 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.



Analysis (a) The properties of air at various states are

( )( )

( ) ( ) ( )

( )

( ) ( )( ) kJ/kg 803.14711.801219.250.821219.25

kJ/kg 711.8059.2815.20071

15.200kJ/kg 1219.25

K 1501

kJ/kg 618.260.75/310.24541.2610.243/

kJ/kg 541.2688.105546.17

5546.1kJ/kg 310.24

K 310

433443

43

43

4

33

121212

12

21

2 == PP

P

1

34

3

12

1

=−−=−−=⎯→⎯−−

=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==

⎯→⎯=

=−+=−+=⎯→⎯−−

=

=⎯→⎯=

==

⎯→⎯=

sTs

T

srr

r

Css

C

srr

r

hhhhhhhh

hPPP

P

Ph

hhhhhhhh

hP

Ph

ηη

ηη

Thus,

T4 = 782.8 K

(b)

)

s

T

1

2s4s

3qin1150 K

310 K

5

6

42

1T

T

( ) ( )( ) (

kJ/kg 108.09=−−−=

−−−=−=

24.31026.61814.80325.12191243inC,outT,net hhhhwww

( )( )(

kJ/kg 738.43618.26803.140.6526.618

242524

25

=−+=

−+=⎯→⎯−−

= hhhhhhhh

εε (c) )

Then,

22.5%===

=−=−=

kJ/kg 480.82kJ/kg 108.09

kJ/kg 480.82738.4319.2512

in

netth

53in

qw

hhq

η


9-83

9-109 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases.

Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas. 3 Kinetic and potential energy changes are negligible.

Properties When assuming constant specific heats, the properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17.

Analysis (a) Assuming constant specific heats,

( )( )( )


( )( )

( )( )

( )( ) kW 39,188===

=−

−−=

−−

−=−

−−=−=

====⎯→⎯=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=⎞⎛

−

−

kW 75,0000.5225

5225.02.6071100

2903.5251111

K 525.3andK 607.2%100

K 607.281K 1100

K 525.3

innet

53

16

53

16

in

outth

2645

0.4/1.4/1

3

434

0.4/1.4/1

QW

TTTT

TTcTTc

qq

TTTT

PP

TT

P

T

p

p

kk

kk

&& η

η

ε

) Assuming variable specific heats,

=⎟⎟⎠

⎜⎜⎝

= 8K 2901

212 P

TT

(b

( )( )

( )

( )( ) kW 40,283===

=−−

−=−−

−=−=

====⎯→⎯=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==

⎯→⎯=

=⎯→⎯===

==

⎯→⎯=

kW 75,0000.5371

5371.037.65107.116116.29012.526111

kJ/kg 526.12andkJ/kg 651.37%100

kJ/kg 651.3789.201.16781

1.167kJ/kg 1161.07

K1100

kJ/kg 526.128488.92311.18

2311.1kJ/kg 90.162

K029

innet

53

16

in

outth

2645

43

4

33

21

2

11

34

3

12

1

QW

hhhh

qq

hhhh

hPPP

P

Ph

T

hPPP

P

Ph

T

T

rr

r

rr

r

&& η

η

ε

s

T

1

2 4

375,000 kW1100 K

290 K

5

qout

6


9-84

9-110 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined.



Analysis (a) The properties at various states are


( )

( )( )(

( ) ( )( ) kJ/kg 192.7=−=−==

−−=−−=⎯→⎯

−−

=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==⎯→⎯==⎯→⎯=

=⎯→⎯=

===

84.65974.90080.0kJ/kg .74900

44.83242.151590.042.1515

kJ/kg .4483206.505.45091

.5450kJ/kg 1515.42K 0014

kJ/kg .84659K 065kJ/kg 10.243K 310

9100/900/

24regen

433443

43

43

4

33

22

11

12

34

3

hhq

hhhhhhhh

hPPP

P

PhThThT

PPr

sTs

T

srr

r

)

ε

ηη

(b)

p

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

40.0%====

=−−=−−=

=−−−=−−−=−=

400.0kJ/kg 662.88kJ/kg 265.08

kJ/kg .886627.192.846591515.42

kJ/kg .08265.24310659.8474.90042.1515

in

netth

regen23in

1243inC,outT,net

qw

qhhq

hhhhwww

η

s

T

1

2s4s

3 qin1400 K

310 K

5

6

4650 K 2


9-85


Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible.



Analysis (a) Using the isentropic relations and turbine efficiency,

( )( )

( )( ) ( )

( )( )

( ) ( ) ( )( )( ) kJ/kg 130.7=−⋅=−=−==

−−=−−=⎯→⎯

−

−=

−−

=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

===

−

K065812.6KkJ/kg 1.0050.80K .6812

3.747140090.01400

K .374791K 1400

9100/900/

2424regen

433443

43

43

43

4.1/4.0/1

3

434

12

TTchhq

TTTTTTcTTc

hhhh

PP

TT

PPr

p

sTsp

p

sT

kk

s

p

εε

ηη s

T

1

2s 4s

3qin1400 K

310 K

5

6

4650 K 2

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

39.9%====

=−−⋅=

−−=−−=

=−−−⋅=

−−−=−=

399.0kJ/kg 623.1kJ/kg 248.7

kJ/kg .1623.7130K0651400KkJ/kg 1.005

kJ/kg .7248K310650.68121400KkJ/kg 1.005

in

netth

regen23regen23in

1243inC,outT,net

qw

qTTcqhhq

TTcTTcwww

p

pp

η

(b)


9-86





Analysis (a) The properties at various states are

( )

( )( )(

( ) ( )( ) kJ/kg 168.6=−=−==

−−=−−=⎯→⎯

−−

=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==⎯→⎯==⎯→⎯=

=⎯→⎯=

===

84.65974.90070.0kJ/kg .74900

44.83242.151590.042.1515

kJ/kg .4483206.505.45091

.5450kJ/kg 1515.42K 0014

kJ/kg .84659K 065kJ/kg 10.243K 310

9100/900/

24regen

433443

43

43

4

33

22

11

12

34

3

hhq

hhhhhhhh

hPPP

P

PhThThT

PPr

sTs

T

srr

r

)

ε

ηη

(b)

p

s

T

1

2s4s

3 qin1400 K

310 K

5

6

4650 K 2

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

38.6%====

=−−=−−=

=−−−=−−−=−=

386.0kJ/kg 687.18kJ/kg 265.08

kJ/kg 18.6876.168.846591515.42

kJ/kg .08265.24310659.8474.90042.1515

in

netth

regen23in

1243inC,outT,net

qw

qhhq

hhhhwww

η


9-87

9-113 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed.


Analysis The expressions for the isentropic compression and expansion processes are


−= 2kk

prTT /)1(1

s

T

1

2 4

3qin

5

6qout

kk

prTT

/)1(

341

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

For an ideal regenerator,

45

TT =

26

TT =

The thermal efficiency of the cycle is

kkp

kkp

kkp

rTT

r

rTT

TTTT

TT

TTTT

TT

TTTT

qq

/)1(

3

1

/)1(

/)1(

3

1

34

12

3

1

35

16

3

1

53

16

in

outth

1

1

11

)/(11)/(

1

)/(11)/(

111

−

−−

−

−=

−

−−=

−−

−=

−−

−=−−

−=−=η


9-88

Brayton Cycle with Intercooling, Reheating, and Regeneration

9-114C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle.

9-115C 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.

9-116C (a) decrease, (b) decrease, and (c) decrease.

9-117C (a) increase, (b) decrease, and (c) decrease.

9-118C (a) increase, (b) decrease, (c) decrease, and (d) increase.

9-119C (a) increase, (b) decrease, (c) increase, and (d) decrease.

9-120C (c) The Carnot (or Ericsson) cycle efficiency.



9-89

9-121 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.


s

T

3

4

1

5qin1200 K

300 K

86

7

10

9

2


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

5

⎞⎛P

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

12

1

=−=−=

=−=−=

==⎯→⎯=⎟⎠

⎜⎝

==

===

⎯→⎯=

==⎯→⎯===

==

⎯→⎯=

hhw

hhw

hhPP

P

Phh

T

hhPPP

P

Ph

T

rr

r

rr

r

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



9-90

9-122 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.



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 3.325813.98679.127722

kJ/kg .4626419.30042.43222

kJ/kg .1398636.94679.127788.079.1277

kJ/kg 946.3633.7923831

238kJ/kg 1277.79K 1200

kJ/kg .42432

/

kJ/kg 411.26158.4386.13

386.1kJ/kg 300.19K 300

65outT,

12inC,

6558665

65

865

6

755

1214212

421

2

11

56

5

12

1

=−=−=

=−=−=

=−−=

−−==⎯→⎯−−

=

==⎯→⎯=⎟⎠⎞

⎜⎝⎛==

===⎯→⎯=

=

−+==⎯→⎯−−

=

=⎯→⎯===

==⎯→⎯=

hhw

hhw

hhhhhhhhh

hhPPP

P

PhhT

hhhhhhhhh

hhPPP

P

PhT

sTs

T

ssrr

r

Css

C

ssrr

r

ηη

ηη

Thus,

s

T

3

4

1

5qin

86

7

1

9

2

86s

84.0/19.30026.41119.30012 −+=

=

45.3%==== 453.0kJ/kg 583.32kJ/kg 264.46

outT,

inC,bw w

wr

4 2s

( ) ( ) ( ) ( )

28.0%====

=−=−=

=−+−=−+−=

280.0kJ/kg 1137.03kJ/kg 318.86

kJ/kg .8631846.26432.583

kJ/kg 1137.03986.131277.79432.421277.79

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

( ) ( )( )

44.2%====

=−=−=

=−=−=

442.0kJ/kg 721.75kJ/kg 318.86

kJ/kg .7572128.41503.1137

kJ/kg 15.28442.43213.98675.0

in

netth

regenoldin,in

48regen

qw

qqq

hhq

η

ε


9-91

9-123 An ideal regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The power produced and consumed by each compression and expansion stage, and the rate of heat rejected are to be determined.



Analysis The pressure ratio for each stage is

464.312 ==pr

s

T

3

4

1

6

288 K

97

8

2

5

1

According to the isentropic process expressions for an ideal gas,

K 7.410K)(3.464) 288( 0.4/1.4/)1(142 ==== − kk

prTTT

Since this is an ideal cycle,

7.410504975


K 7.43828 =+

For the isentropic expansion processes,

==== − kkprTTT

he heat i is

=+=== TTTT

K 7.625K)(3.464) 7.438( 0.4/1.4/)1(786

T nput

(2 56in kJ/kg 8.375K )7.438K)(625.7kJ/kg (1.0052) =−⋅=

e mass flow rate is then

−= TTcq p

Th

kg/s .3311kJ/kg 8.375inqkJ/s 500in ===

Qm

&&

pplication of the first law to the expansion process 6-7 gives

K )7.438K)(625.7kJ/kg 5kg/s)(1.00 331.1(

he same amount of power is produced in process 8-9. When e first law is adapted to the compression process 1-2 it becomes

K )288K)(410.7kJ/kg 5kg/s)(1.00 331.1(

Compression process 3-4 uses the same amount of power. The rate of heat rejection from the cycle is

A

−= )( 76out7,-6 TTcmW p&&

kW 250.1=−⋅=

T th

−= )( 1212,in TTcmW p&&

kW 164.1=

−⋅=

kW 328.3=−⋅=

−=

K )288K)(410.7kJ/kg 5kg/s)(1.00 331.1(2

)(2 32out TTcmQ p&&


9-92

9-124 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The thermal efficiency of the cycle is to be determined.



Analysis The temperatures at various states are obtained as follows

K 9.430K)(4) 290( 0.4/1.4/)1(142 ==== − kk

prTTT


=+=+= TT 5 K 9.450209.430204

s

T

3

4

1

6

290 K

97

8

2 5

10

K 4.749


K 9.450

)(

in56

56in

=⋅

+=+=

−=

p

p

cq

TT

TTcq

K 3.50441K) 4.749(1 0.4/1.4/)1(

67 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

K 8.802KkJ/kg 1.005

kJ/kg 300K 3.504in78 =

⋅+=+=

pcq

TT

K 2.54041K) 8.802(1 0.4/1.4/)1(

89⎜⎛

= TT =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎝

− kk

pr

=−=

The heat input is


=−+−⋅=

pp


20910 −= TT K 2.520202.540

kJ/kg 600300300in =+=q

kJ/kg 373.0R )2909.430290K)(520.2kJ/kg (1.005

)()( 32110

out −+−= TTcTTcq

0.378=−=−=600

0.37311in

outth q

qη


9-93

9-125 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined.



Analysis The temperatures at various states are obtained as follows

K 9.430K)(4) 290( 0.4/1.4/)1(1642 ===== − kk

prTTTT


=+=+= TT

s

T

3

4

1290 K

9

8

2

5

1

11

67

12

13

14

7 K 9.450209.430206

K 4.749


K 9.450

)(

in78

78in

=⋅

+=+=

−=

p

p

cq

TT

TTcq

K 3.50441K) 4.749(1 0.4/1.4/)1(

89 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

K 8.802KkJ/kg 1.005

kJ/kg 300K 3.504in910 =

⋅+=+=

pcq

TT

K 2.54041K) 8.802(1 0.4/1.4/)1(

1011 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

K 7.838KkJ/kg 1.005

kJ/kg 300K 2.540in

1112 =⋅

+=+=pc

qTT

K 4.56441K) 7.838(1 0.4/1.4/)1(

1213 ⎜⎜⎝

=r

TT =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎛− kk

p

The heat input is

he heat r ected is


K 4.544204.564201314 =−=−= TT

kJ/kg 900300300300in =++=q

T ej

kJ/kg .9538R )2909.4302909.430290K)(544.4kJ/kg (1.005

)()()( 5432114out

=−+−+−⋅=

−+−+−= TTcTTcTTcq ppp

40.1%==−=−= 0.401900

9.53811in

outth q

qη


9-94

9-126 A regenerative gas-turbine cycle with three stages of compression and three stages of expansion is considered. The thermal efficiency of the cycle is to be determined.



Analysis Since all compressors share the same compression ratio and begin at the same temperature,

K 9.430K)(4) 290( 0.4/1.4/)1(1642 ===== − kk

prTTTT


−= TT

s

T

3

4

1290 K

9

8

2

5

1

11

67

12

13

14

From the problem statement,

7 6513

The relations for heat input and expansion processes are

p

p cq

TTTTcq in7878in )( +=⎯→⎯−=

kk

prTT

/)1(

891

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

kk

prT

pcq

TT in910 += , T 1

/)1(

1011

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

pcq inTT 1112 += ,

kk /)1(1

−

⎟⎞

⎜⎛

prTT 1213 ⎟

⎠⎜⎝

=

The simultaneous solution of above equations using EES software gives the following results

131211

9

=====

TTTT

lance on the regenerator,

61414136

141367

=+=+=⎯→⎯−=

−=−

TTTTT

TTTT

he heat i is

he heat rejected is


K 7.585 K, 4.870 K, 9.571 K, 8.849K 3.551

10

K, 2.819 K, 7.520 87 == TTT

From an energy ba

)40( 13 −−T K 9.495659.43065

T nput

kJ/kg 900300300300in =++=q

T

kJ/kg 1.490R )2909.4302909.430290K)(495.9kJ/kg (1.005

)()()( 5432114out

=−+−+−⋅=

−+−+−= TTcTTcTTcq ppp

45.5%==−=−= 455.0900

1.49011in

outth q

qη


9-95

Jet-Propulsion Cycles

9-127C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity.

9-128C 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.

9-129C It reduces the exit velocity, and thus the thrust.



9-96

9-130 A turboprop engine operating on an ideal cycle is considered. The thrust force generated is 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 The turbine work output is equal to the compressor work input.

Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K (Table A-1), cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).

Analysis Working across the two isentropic processes of the cycle yields


=== − kkprTT 2 K 7.482K)(10) 250( 0.4/1.4/)1(

1

s

T

1

2 4

3qin

5qout

K 0.403101K) 778(1 0.4/1.4/)1(

35 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

K 3.545)2507.482(778)( 1234 =−−=−−= TTTT

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller,

2

)(2

inlet2

exit54

VVmTTcm ppe

−=− &&

ich when solved for the velocity at which the air leaves the propeller gives wh

m/s 1.216

)m/s 180(kJ/kg 1

/sm 1000K)0.4033.545)(KkJ/kg 005.1(2012

)(21

2 ⎥⎤

⎢⎡

+−= VTTcm

V e&

2/12

22

2/

inlet54exit

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎠⎜⎜⎝

⎛−⋅=

⎥⎦⎢⎣ m pp&

⎟⎟⎞

The mass flow rate through the propeller is

kg/s 0.975/kgm 305.1

m/s 1804m) 3(

4

/kgm 305.1kPa 55

K) 250()mkPa

287.0(RT

3

2

1

12

1

1

33

1

====

=⋅

==

ππvv

v

VDAVm

P

p&

The thrust force generated by this propeller is then

kN 35.2=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−=

2inletexitm/skg 1000

kN 10)m/s181kg/s)(216. (975.0)( VVmF p&


9-97

9-131 A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined.




K 7.482K)(10) 250( 0.4/1.4/)1(12 === − kk

prTT


K 0.40310

K) 778(35 =⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=pr

TT 11 0.4/1.4/)1(⎞⎛⎞⎛

− kk

ince the work produced by expansion 3-4 equals that used by pression 1-2, an energy balance gives

123s

T

1

24

3 qin

5 qout

Scom

7.482(778)( K 3.545)2504 −−=−−= TTTT =

ler is The mass flow rate through the propel

kg/s 0.624

/kgm 305.144 311

====vv

m p&m/s 180m) 4.2(

/kgm 305.1kPa 55

K) 250()mkPa 287.0(

21

21

33

1 =⋅

==

ππ

v

VDAVP

RT

ccording the previous problem,

A to

kg/s 75.4820

kg/s 0.975m&

20=== p

em&

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller,

2)(

2inlet

2exit

54VV

mTTcm ppe−

=− &&

which when solved for the velocity at which the air leaves the propeller gives

m/s 0.234

)m/s 180(kJ/kg 1

/sm 1000K)0.4033.545)(KkJ/kg 005.1(kg/s 0.624kg/s 75.48

2

)(2

2/12

22

2/12 ⎤⎡ me&

inlet54it

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

⎛−⋅=

⎥⎥⎦⎢

⎢⎣

+−= VTTcm

V pp&

The thrust force generated by this propeller is then

ex

kN 33.7=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−=

2inletexitm/skg 1000

kN 10)m/s180kg/s)(234. (624.0)( VVmF p&


9-98

9-132 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is to be determined.


Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K and k = 1.4 (Table A-2a).

Analysis The total mass flow rate is

kg/s 1.676

/kgm 452.1m/s 200

4m) 5.2(

4

/kgm 452.1kPa 50

K) 253()mkPa 287.0(

3

2

1

12

1

1

33

1

====

=⋅

==

ππvv

v

VDAVm

PRT

&

s

T

1

2 4

3qin

5

qoutNow,

kg/s 51.8488e

kg/s 1.676===

mm&

&

he mass flow rate through the fan is

order to roduce the specified thrust force, the velocity at the fan exit will be

T

kg/s 6.59151.841.676 =−=−= ef mmm &&&

In p

m/s 284.5N 1m/skg 1

kg/s 591.6N 50,000m/s) (200

)( inletexit −= f VVmF &

2

inletexit =⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅+=+=

fmFVV&

balance on the stream passing through the fan gives

An energy

K 232.6=

⎟⎠

⎞⎜⎝

⎛⋅

−−=

−−=

−=−

22

22

2inlet

2exit

45

2inlet

2exit

54

/sm 1000kJ/kg 1

)KkJ/kg 005.1(2)m/s 200()m/s 5.284(K 253

2

2)(

p

p

cVV

TT

VVTTc



9-99

9-133 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced 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 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, 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 = 240 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 ≅ 0).


Diffuser:

( )( )

( )( )( )

( ) kPa 88.64K 260K 288.7kPa 45

K 7.288/sm 1000

kJ/kg 1KkJ/kg 1.0052

m/s 240K 2602

2/02

02/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

122

222

11

outin(steady) 0

systemoutinE

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅+=+=

−−=

−+−=⎯→⎯+=+

=⎯→⎯∆=−

−kk

p

p

TT

PP

cV

TT

VTTc

VVhhVhVh

EEEE &&&&

pressor:

&

Com( )( ) ( )( )

( )( )(1K 288.73

23 =⎟⎟⎞

⎜⎜⎛

=PP

TT ) K 7.6003

kPa 5.843kPa 64.8813

0.4/1.4/1

2

243

=⎠⎝

====

− kk

p PrPP

Turbine:

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

or K 0.5187.2887.6008302345 =+−=+−= TTTT Nozzle:

( )( )

( ) 2/02

0

2/2/

K 3.359kPa 843.5

kPa 45K 830

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

+−=⎯→⎯−

+−=

+=+

=⎯→⎯∆=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

&&&&&

( )( )( ) m/s 564.8=⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅==

kJ/kg 1/sm 1000K3.359518.0KkJ/kg 1.0052

22

6 exitVV or

The mass flow rate through the engine is

kg/s 0.291

/kgm 658.1m/s 240

4m) 6.1(

4

/kgm 658.1kPa 45

K) 260()mkPa 287.0(

3

2

1

12

1

1

33

1

====

=⋅

==

ππvv

v

VDAVm

PRT

&

The thrust force generated is then

N 94,520=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−= 2inletexit m/skg 1

N 10)m/s248kg/s)(564. (291.0)( VVmF &

qin

qouts

T

1

2

4

3 5

6


9-100

9-134 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


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 &&&&

ompressor:

&

s

T

1

2

4

3

Qi

5

6

·

C( )( ) ( )( )

( )( )( ) K 593.712K 291.9

kPa 751.2kPa 62.612

0.4/1.4/1

2

32

3

243

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

====

− kk

p

PP

TT

PrPP

ine:

Turb

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

K1098.2291.9593.714002345 =+−=+−= TTTT or

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

+−=⎯→⎯−

+−=

+=+

=⎯→⎯∆=−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

&&&&&

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

/sm 1000K568.21098.2KkJ/kg 1.0052

22

6 =⎟⎟⎠

⎞⎜⎜⎝

⎛−⋅=V or

( ) ( )( ) ( ) kW 13,670=⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

22aircraftinletexit/sm 1000

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

( ) ( ) ( )( )( )

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

&&

&&& (c)


9-101

9-135 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.


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


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

(steady) 0systemoutin

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅+=+=

−−=

−+−=

+=+

=

∆=−

−kk

p

p

TT

PP

cV

TT

VTTc

VVhh

VhVh

EE

EEE&&

&&&

s

T

1

2

4

3

Qin

5s

6

·

5

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


9-102

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

( )( )( ) ( )

kW 11,390=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=

22

aircraftinletexit

/sm 1000kJ/kg 1

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

VVVmW p && (b)

(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

&&

&&&



9-103

9-136 A turbojet aircraft that has a pressure rate of 9 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.


Analysis (a) Using variable specific heats for air,


Compressor:

2311.1kJ/kg 0.1629K 029

1

11

==⎯→⎯=

rPhT

( )( )

( )( )

5.568kJ/kg 1611.65.1067.07544

kJ/kg 5.1067kg/s 20

kJ/s 21,350

kJ/s 350,21kJ/kg 42,700kg/s 0.5HV

kJ/kg .0754408.112311.19

3

12

2323in

inin

fuelin

21

2

=⎯→⎯=+=+=⎯→⎯−=

===

==×=

=⎯→⎯===

rin

rr

Pqhhhhq

mQ

q

mQ

hPPP

P

&

&

&&

s

T

1

3

2

qin

4

5

Turbine:

kJ/kg 7.1357.16290.07544.616111234

4312outturb,incomp,

=+−=+−=

−=−⎯→⎯=

hhhh

hhhhww or

Nozzle:

( )

20

2/2/

kJ/kg .5688817.63915.568

024

25

45

255

244

outin


53

535

VVhh

VhVh

EE

EEE

hPP

PP rr

−+−=

+=+

=

∆=−

=⎯→⎯=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

&&

&&&

or

( ) ( )( ) m/s .6968kJ/kg 1

/sm 1000kJ/kg.568881357.72222

545 =⎟⎟⎠

⎞⎜⎜⎝

⎛−=−= hhV

( ) ( )( ) N 19,370=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−

2inletexit m/skg 1N 1m/s0968.6kg/s 20VVm& Brake force = Thrust =


9-104

9-137 Problem 9-136 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.


P_ratio =9 T_1 = 7 [C] T[1] = T_1+273 "[K]" P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 18.1 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.5 [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] "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))



9-105

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]

21232 21007 20788 20576 20369 20168 19972 19782 19596 19415 19238

23.68 23.22 22.78 22.35 21.94 21.55 21.17 20.8

20.45 20.1

19.77

1284 1307 1330 1352 1375 1398 1420 1443 1466 1488 1510

-20 -15 -10 -5 0 5

10 15 20 25 30

-20 -10 0 10 20 3019000

19450

19900

20350

20800

21250

T1 [C]

Bra

keFo

rce

[N]

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50

500

1000

1500

0x100

5x102

103

103

s [kJ/kg-K]

T [K

]

Air

855 kPa

95 kPa

1

2s

3

4s

5



9-106

9-138 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.


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


J / kg

J / kg

T h

T h

1 1

2 2

280 28013

700 713 27

= ⎯ →⎯ =

= ⎯ →⎯ =

K k

K k

.

.

( ) ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−=

+=++ 211in ( VhmQ &&

=

∆=−

22

222

21

22

12in

222

outin


/sm 1000kJ/kg 1

2m/s 300

13.28027.713kg/s 16kJ/s 000,15

2

)2/()2/

V

VVhhmQ

Vhm

EE

EEE

&&

&

&&

&&&

It gives

V 2 = 1048 m/s

Thus,

15,000 kJ/s

1

7°C 300 m/s16 kg/s

2

427°C

( ) ( )( ) N 11,968=−=−= m/s3001048kg/s 1612 VVmFp &


9-107

Second-Law Analysis of Gas Power Cycles

9-139 The process with the highest exergy destruction for an ideal Otto cycle described in Prob. 9-33 is to be determined.

Analysis From Prob. 9-33, qin = 582.5 kJ/kg, qout = 253.6 kJ/kg, T1 = 288 K, T2 = 661.7 K, T3 = 1473 K, and T4 = 641.2 K. The exergy destruction during a process of the cycle is


⎟⎠

⎜⎝

+−∆==sinksource

0gen0dest TTsTsTx ⎟

⎞⎜⎛ outin qq

ss of the cycle gives

st,1x (isentropic process)

v

P

4

1

3

2

s

T

1

2

4

3 qin

qout

Application of this equation for each proce

de 0=2-

Kk J/kg5746.00K 7.661K 1473ln)KkJ/kg 718.0(

lnln2

3

2

31423

⋅=+⋅=

+=−=−v

vv R

TT

cssss

kJ/kg 51.59=

⎟⎠

⎞⎜⎝

⎛ −⋅=

⎟⎟⎠

⎞⎜⎝

2303-dest,2 TssT ⎜

⎛−−=

K 1473kJ/kg 582.5KkJ/kg 0.5746K) 288(

source

inqx

(isentropic process)

0=4-dest,3x

kJ/kg 88.12=

⎟⎠

⎞⎜⎝

⎛ +⋅−=

⎟⎟⎠

⎜⎜⎝

+−=sink

out4101-dest,4 T

ssTx⎞⎛

K 288kJ/kg 253.6KkJ/kg 0.5746K) 288(

q

The largest exergy destruction in the cycle occurs during the heat-rejection process.


9-108

9-140 The exergy loss of each process for an ideal dual cycle described in Prob. 9-58 is to be determined.

Analysis From Prob. 9-58, qin,x-3 = 114.6 kJ/kg, T1 = 291 K, T2 = 1037 K, Tx = 1141 K, T3 = 1255 K, and T4 = 494.8 K. Also,

kJ/kg 74.67K)10371141)(KkJ/kg 718.0()( 22,in =−⋅=−=− TTcq xx v


v

P

4

1

2

3

qout

x

qinThe exergy destruction during a process of the cycle is


⎟⎟⎠

⎜⎜⎝

+−∆==sink

out

source

in0gen0dest TT

sTsTx ⎞⎛ qq

Application of this equation for each process of the cycle gives

KkJ/kg 1158.0kPa 90kPa 5148ln)KkJ/kg 287.0(

K 291K 1037ln)KkJ/kg 005.1(

lnln1

2

1

212

⋅=

⋅−⋅=

−=−PP

RTT

css p

kJ/kg 33.7=⋅=−= K)kJ/kg K)(0.1158 291()( 1202-1 dest, ssTx

KkJ/kg 06862.00K 1037K 1141ln)KkJ/kg 718.0(

lnln22

2

⋅=+⋅=

+=−v

vv

xxx R

TT

css

kJ/kg 2.65=⎟⎠

⎞⎜⎝

⎛ −⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

K 1255kJ/kg 74.67KkJ/kg 0.06862K) 291(

source

-2 in,20-2 dest, T

qssTx x

xx

KkJ/kg 09571.00K 1141K 1255ln)KkJ/kg 005.1(lnln 33

3 ⋅=−⋅=−=−xx

px PP

RTT

css

kJ/kg 1.28=⎟⎠

⎞⎜⎝

⎛ −⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

K 1255kJ/kg 114.6KkJ/kg 0.09571K) 291(

source

3- in,303- dest, T

qssTx x

xx

KkJ/kg 1339.01.1

18ln)KkJ/kg 287.0(K 1255K 8.494ln)KkJ/kg 718.0(

lnlnlnln3

4

3

4

3

434

⋅=⋅+⋅=

+=+=−crrR

TT

cRTT

css vv v

v

kJ/kg 39.0=⋅=−= K)kJ/kg K)(0.1339 291()( 3404-3 dest, ssTx

KkJ/kg 3811.00K 8.494

K 291ln)KkJ/kg 718.0(lnln4

1

4

141 ⋅−=+⋅=+=−

v

vv R

TT

css

kJ/kg 35.4=⎟⎠

⎞⎜⎝

⎛ +⋅−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

K 291kJ/kg 146.3KkJ/kg 0.3811K) 291(

sink

out4101-dest,4 T

qssTx


9-109

9-141 The exergy loss of each process for an air-standard Stirling cycle described in Prob. 9-74 is to be determined.

Analysis From Prob. 9-74, qin = 1275 kJ/kg, qout = 212.5 kJ/kg, T1 = T2 = 1788 K, T3 = T4 = 298 K. The exergy destruction during a process of the cycle is

⎟⎟⎠

⎞⎜⎜⎝

⎛+−∆==

sink

out

source

in0gen0dest T

qT

qsTsTx



KkJ/kg 7132.0)12ln()KkJ/kg 287.0(0

lnln1

2

1

212

⋅=⋅+=

+=−v

vv R

TT

css

0kJ/kg 0.034 ≈=⎟⎠

⎞⎜⎝

⎛ −⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

K 1788kJ/kg 1275

KkJ/kg 0.7132K) 298(

source

in1202-1 dest, T

qssTx

s

T

3

2 qin

qout

4

1

KkJ/kg 7132.012 ⎠⎝

1ln)KkJ/kg 287.0(0

lnln3

4

3

434

⋅−=⎟⎞

⎜⎛⋅+=

+=−v

vv R

TT

css

0kJ/kg 0.034 ≈−=⎟⎠

⎞⎜⎝

⎛ +⋅−=

⎟⎟⎠

⎞⎛ outq⎜⎜⎝

+−=

K 298kJ/kg 212.5

KkJ/kg 0.7132K) 298(

sink1202-1 dest, T

ssTx

These results are not surprising since Stirling cycle is totally reversible. Exergy destructions are not calculated for processes 2-3 and 4-1 because there is no interaction with the surroundings during these processes to alter the exergy destruction.


9-110

9-142 The exergy destruction associated with each of the processes of the Brayton cycle described in Prob. 9-82 is to be determined.

Analysis From Prob. 9-82, qin = 698.3 kJ/kg, qout = 487.9 kJ/kg, and


Thus,

KkJ/kg 2.66807kJ/kg 783.04

KkJ/kg 3.21751K 1240

KkJ/kg 2.44117kJ/kg .60626

KkJ/kg .685151K 295

44

33

22

11

⋅=⎯→⎯=

⋅=⎯→⎯=

⋅=⎯→⎯=

⋅=⎯→⎯=

o

o

o

o

sh

sT

sh

sT

( )

( ) ( ) ( )( )

( )

( )

( ) ( ) ( )( )

( ) kJ/kg 183.2

kJ/kg 34.53

kJ/kg 105.4

kJ/kg 29.51

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+−==

=⋅−−=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−==

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+−==

=⋅−−=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−==

K 310kJ/kg 487.9

2.668071.68515K 310

ln

1/10lnKkJ/kg 0.2873.217512.66807K 310

ln

K 1600kJ/kg 698.32.441173.21751K 310

ln

10lnKkJ/kg 0.2871.685152.44117K 310

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


9-111

9-143 Exergy analysis is to be used to answer the question in Prob. 9-87.

Analysis From Prob. 9-87, T1 = 288 K, T2s = 585.8 K, T2 = 618.9 K, T3 = 873 K, T4s = 429.2 K, T4 = 473.6 K, rp = 12. The exergy change of a flow stream between an inlet and exit state is given by

)(0 ieie ssThh −−−=∆ψ

4s

s

T

1

2s

4

3 qin

qout

873 K

288 K

2

This is also the expression for reversible work. Application of this equation for isentropic and actual compression processes gives

KkJ/kg 0003998.0

)12ln()KkJ/kg 287.0(K 288K 8.585ln)KkJ/kg 005.1(

lnln1

2

1

212

⋅=

⋅−⋅=

−=−PP

RTT

css sps

kJ/kg 2.299)KkJ/kg 0003998.0)(K 288(K)2888.585)(KkJ/kg 005.1(

)()( 12s01221 rev,

=⋅−−⋅=

−−−=− ssTTTcw sps

KkJ/kg 05564.0)12ln()KkJ/kg 287.0(K 288K 9.618ln)KkJ/kg 005.1(

lnln1

2

1

212

⋅=⋅−⋅=

−=−PP

RTT

css p

)()( 1201221 rev, −−−=− ssTTTcw p


kJ/kg 5.316)KkJ/kg 05564.0)(K 288(K)2889.618)(KkJ/kg 005.1( =⋅−−⋅=

inimum work that must be supplied to the compressor by The irreversibilities therefore increase the m

kJ/kg 17.3=−=−=∆ −− 2.2995.31621 rev,21 rev,Crev, swww

Repeating the calculations for the turbine,

KkJ/kg 0003944.0)12ln()KkJ/kg 287.0(K2.429

K 873ln)KkJ/kg 005.1(

lnln4

3

4

343

⋅=⋅−⋅=

−=−PP

RTT

csss

ps

v, −−−=− ssps ssTTTcw

)()( 4304343 re

kJ/kg 9.445)KkJ/kg 0003944.0)(K 288(K)2.429873)(KkJ/kg 005.1( =⋅−−⋅=

KkJ/kg 09854.0)12ln()KkJ/kg 287.0(K6.473

K 873ln)KkJ/kg 005.1(

lnln4

3

4

343

⋅−=⋅−⋅=

−=−PP

RTT

css p

kJ/kg 8.429)KkJ/kg 09854.0)(K 288(K)6.473873)(KkJ/kg 005.1(

)()( 4304343 rev,

=⋅−−−⋅=

−−−=− ssp ssTTTcw

kJ/kg 16.1=−=−=∆ −− 8.4299.44543 rev,43 rev,Trev, www s

Hence, it is clear that the compressor is a little more sensitive to the irreversibilities than the turbine.


9-112

9-144 The total exergy destruction associated with the Brayton cycle described in Prob. 9-108 and the exergy at the exhaust gases at the turbine exit are to be determined.

Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).

Analysis From Prob. 9-108, qin = 480.82, qout = 372.73 kJ/kg, and

s

T

1

2s 4s

3qin1150 K

310 K

5

6

42

KkJ/kg 08156.2kJ/kg 738.43

KkJ/kg 69407.2kJ/kg 14.803

KkJ/kg 12900.3K 1150

KkJ/kg 42763.2kJ/kg 26.618

KkJ/kg 1.73498K 310

55

44

33

22

11

⋅=⎯→⎯=

⋅=⎯→⎯=

⋅=⎯→⎯=

⋅=⎯→⎯=

⋅=⎯→⎯=

o

o

o

o

o

sh

sh

sT

sh

sT

and, from an energy balance on the heat exchanger,


⋅=⎯→⎯

=−−=⎯→⎯−=−os

hhhhh

Thus,

KkJ/kg 2.52861

kJ/kg 97.682)26.61843.738(14.803

6

66425

( )

( ) ( ) ( )( )

( )

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

( )( )

( )

( ) kJ/kg 142.6=⎟⎟⎠

⎜⎜⎝

+−K 290

2.528611.73498K 290

kJ/kg 58.09

kJ/kg 4.37

kJ/kg 35.83

kJ/kg 38.91

⎞⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+−==

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−−==

=−+−=

−+−=−+−==

=⋅−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−==

=⋅−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−==

kJ/kg 372.73

ln

K 1500kJ/kg 480.822.608153.12900K 290

ln

2.694072.528612.427632.60815K 290

1/7lnKkJ/kg0.2873.129002.69407K 290

ln

7lnKkJ/kg 0.2871.734982.42763K 290

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

oting tha 0 = h@ 290 K = 290.16 kJ/kg, the stream exergy at the exit of the regenerator (state 6) is determined from

N t h

( ) ( ) 06

026

060066 2gzVssThh ++−−−=φ

where

KkJ/kg 0.793631.734982.52861ln0

1

6161606 ⋅=−=−−=−=−

PP

Rssssss oo

Thus,

( )( ) kJ/kg 162.7=⋅−−= KkJ/kg 0.79363K 2900.162997.6826φ


9-113

9-145 Prob. 9-144 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.


"Given" T[1]=310 [K] P[1]=100 [kPa] Ratio_P=7 P[2]=Ratio_P*P[1] T[3]=1150 [K] eta_C=0.75 eta_T=0.82 epsilon=0.65 T_H=1500 [K] T0=290 [K] P0=100 [kPa] "Analysis for Problem 9-154" q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid$, P=P[2], h=h[2]) s[4]=entropy(Fluid$, h=h[4], P=P[4]) s[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x_total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_destroyed_53+x_destroyed_61 x6=h[6]-h[0]-T0*(s[6]-s[0]) "since state 0 and state 1 are identical" "Analysis for Problem 9-116" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) "(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4]



9-114

w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in

Ratio_P xtotal

[kJ/kg] x6

[kJ/kg] 6 7 8 9

10 11 12 13 14

270.1 280

289.9 299.5 308.8 317.8 326.6 335.1 343.3

137.2 143.5 149.6 155.5 161.1 166.6 171.9 177.1 182.1

6 7 8 9 10 11 12 13 14130

150

170

190

210

230

250

270

290

310

330

350

RatioP

x [k

J/kg

]

xtotal,dest

x6



9-115

9-146 The exergy loss of each process for a regenerative Brayton cycle with three stages of reheating and intercooling described in Prob. 9-126 is to be determined.

Analysis From Prob. 9-126,

s

T

3

4

1

9

8

2

5

10

11

67

12

13

14

rp = 4, qin,7-8 = qin,9-10 = qin,11-12 = 300 kJ/kg,

qout,14-1 = 206.9 kJ/kg, qout,2-3 = qout,4-5 = 141.6 kJ/kg,

T1 = T3 = T5 = 290 K , T2 = T4 = T6 = 430.9 K

K 9.495 K, 7.585 K, 4.870 K, 9.571 K, 8.849

K 3.551 K, 2.819 K, 7.520

1413

121110

987

=====

===

TTTTT

TTT

The exergy destruction during a process of a stream from an inlet state to exit state is given by


⎟⎠

⎜⎝

+−−==sinksource

0gen0dest TTssTsTx ie ⎟

⎞⎜⎛ outin qq


0kJ/kg 0.03 ≈=⎥⎦⎤

⎢⎣⎡ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛−===

)4ln()287.0(290

430.9(1.005)ln)290(

lnln1

2

1

206-5 dest,4-3 dest,2-1 dest, P

PR

TT

cTxxx p

kJ/kg 32.1=⎥⎦⎤

⎢⎣⎡ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

4.8703000

520.7819.2(1.005)ln)290(lnln

source

8-in,7

7

8

7

808-7 dest, T

qPP

RTT

cTx p

kJ/kg 26.2=⎥⎦⎤

⎢⎣⎡ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

4.8703000

551.3849.8(1.005)ln)290(lnln

source

10-in,9

7

8

9

10010-9 dest, T

qPP

RTT

cTx p

kJ/kg 22.5=⎥⎦⎤

⎢⎣⎡ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

4.8703000

571.9870.4(1.005)ln)290(lnln

source

12-in,11

11

12

11

12012-11 dest, T

qPP

RTT

cTx p

0kJ/kg 0.05 ≈−=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

41ln)287.0(

819.2551.3(1.005)ln)290(lnln

8

9

8

9098- dest, P

PR

TT

cTx p

0kJ/kg 0.04 ≈−=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

41ln)287.0(

849.8571.9(1.005)ln)290(lnln

10

11

10

11011-10 dest, P

PR

TT

cTx p

0kJ/kg 0.08 ≈−=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

41ln)287.0(

870.4585.7(1.005)ln)290(lnln

12

13

12

13013-12 dest, P

PR

TT

cTx p

kJ/kg 50.6=⎥⎦⎤

⎢⎣⎡ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

2909.2060

495.9290(1.005)ln)290(lnln

sink

1-out,14

14

1

14

101-14 dest, T

qPP

RTT

cTx p

kJ/kg 26.2=⎥⎦⎤

⎢⎣⎡ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−==

2906.1410

430.9290(1.005)ln)290(lnln

sink

3-out,2

2

3

2

305-4 dest,3-2 dest, T

qPP

RTT

cTxx p

kJ/kg 6.66=⎥⎦⎤

⎢⎣⎡ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=∆+∆= −−

585.7495.9(1.005)ln

430.9520.7(1.005)ln)290(

lnln)(13

14

6

701413760regendest, T

Tc

TT

cTssTx pp


9-116

9-147 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. 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

1

Combustion chamber

Turbine

23

4

Compress.

100 kPa30°C

Diesel fuel

700 kPa260°C

0.881=−

=−

=K)303533(12 TTCη −− K)3036.505(12 TT s

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

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

kJ/kg)(0.9 00kg/s)(42,0 21.0(== qmQ η&&

kg/s 6.12kg/s 6.12 =+=+=+=+= aafa

mmmmm

&&&&&

HVin =cf

perature at the exit of combustion chamber is

3323 =⎯→⎯ Tp

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

t

kW 85557)

The tem

)K533kJ/kg.K)( 3kg/s)(1.09 81.12(kJ/s 8555)( −=⎯→⎯−= TTTcmQ && K 1144in

( ) K 7.685kPa 70

K 11443

34 ⎜⎝

=⎟⎠

⎜⎝

=s PTT

0kPa 100 1)/1.357-(1.357/)1(

4 =⎟⎠⎞⎛⎟

⎞⎜⎛

− kkP

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

=−=−= TTcmW &&

t WWW

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, p

&&ne& kW 2287=−=−= 31685455inC,outT,

0.581===kW 5455kW 3168inC,

bw WW

r &

&

outT,

(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

0.364===735.0267.0

max

thII η

ηη and


9-117

9-148 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.



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

V

V

VV===⎯→⎯

+=⎯→⎯

+= 2

33

m 00012.0m 0018.0

16

43

1 m 00192.00018.000012.0 VVVV ==+=+= dc

1

Qin2

3

4

P

V

Qout

xProcess 1-2: Isentropic compression

( )( )

( )( ) kPa 385916kPa 95

K 7.87016K 343

1.336

2

112

1-1.3361

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 1692kPa 3859kPa 7500K) 7.870(

22 ===

PP

TT xx

kJ/kg 6.702K)7.8701692(kJ/kg.K) (0.855)( 2-2 =−=−= TTcq xx v

K)1692(kJ/kg.K) (0.855kJ/kg 6.702)( TTTTcqq xpxx

=

K 2308=⎯→⎯−=⎯→⎯−==− 3333-2

(b) 6.7023-2in +=+= − xx qqq kJ/kg 14056.702

3333 m 0001636.0

K 1692K 2308)m 00012.0( ===

xx T

TVV


( )

( ) kPa 4.279


m 0.001924 ⎝⎠⎝Vm 0.0001636kPa 7500

K 1009m 92m 0.0001636K 2308

1.336

3

33

3

1-1.336

3

313

34

=⎟⎟⎠

⎞⎜⎜⎛

=⎟⎟⎞

⎜⎜⎛

=

=⎟⎟⎠

⎞⎜⎜⎛

=⎟⎟⎞

⎜⎜⎛

=−k

PP

TT

V

V

rocess 4-1: constant voume heat rejection.

0.0014 ⎝⎠⎝k

V

4

P

( ) ( )( ) kJ/kg 3.569K3431009KkJ/kg 0.85514out =−⋅=−= TTcq v

The net work output and the thermal efficiency are

kJ/kg 835.8=−=−= 3.5691405outinoutnet, qqw

59.5%==== 5948.0kJ/kg 1405kJ/kg 835.8

in

outnet,th q

wη


9-118

(c) The mean effective pressure is determined to be

( )( )kg 0.001853

K 343K/kgmkPa 0.287)m 92kPa)(0.001 95(

3

3

1

11 =⋅⋅

==RTP

mV

kPa 860.4=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

0001.000192.0(MEP

21 VV kJmkPa

m)2kJ/kg)kg)(835.8 (0.001853 3

3outnet,mw

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

kW 28.39=⎟⎠⎞

⎜⎝

==60rev/cycle) 2(

kJ/kg) kg)(835.8 001853.0(2netnet mwW& ⎛

s min 1(rev/min) 2200n&

(e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal ef ency (Carnot efficiency). We take the dead state temperature and pressure to be 25ºC and 100 kPa.

Note that there are two revolutions in one cycle in four-stroke engines.

fici

8709.0K 2308

K )27325(113

0max =

+−=−=

TT

η

and

68.3%==== 683.08709.05948.0η

max

th

ηη

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

II

( )

( )( ) kJ/kg 0.285100

4.279ln)kJ/kg.K 287.0(298

1009kJ/kg.K)ln 142.1()298(29810090.855

lnln)(0

4

0

4004040044

=⎥⎦⎤

⎢⎣⎡ −−−=

⎥⎦

⎤⎢⎣

⎡−−−=−−−=

PP

RTT

cTTTcssTuux pv

kW 9.683=⎟⎠⎞

⎜⎝⎛==

s 60min 1

rev/cycle) 2((rev/min) 2200kJ/kg) kg)(285.0 001853.0(

244nmxX&&



9-119

Review Problems

9-149 An Otto cycle with a compression ratio of 7 is considered. The thermal efficiency is to be determined using constant and variable specific heats.

Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible.


Analysis (a) Constant specific heats:


54.1%==−=−=−− 7 11.41th kr

0.54081111η

properties from Table A-17)

Process 1-2: isentropic compression.

v

P

4

1

3

2(b) Variable specific heats: (using air

1.688kJ/kg 48.205

K 2881

11 =

=⎯→⎯=

r

uT

v

kJ/kg 62.4473.98)1.688(112 === vvv

v7 222

12 =⎯→⎯= u

r rrr v

rocess 2-3: v = constant heat addition.

2kJ/kg 51.998

23

3

=−=−=

=

uuq

u

P

1K 1273

33 =

⎯→⎯=Trv

kJ/kg .8955062.44751.998045.

in


kJ/kg 54.47532.84)045.12)(7( 4333

44 =⎯→⎯==== u r rrr vv

v

vv

rocess 4-1: v = constant heat rejection.

P

kJ/kg 06.27048.20554.47514out =−=−= uuq

51.0%==−=−= 0.5098kJ/kg 550.89kJ/kg 270.0611

in

outth q

qη


9-120

9-150 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limits. The net work is to be determined using constant and variable specific heats.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible.


Analysis (a) Constant specific heats:

K 2.549K)(12) 270( 0.4/1.4/)1(12 === − kk

prTT

s

T

1

2

4

3 qin

qout

800 K

270 K

K 3.393121K) 800(1 0.4/1.4/)1(

34 =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

prTT

)(

)()(

2143

1243

compturbnet

TTTTc

TTcTTc

www

p

pp

(b) Variable specific heats: (using air properties from Table A-17)

kJ/kg 128.1=K)2.5492703.393800)(kJ/kg·K 005.1( −+−=

−+−=

−−−=

−=

9590.0kJ/kg 11.270

K 2701

11 =

=⎯→⎯=

rPh

T

kJ/kg 46.55051.11)9590.0)(12( 211

22 =⎯→⎯=== hP

PP

P rr

kJ/kg 04.406979.3)75.47(121

75.47kJ/kg 95.821

K 800

433

44

3

33

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==

⎯→⎯=

hPPP

P

Ph

T

rr

r

kJ/kg 135.6=−−−=

−−−=

−=

)11.27046.550()04.40695.821()()( 1243

compturbnet

hhhh

www



9-121

9-151 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount of power produced per cylinder per mechanical and per thermodynamic cycle is to be determined.

Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have

(a)

cycle)mech kJ/cyl 8.16(= hp 1Btu/min 42.41

rev/min) (1200cylinders) (16hp 3500

cycles) mechanical of (No.cylinders) of (No.producedpower Total

mechanical

⋅⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

cycle mechBtu/cyl 7.73

w

(b)

cycle) thermkJ/cyl 16.31(= cycle thermBtu/cyl 15.46

hp 1Btu/min 42.41

rev/min) (1200/2cylinders) (16hp 3500

cycles) amic thermodynof (No.cylinders) of (No.producedpower Total

micthermodyna

⋅⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=w

ified temperature limits is considered. The pressure ratio for

2 The air-standard assumptions are applicable. 3 Kinetic and potential nergy cha ges are negligible. 4 Air is an ideal gas with constant specific heats.

is k =1.4 (Table A-2).

Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the ompressor and turbine exit can be expressed as

9-152 A simple ideal Brayton cycle operating between the specwhich the compressor and the turbine exit temperature of air are equal is to be determined.

Assumptions 1 Steady operating conditions exist.e n

Properties The specific heat ratio of air

c( )


( )( )

( ) ( ) kk

p

kk

kk

rT

PPTT

/1

3

/1

3

434

/1

1

1−−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎠⎝

Setting T2 = T4 and solving for rp gives

TkkprT

PPTT /1

12

12−=⎟⎟

⎞⎜⎜⎛

=

s1

2

4

3 qin

qout

T3

T1( )16.7=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− 1.4/0.812/

1

3

K 300K 1500

kk

p TT

r

Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 16.7.


9-122

9-153 A four-cylinder spark-ignition engine with a compression ratio of 8 is considered. The amount of heat supplied

ons are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is

e A-1). The properties of air are given in Table A-17.


per cylinder, the thermal efficiency, and the rpm for a net power output of 60 kW are to be determined.

Assumptions 1 The air-standard assumptian ideal gas with variable specific heats.

Properties The gas constant of air is R = 0.287 kJ/kg·K (Tabl


( )

kJ/kg .295642 =⎯→⎯ u

50.543.5725.10

11

3.572kJ/kg 221.25K 031

1

2

11

112

1

====

==⎯→⎯=

r

uT

rrr

r

vvv

vv

v

rocess 2- v = constant heat addition. P 3:

( )( )( )( )

( ) ( )( ) kJ 0.5336=−×=−=

×=

=

−

−

kJ/kg.295641775.3kg 104.406

kg 10.4064K 310K

m 0.0004kPa 98

356.2kJ/kg .3

423in

43

3

uumQ

P

r

V

v

rocess 3-4: isentropic expansion.

v

P

4

1

3

2

Qin

Qout

1800 K

=⎯→⎯= 1775K 0021 33 uT

⋅⋅==

/kgmkPa 0.287 31

11

RTm

(b) P

( )( ) 05.76474.24356.25.10 44

334=⎯→⎯==== ur rrr vv

vv kJ/kg

3v


( ) ( )( )

55.2%====

=−=−=

=−×=−=

5517.0kJ 0.5336kJ 0.2944

kJ 0.29442392.05336.0

kJ 0.2392kJ/kg.25221764.05kg 104.406

in

netth

outinnet

-414out

QW

QQW

uumQ

η

rpm 4586=⎟⎟⎠

⎞⎜⎜⎝

⎛×

==min 1

s 60kJ/cycle) 0.2944(4

kJ/s 45rev/cycle) 2(2cylnet,cyl

net

WnW

n&

& (c)

Note that for four-stroke cycles, there are two revolutions per cycle.


9-123

9-154 Problem 9-153 is reconsidered. The effect of the compression ratio net work done and the efficiency of the s for the cycle are to be plotted.


K]

ber of cyclinders"

nit mass." ession"

2]) v[1]/T[1]

fer (s=const.) with work input"

T=T[1]) eat addition"

3]) 2]*v[2]/T[2]}

ro for v=const, heat is added"

nergy(air,T=T[2]) ion"

fer (s=const) with work output"

T=T[3])

ro for v=const; heat is rejected"

der is found from the volume of the cylinder:"

et=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermal cycle"/2"mechanical cycles"

cycle is to be investigated. Also, the T-s and P-v diagram

A

"Input Data" T[1]=(37+273) [P[1]=98 [kPa] T[3]= 2100 [K]

(L, m^3) V_cyl=0.4 [L]*Convertr_v=10.5 "Compression ratio"

45 [kW] W_dot_net =_cyl=4 "numN

v[1]/v[2]=r_v

one per u"The first part of the solution is d1-2 is isentropic compr"Process

s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]

T=T[2], v=v[s[2]=entropy(air, P[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] "Conservation of energy for process 1 to 2: no heat trans w_in = DELTAu_12

)-intenergy(air,DELTAu_12=intenergy(air,T=T[2]e h"Process 2-3 is constant volum

T=T[3], P=P[s[3]=entropy(air, /T[3]=P[{P[3]*v[3]

P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3: the work is zeq_in = DELTAu_23

])-inteDELTAu_23=intenergy(air,T=T[33-4 is isentropic expans"Process

s[4]=entropy(air,T=T[4],P=P[4]) s[4]=s[3] P[4]*v[4]/T[4]=P[3]*v[3]/T[3] {P[4]*v[4]=R*T[4]} "Conservation of energy for process 3 to 4: no heat trans - w_out = DELTAu_34

34=intenergy(air,T=T[4])-intenergy(air,DELTAu_"Process 4-1 is constant volume heat rejection" v[4]=v[1] "Conservation of energy for process 2 to 3: the work is ze- q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) w_net = w_out - w_in

/q_in*Convert(, %) "Thermal efficiency, in percent" Eta_th=w_net"The mass contained in each cylinV_cyl=m*v[1] "The net work done per cycle is:" W_dot_n



9-124

rv ηth

[%] wnet

[kJ/kg] 5 6 7 8 9 10 11

42 45.55 48.39 50.74 52.73 54.44 55.94

568.3 601.9 625.7 642.9 655.5 664.6 671.2

10-2 10-1 100 101 102101

102

103

104

v [m3/kg]P

[kPa

]

310 K 2100 K

1

2

3

4

s = const

Air Otto Cycle P-v Diagram

4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50

500

1000

1500

2000

2500

3000

s [kJ/kg-K]

T [K

]

98 kPa

6971 kPa

Air Otto Cycle T-s Diagram

1

2

3

4

v = const

5 6 7 8 9 10 1142

44

46

48

50

52

54

56

rv

ηth

[%

]

5 6 7 8 9 10 11

560

580

600

620

640

660

680

rv

wne

t [k

J/kg

]



9-125

9-155 An ideal gas Carnot cycle with helium as the working fluid is considered. The pressure ratio, compression ratio, and minimum temperature of the energy source are to be determined.

Assumptions 1 Kinetic and potential energy changes are negligible. 2 Helium is an ideal gas with constant specific heats.

Properties The specific heat ratio of helium is k = 1.667 (Table A-2a).

Analysis From the definition of the thermal efficiency of a Carnot heat engine,

s

T

3

2qin

qout4

1TH

288 K

K 576=−+

=−

=⎯→⎯−=0.501

K 273)(151

1Carnotth,

Carnotth, ηη L

HH

L TT

TT

An isentropic process for an ideal gas is one in which Pvk remains constant. Then, the pressure ratio is


5.6=⎟⎠

⎜⎝

=⎟⎟⎠

⎜⎜⎝

=11 K 288TP

5⎞⎛⎞⎛ −− )1667.1/(667.1)1/(2 K 576

kkTP

ased on the process equation, the compression ratio is

2

B

2.83==⎟⎟⎠

⎞⎜⎜⎝

⎛= 667.1/1

/1

1

2

2

1 )65.5(k

PP

v

v

9-156 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined.

Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air isan ideal gas with constant specific heats. 4 The combustion efficiency is 100 percent.

cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-

Analysis The heat input to the cycle for 0.043 grams of fuel consumption is 3

HVfuelin =× −

he thermal efficiency is then

Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K,2).

kJ 505.1kJ/kg) kg)(43,000 10035.0(== qmQ

T

6645.0kJ 1.505

kJ 1netth ===

QW

η in

rom the definition of thermal efficiency, we obtain the required compression ratio to be v

P

4

1

3

2 qout

qin

F

15.3=−

=−

=⎯→⎯−=−−− )14.1/(1)1/(1

th1th

)6645.01(1

)1(111

kkr

r ηη


9-126

9-157 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined.


Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given in Table A-17.


v

P

4

1

3

2

qin

qout

( )

( ) ( ) kPa 2129kPa 98K 300K 708.3

9.2

kJ/kg 9.518K 3.70852.672.621

2.911

2.621kJ/kg 214.07K 300

11

2

2

12

1

11

2

22

2

21

2

11

112

1

=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯=

==⎯→⎯====

==⎯→⎯=

PTT

PT

PT

P

uT

r

uT

rrr

r

v

vvv

vvv

vv

v


( )( )

kJ/kg609.8 9.5187.1128

593.8kJ/kg 1128.7K 1416.63.70822

23

3222

323 PTT

3223

3

=−=−=

=

=⎯→⎯====⎯→⎯=

uuq

uTTP

TPP

in

rv

vv

(b) Process 3-4: isentropic expansion.

3

( )( ) kJ/kg 487.7506.79593.82.9 43

433


4r =⎯→⎯==== ur rr vv

vv

ess 4-1: v = constant heat rejection.

14t

=−=−==−=−=

qqwuuq

(c)

v

Proc

kJ/kg336.1 7.2738.609kJ/kg 273.707.21475.487

outin

net

ou

55.1%===kJ/kg 609.8kJ/kg 336.1

in

netth q

wη

( )( )

( ) ( )( )kPa429

kJ 1mkPa 1

1/9.21/kgm 0.879kJ/kg 336.1

/11MEP

/kgm 0.879kPa 98

K 300K/kgmkPa 0.287

3

31

net

21

net

max2min

33

1

11max

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

=⋅⋅

===

rww

r

PRT

vvv

vvv

vv (d)


9-127

9-158 An ideal Otto cycle with air as the working fluid with a compression ratio of 9.2 is considered. The amount of heat transferred to the air, the net work output, the thermal efficiency, and the mean effective pressure are to be determined.




Analysis (a) Process 1-2 is isentropic compression:

( )( )

( ) ( ) kPa 2190kPa 98K 300K 728.8

9.2

K 728.89.2K 300

11

2

2

12

1

11

2

22

0.41

2

112

=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

PTT

PT

PT

P

TTk

v

vvv

v

v

v

P

4

1

3

2

qin

qout


( )( )

( ) ( )( ) kJ/kg523.3 K728.81457.6KkJ/kg 0.718

K 457.618.72822

2323

222

323 PTT

3223

=−⋅=−=−=

====⎯→⎯=

TTcuuq

TTP

TPP

in v

vv

(b) Process 3-4: isentropic expansion.

3

( ) K 600.09.21K 1457.6

0.413 ⎛⎞⎛

−kv

43 =⎟

⎠⎞

⎜⎝

=⎟⎟⎠

⎜⎜⎝

= TTv

ess 4-1: v = constant heat rejection.

4

Proc

( ) ( )( )

kJ/kg307.9 4.2153.523

kJ/kg 215.4K300600KkJ/kg 0.718

outinnet

1414out

=−=−=

=−⋅=−=−=

qqw

TTcuuq v

(c) 58.8%===kJ/kg 523.3kJ/kg 307.9

in

netth q

wη

( )( )

( ) ( )( )kPa393

kJ 1mkPa 1

1/9.21/kgm 0.879kJ/kg 307.9

/11MEP

/kgm 0.879kPa 98

K 300K/kgmkPa 0.287

3

31

net

21

net

max2min

33

1

11max

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅

−=

−=

−=

==

=⋅⋅

===

rww

r

PRT

vvv

vvv

vv (d)


9-128

9-159 An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K, and k = 1.4 (Table A-2).

Analysis The mass of air is

( )( )( )( )

kg 001726.0K 323K/kgmkPa 0.287

m 101600kPa 1003

3-6

1

11 =⋅⋅

×==

RTP

mV


Process 1-2: isentropic compression.

( )( ) K 2.92814K 3232

12 =⎟⎟⎠

⎜⎜⎝

= TTV

0.41

1 =⎞⎛

−kV

rocess 2-x: v = constant heat addition,

)( ) K 1412K2.928KJ/kg x

22in,2

=⎯→⎯−⋅

−=−=−

x

xxx

TT

TTmcuumQ v

rocess x- P = constant heat addition.

v

P

4

1

2

31.1 kJ

Qout

x

0.6 kJ

P

( )( k 0.718kg 0.001726kJ 0.6 =

( ) ( )

P 3:

( ) ( )( )( )( )

1.449K 1412K 2046

K 2046K1412KkJ/kg 1.005kg 0.001726kJ 1.1

3

3

3

33 ==⎯→⎯=x

cx

xx rTT V

33

33in,3

==

=⎯→⎯−⋅=

−−

x

xpxx

TTPP

TT

TT

VVV

rocess 3- isentropic expansion.

=−= mchhmQ

P 4:

( ) K 8.82514

1.449K 2046449.1449.1 0.41

3

1

4

13

1

4

334 =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−− kkk

rTTTT

VV

VV


( ) ( )

( )( )( )

63.3%==−=−=

=−⋅=−=−=

633.0kJ 1.7

kJ 0.623111

kJ 0.6231K323825.8KkJ/kg 0.718kg 0.001726

in

outth

1414out

QQ

TTmcuumQ

η

v


9-129

9-160 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined.


s

T

3

2

qin = 900 kJ/kg

qout

4

1 1800 K

350 K

Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

Analysis (a) The entropy change during process 1-2 is

KkJ/kg 0.5K 1800

kJ/kg 9001212 ⋅===−

HTq

ss

and

( )

( )( ) kPa 5873=⎟⎟⎠

⎞⎜⎜⎝

⎛===⎯→⎯=

=⎯→⎯⋅=⋅⎯→⎯+=− ln1

12 Tcss v

K 350K 800

5.710kPa 200

710.5lnKkJ/kg 0.287KkJ/kg 0.5ln

3

1

1

23

3

1

1

331

1

11

3

33

1

2

1

2

1

20

2

TT

PTT

PPT

PT

P

RT

v

v

v

vvv

v

v

v

v

v

v

(b) The net work output is

1

( ) kJ/kg 725=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−== kJ/kg 900

K 1800K 350

11 ininthnet qTT

qwH

Lη



9-130

9-161 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined.




For rp = 6,

s

T

1

2

4

3 qin

qout

2′

3′ Analysis The properties at various states are

T h

P

T h

P

r

r

1 1

3 3

1

3

1 386

330 9

= ⎯ →⎯ ==

= ⎯ →⎯ ==

300 K 300.19 kJ / kg

1300 K 1395.97 kJ / kg

.

.

( )( )

( )

%9.37kJ/kg 894.57kJ/kg 339.46

kJ/kg 339.4611.55557.894kJ/kg 555.1119.3003.855

kJ/kg 894.5740.50197.1395

kJ/kg 855.315.559.33061

kJ/kg 501.40316.8386.16

in

netth

outinnet

outw

14

23in

43

4

21

2

34

12

===

=−=−==−=−=

=−=−=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

=⎯→⎯===

qw

qqhhqhhq

hPPP

P

hPPP

P

rr

rr

η

rp = 12,

For

( )( )

( )

%5.48kJ/kg 785.37kJ/kg 380.96

kJ/kg 380.9641.40437.785kJ/kg 404.4119.3006.704

kJ/kg 785.3760.61097.1395

kJ/kg 704.658.279.330121

kJ/kg 610.663.16386.112

in

netth

outinnet

14out

23in

43

4

21

2

34

12

===

=−=−==−=−=

=−=−=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

=⎯→⎯===

qw

qqwhhqhhq

hPPP

P

hPPP

P

rr

rr

η

Thus,

(a) ( )increase46.33996.380net kJ/kg 41.5=−=∆w

( )increase%9.37%5.48th 10.6%=−=∆η (b)


9-131

9-162 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined.


Properties The properties of air at room temperature are R = 0.287 kJ/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k = 1.4 (Table A-2).

Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6,

( )( )( )

( )( )

( )( )( )

( )( )( )

%1.40kJ/kg 803.4kJ/kg 321.9

kJ/kg 321.95.4814.803

kJ/kg 481.5K300779.1KkJ/kg 1.005

kJ/kg 803.4K500.61300KkJ/kg 1.005

K 779.161K 1300

K 500.66K 300

in

netth

outinnet

1414out

2323in

0.4/1.4/1

3

434

0.4/1.4/1

1

212

===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

qw

qqw

TTchhq

TTchhq

PP

TT

PP

TT

p

p

kk

kk

η

s

T

1

24

3 qin

qout

2

3′

For rp = 12, ( )

( )( )

( )( )

( )( )( )

( )( )( )

%8.50kJ/kg 693.2kJ/kg 352.3

kJ/kg 352.39.3402.693

kJ/kg 340.9K300639.2KkJ/kg 1.005

kJ/kg 693.2K610.21300KkJ/kg 1.005

K 639.2121K 1300

K 610.212K 300

in

netth

outinnet

1414out

2323in

0.4/1.4/1

3

434

0.4/1.4/1

1

212

===

=−=−=

=−⋅=−=−=

=−⋅=−=−=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

qw

qqw

TTchhq

TTchhq

PP

TT

PP

TT

p

p

kk

kk

η

Thus,

(a) ( )increase9.3213.352net kJ/kg 30.4=−=∆w

( )increase%1.40%8.50th 10.7%=−=∆η (b)



9-132

9-163 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined.




Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.

( )( )( )

( )( ) ( )

( ) ( )

( )( )

( )( ) ( )

( )( )

( )( ) ( )

( )( )

( ) ( ) ( )( )

( ) ( ) ( )( ) kJ/kg .5791K10061400KkJ/kg 1.005222

kJ/kg 375.7K300486.9KkJ/kg 1.005222

K .48769.486100675.09.486

K 10061.942140086.01400

K .194241K 1400

K 486.978.0/3008.445300

/

K 445.84K 300

7676outT,

1212inC,

494549

45

49

45

7667976

76

76

76

0.4/1.4/1

6

7679

1212412

12

12

12

0.4/1.4/1

1

2124

=−⋅=−=−=

=−⋅=−=−=

=−+=

−+=⎯→⎯−

−=

−−

=

=−−=

−−==⎯→⎯−

−=

−−

=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛==

=−+=

−+==⎯→⎯−

−=

−−

=

==⎟⎟⎠

⎞⎜⎜⎝

⎛==

−

−

TTchhw

TTchhw

TTTTTTcTTc

hhhh

TTTTTTTcTTc

hhhh

PP

TTT

TTTTTTTcTTc

hhhh

PP

TTT

p

p

p

p

sTsp

p

sT

kk

ss

Csp

spsC

kk

ss

εε

ηη

ηη

s

T

3

4

1

6

97

8

5

2

97s

4 2s

Thus,

0.475===kJ/kg 791.5kJ/kg 375.7

outT,

inC,bw w

wr

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

45.1%====

=−=−=

=−+−⋅=−+−=−+−=

451.0kJ/kg 922.0kJ/kg 415.8

kJ/kg .84157.3755.791

kJ/kg .0922K10061400.48761400KkJ/kg 1.005

in

netth

inC,outT,net

78567856in

qw

www

TTTTchhhhq p

η


9-133

9-164 Problem 9-163 is reconsidered. The effect of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for


[K]

cy" sentropic efficiency"

ies are constant across the compressor"

ssor for the isentropic case: dy-flow"

])

T[2], P=P[2])

ideal case the entropies are constant across the HP compressor"

r for the isentropic case: dy-flow"

])

P[4])

the cycle is to be plotted.

A

"Input data" T[6] = 1400 T[8] = T[6] Pratio = 4 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1] Eta_reg = 0.75 "Regenerator effectiveness

ta_c =0.78 "Compressor isentorpic efficien"

EEta_t =0.86 "Turbien i "LP Compressor:" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropP[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen_LP/w_comp_LP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the LP compree_in - e_out = DELTAe=0 for steah[1] + w_compisen_LP = h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2 "Actual compressor analysis:"h[1] + w_comp_LP = h[2]

[2]=ENTHALPY(Air,T=T[2]) hs[2]=ENTROPY(Air,T= "HP Compressor:" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For theP[4] = Pratio*P[3] P[3] = P[2] s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at compressor exit" Eta_c = w_compisen_HP/w_comp_HP "compressor adiabatic efficiency, W_comp > W_compisen" "Conservation of energy for the compressoe_in - e_out = DELTAe=0 for steah[3] + w_compisen_HP = h_s[4] h[3]=ENTHALPY(Air,T=T[3]) h_s[4]=ENTHALPY(Air,T=T_s[4 "Actual compressor analysis:"h[3] + w_comp_HP = h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=



9-134


ntercooling heat loss:"

eat exchanger, assuming W=0, ke=pe=0 _in - e_out =DELTAe_cv =0 for steady flow"

[6]=ENTHALPY(Air,T=T[6]) s SSSF constant pressure"

r,T=T[6],P=P[6])

_s[7]=ENTROPY(Air,T=T_s[7],P=P[7])"T_s[7] is the isentropic value of T[7] at HP turbine exit"

ic turbine, assuming: adiabatic, ke=pe=0 teady-flow"

s[7] =T_s[7])

[7]=ENTHALPY(Air,T=T[7]) (Air,T=T[7], P=P[7])

[7] + q_in_reheat = h[8] T[8])

e analysis"

r,T=T[8],P=P[8])

_s[9]=ENTROPY(Air,T=T_s[9],P=P[9])"T_s[9] is the isentropic value of T[9] at LP turbine exit" n > w_turb"

ic turbine, assuming: adiabatic, ke=pe=0 teady-flow"

[9] =T_s[9])

[9]=ENTHALPY(Air,T=T[9]) ir,T=T[9], P=P[9])

ta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency" atio"

t added in the external heat exchanger is"

ALPY(Air, T=T[5]) [5]=ENTROPY(Air,T=T[5], P=P[5])

"Ih[2] = q_out_intercool + h[3] "External heat exchanger analysis" "SSSF First Law for the heh[4] + q_in_noreg = h[6] hP[6]=P[4]"process 4-6 i "HP Turbine analysis" s[6]=ENTROPY(Ais_s[7]=s[6] "For the ideal case the entropies are constant across the turbine" P[7] = P[6] /Pratio sEta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb" "SSSF First Law for the isentrope_in -e_out = DELTAe_cv = 0 for sh[6] = w_turbisen_HP + h_h_s[7]=ENTHALPY(Air,T"Actual Turbine analysis:" h[6] = w_turb_HP + h[7] hs[7]=ENTROPY "Reheat Q_in:" hh[8]=ENTHALPY(Air,T= "HL Turbin P[8]=P[7] s[8]=ENTROPY(Ais_s[9]=s[8] "For the ideal case the entropies are constant across the turbine" P[9] = P[8] /Pratio sEta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbise "SSSF First Law for the isentrope_in -e_out = DELTAe_cv = 0 for sh[8] = w_turbisen_LP + h_sh_s[9]=ENTHALPY(Air,T"Actual Turbine analysis:" h[8] = w_turb_LP + h[9] hs[9]=ENTROPY(A "Cycle analysis" w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LP q_in_total_noreg=q_in_noreg+q_in_reheat EBwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work r "With the regenerator, the heah[5] + q_in_withreg = h[6] h[5]=ENTHsP[5]=P[4]


9-135


]-h[4]) ives h[10] and thus T[10] as:"

ALPY(Air, T=T[10]) [10]=ENTROPY(Air,T=T[10], P=P[10])

_in_total_withreg=q_in_withreg+q_in_reheat

g data is used to complete the Array Table for plotting purposes."

OPY(Air,T=T[5],P=P[5])

_s[10]=s[10] _s[10]=T[10]

η qi g qin eg

"The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[4])/(h[9"Energy balance on regenerator gh[4] + h[9]=h[5] + h[10] h[10]=ENTHsP[10]=P[9] "Cycle thermal efficiency with regenerator" qEta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]" "The followins_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTRT_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6] s_s[8]=s[8] T_s[8]=T[8] sT

ηreg ηC ηt ηth,noreg [%]

th,withreg

[%] n,total,nore

[kJ/kg] ,total,withr

[kJ/kg] wnet

[kJ/kg] 0.6

0.65 0.7

0.75 0.8

0.85 0.9

0.95 1

0.78 0.78

0.86 0.86

30.57 30.57

51.89 53.87

1434 1434

845.1 814.1

438.5 438.5

0.78 0.78 0.78 0.78 0.78 0.78 0.78

0.86 0.86 0.86 0.86 0.86 0.86 0.86

30.57 30.57 30.57 30.57 30.57 30.57 30.57

41.29 42.53 43.85 45.25 46.75 48.34 50.06

1434 1434 1434 1434 1434 1434 1434

1062 1031 1000 969.1 938.1 907.1 876.1

438.5 438.5 438.5 438.5 438.5 438.5 438.5

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50

400

800

1200

1600

s [kJ/kg-K]

T [K

]

100 k

Pa

400 k

Pa

160

0 kP

a

Air

1

2

3

4

5

6

7

8

9

10


9-136

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 128

32

36

40

44

48

52

ηreg

ηth

[%

]

No regeneration

With regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 110

15

20

25

30

35

40

45

50

55

η t

ηth

[%

]

With regeneration

No regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1150

200

250

300

350

400

450

500

550

600

η t

wne

t [k

J/kg

]



9-137

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

800

900

1000

1100

1200

1300

1400

1500

η t

q in,

tota

l [kJ

/kg]

With regeneration

No regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

25

30

35

40

45

50

55

ηc

ηth

[%

] With regeneration

No regeneration



9-138

9-165 A regenerative gas-turbine engine operating with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency are to be determined.

Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.


Properties The properties of helium at room temperature are cp = 5.1926 kJ/kg.K and k = 1.667 (Table A-2).

Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.

( )( )( )

( )( ) ( )

( ) ( )

( )( )

( )( ) ( )

( )( )

( )( ) ( )

( )( )

( ) ( ) ( )( )

( ) ( ) ( )( ) kJ/kg 5323K.48871400KkJ/kg 5.1926222

kJ/kg 2961K300585.2KkJ/kg 5.1926222

K .88112.5854.88775.02.585

K .48870.804140086.01400

K 0.80441K 1400

K .258578.0/3004.522300

/

K .45224K 300

7676outT,

1212inC,

494549

45

49

45

7667976

76

76

76

70.667/1.66/1

6

7679

1212412

12

12

12

70.667/1.66/1

1

2124

=−⋅=−=−=

=−⋅=−=−=

=−+=

−+=⎯→⎯−

−=

−−

=

=−−=

−−==⎯→⎯−

−=

−−

=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛==

=−+=

−+==⎯→⎯−

−=

−−

=

==⎟⎟⎠

⎞⎜⎜⎝

⎛==

−

−

TTchhw

TTchhw

TTTTTTcTTc

hhhh

TTTTTTTcTTc

hhhh

PP

TTT

TTTTTTTcTTc

hhhh

PP

TTT

p

p

p

p

sTsp

p

sT

kk

ss

Csp

spsC

kk

ss

εε

ηη

ηη

s

T

3

4

1

6qin

97

8

5

2

97s

4 2s

Thus,

0.556===kJ/kg 5323kJ/kg 2961

outT,

inC,bw w

wr

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

41.3%====

=−=−=

=−+−⋅=−+−=−+−=

4133.0kJ/kg 5716kJ/kg 2362

kJ/kg 236229615323

kJ/kg 5716K.48871400.88111400KkJ/kg 5.1926

in

netth

inC,outT,net

78567856in

qw

www

TTTTchhhhq p

η


9-139

9-166 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle.

Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ rp. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as

( )( )( )

( ) ( )( )

( ) ( )( ) ( ) ( ) ( ) kk

pkk

pkk

pkk

p

kk

p

kk

kkp

kk

p

kk

kkp

kk

rTrrTrTr

TPPTT

rTr

TPPTTT

rTPPTTT

2/11

2/1/11

2/12

/1

5

/16

5

2/13

/1

3

/1

3

4347

/11

/1

1

2125

1

1

−−−−

−−

−

−−

−−

===⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎛

=

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛==

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

Then,

65⎝

( ) ( )( )( ) ( )( )12/1 −=−=−= − kkrTcTTchhq

1

1161

2/137373in −=−=−= −

ppp

kkppp rTcTTchhq

6out

and thus ( )( )

( )( )kkpp rTcq 2/1

3inth

111

−−−=−=η

which simplifies to

kkpp rTcq 2/1

1out 1− −

( ) kkpr

TT 2/1

3

1th 1 −−=η

The thermal efficiency of the single stage ideal regenerative cycle is given as

( ) kkpr

TT /1

3

1th 1 −−=η

Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio rp.

s

T

1

2

7

qin

5

6

4

3

qout



9-140

9-167 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined.


Properties The gas constant of air is R = 0.287 kJ/kg·K.

Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Optimum intercooling and reheating pressure is


kPa 4.346)1200)(100(412 === PPP

Process 1-2, 3-4: Compression

KkJ/kg 7054.5kPa 100K 300

kJ/kg 43.300K 300

11

1

11

⋅=⎭⎬⎫

==

=⎯→⎯=

sPT

hT

=⎭⎬⎫

===

shss

P kJ/kg 79.428

kJ/kg.K 7054.5kPa 4.346

212

2

kJ/kg 88.46043.300

43.30079.42880.0 2212

12C =⎯→⎯

−−

=⎯→⎯−−

= hhhh

hh sη

s

T

3

4

1

5 qin

8 6

7

10

9

2

8s 6s

4s 2s

KkJ/kg 5040.5kPa 4.346

K 350kJ/kg 78.350K 350

33

3

33

⋅=⎭⎬⎫

==

=⎯→⎯=

sPT

hT

43

=⎭⎬⎫

== shss

kJ/kg 42.500kJ/kg.K 5040.5

kPa 12004 =P

4

kJ/kg 83.53778.350

78.35042.50080.0 4434

34C =⎯→⎯

−−

=⎯→⎯−−

= hhhh

hh sη

Process 6-7, 8-9: Expansion

=⎭⎬⎫

===

shss

P

KkJ/kg 6514.6kPa 1200K 1400

kJ/kg 9.1514K 1400

66

6

66

⋅=⎭⎬⎫

==

=⎯→⎯=

sPT

hT

kJ/kg 9.1083kJ/kg.K 6514.6

kPa 4.3467

67

7

kJ/kg 1.11709.10839.1514

9.151480.0 7

7

76T =

hη 76 =⎯→⎯

−−

=⎯→⎯−−

hh

hhh

s

8

88

⋅=⎭⎬⎫

==

=⎯→⎯=

sPT

h

96kPa 100

99 =

⎭⎬⎫=

shP

T

KkJ/kg 9196.6kPa 4.346

K 1300kJ/kg 6.1395K 1300

88

kJ/kg 00.996kJ/kg.K 91.689 == ss


9-141

kJ/kg 9.107500.9966.1395

6.139580.0 9

9

98

98T =⎯→⎯

−−

=⎯→⎯−−

= hh

hhhh

sη

C ysycle anal is:

kJ/kg 50.34778.35083.53743.30088.4603412inC, =−+−=−+−= hhhhw

kJ/kg 50.6649.10756.13951.11709.15149876outT, =−+−=−+−= hhhhw

0.523===50.66450.347

outT,

inC,bw w

wr

kJ/kg 317.0=−=−= 50.34750.664inC,outT,net www

Regenerator analysis:

kJ/kg 36.67283.5379.1075

9.107575.0 10

10

49

109regen =⎯→⎯

−−

=⎯→⎯−−

= hh

hhhh

ε

1010

⋅=⎭⎬=

sP

(b) =−=−= hhq

KkJ/kg 5157.6kPa 100

K 36.67210 ⎫=h

kJ/kg 40.94183.53736.6729.1075 5545109regen =⎯→⎯−=−⎯→⎯−=−= hhhhhhq

kJ/kg 54.57340.9419.151456in

0.553===54.5730.317

in

netth q

wη

(c) 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,

786.0K 1400K 30011

6

1max =−=−=

TT

η

and

0.704===786.0553.0

maxηη

η thII

(d) The exergies at the combustion chamber exit and the regenerator exit are

kJ/kg 930.7=−−−=

−−−=kJ/kg.K)7054.56514.6)(K 300(kJ/kg)43.3009.1514(

)( 060066 ssThhx

kJ/kg 128.8=−−−=

−−−=kJ/kg.K)7054.55157.6)(K 300(kJ/kg)43.30036.672(

)( 010001010 ssThhx



9-142

9-168 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a three-stage gas turbine is to be compared.



Analysis

Two Stages:


The pressure ratio across each stage is

416 ==pr

The temperatures at the end of compression and expansion are

c K

s

T

3

4

1

6 873 K

283 K

97

8

2

5

10 5.420K)(4) 283( 0.4/1.4/)1(min === − kk

prTT

K 5.58741K) 873(1 0.4/1.4/)1(

max =⎟⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

− kk

pe r

TT

The heat input and heat output are

kJ/kg 9.573K )5.587K)(873kJ/kg 2(1.005)(2 maxin =−⋅=−= ep TTcq

kJ/kg 4.276K )283K)(420.5kJ/kg 2(1.005)(2 minout =−⋅=−= TTcq

e therm of the cycle is then

cp

Th al efficiency

0.518=−=−= 1thη9.573inq4.2761outq

hree Stages:

The pressure ratio across each stage is

==r

The temperatures at the end of compression and expansion are

T

p 520.216 3/1

K 5.368K)(2.520) 283( 0.4/1.4/)1(min === − kk

pc rTT

K 4.6702.520

1K) 873(1/)1(

=⎟⎞

⎜⎛

=− kk

TT0.4/1.4

max =⎟⎠⎞

⎜⎝⎛

⎟⎠

⎜⎝ pr

he heat input and heat output are

kg =−⋅

s

T

3

4

1

9

8

2

5

10

11

6

7

12

13

14

e

T

kJ/ 3(1.005)(3 maxin =−= ep TTcq kJ/kg 8.610K )4.670K)(873

out kJ/kg 8.257K )283K)(368.5kJ/kg (1.0053)(3 min =−⋅=−= TTcq cp


0.578=−=−=8.6108.25711

in

outth q

qη


9-143

9-169 A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is to be determined.




K 6.509K)(9) 272( 0.4/1.4/)1(12 === − kk

prTT

s

T

1

2 4

3qin

5

qout

K 2.34391K 643(35 =⎟

⎟⎠

⎜⎜⎝

=pr

TT )1 0.4/1.4/)1(

=⎟⎠⎞

⎜⎝⎛⎞⎛

− kk

pressor is

The work input to the com

K)(509.6kJ/kg 005.1()( 12C ⋅=−= TTcw p kJ/kg 8.238272)K- =

kJ/kg 005.1(C53

=−−=

p

engine exit is determined from

An energy balance gives the excess enthalpy to be

)( −−=∆ wTTch

kJ/kg 50.62kJ/kg 8.238)K2.343K)(643⋅

The velocity of the air at the

2

2inletexit VV

h−

=∆ 2

earranging,

R

( )

m/s 6.504

)m/s 360(kJ/kg 1

/sm 1000kJ/kg) 50.62(2

22/1

222

2/12inletexit

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

+∆= VhV

The specific impulse is then

m/s 145=−=−= 3606.504inletexit VVmF&



9-144

9-170 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined.



⋅=⎭⎬⎫

=°=

=⎯→⎯°=

sPT

hT

kPa 10002 =

⎭⎬⎫=

shP

1

Combustion chamber

Turbine

23

4

Compress.

100 kPa20°C

450°C1 MPa

325°C 15°C

Sat. vap. 200°C

Heat exchanger

5

Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Process 1-2: Compression

KkJ/kg 682.5kPa 100

C20kJ/kg 5.293C20

11

1

11

kJ/kg 2.567kJ/kg.K 682.512 == ss

2

kJ/kg 8.6115.293

5.2932.56786.0 2212

12C =⎯→⎯

−−

=⎯→⎯−−

= hhhh

hh sη

Process 3-4: Expansion

kJ/kg 5.738C4504 =⎯→⎯°= hT 4

ss hhh 343T −− h

h

4

3 5.738−

We cann ly. However, using the following lines in EES together with the isentropic c relation, we find h3 = 1262 kJ/kg, T3 = 913.2ºC, s3 = 6.507 kJ/kg.K. The solution by hand would require a trial-

h_3=enthalpy(Air, T=T_3)

5 =→h

The inlet water is compressed liquid at 15ºC and at the saturation pressure of steam at 200ºC (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature.

C15

22

2 =⎭⎬⎫

=°=

=⎫°=

ww hx

T

T

12in, =−=−= hhw

hh 43 88.0 =⎯→⎯−

=η

ot find the enthalpy at state 3 directefficien yerror approach.

s_3=entropy(Air, T=T_3, P=P_2)

h_4s=enthalpy(Air, P=P_1, s=s_3)

Also,

C3255 ⎯⎯°=T kJ/kg 4.605

kJ/kg 27921

C200

kJ/kg 47.64kPa 1555 1

1 ⎭⎬= wh

P1w

The net work output is

kJ/kg 2.3185.2938.611C

kJ/kg 4.5235.738126243outT, =−=−= hhw


9-145

kJ/kg 2.2052.3184.523inC,outT,net =−=−= www

The mass flow rate of air is


kg/s 7.311===kJ/s 1500netW

m&

& kJ/kg 2.205netwa

(b) The back work ratio is

0.608===2.318inC,

bww

r 4.523outT,w

The rate of heat input and the thermal efficiency are

kW 4753)kJ/kg8.611kg/s)(1262 311.7()( 23in =−=−= hhmQ a&&

31.6%==== 3156.0kW 4753kW 1500

in

netth Q

W&

&η

(c) An energy balance on the heat exchanger gives

(d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are

kg/s 0.3569=⎯→⎯−=−

−=−

ww

wwwa

mm

hhmhhm

&&

&&

)kJ/kg47.64(2792)kJ/kg4.6055kg/s)(738. 311.7(

)()( 1254

kW 5.973)kJ/kg47.64kg/s)(2792 3569.0()( 12p =−=−= www hhmQ &&

52.0%==+

=+

= 5204.0kW 4753

5.9731500

in

pnet

Q

QWu &

&&ε


9-146

9-171 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flow rate of the air through the compressor, the velocity of the gases at the nozzle exit, the propulsive power, and the propulsive efficiency of the cycle are to be determined.

Assumptions 1 The air-standard assumptions are applicable. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1).


Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Diffuser, Process 1-2:

kJ/kg 23.238C35 11 =⎯→⎯°−= hT

kJ/kg 37.269/sm 1000

kJ/kg 12m/s) (15

/sm 1000kJ/kg 1

2m/s) (900/3.6kJ/kg) 23.238(

22

222

2

222

2

22

2

21

1

=⎯→⎯⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛+

+=+

hh

VhVh

s

T

1

2

4

3

qin

5

qout

6

kPa 50kJ/kg 37.2692

=Ph

KkJ/kg 7951.522

⋅=⎭⎬⎫=

s

Compressor, Process 2-3:

=⎭⎬⎫

===

shss

P kJ/kg 19.505

kJ/kg.K 7951.5kPa 450

323

3

kJ/kg 50.55337.269

37.26919.50583.0 3323

23C =⎯→⎯

−−

=⎯→⎯−−

= hhhh

hh sη

urbine, P cess 3-4:

1028.130437.26950.553 555423 =⎯→⎯−=−⎯→⎯−=− hhhhhh

here the ass flow rates through the compressor and the turbine are assumed equal.

T ro

kJ/kg 8.1304C950 =⎯→⎯°= hT 44

.0 kJ/kg 6

w m

kJ/kg 45.9628.1304

6.10208.130483.0 5554

54T =⎯ →⎯

−−

=⎯→⎯−−

= sss

hhhh

hhη

4⎭=P

545 KkJ/kg 7725.6

kgP

ss

(b) The mass flow rate of the air through the compressor is

KkJ/kg 7725.6C9504 ⋅=⎬

⎫°=s

T

kPa 4504

=5 kJ/ 45.962h s kPa 147.4=

⎭⎬⎫

⋅==

kg/s 1.760=−

=−

=kJ/kg )37.26950.553(

kJ/s 500

23

C

hhW

m&

&

(c) Nozzle, Process 5-6:

⎫=h KkJ/kg 8336.6

kPa 4.147 55

⋅=⎭⎬=

sP

kJ/kg 6.10205


9-147

kJ/kg 66.709kJ/kg.K 8336.6

kPa 406

56

6 =⎭⎬⎫

===

shss

P


kJ/kg 52.76266.7096.1020

6.102083.0 6

6

65

65N =⎯→⎯

−−

=⎯→⎯−−

= hh

hhhh

sη

m/s 718.5=⎯→⎯⎟⎠⎞

⎜⎝⎛+=+

+=+

622

26

26

6

25

5

/sm 1000kJ/kg 1

2kJ/kg 52.7620kJ/kg) 6.1020(

22

VV

VhVh

where the velocity at nozzle inlet is assumed zero.

(d) The propulsive power and the propulsive efficiency are

kW 206.1=⎟⎠⎞

⎜⎝⎛−=−= 22116 /sm 1000

kJ/kg 1m/s) 250)(m/s 250m/s 5.718(kg/s) 76.1()( VVVmWp &&

kW 1322kJ/kg)50.5538.1304(kg/s) 76.1()( 34in =−=−= hhmQ &&

0.156===kW 1322kW 1.206

inQ

W pp &

&η


9-148

9-172 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained.



(b) The work of compression is found by the first law for process 1-2:


2 1 2 1 2

1 2 1 2 2 1

1 2 2 1

)

v

comp v

q w urocess

w u C T T

w w C T T

−

− −

−

− = ∆

= −∆ = − −

= − = −

ound by

)3 2 3 22

v v R T T− = −

The back work ratio is

( )( )

1− −

1 2 0(q isentropic p− =

The expansion work is f

( )exp 2 3w w Pdv P−= = =∫ (3

( )( )

( ))(

3 1 3 11

3 2

/ 1/ 1

comp v vw C T T T TC TT

− −−

Process 1-2 is isentropic; therefore,

s

T

3 2

1

v

P

32

1

exp 3 2 2w R T T R T T= =

−

1

1 21

1k

T VT V r −= =

2 1

k −⎛ ⎞⎜ ⎟⎝ ⎠

and 2P VP V

= 1

1 2

kkr

⎛ ⎞=⎜ ⎟

⎝ ⎠


3 3 3 32 2 1

3 2 2 2 2

PV T VPV V rT T T V V

= ⇒ = = =


3 3 3 31 1 2 kPV T PPV P r= ⇒ = = =

The back work ratio becomes (Cv=R/(k-1))

3 1 1 1 1T T T P P

1

1exp 1w k − −

1 1 11

kcomp

k

w rr r

−

−

−=

(c) Apply first law to the closed system for processes 2-3 and 3-1 to show:

3 1out vq C T T= −


( )3 2in pq C T T= −

( )

( )( )

( )( )

3 1 1 3 1

3 2 2 3 2

/ 111 1 1/ 1

voutth

in p

C T T T T Tqq C T T k T T T

η− −

= − = − = −− −



9-149

111th kk r r

η −= −−

1 1 1kr −

ine the value of the back work ratio and efficiency as r goes to unity.

(d) Determ

( )( )

1

1exp

21 1

1 1 11 1exp 1

1exp

1 1 11 1

11 1 1 1 1 1lim lim lim lim1 1 1 1 1

1 1 1 1lim 11 1 1 1

kcomp

k

kk kcomp

k k k kr r rr

comp

r

w rw k r r

w k rr rw k r r k r r k kr k r

w k kw k k k k

−

−

−− −

− − −→ → →→

→

−=

− −

⎧ ⎫21 k −

⎧ ⎫−⎧ ⎫− −⎪ ⎪ ⎪= = =⎨ ⎬ ⎨ ⎬ ⎨− − − − − − −⎪⎬

⎪ ⎪⎩ ⎭⎪ ⎪ ⎩ ⎭⎩ ⎭− −⎧ ⎫ ⎧ ⎫= = =⎨ ⎬ ⎨ ⎬− − + −⎩ ⎭ ⎩ ⎭

( )

1

1

1 1 11 1 1 1

1

1 1 111

1 1 1 1 1 1lim 1 lim 1 lim 1 lim1 1

1 1lim 1 1 01 1

k

th k

k k k

th k k k kr r r r

thr

rk r r

r r krk r r k r r k kr k r

k kk k k k

η

η

η

−

−

− − −→ → → →

→

−= −

−

2k −

⎧ ⎫⎧ ⎫ ⎧ ⎫− − ⎪ ⎪= − = − = −⎨ ⎬ ⎨ ⎬ ⎨− − − − ⎬⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭

⎧ ⎫ ⎧ ⎫= − = − =⎨ ⎬ ⎨ ⎬− +⎩ ⎭ ⎩ ⎭

These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.



9-150

9-173 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained.



(b) The work of expansion is found by the first law for process 2-3:


T

)1 3 3 13

Pdv P v v R T T= − = − − = −∫


( )( )

2 3 2 3 2 3

2 3

2 3 2 3 3 2

exp 2 3 2 3

0( )

v

v

q w uq isentropic processw u C T

w w C T T

− − −

−

− −

−

− = ∆=

= −∆ = − −

= = −

The compression work is found by

( ) (1

3 1compw w −= −

( )( )

( )( )

( )( )

3 1 1 3 1 3

2 3 2 3

1 / 1 // 1 / 1

comp vw C T T T T T TC T CT R T T

− − −=

−


3

exp 2 3 3

v v

w R T T R T T= =

− −

v

P

3

2

1

s

T

3

2

1

3 3 1 1 1 1 2

3 1 3 3V V3

1PV PV T V VT T T r

= ⇒ = = =

Process 2-3 is isentropic; therefore, 1k

T V−

⎛ ⎞ 32

3 2

kkVP12 2

3 2

krT V

−= =⎜ ⎟⎝ ⎠

and rP V

⎛ ⎞= =⎜ ⎟

⎝ ⎠

The back work ratio becomes (Cv=R/(k-1))

( ) 1 1

11w − − −

exp

1 111 1

mpk k

k rrkw r r r− −= − =

− −

(c) Apply first law to the closed system for processes 1-2 and 3-1 to show:

2 1in vq C T T= −


co

( )( )3 1out pq C T T= −

( )( )

( )( )

3 1 1 3 1

2 1 1 2 1

/ 11 1 1

/ 1poutqηth

in v

C T T T T Tk

q C T T T T T− −

= − = − = −− −


2 2

T1 1 2 2 2

2 1 1 1 3

krT T P P

= ⇒ = = =


PV PV T P P


9-151

111th k

rkr

η −= −

−

(d) Determine the value of the back work ratio and efficiency as r goes to unity.

( )

( ) ( )

( )

1 1exp

11

1exp

11 1 111 1

1

1lim 1 11

compk k

kr

comp

r

w k rrkw r r r

r kr

wk

w k

− −

−→

→

− − −= − =

− −

−⎭ ⎩ ⎭

⎧ ⎫= − =⎨ ⎬−⎩ ⎭

1 1exp

1 1lim 1 lim 1 limcompkr r

w rk kw r→ →

⎧ ⎫ ⎧ ⎫−= − = −⎨ ⎬ ⎨ ⎬−⎩

11 1 1

1

111

1 1lim 1 lim 1 lim1

1lim 1 0

th k

th k kr r r

thr

rkr

rk kr k

kk

η

η

η

−→ → →

→

r

−= −

−⎧ ⎫ ⎧−

= − = −⎨ ⎬ ⎨−⎩ ⎭ ⎩⎧ ⎫= − =⎨ ⎬⎩ ⎭

⎫⎬⎭

These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.



9-152

9-174 The four processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams; an expression for the cycle thermal efficiency is to be obtained; and the limit of the efficiency as the volume ratio during heat rejection approaches unity is to be evaluated.


(b) Apply first law to the closed system for processes 2-3 and 4-1 to show:


3 2

4 1

v

out p

q C T T= −


( )( )q C T T= −

in

( )( )

( )( )

4 11 1 poutth q C T T

η 1 4 1

3 2 2 3 2

/ 11

/ 1in v

C T T T T Tq kT T T

− −= − = − = −

− −


1

1 21

s

T

42

1

3

v

P3

2

1

4

2 1

1k

k

T VT V r

−

−

⎛ ⎞= =⎜ ⎟

⎝ ⎠


113 4

4 3

kk

eT V rT V

−

−⎛ ⎞= =⎜ ⎟

⎝ ⎠


4 4 1 1 4 4p

PV PV T V r= ⇒ = = 4 1 1 1T T VT

113 3 4 1

12 4 1 2

1 kk e

e p k

T T rT T r r rT T T T r r

−−

−

⎛ ⎞= = = ⎜ ⎟⎝ ⎠

p

tant volume and V3 = V2, Since process 2-3 is cons

4 4 4 1

3 2 1 2e pV V V V

V V V Vr r r= = = =

1

3

2

kp k

p p

r rT r rT r

−⎛ ⎞

= =⎜ ⎟⎝ ⎠


1

1111

pth k k

p

rk

r rη −

−= −

−

(c) In the limit as rp approaches unity, the cycle thermal efficiency becomes

1 11 1

1 11

11 1lim 1 lim 1 lim1

1 1 1lim 1 1

p p p

p

pth k k k kr r r

p p

th th Ottok kr

rk k

r r r kr

kr k r

η

η η

− −→ → →

− −→

⎧ ⎫ ⎧−⎪ ⎪ ⎪= − = −⎨ ⎬ ⎨−⎪ ⎪ ⎪⎩ ⎭ ⎩⎧ ⎫= − = − =⎨ ⎬⎩ ⎭

11

1−

⎫⎪⎬⎪⎭


9-153

9-175 The four processes of an air-standard cycle are described. The back work ratio and its limit as rp goes to unity are to be determined, and the result is to be compared to the expression for the Otto cycle.

Analysis The work of compression for process 1-2 is found by the first law:


2 1 2 1 2

1 2 1 2 2 1

,1 2 1 2 2 1

v

comp v

q w u

w u C T T

w w C T T

−

− −

− −

− = ∆

= −∆ = − −

= − = −

)4 1 1 4 4 14

Pdv P v v R T T− = − = − − = −∫

The work of expansion for process 3-4 is found by the first law:

3 4 4 3

exp,3 4 3 4 3 4

v

v

q w u

w u C T T

w w C T T

− − −

− −

− −

− = ∆

= −∆ = − −

= − = −


( )( )

1− −

1 2 0( )q isentropic process− =

The work of compression for process 4-1 is found by

( ) (1

,4 1compw w− = −

( )( )

3 4 3 4 3 4

3 4 0( )q isentropic process− =

3 4

( ) ( )( )

( ) ( )

( )4 1 2 1

4 1 2 1 1

exp 3 4 3 4 3

/ 1 / 1

1 /comp v v

v

R T T T Tw R T T C T T CTw C T T T T T

− + −− + −

= =− −

Using data from the previous problem and Cv = R/(k-1)

( ) ( )11

exp 31 1

( 1) 1 1

11

kpcomp

k kpr r⎝ ⎠

k r rw Tw T

−

− −

− − + −=

⎛ ⎞−⎜ ⎟⎜ ⎟

( ) ( ) ( )( ) ( )1 11 1

1 1exp 3 3

11 1

11

1exp 3

1

( 1) 1 1 1 0 1lim lim 11 11

1lim 11

p p

p

k kpcomp

r r

kk kp

kcomp

r

k

k r r k rw T Tw T T

rr r

w T rw T

r

− −

→ →

−− −

−

→

−

⎧ ⎫⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪− − + − − + −⎪ ⎪ ⎪= =⎨ ⎬ ⎨

⎛ ⎞⎪ ⎪ ⎪ −−⎜ ⎟⎪ ⎪ ⎪⎜ ⎟ ⎩ ⎭⎝ ⎠⎪ ⎪⎩ ⎭⎧ ⎫⎪ ⎪−

= ⎨ ⎬⎪ ⎪−⎩ ⎭

⎪⎬⎪⎪

This result is the same expression for the back work ratio for the Otto cycle.


9-154

9-176 The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given


]

ession"

P=P[2]) v[1]/T[1]

rocess 1 to 2"

T=T[1]) eat addition"

2 to 3"

nergy(air,T=T[2]) ion"

rocess 3 to 4"

T=T[3]) n"

to 1"

nergy(air,T=T[1])-intenergy(air,T=T[4])

Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"

operating conditions is to be investigated.

A

"Input Data" T[1]=300 [K]

] P[1]=100 [kPa[3] = 2000 [KT

r_comp = 12

1-2 is isentropic compr"Process s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]

(air, s=s[2],T[2]=temperatureP[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for pq_12 - w_12 = DELTAu_12 q_12 =0"isentropic process"

12=intenergy(air,T=T[2])-intenergy(air,DELTAu_"Process 2-3 is constant volume hv[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 q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process"

23=intenergy(air,T=T[3])-inteDELTAu_"Process 3-4 is isentropic expanss[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] "Conservation of energy for pq_34 -w_34 = DELTAu_34 q_34 =0"isentropic process"

=intenergy(air,T=T[4])-intenergy(air,DELTAu_34"Process 4-1 is constant volume heat rejectioV[4] = V[1] "Conservation of energy for process 4q_41 - w_41 = DELTAu_41

_41 =0 "constant volume process" wDELTAu_41=inte q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41



9-155

ηth[%]

rcomp wnet[kJ/kg]

45.83 6 567.4 48.67 7 589.3 51.03 8 604.9 53.02 9 616.2 54.74 10 624.3 56.24 11 630 57.57 12 633.8 58.75 13 636.3 59.83 14 637.5 60.8 15 637.9

6 7 8 9 10 11 12 13 14 15560

570

580

590

600

610

620

630

640

rcomp

wne

t [k

J/kg

]

6 7 8 9 10 11 12 13 14 1545

48.5

52

55.5

59

62.5

rcomp

ηth

[%

]



9-156

9-177 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined.


P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 100/100 Eta_t = 100/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[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]) 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,s=s_s[4],P=P[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" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])



9-157

Bwr η Pratio Wc

[kW] Wnet[kW]

Wt[kW]

Qin[kW]

0.254 0.3383 5 175.8 516.3 692.1 1526 0.2665 0.3689 6 201.2 553.7 754.9 1501 0.2776 0.3938 7 223.7 582.2 805.9 1478 0.2876 0.4146 8 244.1 604.5 848.5 1458 0.2968 0.4324 9 262.6 622.4 885 1439 0.3052 0.4478 10 279.7 637 916.7 1422 0.313 0.4615 11 295.7 649 944.7 1406 0.3203 0.4736 12 310.6 659.1 969.6 1392 0.3272 0.4846 13 324.6 667.5 992.1 1378 0.3337 0.4945 14 337.8 674.7 1013 1364 0.3398 0.5036 15 350.4 680.8 1031 1352 0.3457 0.512 16 362.4 685.9 1048 1340 0.3513 0.5197 17 373.9 690.3 1064 1328 0.3567 0.5269 18 384.8 694.1 1079 1317 0.3618 0.5336 19 395.4 697.3 1093 1307 0.3668 0.5399 20 405.5 700 1106 1297 0.3716 0.5458 21 415.3 702.3 1118 1287 0.3762 0.5513 22 424.7 704.3 1129 1277 0.3806 0.5566 23 433.8 705.9 1140 1268 0.385 0.5615 24 442.7 707.2 1150 1259 0.3892 0.5663 25 451.2 708.3 1160 1251 0.3932 0.5707 26 459.6 709.2 1169 1243 0.3972 0.575 27 467.7 709.8 1177 1234 0.401 0.5791 28 475.5 710.3 1186 1227 0.4048 0.583 29 483.2 710.6 1194 1219 0.4084 0.5867 30 490.7 710.7 1201 1211 0.412 0.5903 31 498 710.8 1209 1204 0.4155 0.5937 32 505.1 710.7 1216 1197 0.4189 0.597 33 512.1 710.4 1223 1190 0.4222 0.6002 34 518.9 710.1 1229 1183

5 10 15 20 25 30 35500

525

550

575

600

625

650

675

700

725

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

Pratio

Wne

t [k

W]

η th



9-158

9-178 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Brayton cycle is to be investigated assuming adiabatic efficiencies of 85 percent for both the turbine and the compressor. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined.


P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 1800 [K] m_dot = 1 [kg/s] Eta_c = 85/100 Eta_t = 85/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[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]) 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,s=s_s[4],P=P[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" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta=W_dot_net/Q_dot_in"Cycle thermal efficiency" Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])



9-159

Bwr η Pratio Wc

[kW] Wnet[kW]

Wt[kW]

Qin[kW]

0.3515 0.2551 5 206.8 381.5 588.3 1495 0.3689 0.2764 6 236.7 405 641.7 1465 0.3843 0.2931 7 263.2 421.8 685 1439 0.3981 0.3068 8 287.1 434.1 721.3 1415 0.4107 0.3182 9 309 443.3 752.2 1393 0.4224 0.3278 10 329.1 450.1 779.2 1373 0.4332 0.3361 11 347.8 455.1 803 1354 0.4433 0.3432 12 365.4 458.8 824.2 1337 0.4528 0.3495 13 381.9 461.4 843.3 1320 0.4618 0.355 14 397.5 463.2 860.6 1305 0.4704 0.3599 15 412.3 464.2 876.5 1290 0.4785 0.3643 16 426.4 464.7 891.1 1276 0.4862 0.3682 17 439.8 464.7 904.6 1262 0.4937 0.3717 18 452.7 464.4 917.1 1249 0.5008 0.3748 19 465.1 463.6 928.8 1237 0.5077 0.3777 20 477.1 462.6 939.7 1225 0.5143 0.3802 21 488.6 461.4 950 1214 0.5207 0.3825 22 499.7 460 959.6 1202 0.5268 0.3846 23 510.4 458.4 968.8 1192 0.5328 0.3865 24 520.8 456.6 977.4 1181

5 9 13 17 21 25380

390

400

410

420

430

440

450

460

470

0.24

0.26

0.28

0.3

0.32

0.34

0.36

0.38

0.4

Pratio

Wne

t [k

W]

η thPratio for

W net,max

W net η th



9-160

9-179 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used.


Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) "For Air:" C_V = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = C_V*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = C_V*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "[%]" "The Easy Way to calculate the constant property Otto cycle efficiency is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) "[%]" END "Input Data" T[1]=300 [K] P[1]=100 [kPa] {T[3] = 1000 [K]} r_comp = 12 "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_12 q_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_23 w_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_34 q_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_41



9-161

w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent" Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) "[%]"

PerCentError [%]

rcomp ηth[%]

ηth,ConstProp[%]

ηth,easy[%]

T3[K]

3.604 12 60.8 62.99 62.99 1000 6.681 12 59.04 62.99 62.99 1500 9.421 12 57.57 62.99 62.99 2000 11.64 12 56.42 62.99 62.99 2500

6 7 8 9 10 11 123.6

3.7

3.8

3.9

4

4.1

4.2

4.3

rcom p

Per

Cen

tErr

or [

%]

P ercent E rror = |η th - η th ,ConstProp | / η th

Tm ax = 1000 K

1000 1200 1400 1600 1800 2000 2200 2400 26004

6.2

8.4

10.6

12.8

15

T[3] [K]

Per

Cen

tErr

or [

%]

rcomp = 6

=12



9-162

9-180 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is

.


m diagram window"

P[1])

for the actual compressor, assuming: adiabatic, ke=pe=0"

ot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0"

ycle work, kW"

determined only to produce a T-s plot"

s[4]=entropy(air,T=T[4],P=P[4])

to be investigated. Variable specific heats are to be used

A

"Input data - fro{P_ratio = 8}

{T[1] = 300 [K]P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s]

ta_c = 75/100 EEta_t = 82/100} "Inlet conditions" [1]=ENTHALPY(Air,T=T[1]) h

s[1]=ENTROPY(Air,T=T[1],P= "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"

sure ratio - to find P[2]" P_ratio=P[2]/P[1]"Definition of presT_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])

mpressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Com_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law

lysis" "External heat exchanger anaP[3]=P[2]"process 2-3 is SSSF constant pressure"

ir,T=T[3]) h[3]=ENTHALPY(Am_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"

,T=T[3],P=P[3]) s[3]=ENTROPY(Airs_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4]

=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" T_s[4]=TEMPERATURE(Air,sh_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t"

ta_t=(h[3]-h[4])/(h[3]-h_s[4]) Em_dot*h[3] = W_d "Cycle analysis"

f the net cW_dot_net=W_dot_t-W_dot_c"Definition oEta=W_dot_net/Q_dot_in"Cycle thermal efficiency"

work ratio" Bwr=W_dot_c/W_dot_t "Back "The following state points are T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy(air,T=T[2],P=P[2])



9-163

Bwr η Pratio Wc

[kW] Wnet[kW]

Wt[kW]

Qin[kW]

0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700

0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241

5.0 5.5 6.0 6.5 7.0 7.50

500

1000

1500

s [kJ/kg-K]

T [K

]

100 kPa

800 kPa

1

2s

2

3

4

4s

Air Standard Brayton Cycle

Pressure ratio = 8 and Tmax = 1160K

2 4 6 8 10 12 14 16 18 200.00

0.05

0.10

0.15

0.20

0.25

0

500

1000

1500

2000

2500

Pratio

Cyc

le e

ffic

ienc

y,

Wne

t [k

W]

η Wnet

Tmax=1160 K

Note Pratio for maximum work and η

ηc = 0.75

ηt = 0.82

η



9-164

9-181 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working fluid is to be investigated.


Function hFunc(WorkFluid$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy." IF WorkFluid$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P) endif END Procedure EtaCheck(Eta_th:EtaError$) If Eta_th < 0 then EtaError$ = 'Why are the net work done and efficiency < 0?' Else EtaError$ = '' END "Input data - from diagram window" {P_ratio = 8} {T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium'} "Inlet conditions" h[1]=hFunc(WorkFluid$,T[1],P[1]) s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[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(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=hFunc(WorkFluid$,T_s[2],P[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]=hFunc(WorkFluid$,T[3],P[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(WorkFluid$,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(WorkFluid$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid$,T_s[4],P[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" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW" Eta_th=W_dot_net/Q_dot_in"Cycle thermal efficiency" Call EtaCheck(Eta_th:EtaError$) Bwr=W_dot_c/W_dot_t "Back work ratio" "The following state points are determined only to produce a T-s plot" T[2]=temperature(air,h=h[2]) T[4]=temperature(air,h=h[4]) s[2]=entropy(air,T=T[2],P=P[2]) s[4]=entropy(air,T=T[4],P=P[4])



9-165

Bwr η Pratio Wc

[kW] Wnet

[kW] Wt

[kW] Qin

[kW] 0.5229 0.1 2 1818 1659 3477 16587 0.6305 0.1644 4 4033 2364 6396 14373 0.7038 0.1814 6 5543 2333 7876 12862 0.7611 0.1806 8 6723 2110 8833 11682 0.8088 0.1702 10 7705 1822 9527 10700

0.85 0.1533 12 8553 1510 10063 9852 0.8864 0.131 14 9304 1192 10496 9102 0.9192 0.1041 16 9980 877.2 10857 8426 0.9491 0.07272 18 10596 567.9 11164 7809 0.9767 0.03675 20 11165 266.1 11431 7241

5.0 5 .5 6 .0 6 .5 7 .0 7 .50

500

1000

1500

s [kJ /kg -K ]

T [K

]

100 kP a

800 kP a

1

2 s

2

3

4

4 s

B rayto n C yc le

P ressu re ra tio = 8 an d T m ax = 1160K

2 4 6 8 10 12 14 16 18 200.00

0.05

0.10

0.15

0.20

0.25

0

500

1000

1500

2000

2500

Pratio

Cyc

le e

ffici

ency

,

Wne

t [k

W]

ηWnet

Tmax=1160 K

Note Pratio for maximum work and η

η c = 0.75η t = 0.82

η

Brayton Cycle using Airm air = 20 kg/s



9-166

9-182 The effect of the number of compression and expansion stages on the thermal efficiency of an iregenerative B

deal rayton cycle with multistage compression and expansion and air as the working fluid is to be


umber of compression and expansion stages"

cy" iency"

ysis"


)

the same pressure ssor is the same. The total compressor work is:"

tween turbines is" ges - 1)*C_P*(T_8 - T_7)

to occur until:"

ming: adiabatic, ke=pe=0 steady-flow"

7s)

tages*w_turb

" mp/w_turb "Back work ratio"

investigated.

A

"Input data for air" C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the n

Nstages = 1T_6 = 1200 [K]Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa]

Eta_reg = 1.0 "regenerator effectiveness"entorpic efficienEta_c =1.0 "Compressor is

Eta_t =1.0 "Turbine isentropic efficR_p = Pratio^(1/Nstages)

ressor ana"Isentropic CompT_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit"

ta_c = w_compisen/w_comp E"compressor adiabatic efficiency, W_comp > W_compisen"

ompresso"Conservation of energy for the cteae_in - e_out = DELTAe=0 for s

_1 w_compisen = C_P*(T_2s-TActual compressor analysis:""

w_comp = C_P*(T_2 - T_1)

ccur such that T_3 = T_1 and the compressors have"Since intercooling is assumed to oratio, the work input to each comprew_comp_total = Nstages*w_comp "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"

t exchanger + the reheat be"The heat added in the external hea C_P*(T_6 - T_5) +(Nstaq_in_total =

"Reheat is assumedT_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit"

) T_7s = T_6*(1/R_p)^((k-1)/k"Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen

c turbine, assu"SSSF First Law for the isentropi= 0 fore_in -e_out = DELTAe_cv

w_turbisen = C_P*(T_6 - T_"Actual Turbine analysis:"

_turb = C_P*(T_6 - T_7) ww_turb_total = Ns "Cycle analysis"

turb_total-w_comp_total "[kJ/kg]w_net=w_Bwr=w_co



9-167

P_4=P_2 P_5=P_4 P_6=P_5


rator effectiveness gives T_5 as:"

as:"

ycle thermal efficiency with regenerator"

mperatures, T_6 and T_1 for this problem." ta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

ηth,Ericksson ηth,Regenerative Nstages

T_4 = T_2 "The regeneEta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10T_4 + T_9=T_5 + T_10 "CEta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min teE

[%] [%] 75 49.15 1 75 64.35 2 75 68.32 3 75 70.14 4 75 72.33 7 75 73.79 15 75 74.05 19 75 74.18 22

0 2 4 6 8 10 12 14 16 18 20 22 2440

50

60

70

80

Nstages

ηth

[%]

Ericsson

Ideal Regenerative Brayton


9-168

9-183 The effect of the number of compression and expansion stages on the thermal efficiency of an idealregenerative B

rayton cycle with multistage compression and expansion and helium as the working fluid is to be


umber of compression and expansion stages"

cy" iency"

ysis"


)

the same pressure ssor is the same. The total compressor work is:"

ing W=0, ke=pe=0

tween turbines is" ges - 1)*C_P*(T_8 - T_7)

to occur until:"

ming: adiabatic, ke=pe=0 steady-flow"

_7s)

tages*w_turb

investigated.

A

"Input data for Helium" C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the n

1} {Nstages = T_6 = 1200 [K]Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa]

Eta_reg = 1.0 "regenerator effectiveness"entorpic efficienEta_c =1.0 "Compressor is

Eta_t =1.0 "Turbine isentropic effic_p = Pratio^(1/Nstages) R

"Isentropic Compressor ana T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 "T_2s is the isentropic value of T_2 at compressor exit" Eta_c = w_compisen/w_comp

> W_compisen" "compressor adiabatic efficiency, W_compompresso"Conservation of energy for the c

teae_in - e_out = DELTAe=0 for s_1 w_compisen = C_P*(T_2s-T

"Actual compressor analysis:" w_comp = C_P*(T_2 - T_1)

ccur such that T_3 = T_1 and the compressors have"Since intercooling is assumed to oratio, the work input to each comprew_comp_total = Nstages*w_comp "External heat exchanger analysis" SSSF First Law for the heat exchanger, assum"

e_in - e_out =DELTAe_cv =0 for steady flow"

t exchanger + the reheat be"The heat added in the external hea C_P*(T_6 - T_5) +(Nstaq_in_total =

"Reheat is assumedT_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p "T_7s is the isentropic value of T_7 at turbine exit"

) T_7s = T_6*(1/R_p)^((k-1)/k"Turbine adiabatic efficiency, w_turbisen > w_turb" Eta_t = w_turb /w_turbisen

c turbine, assu"SSSF First Law for the isentropi_in -e_out = DELTAe_cv = 0 fore

w_turbisen = C_P*(T_6 - T "Actual Turbine analysis:"

_turb = C_P*(T_6 - T_7) ww_turb_total = Ns "Cycle analysis"

_net=w_turb_total-w_comp_total w



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mp/w_turb "Back work ratio"

rator effectiveness gives T_5 as:"

nergy balance on regenerator gives T_10 as:"

ycle thermal efficiency with regenerator"

mperatures, T_6 and T_1 for this problem." ta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

ηth,Ericksson ηth,Regenerative Nstages

Bwr=w_co P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 "The regeneEta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "ET_4 + T_9=T_5 + T_10 "CEta_th_regenerative=w_net/q_in_total*Convert(, %) "[%]" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min teE

[%] [%] 75 32.43 1 75 58.9 2 75 65.18 3 75 67.95 4 75 71.18 7 75 73.29 15 75 73.66 19 75 73.84 22

0 2 4 6 8 10 12 14 16 18 20 22 2430

40

50

60

70

80

Nstages

ηth

[%]

EricssonIdeal Regenerative Brayton


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Fundamentals of Engineering (FE) Exam Problems

9-184 An Otto cycle with air as the working fluid has a compression ratio of 10.4. Under cold air standard conditions, the thermal efficiency of this cycle is

(a) 10% (b) 39% (c) 61% (d) 79% (e) 82%

Answer (c) 61%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).

r=10.4 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value"

9-185 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is

(a) Carnot (b) Stirling (c) Ericsson (d) Otto (e) All are the same

Answer (d) Otto

9-186 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 600 kW of net power. The rate of entropy change of the working fluid during the heat addition process is

(a) 0 (b) 0.300 kW/K (c) 0.353 kW/K (d) 0.261 kW/K (e) 2.0 kW/K

Answer (c) 0.353 kW/K


TL=300 "K" TH=2000 "K" Wnet=600 "kJ/s" Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"



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9-187 Air in an ideal Diesel cycle is compressed from 2 L to 0.13 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is

(a) 41% (b) 59% (c) 66% (d) 70% (e) 78%

Answer (b) 59%


V1=2 "L" V2= 0.13 "L" V3= 0.30 "L" r=V1/V2 rc=V3/V2 k=1.4 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency"

9-188 Helium gas in an ideal Otto cycle is compressed from 20°C and 2.5 L to 0.25 L, and its temperature increases by an additional 700°C during the heat addition process. The temperature of helium before the expansion process is

(a) 1790°C (b) 2060°C (c) 1240°C (d) 620°C (e) 820°C

Answer (a) 1790°C


k=1.667 V1=2.5 V2=0.25 r=V1/V2 T1=20+273 "K" T2=T1*r^(k-1) T3=T2+700-273 "C" "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K"



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9-189 In an ideal Otto cycle, air is compressed from 1.20 kg/m3 and 2.2 L to 0.26 L, and the net work output of the cycle is 440 kJ/kg. The mean effective pressure (MEP) for this cycle is

(a) 612 kPa (b) 599 kPa (c) 528 kPa (d) 416 kPa (e) 367 kPa

Answer (b) 599 kPa


rho1=1.20 "kg/m^3" k=1.4 V1=2.2 V2=0.26 m=rho1*V1/1000 "kg" w_net=440 "kJ/kg" Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"

9-190 In an ideal Brayton cycle, air is compressed from 95 kPa and 25°C to 1100 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is

(a) 45% (b) 50% (c) 62% (d) 73% (e) 86%

Answer (b) 50%


P1=95 "kPa" P2=1100 "kPa" T1=25+273 "K" rp=P2/P1 k=1.4 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"



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9-191 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20°C and 1000°C with argon as the working fluid. The net work output of the cycle is

(a) 68 kJ/kg (b) 93 kJ/kg (c) 158 kJ/kg (d) 186 kJ/kg (e) 310 kJ/kg

Answer (c) 158 kJ/kg


P1=100 "kPa" P2=1200 "kPa" T1=20+273 "K" T3=1000+273 "K" rp=P2/P1 k=1.667 Cp=0.5203 "kJ/kg.K" Cv=0.3122 "kJ/kg.K" T2=T1*rp^((k-1)/k) q_in=Cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=Cv*(T3-T2) "Using Cv instead of Cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using Cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*Cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*Cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"

9-192 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be

(a) 74 kJ/kg (b) 95 kJ/kg (c) 109 kJ/kg (d) 128 kJ/kg (e) 177 kJ/kg

Answer (b) 95 kJ/kg


wcomp/wturb=0.4 wturb-wcomp=150 "kJ/kg" Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"



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9-193 In an ideal Brayton cycle, air is compressed from 100 kPa and 25°C to 1 MPa, and then heated to 927°C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is

(a) 349°C (b) 426°C (c) 622°C (d) 733°C (e) 825°C

Answer (a) 349°C


P1=100 "kPa" P2=1000 "kPa" T1=25+273 "K" T3=900+273 "K" rp=P2/P1 k=1.4 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"

9-194 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25°C to 400 kPa, and then heated to 1200°C before entering the turbine. The highest temperature that argon can be heated in the regenerator is

(a) 246°C (b) 846°C (c) 689°C (d) 368°C (e) 573°C

Answer (e) 573°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=0.5203 "kJ/kg.K" P1=100 "kPa" P2=400 "kPa" T1=25+273 "K" T3=1200+273 "K" "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer"



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9-195 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10°C to 400 kPa and 175°C, is heated to 450°C in the regenerator, and then further heated to 1000°C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is

(a) 33% (b) 44% (c) 62% (d) 77% (e) 89%

Answer (d) 77%


k=1.4 Cp=1.005 "kJ/kg.K" P1=80 "kPa" P2=400 "kPa" T1=10+273 "K" T2=175+273 "K" T3=1000+273 "K" T5=450+273 "K" "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"

9-196 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20°C and 900°C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is

(a) 38% (b) 46% (c) 62% (d) 58% (e) 97%

Answer (c) 62%


k=1.3 rp=6 T1=20+273 "K" T3=900+273 "K"


"Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)"

Eta_regen=1-(T1/T3)*rp^((k-1)/k)


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9-197 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is

(a) 36% (b) 40% (c) 52% (d) 64% (e) 76%

Answer (e) 76%


k=1.4 rp=10 T1=290 "K" T3=1200 "K" Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" W1_Eta = 100 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation"

9-198 Air enters a turbojet engine at 320 m/s at a rate of 30 kg/s, and exits at 650 m/s relative to the aircraft. The thrust developed by the engine is

(a) 5 kN (b) 10 kN (c) 15 kN (d) 20 kN (e) 26 kN

Answer (b) 10 kN


Vel1=320 "m/s" Vel2=650 "m/s" Thrust=m*(Vel2-Vel1)/1000 "kN" m= 30 "kg/s" "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"

9-199 ··· 9-207 Design and Essay Problems.


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