ThermoSolutions CHAPTER09

9-1

Chapter 9 GAS POWER CYCLES

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

9-1C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannot be approximated using the hardware of actual power producing devices.

9-2C It is less than the thermal efficiency of a Carnot cycle.

9-3C It represents the net work on both diagrams.

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 Under the air standard assumptions, the combustion process is modeled as a heat addition process, and the exhaust process as a heat rejection process.

9-6C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) allthe 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.

9-7C The clearance volume is the minimum volume formed in the cylinder whereas the displacementvolume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center.

9-8C It is the ratio of the maximum to minimum volumes in the cylinder.

9-9C 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.

9-10C Yes.

9-11C Assuming no accumulation of carbon deposits on the piston face, the compression ratio will remainthe same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a cargets older as a result of wear and tear.

9-12C 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.

9-13C 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.

9-2

9-14 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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

P

qout

qin

3

4

2

1

qout

qin

3

4

2

1

kJ/kg828.149.101310kPa2668

kPa100

kPa2668kPa800K539.8K1800

1310kJ/kg1487.2

K1800

K539.8kJ/kg389.22

088.111.386kPa100kPa800

386.1kJ/kg300.19

K300

43

4

22

33

2

22

3

33

33

2

2

1

2

11

34

3

12

1

hPPP

P

PTT

PT

PT

P

Pu

Tu

PPP

P

Ph

T

rr

r

rr

r

vv

Tv

T

From energy balances,

kJ/kg570.19.5270.1098

kJ/kg527.919.3001.828

kJ/kg1098.02.3892.1487

outinoutnet,

14out

23in

qqw

hhq

uuq s

(c) Then the thermal efficiency becomes

51.9%kJ/kg1098.0

kJ/kg570.1

in

outnet,th q

w

9-3

9-15 EES Problem 9-14 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 beplotted.Analysis Using EES, the problem is solved as follows:

"Input Data"T[1]=300 [K] P[1]=100 [kPa]P[2] = 800 [kPa] T[3]=1800 [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"

9-4

T3 [K] th qin,total [kJ/kg] Wnet [kJ/kg]1500 50.91 815.4 415.11700 51.58 1002 516.81900 52.17 1192 621.72100 52.69 1384 729.22300 53.16 1579 839.12500 53.58 1775 951.2

5.0 5.3 5.5 5.8 6.0 6.3 6.5 6.8 7.0 7.3 7.5200

400

600

800

1000

1200

1400

1600

1800

2000

s [kJ/kg-K]

T[K]

100 kPa

800 kPa

Air

1

2

3

4

10-2 10-1 100 101 102101

102

103

4x103

v [m3/kg]

P[kPa]

300 K

1800 K

Air

1

2

3

4

9-5

1500 1700 1900 2100 2300 250050.5

51

51.5

52

52.5

53

53.5

54

T[3] [K]

th

1500 1700 1900 2100 2300 2500800

1000

1200

1400

1600

1800

T[3] [K]

qin,total

[kJ/kg]

1500 1700 1900 2100 2300 2500400

500

600

700

800

900

1000

T[3] [K]

wnet

[kJ/kg]

9-6

9-16 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the maximum temperature in the cycle 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 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) From the ideal gas isentropic relations and energy balance, P

qin

q41

q34

3

4

2

1

K579.2kPa100kPa1000K300

0.4/1.4/1

1

212

kk

PPTT

K33603max3

2323in

579.2KkJ/kg1.005kJ/kg2800 TTT

TTchhq p

v (c) K336K3360

kPa1000kPa100

33

44

4

44

3

33 TPPT

TP

TP vv

21.0%kJ/kg2800kJ/kg2212

11

kJ/kg2212K300336KkJ/kg1.005K3363360KkJ/kg0.718

in

outth

1443

1443out41,out34,out

qq

TTcTTchhuuqqq

pv

T

q34

q41

qin

3

4

2

1

s

Discussion The assumption of constant specific heats at room temperature is not realistic in this case thetemperature changes involved are too large.

9-17E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the total heat input 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-17E. Analysis (b) The properties of air at various states are

PT Btu/lbm129.06Btu/lbm,92.04R540 111 hu

q12

qout1

2q23

4

3

s

qout

q23

q12

1

2

4

3

v

Btu/lbm593.22317.01242psia57.6psia14.7

1242Btu/lbm48.849

R3200

psia57.6psia14.7R540R2116

Btu/lbm1.537,R2116Btu/lbm04.39230004.92

43

4

33

11

22

1

11

2

22

22

in,121212in,12

34

3

hPPP

P

Ph

T

PTT

PT

PT

P

hTquu

uuq

rr

r

vv

T

From energy balance,

Btu/lbm464.1606.12922.59338.312300

Btu/lbm312.381.53748.849

14out

in23,in12,in

23in23,

hhqqqq

hhqBtu/lbm612.38

(c) Then the thermal efficiency becomes

24.2%Btu/lbm612.38Btu/lbm464.16

11in

outth q

q

9-7

9-18E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the total heat input 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 at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R,and k = 1.4 (Table A-2E).

PAnalysis (b)

Btu/lbm217.4R22943200RBtu/lbm0.24

psia62.46psia14.7R540R2294

R2294R540Btu/lbm.R0.171Btu/lbm300

2323in,23

11

22

1

11

2

22

2

2

1212in,12

TTchhq

PTT

PT

PT

PT

TTTcuuq

P

vv

v

qout

q23

q12

1

2

4

3

v

Process 3-4 is isentropic:

Btu/lbm378.55402117Btu/lbm.R0.240

4.217300

R2117psia62.46

psia14.7R3200

1414out

in,23in,12in

0.4/1.4/1

3

434

TTchhq

qqq

PP

TT

p

kk

Btu/lbm517.4

T

q12

qout1

2q23

4

3

(c) 26.8%Btu/lbm517.4Btu/lbm378.5

11in

outth q

q s

9-19 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the heat rejected 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 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) K579.2kPa100kPa1000

K3000.4/1.4/1

1

212

kk

PP

TTP

qout

qin

1

2 3

v

K12662.579KkJ/kg1.005kg0.004kJ2.76 33

2323in

TT

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/kg7.273KkJ/kg0.287kPa1000K1266

kPa100K300

2kPa1001000

22area

3

3

1

11331

1331 P

RTP

RTPPPPw vv T

Energy balance for process 3-1 gives

kJ1.679K1266-300KkJ/kg0.718273.7kg0.004)(

)(

31out31,31out31,out31,

31out31,out31,systemoutin

TTcwmTTmcmwQ

uumWQEEE

vv

qin

qout1

23

s

(c) 39.2%kJ2.76kJ1.679

11in

outth Q

Q

9-8

9-20 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the net work per cycle 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 P

qout

qin

1

2

3

kJ0.422651.1073.2

kJ1.651kJ/kg290.16840.38kg0.003

kJ2.073kJ/kg206.91897.91kg0.003

kJ/kg840.3851.8207.2kPa380

kPa95

2.207kJ/kg,897.91

K1160K290kPa95kPa380

kJ/kg290.16kJ/kg206.91

K029

outinoutnet,

13out

12in

32

3

2

11

22

1

11

2

22

1

11

23

2

QQW

hhmQ

uumQ

hPPP

P

Pu

TPP

TT

PT

P

hu

T

rr

r

vv

v

T

qout

qin

2

31

s(c) 20.4%

kJ2.073kJ0.422

in

outnet,th Q

W

9-21 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the net work per cycle 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 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) From the isentropic relations and energy balance,

kJ0.3948.187.1

kJ1.48K290780.6KkJ/kg1.005kg0.003

kJ1.87K2901160KkJ/kg0.718kg0.003

K780.6kPa380

kPa95K1160


outinoutnet,

1313out

1212in

0.4/1.4/1

2

323

11

22

1

11

2

22

QQW

TTmchhmQ

TTmcuumQ

PP

TT

TPP

TT

PT

P

p

kk

v

vv P

qout

qin

2

31

s

qout

qin

1

2

3

v

T

(c) 20.9%kJ1.87kJ0.39

in

netth Q

W

9-9

9-22 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle isto 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,

kPa3.935K300K900

kPa201.4/0.41/

3

232

/1

3

2

3

2

kk

kk

TT

PP

PP

TT T

The heat input is determined from

Then,

kJ0.393kJ0.58890.667

.7%66K900K300

11

kJ0.5889KkJ/kg0.2181K900kg0.003

KkJ/kg0.2181kPa2000kPa935.3

lnKkJ/kg0.287lnln

inthoutnet,

th

12in

1

20

1

212

QW

TT

ssmTQ

PP

RTT

css

H

L

H

p

1

4qout

qin2

3

900

300

s

9-23 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle, the heat transfer to the working fluid, and the mass of the working fluid are to be determined.Assumptions Air is an ideal gas with variable specific heats. Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process,which is state 1.

(Table A-17)T P

T P

r

r

1

4

1

4

238

2 379

1200 K

350 K .T

Wnet = 0.5 kJ

1

4Qout

Qin2

3

1200

max41 kPa300379.2

238

4

1 PPP

PP

r

rkPa30,013

350(b) The heat input is determined from

s

kJ0.7060.7083/kJ0.5/

%83.70K1200K350

11

thoutnet,in

th

WQ

TT

H

L

(c) The mass of air is

kg0.00296kJ/kg169.15kJ0.5

kJ/kg169.15K3501200KkJ/kg0.199

KkJ/kg0.199

kPa150kPa300

lnKkJ/kg0.287ln

outnet,

outnet,

12outnet,

21

3

403434

wW

m

TTssw

ss

PP

Rssss

LH

9-10

9-24 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle, the heat transfer to the working fluid, and the mass of the working fluid are to be determined.Assumptions Helium is an ideal gas with constant specific heats. Properties The properties of helium at room temperature are R = 2.0769 kJ/kg.K and k = 1.667 (Table A-2).Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process,which is state 1.

or,

kPa652471.667/0.661/

4

141

/1

4

1

4

1

K350K1200

kPa300kk

kk

TT

PP

PP

TT

Wnet = 0.5 kJ

350

1200 1

4

Qin2

3

T

(b) The heat input is determined from s

kJ0.7060.7083/kJ0.5/

70.83%K1200K350

11

thoutnet,in

th

WQ

TT

H

L

(c) The mass of helium is determined from

kg0.000409kJ/kg1223.7kJ0.5

kJ/kg1223.7K3501200KkJ/kg1.4396

KkJ/kg1.4396

kPa150kPa300

lnKkJ/kg2.0769lnln

outnet,

outnet,

12outnet,

21

3

40

3

434

wW

m

TTssw

ss

PP

RTT

css

LH

p

9-11

9-25 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimumpressure 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.Assumptions Air is an ideal gas with constant specific heats. 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

K350K750KkJ/kg0.25kJ/kg10012net LLLH TTTTssw

The pressure at state 4 is determined from

or

kPa1.110K350K750kPa800 4

1.4/0.4

4

1/

4

141

/1

4

1

4

1

PP

TT

PP

PP

TT

kk

kk

qout

wnet=100 kJ/kg

750 K 1

4

qin2

3

T

s

The minimum pressure in the cycle is determined from

kPa46.133

3

40

3

43412

kPa110.1lnKkJ/kg0.287KkJ/kg25.0

lnln

PP

PP

RTT

css p

(b) The heat rejection from the cycle is kgkJ/87.5kJ/kg.K)K)(0.25350(12out sTq L

(c) The thermal efficiency is determined from

0.533K750K350

11thH

L

TT

(d) The power output for the Carnot cycle is

kW9000kJ/kg)kg/s)(10090(netCarnot wmW

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

0.578kW9000kW5200

Carnot

actualII W

W

9-12

Otto Cycle

9-26C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heataddition, (3) isentropic expansion, and (4) v = constant heat rejection.

9-27C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.

9-28C 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.

9-29C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.

9-30C It increases with both of them.

9-31C Because high compression ratios cause engine knock.

9-32C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k= 1.667.

9-33C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasolineengines.

9-13

9-34 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle 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. The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.

621.2kJ/kg214.07

K3001

11

r

uT

vP

750 kJ/kg 2

3

1

4

vkPa1705kPa95K300K673.1

8

kJ/kg491.2K673.1

65.772.621811

11

2

2

12

1

11

2

22

2

2

1

2112

PTT

PT

PT

P

uT

r rrr

v

vvv

vvv

vv

Process 2-3: v = constant heat addition.

588.6kJ/kg241.217502.4913

3in23,2323in23,

r

Tquuuuq

v

K1539

kPa3898kPa1705K673.1K1539

22

33

2

22

3

33 PTT

PT

PT

P vv

(b) Process 3-4: isentropic expansion.

kJ/kg571.69K774.5

70.52588.684

4

2

1334 u

Tr rrr vv

v

vv

Process 4-1: v = constant heat rejection.q u uout 571.69 214.07 357.62 kJ / kg4 1

kJ/kg392.462.357750outinoutnet, qqw

(c) 52.3%kJ/kg750kJ/kg392.4

in

outnet,th q

w

(d)

kPa495.0kJ

mkPa1/81/kgm0.906

kJ/kg392.4)/11(

MEP

/kgm0.906kPa95

K300K/kgmkPa0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv

9-14

9-35 EES Problem 9-34 is reconsidered. The effect of the compression ratio on the net work output and thermal efficiency is to be investigated. Also, T-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]=300 [K] P[1]=95 [kPa]q_23 = 750 [kJ/kg] {r_comp = 8}

"Process 1-2 is isentropic compression"s[1]=entropy(air,T=T[1],P=P[1])s[2]=s[1]T[2]=temperature(air, s=s[2], P=P[2])P[2]*v[2]/T[2]=P[1]*v[1]/T[1]P[1]*v[1]=R*T[1]R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp"Conservation of energy for process 1 to 2"q_12 - w_12 = DELTAu_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 w_41 =0 "constant volume process"DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4])q_in_total=q_23q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 "Thermal efficiency, in percent""The mean effective pressure is:"MEP = w_net/(V[1]-V[2])"[kPa]"

9-15

rcomp th MEP [kPa] wnet [kJ/kg]5 43.78 452.9 328.46 47.29 469.6 354.77 50.08 483.5 375.68 52.36 495.2 392.79 54.28 505.3 407.1

10 55.93 514.2 419.5

4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T [K

]

95kP

a

3900 kPa0.

9

0.11 m3/kg

Air

1

2

3

4

10-2 10-1 100 101 102101

102

103

104

v [m3/kg]

P [k

Pa]

300 K

1500 K

Air

1

2

3

4

s2 = s1 = 5.716 kJ/kg-K

s4 = 33 = 6.424 kJ/kg-K

9-16

5 6 7 8 9 10320

340

360

380

400

420

rcomp

wnet

[kJ/kg]

5 6 7 8 9 10450

460

470

480

490

500

510

520

rcomp

MEP[kPa]

5 6 7 8 9 1042

44

46

48

50

52

54

56

rcomp

th

9-17

9-36 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle 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.

kPa1745kPa95K300K689

8

K6898K300

11

2

2

12

1

11

2

22

0.41

2

112

PTT

PT

PT

P

TTk

v

vvv

v

v P

750 kJ/kg 2

3

1

4

vProcess 2-3: v = constant heat addition.

K1734K689KkJ/kg0.718kJ/kg750

3

3

2323in23,

TTTTcuuq v

kPa4392kPa1745K689K1734

22

33

2

22

3

33 PTT

PT

PT

P vv


K75581K1734

0.41

4

334

k

TTv

v

Process 4-1: v = constant heat rejection.kJ/kg327K300755KkJ/kg0.7181414out TTcuuq v

kJ/kg423327750outinoutnet, qqw

(c) 56.4%kJ/kg750kJ/kg423

in

outnet,th q

w

(d)

kPa534kJ

mkPa1/81/kgm0.906

kJ/kg423)/11(

MEP

/kgm0.906kPa95

K300K/kgmkPa0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv

9-18

9-37 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 effectivepressure 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.

P

kPa2338kPa100K308K757.9

9.5

K757.99.5K308

11

2

2

12

1

11

2

22

0.41

2

112

PTT

PT

PT

P

TTk

v

vvv

v

v

QoutQin

2

3

1

4

vProcess 3-4: isentropic expansion.

K19690.41

3

443 9.5K800

k

TTv

v


kPa6072kPa2338K757.9K1969

22

33

2

22

3

33 PTT

PT

PT

P vv

(b) kg10788.6K308K/kgmkPa0.287

m0.0006kPa100 43

3

1

11

RTP

mV

kJ0.590K757.91969KkJ/kg0.718kg106.788 42323in TTmcuumQ v

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

kJ0.240K308800KkJ/kg0.718kg106.788)( 41414out TTmcuumQ v

kJ0.350240.0590.0outinnet QQW

59.4%kJ0.590kJ0.350

in

outnet,th Q

W

(d)

kPa652kJ

mkPa1/9.51m0.0006

kJ0.350)/11(

MEP3

31

outnet,

21

outnet,

max2min

rWW

r

VVV

VVV

9-19

9-38 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 effectivepressure 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.

P

kPa2338kPa100K308K757.9

9.5

K757.99.5K308

11

2

2

12

1

11

2

22

0.41

2

112

PTT

PT

PT

P

TTk

v

vvv

v

v

800 K

Polytropic

Qout

Qin

2

3

1308 K

4

vProcess 3-4: polytropic expansion.

kg10788.6K308K/kgmkPa0.287

m0.0006kPa100 43

3

1

11

RTP

mV

kJ0.53381.351

K1759800KkJ/kg0.287106.7881

9.5K800

434

34

35.01

3

443

nTTmR

W

TTn

K1759v

v

Then energy balance for process 3-4 gives

kJ0.0664kJ0.5338K1759800KkJ/kg0.718kg106.788 4in34,

out34,34out34,34in34,

34out34,in34,

outin

QWTTmcWuumQ

uumWQ

EEE system

v

That is, 0.066 kJ of heat is added to the air during the expansion process (This is not realistic, and probablyis due to assuming constant specific heats at room temperature).(b) Process 2-3: v = constant heat addition.

kPa5426kPa2338K757.9K1759

22

33

2

22

3

33 PTT

PT

PT

P vv

kJ0.4879K757.91759KkJ/kg0.718kg106.788 4in23,

2323in23,

QTTmcuumQ v

Therefore,kJ0.55430664.04879.0in34,in23,in QQQ

(c) Process 4-1: v = constant heat rejection.kJ0.2398K308800KkJ/kg0.718kg106.788 4

1414out TTmcuumQ v

kJ0.31452398.05543.0outinoutnet, QQW

56.7%kJ0.5543kJ0.3145

in

outnet,th Q

W

(d)

kPa586kJ

mkPa1/9.51m0.0006

kJ0.3145)/11(

MEP3

31

outnet,

21

outnet,

max2min

rWW

rV

VVV

VV

9-20

9-39E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heattransferred 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.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-17E. Analysis (a) Process 1-2: isentropic compression. P

540

2400 R

qout

qin

2

3

1

4

v

T 32.144Btu/lbm92.04

R5401

11

r

uv

Btu/lbm11.28204.1832.144811

21

2222

ur rrr vv

v

vv


Btu/lbm241.4228.21170.452

419.2Btu/lbm452.70

R2400

23

33

3

uuq

uT

in

rv


Btu/lbm205.5435.19419.28 43

4334

ur rrr vvv

vv

Process 4-1: v = constant heat rejection.Btu/lbm50.11304.9254.20514out uuq

53.0%Btu/lbm241.42Btu/lbm113.50

11in

outth q

q

(c) 77.5%R2400

R54011Cth,

H

L

TT

9-40E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount ofheat transferred to the argon during the heat addition process, the thermal efficiency, and the thermalefficiency 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 andpotential energy changes are negligible. 3 Argon is an ideal gas with constant specific heats.Properties The properties of argon are cp = 0.1253 Btu/lbm.R, cv = 0.0756 Btu/lbm.R, and k = 1.667 (Table A-2E). Analysis (a) Process 1-2: isentropic compression.

P

R21618R540 0.6671

2

112

k

TTv

v

Process 2-3: v = constant heat addition.Btu/lbm18.07R21612400Btu/lbm.R0.07562323in TTcuuq v

qout

qin

2

3

1

4

v(b) Process 3-4: isentropic expansion.

R60081R2400

0.6671

4

334

k

TTv

v

Process 4-1: v = constant heat rejection.Btu/lbm4.536R540600Btu/lbm.R0.07561414out TTcuuq v

74.9%Btu/lbm18.07Btu/lbm4.536

11in

outth q

q

(c) 77.5%R2400R540

11Cth,H

L

TT

9-21

9-41 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeledas 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.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K,and k = 1.349 (Table A-2b).Analysis (a) Process 1-2: polytropic compression

Qout

4

3

2

Qin

1

kPa199510kPa100

K4.66410K333

1.3

2

112

1-1.31

2

112

n

n

PP

TT

v

v

v

v

Process 2-3: constant volume heat addition

K2664kPa1995kPa8000K664.4

2

323 P

PTT

kJ/kg1646K664.42664KkJ/kg0.8232323in TTcuuq v

Process 3-4: polytropic expansion.

K13351-1.31

4

334 10

1K2664n

TTv

v

kPa9.400101kPa8000

1.3

1

234

n

PPv

v

Process 4-1: constant voume heat rejection.kJ/kg8.824K3331335KkJ/kg0.8231414out TTcuuq v

(b) The net work output and the thermal efficiency are kJ/kg820.98.8241646outinoutnet, qqw

0.499kJ/kg1646kJ/kg820.9

in

outnet,th q

w

(c) The mean effective pressure is determined as follows

kPa954.3kJ

mkPa1/101/kgm0.9557

kJ/kg820.9)/11(

MEP

/kgm0.9557kPa100

K333K/kgmkPa0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv

(d) The clearance volume and the total volume of the engine at the beginning of compression process (state1) are

9-22

33

m0002444.0m0022.0

10 cc

c

c

dc VV

VV

VVr

31 m002444.00022.00002444.0dc VVV

The total mass contained in the cylinder is

kg0.002558K333K/kgmkPa0.287

)m444kPa)/0.002100(3

3

1

11

RTVP

mt

The engine speed for a net power output of 70 kW is

rev/min4001min1

s60cycle)kJ/kgkg)(820.9002558.0(

kJ/s70rev/cycle)2(2net

net

wmW

nt

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

(e) The mass of fuel burned during one cycle is

kg0001505.0kg)002558.0(

16AF ff

f

f

ft

f

a mm

mm

mmmm

Finally, the specific fuel consumption is

g/kWh258.0kWh1

kJ3600kg1

g1000kJ/kg)kg)(820.9002558.0(

kg0001505.0sfc

netwmm

t

f

Diesel Cycle

9-42C A diesel engine differs from the gasoline engine in the way combustion is initiated. In dieselengines 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 thecutoff ratio decreases, the efficiency of the diesel cycle increases.

9-23

9-47 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. Thetemperature after the heat addition process, 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 variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K.The properties of air are given in Table A-17.

P

qout

qin 32

1

4

v

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

2.621kJ/kg214.07

K3001

11

r

uT

v

kJ/kg90.98K862.4

825.382.6211611

2

2

1

2112 h

Tr rrr vv

v

vv

Process 2-3: P = constant heat addition.

546.4kJ/kg1910.6

K862.4223

322

2

33

2

22

3

33

r

hTTT

TP

TP

vv

vvvK1724.8

(b) kJ/kg019.719.8906.191023in hhq

Process 3-4: isentropic expansion.

kJ/kg659.737.36546.42

1622 4

2

4

3

43334

urrrrr vv

v

vv

v

vv

Process 4-1: v = constant heat rejection.

56.3%kJ/kg1019.7kJ/kg445.63

11

kJ/kg445.6307.2147.659

in

outth

14out

qq

uuq

(c)

kPa675.9kJ

mkPa1/161/kgm0.906

kJ/kg574.07/11

MEP

/kgm0.906kPa95

K300K/kgmkPa0.287

kJ/kg574.0763.4457.1019

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

outinoutnet,

rww

r

PRT

qqw

vvv

vvv

vv

9-24

9-48 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. Thetemperature after the heat addition process, 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. P

qout

qin 32

1

4

v

K909.416K300 0.41

2

112

k

TTv

v


K1818.8K909.422 222

33

2

22

3

33 TTTT

PT

Pv

vvv

(b) kJ/kg913.9K909.41818.8KkJ/kg1.0052323in TTchhq p


K791.7162K1818.8

2 0.41

4

23

1

4

334

kk

TTTv

v

v

v


61.4%kJ/kg913.9

kJ/kg35311

kJ/kg353K300791.7KkJ/kg0.718

in

outth

1414out

qq

TTcuuq v

(c)

kPa660.4kJ

mkPa1/161/kgm0.906

kJ/kg560.9/11

MEP

/kgm0.906kPa95

K300K/kgmkPa0.287

kJ/kg560.93539.913

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

outinnet.out

rww

r

PRT

qqw

vvv

vvv

vv

9-25

9-49E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, theheat rejection per unit mass, 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-17E.

PAnalysis (a) Process 1-2: isentropic compression.3000 R

qout

qin 32

1

4

v

T 32.144Btu/lbm40.92

R5401

11

r

uv

Btu/lbm402.05R1623.6

93.732.1442.18

11

2

2

1

2112 h

Tr rrr vv

v

vv


1.848R1623.6

R3000

2

3

2

3

2

22

3

33

TT

TP

TP

v

vvv

(b)

Btu/lbm388.6305.40268.790

180.1Btu/lbm790.68

R3000

23in

33

3

hhq

hT

rv


Btu/lbm91.250621.11180.1848.1

2.18848.1848.1 4

2

4

3

43334

urrrrr vv

v

vv

v

vv


(c) 59.1%

Btu/lbm158.87

Btu/lbm388.63Btu/lbm158.87

11

04.9291.250

in

outth

14out

qq

uuq

9-26

9-50E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, theheat rejection per unit mass, 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 at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R,and k = 1.4 (Table A-2E).Analysis (a) Process 1-2: isentropic compression. P qin 32

1

4

v

R172418.2R540 0.41

2

112

k

TTv

v


1.741R1724R3000

2

3

2

3

2

22

3

33

TT

TP

TP

v

vvv

(b) Btu/lbm306R17243000Btu/lbm.R0.2402323in TTchhq p


R117318.21.741R3000

741.1 0.41

4

23

1

4

334

kk

vTTT

v

v

v


(c) 64.6%

Btu/lbm108

Btu/lbm306Btu/lbm108

11

R0541173Btu/lbm.R0.171

in

outth

1414out

qq

TTcuuq v

9-27

9-51 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermalefficiency 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.

P

qout

qin 32

1

4

v

K971.120K293 0.41

2

112

k

TTV

V


2.265K971.1K2200

2

3

2

3

2

22

3

33

TT

TP

TP

V

VVV


63.5%kJ/kg1235kJ/kg784.4

kJ/kg784.46.4501235

kJ/kg450.6K293920.6KkJ/kg0.718

kJ/kg1235K971.12200KkJ/kg1.005

K920.620

2.265K2200265.2265.2

in

outnet,th

outinoutnet,

1414out

2323in

0.41

3

1

4

23

1

4

334

qw

qqw

TTcuuq

TTchhq

rTTTT

p

kkk

v

V

V

V

V

(b)

kPa933kJ

mkPa1/201/kgm0.885

kJ/kg784.4/11

MEP

/kgm0.885kPa95

K293K/kgmkPa0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv

9-28

9-52 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. P

Polytropic

qout

qin 32

1

4

v

K971.120K293 0.41

2

112

k

TTV

V


2.265K971.1K2200

2

3

2

3

2

22

3

33

TT

TP

TP

V

VVV

Process 3-4: polytropic expansion.

kJ/kg526.3K2931026KkJ/kg0.718

kJ/kg1235K971.12200KkJ/kg1.005

K102620

2.265K2200r

2.2652.265

1414out

2323in

0.351n

3

1n

4

23

1

4

334

TTcuuq

TTchhq

TTTT

p

n

v

V

V

V

V

Note that qout in this case does not represent the entire heat rejected since some heat is also rejected during the polytropic process, which is determined from an energy balance on process 3-4:

kJ/kg120.1K22001026KkJ/kg0.718kJ/kg963

kJ/kg9631.351

K22001026KkJ/kg0.2871

34out34,in34,34out34,in34,

systemoutin

34out34,

TTcwquuwq

EEEnTTR

w

v

which 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 losesheat during polytropic expansion. The cause of this unrealistic result is the constant specific heatassumption. If we were to use u data from the air table, we would obtain

kJ/kg1.128)4.18723.781(96334out34,in34, uuwq

which is a heat loss as expected. Then qout becomeskJ/kg654.43.5261.128out41,out34,out qqq

and

47.0%kJ/kg1235kJ/kg580.6

kJ/kg580.64.6541235

in

outnet,th

outinoutnet,

qw

qqw

(b)

kPa691kJ

mkPa11/201/kgm0.885

kJ/kg580.6/11

MEP

/kgm0.885kPa95

K293K/kgmkPa0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv

9-29

9-53 EES Problem 9-52 is reconsidered. The effect of the compression ratio on the net work output, meaneffective pressure, and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle areto be plotted.Analysis Using EES, the problem is solved as follows:

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])^ns[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])

9-30

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 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 th MEP [kPa] wnet [kJ/kg]14 47.69 970.8 797.916 50.14 985 817.418 52.16 992.6 829.820 53.85 995.4 837.022 55.29 994.9 840.624 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]

95kP

a

340.1

kPa

5920

kPa

0.044

0.1

0.88m3/kg

Air

2

1

3

4

10-2 10-1 100 101 102101

102

103

104

101

102

103

104

v [m3/kg]

P[kPa] 293 K

1049 K

2200 K

5.69

6.74kJ/kg-K

Air

9-31

14 16 18 20 22 24790

800

810

820

830

840

850

rcomp

wnet

[kJ/kg]

14 16 18 20 22 2447

49

51

53

55

57

rcomp

th

14 16 18 20 22 24970

975

980

985

990

995

1000

rcomp

MEP[kPa]

9-32

9-54 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 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. 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 Process 1-2: isentropic compression.

P

Qout

Qin 32

1

4

v

K101917K328 0.41

2

112

k

TTV

V


K2241K10192.22.2 222

33

2

22

3

33 TTTT

PT

Pv

vvv


kW46.6kJ/rev1.864rev/s1500/60

kJ/rev864.1174.1038.3

kJ174.1K328989.2KkJ/kg0.718kg10473.2

kJ3.038K)10192241)(KkJ/kg1.005)(kg10.4732(

kg10473.2)K328)(K/kgmkPa0.287(

)m0.0024)(kPa97(

K989.217

2.2K22412.22.2

outnet,outnet,

outinoutnet,

31414out

32323in

33

3

1

11

4.01

3

1

4

23

1

4

334

WnW

QQW

TTmcuumQ

TTmchhmQ

RTP

m

rTTTT

v

p

kkk

V

V

V

V

V

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

9-33

9-55 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined.Assumptions 1 The air-standard assumptions are applicable with nitrogen as the working fluid. 2 Kineticand 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).

PAnalysis Process 1-2: isentropic compression.

Qout

Qin 32

1

4

v

K101917K328 0.41

2

112

k

TTV

V


K2241K10192.22.2 222

33

2

22

3

33 TTTT

PT

Pv

vvv


kW46.6kJ/rev1.863rev/s1500/60

kJ/rev.8631175.1037.3

kJ1.175K328989.2KkJ/kg0.743kg102.391

kJ3.037K10192241KkJ/kg1.039kg102.391

kg10391.2K328K/kgmkPa0.2968

m0.0024kPa97

K989.2172.2K22412.22.2

outnet,outnet,

outinoutnet,

31414out

32323in

33

3

1

11

0.41

3

1

4

23

1

4

334

WnW

QQW

TTmcuumQ

TTmchhmQ

RTP

m

rTTTT

p

kkk

v

V

V

V

V

V

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

9-34

9-56 [Also solved by EES on enclosed CD] An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heat transferred at constant volume and the thermal efficiency of the cycle 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. P

x

Qout

1520.4kJ/kg

3

2

1

4

v

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

621.2kJ/kg214.07

K3001

11

r

uT

v

kJ/kg611.2K823.1

37.442.6211411

2

2

1

2112 u

Tr rrr vv

v

vv

Process 2-x, x-3: heat addition,

xx

xx

r

huhhuuqqq

hT

2.25032.6114.1520

012.2kJ/kg2503.2

K2200

32inx,3in2,xin

33

3v

By trial and error, we get Tx = 1300 K and hx = 1395.97, ux = 1022.82 kJ /kg.Thus,

and

27.1%kJ/kg1520.4kJ/kg411.62

ratio

kJ/kg411.622.61182.1022

in

in,2

2in,2

qq

uuq

x

xx

(b) cxxx

xx rTT

TP

TP

1.692K1300K220033

3

33

v

vvv

kJ/kg886.3648.16012.2692.114

692.1692.1 42

4

3

43334

urrrrr vv

v

vv

v

vv


55.8%kJ/kg1520.4kJ/kg672.23

11

kJ/kg72.23607.2143.886

in

outth

14out

qq

uuq

9-35

9-57 EES Problem 9-56 is reconsidered. The effect of the compression ratio 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]=300 [K] P[1]=100 [kPa] T[4]=2200 [K] q_in_total=1520 [kJ/kg] r_v = 14 v[1]/v[2]=r_v "Compression ratio"

"Process 1-2 is isentropic compression"s[1]=entropy(air,T=T[1],P=P[1])s[2]=s[1]s[2]=entropy(air, T=T[2], v=v[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 [kJ/kg]"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 constant pressure heat addition"s[4]=entropy(air, T=T[4], P=P[4]) {P[4]*v[4]/T[4]=P[3]*v[3]/T[3]}P[4]*v[4]=R*T[4]P[4]=P[3]"Conservation of energy for process 3 to4"q_34 -w_34 = DELTAu_34 w_34 =P[3]*(v[4]-v[3]) "constant pressure process"DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3])q_in_total=q_23+q_34"Process 4-5 is isentropic expansion"s[5]=entropy(air,T=T[5],P=P[5])s[5]=s[4]P[5]*v[5]/T[5]=P[4]*v[4]/T[4]{P[5]*v[5]=0.287*T[5]}"Conservation of energy for process 4 to 5"q_45 -w_45 = DELTAu_45 q_45 =0 [kJ/kg] "isentropic process"DELTAu_45=intenergy(air,T=T[5])-intenergy(air,T=T[4])"Process 5-1 is constant volume heat rejection"v[5]=v[1]"Conservation of energy for process 2 to 3"q_51 -w_51 = DELTAu_51

9-36

w_51 =0 [kJ/kg] "constant volume process"DELTAu_51=intenergy(air,T=T[1])-intenergy(air,T=T[5])

w_net = w_12+w_23+w_34+w_45+w_51 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"

rv th [%] wnet [kJ/kg]10 52.33 795.411 53.43 812.112 54.34 82613 55.09 837.414 55.72 846.915 56.22 854.616 56.63 860.717 56.94 865.518 57.17 869

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

3500

s [kJ/kg-K]

T[K]

100 kPa

382.7 kPa

3842 kPa

6025 kPa

T-s Diagram for Air Dual Cycle

1

2

3

4

5

v=const

p=const

1

2

3 4

5

s=const

10-2 10-1 100 101 102101

102

103

8x103

v [m3/kg]

P[kPa]

300 K

2200 K

P-v Diagram for Air Dual Cycle

9-37

10 11 12 13 14 15 16 17 1852

53

54

55

56

57

58

rv

th[%]

10 11 12 13 14 15 16 17 18790

800

810

820

830

840

850

860

870

rv

wnet

[kJ/kg]

9-58 An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heattransferred at constant volume and the thermal efficiency of the cycle 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.

K86214K300 0.41

2

112

k

TTv

v

Process 2-x, x-3: heat addition,

xx

xpx

xxxx

TT

TTcTTchhuuqqq

2200KkJ/kg1.005862KkJ/kg0.718kJ/kg1520.432

32in,3in,2in

v

P

4

1

2

3

1520.4kJ/kg

x

Qout

v

Solving for Tx we get Tx = 250 K which is impossible. Therefore, constant specific heats at roomtemperature turned out to be an unreasonable assumption in this case because of the very high temperaturesinvolved.

9-38

9-59 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximumtemperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the meaneffective pressure, the net power output, and the specific fuel consumption are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K,and k = 1.349 (Table A-2b).Analysis (a) Process 1-2: Isentropic compression

Qout

4

32

Qin

1

kPa434117kPa95

K7.88117K328

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

m0002813.0

m0045.017

c

c

c

c

dcr

V

V

V

V

VV

31 m004781.00045.00002813.0dc VVV

The total mass contained in the cylinder is

kg0.004825K328K/kgmkPa0.287

)m781kPa)(0.00495(3

3

1

11

RTP

mV

The mass of fuel burned during one cycle is

kg000193.0kg)004825.0(

24 ff

f

f

f

f

a mm

mm

mmmm

AF

Process 2-3: constant pressure heat additionkJ039.88)kJ/kg)(0.9kg)(42,500000193.0(HVin cf qmQ

K23833323in K)7.881(kJ/kg.K)kg)(0.823004825.0(kJ039.8)( TTTTmcQ v

The cutoff ratio is

2.7K881.7K2383

2

3

TT

(b) 33

12 m0002813.0

17m0.004781

rV

V

23

14

3323 m00076.0)m0002813.0)(70.2(

PPVV

VV

9-39


kPa2.363m0.004781m0.00076kPa4341

K1254m0.004781m0.00076K2383

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.kJ677.3K3281254KkJ/kg0.823kg)004825.0(14out TTmcQ v

The net work output and the thermal efficiency are kJ4.361677.3039.8outinoutnet, QQW

0.543kJ8.039kJ4.361

in

outnet,th Q

W

(c) The mean effective pressure is determined to be

kPa969.2kJ

mkPam)0002813.0004781.0(

kJ4.361MEP

3

321

outnet,

VV

W

(d) The power for engine speed of 2000 rpm is

kW72.7s60

min1rev/cycle)2(

(rev/min)2000kJ/cycle)(4.3612netnetnWW

Note that there are two revolutions in one cycle in four-stroke engines.(e) Finally, the specific fuel consumption is

g/kWh159.3kWh1

kJ3600kg1

g1000kJ/kg4.361

kg000193.0sfcnetW

mf

9-40

Stirling and Ericsson Cycles

9-60C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.

9-61C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.

9-62C The Stirling cycle.

9-63C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.

9-64E An ideal Ericsson engine with helium as the working fluid operates between the specifiedtemperature and pressure limits. The thermal efficiency of the cycle, the heat transfer rate in theregenerator, and the power delivered are to be determined.Assumptions Helium is an ideal gas with constant specific heats. Properties The gas constant and the specific heat of helium at room temperature are R = 0.4961 Btu/lbm.Rand cp = 1.25 Btu/lbm·R (Table A-2E).Analysis (a) The thermal efficiency of this totally reversible cycle is determined from

81.67%R3000

R55011th

H

L

TT

(b) The amount of heat transferred in the regenerator is

Btu/s42,875R5503000RBtu/lbm1.25lbm/s14

4141in41,regen TTcmhhmQQ p

(c) The net power output is determined from

2

T

550 R

3000 R 1

4Qout·

Qin·

3

s

Btu/s35,384328,438167.0

Btu/s,32843RBtu/lbm1.0316R3000lbm/s14

RBtu/lbm1.0316psia200

psia25lnRBtu/lbm0.4961lnln

inthoutnet,

12in

1

20

1

212

QW

ssTmQ

PP

RTT

css

H

p

9-41

9-65 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximumpressure 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

KkJ/kg0.5K300

kJ/kg150

0

out34,34 T

qss

300 K

1200 K 1

4qout

qin2

3

T

andKkJ/kg0.5

kPa120P

lnKkJ/kg0.287

lnln

4

3

40

3

434 P

PR

TT

css p

s

It yields P4 = 685.2 kPa

(b) For reversible cycles, kJ/kg600kJ/kg150K300K1200

outinin

out qTT

qTT

qq

L

H

H

L

Thus, kJ/kg450150600outinoutnet, qqw

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

75.0%K1200K300

11thH

L

TT

9-66 An ideal Stirling engine with helium as the working fluid operates between the specified temperatureand pressure limits. The thermal efficiency of the cycle, the amount of heat transfer in the regenerator, andthe work output per cycle are to be determined.Assumptions Helium is an ideal gas with constant specific heats. Properties The gas constant and the specific heat of helium at room temperature are R = 2.0769 kJ/kg.K,cv = 3.1156 kJ/kg.K and cp = 5.1926 kJ/kg.K (Table A-2).Analysis (a) The thermal efficiency of this totally reversible cycle is determined from

85.0%K2000

K30011th

H

L

TT

300 K

2000 K 1

4qout

qin2

3

T(b) The amount of heat transferred in the regenerator is

kJ635.6K3002000KkJ/kg3.1156kg0.12

4141in41,regen TTmcuumQQ v

(c) The net work output is determined from s

kJ465.5kJ6.54785.0

kJ547.6KkJ/kg2.282K2000kg0.12

KkJ/kg282.23lnKkJ/kg2.0769lnln

3kPa150K2000kPa3000K300

inthoutnet,

12in

1

20

1

212

1

2

31

13

1

3

1

11

3

33

QW

ssmTQ

RTT

css

PTPT

TP

TP

H

v

v

v

v

v

vvv

v

9-42

Ideal and Actual Gas-Turbine (Brayton) Cycles

9-67C In gas turbine engines a gas is compressed, and thus the compression work requirements are verylarge since the steady-flow work is proportional to the specific volume.

9-68C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropicexpansion (in a turbine), and (4) P = constant heat rejection.

9-69C 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.

9-70C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It isusually between 0.40 and 0.6 for gas turbine engines.

9-71C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.

9-72E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The airtemperature 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 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Noting that process 1-2 is isentropic,

520 R

2000 R

qout

qin

3

4

2

1

T

ThPr

11

112147520 R

124.27 Btu / lbm.

Btu/lbm240.11147.122147.110

2

2

1

212 h

TP

PP

P rrR996.5

(b) Process 3-4 is isentropic, and thuss

Btu/lbm38.88283.26571.504

Btu/lbm115.8427.12411.240

Btu/lbm265.834.170.174101

0.174Btu/lbm504.71

R2000

43outT,

12inC,

43

4

33

34

3

hhw

hhw

hPPP

P

Ph

T

rr

r

Then the back-work ratio becomes

48.5%Btu/lbm238.88Btu/lbm115.84

outT,

inC,bw w

wr

(c)

46.5%Btu/lbm264.60Btu/lbm123.04

Btu/lbm123.0484.11588.238

Btu/lbm264.6011.24071.504

in

outnet,th

inC,outT,outnet,

23in

qw

www

hhq

9-43

9-73 [Also solved by EES on enclosed CD] A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. 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 andpotential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.

4

2

310 K

1160 K

qout

qin

3

4s

2s

1

T

Properties The properties of air are given in Table A-17. Analysis (a) Noting that process 1-2s is isentropic,

ThPr

11

115546310 K

310.24 kJ / kg.

s

kJ/kg89.16719.69292.123082.092.1230

K3.680andkJ/kg692.1990.252.20781

2.207kJ/kg1230.92

K1160

kJ/kg646.775.0

24.31058.56224.310

K557.25andkJ/kg562.5844.125546.18

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

Thus, T4 = 770.1 K(b)

kJ/kg105.392.4782.584

kJ/kg478.9224.31016.789

kJ/kg584.27.64692.1230

outinoutnet,

14out

23in

qqw

hhq

hhq

(c) 18.0%kJ/kg584.2kJ/kg105.3

in

outnet,th q

w

9-44

9-74 EES Problem 9-73 is reconsidered. The mass flow rate, pressure ratio, turbine inlet temperature, andthe isentropic efficiencies of the turbine and compressor are to be varied and a general solution for theproblem by taking advantage of the diagram window method for supplying data to EES is to be developed.Analysis Using EES, the problem is solved as follows:

"Input data - from diagram window"{P_ratio = 8}{T[1] = 310 [K] P[1]= 100 [kPa] T[3] = 1160 [K] m_dot = 20 [kg/s] Eta_c = 75/100 Eta_t = 82/100}

"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])

9-45

Bwr Pratio Wc [kW] Wnet [kW] Wt [kW] Qin [kW]0.5229 0.1 2 1818 1659 3477 165870.6305 0.1644 4 4033 2364 6396 143730.7038 0.1814 6 5543 2333 7876 128620.7611 0.1806 8 6723 2110 8833 116820.8088 0.1702 10 7705 1822 9527 10700

0.85 0.1533 12 8553 1510 10063 98520.8864 0.131 14 9304 1192 10496 91020.9192 0.1041 16 9980 877.2 10857 84260.9491 0.07272 18 10596 567.9 11164 78090.9767 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

Cycleefficiency,

Wnet

[kW]

Wnet

Tmax=1160 K

Note Pratio for maximum work and

c = 0.75t = 0.82

9-46

9-75 A simple Brayton cycle with air as the working fluid has a pressure ratio of 8. 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 Kineticand 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,

K733.94.640116082.01160

K645.375.0

3105.561310

K640.481K1160

K561.58K310

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

4

2

310 K

1160 K

qout

qin

3

4s

2s

1

T

s

(b)

kJ/kg91.30.4263.517

kJ/kg426.0K310733.9KkJ/kg1.005

kJ/kg517.3K645.31160KkJ/kg1.005

outinoutnet,

1414out

2323in

qqw

TTchhq

TTchhq

p

p

(c) 17.6%kJ/kg517.3

kJ/kg91.3

in

outnet,th q

w

9-47

9-76 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid hasa specified pressure ratio. The required mass flow rate of air is to be determined for two cases. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature arecp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).Analysis (a) Using the isentropic relations,

kg/s352kJ/kg199.1

kJ/s70,000

kJ/kg99.1175.31184.510

kJ/kg510.84K491.71000KkJ/kg1.005

kJ/kg311.75K300610.2KkJ/kg1.005

K491.7121K1000

K610.212K300

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

4

2

300 K

1000 K 3

4s

2s

1

T

s

(b) The net work output is determined to be

kg/s1037kJ/kg67.5

kJ/s70,000

kJ/kg67.50.85311.7584.51085.0

/

outnet,a,

outnet,

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

wW

m

wwwww

a

CT

9-48

9-77 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The power delivered by this plant is to be determined assuming constant and variable specific heats.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas. Analysis (a) Assuming constant specific heats,

kW15,680kW35,0000.448

448.03.5251100

2902.6071111

K607.281K1100

K525.38K290

inthoutnet,

23

14

23

14

in

outth

0.4/1.4/1

3

434

0.4/1.4/1

1

212

QW

TTTT

TTcTTc

qq

PP

TT

PP

TT

p

p

kk

s

kk

s

290 K

1100 K

qout

qin

3

4

2

1

T

s

(b) Assuming variable specific heats (Table A-17),

kW15,085kW35,0000.431

431.011.52607.116116.29037.651111

kJ/kg651.3789.201.16781

1.167kJ/kg1161.07

K1100

kJ/kg526.128488.92311.18

2311.1kJ/kg290.16

K290

inoutnet,

23

14

in

outth

43

4

33

21

2

11

34

3

12

1

QW

hhhh

qq

hPPP

P

Ph

T

hPPP

P

Ph

T

T

rr

r

rr

r

9-49

9-78 An actual gas-turbine power plant operates at specified conditions. The fraction of the turbine workoutput used to drive the compressor and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Using the isentropic relations,

T h

T h1 1

2 2

300 K 300.19 kJ / kg

580 K 586.04 kJ / kg

kJ/kg542.0905.831536.0486.0

kJ/kg285.8519.30004.586

kJ/kg05.83973.6711.47471

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

r4

2

300 K

580 K

950 kJ/kg 3

4s

2s

1

T

s

Thus, 52.7%kJ/kg542.0kJ/kg285.85

outT,

inC,bw w

wr

(b)

27.0%kJ/kg950

kJ/kg256.15

kJ/kg256.15285.85.0542

in

outnet,th

inC,outT,net.out

qw

www

9-50

9-79 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work outputused to drive the compressor 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 constant specific heats,

kJ/kg562.2K874.81525.3KkJ/kg1.00586.0

kJ/kg281.4K300580KkJ/kg1.005

K874.871K1525.3

K1525.3KkJ/kg1.005/kJ/kg950K580

/

7100700

4343outT,

1212inC,

0.4/1.4/k1k

3

434s

in232323in

1

2

spTsT

p

pp

p

TTchhw

TTchhw

PP

T

cqTTTTchhq

PP

r

T

4

2

300 K

580 K

950 kJ/kg 3

4s

2s

1

T

s

Thus, 50.1%kJ/kg562.2kJ/kg281.4

outT,

inC,bw w

wr

(b)

29.6%kJ/kg950kJ/kg280.8

kJ/kg280.8281.42.562

in

outnet,th

inC,outT,outnet,

qw

www

9-80E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The netpower output of the plant is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis Using variable specific heats for air,

T h

T h3 3

4 4

2000

1200 291.30

R 504.71 Btu / lbm

R Btu / lbm

kW3373Btu/s197353398536

Btu/s8536Btu/lbm291.30504.71lbm/s40

Btu/s53390.80/131.30238.07lbm/s40/

Btu/lbm8.072379.11474.18

474.1Btu/lbm131.306400/40291.30

815

120

inC,outT,outnet,

43outT,

12inC,

21

2

114out

1

2

12

1

WWW

hhmW

hhmW

hPPP

P

PhhhmQ

PP

r

Cs

srr

r

p

4

21200 R

2000 R

6400 Btu/s

3

4s

2s

1

T

s

9-51

9-81E A gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The compressor efficiency for which the power plant produces zero net work is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standardassumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats.

TProperties The properties of air are given in Table A-17E. Analysis Using variable specific heats,

T h

T h3 3

4 4

2000

1200 291.30

R 504.71 Btu / lbm

R Btu / lbm

Btu/lbm8.072379.11474.18

474.1Btu/lbm131.306400/40291.30

815

120

21

2

114out

1

2

12

1

srr

r

p

hPPP

P

PhhhmQ

PP

r

4

2Wnet = 0 ·

3

4s

2s

1

2000 R

1200 R

s

Then,

50.0%30.29171.50430.13107.238

/

43

12

4312outT,inC,

hhhh

hhmhhmWW

sC

Cs

9-82 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid.The mass flow rate of air through the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis Using variable specific heats,

T 5546.1kJ/kg310.24

K3101

11

rPh

kJ/kg40.6880.0/24.31026.56232.51993.93286.0

/

kJ/kg519.32411.929.7581

29.75kJ/kg32.939

K900

kJ/kg62.26544.125546.18

1243inC,outT,outnet,

43

4

33

21

2

34

3

12

CssT

srr

r

srr

hhhhwww

hPPP

P

Ph

T

hPPP

P4

2

310 K

900 K

Wnet =32 MW

·

3

4s

2s

1

T

s

and kg/s786.6kJ/kg40.68

kJ/s32,000

outnet,

outnet,

wW

m

9-52

9-83 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid.The mass flow rate of air through the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand 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 Using constant specific heats,

kJ/kg32.5

K0.80/310561.5496.89000.86KkJ/kg1.005

/

K496.881K900

K561.58K310

1243CoutT,outnet,

0.4/1.4/1

3

434

0.4/1.4/1

1

212

in, CspspT

kk

s

kk

s

TTcTTcwww

PP

TT

PP

TT

4

2

310 K

900 K

Wnet =32 MW

·

3

4s

2s

1

T

s

and

kg/s984.6kJ/kg32.5

kJ/s32,000

outnet,

outnet,

wW

m

9-53

9-84 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 theproperties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions ofboth temperature and pressure.

1.2 MPa

500°C100 kPa 30°C

Compress.

4

32

Turbine

Combustionchamber

1

Process 1-2: Compression

KkJ/kg7159.5kPa100

C30kJ/kg60.303C30

11

1

11

sPT

hT

kJ/kg37.617kJ/kg.K7159.5

kPa12002

12

2sh

ssP

kJ/kg24.68660.303

60.30337.61782.0 2212

12C h

hhhhh s

Process 3-4: Expansion

kJ/kg62.792C500 44 hT

ss hhh

hhhh

43

3

43

43T

62.79288.0

We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together withthe isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach.

h_3=enthalpy(Air, T=T_3)s_3=entropy(Air, T=T_3, P=P_2)h_4s=enthalpy(Air, P=P_1, s=s_3)

The mass flow rate is determined from

kg/s875.2K27330K/kgmkPa0.287

)s/m0kPa)(150/6100(3

3

1

11

RTP

mV

The net power output iskW1100)kJ/kg60.30324kg/s)(686.875.2()( 12inC, hhmW

W kW1759)kJ/kg62.792.7kg/s)(1404875.2()( 43outT, hhm

kW65911001759inC,outT,net WWW

(b) The back work ratio is

0.625kW1759kW1100

outT,

inC,bw W

Wr

(c) The rate of heat input and the thermal efficiency are kW2065)kJ/kg24.686.7kg/s)(1404875.2()( 23in hhmQ

0.319kW2065

kW659

in

net

QW

th

9-54

Brayton Cycle with Regeneration

9-85C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the wasteheat from the exhaust gases and preheating the air before it enters the combustion chamber.

9-86C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than thetemperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in aregenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermalefficiency will decrease.

9-87C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness , and is defined as = qregen, act /qregen, max.

9-88C (b) turbine exit.

9-89C 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 and W increases while Qin remainsconstant. Steam can be obtained by utilizing the hot exhaust gases.

9-55

9-90E 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 95 hp are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Theambient air is 540 R and 14.5 psia. 4 The effectiveness of the regenerator is 0.9, and the isentropicefficiencies 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 fromTable A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows:

Btu/lbm372.253.5712.23041

12.230Btu/lbm549.35

R2160

Btu/lbm0.192544.5386.14

386.1Btu/lbm129.06

R054

43

4

33

21

2

11

34

3

12

1

srr

r

srr

r

hPPP

P

Ph

T

hPPP

P

Ph

T

24

5

540 R

2160 R qin 3

4s2s

1

T

s

and

Btu/lbm63.4072.37235.549

35.5490.80

Btu/lbm74.20706.129

06.1290.1920.80

44

43

43turb

2212

12comp

hh

hhhh

hhhh

hh

s

s

Then the thermal efficiency of the gas turbine cycle becomesq Btu/lbm179.9)74.20763.407(9.0)( 24regen hh

Btu/lbm63.0)06.12974.207()63.40735.549()()(

Btu/lbm161.7=9.179)74.20735.549()(

1243inC,outT,outnet,

regen23in

hhhhwww

qhhq

0.39Btu/lbm161.7Btu/lbm63.0

in

outnet,th %39

qw

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

lbm/s1.07hp1

Btu/s7068.0Btu/lbm63.0

hp95

net

netair w

Wm

9-56

9-91 [Also solved by EES on enclosed CD] The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 3 The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17. Analysis The properties at various states are

24

5

293 K

1561 K qin 3

4s2s

1

T

kJ/kg23.82547.485.7127.14

1

5.712kJ/kg1710.0

K1561C1288

kJ/kg3.643765.182765.17.14

2765.1kJ/kg293.2

K293=C20

43

4

33

21

2

11

34

3

12

1

srr

r

srr

r

hPPP

P

Ph

T

hPPP

P

Ph

T

s

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

C650kJ/kg54.9582.29334.665kJ/kg34.66566.3720.1038

kJ/kg0.67210381710

kJ/kg0.10380.359

kJ/kg372.66

kJ/kg66.372h1

s3600kg/h1,536,000kW159,000

41out414out

netinout

3223in

th

netin

netnet

Thqhhhqwqq

qhhhhq

wq

mW

w

in

Then the compressor and turbine efficiencies become

0.924

0.849

2.2936722.2933.643

23.825171054.9581710

12

12

43

43

hhhh

hhhh

sC

sT

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

0.496kJ/kg751.46kJ/kg372.66

kJ/kg751.46=54.2861038

kJ/kg286.54=672.0)-.54(0.80)(958=)(

newin,

netnewth,

regeninnewin,

24regen

qw

qqq

hhq

Discussion Note an 80% efficient regenerator would increase the thermal efficiency of this gas turbinefrom 35.9% to 49.6%.

9-57

9-92 EES Problem 9-91 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. Analysis Using EES, the problem is solved as follows:

"Input data"T[3] = 1288 [C] Pratio = 14.7 T[1] = 20 [C] P[1]= 100 [kPa] {T[4]=589 [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 = 0.80 Eta_c = 0.892 "Compressor isentorpic efficiency"Eta_t = 0.926 "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_doth[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] /Pratios_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"

9-58

m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])"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]

t c th,noreg th,withreg Qinnoreg

[kW]Qinwithreg

[kW]Wnet [kW]

0.7 0.892 0.2309 0.3405 442063 299766 102.10.75 0.892 0.2736 0.3841 442063 314863 120.90.8 0.892 0.3163 0.4237 442063 329960 139.8

0.85 0.892 0.359 0.4599 442063 345056 158.70.9 0.892 0.4016 0.493 442063 360153 177.6

0.95 0.892 0.4443 0.5234 442063 375250 196.41 0.892 0.487 0.5515 442063 390346 215.3

9-59

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 1100

120

140

160

180

200

220

t

Wnet

[kW]

0.7 0.75 0.8 0.85 0.9 0.95 1275000

310000

345000

380000

415000

450000

t

Qdot,in

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

0.6

t

Etath

with regeneration

no regeneration

9-60

9-93 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle 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.Properties The properties of air are given in Table A-17. Analysis Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic,we have

kJ/kg601.9485.67579.1277

kJ/kg279.6819.30087.579

kJ/kg675.858.23238101

238kJ/kg1277.79

K1200

kJ/kg579.8786.13386.110

386.1kJ/kg300.19

K300

43outT,

12inC,

43

4

33

21

2

11

34

3

12

1

hhw

hhw

hPPP

P

Ph

T

hPPP

P

Ph

T

rr

r

rr

r

5

300 K

1200 K qin3

42

1

T

s

Thus,kJ/kg322.2668.27994.601inC,outT,net www

Also, 100% 5 4h h 675.85 kJ / kg

and

53.5%kJ/kg601.94kJ/kg322.26

kJ/kg601.9485.67579.1277

in

netth

53in

qw

hhq

9-61

9-94 EES Problem 9-93 is reconsidered. The effects of the isentropic efficiencies for the compressor and turbine and regenerator effectiveness on net work done and the heat supplied to the cycle are to be studied.Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows:

"Input data"T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency"Eta_t =0.9 "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_compisen/w_comp"compressor adiabatic efficiency, W_comp > W_compisen""Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"h[1] + w_compisen = 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 = 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_in - e_out =DELTAe_cv =0 for steady flow"h[2] + q_in_noreg = h[3] 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] /Pratios_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_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb""SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])"Actual Turbine analysis:"h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,T=T[4], P=P[4])"Cycle analysis"w_net=w_turb-w_compEta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb"Back work ratio"

9-62

"With the regenerator the heat added in the external heat exchanger is"h[5] + q_in_withreg = h[3] 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:"h[2] + h[4]=h[5] + 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_net/q_in_withreg*Convert(, %) "[%]""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]

c t th,noreg th,withreg qinnoreg

[kJ/kg]qinwithreg

[kJ/kg]wnet

[kJ/kg]0.6 0.9 14.76 13.92 510.9 541.6 75.4

0.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 142

0.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 192

0.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8

4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T[K] 10

0kPa

1000

kPa

Air

2s

1

2 5

3

44s

6

9-63

0.7 0.75 0.8 0.85 0.9 0.95 110

15

20

25

30

35

40

45

t

th

c = 0.8

With regeneration

No regeneration

0.7 0.75 0.8 0.85 0.9 0.95 150

95

140

185

230

275

t

wnet

[kJ/kg]

c = 0.8

0.7 0.75 0.8 0.85 0.9 0.95 1400

450

500

550

600

650

t

qin

c = 0.8

No regeneration

With regeneration

9-64

0.6 0.65 0.7 0.75 0.8 0.85 0.910

15

20

25

30

35

40

45

c

th

t = 0.9

With regeneration

No regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.975

110

145

180

215

250

c

wnet

[kJ/kg]

t = 0.9

0.6 0.65 0.7 0.75 0.8 0.85 0.9500

520

540

560

580

600

620

640

660

680

c

qin

t = 0.9

No regeneration

With regeneration

9-65

9-95 An ideal Brayton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle 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 Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic, we have

517.0)10(120030011or

5.62112003002.5791111

K579.2andK621.5%100

K621.5101K1200

K579.210K300

1)/1.4(1.4/)1(

3

1th

53

16

53

16

in

outth

2645

0.4/1.4/1

3

434

0.4/1.4/1

1

212

kkp

p

p

kk

kk

rTT

TTTT

TTcTTc

qq

TTTT

PP

TT

PP

TT

0.517

5

300 K

1200 K qin3

42

1

T

s

Then,

kJ/kg300.8300)]K-(579.2-621.5)-1200kJ/kg.K)[(005.1(

)]()[(

)()(

1243

1243incomp,outturb,net

TTTTc

hhhhwww

p

or,

kgkJ6300=621.5)-200kJ/kg.K)(1005.1)(517.0(

)()(

53th

53th

inthnet

/.

TTchh

qw

p

9-66

9-96 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 areapplicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.Properties The properties of air are given in Table A-17. Analysis (a) The properties of air at various states are

kJ/kg803.14711.801219.250.821219.25

kJ/kg711.8059.2815.20071

15.200kJ/kg1219.25

K1501

kJ/kg618.260.75/310.24541.2610.243/

kJ/kg541.2688.105546.17

5546.1kJ/kg310.24

K310

433443

43

43

4

33

121212

12

21

2

11

34

3

12

1

sTs

T

srr

r

Css

C

srr

r

hhhhhhhh

hPPP

P

Ph

T

hhhhhhhh

hPPP

P

Ph

T2

4

6

5

310 K

1150 K qin 3

4s2s

1

T

s

Thus, T4 = 782.8 K(b)

kJ/kg108.0924.31026.61814.80325.1219

1243inC,outT,net hhhhwww

(c)

kJ/kg738.43618.26803.140.6526.618

242524

25 hhhhhhhh

Then,

22.5%kJ/kg480.82kJ/kg108.09

kJ/kg480.82738.4319.2512

in

netth

53in

qw

hhq

9-67

9-97 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 potentialenergy 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 obtainedfrom Table A-17. Analysis (a) Assuming constant specific heats,

kW39,188kW75,0000.5225

5225.02.6071100

2903.5251111

K525.3andK607.2%100

K607.281K1100

K525.38K290

innet

53

16

53

16

in

outth

2645

0.4/1.4/1

3

434

0.4/1.4/1

1

212

QW

TTTT

TTcTTc

qq

TTTT

PP

TT

PP

TT

T

p

p

kk

kk

6

qout

5

290 K

1100 K 75,000 kW 3

42

1

T

s

(b) Assuming variable specific heats,

kW40,283kW75,0000.5371

5371.037.65107.116116.29012.526111

kJ/kg526.12andkJ/kg651.37%100

kJ/kg651.3789.201.16781

1.167kJ/kg1161.07

K1100

kJ/kg526.128488.92311.18

2311.1kJ/kg90.162

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

9-68

9-98 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heattransfer in the regenerator 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.Properties The properties of air are given in Table A-17. Analysis (a) The properties at various states are

kJ/kg152.504.58688.79772.0kJ/kg797.88

75.71979.127786.079.1277

kJ/kg719.7575.290.23881

238.0kJ/kg1277.79K2001

kJ/kg86.045K580kJ/kg00.193K300

8100/800/

24regen

433443

43

43

4

33

22

11

12

34

3

hhq

hhhhhhhh

hPPP

P

PhThThT

PPr

sTs

T

srr

r

p

2580 K

4

6

5

300 K

1200 K qin 3

4s2s

1

T

s

(b)

36.0%kJ/kg539.23kJ/kg194.06

kJ/kg539.23152.52586.041277.79

kJ/kg194.06300.19586.0488.79779.1277

in

netth

regen23in

1243inC,outT,net

qw

qhhq

hhhhwww

9-69

9-99 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heattransfer in the regenerator and the thermal efficiency are to be determined.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.Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2a).Analysis (a) Using the isentropic relations and turbine efficiency,

kJ/kg114.2K580737.8KkJ/kg1.0050.72K737.8

5.662120086.01200

K662.581K1200

8100/800/

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

2580 K

4

6

5

300 K

1200 K qin 3

4s2s

1

T

s

(b)

36.0%kJ/kg508.9kJ/kg183.1

kJ/kg508.9114.2K5801200KkJ/kg1.005

kJ/kg183.1K300580737.81200KkJ/kg1.005

in

netth

regen23regen23in

1243inC,outT,net

qw

qTTcqhhq

TTcTTcwww

p

pp

9-100 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heattransfer in the regenerator 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.Properties The properties of air are given in Table A-17. Analysis (a) The properties of air at various states are

kJ/kg148.304.58688.79770.0

kJ/kg797.8875.71979.127786.079.1277

kJ/kg719.7575.290.23881

238.0kJ/kg1277.79K2001

kJ/kg86.045K580kJ/kg00.193K300

8100/800/

23regen

433443

43

43

4

33

22

11

12

34

3

hhq

hhhhhhhh

hPPP

P

PhThThT

PPr

sTs

T

srr

r

p

2580 K

4

6

5

300 K

1200 K qin 3

4s2s

1

T

s

(b)

35.7%kJ/kg5543.kJ/kg194.06

kJ/kg543.5148.3586.041277.79

kJ/kg194.06300.19586.0488.79779.1277

in

netth

regen23in

1243inC,outT,net

qw

qhhq

hhhhwww

9-70

Brayton Cycle with Intercooling, Reheating, and Regeneration

9-101C As the number of compression and expansion stages are increased and regeneration is employed,the ideal Brayton cycle will approach the Ericsson cycle.

9-102C (a) decrease, (b) decrease, and (c) decrease.

9-103C (a) increase, (b) decrease, and (c) decrease.

9-104C (a) increase, (b) decrease, (c) decrease, and (d) increase.

9-105C (a) increase, (b) decrease, (c) increase, and (d) decrease.

9-106C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output.

9-107C (c) The Carnot (or Ericsson) cycle efficiency.

9-71

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

kJ/kg62.86636.94679.127722

kJ/kg2.142219.30026.41122

kJ/kg946.3633.7923831

238kJ/kg77.7912

K2001

kJ/kg411.26158.4386.13

386.1kJ/kg300.19

K300

65outT,

12inC,

865

6

755

421

2

11

56

5

12

1

hhw

hhw

hhPPP

P

Phh

T

hhPPP

P

Ph

T

rr

r

rr

r

2

9

10

7

6 8

300 K

1200 K qin5

1

4

3

T

s

Thus, 33.5%kJ/kg662.86kJ/kg222.14

outT,

inC,bw w

wr

36.8%kJ/kg1197.96kJ/kg440.72

kJ/kg440.72222.1486.662

kJ/kg1197.9636.94679.127726.41179.1277

in

netth

inC,outT,net

6745in

qw

www

hhhhq

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

55.3%kJ/kg796.63kJ/kg440.72

kJ/kg796.6333.40196.1197

kJ/kg33.40126.41136.94675.0

in

netth

regenoldin,in

48regen

qw

qqq

hhq

9-72

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

kJ/kg63.44507.99679.127722

kJ/kg77.68219.30003.43922

kJ/kg96.07936.94679.127785.079.1277

kJ/kg946.3633.7923831

238kJ/kg1277.79K1200

kJ/kg9.034380.0/19.30026.41119.300

/

kJ/kg411.26158.4386.13

386.1kJ/kg300.19K300

65outT,

12inC,

6558665

65

865

6

755

1214212

12

421

2

11

56

5

12

1

hhw

hhw

hhhhhhhhh

hhPPP

P

PhhT

hhhhhhhhh

hhPPP

P

PhT

sTs

T

rr

r

Css

C

ssrr

r T

4

8

22s

9

10

7

66s

8qin

5

1

4

3s

Thus, 49.3%kJ/kg563.44kJ/kg277.68

outT,

inC,bw w

wr

25.5%kJ/kg1120.48kJ/kg285.76

kJ/kg5.762868.27744.563

kJ/kg1120.48996.071277.79439.031277.79

in

netth

inC,outT,net

6745in

qw

www

hhhhq

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

40.7%kJ/kg702.70kJ/kg285.76

kJ/kg.7070278.41748.1120

kJ/kg17.78403.43907.99675.0

in

netth

regenoldin,in

48regen

qw

qqq

hhq

9-73

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

kg/s249.6kJ/kg440.72

kJ/s110,000

kJ/kg440.72222.1468.662

kJ/kg662.86946.361277.7922

kJ/kg222.1419.30026.41122

kJ/kg6.369433.7923831

238kJ/kg,1277.79K0120

kJ/kg11.264158.4386.13

386.1kJ/kg,00.193K003

net

net

inC,outT,net

65outT,

12inC,

865

6

755

421

2

11

56

5

12

1

wW

m

www

hhw

hhw

hhPPP

P

PhhT

hhPPP

P

PhT

rr

r

rr

r

2

7

6 8

300 K

1200 K 5

1

4

3

T

s

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

kg/s404.7kJ/kg271.8

kJ/s110,000

kJ/kg271.83.1721.444

kJ/kg444.1K773.21200KkJ/kg0.5203222

kJ/kg172.3K300465.6KkJ/kg0.5203222

K773.231K1200

K465.63K300

net

net

inC,outT,net

6565outT,

1212inC,

70.667/1.66/1

5

656

70.667/1.66/1

1

212

wW

m

www

TTchhw

TTchhw

PP

TT

PP

TT

p

p

kk

kk

2

7

6 8

300 K

1200 K 5

1

4

3

T

s

9-74

Jet-Propulsion Cycles

9-112C The power developed from the thrust of the engine is called the propulsive power. It is equal tothrust times the aircraft velocity.

9-113C The ratio of the propulsive power developed and the rate of heat input is called the propulsiveefficiency. It is determined by calculating these two quantities separately, and taking their ratio.

9-114C It reduces the exit velocity, and thus the thrust.

9-115E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure atthe turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible,except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor workinput.Properties The properties of air at room temperature are cp = 0.24 Btu/lbm.R and k = 1.4 (Table A-2Ea). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards theaircraft at a velocity of V1 = 900 ft/s. Ideally, the air will leave the diffuser with a negligible velocity (V20).Diffuser:

psia11.19R470R537.3

psia7

R537.4/sft25,037

Btu/lbm1RBtu/lbm0.242

ft/s900470

2

2/02

0

2/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

12

222

211

outin

(steady)0systemoutin

kk

p

p

TT

PP

cV

TT

VTTc

VVhh

VhVh

EE

EEE T

6

qout

5

qin

3

4

2

1s

Compressor:

R1118.313R537.4

psia145.5psia11.1913

0.4/1.4/1

2

323

243

kk

p

PP

TT

PrPP

9-75

Turbine:

54235423outturb,incomp, TTcTTchhhhww pp

or,

psia55.21.4/0.41/

4

545

2345

R2400R1819.1

psia145.5

R1819.14.5371118.34002kk

TT

PP

TTTT

(b) Nozzle:

2/02

0

2/2/

R1008.6psia55.2

psia7R1819.1

2656

025

26

56

266

255

outin


0.4/1.4/1

5

656

VTTc

VVhh

VhVh

EE

EEE

PP

TT

p

kk

or,

s/ft3121Btu/lbm1

/sft25,037R1008.61819.1RBtu/lbm0.2402

22

6V

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

%25.9Btu/lbm307.6

Btu/lbm79.8

Btu/lbm307.6R1118.32400RBtu/lbm0.24

Btu/lbm79.8/sft25,037

Btu/lbm1ft/s900ft/s9003121

in

3434in

22

aircraftinletexit

qw

TTchhq

VVVw

pp

p

p

9-76

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

Diffuser:8548.0

Btu/lbm112.20R470

1

11

rPhT

T

psia11.220.85481.3698psia7

3698.1Btu/lbm128.48/sft25,037

Btu/lbm12ft/s900

20.1122

20

2/2/

1

2

2

12

r22

221

12

21

022

12

222

211

outin


r

r

P

PPP

PV

hh

VVhh

VhVh

EE

EEE

6

qout

5

qin

3

4

2

1s

Compressor:

Btu/lbm267.5680.171.36811.22145.8

psia145.8psia11.2213

32

3

243

23hP

PP

P

PrPP

rr

p

Turbine: T 6.367Btu/lbm22.617

R24004

44

rPh

5423

outturb,incomp,

hhhh

ww

or,

psia56.6367.6142.7psia145.8

7.142Btu/lbm14.47848.12856.26722.617

4

5

5

45

2345

r

r

r

P

PPP

Phhhh

9-77

(b) Nozzle:

20

2/2/

Btu/lbm266.9317.66psia56.6

psia7)7.142(

025

26

56

266

255

outin


65

656

VVhh

VhVh

EE

EEE

hPP

PP rr

or,

ft/s3252Btu/lbm1

/sft25,037Btu/lbm)93.26614.478(22

22

656 hhV

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

24.2%Btu/lbm349.66

Btu/lbm84.55

Btu/lbm349.6656.26722.617

Btu/lbm84.55/sft25,037

Btu/lbm1ft/s)(900ft/s)900)(3252

in

34

22

aircraftinletexit

qw

hhq

VVVw

pp

in

p

9-78

9-117 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible,except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor workinput.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:

kPa62.6K241K291.9

kPa32

K291.9/sm1000

kJ/kg1KkJ/kg1.0052

m/s320K241

2

2/02

02/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

122

222

11

outin(steady)0

systemoutin

kk

p

p

TT

PP

cV

TT

VTTc

VVhhVhVh

EEEEE

T

6

5

Qi·

3

4

2

1s

Compressor:

K593.712K291.9

kPa751.2kPa62.612

0.4/1.4/1

2

323

243

kk

p

PP

TT

PrPP

Turbine:

or,K1098.2291.9593.714002345

54235423outturb,incomp,

TTTT

TTcTTchhhhww pp

Nozzle:

2/02

0

2/2/

K568.2kPa751.2

kPa32K1400

2656

025

26

56

266

255

outin(steady)0

systemoutin

0.4/1.4/1

4

646

VTTcVV

hh

VhVh

EEEEE

PP

TT

p

kk

or, m/s1032kJ/kg1

/sm1000K568.21098.2KkJ/kg1.0052

22

6V

(b) kW13,67022aircraftinletexit

/sm1000kJ/kg1

m/s320m/s3201032kg/s60VVVmW p

(c)

kg/s1.14kJ/kg42,700kJ/s48,620

HV

kJ/s48,620K593.71400KkJ/kg1.005kg/s60

infuel

3434in

Qm

TTcmhhmQ p

9-79

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

0). Diffuser:

kPa62.6K241K291.9

kPa32

K291.9/sm1000

kJ/kg1KkJ/kg1.0052

m/s320K241

2

2/02

0

2/2/

1.4/0.41/

1

212

22

221

12

2112

21

022

12

222

211

outin


kk

p

p

TT

PP

cV

TT

VTTc

VVhh

VhVh

EE

EEE T

5

6

5s

Qi·

3

4

2

1s

Compressor:

K593.712K291.9

kPa751.2kPa62.612

0.4/1.4/1

2

323

243

kk

s

p

PP

TT

PrPP

K669.20.80/291.9593.71.929/2323

23

23

23

23

Cs

p

spsC

TTTT

TTcTTc

hhhh

Turbine:

or,K1022.7291.9669.214002345

54235423outturb,incomp,

TTTT

TTcTTchhhhww pp

kPa197.7K1400K956.1

kPa751.2

K956.185.0/1022.714001400/1.4/0.41/

4

545

5445

54

54

54

54

kks

Ts

sp

p

sT

TT

PP

TTTT

TTcTTc

hhhh

9-80

Nozzle:

2/02

0

2/2/

K607.8kPa197.7

kPa32K1022.7

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/s913.2kJ/kg1

/sm1000K607.81022.7KkJ/kg1.0052

22

6V

(b)

kW11,39022

aircraftinletexit

/sm1000kJ/kg1

m/s320m/s320913.2kg/s60

VVVmW p

(c)

kg/s1.03kJ/kg42,700kJ/s44,067

HV

kJ/s44,067K669.21400KkJ/kg1.005kg/s60

infuel

3434in

Qm

TTcmhhmQ p

9-81

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

TAnalysis (a) Using variable specific heats for air,

5

4

qin

2

3

1

Compressor: T386.1

kJ/kg300.19K300

1

11

rPh

27.396kJ/kg1464.65854610.65

kJ/kg854kg/s10

kJ/s8540

kJ/s8540kJ/kg42,700kg/s0.2HV

kJ/kg610.6563.16386.112

3

12

2323in

inin

fuelin

21

2

r

in

rr

Pqhhhhq

mQ

q

mQ

hPPP

P

s

Turbine:

or,kJ/kg741.17300.19610.65464.6511234

4312outturb,incomp,

hhhh

hhhhww

Nozzle:

20

2/2/

kJ/kg41.79702.3312127.396

024

25

45

255

244

outin


53

535

VVhh

VhVh

EE

EEE

hPP

PP rr

or,

m/s908.9kJ/kg1

/sm1000kJ/kg741.171154.1922

22

545 hhV

Brake force = Thrust = N90892inletexit

m/skg1N1

m/s0908.9kg/s10VVm

9-82

9-120 EES Problem 9-119 is reconsidered. The effect of compressor inlet temperature on the force thatmust be applied to the brakes to hold the plane stationary is to be investigated.Analysis Using EES, the problem is solved as follows:

P_ratio = 12 T_1 = 27 [C] T[1] = T_1+273 "[K]"P[1]= 95 [kPa] P[5]=P[1]Vel[1]=0 [m/s]V_dot[1] = 9.063 [m^3/s] HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.2 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0

"Inlet conditions"h[1]=ENTHALPY(Air,T=T[1])s[1]=ENTROPY(Air,T=T[1],P=P[1])v[1]=volume(Air,T=T[1],P=P[1])m_dot = V_dot[1]/v[1]"Compressor anaysis"s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor"P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]"T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit"h_s[2]=ENTHALPY(Air,T=T_s[2])Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. "m_dot*h[1] +W_dot_c=m_dot*h[2] "SSSF First Law for the actual compressor, assuming:adiabatic, ke=pe=0"

"External heat exchanger analysis"P[3]=P[2]"process 2-3 is SSSF constant pressure"h[3]=ENTHALPY(Air,T=T[3])Q_dot_in = m_dot_fuel*HV_fuelm_dot*h[2] + Q_dot_in= m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0"

"Turbine analysis"s[3]=ENTROPY(Air,T=T[3],P=P[3])s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine"{P_ratio= P[3] /P[4]}T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit"{h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency,Wts_dot > W_dot_t"Eta_t=(h[3]-h[4])/(h[3]-h_s[4])m_dot*h[3] = W_dot_t + m_dot*h[4] "SSSF First Law for the actual compressor, assuming:adiabatic, ke=pe=0"T[4]=TEMPERATURE(Air,h=h[4])P[4]=pressure(Air,s=s_s[4],h=h_s[4])"Cycle analysis"W_dot_net=W_dot_t-W_dot_c"Definition of the net cycle work, kW"W_dot_net = 0 [kW]

9-83

"Exit nozzle analysis:"s[4]=entropy('air',T=T[4],P=P[4])s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle"

T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzleexit"h_s[5]=ENTHALPY(Air,T=T_s[5])Eta_N=(h[4]-h[5])/(h[4]-h_s[5])m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg))m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg))T[5]=TEMPERATURE(Air,h=h[5])s[5]=entropy('air',T=T[5],P=P[5])

"Brake Force to hold the aircraft:"Thrust = m_dot*(Vel[5] - Vel[1]) "[N]"BrakeForce = Thrust "[N]""The following state points are determined only to produce a T-s plot"T[2]=temperature('air',h=h[2])s[2]=entropy('air',T=T[2],P=P[2])

BrakeForce

[N]

m[kg/s]

T3

[K]T1

[C]

9971 11.86 1164 -209764 11.41 1206 -109568 10.99 1247 09383 10.6 1289 109207 10.24 1330 209040 9.9 1371 30

4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T [K

]

95kP

a

114

0kP

a

Air

1

2s

3

4s

5s

-20 -10 0 10 20 309000

9200

9400

9600

9800

10000

T1 [C]

Bra

keFo

rce

[N]

9-84

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

T h

T h1 1

2 2

280 28013

700 713 27

K k

K k

.

.

J / kg

J / kg 15,000 kJ/s

22

222

21

22

12in

222

211in

outin


/sm1000kJ/kg1

2m/s300

13.28027.713kg/s16kJ/s000,15

2

)2/()2/(

V

VVhhmQ

VhmVhmQ

EE

EEE 7 C300 m/s16 kg/s

427 C

1 2

It gives V 2 = 1048 m/sThus,

N11,968m/s3001048kg/s1612 VVmFp

9-85

Second-Law Analysis of Gas Power Cycles

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

kJ/kg245.12K2000

kJ/kg750K300kJ/kg357.62

K300inout0destroyed

HL Tq

Tq

Tx

Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke(state 4) is determined from

040040044 vvPssTuu

where

KkJ/kg0.7081K300K774.5lnKkJ/kg0.2871.70203.68232

lnlnln

0kJ/kg357.62

1

414

41

1414

1

4141404

1404

out1404

TT

RssTT

RssPP

Rssssss

quuuu

v

v

vvvv

Thus,kJ/kg145.20KkJ/kg0.7081K300kJ/kg357.624

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

kJ/kg292.7K2000

kJ/kg1019.7K300kJ/kg445.63

K300inout0destroyed

HL Tq

Tq

Tx

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

kJ/kg348.63

312012012

020020022

mkPa1kJ1

/kgm0.9060.0566kPa950214.07643.3

vv

vv

PssTuuPssTuu

9-86

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

RBtu/lbm0.1741R1420.6

R540lnRBtu/lbm0.180lnln0

4

1

4

141 v

vRTTcss v

Thus,

Btu/lbm64.9R540

Btu/lbm158.9RBtu/lbm0.1741R54041,41041destroyed,

R

R

Tq

ssTx

Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke(state 4) is determined from

040040044 vvPssTuu

where

RBtu/lbm0.17410

RBtu/lbm9.581

1404

1404

out1404

ssss

quuuuvvvv

Thus,Btu/lbm64.90RBtu/lbm0.1741R540Btu/lbm158.94

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

9-87

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

KkJ/kg2.67602kJ/kg789.16

KkJ/kg3.13916K1160

KkJ/kg2.47256kJ/kg6.364

KkJ/kg.734981K310

44

33

22

11

sh

sT

sh

sT

Thus,

kJ/kg206.0

kJ/kg38.76

kJ/kg87.35

kJ/kg40.83

K310kJ/kg478.92

2.676021.73498K290

ln

1/8lnKkJ/kg0.2873.139162.67602K290

ln

K1600kJ/kg584.62

2.472563.13916K290

ln

8lnKkJ/kg0.2871.734982.47256K290

ln

out0

4

1410

41,41041gen,041destroyed,

3

434034034gen,034destroyed,

in0

2

3230


1


LR

R

HR

R

Tq

PP

RssTT

qssTsTx

PP

RssTssTsTx

Tq

PP

RssTT

qssTsTx

PP

RssTssTsTx

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

kJ/kg179.4K1800

kJ/kg601.94K300kJ/kg279.68

K300inout0destroyed

HL Tq

Tq

Tx

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

06

026

060066 2gz

VssThh

where s s s s s s R PP6 0 6 1 6 1

6

1

0

ln 2.36275 1.70203 0.66072 kJ / kg K

Thus, kJ/kg81.5KkJ/kg0.66072K300300.1979.8756

9-88

9-127 EES Problem 9-126 is reconsidered. The effect of the cycle pressure on the total irreversibility forthe cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated.Analysis Using EES, the problem is solved as follows:

"Input data"T_o = 300 [K] T_L = 300 [K] T_H = 1400 [K] T[3] = 1200 [K] {Pratio = 10}T[1] = 300 [K] C_P=1.005 [kJ/kg-K] P[1]= 100 [kPa] P_o=P[1]Eta_reg = 1.0 Eta_c =1.0"Compressor isentorpic efficiency"Eta_t =1.0"Turbien isentropic efficiency"MM=MOLARMASS(Air)R=R_u/MMR_u=8.314 [kJ/kmol-K]C_V=C_P - R k=C_P/C_V"Isentropic Compressor anaysis""For the ideal case the entropies are constant across the compressor"P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit"T_s[2]=T[1]*(Pratio)^((k-1)/k)Eta_c = w_compisen/w_comp"compressor adiabatic efficiency, W_comp > W_compisen""Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"w_compisen = C_P*(T_s[2]-T[1]) "Actual compressor analysis:"w_comp = C_P*(T[2]-T[1]) "External heat exchanger analysis""SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow"q_in_noreg = C_P*(T[3]-T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure""Turbine analysis""For the ideal case the entropies are constant across the turbine"P[4] = P[3] /PratioT_s[4]=T[3]*(1/Pratio)^((k-1)/k)"T_s[4] is the isentropic value of T[4] at turbine exit"Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb""SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen=C_P*(T[3] - T_s[4]) "Actual Turbine analysis:"w_turb= C_P*(T[3]-T[4])

"Cycle analysis"w_net=w_turb-w_comp "[kJ/kg]"Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb "Back work ratio"

9-89

"With the regenerator the heat added in the external heat exchanger is"q_in_withreg = C_P*(T[3]-T[5]) P[5]=P[2]"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:""h[2] + h[4]=h[5] + h[6]"T[2] + T[4]=T[5] + T[6]P[6]=P[4]"Cycle thermal efficiency with regenerator"Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]"

"Irreversibility associated with the Brayton cycle is determined from:"q_out_withreg = q_in_withreg - w_net i_withreg = T_o*(q_out_withreg/T_L - q_in_withreg/T_H)q_out_noreg = q_in_noreg - w_net i_noreg = T_o*(q_out_noreg/T_L - q_in_noreg/T_H)"Neglecting the ke and pe of the exhaust gases, the exergy of the exhaust gases at the exit of the regenerator is:""Psi_6 = (h[6] - h_o) - T_o(s[6] - s_o)"Psi_exit_withreg = C_P*(T[6] - T_o) - T_o*(C_P*ln(T[6]/T_o)-R*ln(P[6]/P_o))Psi_exit_noreg = C_P*(T[4] - T_o) - T_o*(C_P*ln(T[4]/T_o)-R*ln(P[4]/P_o))

inoreg iwithreg Pratio exit,noreg

[kJ/kg]exit,withreg

[kJ/kg]th,noreg

[%]th,withreg

[%]270.8 97.94 6 157.8 47.16 40.05 58.3244.5 113.9 7 139.9 56.53 42.63 56.42223 128.8 8 125.5 65.46 44.78 54.73205 142.7 9 113.6 74 46.61 53.18

189.6 155.9 10 103.6 82.17 48.19 51.75176.3 168.4 11 95.05 90.02 49.58 50.41164.6 180.2 12 87.62 97.58 50.82 49.17154.2 191.5 13 81.11 104.9 51.93 47.99144.9 202.3 14 75.35 111.9 52.94 46.88

4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T [K

]

100 kPa

1000 kPa

Air

2s

1

2 5

3

44s

6

9-90

6 7 8 9 10 11 12 13 1475

115

155

195

235

275

Pratio

i [kJ

/kg]

No regeneration

With regeneration

6 7 8 9 10 11 12 13 1440

60

80

100

120

140

160

Pratio

Psi

exit

[kJ/

kg]

No regeneration

With regeneration

6 7 8 9 10 11 12 13 1440

44

48

52

56

60

Pratio

th[%

]

With regeneration

No regeneration

9-91

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

KkJ/kg08336.2kJ/kg738.56

KkJ/kg68737.2kJ/kg88.797

KkJ/kg17888.3K1200

KkJ/kg.373482kJ/kg04.586

KkJ/kg1.70203K300

55

44

33

22

11

sh

sh

sT

sh

sT

and, from an energy balance on the heat exchanger,

KkJ/kg2.47108

kJ/kg645.36.04586738.5688.797

6

66425

s

hhhhh

Thus,

kJ/kg114.5

kJ/kg42.78

kJ/kg5.57

kJ/kg31.59

kJ/kg22.40

K300kJ/kg345.17

2.471081.70203K300

ln

K1260kJ/kg539.23

2.608333.17888K300

ln

2.687372.471082.373482.60833K300

1/8lnKkJ/kg0.2873.178882.68737K300

ln

8lnKkJ/kg0.2871.702032.37348K300

ln

out0

6

1610


0

5

3530


4625046250regengen,0regendestroyed,

3


1


LR

R

H

in

R

R

Tq

PP

RssTT

qssTsTx

Tq

PP

RssTT

qssTsTx

ssssTssssTsTx

PP

RssTssTsTx

PP

RssTssTsTx

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

06

026

060066 2gzVssThh

where KkJ/kg0.769051.702032.47108ln0

1

6161606 P

PRssssss

Thus, kJ/kg114.5KkJ/kg0.76905K300300.1936.6456

9-92

9-129 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-lawefficiency are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energychanges are negligible. 3 Air is an ideal gas withconstant specific heats.

700kPa

Diesel fuel

100kPa

Compres

4

32

Turbine

Combustionchamber

1

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

K6.505kPa100kPa700K303

1)/1.357-(1.357/)1(

1

212

kk

s PP

TT

0.881K)303533(K)3036.505(

12

12

TTTT s

C

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

kg/s81.12kg/s21.0kg/s6.1260

kg/s6.12kg/s6.12AF

aafat

mmmmm

kW85557)kJ/kg)(0.900kg/s)(42,021.0(HVin cf qmQThe temperature at the exit of combustion chamber is

K1144)K533kJ/kg.K)(3kg/s)(1.0981.12(kJ/s8555)( 3323in TTTTcmQ p

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

K7.685kPa700kPa100K1144

1)/1.357-(1.357/)1(

3

434

kk

s PP

TT

K4.754K)7.6851144(

K)1144(85.0 4

4

43

43 TT

TTTT

sT

The net power and the back work ratio are kW3168)K30333kJ/kg.K)(53kg/s)(1.096.12()( 12inC, TTcmW pa

kW5455)K4.754144kJ/kg.K)(13kg/s)(1.0981.12()( 43outT, TTcmW p


0.581kW5455kW3168

outT,

inC,bw W

Wr

(c) The thermal efficiency is 0.267kW8555kW2287

in

netth Q

W

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

735.0K1144K30311

3

1max T

T

and 0.364735.0267.0

max

thII

9-93

9-130 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperaturein the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output,the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are cp = 1.110 kJ/kg·K, cv = 0.823 kJ/kg·K, R = 0.287 kJ/kg·K,and k = 1.349 (Table A-2b).Analysis (a) The clearance volume and the total volume of the engine at the beginning of compressionprocess (state 1) are

xcc

c

c

dcr VVVV

VV

VV2

33

m0002154.0m0028.014

43

1 m003015.00028.00002154.0 VVVV dc

Process 1-2: Isentropic compression

kPa334114kPa95

K9.82314K328

1.349

2

112

1-1.3491

2

112

k

k

PP

TT

v

v

v

v

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

K2220kPa3341kPa9000K)9.823(

22 P

PTT x

x

kJ/kg1149K)9.8232220(kJ/kg.K)(0.823)( 2-2 TTcq xx v

x

Qout

V

P

4

3

2Qin

1

K32543333-2 K)2220(kJ/kg.K)(0.823kJ/kg1149)( TTTTcqq xpxx

(b) kJ/kg2298114911493-2in xx qqq

3333 m0003158.0

K2220K3254)m0002154.0(

xx T

TVV


kPa9.428m0.003015m0.0003158kPa9000

K1481m0.003015m0.0003158K3254

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/kg7.948K3281481KkJ/kg0.82314out TTcq v

The net work output and the thermal efficiency are kJ/kg13497.9482298outinoutnet, qqw

0.587kJ/kg2298kJ/kg1349

in

outnet,th q

w

9-94


kg0.003043K328K/kgmkPa0.287

)m015kPa)(0.00395(3

3

1

11

RTP

mV

kPa1466kJ

mkPam)0002154.0003015.0(

kJ/kg)kg)(1349(0.003043MEP

3

321

outnet,

VV

mw


kW120s60

min1rev/cycle)2(

(rev/min)3500kJ/kg)kg)(1349003043.0(2netnetnmwW

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

908.0K3254

K)27325(113

0max T

T

and 0.646908.0587.0

max

thII

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

kJ/kg6.567100

9.428ln287.0298

1481kJ/kg.Kln110.1()298(29814810.823

lnln)(0

4

0

4004040044 P

PR

TT

cTTTcssTuux pv

kW50.4s60

min1rev/cycle)2(

(rev/min)3500kJ/kg)kg)(567.6003043.0(244nmxX

9-95

9-131 A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion chamber, the net power output, the back work ratio, the thermal efficiency, the second law efficiency, the exergy efficienciesof the compressor, the turbine, and the regenerator, and the rate of the exergy of the combustion gases atthe regenerator exit are to be determined.Assumptions 1 The air-standard assumptions areapplicable. 2 Kinetic and potential energychanges are negligible. 3 Air is an ideal gas withconstant specific heats. Properties The properties of air at 500ºC = 773 K are cp = 1.093 kJ/kg·K, cv = 0.806 kJ/kg·K,R = 0.287 kJ/kg·K, and k = 1.357 (Table A-2b).Analysis (a) For the compressor and the turbine:

K6.505kPa100kPa700K303

1.3571-1.3571

1

212

kk

s PPTT

65

400°C700 kPa 260°C

Regenerator

871°C

Compress.

432

Turbine

Combustionchamber

1100 kPa 30°C

0.881K)303533(K)3036.505(

12

12

TTTT s

C

K6.685kPa700kPa100K1144

1)/1.357-(1.357/)1(

3

434

kk

s PP

TT

K4.754K)6.6851144(

K)1144(85.0 4

4

43

43 TT

TTTT

sT

(b) The effectiveness of the regenerator is

0.632K)5334.754(

K)533673(

24

25regen TT

TT

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

kg/s1613.0673)K144kJ/kg.K)(1093.1)(6.12(7)kJ/kg)(0.9000,42(

)()( 53HVin

fff

pafcf

mmm

TTcmmqmQ

78.141613.0

6.12AFf

a

mm

Also, kg/s76.121613.06.12fa mmm

kW65707)kJ/kg)(0.900kg/s)(42,01613.0(HVin cf qmQ(d) The net power and the back work ratio are

kW3168)K30333kJ/kg.K)(53kg/s)(1.096.12()( 12inC, TTcmW pa

kW5434)K4.754144kJ/kg.K)(13kg/s)(1.0976.12()( 43outT, TTcmW p


0.583kW5434kW3168

outT,

inC,bw W

Wr

(e) The thermal efficiency is

0.345kW6570kW2267

in

net

QW

th

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

9-96

735.0K1144K30311

3

1max T

T

and 0.469735.0345.0

max

thII

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

kW2943100700ln287.0

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

lnln)(1

2

1

201212012C P

PR

TT

cTTTcmssThhmX ppaa

0.929kW3168kW2943

inC,

CCII, W

X

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

kW5834100700ln287.0

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

lnln4

3

4

3043T P

PR

TT

cTTTcmX pp

0.932kW5834kW5434

T

inT,, X

WTII

An energy balance on the regenerator gives

K2.616)4.754)(093.1)(76.12()53373)(1.093)(66.12(

)()(

66

6425

TT

TTcmTTcm ppa

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

kW10730616.2754.4ln)093.1()303()2.6164.754)(093.1()76.12(

0ln6

4064hotregen, T

TcTTTcmX pp

kW8.9540533673ln)093.1()303()533673)(093.1()76.12(

0ln2

5025coldregen, T

TcTTTcmX pp

0.890kW1073kW8.954

hotregen,

coldregen,, X

XTII

The exergy of the combustion gases at the regenerator exit:

kW13510303

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

0ln0

60066 T

TcTTTcmX pp

9-97

Review Problems

9-132 A turbocharged four-stroke V-16 diesel engine produces 3500 hp at 1200 rpm. The amount ofpower 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 mechanicalcycles, we have (a)

cycle)mechkJ/cyl8.16(=hp1Btu/min42.41

rev/min)(1200cylinders)(16hp3500

cycles)mechanicalof(No.cylinders)of(No.producedpowerTotal

mechanical

cyclemechBtu/cyl7.73

w

(b)

cycle)thermkJ/cyl16.31(=hp1Btu/min42.41

rev/min)(1200/2cylinders)(16hp3500

cycles)amic thermodynof(No.cylinders)of(No.producedpowerTotal

micthermodyna

cyclethermBtu/cyl15.46

w

9-133 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to bedetermined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kineticand potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.Properties The specific heat ratio of air 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 compressor and turbine exit can be expressed as

kk

p

kk

kkp

kk

rT

PPTT

rTPPTT

/1

3

/1

3

434

/11

/1

1

212

1

T1

T3

qout

qin

3

4

2

1

T

Setting T2 = T4 and solving for rp givess

16.71.4/0.812/

1

3

K300K1500

kk

p TT

r

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

9-98

9-134 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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.

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

qout

q23

q12

1

2

4

3Analysis (b) We treat air as an ideal gas with variable specific heats,

kJ/kg187.2827.73655.923

kJ/kg736.2719.30046.1036

kJ/kg923.5593.93297.139507.21458.674

kJ/kg1036.46

3.110330.9kPa300kPa100

9.330kJ/kg,1395.97kJ/kg1022.82K0130

kJ/kg932.93kJ/kg674.58K900

K300kPa100kPa300

kJ/kg300.19kJ/kg214.07K300

outinnet

14out

2312in23,in12,in

4

3

4

33

33

2

2

11

22

1

11

2

22

1

11

34

qqw

hhq

hhuuqqq

h

PPP

P

PhuT

hu

TPP

TT

PT

P

huT

rr

r

vvv

T

q12

qout1

2q23

4

3

s

(c) 20.3%kJ/kg923.55kJ/kg187.28

in

netth q

w

9-99

9-135 All four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, 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 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).

PAnalysis (b) Process 3-4 is isentropic:

6538.832

kJ/kg653K300949.8KkJ/kg1.005

kJ/kg832.8K9001300KkJ/kg1.005K300900KkJ/kg0.718


K949.831K1300

outinnet

1414out

23122312in23,in12,in

11

22

1

11

2

22

0.4/1.4/1

3

434

kJ/kg179.8qqw

TTchhq

TTcTTchhuuqqq

TPP

TT

PT

P

PP

TT

p

pv

kk

vvqout

q23

q12

1

2

4

3

v

T

q12

qout1

2q23

4

3

(c) 21.6%kJ/kg832.8kJ/kg179.8

in

netth q

w s

9-136 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the maximum temperature in the cycle 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. P

qout

qin

1

2 3Analysis (b) We treat air as an ideal gas with variable specific heats,

kJ/kg2377.7kJ/kg1775.3K2100

K300kPa100kPa700

kJ/kg523.90702.9386.1kPa100kPa700

386.1kJ/kg214.07K300

3

33

11

33max

1

11

3

33

21

2

11

12

1

huT

TPP

TTT

PT

P

hPPP

P

PuT

rr

r

K2100vv

v

T

qin

qout1

23

(c)

kJ/kg1853.8kJ/kg1561.23

11

kJ/kg1561.2307.2143.1775

kJ/kg1853.89.5237.2377

in

outth

13out

23in

15.8%qq

uuq

hhqs

9-100

9-137 All three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-sdiagrams, and the maximum temperature in the cycle 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 at room temperature are cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg·K, and k= 1.4 (Table A-2).

P

qout

qin

1

2 3

Analysis (b) We treat air as an ideal gas with constant specific heats. Process 1-2 is isentropic:

K2100K300kPa100kPa700

K523.1kPa100kPa700K300

11

33max

1

11

3

33

0.4/1.4/1

1

212

TPPTT

TP

TP

PPTT

kk

vv vT

qin

qout1

23(c)

18.5%kJ/kg1584.8kJ/kg1292.4

11

kJ/kg1292.4K3002100KkJ/kg0.718

kJ/kg1584.8K523.12100KkJ/kg1.005

in

outth

1313out

2323in

qq

TTcuuq

TTchhq

v

p

s

9-138 A Carnot cycle executed in a closed system uses air as the working fluid. The net work output per cycle is to be determined.Assumptions 1 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).

th = 60%

300 K

1 MPa1

4

2700 kPa

3

T

s

Analysis (a) The maximum temperature is determined from

kJ0.115K300750KkJ/kg0.1024kg0.0025

KkJ/kg0.1204kPa1000kPa700

lnKkJ/kg0.287ln

K750K300

160.01

12

1

21

0

212

LHnet

HHH

Lth

TTssmW

PP

Rssss

TTT

T

9-101

9-139 [Also solved by EES on enclosed CD] A four-cylinder spark-ignition engine with a compression ratioof 8 is considered. The amount of heat supplied per cylinder, the thermal efficiency, and the rpm for a netpower output of 60 kW 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). The properties of air are given inTable A-17.Analysis (a) Process 1-2: isentropic compression.

P

1800 K

Qout

Qin

2

3

1

4

v

kJ/kg475.11

51.841.676811

1.676kJ/kg206.91K029

2

1

2

11

112

1

ur

uT

rrr

r

vvv

vv

v


kJ0.715kJ/kg475.111487.2kg107.065

kg107.065K290K/kgmkPa0.287

m0.0006kPa98

994.3kJ/kg1487.2K1800

423in

43

3

1

11

33

3

uumQ

RTP

m

uT

r

V

v


kJ/kg693.2395.31994.38 43

4334

ur rrr vvv

vv


51.9%kJ0.715kJ0.371

kJ0.371344.0715.0

kJ/kg206.91693.23kg107.065

in

netth

outinnet

-414out

QW

QQW

uumQ kJ0.344

(c) rpm4852min1

s60kJ/cycle)0.371(4

kJ/s60rev/cycle)2(2

cylnet,cyl

net

WnW

n

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

9-102

9-140 EES Problem 9-139 is reconsidered. The effect of the compression ratio net work done and theefficiency of the cycle is to be investigated. Also, the 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]=(17+273) [K] P[1]=98 [kPa]T[3]=1800 [K] V_cyl=0.6 [L]*Convert(L, m^3) r_v=8 "Compression ratio"W_dot_net = 60 [kW] N_cyl=4 "number of cyclinders"v[1]/v[2]=r_v

"The first part of the solution is done per unit mass.""Process 1-2 is isentropic compression"s[1]=entropy(air,T=T[1],P=P[1])s[2]=s[1]s[2]=entropy(air, T=T[2], v=v[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: no heat transfer (s=const.) with work input" w_in = DELTAu_12 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: the work is zero for v=const, heat is added"q_in = DELTAu_23 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]=R*T[4]}"Conservation of energy for process 3 to 4: no heat transfer (s=const) with work output" - w_out = DELTAu_34 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 2 to 3: the work is zero for v=const; heat is rejected"- q_out = DELTAu_41 DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4])w_net = w_out - w_in Eta_th=w_net/q_in*Convert(, %) "Thermal efficiency, in percent""The mass contained in each cylinder is found from the volume of the cylinder:"V_cyl=m*v[1]"The net work done per cycle is:"W_dot_net=m*w_net"kJ/cyl"*N_cyl*N_dot"mechanical cycles/min"*1"min"/60"s"*1"thermalcycle"/2"mechanical cycles"

9-103

th[%]

rv wnet[kJ/kg]

42.81 5 467.146.39 6 492.549.26 7 509.851.63 8 521.753.63 9 529.855.35 10 535.256.85 11 538.5

10-2 10-1 100 101 10250

100

1000

8000

v [m3/kg]

P[kPa]

290 K

1800 K

Air Otto Cycle P-v Diagram

s = const

1

2

3

4

1

2

3

4

v = const

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

s [kJ/kg-K]

T[K] 98 kPa

4866 kPa

Air Otto Cycle T-s Diagram

9-104

5 6 7 8 9 10 11460

470

480

490

500

510

520

530

540

rv

wne

t [k

J/kg

]

5 6 7 8 9 10 1142

44

46

48

50

52

54

56

58

rv

th [

%]

9-105

9-141 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 effectivepressure 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). The properties of air are given inTable A-17. Analysis (a) Process 1-2: isentropic compression.

kPa2129kPa98K300K708.3

9.2

kJ/kg9.518K3.70852.672.621

2.911

2.621kJ/kg214.07K300

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

P

qout

qin

2

3

1

4

v


kJ/kg609.89.5187.1128

593.8kJ/kg1128.7K1416.63.70822

23

3222

33

2

22

3

33

3

uuq

uTTPP

TT

PT

P

in

rv

vv


kJ/kg487.7506.79593.82.9 43

4334

ur rrr vvv

vv


kJ/kg336.17.2738.609kJ/kg273.707.21475.487

outinnet

14out

qqwuuq

(c) 55.1%kJ/kg609.8kJ/kg336.1

in

netth q

w

(d)

kPa429kJ1

mkPa11/9.21/kgm0.879

kJ/kg336.1/11

MEP

/kgm0.879kPa98

K300K/kgmkPa0.287

3

31

net

21

net

max2min

33

1

11max

rww

r

PRT

vvv

vvv

vv

9-106

9-142 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 effectivepressure 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 is isentropic compression:

kPa2190kPa98K300K728.8

9.2

K728.89.2K300

11

2

2

12

1

11

2

22

0.41

2

112

PTT

PT

PT

P

TTk

v

vvv

v

v P


kJ/kg523.3K728.81457.6KkJ/kg0.718

K457.618.72822

2323

222

33

2

22

3

33

TTcuuq

TTPP

TT

PT

P

in v

vv

qout

qin

2

3

1

4

v


K600.09.21K1457.6

0.41

4

334

k

TTv

v


kJ/kg307.94.2153.523

kJ/kg215.4K300600KkJ/kg0.718

outinnet

1414out

qqw

TTcuuq v

(c) 58.8%kJ/kg523.3kJ/kg307.9

in

netth q

w

(d)

kPa393kJ1

mkPa11/9.21/kgm0.879

kJ/kg307.9/11

MEP

/kgm0.879kPa98

K300K/kgmkPa0.287

3

31

net

21

net

max2min

33

1

11max

rww

r

PRT

vvv

vvv

vv

9-107

9-143 An engine operating on the ideal diesel cycle with air as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, 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 variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The properties of air are given inTable A-17.Analysis (a) The compression and the cutoff ratios are

P2

cm75cm15016

cm75cm1200

3

3

2

33

3

2

1

VV

VV

crr

Process 1-2: isentropic compression.

kJ/kg863.03K837.3256.421.676

1611

1.676kJ/kg206.91K029

2

21

2

11

112

1

hT

r

uT

rrr

r

vvv

vv

v

qout

qin 32

1

4

v


002.5kJ/kg1848.9

K1674.63.83722

3

3

222

33

2

22

3

33

r

h

TTTT

PT

P

v

v

vvv


kJ/kg636.00K853.4016.40002.5

216

224

42

4

3

43334

uTr

rrrr vvv

vv

vv

v


kPa294.3kPa100K290K853.4

11

44

1

11

4

44 PTTP

TP

TP vv

(b) kg101.442K290K/kgmkPa0.287

m0.0012kPa100 33

3

1

11

RTP

mV

kJ0.803619.0422.1

kJ0.619kJ/kg206.91636.00kg101.442

kJ1.422863.081848.9kg101.442

outinnet

3-14out

-323in

QQW

uumQ

hhmQ

(c) kPa714kJ1

mkPa11/161m0.0012

kJ0.803/11

MEP3

31

net

21

net

rWW

VVV

9-108

9-144 An engine operating on the ideal diesel cycle with argon as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, 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 Argon is an ideal gas with constant specific heats.Properties The properties of argon at room temperature are cp = 0.5203 kJ/kg.K, cv = 0.3122 kJ/kg·K, R= 0.2081 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The compression and the cutoff ratios are

2cm75cm15016

cm75cm1200

3

3

2

33

3

2

1

VV

VV

crr P

Qout

Qin 32

1

4

v


K184316K290 0.6671

1

212

k

TTVV


K3686184322 222

33

2

22

3

33 TTTT

PT

Pvvvv


K920.9162K368622 0.6671

3

1

4

23

1

4

334

kkk

rTTTT

VV

VV


kPa317.6kPa100K290K920.9

11

44

1

11

4

44 PTTP

TP

TP vv

(b) kg101.988K290K/kgmkPa0.2081

m0.0012kPa100 33

3

1

11

RTPm V

kJ1.514392.0906.1

kJ0.392K290920.9KkJ/kg0.3122kg101.988

kJ1.906K18433686KkJ/kg0.5203kg101.988

outinnet

3-1414out

-32323in

QQW

TTmcuum

TTmchhmQ p

vQ

(c) kPa1346kJ1

mkPa11/161m0.0012

kJ1.514/11

MEP3

31

net

21

net

rWW

VVV

9-109

9-145E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. Thethermal efficiency of the cycle is 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 = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R,and k = 1.4 (Table A-2E). Analysis The mass of air is

lbm103.132R550R/lbmftpsia0.3704

ft75/1728psia14.7 33

3

1

11

RTPm V P

0.3 Btu

x

Qout

1.1 Btu 3

2

1

4

v


R148612R550 0.41

2

112

k

TTVV

Process 2-x: v = constant heat addition,

R2046R1486RBtu/lbm0.171lbm103.132Btu0.3 x3

22in,2

x

xxx

TT

TTmcuumQ v

Process x-3: P = constant heat addition.

1.715R2046R3509

R3509R2046RBtu/lbm0.240lbm103.132Btu1.1

33

3

33

333

33in,3

xxc

x

xx

xpxx

TT

rT

PT

P

TT

TTmchhmQ

V

VVV


R161112

1.715R3509715.1715.1 0.41

3

1

4

13

1

4

334

kkk

rTTTT

V

V

V

V


59.4%Btu1.4

Btu0.56811

Btu0.568R5501611RBtu/lbm0.171lbm103.132

in

outth

31414out

QQ

TTmcuumQ v

9-110

9-146 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in thecycle and the net work output are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

350 K

1800 K 1

4qout

qin = 900 kJ/kg

2

3

T

Properties The properties of air at room temperature areR = 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

sKkJ/kg0.5

K1800kJ/kg90012

12HT

qss

and

kPa5873K350K1800

5.710kPa200

710.5lnKkJ/kg0.287KkJ/kg0.5lnln

3

1

1

23

3

1

1

331

1

11

3

33

1

2

1

2

1

20

1

212

TT

PTT

PPT

PT

P

RTT

css

v

v

v

vvv

v

v

v

v

v

vv

(b) kJ/kg725kJ/kg900K1800K350

11 ininthnet qTT

qwH

L

9-111

9-147 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the network output per unit mass 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 The properties at various states are T

3

2qin 3

4qout

2

1

T hP

T hP

r

r

1 1

3 3

1

3

1 386

330 9

300 K 300.19 kJ / kg

1300 K 1395.97 kJ / kg

.

.

For rp = 6, s

%9.37kJ/kg894.57kJ/kg339.46

kJ/kg339.4611.55557.894kJ/kg555.1119.3003.855

kJ/kg894.5740.50197.1395

kJ/kg855.315.559.33061

kJ/kg501.40316.8386.16

in

netth

outinnet

14out

23in

43

4

21

2

34

12

qw

qqwhhqhhq

hPPP

P

hPPP

P

rr

rr

For rp = 12,

%5.48kJ/kg785.37kJ/kg380.96

kJ/kg380.9641.40437.785kJ/kg404.4119.3006.704

kJ/kg785.3760.61097.1395

kJ/kg704.658.279.330121

kJ/kg610.663.16386.112

in

netth

outinnet

14out

23in

43

4

21

2

34

12

qw

qqwhhqhhq

hPPP

P

hPPP

P

rr

rr

Thus,(a) increase46.33996.380net kJ/kg41.5w

(b) increase%9.37%5.48th 10.6%

9-112

9-148 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the network output per unit mass 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 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/kg803.4kJ/kg321.9

kJ/kg321.95.4814.803

kJ/kg481.5K300779.1KkJ/kg1.005

kJ/kg803.4K500.61300KkJ/kg1.005

K779.161K1300

K500.66K300

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

T3

2qin 3

4qout

2

1s

For rp = 12,

%8.50kJ/kg693.2kJ/kg352.3

kJ/kg352.39.3402.693

kJ/kg340.9K300639.2KkJ/kg1.005

kJ/kg693.2K610.21300KkJ/kg1.005

K639.2121K1300

K610.212K300

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

Thus,(a) increase9.3213.352net kJ/kg30.4w

(b) increase%1.40%8.50th 10.7%

9-113

9-149 A regenerative Brayton cycle with helium as the working fluid is considered. The thermal efficiency and the required mass flow rate of helium are to be determined for 100 percent and 80 percent isentropicefficiencies for both the compressor and the turbine. 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 are cp = 5.1926 kJ/kg.Kand k = 1.667 (Table A-2). T

Analysis (a) Assuming T = C = 100%,

K759.84.6893.78375.04.689

K783.381K1800

K689.48K300

242524

25

24

25

70.667/1.66/1

3

434

70.667/1.66/1

1

212

TTTTTTcTTc

hhhh

PP

T

PP

TT

p

p

kk

s

kk

s

T

24

6

5

qin 3

4s2s

1

1800 K

300 Ks

60.3%kJ/kg5401.3kJ/kg3257.3

kJ/kg5401.3K759.81800KkJ/kg5.1926

kJ/kg3257.3kJ/s60,000

kJ/kg3257.3K300689.4783.31800KkJ/kg5.1926

in

netth

5353in

net

net

12431243inC,outT,net

qw

TTchhq

wW

m

TTTTchhhhwww

p

p

kg/s18.42

(b) Assuming T = C = 80%,

K986.63.783180080.01800

K783.381K1800

K786.880.0/3004.689300/

K689.48K300

433443

43

43

43

70.667/1.66/1

3

434

121212

12

12

12

70.667/1.66/1

1

212

sTsp

p

sT

kk

s

Csp

spsC

kk

s

TTTTTTcTTc

hhhh

PP

TT

TTTTTTcTTc

hhhh

PP

TT

37.8%kJ/kg4482.8kJ/kg1695.9

kJ/kg4482.8K936.71800KkJ/kg5.1926

kJ/kg1695.9kJ/s60,000

kJ/kg1695.9K300786.8986.61800KkJ/kg5.1926

K936.78.7866.98675.08.786

in

netth

5353in

net

net

12431243inC,outT,net

242524

25

24

25

qw

TTchhq

wWm

TTTTchhhhwww

TTTTTTcTTc

hhhh

p

p

p

p

kg/s35.4

9-114

9-150 A regenerative gas-turbine engine operating with two stages of compression and two stages ofexpansion 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 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 The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.

kJ/kg624.1K889.51200KkJ/kg1.005222

kJ/kg332.7K300465.5KkJ/kg1.005222

K770.85.4655.88972.05.465

K889.59.838120086.01200

K838.93.51K1200

K465.578.0/3001.429300

/

K429.13.5K300

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 T

4

9

22s

5

8

77s

9

6

1

4

3s

Thus, 53.3%kJ/kg624.1kJ/kg332.7

outT,

inC,bw w

wr

39.2%kJ/kg743.4kJ/kg291.4

kJ/kg291.47.3321.624

kJ/kg743.4K889.51200770.81200KkJ/kg1.005

in

netth

inC,outT,net

78567856in

qw

www

TTTTchhhhq p

9-115

9-151 EES Problem 9-150 is reconsidered. The effect of the isentropic efficiencies for the compressor andturbine and regenerator effectiveness on net work done and the heat supplied to the cycle is to be investigated. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows:

"Input data"T[6] = 1200 [K] T[8] = T[6]Pratio = 3.5 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = T[1]Eta_reg = 0.72 "Regenerator effectiveness"Eta_c =0.78 "Compressor isentorpic efficiency"Eta_t =0.86 "Turbien isentropic efficiency"

"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 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_compisen_LP/w_comp_LP"compressor adiabatic efficiency, W_comp > W_compisen"

"Conservation of energy for the LP compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"h[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] h[2]=ENTHALPY(Air,T=T[2])s[2]=ENTROPY(Air,T=T[2], P=P[2])

"HP Compressor:"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 HP compressor"P[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 compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"h[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]

9-116

h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,T=T[4], P=P[4])

"Intercooling heat loss:"h[2] = q_out_intercool + h[3]

"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"h[4] + q_in_noreg = h[6]

h[6]=ENTHALPY(Air,T=T[6])P[6]=P[4]"process 4-6 is SSSF constant pressure"

"HP Turbine analysis"

s[6]=ENTROPY(Air,T=T[6],P=P[6])s_s[7]=s[6] "For the ideal case the entropies are constant across the turbine"P[7] = P[6] /Pratios_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"Eta_t = w_turb_HP /w_turbisen_HP "turbine adiabatic efficiency, w_turbisen > w_turb"

"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"h[6] = w_turbisen_HP + h_s[7] h_s[7]=ENTHALPY(Air,T=T_s[7])"Actual Turbine analysis:"h[6] = w_turb_HP + h[7] h[7]=ENTHALPY(Air,T=T[7])s[7]=ENTROPY(Air,T=T[7], P=P[7])

"Reheat Q_in:"h[7] + q_in_reheat = h[8] h[8]=ENTHALPY(Air,T=T[8])

"HL Turbine analysis"

P[8]=P[7]s[8]=ENTROPY(Air,T=T[8],P=P[8])s_s[9]=s[8] "For the ideal case the entropies are constant across the turbine"P[9] = P[8] /Pratios_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"Eta_t = w_turb_LP /w_turbisen_LP "turbine adiabatic efficiency, w_turbisen > w_turb"

"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"h[8] = w_turbisen_LP + h_s[9] h_s[9]=ENTHALPY(Air,T=T_s[9])"Actual Turbine analysis:"h[8] = w_turb_LP + h[9] h[9]=ENTHALPY(Air,T=T[9])s[9]=ENTROPY(Air,T=T[9], P=P[9])

"Cycle analysis"w_net=w_turb_HP+w_turb_LP - w_comp_HP - w_comp_LPq_in_total_noreg=q_in_noreg+q_in_reheatEta_th_noreg=w_net/(q_in_total_noreg)*Convert(, %) "[%]" "Cycle thermal efficiency"

9-117

Bwr=(w_comp_HP + w_comp_LP)/(w_turb_HP+w_turb_LP)"Back work ratio"

"With the regenerator, the heat added in the external heat exchanger is"h[5] + q_in_withreg = h[6] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5])P[5]=P[4]

"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (h[5]-h[4])/(h[9]-h[4]) "Energy balance on regenerator gives h[10] and thus T[10] as:"h[4] + h[9]=h[5] + h[10] h[10]=ENTHALPY(Air, T=T[10]) s[10]=ENTROPY(Air,T=T[10], P=P[10])P[10]=P[9]

"Cycle thermal efficiency with regenerator"q_in_total_withreg=q_in_withreg+q_in_reheatEta_th_withreg=w_net/(q_in_total_withreg)*Convert(, %) "[%]"

"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]s_s[8]=s[8]T_s[8]=T[8]s_s[10]=s[10]T_s[10]=T[10]

C reg t th,noreg

[%]th,withreg

[%]qin,total,noreg

[kJ/kg]qin,total,withreg

[kJ/kg]wnet

[kJ/kg]0.78 0.6 0.86 27.03 36.59 1130 834.6 305.40.78 0.65 0.86 27.03 37.7 1130 810 305.40.78 0.7 0.86 27.03 38.88 1130 785.4 305.40.78 0.75 0.86 27.03 40.14 1130 760.8 305.40.78 0.8 0.86 27.03 41.48 1130 736.2 305.40.78 0.85 0.86 27.03 42.92 1130 711.6 305.40.78 0.9 0.86 27.03 44.45 1130 687 305.40.78 0.95 0.86 27.03 46.11 1130 662.4 305.40.78 1 0.86 27.03 47.88 1130 637.8 305.4

9-118

4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T[K] 10

0kPa

350kP

a

1225

kPa

Air

1

2

3

4

5

6

7

8

9

10

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

t

th

reg = 0.72

c = 0.78

With regeneration

No regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1600

700

800

900

1000

1100

1200

t

q in,

tota

l

reg = 0.72

c = 0.78No regenerationWith regeneration

9-119

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1100

150

200

250

300

350

400

450

t

wne

t [k

J/kg

]reg = 0.72

c = 0.78

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 125

30

35

40

45

50

reg

th

c = 0.78

t = 0.86

With regenertion

No regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 115

20

25

30

35

40

45

50

c

th

reg = 0.72

t = 0.86

With regeneration

No regeneration

9-120

9-152 A regenerative gas-turbine engine operating with two stages of compression and two stages ofexpansion 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 (TableA-2).Analysis The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine.

kJ/kg4225.7K793.11200KkJ/kg5.1926222

kJ/kg2599.4K300550.3KkJ/kg5.1926222

K725.13.5501.79372.03.550

K793.19.726120086.01200

K726.93.51K1200

K550.378.0/3002.495300

/

K495.23.5K300

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 T

4

9

22s

5

8

77s

9qin

6

1

4

3s

Thus, 61.5%kJ/kg4225.7kJ/kg2599.4

outT,

inC,bw w

wr

35.5%kJ/kg4578.8kJ/kg1626.3

kJ/kg1626.34.25997.4225

kJ/kg4578.8K793.11200725.11200KkJ/kg5.1926

in

netth

inC,outT,net

78567856in

qw

www

TTTTchhhhq p

9-121

9-153 An ideal regenerative Brayton cycle is considered. The pressure ratio that maximizes the thermalefficiency of the cycle is to be determined, and to be compared with the pressure ratio that maximizes the cycle net work. Analysis Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as

kkp

kk

p

kk

kkp

kk

rTr

TPP

TT

rTPP

TT

/13

/1

3

/1

3

434

/11

/1

1

212

1

T

Then,

1/1

1/1

33outinnet

/11121616out

/13435353in

1

1

TrTrTTcqqw

rTcTTcTTchhq

rTcTTcTTchhq

kkp

kkpp

kkpppp

kkpppp

6

5

qin3

42

1s

To maximize the net work, we must have

01111 /11

/13

net kkp

kkpp

prT

kkrT

kkc

rw

Solving for rp giveskk

p TTr

12/

3

1

Similarly,

kkpp

kkpp

rTc

rTcqq

/13

/11

in

outth

1

111

which simplifies to

kkpr

TT /1

3

1th 1

When rp = 1, the thermal efficiency becomes th = 1 - T1/T3, which is the Carnot efficiency. Therefore, the efficiency is a maximum when rp = 1, and must decrease as rp increases for the fixed values of T1 and T3.Note that the compression ratio cannot be less than 1, and the factor

kkpr /1

is always greater than 1 for rp > 1. Also note that the net work wnet = 0 for rp = 1. This being the case, thepressure ratio for maximum thermal efficiency, which is rp = 1, is always less than the pressure ratio formaximum work.

9-122

9-154 An ideal gas-turbine cycle with one stage of compression and two stages of expansion andregeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressureratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to becompared with the efficiency of the standard regenerative cycle.Analysis The T-s diagram of the cycle is as shown in thefigure. If the overall pressure ratio of the cycle is rp, which is the pressure ratio across the compressor, then the pressure ratioacross each turbine stage in the ideal case becomes rp. Usingthe isentropic relations, the temperatures at the compressor and turbine exit can be expressed as

T

kkp

kkp

kkp

kkp

kk

p

kk

kkp

kk

p

kk

kkp

kk

rTrrTrTr

TPPTT

rTr

TPPTTT

rTPPTTT

2/11

2/1/11

2/12

/1

5

/1

5

656

2/13

/1

3

/1

3

4347

/11

/1

1

2125

1

1 qout

3

4

6

5

qin

7

2

1s

Then,

1

1

2/111616out

2/137373in

kkppp

kkppp

rTcTTchhq

rTcTTchhq

and thus

kkpp

kkpp

rTc

rTcqq

2/13

2/11

in

outth

1

111

which simplifies to

kkpr

TT 2/1

3

1th 1

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

kkpr

TT /1

3

1th 1

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.

9-123

9-155 A spark-ignition engine operates on an ideal Otto cycle with a compression ratio of 11. The maximum temperature and pressure in the cycle, the net work per cycle and per cylinder, the thermalefficiency, the mean effective pressure, and the power output for a specified engine speed 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) First the mass in one cylinder is determined as follows

cylinderoneform000045.04/)0018.0(

11 3c

c

c

c

dcr VV

V

V

VV

31 m000495.000045.0000045.0dc VVV

kg0.0004805K323K/kgmkPa0.287

)m495kPa)(0.00090(3

3

1

11

RTPm V

For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is afunction of temperature only whereas entropy is functions of both temperature and pressure. Process 1-2: Isentropic compression

KkJ/kg8100.5kPa90

K233C50kJ/kg88.230K233C50

11

1

11

sPT

uT

Qout

V

P

4

3

2

Qin

1/kgm09364.0kg0004805.0

m000045.0

m000045.0

/kgm0302.1kg0004805.0

m000495.0

33

22

32

33

11

m

m

c

Vv

VV

Vv

K3.807kg/m09364.0

kJ/kg.K8100.523

2

12 Tss

v

kJ/kg33.598K3.807 22 uT

kPa2474K323K807.3

m0.000045m0.000495kPa)90(

3

3

1

2

2

112 T

TPP

V

V

Process 2-3: constant volume heat addition

kJ/kg8.3719kJ/kg)33.598kg)(0004805.0(kJ5.1)( 3323in uuuumQ

K403732 kJ/kg33.598 Tu

kPa12,375K807.3K4037kPa)2474(

2

323 T

TPP

KkJ/kg3218.7kPa375,12

K40373

3

3 sPT

9-124


K2028kg/m0302.1

kJ/kg.K3218.743

14

34 Tssvv

kJ/kg6.1703K2028 44 uT

kPa565K4037K2028

111kPa)375,12(

3

4

4

334 T

TPP

V

V

Process 4-1: constant voume heat rejectionkJ7077.0kJ/kg88.2301703.6kg)0004805.0()( 14out uumQ

The net work output and the thermal efficiency are cylinder)percycle(per7077.05.1outinoutnet, kJ0.792QQW

0.528kJ1.5

kJ0.792

in

outnet,th Q

W


kPa1761kJ

mkPam)000045.0000495.0(

kJ0.7923MEP

3

321

outnet,

VV

W


kW79.2s60

min1rev/cycle)2(

rev/min)3000(cycle)-rkJ/cylinde0.792cylinder)(4(2netcylnetnWnW

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

9-125

9-156 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. Theback work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies atthe exits of the combustion chamber and the regenerator 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.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: kPa4.346)1200)(100(412 PPP

Process 1-2, 3-4: Compression

T

KkJ/kg7054.5kPa100K300

kJ/kg43.300K300

11

1

11

sPT

hT

4

8

22s

9

10

7

66s

8qin

5

1

4

3

kJ/kg79.428kJ/kg.K7054.5

kPa4.3462

12

2sh

ssP

kJ/kg88.46043.300

43.30079.42880.0 2212

12C h

hhhhh s s

KkJ/kg5040.5kPa4.346

K350kJ/kg78.350K350

33

3

33

sPT

hT

kJ/kg42.500kJ/kg.K5040.5

kPa12004

34

4sh

ssP

kJ/kg83.53778.350

78.35042.50080.0 4434

34C h

hhhhh s

Process 6-7, 8-9: Expansion

KkJ/kg6514.6kPa1200K1400

kJ/kg9.1514K1400

66

6

66

sPT

hT

kJ/kg9.1083kJ/kg.K6514.6

kPa4.3467

67

7sh

ssP

kJ/kg1.11709.10839.1514

9.151480.0 7

7

76

76T h

hhhhh

s

KkJ/kg9196.6kPa4.346

K1300kJ/kg6.1395K1300

88

8

88

sPT

hT

kJ/kg00.996kJ/kg.K9196.6

kPa1009

89

9sh

ssP

9-126

kJ/kg9.107500.9966.1395

6.139580.0 9

9

98

98T h

hhhhh

s

Cycle analysis: kJ/kg50.34778.35083.53743.30088.4603412inC, hhhhw

kJ/kg50.6649.10756.13951.11709.15149876outT, hhhhw

0.52350.66450.347

outT,

inC,bw w

wr

kJ/kg317.050.34750.664inC,outT,net www

Regenerator analysis:

kJ/kg36.67283.5379.1075

9.107575.0 10

10

49

109regen h

hhhhh

KkJ/kg5157.6kPa100

K36.67210

10

10 sPh

kJ/kg40.94183.53736.6729.1075 5545109regen hhhhhhq

(b) kJ/kg54.57340.9419.151456in hhq

0.55354.5730.317

in

netth q

w

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

786.0K1400K30011

6

1max T

T

and 0.704786.0553.0

max

thII

(d) The exergies at the combustion chamber exit and the regenerator exit are kJ/kg930.7kJ/kg.K)7054.56514.6)(K300(kJ/kg)43.3009.1514()( 060066 ssThhx

kJ/kg128.8kJ/kg.K)7054.55157.6)(K300(kJ/kg)43.30036.672()( 010001010 ssThhx

9-127

9-157 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 flowrate 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.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. Analysis (a) For this problem, we use theproperties of air from EES software. Rememberthat for an ideal gas, enthalpy is a function oftemperature only whereas entropy is functions ofboth temperature and pressure.

5Heatexchanger

Sat. vap.200°C

25°C350°C

1.2 MPa 500°C

100 kPa30°C

Compress.

43

2

Turbine

Combustionchamber

1

Process 1-2: Compression

KkJ/kg7159.5kPa100

C30kJ/kg60.303C30

11

1

11

sPT

hT

kJ/kg37.617kJ/kg.K7159.5

kPa12002

12

2sh

ssP

kJ/kg24.68660.303

60.30337.61782.0 2212

12C h

hhhhh s

Process 3-4: Expansion

kJ/kg62.792C500 44 hT

ss hhh

hhhh

43

3

43

43T

62.79282.0

We cannot find the enthalpy at state 3 directly. However, using the following lines in EES together withthe isentropic efficiency relation, we find h3 = 1404.7 kJ/kg, T3 = 1034ºC, s3 = 6.5699 kJ/kg.K. The solution by hand would require a trial-error approach.

h_3=enthalpy(Air, T=T_3)s_3=entropy(Air, T=T_3, P=P_2)h_4s=enthalpy(Air, P=P_1, s=s_3)

Also, kJ/kg44.631C350 55 hT

The inlet water is compressed liquid at 25ºC 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 liquidenthalpy at the given temperature.

kJ/kg0.27921

C200

kJ/kg27.106kPa1555

C25

22

2

11

1

ww

ww

hx

T

hP

T

The net work output is

kJ/kg03.61262.7927.1404kJ/kg64.38260.30324.686

43outT,

12inC,

hhwhhw

kJ/kg39.22964.38203.612inC,outT,net www

9-128

The mass flow rate of air is

kg/s3.487kJ/kg39.229

kJ/s800

net

net

wW

ma

(b) The back work ratio is

0.62503.61264.382

outT,

inC,bw w

wr

The rate of heat input and the thermal efficiency are

kW2505)kJ/kg24.686.7kg/s)(1404487.3()( 23in hhmQ a

0.319kW2505

kW800

in

netth Q

W

(c) An energy balance on the heat exchanger gives

kg/s0.2093ww

wwwa

mm

hhmhhm

)kJ/kg27.106(2792.0)kJ/kg44.63162kg/s)(792.487.3(

)()( 1254

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

kW1.562)kJ/kg27.106.0kg/s)(27922093.0()( 12p www hhmQ

0.544kW2505

1.562800

in

pnet

Q

QWu

9-129

9-158 A turbojet aircraft flying is considered. The pressure of the gases at the turbine exit, the mass flowrate 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. 3Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). TAnalysis (a) For this problem, we use the properties from EES software.Remember that for an ideal gas, enthalpy is a function of temperatureonly whereas entropy is functions of both temperature and pressure. Diffuser, Process 1-2:

kJ/kg23.238C35 11 hT

kJ/kg37.269/sm1000

kJ/kg12m/s)(15

/sm1000kJ/kg1

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

22

222

2

222

2

22

2

21

1

hh

VhVh

6

qout

5

qin

3

4

2

1s

KkJ/kg7951.5kPa50

kJ/kg37.2692

2

2 sPh

Compressor, Process 2-3:

kJ/kg19.505kJ/kg.K7951.5

kPa4503

23

3sh

ssP

kJ/kg50.55337.269

37.26919.50583.0 3323

23C h

hhhhh s

Turbine, Process 3-4:

kJ/kg8.1304C950 44 hT

kJ/kg6.10208.130437.26950.553 555423 hhhhhh

where the mass flow rates through the compressor and the turbine are assumed equal.

kJ/kg45.9628.1304

6.10208.130483.0 5554

54T s

ssh

hhhhh

KkJ/kg7725.6kPa450C950

44

4 sPT

kPa147.4545

5

KkJ/kg7725.6kJ/kg45.962

Pss

h s

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

kg/s1.760kJ/kg)37.26950.553(

kJ/s500

23

C

hhW

m

(c) Nozzle, Process 5-6:

KkJ/kg8336.6kPa4.147kJ/kg6.1020

55

5 sPh

9-130

kJ/kg66.709kJ/kg.K8336.6

kPa406

56

6sh

ssP

kJ/kg52.76266.7096.1020

6.102083.0 6

6

65

65N h

hhhhh

s

m/s718.5622

26

26

6

25

5

/sm1000kJ/kg1

2kJ/kg52.7620kJ/kg)6.1020(

22

VV

VhVh

where the velocity at nozzle inlet is assumed zero.(d) The propulsive power and the propulsive efficiency are

kW206.122116 /sm1000kJ/kg1m/s)250)(m/s250m/s5.718(kg/s)76.1()( VVVmWp

kW1322kJ/kg)50.5538.1304(kg/s)76.1()( 34in hhmQ

0.156kW1322kW1.206

inQ

W pp

9-131

9-159 EES The effect of variable specific heats on the thermal efficiency of the ideal Otto cycle using air as the working fluid is to be investigated. The percentage of error involved in using constant specific heatvalues at room temperature for different combinations of compression ratio and maximum cycletemperature is to be determined.Analysis Using EES, the problem is solved as follows:

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"

9-132

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])q_in_total=q_23q_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 th

[%]th,ConstProp

[%]th,easy

[%]T3

[K

3.604 12 60.8 62.99 62.99 10006.681 12 59.04 62.99 62.99 15009.421 12 57.57 62.99 62.99 200011.64 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

rcomp

Per

Cen

tErr

or [

%]

Percent Error = | th - th,ConstProp | / th

Tmax = 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

9-133

9-160 EES The effects of compression ratio on the net work output and the thermal efficiency of the Otto cycle for given operating conditions is to be investigated.Analysis Using EES, the problem is solved as follows:

"Input Data"T[1]=300 [K] P[1]=100 [kPa]T[3] = 2000 [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 w_41 =0 "constant volume process"DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4])

q_in_total=q_23q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) "Thermal efficiency, in percent"

9-134

th[%]

rcomp wnet[kJ/kg]

45.83 6 567.448.67 7 589.351.03 8 604.953.02 9 616.254.74 10 624.356.24 11 63057.57 12 633.858.75 13 636.359.83 14 637.560.8 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

th [

%]

9-135

9-161 The effects of pressure ratio on the net work output and the thermal efficiency of a simple Braytoncycle is to be investigated. The pressure ratios at which the net work output and the thermal efficiency are maximum are to be determined.Analysis Using EES, the problem is solved as follows:

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


"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"


"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])

9-136

Bwr Pratio Wc[kW]

Wnet[kW]

Wt[kW]

Qin[kW]

0.254 0.3383 5 175.8 516.3 692.1 15260.2665 0.3689 6 201.2 553.7 754.9 15010.2776 0.3938 7 223.7 582.2 805.9 14780.2876 0.4146 8 244.1 604.5 848.5 14580.2968 0.4324 9 262.6 622.4 885 14390.3052 0.4478 10 279.7 637 916.7 14220.313 0.4615 11 295.7 649 944.7 14060.3203 0.4736 12 310.6 659.1 969.6 13920.3272 0.4846 13 324.6 667.5 992.1 13780.3337 0.4945 14 337.8 674.7 1013 13640.3398 0.5036 15 350.4 680.8 1031 13520.3457 0.512 16 362.4 685.9 1048 13400.3513 0.5197 17 373.9 690.3 1064 13280.3567 0.5269 18 384.8 694.1 1079 13170.3618 0.5336 19 395.4 697.3 1093 13070.3668 0.5399 20 405.5 700 1106 12970.3716 0.5458 21 415.3 702.3 1118 12870.3762 0.5513 22 424.7 704.3 1129 12770.3806 0.5566 23 433.8 705.9 1140 12680.385 0.5615 24 442.7 707.2 1150 12590.3892 0.5663 25 451.2 708.3 1160 12510.3932 0.5707 26 459.6 709.2 1169 12430.3972 0.575 27 467.7 709.8 1177 12340.401 0.5791 28 475.5 710.3 1186 12270.4048 0.583 29 483.2 710.6 1194 12190.4084 0.5867 30 490.7 710.7 1201 12110.412 0.5903 31 498 710.8 1209 12040.4155 0.5937 32 505.1 710.7 1216 11970.4189 0.597 33 512.1 710.4 1223 11900.4222 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

Wnet

[kW]

th

9-137

9-162 EES The effects of pressure ratio on the net work output and the thermal efficiency of a simpleBrayton 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 maximumare to be determined.Analysis Using EES, the problem is solved as follows:

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






"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])

9-138

s[4]=entropy('air',T=T[4],P=P[4])

9-139

Bwr Pratio Wc[kW]

Wnet[kW]

Wt[kW]

Qin[kW]

0.3515 0.2551 5 206.8 381.5 588.3 14950.3689 0.2764 6 236.7 405 641.7 14650.3843 0.2931 7 263.2 421.8 685 14390.3981 0.3068 8 287.1 434.1 721.3 14150.4107 0.3182 9 309 443.3 752.2 13930.4224 0.3278 10 329.1 450.1 779.2 13730.4332 0.3361 11 347.8 455.1 803 13540.4433 0.3432 12 365.4 458.8 824.2 13370.4528 0.3495 13 381.9 461.4 843.3 13200.4618 0.355 14 397.5 463.2 860.6 13050.4704 0.3599 15 412.3 464.2 876.5 12900.4785 0.3643 16 426.4 464.7 891.1 12760.4862 0.3682 17 439.8 464.7 904.6 12620.4937 0.3717 18 452.7 464.4 917.1 12490.5008 0.3748 19 465.1 463.6 928.8 12370.5077 0.3777 20 477.1 462.6 939.7 12250.5143 0.3802 21 488.6 461.4 950 12140.5207 0.3825 22 499.7 460 959.6 12020.5268 0.3846 23 510.4 458.4 968.8 11920.5328 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]

thPratio for

W net,max

W netth

9-140

9-163 EES 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 cyclewith air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used.Analysis Using EES, the problem is solved as follows:

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])

9-141

"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])q_in_total=q_23q_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 th[%]

th,ConstProp[%]

th,easy[%]

T3[K]

3.604 12 60.8 62.99 62.99 10006.681 12 59.04 62.99 62.99 15009.421 12 57.57 62.99 62.99 200011.64 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 Error = | th - th ,ConstProp | / th

Tm ax = 1000 K

9-142

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

9-143

9-164 EES 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 to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows:

"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 = 75/100 Eta_t = 82/100}






"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])

9-144

s[4]=entropy('air',T=T[4],P=P[4])

9-145

Bwr Pratio Wc[kW]

Wnet[kW]

Wt[kW]

Qin[kW]

0.5229 0.1 2 1818 1659 3477 165870.6305 0.1644 4 4033 2364 6396 143730.7038 0.1814 6 5543 2333 7876 128620.7611 0.1806 8 6723 2110 8833 116820.8088 0.1702 10 7705 1822 9527 10700

0.85 0.1533 12 8553 1510 10063 98520.8864 0.131 14 9304 1192 10496 91020.9192 0.1041 16 9980 877.2 10857 84260.9491 0.07272 18 10596 567.9 11164 78090.9767 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


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]

W net

Tmax=1160 K


c = 0.75t = 0.82

9-146

9-165 EES 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.Analysis Using EES, the problem is solved as follows:

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) endifENDProcedure 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"

9-147

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])

Bwr Pratio Wc

[kW]Wnet

[kW]Wt

[kW]Qin

[kW]0.5229 0.1 2 1818 1659 3477 165870.6305 0.1644 4 4033 2364 6396 143730.7038 0.1814 6 5543 2333 7876 128620.7611 0.1806 8 6723 2110 8833 116820.8088 0.1702 10 7705 1822 9527 10700

0.85 0.1533 12 8553 1510 10063 98520.8864 0.131 14 9304 1192 10496 91020.9192 0.1041 16 9980 877.2 10857 84260.9491 0.07272 18 10596 567.9 11164 78090.9767 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

Brayton Cycle


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

Cycleefficiency,

Wnet

[kW]

Wnet

Tmax=1160 K


c = 0.75t = 0.82

Brayton Cycle using Airm air = 20 kg/s

9-148

9-166 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermalefficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows:

"Input data for air"C_P = 1.005 [kJ/kg-K] k = 1.4 "Other Input data from the diagram window"{T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8"Compressor isentorpic efficiency" Eta_t =0.9"Turbien isentropic efficiency"}

"Isentropic Compressor anaysis"T_s[2] = T[1]*Pratio^((k-1)/k)P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit"Eta_c = w_compisen/w_comp"compressor adiabatic efficiency, W_comp > W_compisen"

"Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1])

"Actual compressor analysis:" w_comp = C_P*(T[2]-T[1])

"External heat exchanger analysis""SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow"q_in_noreg = C_P*(T[3] - T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure"

"Turbine analysis"P[4] = P[3] /Pratio T_s[4] = T[3]*(1/Pratio)^((k-1)/k)"T_s[4] is the isentropic value of T[4] at turbine exit"Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb"

"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen = C_P*(T[3] - T_s[4])

"Actual Turbine analysis:"w_turb = C_P*(T[3] - T[4])

"Cycle analysis"w_net=w_turb-w_compEta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb "Back work ratio"

9-149

"With the regenerator the heat added in the external heat exchanger is"q_in_withreg = C_P*(T[3] - T[5]) P[5]=P[2]

"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:"T[2] + T[4]=T[5] + T[6] P[6]=P[4]

"Cycle thermal efficiency with regenerator"Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "[%]"

c t th,noreg[%]

th,withreg[%]

qin,noreg[kJ/kg]

qin,withreg[kJ/kg]

wnet[kJ/kg]

0.6 0.9 14.76 13.92 510.9 541.6 75.40.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 1420.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 1920.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8

9-150

9-167 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are tobe used. Analysis Using EES, the problem is solved as follows:

"Input data""Input data from the diagram window"{T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "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_compisen/w_comp"compressor adiabatic efficiency, W_comp > W_compisen"

"Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow"h[1] + w_compisen = 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 = 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_in - e_out =DELTAe_cv =0 for steady flow"h[2] + q_in_noreg = h[3]

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_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb"

"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"

9-151

h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4])"Actual Turbine analysis:"h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,T=T[4], P=P[4])"Cycle analysis"w_net=w_turb-w_compEta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb"Back work ratio"

"With the regenerator the heat added in the external heat exchanger is"h[5] + q_in_withreg = h[3]

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:"h[2] + h[4]=h[5] + h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6])P[6]=P[4]


4.5 5.0 5.5 6.0 6.5 7.0 7.5200

400

600

800

1000

1200

1400

1600

s [kJ/kg-K]

T [K

]

100 kPa

1000 kPa

Air

2s

1

2 5

3

44s

6

"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]

c t th,noreg[%]

th,withreg[%]

qin,noreg[kJ/kg]

qin,withreg[kJ/kg]

wnet[kJ/kg]

0.6 0.9 14.76 13.92 510.9 541.6 75.40.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 1420.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 1920.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8

9-152

0.7 0.75 0.8 0.85 0.9 0.95 110

15

20

25

30

35

40

45

t

thc = 0.8

With regeneration

No regeneration

0.7 0.75 0.8 0.85 0.9 0.95 150

95

140

185

230

275

t

wne

t [k

J/kg

]

c = 0.8

0.7 0.75 0.8 0.85 0.9 0.95 1400

450

500

550

600

650

t

q in

c = 0.8

No regeneration

With regeneration

9-153

0.6 0.65 0.7 0.75 0.8 0.85 0.910

15

20

25

30

35

40

45

c

tht = 0.9

With regeneration

No regeneration

0.6 0.65 0.7 0.75 0.8 0.85 0.975

110

145

180

215

250

c

wne

t[k

J/kg

]

t = 0.9

0.6 0.65 0.7 0.75 0.8 0.85 0.9500

520

540

560

580

600

620

640

660

680

c

q in

t = 0.9

No regeneration

With regeneration

9-154

9-168 EES The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with helium as the working fluid is to be investigated.Analysis Using EES, the problem is solved as follows:

"Input data for helium"C_P = 5.1926 [kJ/kg-K] k = 1.667

"Other Input data from the diagram window"{T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "Turbien isentropic efficiency"}

"Isentropic Compressor anaysis"

T_s[2] = T[1]*Pratio^((k-1)/k)P[2] = Pratio*P[1] "T_s[2] is the isentropic value of T[2] at compressor exit"Eta_c = w_compisen/w_comp"compressor adiabatic efficiency, W_comp > W_compisen"

"Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_s[2]-T[1])

"Actual compressor analysis:" w_comp = C_P*(T[2]-T[1])

"External heat exchanger analysis""SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 e_in - e_out =DELTAe_cv =0 for steady flow"q_in_noreg = C_P*(T[3] - T[2]) P[3]=P[2]"process 2-3 is SSSF constant pressure"

"Turbine analysis"P[4] = P[3] /Pratio

T_s[4] = T[3]*(1/Pratio)^((k-1)/k)"T_s[4] is the isentropic value of T[4] at turbine exit"Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb"

"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen = C_P*(T[3] - T_s[4])

"Actual Turbine analysis:"w_turb = C_P*(T[3] - T[4])

"Cycle analysis"w_net=w_turb-w_comp

9-155

Eta_th_noreg=w_net/q_in_noreg*Convert(, %) "[%]" "Cycle thermal efficiency"Bwr=w_comp/w_turb "Back work ratio"

"With the regenerator the heat added in the external heat exchanger is"q_in_withreg = C_P*(T[3] - T[5]) P[5]=P[2]

"The regenerator effectiveness gives h[5] and thus T[5] as:"Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:"T[2] + T[4]=T[5] + T[6] P[6]=P[4]


c t th,noreg[%]

th,withreg[%]

qin,noreg[kJ/kg]

qin,withreg[kJ/kg]

wnet[kJ/kg]

0.6 0.9 14.76 13.92 510.9 541.6 75.40.65 0.9 20.35 20.54 546.8 541.6 111.30.7 0.9 24.59 26.22 577.5 541.6 1420.75 0.9 27.91 31.14 604.2 541.6 168.60.8 0.9 30.59 35.44 627.5 541.6 1920.85 0.9 32.79 39.24 648 541.6 212.50.9 0.9 34.64 42.61 666.3 541.6 230.8

9-156

9-169 EES The effect of the number of compression and expansion stages on the thermal efficiency of anideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated.Analysis Using EES, the problem is solved as follows:

"Input data for air"C_P = 1.005 [kJ/kg-K] k = 1.4 "Nstages is the number of compression and expansion stages"Nstages = 1 T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness"Eta_c =1.0 "Compressor isentorpic efficiency"Eta_t =1.0 "Turbine isentropic efficiency"R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis"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"compressor adiabatic efficiency, W_comp > W_compisen"

"Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1)

"Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:"w_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""The heat added in the external heat exchanger + the reheat between turbines is"q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:"T_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"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen = C_P*(T_6 - T_7s) "Actual Turbine analysis:"w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb

"Cycle analysis"

9-157

w_net=w_turb_total-w_comp_total "[kJ/kg]"Bwr=w_comp/w_turb "Back work ratio"P_4=P_2P_5=P_4P_6=P_5T_4 = T_2 "The regenerator effectiveness gives T_5 as:"Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 "Energy balance on regenerator gives T_10 as:"T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator"Eta_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 temperatures, T_6 and T_1 for this problem."Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

th,Ericksson[%]

th,Regenerative[%]

Nstages

75 49.15 175 64.35 275 68.32 375 70.14 475 72.33 775 73.79 1575 74.05 1975 74.18 22

0 2 4 6 8 10 12 14 16 18 20 22 2440

50

60

70

80

Nstages

th [%

]

Ericsson

Ideal Regenerative Brayton

9-158

9-170 EES The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the workingfluid is to be investigated.Analysis Using EES, the problem is solved as follows:

"Input data for Helium"C_P = 5.1926 [kJ/kg-K] k = 1.667 "Nstages is the number of compression and expansion stages"{Nstages = 1}T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness"Eta_c =1.0 "Compressor isentorpic efficiency"Eta_t =1.0 "Turbine isentropic efficiency"R_p = Pratio^(1/Nstages) "Isentropic Compressor anaysis"

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"compressor adiabatic efficiency, W_comp > W_compisen""Conservation of energy for the compressor for the isentropic case: e_in - e_out = DELTAe=0 for steady-flow" w_compisen = C_P*(T_2s-T_1) "Actual compressor analysis:" w_comp = C_P*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:"w_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"

"The heat added in the external heat exchanger + the reheat between turbines is"q_in_total = C_P*(T_6 - T_5) +(Nstages - 1)*C_P*(T_8 - T_7) "Reheat is assumed to occur until:"T_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"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 e_in -e_out = DELTAe_cv = 0 for steady-flow"w_turbisen = C_P*(T_6 - T_7s)

"Actual Turbine analysis:"w_turb = C_P*(T_6 - T_7) w_turb_total = Nstages*w_turb

9-159

"Cycle analysis"w_net=w_turb_total-w_comp_totalBwr=w_comp/w_turb "Back work ratio"

P_4=P_2P_5=P_4P_6=P_5T_4 = T_2 "The regenerator effectiveness gives T_5 as:"Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7"Energy balance on regenerator gives T_10 as:"T_4 + T_9=T_5 + T_10

"Cycle thermal efficiency with regenerator"Eta_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 temperatures, T_6 and T_1 for this problem."Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %) "[%]"

th,Ericksson[%]

th,Regenerative[%]

Nstages

75 32.43 175 58.9 275 65.18 375 67.95 475 71.18 775 73.29 1575 73.66 1975 73.84 22

0 2 4 6 8 10 12 14 16 18 20 22 2430

40

50

60

70

80

Nstages

th [%

]

EricssonIdeal Regenerative Brayton

9-160

Fundamentals of Engineering (FE) Exam Problems

9-171 An Otto cycle with air as the working fluid has a compression ratio of 8.2. Under cold air standardconditions, the thermal efficiency of this cycle is(a) 24% (b) 43% (c) 52% (d) 57% (e) 75%

Answer (d) 57%

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 numericalvalues).

r=8.2k=1.4Eta_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"

9-172 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

9-173 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"

9-161

9-174 Air in an ideal Diesel cycle is compressed from 3 L to 0.15 L, and then it expands during theconstant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermalefficiency of this cycle is(a) 35% (b) 44% (c) 65% (d) 70% (e) 82%

Answer (c) 65%


V1=3 "L"V2= 0.15 "L"V3= 0.30 "L"r=V1/V2rc=V3/V2k=1.4Eta_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"

9-175 Helium gas in an ideal Otto cycle is compressed from 20 C and 2.5 L to 0.25 L, and its temperatureincreases by an additional 700 C during the heat addition process. The temperature of helium before theexpansion process is(a) 1790 C (b) 2060 C (c) 1240 C (d) 620 C (e) 820 C

Answer (a) 1790 C


k=1.667V1=2.5V2=0.25r=V1/V2T1=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"

9-162

9-176 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=1.4V1=2.2V2=0.26m=rho1*V1/1000 "kg"w_net=440 "kJ/kg"Wtotal=m*w_netMEP=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"

9-177 In an ideal Brayton cycle, air is compressed from 95 kPa and 25 C to 800 kPa. Under cold airstandard conditions, the thermal efficiency of this cycle is(a) 46% (b) 54% (c) 57% (d) 39% (e) 61%

Answer (a) 46%


P1=95 "kPa"P2=800 "kPa"T1=25+273 "K"rp=P2/P1k=1.4Eta_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"

9-163

9-178 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa andtemperature 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/P1k=1.667Cp=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"

9-179 An ideal Brayton cycle has a net work output of 150 kJ/kg and a backwork ratio of 0.4. If both theturbine 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=0.4wturb-wcomp=150 "kJ/kg"Eff=0.85w_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"

9-164

9-180 In an ideal Brayton cycle, air is compressed from 100 kPa and 25 C to 1 MPa, and then heated to1200 C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is(a) 490 C (b) 515 C (c) 622 C (d) 763 C (e) 895 C

Answer (a) 490 C


P1=100 "kPa"P2=1000 "kPa"T1=25+273 "K"T3=1200+273 "K"rp=P2/P1k=1.4T4=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"

9-181 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


k=1.667Cp=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/P1T2=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"

9-165

9-182 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=1.4Cp=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/P1T2check=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"

9-183 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle withregeneration between the temperature limits of 20 C and 900 C. If the specific heat ratio of the workingfluid 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=1.3rp=6T1=20+273 "K"T3=900+273 "K"Eta_regen=1-(T1/T3)*rp^((k-1)/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)"

9-166

9-184 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=1.4rp=10T1=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"

9-185 Air enters a turbojet engine at 260 m/s at a rate of 30 kg/s, and exits at 800 m/s relative to theaircraft. The thrust developed by the engine is(a) 8 kN (b) 16 kN (c) 24 kN (d) 20 kN (e) 32 kN

Answer (b) 16 kN


Vel1=260 "m/s"Vel2=800 "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"

9-186 ··· 9-191 Design and Essay Problems.

ThermoSolutions CHAPTER09

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