Thermo 7e SM Chap09-1

9-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition

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

Chapter 9 GAS POWER CYCLES

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

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

9-2

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

9-1C It is less than the thermal efficiency of a Carnot cycle.

9-2C It represents the net work on both diagrams.

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

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

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

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

9-7C 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.

9-8C Yes.

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

9-10C 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.



9-3

9-11C 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. 9-12E The maximum possible thermal efficiency of a gas power cycle with specified reservoirs is to be determined.

Analysis The maximum efficiency this cycle can have is

0.643=++== 1 LT

R )460(940R )460(401Carnotth,

HT



9-4

9-13 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams and the back work ratio are to be determined.

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

Properties The properties of air are given as R = 0.287 kPam3/kgK, cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, and k = 1.4.

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

(b) Process 1-2: Isentropic compression

)( 12,21 TTmcw in = v

s

T

3 2

1

v

P

32

1

11

212 =

= rTTTv

1

1 kkv

rocess 2-3: Constant pressure heat addition

TTmRP == VVv

he back w rk ratio is

P

2

,32 Pdw out = )()( 232323T o

)()(

23

12

,32

,21

TTmRTTmc

ww

rout

inbw

== v

Noting that

1

thus,and and === kRc

cc

kccR pp vv

v

From ideal gas relation,

rTT ===

2

1

2

3

2

3

vv

vv

Substituting these into back work relation,

( )( )

0.256=

=

=

=

=

1661

14.11

11

11

1

11

11

)1/()/1(1

1

4.0

11

23

21

2

2

rr

krr

k

TTTT

TT

RkRr

kk

bw



9-5

9-14 The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the back work ratio and the thermal efficiency are to be determined.


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

k = 1.4.


(b) The temperature at state 2 is

K 2100kPa 100kPa 700K) 300(

1

212 === P

PTT

K 210023 == TT During process 1-3, we have

3

,13

==

=== TTRPPdw in VVv During process 2-3, we have

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

)()( 313111

s

T

32

1

v

P

3

2

1

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

kJ/kg 8.1172n7K)(2100)KlkJ/kg 287.0(22

==

7ln7

lnln2

2

2

33

,3 ===== RTRTRTdRTPdw out VVVVvVv The back work ratio is then

3

2

0.440===

kJ/kg 8.1172kJ/kg 6.516

,32

,13

out

inbw w

wr

Heat input is determined from an energy balance on the cycle during process 1-3,

) ,321,3231,31

31

=+

++=

=

out

outin

ut

wTwuq

u

The net work output is

,32,31 oin wq

718.0(( 3

== v Tc

kJ/kg 2465kJ/kg 1172.8300)K)(2100kJ/kg

kJ/kg 2.6566.5168.1172,13,32 === inoutnet www (c) The thermal efficiency is then

26.6%==== 266.0kJ 2465kJ 656.2

in

netth q

w

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

9-6

9-15 The three processes of an ideal gas power cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the maximum temperature, expansion and compression works, and thermal efficiency are to be determined.

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

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

k = 1.5.


(b) The maximum temperature is determined from


K 734.8=+==

== 112

112max 27(rTTTT V

11.51 K)(6) 273kk

V

) An energy balance during process 2-3 gives

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

Then, the work of compression is

s

T

32

1

v

P

3

2

1

(c

ouin wq ,32,32 = t

kJ/kg 395.0==

=== K)ln6 K)(734.8kJ/kg 3.0(

lnln 22

32

22,32,32 rRTRTd

RTPdoutin VV

vV

v

(d) The work during isentropic compression is determined from an energy 1-2:

== 300)

)( 1221,2 TTcuw vin

) Net work output is

== 33wq

balance during process

1

kJ/kg 260.9== K)(734.8kJ/kg 6.0(

(e

kJ/kg 1.1349.2600.395,21,32 === inoutnet www The thermal efficiency is then

33.9%==== 339.0kJ 395.0kJ 134.1

in

netth q

w


9-7

9-16 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined.

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

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

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

v

P

1

2

4

3qin

qout

s

T

1

24

3qin

qout

( )

( )( ) kJ/kg 71.73979.32601.9

kPa 1835.6kPa 100

kPa 6.1835kPa 600K 490.3K 1500

9.601kJ/kg 41.1205

K 1500

K 3.490kJ/kg 352.29

841.71.3068kPa 100kPa 600

3068.1kJ/kg .17295

K 2951 =T

43

4

22

323

33

2

2

1

2

1

34

3

12

1

====

=

===

=====

==

hPPP

P

TT

Pu

T

Tu

PPP

P

Ph

rr

r

rr

r

From energy balances,

(c) Then the thermal efficiency becomes

32233 === PTPPP vvT

kJ/kg 408.6=========

5.4441.853

kJ/kg .544417.29571.739

kJ/kg 1.85329.35241.1205

outinoutnet,

14out

23in

qqw

hhq

uuq

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

in

outnet,th q

w



9-8

9-17 Problem 9-16 is reconsidered. The effect of the maximum temperature of the cycle on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted.

Analysis Using EES, the problem is solved as follows:

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



9-9

T3 [K]

th qin,total [kJ/kg]

Wnet [kJ/kg]

1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500

47.91 48.31 48.68 49.03 49.35 49.66 49.95 50.22 50.48 50.72 50.95

852.9 945.7 1040 1134 1229 1325 1422 1519 1617 1715 1813

408.6 456.9 506.1 556

606.7 658.1 710.5 763

816.1 869.8 924

1500 1700 1900 2100 2300 250047.5

48

48.5

49

49.5

50

50.5

51

T[3] [K]

th

1500 1700 1900 2100 2300 2500800

1020

1240

1460

1680

1900

T[3


] [K]

q in,

tota

l [k

J/k

g]


9-10

1500 1700 1900 2100 2300 2500400

500

600

700

800

900

1000

T[3] [K]

wne

t [k

J/kg

]

5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.50

500

1000

1500

2000

s [kJ/kg-K]

T [K

]

100 kPa

600 kPa

Air

1

2

3

4

10-2 10-1 100 101 102101

102

103

104

v [m3/kg]

P [k

Pa]

295 K 1500 K

Air

1

2

3

4



9-11

9-18 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the heat rejected and the thermal efficiency are to be determined.


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

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


( )( ) K 579.2

kPa 100kPa 1000

K 3000.4/1.4/1

1

212 =

=

=

kk

PP

TT

( ) ( )( )( )( )2.579KkJ/kg 1.005kg 0.004kJ 2.76 3 = T

P

K 12663

2323in

===

T

TTmchhmQ p

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

( )( )

v

P

32

1

qin

qout

s

T

32

1 qout

qin

kJ/kg 7.273=KkJ/kg 0.287

kPa 1000K 1266

kPa 100K 300

2kPa 1001000

22area

3

3

1

11331

1331

+=

+=+==

PRT

PRTPPPP

w vv

Energy balance for process 3-1 gives

==

K1266-300KkJ/kg 0.718273.7kg 0.004)(

)(

31out31,31out31,out31,

31out31,out31,systemoutin

TTcwmTTmcmwQ

uumWQ

vv

) The thermal efficiency is then

( )[ ]( ) ( )( )[ ] kJ 1.679=+=

+= EEE

=

(c

39.2%===kJ 2.76kJ 1.679

11in

outth Q

Q


9-12

9-19E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined.


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

v

P

3

4

2

1

q12

q23

qout

s

T3

42

1 qout

q23

q12

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

Btu/lbm 129.06 Btu/lbm, 92.04R 540 111 === huT

( )

( ) Btu/lbm 593.22317.01242psia 57.6psia 14.7

1242Btu/lbm 48.849

R 3200

psia 57.6psia 14.7R 540R 2116

Btu/lbm 1.537,R2116Btu/lbm 04.39230004.92

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

From energy balance,

Btu/lbm 464.1606.12922.593 38.31300in23,in12,in

23in23,

+=+= qqq 2Btu/lbm 312.381.53748.849

14out ====

===

hhq

hhqBtu/lbm612.38

(c) Then the thermal efficiency becomes

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

11in

outth q

q



9-13

9-20E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined.


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


)( )( )

(

( )( ) ( )( ) Btu/lbm 217.4R22943200RBtu/lbm 0.24

psia 62.46psia 14.7R 540R 2294

R 2294R540Btu/lbm.R .171

2323in,23

11

22

1

11

2

22

2

2

1212in,12

====

=====

==

TTchhq

PTT

PT

PT

PT

TTTcuuq

P

vv

v

Process 3-4 is isentropic:

0Btu/lbm 300 =

( )( )

( ) ( )( ) Btu/lbm 378.55402117Btu/lbm.R 0.240 4.217300

R 2117psia 62.46

psia 14.7R 3200

1414out

in,23in,12in

0.4/1.4

3

434

=====+=+=

=

=

=

TTchhq

qqq

PP

T

p

Btu/lbm517.4

(c) The thermal efficiency is then

/1 kkT

26.8%===Btu/lbm 517.4Btu/lbm 378.5

11in

outth q

q

v

P

3

4

2

1

q12

q23

qout

s

T

3

42

1 qout

q23

q12


9-14

9-21 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is to be determined.

Assumptions Air is an ideal gas with constant specific heats.

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

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

or

( )

( )( ) kPa 1888

K 300K 1100

kPa 201.4/0.41/

3

232

/1

3

2

3

2

=

=

=

=

kk

kk

TT

PP

PP

TT

s

T

3

2qin

qout4

11100

300 The heat input is determined from

Then,

( )( ) ( )( )( )

( )( ) kJ 63.8=======

===

===

kJ 87.730.7273

.7%727273.0K 1100K 30011

kJ 73.87KkJ/kg 0.1329K 1100kg 0.6

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

inthoutnet,

th

12in

1

20

1

212

QW

TT

ssmTQ

PP

RTT

css

H

L

H

p



9-15

9-22 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimum pressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the second-law efficiency of an actual cycle operating between the same temperature limits are to be determined.

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

( )( ) ( )( ) K 350K750KkJ/kg 0.25kJ/kg 10012net === LLLH TTTTssw The pressure at state 4 is determined from

( )

( )

kPa 1.110


or

K 350 K 750

kPa 800 41.4/0.4

4

1/

4

141

/1

=

=

=

PP

TT

PP

PT

kk

kk

he minim m pressure in the cycle is determined from s

T

3

2qin

4

1750 K

wnet=100 kJ/kg

qout

4

1

4

1

=PT

T u

( ) kPa 46.1====

04ln

Tcss

33

3

43412

kPa 110.1lnKkJ/kg 0.287KkJ/kg 25.0

ln

PP

PP

RTp

) The heat rejection from the cycle is

3

(b

kgkJ/ 87.5=== kJ/kg.K) K)(0.25 350(12out sTq L (c) The thermal efficiency is determined from

0.533===K 750K 350

11thH

L

TT

(d) The power output for the Carnot cycle is

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

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

0.578===kW 9000kW 5200

Carnot

actualII W

W&&


9-16

9-23 An ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature of the low-temperature energy reservoir, the cycle's thermal efficiency, and the amount of heat that must be supplied per cycle are to be determined.


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

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


K 481.1=+=

=1

21 12K) 2731027(TT

v

14.112 1

kv

ince the Carnot engine is completely reversible, its efficiency is

s

T

3

2qin

qout4

1 1300 K S

0.630=+== 11Carnotth, HL

T

K 273)(1027K 481.1T

he work tput per cycle is

T ou

kJ/cycle 20min 1

s 60cycle/min 1500

kJ/s 500netnet =

==

nW&

W &

According to the definition of the cycle efficiency,

kJ/cycle 31.75====0.63kJ/cycle 20

Carnotth,

netin

in

netCarnotth,

WQ

QW


9-17

9-24 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to be determined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained.


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


(b) Noting that


1.4290.1 ===

7.0

K kJ/kg 0.13.07.0 =+=+=

v

v

c

Rcc

Process 1-2: Isentropic compression

s

T

32

1

v

P

32

1

p

ck p

K 4.584)5)(K 293( 429.0111

2

112 ===

=

k

k

rTTTv

v

= ( 2in2,1 Tcw v kJ/kg 204.0== K )2934.584)(KkJ/kg 7.0()1T 2q 0=1

From ideal gas relation,

2922)5)(4.584(32

1

2

3

2

3 ===== TrTT

v

v

v

v

Process 2-3: Constant pressure heat addition

K )4.5842922)(KkJ/kg 3.0(

)233

2

TT

K )4.5842922)(KkJ/kg 1()( 233232,32in3,

TTchuwq

p

out

Process 3-1: Constant volume heat rejection

=== ()( 232out3,2 RPPdw vvv

kJ/kg 701.3==

kJ/kg 2338====+= 2

== (31out1,3 Tcuq v kJ/kg 1840.3== K 293)-K)(2922kJ/kg 7.0()13 T w

(c) Net work is

0=13

KkJ/kg 3.4970.2043.701in2,1out3,2net === www The thermal efficiency is then

21.3%==== 213.0kJ 2338kJ 497.3

in

netth q

w


9-18

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

=

=

=

=

=

=

==

1

1

11

1

11

11

111

11

11

11

1

23

1223

in

in2,1out3,2

in

netth

11)1(

111

11)1(

1

1)1(

1

)1(

1

)()(

)()()(

k

kp

kp

kp

kk

v

p

kkp

kv

p

p

v

rrkk

rrkcR

rTT

rkcR

rrTc

rTT

rTc

cR

rTrrTcTrTc

cR

TTcTTcTTR

qww

qw

since

111kc

cc

cccR

p

v

p

vp

p===



9-19

Otto Cycle

9-25C 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.

9-26C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency.

9-27C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heat addition, (3) isentropic expansion, and (4) v = constant heat rejection.

9-28C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.

9-29C It increases with both of them.

9-30C Because high compression ratios cause engine knock.

9-31C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k = 1.667.

9-32C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline engines.



9-20

9-33 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.


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

Analysis The definition of cycle thermal efficiency reduces to


61.0%==== 6096.010.51111 11.41th kr

The rate of heat addition is then

kW 148===0.6096

kW 90

th

netin

WQ

&&

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

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

nalysis The definition of cycle thermal efficiency reduces to

v

P

4

1

3

2

qinqout

9-34 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.

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

Propert p2a).

A

v

P

4

1

3

2

qinqout

57.5%==== 8.511 1.1th kr 5752.011

14

he rate o eat addition is then T f h

kW 157===0.5752

kW 90

th

netin

WQ

&&


9-21

9-35 The two isentropic processes in an Otto cycle are replaced with polytropic processes. The heat added to and rejected from this cycle, and the cycles thermal efficiency are to be determined.


Properties The properties of air at room temperature are R = 0.287 kPam3/kgK, cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, and k = 1.4 (Table A-2a).

Analysis The temperature at the end of the compression is

v

P

4

1

3

2

K 4.537K)(8) 288( 13.1111

2

112 ===

=

n

n

rTTTv

v

And the temperature at the end of the expansion is

K 4.7898

K) 1473(34

34 ===

=r

TTTv

1113.111

3 nnv

for the polytropic compression gives The integral of the work expression


kJ/kg 6.238)18(13.1

11 2

21 n

wv

K) K)(288kJ/kg 287.0( 13.11

11 == =

nRT v

imilarly, the work produced during the expansion is S

kJ/kg 0.654181

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

11

13.11 n4

3343 =

=

=

nRT

wv

v

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

A

1 kJ/kg 53.59K)2884.537)(KkJ/kg 718.0(kJ/kg 6.238)( 12212 === TTcwq v 1473)(KkJ/kg 718.0()( 2332 == TTcq v kJ/kg 8.671K)4.537 = kJ/kg 2.163K)4.7891473)(KkJ/kg 718.0(kJ/kg 0.654)( 434343 === TTcwq v

kJ/kg 0.360K)2884.789)(KkJ/kg 718.0()( 1414 === TTcq v The head added and rejected from the cycle are

The thermal efficiency of this cycle is then

kJ/kg 419.5kJ/kg 835.0

=+=+==+=+=

0.36053.592.1638.671

1421out

4332in

qqqqqq

0.498===0.8355.41911

in

outth q

q


9-22

9-36 An ideal Otto cycle is considered. The heat added to and rejected from this cycle, and the cycles thermal efficiency are to be determined.


Properties The properties of air at room temperature are R = 0.287 kPam3/kgK, cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, and k = 1.4 (Table A-2a).

Analysis The temperature at the end of the compression is

v

P

4

1

3

2

K 7.661K)(8) 288( 14.1111

2

112 ===

=

k

k

rTTTv

v

and the temperature at the end of the expansion is

K 2.648

K) 1473(34

34 ===

=r

TTTv

11114.111

3 kkv

pplication of the first law to the heat addition process gives A

== ()( 23in TTcq v


kJ/kg 582.5= K)7.6611473)(KkJ/kg 718.0 imilarly, the heat rejected is S

== KkJ/kg 718.0()( TTcq kJ/kg 253.6= K)2882.641)(14out v he thermal efficiency of this cycle is then T

0.565===5.5826.25311

in

outth q

q


9-23

9-37E A six-cylinder, four-stroke, spark-ignition engine operating on the ideal Otto cycle is considered. The power produced by the engine is to be determined.


Properties The properties of air at room temperature are R = 0.3704 psiaft3/lbm.R (Table A-1E), cp = 0.240 Btu/lbmR, cv = 0.171 Btu/lbmR, and k = 1.4 (Table A-2Ea).

Analysis From the data specified in the problem statement,

v

P

4

1

3

2

143.7


140 12 vv .11 == vvr

Since the compression and expansion processes are isentropic,

=

( ) R 1153143.7R) 525( 14.1111

2

112 ===

=

k

k

rTTTvv

R 2.9387.143

11 14.111

3 kkv R) 2060(34

34 ===

=r

TTTv

mpression and expansion processes gives

=

Application of the first law to the co

171.0(R)2.9382060)(RBtu/lbm 171.0()()( 1243net

== TTcTTcw vv

Btu/lbm 44.84R)5251153)(RBtu/lbm

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

/lbmft 89.13psia 14

R) 525)(R/lbmftpsia 3704.0( 33

1

11 === P

RTv

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

lbm 009380.0/lbmft 89.13

ft)/4 12/9.3(ft) 12/5.3()6(4/cylcyl === vvV SBNNm

3

2

1

2

1=

he net work produced per cycle is

T

Btu/cycle 7920.0Btu/lbm) lbm)(84.44 (0.009380netnet === mwW he power produced is determined from

T

hp 23.3=

==Btu/s 7068.0

hp 1rev/cycle 2

rev/s) (2500/60Btu/cycle) (0.7920

rev

netnet N

nWW

&&

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


9-24

9-38E An Otto cycle with non-isentropic compression and expansion processes is considered. The thermal efficiency, the heat addition, and the mean effective pressure are to be determined.



Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then

v

P

4

1

3

2 qoutqin


( ) R 11958R) 520(12

12 ===

=s rTTT v 14.11

11 kkv

With the isentropic compression efficiency, the actual temperature at e end of the compression is th

R 13140.85

R 520)(1195R) 520(121212

12 = TTTs =+=+=

TTT

TT s

imilarly for the expansion,

S

R 120181R) 4602300(1

14.11

3

1

4 3

34 =

+=

=

=

kks r

TTTv

v

R 1279R )1201(2760)95.0(R) 2760()( 433443 sTT43 ===

sTTTT

TT

The specific heat addition is that of process 2-3,

=

Btu/lbm 247.3=== R)13142760)(RBtu/lbm 171.0()( 23in TTcq v The net work production is the difference between the work produced by the expansion and that used by the compression,

=

The thermal efficiency of this cycle is then

)()( 1243net = TTcTTcw vv R)5201314)(RBtu/lbm 171.0(R)12792760)(RBtu/lbm 171.0(

Btu/lbm 5.117=

0.475===Btu/lbm 247.3inqBtu/lbm 117.5net

thw

t the beg ning of compression, the maximum specific volume of this cycle is A in

/lbmft 82.14psia 13

R) 520)(R/lbmftpsia 3704.0( 311

== RTv 31

=P

hile the minimum specific volume of the cycle occurs at the end of the compression

w

/lbmft 852.18

/lbmft 82.14 3312 === r

vv

The engines mean effective pressure is then

psia 49.0=

== Btu 1ftpsia 404.5

/lbmft )852.182.14(Btu/lbm 5.117MEP

3

321

net

vv

w


9-25

9-39 An ideal Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined.


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

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

v

P

4

1

3

2Qin Qout

( )( )( ) ( ) kPa 2338kPa 100

K


308K 757.9

9.5

K 757.99.5K

11

2

2

12

1

11

2

22

0.41

=

===

=

PTT

PT

PT

P

k

v

vvv

v

308

2

112 =

= TTv

Process 3-4: isentropic expansion.

( )( ) K 1969==

=

0.4

1

3

443 9.5K 800

k

TTv

v

Process 2-3: v = constant heat addition.

( ) kPa 6072=

=== kPa 2338

K 757.9K 1969

22

323 TTT

3223 PT

PPP vv

(b)

3

( )( )( )( ) kg10788.6K 308K/kgmkPa 0.287 m 0.0006kPa 100 433

1

11 === RTP

mV

( ) ( ) ( )( )( ) kJ 0.590==== K757.91969KkJ/kg 0.718kg106.788 42323in TTmcuumQ v (c) Process 4-1: v = constant heat rejection.

( ) ( )( )( ) kJ 0.240K308800KkJ/kg 0.718kg106.788)( 41414out ==== TTmcuumQ v kJ 0.350240.0590.0outinnet === QQW

59.4%===kJ 0.590kJ 0.350

in

outnet,th Q

W

( )( ) kPa 652=

===

==

kJmkPa

1/9.51m 0.0006kJ 0.350

)/11(MEP

3

31

outnet,

21

outnet,

max2min

rWW

r

VVV

VVV (d)


9-26

9-40 An Otto cycle with air as the working fluid has a compression ratio of 9.5. The highest pressure and temperature in the cycle, the amount of heat transferred, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.


Properties The properties of air at room temperature are cp = 1.005 kJ/kgK, cv = 0.718 kJ/kgK, R = 0.287 kJ/kgK, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression.

( )( )( ) ( ) kPa 2338kPa 100

K 308K 757.9

9.5

K 757.99.5K 308

11

2

2

12

1

11

2

22

0.41

2

112

=

===

==

=

PTT

PT

PT

P

TTk

v

vvv

v

v

v

P

4

1

3

2

Qin

Qout

Polytropic

800 K

308 K

Process 3-4: polytropic expansion.

( )( )( )( ) kg10788.6K 308K/kgmkPa 0.287 m 0.0006kPa 100 433

1

11 === RTP

mV

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

kJ 0.53381.351

K1759800KkJ/kg 0.287

106.7881

3434

==n

TTmRW

9.5K 800

4

35.01

3

443

=

==

=

TT

n

K 1759v

v

=+=+=+=

QWTTmcWuumQ v

That is, 0.066 kJ of heat is added to the air during the expansion process (This is not realistic, and probably is due to nstant specific heats at room temperature).

v = constant heat addition.

T gyhen ener balance for process 3-4 gives

outin = EEE system

( )34out34,in34, = uumWQ( ) ( )( )( )( ) kJ 0.0664kJ 0.5338K1759800KkJ/kg 0.718kg106.788 4in34, out34,34out34,34in34,

assuming co(b) Process 2-3:

( ) kPa 5426=

TP vv === kPa 2338

K 757.9K 1759

22

33

2

22

3

3 PT

PT

P

====

QTTmcuumQ v

efore,

3

T

( ) ( )( )( )( ) kJ 0.4879K757.91759KkJ/kg 0.718kg106.788 4in23, 2323in23, kJ 0.5543=+=+= 0664.04879.0in34,in23,in QQQ Ther

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

( ) ( ) ( )( )( ) kJ 0.2398K308800KkJ/kg 0.718kg 106.788 41414out ==== TTmcuumQ v kJ 0.31452398.05543.0outinoutnet, === QQW

56.7%===kJ 0.5543kJ 0.3145

in

outnet,th Q

W

( )( ) kPa 586=

===

==

kJmkPa

1/9.51m 0.0006kJ 0.3145

)/11(MEP

3

31

outnet,

21

outnet,

max2min

rWW

r

VVV

VVV (d)


9-27

9-41E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined.



v

P

4

1

3

2

qinqout

2400 R

540 R



32.144Btu/lbm92.04

R5401

11 =

==r

uT

v

( ) Btu/lbm 11.28204.1832.14481 222

===r rrr

vvv

v11

22 == uv

rocess 2- v = constant heat addition.

=28.21170.452

419.2Btu/lbm 0

R2400

23

33

uuq

T

in

rv

P 3:

= 452.73u

Btu/lbm 241.42====

(b) Process 3-4: isentropic expansion.

( )( ) Btu/lbm 205.5435.19419.28 43

4334

===== ur rrr vvvv

v

Process 4-1: v = constant heat rejection.

Btu/lbm 50.11304.9254.20514out === uuq

53.0%===Btu/lbm 241.42Btu/lbm 113.50

11in

outth q

q

(c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is

77.5%===R 2400

R 54011Cth,

H

L

TT


9-28

9-42E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount of heat transferred to the argon during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined.

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

Properties The properties of argon are cp = 0.1253 Btu/lbm.R, cv = 0.0756 Btu/lbm.R, and k = 1.667 (Table A-2E).


( )( ) R 21618R 540 0.66712

112 ==

=

kTT

v

v

v

P

4

1

3

2

qinqout

Process 2-3: v = constant heat addition.

( )


( )(= 2400Btu/lbm.R 0.0756 )Btu/lbm 18.07=

==

R 21612323 TTcuuq v

) Process 3-4: isentropic expansion.

in

(b

( ) R 6008

0.667

4=v


1R 24001

334 =

=

kTT

v

( ) ( )( ) Btu/lbm 4.536R540600Btu/lbm.R 0.07561414out ==== TTcuuq v 74.9%=== Btu/lbm 4.53611 outq

Btu/lbm 18.07inth q

(c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is

77.5%===R 2400R 540

11Cth,H

L

TT


9-29

9-43 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeled as polytropic. The temperature at the end of expansion process, the net work output, the thermal efficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuel consumption are to be determined.


Properties The properties of air at 850 K are cp = 1.110 kJ/kgK, cv = 0.823 kJ/kgK, R = 0.287 kJ/kgK, and k = 1.349 (Table A-2b).

Analysis (a) Process 1-2: polytropic compression

1

Qin

2

3

4

P

V

Qout

( )( )

( )( ) kPa 225811kPa 100

K 5.63611K 310

1.3

2

112

1-1.31

2

112

==

=

==

=

n

n

PP

TT

v

v

v

v

kJ/kg 3.3121.31112 n

310)KK)(636.5kJ/kg (0.287)( 12 === TTRw

Process 2-3: constant volume heat addition

( ) K 2255kPa 2258kPa 8000K 636.5

2

323 =

=

=

PP

TT

( 2323in == TTcuuq v )( )( ) kJ/kg 1332K636.52255KkJ/kg 0.823 == rocess 3-4: polytropic expansion.

P

( ) K 1098=

=

=

1-1.31

4

334 11

1K 2255n

TTv

v

( ) kPa 2.3541 1.3 nv 11

kPa 80001

234 ==

= PPv

kJ/kg 11061.31

2255)KK)(1098kJ/kg (0.287)( TTR1

34 === nw

Process 4-1: constant volume heat rejection.

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

34

kJ/kg 794=== 3.31211061234outnet, www

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

in

outnet,th q

w

(c) The mean effective pressure is determined as follows

( )( )

( )( ) kPa 982=

===

==

====

kJmkPa

1/111/kgm 0.8897kJ/kg 794

)/11(MEP

/kgm 0.8897kPa 100

K 310K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv



9-30

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

33

m 00016.0m 0016.0

11 =+=+= cc

c

c

dcr VV

V

V

VV

31 m 00176.00016.000016.0 =+=+= dc VVV

The total mass contained in the cylinder is

( )( ) kg 0.001978K 310K/kgmkPa 0.287 )m 76kPa)/0.001 100( 33

1

11 === RTP

mtV

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

rev/min 3820=== min 1cycle)kJ/kg kg)(794 001978.0(rev/cycle) 2(2 net

net

wmn

t& s 60kJ/s 50W&

n four-stroke engines.

) The mass of fuel burned during one cycle is

Note that there are two revolutions in one cycle i

(e

kg 0001164.0kg) 001978.0( =16AF ===

ff

a

mm

f

f

fft mm

mmmm

inally, the specific fuel consumption is

F

g/kWh 267=

==

kWh 1kJ 3600

kg 1g 1000

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

netwmm

t

f



9-31

9-44 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled.


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

111

2

112

=

= k

k

rTTTvv

v

P

4

1

3

2 qoutqinsince T1 is fixed. The temperature rise during the combustion remains constant since the

amount of heat addition is fixed. Then, the maximum cycle temperature is given by

11in2in3 //+=+= krTcqTcqT vv

The smallest gas specific volume during the cycle is

r1

3v

v =

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

( )11in13

33 /

+== krTcqRrRTP vvv

9-45 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent.


nalysis During a polytropic process,

=T

=Pv

is lost during the expansion of the gas,

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

when

A

constant/)1(

= nn

nPv P

4

1

3

2 qoutqin

P constant

and for an isentropic process,

k constant

constant/)1( = kkTPIf heat

v

4 sTT 4

w

kn



9-32

Diesel Cycle

9-46C A diesel engine differs from the gasoline engine in the way combustion is initiated. In diesel engines combustion is initiated by compressing the air above the self-ignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine.

9-47C 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.

9-48C The gasoline engine.

9-49C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem.

9-50C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process. As the cutoff ratio decreases, the efficiency of the diesel cycle increases.

9-51 An ideal diesel cycle has a compression ratio of 20 and a cutoff ratio of 1.3. The maximum temperature of the air and the rate of heat addition are to be determined.


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

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


( ) K 6.95420K) 288( 14.1111

2

112 ===

=

k

k

rTTTv

v

K 1241==== K)(1.3) 6.954(2

223 crTTT v

3v

ombining the first law as applied to the various processes with the process equations gives v

P

4

1

2 3qin

qout

C

6812.0)13.1(4.120

1)1(

111.41th == ck rkr

13.11114.1

=k

cr

ccording the definition of the thermal efficiency,

A to

kW 367===0.6812

kW 250

th

netin

WQ

&&


9-33

9-52E An ideal diesel cycle has a a cutoff ratio of 1.4. The power produced is to be determined.



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

v

P

4

1

2 3qin

qout

/lbmft 12.13psia 4.14

R) 510)(R/lbmftpsia 3704.0( 33

1

11 === P

RTv

The total air mass taken by all 8 cylinders when they are charged is

lbm 01774.0/lbmft 12.13

)8(3

1cyl

1cyl === vv NNm

ft)/4 12/4(ft) 12/4(4/ 22 = V SB

he rate at which air is processed by the engine is determined from T


lbm/h 0.958lbm/s 2661.0rev/cycle 2rev

====N

m& rev/s) (1800/60lbm/cycle) (0.01774nm &

nce there re two revolutions per cycle in a four-stroke engine. The compression ratio is

si a

22.22045.01 ==r

=== krTT work integral to the constant pressure heat addition gives

At the end of the compression, the air temperature is

( ) R 176322.22R) 510( 14.111 2Application of the first law and

Btu/lbm 239.3R)17632760)(RBtu/lbm 240.0()( 23in === TTcq p while the thermal efficiency is

6892.0)14.1(4.1

14.122.22

111 kr 1)1(

14.1

11.41=

== cc

k rkr

he power produced by this engine is then

th

T

hp 62.1=

===

Btu/h 5.2544hp 1Btu/lbm) 892)(239.3lbm/h)(0.6 (958.0

inthnetnet qmwmW &&&


9-34

9-53 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined.




v

P

4

1

2

3

qout

x

qin/kgm 051.1

kPa 80K) 293)(K/kgmkPa 287.0( 3

3

1

11 === P

RTv

and the specific volume at the end of the compression is


/kgm 07508.0142

=== /kgm 051.1 33

1vvr

The pressure at the end of the compression is

( ) kPa 321914kPa) 80( 4.112

12 v1 ==== k

k

rPPPv

nd the ma imum pressure is

he temperature at the end of the compression is

a x

4829==== kPa) 3219)(5.1(23 PrPP px kPa T

( ) K 0.84214K) 293( 14.1111

2

11 2 ===

=

k

k

rTTTv

v

nd K 1263kPa 3219kPa 4829K) 0.842(

2

32 =

=

P

=P

TTx

he remaining state temperatures are then

a

From the definition of cutoff ratio

/kgm 09010.0)/kgm 07508.0)(2.1( 3323 ==== vvv cxc rr T

K 1516==

=0.07508

K) 1263(33x

xTT v 0.09010v

K 5.5671.051

0.09010K) 1516(14.11

4

334 = TT v =

=

k

v

ng the first law and work expression to the heat addition processes gives

The heat rejected is

Applyi

kJ/kg 556.5=+=

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

)()( 32in xpx TTcTTcq v

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

0.646===kJ/kg 556.5kJ/kg 197.1

11in

outth q

q Then,


9-35

9-54 An ideal dual cycle has a compression ratio of 14 and cutoff ratio of 1.2. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined.




/kgm 9076.0kPa 80

K) 253)(K/kgmkPa 287.0( 33

1

11 === P

RTv

v

P

4

1

2

3

qout

x

qinand the specific volume at the end of the compression is


/kgm 06483.0142

=== /kgm 9076.0 33

1vvr

The pressure at the end of the compression is

( ) kPa 321914kPa) 80( 4.112

12 v1 ==== k

k

rPPPv

nd the ma imum pressure is

he temperature at the end of the compression is

a x

4829==== kPa) 3219)(5.1(23 PrPP px kPa T

( ) K 1.72714K) 253( 14.1111

2

11 2 ===

=

k

k

rTTTv

v

nd K 1091kPa 3219kPa 4829K) 1.727(

2

32 =

=

P

=P

TTx

he remaining state temperatures are then

a

From the definition of cutoff ratio

/kgm 07780.0)/kgm 06483.0)(2.1( 3323 ==== vvv cxc rr T

K 1309==

=0.06483

K) 1091(33x

xTT v 0.07780v

K 0.4900.9076

0.07780K) 1309(14.11

4

334 = TT v =

=

k

v

ng the first law and work expression to the heat addition processes gives

The heat rejected is

Applyi

kJ/kg 480.4=+=

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

)()( 32in xpx TTcTTcq v

kJ/kg 2.170K)2530.490)(KkJ/kg 718.0()( 14out === TTcq v

0.646===kJ/kg 480.4kJ/kg 170.2

11in

outth q

q Then,


9-36

9-55E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined.



v

P

4

1

2 3qin

qout

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

32.144Btu/lbm 40.92

R 5401

11 =

==r

uT

v

( )Btu/lbm 402.05R 1623.6

93.732.1442.181 112 r

rrr v11

2

22

======

hT

vvv

v

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

1.848====R 1623.6

R 3000

2

3

2

3

2

22

3

33

TT

TP

T

Pv

vvv

(b) R 3000

23in

33

=====

hhq

Trv

Btu/lbm 388.6305.40268.790

180.1Btu/lbm 790.683 =h


( ) Btu/lbm 91.250621.11180.1848.1

2.18848.1848.1 42

4

3

43334

====== ur rrrr vvvv

vv

vv


(c) 59.1%

Btu/lbm 158.87

======

Btu/lbm 388.63Btu/lbm 158.87

11

04.9291.250

in

outth

14out

qq

uuq



9-37

9-56E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined.


Properties The properties of air at room temperature are cp = 0.240 Btu/lbm.R, cv = 0.171 Btu/lbm.R, and k = 1.4 (Table A-2E).


v

P

4

1

2 3qin

( )( ) R172418.2R540 0.41

2

112 ==

=

kTT

v

v

Process 2-3: P = constant heat addition.

1.741====R 1724R 3000

2

3

2

3

2

22

3

33

TT

TP

TP

v

vvv

(b) ( ) ( )( ) Btu/lbm 306R17243000Btu/lbm.R 0.2402323in == rocess 3- isentropic expansion.

== TTchhq pP 4:

( ) R 117318.21.741R 3000

741.11

3 k vv 0.414

23

434 =

=

=

=k

vTTT

v


(c)

( )( )( )

64.6%

Btu/lbm 108

=====

==

Btu/lbm 306Btu/lbm 108

11

R0541173Btu/lbm.R 0.171

in

outth

1414out

qq

TTcuuq

v



9-38

9-57 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined.




v

P

4

1

2 3 qin

qout

( )( ) K 971.120K 2932

112 =

= TTV

0.41

=k

V

rocess 2-3: P = constant heat addition.

P

2.265K971.1K2200

2

3

2

3

2

22

3

33 ==TT V

==TTPP VVV

rocess 3- isentropic expansion.

P 4:

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

63.5%===

===========

=

=

=

=

=

kJ/kg 1235kJ/kg 784.4

kJ/kg 784.46.4501235

kJ/kg 450.6K293920.6KkJ/kg 0.718

kJ/kg 1235K971.12200KkJ/kg 1.005

K 920.620

2.265K 2200265.2265.2

in

outnet,th

outinoutnet,

1414out

2323in

0.41

3

1

4

23

1

4

334

qw

qqw

TTcuuq

TTchhq

rTTTT

p

kkk

v

V

V

V

V

(b) ( )( )

( ) ( )( ) kPa933 kJmkPa1/201/kgm 0.885 kJ/kg 784.4/11MEP

/kgm 0.885kPa 95

K 293K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

=

===

==

====

rww

r

PRT

vvv

vvv

vv



9-39

9-58 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/kgK, cv = 0.718 kJ/kgK, R = 0.287 kJ/kgK, and k = 1.4 (Table A-2). Analysis (a) Process 1-2: isentropic compression.

v

P

4

1

2 3 qin

qout

Polytropic


( )( )20K 293 0.42

112 =

= TTV

K 971.11

=k

V

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

2.265K 971.1K 2200

2

3

2

32233 ==T

PT

PV

VVV

23==

TT

Process 3-4: polytropic expansion.

( )( ) ( )( )( ) ( )( ) kJ/kg 526.3K 2931026KkJ/kg 0.718

kJ/kg 1235K 971.12200KkJ/kg 1.005

K 102620

2.265K 2200r

1414out

2323in

0.35

34

34

34

========

=

===

=

TTcuuq

TTchhq

TTTT

p

v

VV

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:

2.2652.265 1n1n

21

3 n VV

( ) ( )( )

( )( )(

kJ/kg )

1K 22001026KkJ/kg 0.718kJ/kg 963

K 22001026KkJ/kg 0.287

34out34,in34,34

+=+=

TTcwqu

TTR

v

hich means that 120.1 kJ/kg of heat is transferred to the comlistic since the gas is at a much higher temperature than the surroundings, and a hot gas loses heat during polytropic

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

kJ/kg 9631.3511

out34,in34,

systemoutin

34out34,

==

===

uwq

EEEn

w

120.=w bustion gases during the expansion process. This is unreae . a e

( ) kJ/kg 1.128)4.18723.781(96334out34,in34, =+=+= uuwq which is a heat loss as expected. Then qout becomes kJ/kg 654.43.5261.128out41,out34,out =+=+= qqq and

47.0%===

===

kJ/kg 1235kJ/kg 580.6

kJ/kg 580.64.6541235

in

outnet,th

outinoutnet,

qw

qqw

( )( )

( ) ( )( ) kPa 691=

===

==

====

kJmkPa 1

1/201/kgm 0.885kJ/kg 580.6

/11MEP

/kgm 0.885kPa 95

K 293K/kgmkPa 0.287

3

31

outnet,

21

outnet,

max2min

max3

3

1

11

rww

r

PRT

vvv

vvv

vv (b)


9-40

9-59 Problem 9-58 is reconsidered. The effect of the compression ratio on the net work output, mean effective lso, T-s and P-v diagrams for the cycle are to be plotted.

nalysis Using EES, the problem is solved as follows:

tal(q_12,q_23,q_34,q_41: q_in_total,q_out_total)

> 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41

ssion"

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

rocess 1 to 2"

=T[1]) heat addition"

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

pic process" )-intenergy(air,T=T[3])

4 to 1"

pressure, and thermal efficiency is to be investigated. A

A

Procedure QToq_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_41END "Input Data" T[1]=293 [K] P[1]=95 [kPa] T[3] = 2200 [K]

=1.35 n{r_comp = 20}

1-2 is isentropic compre"Process s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1]

(air, s=s[2],T[2]=temperatureP[2]*v[2]/T[2]=P[1]*P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp "Conservation of energy for pq_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T"Process 2-3 is constant pressureP[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

sure process" w_23 =P[2]*(V[3] - V[2])"constant presDELTAu_23=intenergy(air,T=T[3])-inte"Process 3-4 is polytropic expans

[3])^n P[3]/P[4] =(V[4]/Vs[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 ploytroDELTAu_34=intenergy(air,T=T[4]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 q_41 - w_41 = DELTAu_41

_41 =0 "constant volume process" wDELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) w_net = w_12+w_23+w_34+w_41



9-41


"Thermal efficiency, in percent" he mean effective pressure is:" EP = w_net/(V[1]-V[2])

r

Eta_th=w_net/q_in_total*100 "TM

comp th MEP [kPa]

wnet [kJ/kg]

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

4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5200400600800

10001200140016001800200022002400

s [kJ/kg-K]

T [K

]

95

kPa

340

.1 kP

a

592

0 kP

a

0.0

44

0.1

0.

88 m

3/kg

Air

2

1

3

4

10-2 10-1 100 101 102101

102

103

104

101

102

103

104

v [m3/kg]

P [k

Pa] 293 K

1049 K

2200 K

5.69 6.74 kJ/kg-K

Air


9-42

14 16 18 20 22 24790

800

810

820

830

840

850

rcomp

wne

t [k

J/kg

]

14 16 18 20 22 2447

49

51

53

55

57

rcomp

th

14 16 18 20 22 24970

975

980

985

990

995

1000

rcomp

MEP

[kP

a]



9-43

9-60 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.

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


Analysis Process 1-2: isentropic compression.

v

P

4

1

2 3 Qin

Qout


( )( )22K 3432

112 =

= TTV

K 11810.41

=k

V


( )( ) K 2126K 11811.88.1 222

33

2

22

3

33 == TTT

=== TTPPv

vvv

rocess 3- isentropic expansion.

P 4:

( )

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

( )( ) kW 48.0======

====

====

===

=

=

=

=

=

4.0111T

kkk

kJ/rev 1.251rev/s 2300/60

kJ/rev 251.16198.0871.1

kJ 6198.0K343781KkJ/kg 0.718kg 001971.0

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

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

)m 0.0020)(kPa 97(

K 78122

8.1K 22162.22.2

outnet,outnet,

outinoutnet,

1414out

2323in

3

3

1

11

34

23

4

334

WnW

QQW

TTmcuumQ

TTmchhmQ

RTP

m

rTTT

v

p

&&

V

VV

V

V

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


9-44

9-61 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at 2300 rpm is to be determined.

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

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

Analysis Process 1-2: isentropic compression.

v

P

4

1

2 3 Qin

Qout

( )( ) K 118122K 343 0.412

112 ==

=

kTT

V

V


( )( ) K 2126K 11811.88.1 222

33

23 TT223 ===== TTTPP

v

vvv


3

( )

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

( )( ) kW 47.9=== kJ/rev 1.251rev/s 2300/60outnet,outnet, WnW &&Discussion No

=====

====

==

===

=

=

=

=

=

kJ/rev 251.16202.0871.1

kJ 6202.0K343781KkJ/kg 0.743kg 001906.0

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

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

)m 0.0020)(kPa 97(

K 78122

8.1K 22162.22.2

outinoutnet,

1414out

2323in

3

3

1

11

4.01

3

1

4

23

1

4

334

QQW

TTmcuumQ

TTmchhmQ

RTP

m

rTTTT

v

p

kkk

V

VV

V

V

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



9-45

9-62 An ideal dual cycle has a compression ratio of 18 and cutoff ratio of 1.1. The power produced by the cycle is to be determined.



Analysis We begin by fixing the temperatures at all states.


( ) K 7.92418K) 291( 14.1111

2

112 ===

=

k

k

rTTTv

v

v

P

4

1

2

3

qout

x

qin

( ) kPa 514818kPa) 90( 4.112

11 2 ===

= k

k

rPPPv

v

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

K 1017kPa 5148kPa 5663K) 7.924(

22 =

=

=

PP

TT x x

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

K 8.365181.K) 1119(

1

33

34 ==

=

kcrTTT

v 1 14.11

4=

k

rv

pplying the first law to each of the processes gives

A

1 kJ/kg 0.455K)2917.924)(KkJ/kg 718.0()( 122 === TTcw v kJ/kg 5.102K)10171119)(KkJ/kg 005.1()( 33 === xpx TTcq

( 333 = xx Tcqw v kJ/kg 26.29K)10171119)(KkJ/kg 718.0(5.102) == xT 3 kJ/kg 8.540K)8.3651119)(KkJ/kg 718.0()( 434 === TTcw v

1.1150.45526.298.54021343net

The net work of the cycle is

k J/kg=+=+= wwww x The mass in the device is given by

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

)m kPa)(0.003 90( 3PV3

1

11 ==RTm

he net power produced by this engine is then

=

T

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


9-46

9-63 A dual cycle with non-isentropic compression and expansion processes is considered. The power produced by the cycle is to be determined.



Analysis We begin by fixing the temperatures at all states.


( ) K 7.92418K) 291( 14.1111

2

112 ===

=

k

k

s rTTT vv

K 10370.85

K )291(924.7K) 291(121212

12 = TT s =+=+= TT

TTTT

s

v

P

4

1

2

3

qout

x

qin

( ) kPa 514818kPa) 90( 4.112

112 ===

= k

k

rPPPv

v

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

K 1141kPa 5148kPa 5663K) 1037(

22 =

=

=

PP

TT x x

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

K 3.410181.1K) 1255(

14.11

3

1

4

33 =

=

=

=

kc

k

rr

TTTv

v 4s

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

43 = === s

sTTTT

TTTT

pplying e first law to each of the processes gives A th

kJ/kg 6.535K)2911037)(KkJ/kg 718.0()( 1221 === TTcw v kJ/kg 6.114K)11411255)(KkJ/kg 005.1()( 33 === TTcq

11411255)(KkJ/kg 718.0(6.114)( 333

xpx

) kJ/kg 75.32K === xxx TTcqw v 8 71.0()( 4343 == TTcw v kJ/kg 8.545K)8.4941255)(KkJ/kg =

he net w k of the cycle is T or

328.54521343net +=+= wwww x kJ/kg 95.426.53575. = he mass in the device is given by

T

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

)m kPa)(0.003 90(3

3

1

11 === RTP

mV

The net power produced by this engine is then

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


9-47

9-64E An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, and the thermal efficiency are to be determined.



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

( ) R 158015R) 535( 14.1111

2

112 ===

=

k

k

rTTTv

v

v

P

4

1

2

3

qout

x

qin( ) psia 2.62915psia) 2.14( 4.112

11


2 ===

= k

k

rPPPv

v

psia 1.692psia) 2.629)(1.1(23 ==== PrPP px

R 1738psia 629.2psia 692.1

R) 1580(2

2 =

=

=

PP

TT xx

R 2433R)(1.4) 1738(33 ===

= cx

xx rTTT v

v

R 2.942151.4R) 2433(

14.11

3

1 k

4

33 =

=

=

=

kc

rr

TTTv

v

pplying e first law to each of the processes gives

4

A th

Btu/lbm 7.178R)5351580)(RBtu/lbm 171.0()( 1221 === TTcw v 2 Btu/lbm 02.27R)15801738)(RBtu/lbm 171.0()( 2 === TTcq xx v )( 33 Btu/lbm 8.166R)17382433)(RBtu/lbm 240.0( ==

2433)(RBtu/lbm 171.0(Btu/lbm 8.166)( 333

= xpx TTcq 8 Btu/lbm 96.47R)173 === xxx TTcqw v )( 4343 Btu/lbm 9.254R)2.9422433)(RBtu/lbm 171.0( ==

he net w k of the cycle is

= TTcw vT or

+= wwww Btu/lbm 124.2=+= 7.17896.479.25421343net x nd the net heat addition is

Hence, the thermal efficiency is

a

Btu/lbm 193.8=+=+= 8.16602.2732in xx qqq

0.641===Btu/lbm 193.8Btu/lbm 124.2

in

netth q

w


9-48

9-65 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined.


Properties The properties of air at 850 K are cp = 1.110 kJ/kgK, cv = 0.823 kJ/kgK, R = 0.287 kJ/kgK, and k = 1.349 (Table A-2b).

Analysis (a) Process 1-2: Isentropic compression

1

Qin

2 3

4

Qout

( )( )

( )( ) kPa 504419kPa 95 1.3492

112

1-1.3491

==

=

k

PPv

v

v

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

K 1.95019K 3402

112 ==

=k

TTv

3

3

m 0001778.0

m 0045.019 ==

c

c

c

dcrVV

=

++

cV

VVV

m 003378.00032.00001778.0 =+=+= VVV 31 dcass contained in the cylinder is The total m

( )( ) kg 0.003288K 340K/kgmkPa 0.287 )3

1=

The mass of fuel burned during one cycle is

m 378kPa)(0.003 95(3

11 ==RTP

mV

kg 0001134.0kg) 003288.0(

28 ==== amm

AF ff

f

f

f

fm

mm

mmm

rocess 2-3: constant pressure heat a

P ddition

kJ 723.48)kJ/kg)(0.9 kg)(42,500 0001134.0(HVin === cf qmQ

The cutoff ratio is

K 2244=== 3323in K)1.950(kJ/kg.K) kg)(0.823 003288.0(kJ 723.4)( TTTTmcQ v

2.362===K 950.1K 2244

2

3

TT

(b) 33

12 m 0001778.019

m 0.003378 ===r

VV

23

14

3323 m 0004199.0)m 0001778.0)(362.2(

PP ==

===VV

VV



9-49


( )

( ) kPa 9.302


m 0.003378kPa 5044 3

Documents

Thermo 7e SM Chap09-1