Internal Combustion Engines - SENS

40
1 Bottom Dead Center BDC Top Dead Center TDC 1 V 2 V 2 1 c V V r Ratio, n Compressio = Internal Combustion Engines

Transcript of Internal Combustion Engines - SENS

Page 1: Internal Combustion Engines - SENS

1

BottomDeadCenterBDC

TopDeadCenterTDC

1V

2V

2

1c V

Vr Ratio,n Compressio =

Internal Combustion Engines

Page 2: Internal Combustion Engines - SENS

2

Intake0-1

p=constant

Combustion1-2

v=constant Exhaust

3-0p=constant

Power2-3

Q=0one revolution

Lenoir Cycle 1860

TDCTDC

0

1

2

3

p

v

Q=0v=c

p=c

Air StandardLenoir Cycle

Q

Q

Page 3: Internal Combustion Engines - SENS

3

0

1

2

3

p Q=0v=c

p=c

Air Standard Lenoir Cycle

Q

Q

)u(uqwhh)v(vp)u(uq

wΔuqcp

ProcessExhaust 13

)u(uwvv

pp

TT

ocessadiabticpr0,qStrokePower Process,Expansion 32

uuq0wconstantv

Process Combustion 32

wΔuqmass ofquantity System micThermodyna Closed

1331

3131331

32

1

1

21k

k

2

3

2

3

12

−−=−=−+−=

+==−

−−=

=

=

=−

−==⇒=

+=−

−−

k

Page 4: Internal Combustion Engines - SENS

4

Compression1-2

Combustion2-3

TDC

Power3-4

Exhaust4-1

BDC

Intake0-1BDC

Exhaust1-0

TDC

0 1

2

3

4

revolution 1 revolution 2

OTTO CYCLE , 2 Revolutions/cycle

p

v

OTTO CYCLE

spark ignition, SI

fuel injection + SI

Page 5: Internal Combustion Engines - SENS

5

4 CYCLE, 2 REVOLUTIONS

Revolution 1 Revolution 2

Page 6: Internal Combustion Engines - SENS

6

intake opensexhaust closes

exhaust opens

intake closes

ignition

Q

W

W

p

v

p

v

Air StandardOtto Cycle Closed System- a quantity of mass

adiabatic andisentropic

atmospheric pressure

W

Page 7: Internal Combustion Engines - SENS

7

inQoutQ

Compression1-2

Combustion2-3

ExpansionPower

3-4

Exhaust4-1

1

2 4

pW

W

inQ

outQ 1

2 4

p

v v

v=const

v=constadiabaticisentropic

adiabaticisentropic

AIR STANDARD OTTO CYCLE, 2 Revolutions/cycle

inW outW

V=c

V=c

Page 8: Internal Combustion Engines - SENS

8

“COLD”

Air Standard Otto Cycle

T

v

2 4

1

.Tat c cconstant COLD""

wΔuq SYSTEM CLOSED

1vp,

⇒+=

( )

( )

( )

( )

( )( )23v

14v

2-3

1-4

in

outcycle

14v1414

43v4343

1k

4

334

k23v2332

12v1212

1k

2

112

k22

k11

k

TTcTTc1

qq1

qq1η

TTcuuΔuq

ΔUq 0,pdv wc, v,1

TTcuuΔu wvvTT

ΔU w0,q c, pv 0,Δs ,

TTcuuΔuq

ΔUq 0,pdv wc,v,

TTcuuΔu wvvTT

vpvp

Δu w0,q c, pv 0,Δs ,

−−

−=−=−=

−=−==

====→

−=−==

=

====→

−=−==

====→

−===

=

=

====→

−−

Cycle

4Exhaust

43 Expansion

32 Combustion

21 nCompressio

3

0,Q c pv

0Δsk

==

=inq

outq

Page 9: Internal Combustion Engines - SENS

9

“COLD”Air Standard Otto Cycle

( )

( )

( )21

net

VVw(mep)

−=

−×=

−=

−=

−=

=−

=

=

=∆

−−−

→→

Pressure EffectiveMean

VVpw

r11

rationcompressio

11

vv

11η

vv

TT

TT

TT

vvlnR

TTlnc

vvlnR

TTlnc

Δss

12meannet

1k1k1k

2

1

1k

1

2

2

1

1

4

2

3

1

4

1

4v

2

3

2

3v

1432

( )( )

−=−−

−=

−−

−=−−

−=

−=−

−=

===

1TT

1TT

TT1

TTTT1η

TTcTTc1

uuuu1η

QQ1

QQQ1η

vv

VVr Ration Compressio

2

3

1

4

2

1

23

14

23v

14v

23

14

in

out

in

outin

2

1

2

1c

T

2

3

4

1

ΔUQ 0,pdv Wc, v3,2 Exhause

ΔU W0,Q 0,Δs 4,3Expansion

ΔUQ 0,pdv Wc,v3,2 Combustion

ΔU W0,Q 0,Δs 2,1n CompressioWΔUQ SYSTEM, CLOSED

====→

===→

====→

===→+=

1V2V

1vp, Tat c cconstant COLD"" ⇒

1.0 1.0

Page 10: Internal Combustion Engines - SENS

10

( )( )

( )

( )

( )

%1.516

11

vv

11η

51.1%361.4

100.3285.2q

b176.5BTU/l559.71591.9.171quuq

b285.2BTU/l1591.93259.7.171wuuw

R1591.9613259.7

vvTT

b361.4BTU/l11463259.7.171quuq

b100.3BTU/l559.71146.171wTTcuuw

R11466559.7vvTT

14.11k

2

1

cycle

in

netcycle

34

3434

43

4343

o.41k

4

334

32

1332

12

12v1212

o.41k

2

112

=−=

−=

=

=

=

=−=

−==−=

−=

=

=

=

=−×=−=

=−×=−==

=×=

=

−−

−−

−What is the thermal efficiencyof a COLD air standard Otto cycleoperating between 2800 F and 100 F with a compression ratio of 6?

p

v

1100F

2

32800 F

4

Q

Q

W

W

ΔUq 0,pdv wc, v3,2Exhaust

Δu w0,q 0,Δs 4,3Expansion

Δuq 0,pdv wc,v3,2 Combustion

Δu w0,q 0,Δs 2,1n CompressioWΔUQ

System Closed

====→

===→

====→

===→+=

Page 11: Internal Combustion Engines - SENS

11

AIR TABLE - Tables A-22, A-22Especific heats are integrated as variables of T- page 113, 230

( )

( ) dTTcu

Energy Internal 4)

dTTch

Enthalpy 3)

)(v)(v

vv

)(p)(p

pp

constantpv Process, Isentropic 2)

C0 F,0 base Table 1)

v

p

2r

1r

2

1

2r

1r

2

1

k

oo

=

=

=

=

=

( ) ( )

( ) ( )

−−=−

+−=−

−=

+=

=−

+

=−

1

21

02

012

1

21

02

012

p

v

1

2

1

2p12

1

2

1

2v12

ppln R)(Ts)(Tsss

vvln R)(Ts)(Tsss

ratio pressurelnRTdTTcs

ratio volumeln RTdTTcs

22E,A Table and 22A Table Using-Entropyppln R

TTln css

vvln R

TTln css

gas ideal -Entropy 5)

Page 12: Internal Combustion Engines - SENS

12

9.1

( ) ( ) g674.25kJ/k214.07888.32uuQkJ/kg 888.32u

16.3near v 22TableA eInterpolat

16.38.51.9176vvvv

1.9176vkJ/kg 1903.6unear 22eTableAInterpolat

kJ/kg 1903.06503.61400ukJ/kg 1400uuQ

.17692.94.52

56,725.7556.7208.73ratioion interpolat

504.45 72.56 ukJ/g 503.06 73.08

496.62 75.5 u vT

73.08621.2/8.5vvvv,

vv

vv

/kgm .858100

300.286p

RTv

214.07u 621.2, vK,300 @

43out

4

r4

3

4r3r4

r3

3

3

23in

3

r

1

2r1r2

1

2

r1

r2

31

1r1o

=−=−==

=−

=×=

=

=

=−=−=

=−=

==−−

=

=

==

==

==

==

psi 958.7

8.511.858

725.75mep

vv1v

WVV

Wmep c)

% 51.81400

725.751QQ1η b)

g725.75kJ/kW674.251400QQ Wa)

1

21

21

in

out

outin

=

−×

==

=−

=

=−=

−=

=

−=−=

1100 kPa300 K

2

3

4

1400kJ/kg

p

v

r=8.5

Page 13: Internal Combustion Engines - SENS

13

( ) ( )

( )( )

1685psia13055983367p

367psia1014.7VVpp

adiabatic, and isentropic is 21 SinceTTpp

mRT,Pv law gas ideal thefrom ,VV SinceR5893T

1305T.17800 TTcQHeat, Combustion

b)b134.2BTU/l5201305.17TTcΔU

a)

130510520VVTT

constantpv with adiabatic ,isentropic 2-1 Process

3

1.4k

2

112

2

323

32

o3

3

23v

12v

o.41k

2

112

k

=

×=

=×=

=

=

==

=

−×=−=

=−×=−=

=×=

×=

=−

The initial temperature and pressure of air inan ideal Cold Otto cycle having acompression ratio of10, is 60 F and 14.7 psiarespectively. Heat isadded in the constant volume process at a rateof 800 BTU/lbm. Consider air to be an idea gas.

Calculate: a) The change in internal energy per pound of air during the compression process. b) The maximum temperature and pressure of the cycle

R.24BT U/lbmcRm.171BT U/lbc

1.4k

p

v

−=−=

=

1 14.7 psia 520 R

2

3

p

V

Compression

CombustionQ=800 BTU/lb

Expansion

Exhaust

1V1V.1×

4

Page 14: Internal Combustion Engines - SENS

14

hrfuel/HP .44lbnConsumptio Fuel SpecificHP 167.3

lb/hr 74.18nConsumptio Fuel Specific

d)74.18lb/hrFuel

strokerev/power 218,900604600cylinders 8inderstroke/cylBTU/power 1.27Fuel

c)

30.4%7781.277

1200)(1500q

b)HP 167.3HP

sec/min 60strokepower rev/ 2lb/HPft 550rpm 4600cylinders 8inderstroke/cylBTU/power 1200)(1500HP

a)

in

net

=

=

×××=

−==

=××−

××−=

265cu in

14.7 psia70 F

p

v

We=1500

Wc=1200

Q=1.27

Intake

Compression

Power

Exhaust

---1st Revolution

---2nd Revolution

A 265 cubic in V8 gasoline engineruns at 4600 rpm. Compression workis 1200 ft-lb, expansion work is 1500 ft-lb and heat input is 1.27 BTUper piston per cycle. The atmosphereis 14.7 psia and 70 F. The air fuelratio is 20:1. The heating value ofgasoline is 18,900 BTU/lb. Find

a) indicated HP b) thermal efficiencyc) gasoline consumption per hrd) specific fuel consumption.

Page 15: Internal Combustion Engines - SENS

15

2 CYCLE, 1 Revolution/cycle

Compression1-2

Ignition2-3

Power3-4

Exhaus4-1

Intake1

p

v1

2

3

4

Page 16: Internal Combustion Engines - SENS

16

Diesel Cycle

p

v

Compression - isentropic, Q=0Combustion -constant pressureExpansion - isentropic, Q=0Exhaust - constant volume

Closed system

2 3

4 ΔUQΔHQΔUW

14

32

21

===

( ) ( )

−−

−=

=

=

− 1rk1r

r11η

VVr ratio, offCut

VVr ratio,n Compressio

c

kc

1kcycle

2

3c

2

11

Page 17: Internal Combustion Engines - SENS

17

( ) ( )

( ) ( )

43.9%518.15

290.7518.15q

qqη

BTU/lb290.7q5352235.171TTcq

BTU/lb518.15q

12103360.241TTcq

R223513.945.0323360

vvTT

/lbft5.032121033601.812

TTvv

/lbft1.812121053513.94

TTvv

/lbft13.9414414.253553.35

pRTv

in

outincycle

out

14vout

in

23pin

o.41k

4

334

3

1

223

32.51k

1

2

112

3

1

11

=−

=−

=

==−=−=

=

−=−=

=

=

=

=

=

=

=

=

=

=××

==

7-21

What is the thermal efficiencyof a COLD air standard diesel

cycle operating on 14.7 psia air at 75 F? The temperature of the air before and after heat addition are 750 F and 2900 F respectively.

p

v

175 F14.7 psia

1210 R2

3360 R3

4

Qin

Qout

Page 18: Internal Combustion Engines - SENS

18

1

2

3 4

5

p

v ( )

1515

545454

433443

233232

1221

1

2

3

4c

uuq 0w0q uuw

hhq vvpw

uuq uuw

0q uuw

VVr ratio,n Compressio

VVr ratio, offCut

15

43

43

21

−===−=

−=−=

−=−=

=−=

=

=

→→

AIR STANDARD DUAL CYCLE

adiabatic andisentropic

( )Δhq

ΔvpwwΔuq

mass ofquantity System Closed

constantp

constantp

=

=+=

=

=

Page 19: Internal Combustion Engines - SENS

19

Stirling and Erickson Cycles

Page 20: Internal Combustion Engines - SENS

20

1 2

4 3

T

v

Δuq constant v

1vvln T R w

decreases v,Tconatant at q

Δuq constant v

vvlnRT w

increases v,Tconatant at q

rregenerato

3

43in

Hout

rregenerato

1

21out

Hin

==

=

==

=

4 Process

43 Process

32 Process

21 ProcessSystem Closed Cycle, Stirling

outW

inW

Page 21: Internal Combustion Engines - SENS

21

Simple Brayton, Gas Turbine, Cycle

compressorW Net

WorkOutput

T

s

combustor

turbine

compressor

1

23

4

1

2

3

4

p=c

reversibleconstantpvadiabatic 0,Qisentropic 0,Δs

k =

==

( ) netWKEhΔQ LawFirst of FormEquation Energy FlowSteady

Systems icThermodyanOpen are Turbine and,Combustion ,Compressor

++=

inQ

Page 22: Internal Combustion Engines - SENS

22

Propulsion Gas Turbine

Page 23: Internal Combustion Engines - SENS

23

inQ

turbinecompressorWW = outputnet

W

Aero-derivative Gas Turbine

PowerTurbine

GasGenerator

Page 24: Internal Combustion Engines - SENS

24

Figure 9.4 Air Standard Brayton Cycle

Page 25: Internal Combustion Engines - SENS

25

Brayton Cycle with Real Compression and Expansion

turbine

compressor

combustor

12

12sc

c

4s3

43t

t

hhhhη

workactual workisentropicη

Efficiency Compressorhhhhη

workisentropic workactualη

Efficiency Turbine

−−

=

=

−−

=

=

Page 26: Internal Combustion Engines - SENS

26

( )( )

( )( )14

43

23

12

shaft

2

hhmQ 0, WProcess, Exhausehhm W0,Q Process, Expansion

hhmQ 0, WProcess, Combustionhhm W0,Q Process,n Compressio

Wρgh)2

Vpv(umQ

EquationEnergy FlowSteady spacein region -SystemOpen Flow,Steady

−==−==

−==−==

++++∆×=

Brayton Air Standard Cycle

Air StandardProcess

Actual Process

Page 27: Internal Combustion Engines - SENS

27

( )

( )( )

( )( )

( ) ( )

( )( )

( ) ( )( )

( ) ( )

33.9%215.26

139.92215.26qwη

33.9%1103.012000

5201113.031hhhh1η

lb142.33BTU/520.-1113.03.241TT.241hhqBTU/lb213.591110.032000.241hhw

R1113.031035.92000.932000TT.92TT

.931029.51959.7

T1959.7TTcTTc

hhhh

wwη

R1035.91012000

ppTT

BTU/lb139.925201103.01.241TTcwhhw

BTU/lb215.261103.012000.241q

TTchhq

R1103.01.83

5201003.9520.83

TTTT

.83TTcTTc

hhhh

WWη

R1003.91060460ppTT

in

netcycle

23

14cycle

141414

4343

o4is334

4

4i3p

43p

4i3

43

ideal

actualturbine

o.2857k

1k

3

434i

12p21

1221

32

23p2332

o12is12

12p

12isp

12

12is

actual

idealcompressor

o.2857k

1k

1

212i

=−

==

=−

−−=

−−

−=

==−=−==−=−=

=−×−=−×−=

=−

−=

−=

−−

==

=

×=

=

=−=−=−=

=−=

−=−=

=−

+=−

+=

=−−

=−−

==

=×+=

=

7-23

In an air COLD standard gasturbine, 60 F and 14.7 psia air is

compressed through a pressure ratio of 10. Air enters at 1540 Fand expands to14.7 psia. If the

isentropic efficiency of the compressor and turbine are 83%and 93% respectively.What is thethermal efficiency of the cycle?

T

s

160 F

14.7 psia

2i2

1540F3

44i

%83=cη

%93=tη

93.7273.6733.14226.215139.92-213.59

qw:Check

=−=

=∑ ∑

Page 28: Internal Combustion Engines - SENS

28

In an air standard gasturbine, 60 F and 14.7 psia air is

compressed through a pressure ratio of 10. Air enters at 1540 Fand expands to 14.7 psia. If the

isentropic efficiency of the compressor and turbine are 83%and 93% respectively.What is thethermal efficiency of the cycle?

160 F

14.7 psia

2i2

1540F3

44i

%83=cη

%93=tη

T

s

( )

35.3%240.42155.431

QQ1η

BTU/lb 155.43124.27282.7hhqBTU/lb 240.42264.28504.7hhq

BTU/lb 282.7hh.92hh

.93hhhh

WWη

BTU/lb 265.99h 1.74,pat ion interpolatby

17.41/10174.p

ppp

504.71h 174.,p R,2000At BTU/lb 264.28h

.83124.27240.48124.27

.83hhhh

.83hhhh

WWη

BTU/lb 240.48h

12.147101.2147p

ppp

BTU/lb 124.27h 1.2147,p R,520at

out

in

14out

23in

4is334

4i3

43

ideal

actualturbine

4isr4is

1

4isr3r4is

3r3o

2

12is12

12

12is

actual

idealcompressor

2is

1

2isr1r2is

1r1o

=−=−=

=−=−==−=−=

=−×−=

=−−

==

==

=×=

=

==

=

−+=

−+=

=−−

==

=

=×=

=

==

.10061.69.17 ratio

240.98 12.30 1000240.48 12.147 998

236.02 10.61 980hp T r

=

Page 29: Internal Combustion Engines - SENS

29

24

64

24

25

hhhhessEffectivenr Regenerato

hhhhessEffectivenr Regenerato

TransferHeatr IdealerHeatTransf ActualessEffectivenr Regenerato

−−

=

−−

=

=

Regenerated Brayton Cycle

Page 30: Internal Combustion Engines - SENS

30

1

2

3

4

6

7

8

9510T

s

Brayton Cycle with:IntercoolingRegenerationReheat

Page 31: Internal Combustion Engines - SENS

31

A gas turbine cycle operates with 2 stages of inter-cooled compression, a regenerator, and two stages of expansion with reheat between them. Air enters the compressor at 100 kpa, 300 K at a volume flow of 5 cubic meters/ sec. The turbine and compressor stages have an equal pressure ratio, isentropic efficiencies of 80 % and the overall cycle pressure ratio is 8. The regenerator effectiveness is 80%. Turbine inlet temperature is 1400 K. Determine: a) the cycle efficiency.

1

2

3

4

7

8

9510T

s 588.05 100 10 1218.9 283 9

73 1144. 159.3 283 1105.9 8s1515.42 450.50 800 1400 8

1218.9 283 773 1144. 159.3 283 1105.9 7s

1515.42 450.50 800 1400 6 1061.2 800 5 430.34 283 4

404.31 3.920 283 403.29 3s300.19 1.386 100 300 3

430.34 283 2404.31 3.920 283 403.29 2s300.19 1.386 100 300 1h )(p p TPt r

inQ inQ

outQ

outQ

Page 32: Internal Combustion Engines - SENS

32

( )

( ) ( )( )

( )

( )( ) ( )

( )( )

( )

( ) kJ/kg 588.05hhεhh 10Point

kJ/kg 1061.2hhεhh 5Point

hhhh

hhhh

QQε

alancer Energy BRecuperatokJ/kg 1218.9/.8hhhh

1105.9T 1144.73,h 159.3,p @ 159.3/ppp

450.5p kJ/kg, 11414.42h 1400,T @ 22TableA 6Point

430.34/.8hhhh 3Point

404.31h 3.92,p @ 3.92ppp

2.83pr 8,ppratio pressure stage 2sPoint

1.386p kJ/kg, 300.19h 300,T @ 22A Table kg/sec 5.81/sec/v5mm

kg/sec .861300/100.287RT/p v1Point

49910

4945

49

109

49

45

max

actual

7s667

r7s

rr6sr7s

r1

12s12

2sr2s

rr1sr2s

rr

r1

3

=−×−=

=−×+=

−−

=−−

==

=−−====

=====−

=−+=

===×=

==×=

===−==

=×==

1

2

3

4

7

8

9510T

s

inQ inQ

outQ

outQ

Page 33: Internal Combustion Engines - SENS

33

( ) ( )( ) ( )

.438.52130.15/296Work RatioBack /WWWork RatioBack

KW1921.6kJ/sec 1921.6W(130.15)][(296.52)25.81W

300.19)](430.341218.9)[(1515.4225.81W)]h(h)h(h)h(h)h[(hmW

Work CompressorWork TurbineW

% 44.3296.52454.22130.15288.051η

ngintercooliexhaustreheatcombustor1η

hhhhhhhh1

QQ1η

turbinecompressor

34129876

cycle

cycle

3256

32110

in

outcycle

==

=

==−××=

−−−−××=−−−−−+−×=

−=

=++

−=

++

−=

−+−−+−

−=−=

1

2

3

4

7

8

9510

T

s

inQ inQ

outQ

outQ

Page 34: Internal Combustion Engines - SENS

3413

42is4is

T

k1k

1

212is

3443

3232

1221

ppTT

hhwhhqhhw

=

−=−=−=

Q

WW

Intercooled Compressionstagesingle2

2

stagesingle2is

inletpressure

dischargepressure

s

T

inletpressure

dischargepressure

s

Page 35: Internal Combustion Engines - SENS

35

1

2

3

42is4isT

s

Q WW

2

stagesingle2is

( )

( )( ) ( )

( )

( ) ( )

( )

( ) ( )

higher 11.4%

BTU/lb 77.68530851.95.24TTcw

R851.95.8

TTTT

R787.564530p/pTT

BTU/lb 68.3133.5134.8w

BTU/lb 33.51510649.61.24TTcw

R649.61.8

510621.69510.8

TTTT

R621.692510ppTT

BTU/lb 32.64530646.24TTcq

lb/BTU80.34)530675(24.TTcw

R675.8

530646530.8

TTTT

R6462530ppTT

34pstagesingle

o1

stage single 2is

1stagesingle

o.2857k1k

1stage single 21

stage single 2is

cooledinter

34p43

o34is34

o.2857k

1k

3

414is

32p32

32p12

o12is12

o.2857k

1k

1

212is

=−×=−=

=−

+=

=×=

=

=+=

=−×=−=

=−

+=−

+=

==

=

=−×=−=

=−×=−=

=−

+=−

+=

==

=

Compare compression overa pressure ratio of 4 from14.7 psia, 530 R withcompression in two stages of equal pressure ratio intercooledto 510 R. Theefficieny ofall stages is 80%

stagesingle2

Page 36: Internal Combustion Engines - SENS

36

INTEGRAL GEARCENTRIFUGAL COMPRESSOR

Page 37: Internal Combustion Engines - SENS

37

Intercooled Supercharger

TurbochargerSupercharger

Q

1

2 3

4

Page 38: Internal Combustion Engines - SENS

38

Turbocharger

W

1

2 3

4

1

2

3

4enginecompressionstroke

combustionv=constant

enginepowerstroke

TP atm

Air Standard Otto Cyclewith a Turbocharger

Page 39: Internal Combustion Engines - SENS

39

ProCharger Supercharger

Page 40: Internal Combustion Engines - SENS

40

TurboCharger

CompressedAir to Engine

EngineExhaust

Air Inlet