Compression1-2

Combustion2-3

Exhaust4-1

Intake0-1

Exhaust1-0

3revolution 1 revolution 2

pBDCBottom Dead Center

OTTO CYCLE

spark ignition, SI

fuel injection + SI

Diesel

fuel injection

ACTUAL 4 CYCLE , 2 Revolutions/cycle

Power3-4

TDCTop Dead Center

42

0 1v

4 CYCLE, 2 REVOLUTIONS

Revolution 1 Revolution 2

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

AIR STANDARD OTTO CYCLE, 2 Revolutions/cycle

inQoutQ

Compression1-2

ExpansionPower3-4

1

2

W

W

inQ

outQ 1

4

p

v v

v=const

v=constadiabaticisentropic

adiabaticisentropic

inW outW

V=c

V=c

Exhaust4-1

Combustion2-3

p

4 2

“COLD”

Air Standard Otto Cycle

4

1

.Tat c cconstant COLD""

W∆UQ SYSTEM CLOSED

1vp,

⇒+=

( )

( )

( )

( )

( )( )23v

14v

2-3

1-4

in

outcycle

14v1414

43v4343

1k

3

434

k23v2332

12v1212

1k

2

112

k22

k11

k

TTcTTc1

QQ1

QQ1η

TTcuu∆UQ

∆UQ 0,pdv Wc, v,1

TTcuu∆U WvvTT

∆U W0,Q c, pv 0,∆s ,

TTcuu∆UQ

∆UQ 0,pdv Wc,v,

TTcuu∆U WvvTT

vpvp

∆U W0,Q c, pv 0,∆s ,

−−

−=−=−=

−=−==

====→

−=−==

=

====→

−=−==

====→

−===

=

=

====→

−

−

−

−

−−

−

∫

∫

Cycle

4Exhaust

43 Expansion

32 Combustion

21 nCompressio

3

0,Q c pv

0∆sk

==

=inQ

outQ

T

2

v

“COLD”Air Standard Otto Cycle

( )

( )

( )21 VVW(mep)−

=

−×=

−=

−=

−=

=−

=

−

=

−

=∆

−−−

−

→→

Pressure EffectiveMean

VVpW

11

rationcompressio

11

vv

11η

vv

TT

TT

TT

lnTTlncln

TTlnc

∆ss

12mean

11k1k

2

1

1k

1

2

2

1

1

4

2

3

1

4

1

4v

2

3

2

3v

1432

kr

vvR

vvR

( )( )

−

−

−=−−

−=

−−

=−−

−=

−=−

−=

===

1TT

1TT

TT1

TTTT1η

TTcTTc

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

4

1

∆UQ 0,pdv Wc, v3,2 Exhause

∆U W0,Q 0,∆s 4,3Expansion

∆UQ 0,pdv Wc,v3,2 Combustion

∆U W0,Q 0,∆s 2,1n CompressioW∆U WSYSTEM, CLOSED

∫

∫

====→

===→

====→

===→+=

1V2V

1vp, Tat c cconstant COLD"" ⇒

1.0 1.0

3

T

2

( )( )

( )

( )

( )

%1.516

1111η

51.1%361.4

100.3285.2QWη

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

14.11

2

1

cycle

in

netcycle

34

3434

43

4343

o.41k

4

334

32

1332

12

12v1212

o.41k

2

112

=−=

−=

=

−

=

=

=−=−=

=−=−=

=

=

=

=−×=−=

=−×=−==

=×=

=

−−

−

−

−

−

−

−

−

−

−−

−

k

vv

What is the thermal efficiency of a COLD air standard Otto cycle operating between 2800 F and 100 F with a compression ratio of 6?

p

1100F

2

32800 F

4

Q

Q

W

W

∆UQ 0,pdv Wc, v3,2Exhaust

∆U W0,Q 0,∆s 4,3Expansion

∆UQ 0,pdv Wc,v3,2 Combustion

∆U W0,Q 0,∆s 2,1n CompressioW∆UW

System Closed

∫

∫

====→

===→

====→

===→+=

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

( ) ( )

( ) ( )

−−=−

+−=−

−=

+=

−

−

=−

+

=−

∫

∫

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 pressurelnRT

dTTcs

ratio volumeln RT

dTTcs

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

TTln css

vvln R

TTln css

gas ideal -Entropy 5)

( )

( ) 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

∫

∫

=

=

=

=

=

( ) ( ) 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

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43out

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1400kJ/kg

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r=8.53

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1100 kPa300 K

psi 958.7

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% 51.81400

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1

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21

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out

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1305T.17800 TTcQHeat, Combustion

b)b134.2BTU/l5201305.17TTc∆U

a)

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constantpv with adiabatic ,isentropic 2-1 Process

3

1.4k

2

112

2

323

32

o3

3

23v

12v

o.41k

2

112

k

=

×=

=×=

=

−

=

===

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=−×=−=

=×=

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=−

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

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)(1500QWη

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.

( ) ( )

( ) ( )

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

=−

=−

=

==−=−=

=

−=−=

=

=

=

=

=

=

=

=

=

=××

==

−

−

What is the thermal efficiency of 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 R 2

3360 R 3

4

Qin

Qout

Stirling and Erickson Cycles

1 2

∆UQ constant v

1vvln T R W

decreases v,Tconatant at Q

∆UQ 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

4 3

T

v

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,∆s

k =

==

( ) netWKEh∆Q LawFirst of FormEquation Energy FlowSteady

Systems icThermodyanOpen are Turbine and,Combustion ,Compressor

++=

inQ

Propulsion Gas Turbine

inQ

turbinecompressorWW = outputnet

W

GasGenerator

PowerTurbine

Aero-derivative Gas Turbine

Figure 9.4 Air Standard Brayton Cycle

Brayton Cycle with Real Compression and Expansion

turbine

compressor

combustor

43

4s3c

c

4s3

43t

t

hhhhη

workactual workisentropicη

Efficiency Compressorhhhhη

workisentropic workactualη

Efficiency Turbine

−−

=

=

−−

=

=

( )( )

( )( )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

−==−==−==−==

++++∆×=

Air StandardProcess

Actual Process

Brayton Air Standard Cycle

( )

( )( )

( )( )

( ) ( )

( )( )

( ) ( )( )( ) ( )

33.9%215.26

142.33215.26QWη

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lb142.33BTU/520.-1113.03.241TT.241hhQBTU/lb213.591110.032000.241hhW

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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

=−

==

=−

−−=

−−

−=

==−=−==−=−=

=−×−=−×−=

=−−

=−

−=

−−

==

=

×=

=

=−=−=−=

=−=

−=−=

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−

−

−

−

−

−

−

7-23

In an air COLD standard gas turbine, 60 F and 14.7 psia air iscompressed through a pressureratio of 10. Air enters at 1500 F and expands to14.7 psia. If theisentropic efficiency of thecompressor and turbine are 83% and 93% respectively.What is the thermal 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

=−=

=∑ ∑

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

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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.83

124.27240.48124.27.83

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BTU/lb 240.48h

12.147101.2147p

ppp

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

out

in

14out

23in

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43

ideal

actualturbine

4isr4is

1

4isr3r4is

3r3o

2

12is12

12

12is

actual

idealcompressor

2is

1

2isr1r2is

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=−=−=

=−=−==−=−=

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.10061.69.17 ratio

240.98 12.30 1000240.48 12.147 998

236.02 10.61 980hp T r

=

Regenerated Brayton Cycle

24

64

24

25

hhhhessEffectivenr Regenerato

hhhhessEffectivenr Regenerato

TransferHeatr IdealerHeatTransf ActualessEffectivenr Regenerato

−−

=

−−

=

=

13

42is4is

Q WW

stagesingle2

2

stagesingle2is Intercooled Compression

k1k

1

212is

3443

3232

1221

ppTT

hhWhhQhhW

−

→

→

→

=

−=−=−=2

T

s

( )

( ) ( )

( )

( ) ( )

( )

( ) ( ) BTU/lb 77.68530851.95.24TTcW

R851.95.8

TTTT

R787.564510p

pTT

BTU/lb 66.1533.5132.64W

BTU/lb 33.51510649.61.24TTcW

R649.61.8

510621.69510.8

TTTT

R621.692510ppTT

BTU/lb 32.64530646.24TTcQ

R675.8

530646530.8

TTTT

R6462530ppTT

34pstagesingle

o1

stage single 2is

1stagesingle

o.2857

k1k

1

stage single 2

1stage single 2is

cooledinter

34p43

o34is34

o.2857k

1k

3

414is

32p32

o12is12

o.2857k

1k

1

212is

=−×=−=

=−

+=

=×=

=

=+=

=−×=−=

=−

+=−

+=

==

=

=−×=−=

=−

+=−

+=

==

=

−

→

−

→

−

1

2

3

42is4is

Q WW

2

stagesingle2is

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

T

s

1

2

3

4

6

7

8

9510T

Brayton Cycle with:IntercoolingRegenerationReheat s