Internal Combustion Engines - SENS
Transcript of Internal Combustion Engines - SENS
1
BottomDeadCenterBDC
TopDeadCenterTDC
1V
2V
2
1c V
Vr Ratio,n Compressio =
Internal Combustion Engines
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
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
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
5
4 CYCLE, 2 REVOLUTIONS
Revolution 1 Revolution 2
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
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
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
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
10
( )( )
( )
( )
( )
%1.516
11
vv
11η
51.1%361.4
100.3285.2q
wη
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
∫
∫
====→
===→
====→
===→+=
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)
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
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
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
wη
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.
15
2 CYCLE, 1 Revolution/cycle
Compression1-2
Ignition2-3
Power3-4
Exhaus4-1
Intake1
p
v1
2
3
4
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
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
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
=
=+=
−
=
=
19
Stirling and Erickson Cycles
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
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
22
Propulsion Gas Turbine
23
inQ
turbinecompressorWW = outputnet
W
Aero-derivative Gas Turbine
PowerTurbine
GasGenerator
24
Figure 9.4 Air Standard Brayton Cycle
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
−−
=
=
−−
=
=
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
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
=−=
=∑ ∑
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
=
29
24
64
24
25
hhhhessEffectivenr Regenerato
hhhhessEffectivenr Regenerato
TransferHeatr IdealerHeatTransf ActualessEffectivenr Regenerato
−−
=
−−
=
=
Regenerated Brayton Cycle
30
1
2
3
4
6
7
8
9510T
s
Brayton Cycle with:IntercoolingRegenerationReheat
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
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
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
3413
42is4is
T
k1k
1
212is
3443
3232
1221
ppTT
hhwhhqhhw
−
→
→
→
=
−=−=−=
Q
WW
Intercooled Compressionstagesingle2
2
stagesingle2is
inletpressure
dischargepressure
s
T
inletpressure
dischargepressure
s
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
36
INTEGRAL GEARCENTRIFUGAL COMPRESSOR
37
Intercooled Supercharger
TurbochargerSupercharger
Q
1
2 3
4
38
Turbocharger
W
1
2 3
4
1
2
3
4enginecompressionstroke
combustionv=constant
enginepowerstroke
TP atm
Air Standard Otto Cyclewith a Turbocharger
39
ProCharger Supercharger
40
TurboCharger
CompressedAir to Engine
EngineExhaust
Air Inlet