thermodynamica 1 - tu delft ocw · march 15, 2010. 1. thermodynamica 1. lecture 11: processtappen....
TRANSCRIPT
March 15, 2010
1
Thermodynamica 1
Lecture 11: ProcesstappenKringprocessen
• Stirling• Otto (2 en 4 slags)
Bendiks Jan BoersmaWiebren de JongThijs VlugtTheo Woudstra
Energy Technology
Energy Technology
Lecture 11, March 15, 2010 2
Recap College 10• Carnot process• working with the entropy
• Carnot process
2 isothermal and 2 adiabatic processesreversible isentropic=adiabatic
Carnotreal
h
cCarnot T
TTT
114
3
Energy Technology
Efficiency Stirling cycle
Lecture 11, March 15, 2010 4
pTpT
pTpT
vvRTTcT
vvRTTcT
vvvvTTTT
vvRTTTcTTcvvRT
QQQQ
QW
h
c
v
v
v
v
in
uit
in
11
)/ln()()/ln(
)(
1
//)isothermenzijn 4-3en 2-(1
)/ln()()()/ln(111
3
1
43
233
21
141
4312
2314
43332
14211
3423
4112
Energy Technology
Stirling met regenerator
Lecture 11, March 15, 2010 5
1 2 1 2 1
3 4 3 4 3
4 1 3 2
2 1 3 4
1
3
| | | | | ln ( / ) |1 1 1ln ( / )
(1 -2 en 3 -4 z ijn iso th e rm en )/ /
1 1 M et reg en e ra to r d u s C a rn o t!
u it
in in
c
h
Q Q R T v vWQ Q Q R T v v
T T T Tv v v v
TTT T
2341 d toegevoersysteemhet aan weer regenrator wordt via QQ
Energy Technology
Lecture 11, March 15, 2010 6
Verbrandingsmotoren bv Otto motor
2 adiabaten en 2 isochoren, dus geen Carnot machine
en Carnot cyclus!
Energy Technology
Lecture 11, March 15, 2010 7
Standaardlucht aannames (air standard):
• Werkmedium: lucht als ideaal gas• Kringproces is gesloten• Deelprocessen zijn inwendig reversibel• Warmtetoevoer i.p.v. verbrandingsprocessen• Warmteafvoer i.p.v. uitdrijfproces
2
2 112
2 1 2 11
; ( )
( )
p
p p p
p p
dh c dT
h h c dT c c T
h h c dT c T T
Gebruik dus tabel A-22
Energy Technology
Lecture 11, March 15, 2010 8
Koude lucht aannames (cold air standard):
• Werkmedium: lucht als ideaal gas, Cp en Cv zijn constant• Kringproces is gesloten• Deelprocessen zijn inwendig reversibel• Warmtetoevoer i.p.v. verbrandingsprocessen• Warmteafvoer i.p.v. uitdrijfproces
2
2 1 2 11
( )
p
p p
dh c dT
h h c dT c T T
Gebruik dus geen tabel
Energy Technology
Lecture 11, March 15, 2010 9
Ideal gas (cold air standard)
1 *
1 11 1 2 2
11
1 2 1
2 1 2
1 **
11 1
2 2
/
/
P V C onstT V C onst
P m R T V
T V T V
T V VT V V
P V C onstP T C onst
V m R T P
P TP T
Energy Technology
Lecture 11, March 15, 2010 10
Isentroop process (example 6.9)
1
1
2
2
2 2 2
1 1 1
2 1/( 1) 1.39 /.39
2 2 2
1 1 1
1300650?
21.86Air Standard: ( 22)1.3860
15.77 15.77
650Cold air Standard: 15.81300
r
r
k k
p barT KT Kp
p p pT Ap p p
p p bar
p T p barp T p
Energy Technology
Lecture 11, March 15, 2010 11
Recap: reversible process steps in closed systems
isotherm (T=const.)
isochor (v=const.)
isobar (p=const.)
adiabatic (dQ=0)
polytropic(pvn=const.)
W12 2
1
v
vdvpmQ12
1
2
1
2 lnlnvvRT
ppRT QW 1212
0
0
121 vvpm
12 uum
11221vpvp
nm
12
2
1
uumdTcmT
Tv
12
2
1
hhmdTcmT
Tp
Wuum 1212
ideal gas ideal gas
Energy Technology
Lecture 11, March 15, 2010 12
Recap: reversible process steps in closed systems
isotherm (T=const.)
isochor (v=const.)
isobar (p=const.)
adiabatic (dQ=0)
polytropic(pvn=const.)
W12Q12p
V
W121
2
T
S
Q121
2 pvp
V
2
1
T
S
1
2
p
V
W121
2
p
V
1
2
for 1>n>kW12T
S
1
2 Q12
for 1>n>k
T
S
Q1212
pv
p
V
21 W12
T
S
Q12
1 2
T1T2
T2
T1
012 Q
012 W
Energy Technology
Lecture 11, March 15, 2010 14
.
NOx- emission and efficiency of combustion engines
Energy Technology
Lecture 11, March 15, 2010 15
Q23
Q41
Recap: Otto cycleT
S
p
V
2
1 T1
3
4 T4 1
2
p
v
4
3
2323 uumQ 023 W
4141 uumQ 041 W
012 Q
034 Q
1212 uumW
3434 uumW
internally reversible processes1-2
adiabatic
compression
of the air as the piston moves from
bottom
dead
center to top dead
center
2-3
isochoric
heating
of air from
external
source
–
to mimic
the ignition
and oxidation
of fuel-air
mixture
3-4
power stroke: adiabatic
expansion
4-1
isochoric
heat transfer
to external
reservoir –
to mimic
the release of hot exhaust
Energy Technology
Lecture 11, March 15, 2010 16
Efficiency Otto cycle (cold air standard)
thermal efficiency
Poisson relation
3241 , VVVV
1 .k
p
v
TV constc
kc
1
2
1
1
2
k
VV
TT
2 1 4 312 34 4 1
23 23 3 2 3 21
cycrev u u u uW W u uW
Q Q u u u u
1
2
1
2
1
1
3
4
4
3
TT
VV
VV
TT
kk
4 1 1 4 1 1
3 2 2 3 2 2
4
3
11 1 1
1
=1-
vrev
v
c T T T T T Tc T T T T T T
TT
1
4
2
3
TT
TT
Energy Technology
Lecture 11, March 15, 2010 17
rev
21 VVr
Efficiency Otto cycle (cold air standard)
thermal efficiency
with compression ratio 1
2
VrV
4 1 1
3 2 2
( )1 .. 1( )
rev v
v
C T T TC T T T
11
1
2
2
1
k
TT
VVrkpV const
.1 constTV k
1
11revkr
1 5 10
Energy Technology
Lecture 11, March 15, 2010 18
The efficiency monotonically increases with the compression ratio
in practice, high compression ratios leads to high temperatures, resulting{1} in NOx emissions (oxidation of air-nitrogen, severe at T>1600 K){2} in “knock”, the spontaneous ignition of air-fuel mixture prior to spark plug{3} incomplete combustion
compression ratios of r10 are therefore common (efficiency <60%)
1
11revkr
Energy Technology
Lecture 11, March 15, 2010 19
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 2 4 6 8 10 12 14 16
Compressie verhouding, r [-]
Ren
dem
ent [
-]
Mas
erat
i Qua
trop
orto
3.2
(ren
d.=0
.41;
247
kW
)
Pors
che
Car
rera
(ren
d.=0
.50;
235
kW
)
Lam
borg
ini M
urci
elag
o (r
end.
=0.4
5; 4
26 k
W)
Lotu
s Es
prit
(ren
d.=0
.52;
260
kW
)
Mer
cede
s SL
(ren
d.=0
.50;
350
kW
)
Ope
l Ast
ra 1
.6 (r
end.
=0.2
3; 5
5 kW
)
Volk
swag
en G
olf 1
.6 (r
end.
=0.2
9; 7
7 kW
)Pe
ugeo
t 306
1.6
(ren
d.=0
.29;
80
kW)
Angulo-Brown et al [1994]
9.3
F. Angulo-Brown et al [1994] “ Compression ratio of an optimized air standard Otto-cycle model”, Eur. J. Physics, Vol. 15, pp. 38-42
irreversible processes: heat transfer processes, friction in cylinders and dissipation in cycle, non-isochoric processes
1
11 krev
r
21 VVr
Efficiency Otto cycle
irrev
Energy Technology
Lecture 11, March 15, 2010 20
1=5
4
3
2
dual cycle
6
4-stroke Otto cycle
Ideal process steps of 4-stroke Otto cycle1-2 adiabatic compression of air (1st stroke)2-3 isochoric heating, to mimic the ignition and oxidation of fuel-air mixture3-4 power stroke: adiabatic expansion (2nd stroke)4-5 isochoric heat transfer, to mimic opening of exhaust valve5-6 exhaust stroke (3rd stroke): isobaric displacement of exhaust gas6-1 intake stroke (4th stroke): isobarically drawing fresh air into the cylinder
p
V
Energy Technology
Lecture 11, March 15, 2010 21
Dual cycle
1=5
4
3
2
dual cycle
6
4 stroke dual cycle
Energy Technology
Lecture 11, March 15, 2010 22
34 45 32 51
23 34 23 34
23 3 2
34 4 3
51 5 1
5 14
3 2 4 3
1 1
( )( )
( )
1( )
/
out
in in
v
p
v
tact
p v
W W W Q QWQ Q Q Q Q Q
Q mc T TQ mc T T
Q mc T TT T
T T k T Tk c c
Efficiency of dual cycle
Energy Technology
Lecture 11, March 15, 2010 23
Tentamen juni 2010We willen een gasmotor gebruiken om 3MW elektriciteit op te wekken. De compressie verhouding van de gasmotor is 12 en het cilinder volume aan het begin van de arbeidsslag is 2 liter. Er is gegeven dat Cp =1.0KJ/kgK en Cv=0.714KJ/kgK en dat de laagste druk in het systeem gelijk is aan 1 bar
• Schets het p-v en T-s diagram voor de Otto cyclus. • Bepaal het volume V1, als gegeven is dat T1=300K bepaal dan T2.• Tijdens het proces 23 neemt de druk in de motor toe met een factor 2.5.• Hoe groot zijn dan T3 en T-4?• Bepaal de warmte toe en afvoer aan het systeem Q23 en Q41?• Bepaal de geleverde arbeid en het rendement van de motor