Getting Ready for an Internal Audit – Cycle 2 A Review of Internal Controls 1.
Lecture 40: Air standard cycle, internal combustion...
Transcript of Lecture 40: Air standard cycle, internal combustion...
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1.1
ME 200 –Thermodynamics I Spring 2016
Lecture 40: Air standard cycle, internal combustion
engines, Otto cycle
Yong Li
Shanghai Jiao Tong University
Institute of Refrigeration and Cryogenics
800 Dong Chuan Road Shanghai, 200240, P. R. China
Email : [email protected]
Phone: 86-21-34206056; Fax: 86-21-34206056
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1.2
Air Standard Cycles
Air standard cycles are idealized cycles based on the
following approximations: A fixed amount of air modeled as an ideal gas (working fluid).
The combustion process is replaced by a heat transfer from an
external source. There are no exhaust and intake processes as in
an actual engine.
The cycle is completed by a constant-volume heat transfer
process taking place while the piston is at the bottom dead center
position.
All processes are internally reversible.
Cold air-standard analysis The specific heats are assumed constant at Ta.
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1.3
Continue Air Standard Cycles
Cycles under consideration:
» Carnot cycle – maximum cycle efficiency
» Otto cycle – Spark-ignition engine (SI engine)
» Diesel cycle – Compression-ignition engine (CI engine)
» Dual cycle – modern CI engine
» Brayton cycle – gas turbines
» Other cycles that we want to analyze:
– Stirling cycle
– Ericsson cycle
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1.4
Ideal Gas Model Review
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1.5
Ideal Gas Model Review
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1.6
Polytropic Processes on p–v and T–s Diagrams
cpvn
cpn 0
cvpn /1
11 cTcpvRTn
cscpvkn k
cvn
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1.7
Continue Air Standard Cycles
Air Standard Carnot Cycle:
s s1=s2 s3=s4
T
TH
TL
1
2 3
4 qL
p3
p2
p4
p1
qH
wnet
H H 23
L L 41
net H L
H Lnetth,Carnot
H H
L L 41
th,Carnot
H H 23
Lth,Carnot
H
q T s
q T s
w q q
q qw
q q
q T s1 1
q T s
T1
T
Question: How to have isothermal heat transfer with air
as the working fluid?
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1.8
Continue Air Standard Carnot Cycle
Answer: Must have work!
For an ideal gas, is T = constant, then
p v = constant!
v
p
1
2
3
4
p3
p2
p4
p1
TL = const.
v2 v1 v3 v4
TH = const.
s = const.
s = const.
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1.9
Continue Air Standard Carnot Cycle
Physical Devices:
2
3 4
wnet
isentropic
compression
isentropic
expansion
isothermal
compression
isothermal
expansion
1
1
2
3
2
4
3
4
1
qL
qL
qH
qH
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1.10
ME 200 –Thermodynamics I Spring 2015
Otto Cycle
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1.11
Otto Cycle--Internal Combustion Engine
The stroke ::: the distance the piston moves in
one direction
Bore ::: cylinder diameter
Top dead center ::: a position when the piston
moves to the minimum cylinder volume
Clearance volume ::: minimum volume
Bottom dead center ::: the position when the
piston moves to the maximum cylinder volume.
Displacement volume ::: The volume swept out
by the piston as it moves from the top dead
center to the bottom dead center position.
Compression ratio r ::: the volume at bottom
dead center divided by the volume at top dead
center max BDC
min TDC
V Vr
V V
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1.12
Otto Cycle--Internal Combustion Engine
Ignition
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1.13
Continue Internal Combustion Engines
Spark ignition (SI):
» combustion initiated by spark
» air and fuel can be added together
» Light and lower in cost, used in
automobiles
Compression ignition (CI):
» combustion initiated by auto
ignition
» requires fuel injection to control
ignition
» large power, heavy trucks,
locomotives, ships
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1.14 An automotive engine with the
combustion chamber exposed
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1.15
Spark ignition (SI):
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1.16
Continue Internal Combustion Engines
Mean effective pressure, MEP
Notes:
» MEP would produce the same net work
with constant pressure as for actual cycle
(includes both expansion and
compression)
» want high MEP (high power density)
net
max min
W net work for one cycleMEP
V V displacement volume
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1.17
Otto Cycle
The ideal cycle for spark ignited engines:
1
2
3
2 3
4
1
4 qL qH win wout
TDC
BDC
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1.18
Continue Otto Cycle
Ideal Otto cycle
» 1-2 Isentropic compression
» 2-3 Constant-volume heat addition
» 3-4 Isentropic expansion
» 4-1 Constant-volume heat rejection
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1.19
Continue Otto Cycle
Otto cycle energy balances:
» Assuming:
– W = 0 during heat transfer processes, KE = PE = 0
– Air is ideal gas, constant specific heats
wnet = wout – |win| = qin – |qout|
qin = u3 – u2 = cv (T3 – T2)
qout = u4 – u1 = cv (T4 – T1)
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1.20
Continue Otto Cycle
Otto cycle efficiency:
net in out out
th
in in in
w q q q1
q q q
pk
v
cFor isentropic process : pv constant, with k
c
k k
1 1 2 2For process 1 2 : p v p v
2
k
1 2 2 12
k12 1 1 2
1
RT
v p T vv
RTv p T v
v
k 1k k 1
1 2 2 1 1
k k 1
2 1 1 2 2
v v T v v
v v T v v
4
1
3
2
v 4 1 1
v 3 2 2
T
T
T
T
1c (T T ) T
1 1c (T T ) T 1
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1.21
Continue Otto Cycle
Continue Otto cycle efficiency:
k 1k 1
k 12 1 BDC
1 2 TDC
T V VFor m = constant: r
T V V
For process 3 4, using the same analysis:k 1 k 1
k 13 4 BDC
4 3 TDC
T V Vr
T V V
2 3 3 4
1 4 2 1
T T T TThen, or
T T T T
th k 1
11
r
max BDC
min TDC
V Vr
V V
Compression ratio
=1
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1.22
Continue Otto Cycle
Question: Why increase in th with increase in r?
Typical r for gasoline
engines
th k 1
11
r
Answer: Increase in r results in:
» an increase in T for heat addition and
» a decrease in T for heat rejection
Note: Need to consider material
limitations as function of T and p
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1.23
Example 1: Air-Standard Otto Cycle
Known: r = 8, state 1: p1 =100 kPa, T1 =290 K, qin = 800 kJ/kg
Find: p, T and v at all state points, wnet and ηth, MEP
Assumptions: 1) Air-Standard assumptions. 2) ΔKE=ΔPE =0.
3)Variable cp
Analysis:
cold-air-standard assumptions
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1.24
net in out outth
in in in
w q q q1
q q q