Gas Power Cycles
Post on 26-Dec-2015
23 Views
Preview:
DESCRIPTION
Transcript
Presented by:Presented by:Damon OgdenDamon Ogden
22--1313--0101
ME 372 Thermodynamics II
Chapter 8Gas Power Cycles
ME 372 Thermodynamics II
Chapter 8Gas Power Cycles
Reciprocating EnginesReciprocating Engines
CH 8
Sections
8.4 - 8.6
Short VideoShort Video
• 3.3 Liter• V 12• 1965 Ferrari 275• Water Brake Dynamometer • Over 300 hp
Discussion TopicsDiscussion Topics
1. - History2. - General Knowledge3. - 4 Stroke Cycle4. - Otto Cycle5. - Diesel Cycle
HistoryHistory• Nikolaus Otto patented the “Otto Cycle”
Engine 1876Spark ignition engines use the Otto Cycle
• Rudolf Diesel received patent for “Diesel Cycle” Engine in 1892
Compression Ignition engines use theDiesel Cycle and hence the name “diesel” engines
How much do you know?How much do you know?
• What are are some differences between gasoline and diesel engines?
Ways to describe an engineWays to describe an engine
• Application• Design Geometry• Working Cycle• Fuel• Method of load control• Method of ignition
Piston -CylinderNomenclature
Otto / Diesel Cycle DifferencesOtto / Diesel Cycle DifferencesFour Stroke Cycle
Otto Cycle Diesel Cycle
Throttle Air Load Control Meter Fuel
Fuel/Air mix Working Fluid Air During Compression
Spark Start Combustion Inject Fuel
Four-Stroke CycleFour-Stroke Cycle
• Intake
http://www.howstuffworks.com/engine.htm
• Compression• Power• Exhaust
SuckSqueezeBangBlow
Thermodynamic Properties
Thermodynamic Properties
• The working fluid is air, and behaves as an ideal gas.
• Combustion processes are modeled as heat addition from an external source.
• Heat reject process to the surroundings is used to restore working fluid to its initial state.
• All processes are internally reversible.
Four Stroke Cycle P-V diagram
Intake Valve Opens
Combustion
Volume
Pressure Exhaust Valve Opens
Intake Valve Closes
Exhaust Valve Closes
TDC BDC
Intake
Exhaust
Power
Spark Ignition, Otto Cycle Engines
Spark Ignition, Otto Cycle Engines
• From < 1 hp to > 3000 hp• Cars• Trucks• Motor Cycles• Small Engines, i.e.;
Lawn Mower, • Mems
Spark Ignition Engine Characteristics
Spark Ignition Engine Characteristics
• Compression Ratio of 6 - 12=• RPM 900 - 8000• Power per unit Volume
3 - 60 kW / LiterFerrari in video @ 68 kW / liter
• Bore = .05 - .45 m
@TDC
@BDC
VV
BBDC
TDC
p
v
1
2
3
4
Otto Cycle
s = constant
A Quick Review of TermsA Quick Review of Terms• s = entropy, h = enthalpy, • u = internal energy• Isometric = Constant Volume• Isobaric = Constant Pressure• Isothermal = Constant Temperature• Isentropic = No irreversibilities,
Ideal process foradiabatic processes(adiabatic = no heat transfer)
Otto Cycle ModelOtto Cycle Model
• Intake and exhaust strokes not modeled
• Compression modeled as isentropic• Heat input modeled as isometric,
occurring at TDC• Expansion modeled as isentropic• Heat rejection modeled as isometric,
occurring at BDC
• Process 1-2: Isentropic Compression
• Apply First Lawq w u− = ∆
rvv
vv
r
r 11
2
1
2 ==
Since ∆s = 0
p
v
1
2
Otto Cycle AnalysisOtto Cycle Analysis
• Process 2-3 Isometric Heating• Apply First Law
q w u− = ∆
q u uin = −3 2
but v = constant, therefore w = 0
p
v
1
2
3
Otto Cycle AnalysisOtto Cycle Analysis
• Process 3-4 Isentropic Expansion• Apply First Law
q w u− = ∆
vv
vv
rr
r
4
3
4
3= =
p
v
1
2
3
4But since ∆s = 0
Otto Cycle AnalysisOtto Cycle Analysis
• Process 4-1 Isometric Cooling• Apply First Law
q w u− = ∆
q u uout = −1 4
p
v
1
2
3
4but v = constant, therefore:
Otto Cycle AnalysisOtto Cycle Analysis
Compression Ignition Diesel Cycle EnginesCompression Ignition Diesel Cycle Engines
• From < 50 hp to > 6000 hp• Cars• Trucks• Trains• Boats• Power Plants• Construction Equipment 1.47 L, 50 hp
Volkswagen
Combustion Ignition Engine Characteristics
Combustion Ignition Engine Characteristics
• Compression Ratios, rc of 12 - 23• Power per unit volume
2 - 26 kW / Liter• Bore .075 - 1 m• RPM 110 - 5000
Caterpillar
BBDC
TDC
@TDC
@BDC
VV
=cr 8 Liter
400 hp
Diesel Cycle ModelDiesel Cycle Model
• Intake and exhaust strokes not modeled
• Compression modeled as isentropic• Heat input modeled as isobaric,
occurring from TDC to appropriate volume
• Expansion modeled as isentropic• Heat rejection modeled as isometric,
occurring at BDC
T
s
1
2
3
4
p=const
v=const
Diesel Cycle
p
v1
23
4s=c
s=c
q w u− = ∆
rvv
vv
r
r 11
2
1
2 ==
p
v1
2
Since ∆s = 0
Diesel Cycle AnalysisDiesel Cycle Analysis• Process 1-2: Isentropic
Compression• Apply First Law
p
v1
2
Diesel Cycle AnalysisDiesel Cycle Analysis
• Process 2-3 Isobaric Heating• Apply First Law• q - w = ∆u• but p = constant, • therefore w = p ∆∆∆∆v• qin = h3 - h2
• Cutoff Ratio = rc
• rc = v3 - v2• Only process where Otto and Diesel Cycle differ
• Process 3-4 Isentropic Expansion
• Apply First Law• q - w = ∆u
Diesel Cycle AnalysisDiesel Cycle Analysis
p
v1
23
4•But since ∆∆∆∆s = 0
cr
r
rr
vv
vv ==
3
4
3
4
q w u− = ∆
Diesel Cycle AnalysisDiesel Cycle Analysis
• Process 4-1 Isometric Cooling• Apply First Law
• but v = constant, therefore:
p
v1
23
4
q u uout = −1 4
Comparison Of the Two Cycles
Comparison Of the Two Cycles
• Otto cycle utilizes an external energy source to initiate Combustion (spark plug)
• Diesel cycle relies on temp and pressure to start combustion
• Diesels only compress air, during compression stroke
• Otto cycle, spark engines compress both air and fuel during compression stroke
Comparison ContinuedComparison ContinuedFuel plays major role in cycle differences
• Spark ignition engines need fuels that are resistant to (knock)
• Diesel engines require fuels that will auto ignite under proper pressure and temperature
• Diesel fuel allows for higher Comp Ratios, = higher efficiencies.
Summary of Reciprocating Engines
Summary of Reciprocating Engines
• Been around for 125 years• Haven’t changed much:
Computers, Lower Emissions• Good for man, bad for the earth:• Not going anywhere soon• Future is Hydrogen, exhaust = H2O
Future = Mems Engines?Future = Mems Engines?
.001 kW
The EndThe End
Any Questions ?
Blank SlideBlank Slide
w u uon = −1 2
w u u p v vby = − + −3 4 2 3 2( )
q u uout = −1 4
q h hin = −3 2
w w w u u p v v u unet by on= + = − + − + −( ) ( ) ( )3 4 2 3 2 1 2
η thnet
in
wq
u u p v v u uh h
= = − + − + −−
3 4 2 3 2 1 2
3 2
( )
Diesel Cycle AnalysisDiesel Cycle Analysis• Combining above equations:
RTpv =uwq ∆=−
For isentropic processes:
rvv
vv
r
r 1
1
2
1
2 ==
in
netth q
w=η
vwMEP net
∆=
Equations:Equations:
Example: 9-2 Diesel CycleAn air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of compression, p1= 14.2 psi, V1 = 0.5 ft3, and T1 = 520oR. Determine:(a) the heat added, [Btu](b) the heat rejected, [Btu](c) the thermal efficiency(d) the Carnot efficiency(e) the mean effective pressure, [psi]
point 1 2 3 4p - psia 14.2
T oR 520v ft3/lb
u - Btu/lbh - Btu/lb
vr
V - ft3 0.5
Summary of Conditions - Diesel Cycle
r = 16, rc = 2
( )lbft
ftin
inlb
RRlb
lbftp
RTv
f
oo
f3
2
2
2
1
11
56.13
11442.14
52097.28
1545
=
==
Solution: (point 1)Solution: (point 1)
(isentropic compression from point 1:
911.916
58.15812
1
2
1
2 ==⇒== rr
r vrv
vvv
from air table, T2 is between 1480 & 1520 o
563.034.10578.934.10911.9
)1480()1520(
)1480(2 =−−=
−−
=rr
rr
vvvv
fr
( )( ) RT o150214801520563.014802 =−+=
Solution cont'd (point 2)Solution cont'd (point 2)
( )( )
lbBtu
u
84.266
44.26226.270563.044.2622
=
−+=
( )( )
lbBtu
h
4.369
89.36347.374563.089.3632
=
−+=
Point 2 cont'dPoint 2 cont'd
( ) 222
1
21
2
1
1
212
1
11
2
22
65616520
15022.14inlb
inlb
p
rTTp
vv
TTpp
Tvp
Tvp
ff =
=
==⇒=
Point 3
( )( ) RRT
rTvv
ppTT
Tvp
Tvp
oo
c
3005215023
22
3
2
323
2
22
3
33
==
==⇒=
Point 2 cont'dPoint 2 cont'd
from Air Table, T3 is between 3000 & 3050oRusing appropriate linear interpolation, determine
( )( )lb
Btuu 16.58604.58528.5961.004.5853 =−+=
( )( )lb
Btuh 15.79268.79034.8051.068.7903 =−+=
( )( ) 174.1180.1118.11.0180.13 =−+=rv
Point 3 cont'dPoint 3 cont'd
Point 4Point 4Isentropic expansion from point 3
( ) 390.90625.0
5.0174.13
434 =
==
VVvv rr
from table, 1520 & 1560oR, interpolation
273.0578.9890.8578.9390.9
)1520()1560(
)1520(4 =−−=
−−
=rr
rr
vvvv
fr
( )( )( )( )
lbBtuu
RT o
4.27226.27013.178273.026.270
153115201560273.01520
4
4
=−+=
=−+=
Point 1 2 3 4Start of Start of Start of Start of
Compressio Heating Expansion Cooling
p - psia 14.2 656T oR 520 1502 3005 1531
v ft3/lb 13.56u - Btu/lb 88.62 266.84 586.16 272.4h - Btu/lb 124.7 369.84 792.15
vr 158.58 9.911 9.39V - ft3 0.5 0.0313 0.0625 0.5
Summary of Conditions - Diesel Cycle
r = 16, rc = 2
Calculations:Calculations:(a)
lb
lbft
ftvVm 2
3
3
1
1 1069.356.13
5.0 −×===
lbBtuhhqin 31.42284.36915.79223 =−=−=
( ) Btulb
BtulbqmQ inin 57.153.4220369.0 =
==
BtuQin 57.15=
Calculations cont'dCalculations cont'd(b)
( ) ( )( )lb
BtulbuumQout 4.27262.880369.041 −=−=
BtuQout 78.6−=(c)
57.1578.611 −=−=
in
outth Q
Qη
564.0=thη
w u uon = −1 2
w u uby = −3 4
q u uout = −1 4
q u uin = −3 2
w w w u u u unet by on= + = − + −( ) ( )3 4 1 2
η thnet
in
wq
u u u uu u
= = − + −−
3 4 1 2
3 2
Otto Cycle AnalysisOtto Cycle Analysis• Combining above equations:
Example: 9-1At the beginning of the compression process in an air-standard Otto Cycle, p1 = 14.7 psi, T1 = 530oR. The compression ratio is 8. Determine for a maximum cycle temperature of 2000oR: thermal efficiency, Carnotefficiency, and mean effective pressure.
Point 1 2 3 4start start start start
of of of ofcompression heating expansion cooling
p-psia 14.7T oR 530 2000
v-ft3/lbu-Btu/lb
vr
Summary of Conditions
r = 8
Point 1 2 3 4start start start start
of of of ofcompression heating expansion cooling
p-psia 14.7T oR 530 2000
v-ft3/lbu-Btu/lb 90.3 367.61
vr 151.38 4.258
Summary of Conditions - data from air table
r = 8
RTpv =uwq ∆=−
For isentropic processes:
rvv
vv
r
r 1
1
2
1
2 ==
in
netth q
w=η
vwMEP net
∆=
Equations:Equations:
( )lbft
inlb
inftR
Rlblbft
pRTv
f
oo
f3
2
2
2
1
11 35.13
7.14
1441530
97.281545
=
==
lbftlbft
rvv
331
2 67.18
/35.13 ===
92.188
38.15112 ===
rvv r
r
Solution:Solution:
From air tables, note that T2 is between1160 and 1200 oR.doing the interpolation:T2 = 1191 oRu2 = 207.33 Btu/lb
Solution cont'dSolution cont'd
Point 1 2 3 4start start start start
of of of ofcompression heating expansion cooling
p-psia 14.7T oR 530 1191 2000 954
v-ft3/lb 13.35 1.67u-Btu/lb 90.3 207.3 367.61 164.2
vr 151.38 18.92 4.258 34.064
Summary of Conditions - calculations
r = 8
( ) ( )
net
net
wlb
Btu
wwwww
=
=+−++−=+++= −−−−
4.86
02.1647.36703.2073.9014433221
( ) in
in
qlb
Btuuuqq
==−
=−== −
3.1603.2076.367
2332
Solution - cont'dSolution - cont'd
thin
netth q
w ηη ==== 54.03.1604.86
Carnoth
cCarnot T
T ηη ==−=−= 74.0200053011
( )
2
2
2321
9.39
14467.135.13
7784.86
inlb
MEP
ftin
lbft
Btulbft
lbBtu
vvwMEP
f
f
net
=
−
=−
=
Solution - cont'dSolution - cont'd
Calculations cont'dCalculations cont'd(d)
300552011 −=−=
h
cCarnot T
Tη
827.0=Carnotη(e)
−
−
=−+=
−=
2
233
2121
144
778
0313.05.078.657.15
ftinBtulbft
ftftBtuBtu
VVQQ
VVWMEP
f
outinnet
2101inlb
MEP f=
Internal Combustion Engine Models
Internal Combustion Engine Models
• Spark Ignition Engine -Otto Cycle
• Compression Ignition Engine -Diesel Cycle
The piston moves the length The piston moves the length of the cylinder four times for one of the cylinder four times for one complete cycle.complete cycle.
Hence the name “four stroke” orHence the name “four stroke” or“four cycle engine”.“four cycle engine”.
Otto Cycle Model
Diesel Cycle Model
Piston-Cylinder NomenclaturePiston-Cylinder Nomenclature• Bore = B, Stroke = S• Bottom dead center = BDC• Top dead center = TDC• Mean Piston Speed = Sp = 2SN ,
N = engine speed in rev/s• Clearance volume = VTDC
• Displacement volume = VBDC - VTDC
• Compression ratio r = VBDC/VTDC
Method of Load ControlMethod of Load Control
• Throttle Air • (Otto cycle, spark ignition)
• Fuel Metering• (Diesel Cycle, compression
ignition)
T
s
1
2
3
4
p=const
v=const
Diesel Cycle
p
v1
23
4s=c
s=c
q w u− = ∆
Diesel Cycle AnalysisDiesel Cycle Analysis• Process 3-4 Isentropic
Expansion• Apply First Law
• But since ∆∆∆∆s = 0
p
v1
23
4
q w u− = ∆
cr
r
rr
vv
vv ==
3
4
3
4
p
v1
23
4
Diesel Cycle AnalysisDiesel Cycle Analysis• Process 3-4 Isentropic
Expansion• Apply First Law
• But since ∆∆∆∆s = 0
q w u− = ∆
q u uout = −1 4
p
v1
23
4
Diesel Cycle AnalysisDiesel Cycle Analysis• Process 4-1 Isometric Cooling• Apply First Law
• but v = constant, therefore:
Intake Stroke(Suck)
Intake Stroke(Suck)
• Intake valve open
• Exhaust valve closed
• Piston moves from TDC to BDC drawing in mixture of fuel and air (Only air for diesel cycle)
Compression Stroke(Squeeze)
Compression Stroke(Squeeze)
• Intake valve closes
• Exhaust valve remains closed
• Piston moves from BDC to TDC compressing fuel/air mixture, or air only
Otto Cycle
• Near top of stroke, spark ignites fuel/air mixture, causing heat input
Diesel Cycle
• Fuel injected into compressed air, in the cylinder
Power Stroke(Bang)
Power Stroke(Bang)
• Both valves closed
• Hot combustion products expand causing piston to move from TDC to BDC, producing work output
Exhaust Stroke(Blow)
Exhaust Stroke(Blow)
• Exhaust valve opens
• Intake valve remains closed
• Piston moves from BDC to TDC, forcing spent combustion products out of cylinder
ApplicationsApplications
• Auto• Semi-Truck• Locomotive• Marine• Aircraft• Stationary Power
Design Geometry Operating CycleDesign Geometry Operating Cycle
• V• Inline• Opposed• Rotary• Radial
• 4 stroke(There are others, but
those will be for a discussion)
FuelsFuels
• Gasoline• Diesel• Natural Gas• Liquid Propane• Methanol• Mixed gases
top related