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33
ME 576
• THERMODYNAMICS PRIMER• THE 3 T’s OF COMBUSTION• GAS POWER CYCLES• INTERNAL COMBUSTION ENGINES• CHEMICAL KINETICS• WASTE INCINERATION• DETONATION WAVES• LAMINAR PRE-MIXED FLAMES• DIFFUSION (NON PRE-MIXED) FLAMES• TURBULENT CHEMICALLY REACTING FLOW
MODELING• NUMERICAL SIMULATION OF COMBUSTION• ABLATION
HISTORICAL JOURNEY OF THERMO
Sir Willard Gibbs (circa 1900)
Prof. Laszlo Tisza (MIT)
Prof. Herb Callen (Univ. Penn.)
Prof. Mel Branch (CU Boulder, 1993)
Prof. K.R. Anderson, Ph.D.,P.E.4
Prof. Gary Koonce,Cal Poly Pomona, 1989
THERMODYNAMICS REVIEW
extensive,,
intensive,,
system of mass
;;
Properties Extensive vs.Intensive
Postulate)Nernst the(aka reached benot can K 0 T Zero,Absolute :Law 3
leirreversib , reversible;:Law 2
dependentpath ) (~
;~
:Law 1
re, temperatusame at the
are then theym,equilibriuin are and bodies if :Law 0
micsThermodyna of Laws
rd
nd
st
th
==
=
===
=
>=≥
=+=
==
SHU
shu
mm
Ss
m
Hh
m
Uu
T
dQdS
dWddQdU
TTT
BA
equilBA
5
IDEAL GAS RELATIONS
vp
vp
CCk
CCR
RTp
v
RTpv
mVv
mRTpV
/
/1
/
==
−===
==
=
γ
ρρ
RTRTupvuh
RTu
MWRRk
RC
Rk
kC
u
v
p
2
52
3
:Gases Ideal Monatomic
/1
1
=+=+=
=
=−
=
−=
6
SPECIFIC HEAT DEFINITIONS
• CONSTANT VOLUME CASE
• CONSTANT PRESSURE CASE
V
VV
CT
U
T
Q
UWQ
≡∂∆∂=
∂∂
∆=−0, NO MOVING BOUNDARIES
P
pp
CT
H
T
Q
HQ
pVUH
UUVpVpQ
UWQ
≡∂∆∂=
∂∂
∆=+=
−+−=∆+=
121122
7
IDEAL GAS ENTROPY CHANGE
v
Rdv
T
dTCds
T
dvp
T
duds
v
RTp
dTCdu
v
v
+=
∴
+=
=
=
Relation1st sGibb'
Law Gas Ideal
C of Definitionv
1
2
2
1
12
2
1
2
1
2
1
ln)(
)(
:affordsn Integratio
v
vR
T
dTTCsss
v
Rdv
T
dTTCds
v
v
+=−=∆
+=
∫
∫∫∫
8
IDEAL GAS ENTROPY CHANGE
∫ −=−=∆
−=
−=
=
=
2
1 1
2
12ln
)(
Relation 2nd sGibb'
/
Law Gas Ideal
of Definition
ely,Alternativ
p
pR
T
dTTCsss
p
dpR
T
dTCds
T
vdp
T
dhds
pdTv
dTCdh
C
p
p
p
p
9
IDEAL GAS ENTROPY CHANGE
• In order to proceed– Assume constant specific heats– Assume variable specific heats
• Use constant specific heats for monatomic (i.e. He) ideal gases where Cp ≠ Cp (T) or when Cp & Cv vary linearly over ∆T range of interest and use average values
• Use variable specific heats when ∆T during the process is large and thus, the Cp & Cv vary non-linearly within the ∆T range of interest
10
IDEAL GAS ENTROPY CHANGE
• Constant Specific Heats
• Average Specific Heats
• Variable Specific Heats
∫
∫
=−
=
+=−=+=∆
−=+=∆
2
1
12
0
21
1
2
1
2,
1
2
1
2,
1
2
1
2
1
2
1
2
)()()(
)()( Define
2;lnlnlnln
lnlnlnln
dTTCTsTs
dTTCTs
TTT
p
pR
T
TC
v
vR
T
TCs
p
pR
T
TC
v
vR
T
TCs
poo
T
p
avgavgpavgv
pv
o
11
IDEAL GAS ENTROPY CHANGE
• Table A-17 “Ideal Gas Properties of Air” tabulates T (K) vs. so(T) [kJ/kg-K]
• Thus, for variable specific heats
1
2
12ln
p
pRsss oo −−=∆
12
IDEAL GAS ISENTROPIC PROCESSES
avgvC
R
avgv
v
avgv
v
v
T
T
v
v
C
R
T
T
v
v
C
R
T
T
v
vR
T
TC
s
,
1
2
1
2
1
2
,1
2
1
2
1
2
1
2
1
2,
lnexplnexp
lnln
lnln0
const.) adiabatic & e(reversibl isentropic heats specific Const.
−
=
−=
−=
+=
=⇒
13
IDEAL GAS ISENTROPIC PROCESSES
1
2
1
1
2
,
,
,
,
,
,,
lyconsequent
11
) k use books (some ratioheat specific
Recall,
−
=
−=−=
∴
==≡
−≡
k
avgv
avgp
avgv
avgv
avgp
avgvavgp
v
v
T
T
kC
C
C
R
C
Ck
CCR
γ
“1st Isentropic ideal gas relationship”
14
IDEAL GAS ISENTROPIC PROCESSES
k
k
avgp
avgvavgpC
R
avgp
avgp
avgp
p
p
T
T
k
k
kC
CCR
p
p
T
T
p
p
C
R
T
T
p
p
C
R
T
T
p
pR
T
TC∆s
avgp
1
1
2
1
2
,
,,
1
2
1
2
1
2
,1
2
1
2
,1
2
1
2
1
2,
111Using,
lnexplnexp
lnln
lnln0
process isentropic andfor Next,
,
−
=
−=−=−
=
=
=
=
−==
“2nd Isentropic ideal gas relationship”15
IDEAL GAS ISENTROPIC PROCESSES
k
k
kk
k
kk
v
v
p
p
v
v
p
p
p
p
v
v
T
T
=
=
=
=
−−
−−
2
1
1
2
11
2
1
1
2
1
1
2
1
2
1
1
2
Finally,
“3rd Isentropic ideal gas relationship”
16
IDEAL GAS ISENTROPIC PROCESSES
• Summary
.
.
.1
1
constpv
constTp
constTv
k
k
k
k
==
=−
−
“Ideal Gas Isentropic Process Relationships”
k
vpvpW
v
v
p
p
T
T
v
v
p
pconstpv
kkk
k
kk
−−=
=
=
=
=
=
−−−
1
.;
1122
)1(
1
2
)1(
2
1
)1(
2
1
2
1
1
2
1
2
1
ρρ
17
VARIABLE SPECIFIC HEATS
• Start with an isentropic ideal gas process
:follows as volume"specific relative"
and pressure" relative" introduce wes,remedy thi To
ratio pressure oflieu in given
is ratio volumeerror when & trialrequires
ln
ln0
1
2
12
1
2
12
p
pRss
p
pRss
oo
oo
+=
−−=
18
VARIABLE SPECIFIC HEATS
2
1
1
2
2
1
1
2
211
1
2
.1
2
1
212
1
2
read4 Step
compute 3 Step
read 2 Step
at enter 1 Step
via findcan then we
process isentropican for
,given 17-A TablesAir
only offunction )/exp(
)/exp(
Pressure Relative
)/exp(
)/exp(exp
Ratio Pressure
T
pp
pp
p
T
pT
, & p, Tp
TRs
Rs
p
pp
Rs
Rs
R
ss
p
p
rr
r
r
o
o
consts
r
o
ooo
⇒
=⇒
⇒
⇒
==≡
=−=
=
cf. Cengel & Boles
19
VARIABLE SPECIFIC HEATS
11
22
1
2
.1
2
11
22
2
1
1
2
21
12
1
2
1212
/
/
Now,
only offunction
a is exp sinceonly offunction /
/
/ Ratio Volume
Thus,
analysis engine automotivein i.e.
of insteadknown is Sometimes
r
r
r
r
consts
o
rr
r
r
r
r
pT
pT
v
v
v
v
T
/R)(sT/T/pTpT
pT
pT
p
p
T
T
pT
pT
v
v
/pp/vv
==
==
====
=
20
VARIABLE SPECIFIC HEATS
air tables ofcolumn in
given is / :NOTE
read 4 Step
compute 3 Step
read 2 Step
with 17-A Tableenter 1 Step
find process, isentropican and ,Given
2
1
1
2
2
1
1
2211
r
rr
rr
r
v
pTv
T
vv
vv
v
T
T, v,Tv
=
⇒
=⇒
⇒
⇒
cf. Cengel & Boles
21
3 T’s of Combustion• Combustion efficiency can be further explained in terms of the three
T’s;– Time– Temperature– Turbulence
• Time (aka Residence Time, or time the gases reside in the combustion chamber) For example, in oil heat systems, the amount of time oil vapor has to combust (or reside in the flame front or burning zone) has been improved dramatically with the advent of flame retention burners. The primary difference between this type burner and the older conventional style burner is that the flame retention burners violently spin the air/fuel mixture resulting in better mixing
• This reduces the amount of excess combustion air necessary to insure each fuel droplet is completely surrounded by oxygen and burns completely
• As the amount of combustion air is reduced, efficiency increases
3 T’s of Combustion
• O2 readings within the manufacturer’s specifications document that the burner’s flame retention properties are operating as designed and engineered
• With gas fired systems, air and fuel mixing occurs inside the burner. Again, increasing the amount of time the fuel and air have the opportunity to thoroughly mix will help insure complete combustion
• Adjustable air intake shutters on atmospheric burners primarily control the velocity at which the fuel/air mixture moves down the burner throat or ‘mixing area’
3 T’s of Combustion
• Increasing the amount of time flue gases are in the heat exchanger also increases the amount of time for heat transfer to occur
• O2 and stack temperature readings within the manufacturer’s specifications document that the flue gases are moving through the heat exchanger at a velocity which allows time for maximum heat transfer while still permitting the introduction of additional combustion air
• As the temperature difference (∆T) between the source of heat and the material being heated increases, so does the rate of heat transfer. This heat transfer rate is measurable in forced air systems and boilers. By reducing the amount of combustion air introduced into the combustion process to the absolute minimum necessary, we increase the DT between the flame/flue gases and the distribution air or boiler water
3 T’s of Combustion
• Radiant heat energy is much more effective in transferring heat than convective/conductive heat transfer. It is primarily put off by the flame itself and is directly related to flame temperature. As excess air is reduced, the higher flame temperature generates more radiant heat energy
• The burn rate in combustion process is very sensitive to temperature. If flame temperature is increased by 10%, the rate of combustion more than doubles. Unfortunately, the same increase in flame temperature also increases production of NOx gases by more than 10 times when sufficient O2 is available
3 T’s of Combustion
• To estimate the actual flame temperature, continue the horizontal line from the top of fuel to the point where it intersects the line corresponding to the percent of excess air. From the point of intersection, draw a vertical line up to the point where it intersects the combustion air temperature line (80 DEG F); Then draw a horizontal line to the vertical axis – temperature rise
3 T’s of Combustion
• For example:– A natural gas burner operating with 25% excess air (4.5%
O2) has an estimated flame temperature of 2930 °°°°F plus 80 °°°°F (combustion air temperature) equals 3010 °°°°F
• Note: Deduct approximately 11% for gas and 4% to 6% for oil due to water vapor in the flue gases
• Note that in this example burning natural gas, a 25% excess air translates to a 4.5% O2 reading and a 20% excess air translates to a 4.0% O2 reading on the analyzer
• O2 readings within the manufacturer’s specifications document that the flame temperature has been maximized for the most efficient radiant heat production and transfer
3 T’s of Combustion• ‘Turbulation’ of the fuel, air and heat
source provides for more complete combustion by keeping these components in contact with each other for a longer period of time
• Agitation of flue gases in a heat exchanger serves to provide a continual circulation of hotter flue gasses in contact with the heat exchanger surfaces. Typically, heat exchanger surfaces have a wide variety of irregular surfaces incorporating bumps, ridges etc. to provide this effect. Boilers and domestic hot water heaters commonly have turbulators that provide for this mixing process. These surfaces also produce eddy currents that recirculate flue gases and increase
3 T’s of Combustion
• These ‘turbulators’ and irregular heat exchanger surfaces are designed to be most effective at the appliance’s full firing rate. Under firing a burner results in the smaller volume of flue gases taking the path of least resistance through the boiler thus reducing the scrubbing and mixing of the flue gases against the fire side of the heat exchanger
• Again, O2 readings within the manufacturer’s specifications verify that the burner is firing at full capacity, maximizing efficient heat transfer
33
THE CARNOT CYCLE
• The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature TH and a sink at temperature TL
• We can never get better than Carnot’s efficiency
H
L
Carnotth T
T−=1,
η
cf. Cengel & Boles
34
Example
Consider a Carnot cycle executed in a closed system with 0.003 kg of air. The temperature limits of the cycle are 300 K and 900 K, while the minimum and maximum pressures are 20 kPa and 2000 kPa, respectively. Determine the net work per cycle.
p
v
2
3
41
2000 kPa
20 kPa
35kJ 393.01960.003(2/3)
kJ/kg 1960
kJ/kg 196935
2000ln)900(287.0
kPa 935
900300
20kPa, 2000
lnlnln
3
2
3
2
900
30011
3232
32
4.0
4.1
1
3
4
4
32
3
2
2
3
2
3
2232
==∴==⇒=−
==
=
=
==
===
==
=−=−=
−−
−
−
−
W
wqwq
w
T
T
ppp
p
pRT
v
vRT
v
vvpw
qqw
T
T
HH
k
k
HH
HHCarnotnet
H
L
Carnot
η
η
36
AIR-STANDARD CYCLE ASSUMPTIONS
• The working fluid is air• Processes are reversible• Combustion process is replaced by a heat
addition process• Exhaust process is replaced by a heat
rejection process
GEOMETRIC PROPERTIES OF RECIPROCATING ENGINES
37speed crankshaft speed;piston mean 2
ratio radiuscrank length / rod/
ratio ebore/strok/
ration compressioolumecylinder v min.
olumecylinder v max.
Center Dead Bottom
Center Dead Top
===
====
==+
=
====
NLNS
alR
LBR
V
VVr
BDCBC
TDCTC
p
bs
c
cdc
44444 344444 21end lowat diesels marine end,high at Automobile
m/s) 15~8( fpm 300 ~1500
engines C.I. speed low large 9~5
engines, medium small, 4~3
engines C.I. speed low 0.5
engines, medium small, 1.2~ 0.8
Engines C.I. 24~12
Engines S.I. 12~8
Typ.
2
=
=
===
=
p
c
c
S
R
B/L
r
r
aL
p
p
S
S
4-STROKE ENGINE OPERATING CYCLES
38
1) INTAKE STROKE – STARTS WITH PISTON AT TDC AND ENDS WITH THE PISTON AT BDC DRAWING MIXTURE INTO THE CYLINDER, TO INDUCE MASS THE INLET VALVE OPENS SHORTLY BEFORE THE STROKE BEGINS, AND CLOSES AFTER IT ENDS
2) COMPRESSION STROKE – WHEN BOTH VALVES ARE CLOSED AND THE MIXTURE INSIDE THE CYLINDER IS COMPRESSED, AT THE END OF THE COMPRESSION STROKE COMBUSTION IS INITIATED AND THE CYLINDER PRESSURE RISES RAPIDLY
3) POWER STROKE (AKA EXPANSION STROKE) – STARTS WITH PISTON AT TDC AND ENDS AT BDC AS THE HIGH TEMP. , HIGH PRESS. GASES PUSH THE PISTON DOWN AND FORCE THE CRANK TO ROTATE, APPROX. 5 TIMES AS MUCH WORK IS DONE ON THE PISTON DURING THE POWER STROKE AS WHAT HAD TO BE DONE DURING COMPRESSION, AS THE PISTON NEARS BDC THE EXHAUST VALVE OPENS TO BEGIN THE EXHAUST PROCESS AND DROP THE CYLINDER PRESSURE CLOSE TO THE EXHAUST PRESSURE
4) EXAUST STROKE – THE REMAINING UNBURNED GASES EXIT THE CYLINDER, JUST AFTER TD THE EXHAUST VALVE CLOSES AND THE CYCLE REPEATS ITSELF
CF. HEYWOOD
2-STROKE ENGINE OPERATING CYCLES
39
1) COMPRESSION STROKE – BOTH INLET AND EXHAUST PORTS ARE CLOSED, THEN THE CYLINDER CONTENTS ARE COMPRESSED AND FRESH CHARGE IS DRAWN INTO THE CRANKCASE, AS THE PISTON NEARS TDC COMBUSTION IS INITIATED
2) SCAVENGING (AKA POWER AKA EXPANSION) STROKE – MOST OF THE BURNT GASES EXIT THE CYLINDER IN AN EXHAUST BLOWDOWN PROCESS, WHEN THE INLET PORTS ARE UNCOVERED THE FRESH CHARGE WHICH HAS BEEN COMPRESSED IN THE CRANKCASE FLOWS INTO THE CYLINDER
*EACH ENGINE CYCLE WITH ONE POWER STROKE IS COMPLETED IN ONE CRANKSHAFT REVOLUTION, BUT IT IS DIFFICULT TO COMPLETELY FILL THE DISPLACE VOLUME WITH FRESH CHARGE, AND SOME OF THE FRESH MIXTURE FLOWS DIRECTLY OUT OF THE CYLINDER DURING THE SCAVENGING PROCESS (THE FIGURE ABOVE SHOWS A CROSS-SCAVENGED DESIGN)
Reed springInlet valve
CF. HEYWOOD
40
COMPRESSION RATIO
FLpALSTROKEAREAMEPW
VV
WMEP
V
V
V
Vr
net
net
TDC
BDC
===−
=
==
**
Pressure EffectiveMean
Ration Compressio
minmax
min
max
cf. Cengel & Boles
cf. Cengel & Boles
MEP
• MEP = PnR/VdN– nR = number of crank revs for each power
stroke (nR = 2 for 4 stroke, 1 for 2 stroke engines)
– Vd=displaced volume– N = speed– P = cylinder pressure
41
MEP
• Formulas for MEP
42
][in
ft]-[lbf 4.75[psi]
][dm
m]-[N 28.6[kPa]
torque,of In terms
]rpm[ ][in
396,000*[hp] [psi]
]rev/s[ ][dm
10*[kW] [kPa]
3
3
3
3
3
d
R
d
R
d
R
d
R
V
TnMEP
V
TnMEP
NV
nPMEP
NV
nPMEP
=
=
=
=
MEP• Typ. Values for MEP
– Naturally aspirated spark-ignition engines 125 ~ 150 psi (850 ~1050 kPa) @ 3000 rpm where engine max. torque is reached
• At max. rated power, MEP values are 10~15% lower than above
– Turbocharged spark-ignition engines180 ~ 250 psi (1250 ~ 1700 kPa)
• At max. rated power 130 ~ 200 psi (900 ~ 1400kPa)– Naturally aspirated 4-stroke diesel engines 100 ~ 130 psi (700 ~
900 kPa)• At max. rated power ~100 psi (~ 700kPa)
– Turbocharged 4-stroke diesel engines 145 ~ 175 psi (1000 ~ 1200 kPa), for aftercooled MEP ~200 psi (~1400 kPa)
• At max. rated power 125~140 psi (850~ 950 kPa)
– Two-stroke diesel engines 225 psi (~1600 kPa)43
Example• A four cylinder automotive spark-ignition engine provides a max.
brake torque of 150 N-m (110 lbf-ft) in the mid-speed range of about 3000 rpm. Estimate the required engine displacement, bore, stroke and the max. brake power the engine delivers
44
hp 94kW 701000*2
87*2*800
1000*
**
kPa 800 power max. @ MEP Assume
rpm 5200rev/sec 87m/s 15 assume 2
mm 86,2 assume
4*4
engine,cylinder -four aFor
dm 2925
150*2*28.628.6
kPa 925 MEP Assume
maxmax
maxmax
3
2
3max
====
=
==⇒==
==⇒=
=
===
=
r
pp
d
r
n
NVMEPP
NSLNS
BBLB
LBV
MEP
TnV
π
π
45
PLAN FORMULA
• Four-stroke cycle– Intake– Combustion– Power– Exhaust
• A four-stroke cycle produces one power stroke for each piston for 2 revolutions of the main shaft
• i.e. a 4000 rpm 4-stroke engine gives N=4000/2 = 2000 power strokes/min
• Four stroke cycle produces one power stroke for each piston for two revolutions of the crankshaft
46
PLAN FORMULA
• Two-stroke cycle• In one revolution of the crankshaft
– Air intake– Compression– Expansion– Exhaust
• Every outward stroke of the piston is a power producing stroke
47
PLAN FORMULA
• PLAN Formula (ENG)
[ ] [ ] [ ] [ ]
[ ][ ][ ][ ] /minstrokespower ofnumber nstrokes/mipower
area sectoinal-crosspiston in
strokeft
psi
cylinder a ofoutput horsepower
minhpblfft
000,33
nstrokes/mipower inftpsi
2
2
==
==
=
−−
=
N
A
L
MEPP
HP
NALPHP
48
PLAN FORMULA
• PLAN Formula (SI)
• Example: compute the HP of a 2.0 L engine that has a MEP = 1 MPa operating @ 4000 rpm. The engine is a four-stroke engine
Solution: Four-stroke →N=4000/2=2000 power strokes/min
HP=1000 kPa(2 L) (2000 power strokes/min)/ (44760 N-m/min-hp)
HP =89.4 hp
[ ] [ ] [ ]
−−
=
hpminmN
760,44
nstrokes/mipower LiterskPa NLAPHP
49
PLAN FORMULA
( )
psia 117
minstrokespower
2000)in ft(42.412 3/1
psiahpin
minstrokespower
hp) 100(000,33
2)by divide engine, stroke-four a isit (becausemin
strokespower 2/4000
in 412.42cyl) 6(cyl 1
in 34ft, 3/1
in/ft 12
inches 4
000,33
:Solution
inches. 4 is stroke theand inches 3 is borecylinder
The rpm. 4000 @output hp 100 a has that engine
cylindersix stroke-four a of MEP theDetermine
Example
2
3
2
2
=
==
=
====
=
MEPP
N
AL
PLANHP
π
51
OTTO CYCLE
1-2: isentropic compression
2-3: isochoric heat addition
3-4: isentropic expansion
4-1: isochoric heat rejection
cf. Cengel & Boles
52
OTTO CYCLE
( )( )1/
1/1
11
)(
)(
0) PEKE system, (closed Energy ofon Conservati
232
141
,
23
14
,
1414
2323
−−−=
−−−=−==
−=−=−=−=
∆=−+−
=∆=∆
TTT
TTT
TT
TT
q
q
q
w
TTCuuq
TTCuuq
uwwqq
Ottoth
in
out
in
net
Ottoth
vout
vin
outinoutin
η
η
53
OTTO CYCLE
1,
3
4
1
4
3
1
1
2
2
1
1423
11
processes isochoric
are 1-4 and 3-2 since & also
process, isentropic are 4-3 & 2-1
−
−−
−=
=
=
=
∴
==
kOttoth
kk
r
T
T
v
v
v
v
T
T
v v vv
η
cf. Cengel & Boles
54
DIESEL CYCLE
• Ideal cycle for compression ignition engines
1-2: isentropic compression
2-3: isobaric heat addition
3-4: isentropic expansion
4-1: isochoric heat rejection
cf. Cengel & Boles
cf. Cengel & Boles
55
DIESEL CYCLE
)1/(
)1/(1
)(
)(1
)(
)(11
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)(
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232
141
23
14
,
23
14
,
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41
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23232
23
−−−=
−−−=
−−−=−==
−=−=−=−
−=−=−+−=
−=−
TTkT
TTT
TTk
TT
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TTC
q
q
q
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TTCuuq
uuq
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uuwq
Dieselth
p
v
in
out
in
net
Dieselth
vout
out
pin
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η
η
56
DIESEL CYCLE
−−−=
−−
==≡
− )1(
111
thusprocesses, gas ideal
isentropic are 43&21
combustion before volume
combustionafter volumeratio off-cut
Define
1,
2
3
c
kc
kDieselth
c
rk
r
r
v
vr
η
cf. Cengel & Boles
57
DUAL CYCLE
• Models the combustion process as a combination of a constant volume and a constant pressure process
cf. Cengel & Boles
58
Example
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17 C, and 800 kJ/kg of heat is transferred to the air during the constant-volume heat addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum T and p that occur during the cycle, (b) the net work output, (c) the thermal efficiency, (d) MEP
cf. Cengel & Boles
59
kPa7.17998290
4.652100
kJ/kg 475.12C, 4.652 :ionInterpolat
481.01 660 1.598
51.84
473.25 650 34.85
:17-A Table
51.848
1.6761
gas idealan ofn compressio isentropic :2-1
676.1kJ/kg, 206.9117-A TableK 290 (a)
21
1212
1
11
2
22
22
22
12
1
2
1
2
111
===⇒=
==
===⇒==
==⇒=
vT
vTpp
T
vp
T
vp
u T
uT
u T v
r
vv
rv
v
v
v
vuT
r
rr
r
r
r
o
60
MPa 345.414.652
14.15757.1799
6.103K, 1575.4 :affordsion Interpolat
6.046 1580 65.1279
12.1275
6.301 1560 99.1260
17-A Table
kJ/kg 12.1275
12.475800
additionheat olumeconstant v :3-2
32
23
23
2
22
3
33
33
33
3
3
23
===⇒=
==
=−=
−=
vT
vTpp
T
vp
T
vp
v T
v T
v T u
u
u
uuq
r
r
r
in
61kJ/kg 38291.20692.588
rejectionheat olumeconstant v :1-4
kJ/kg 588.92 K, 795.16:yieldsion Interpolat
592.30 800 48.08
8.8244
576.12 780 64.51
17-A Table
824.48)103.6(8
gas ideal of
expansion isentropic 4-3 cycle, during done Work (b)
14
44
44
343
4
3
4
=−=−=
==
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uuq
uT
u T
u T v
rvvrv
v
v
v
out
r
rrr
r
62
( )
kPa 2.574kJ 1
m-kPa 1)8/11(/kgm 832./kJ/kg 418
/kgm 832.100/290*287./
11
(d)
cycle Otto idealan smaller th is efficiency
i.e. realistic, more is assumptionheat specific variable
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1
sassumptionheat specificconstant under versus
%52800
418
kJ/kg 418382800
3
3
3
111
1
1
121
4.11
1,
=−=
===
−=
−=
−=
>=−=−=
===
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−
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pRTv
rv
w
r
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r
q
w
qqqw
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η
η
63
Example
An air standard Diesel cycle has a compression ratio of
18.2. The air is at 80 °F and 14.7 psia at the beginning of
the compression process and at 3000 R at the end of the
heat addition process. Accounting for variable specific
heats with temperature, determine:
(a) cut-off ratio
(b) Heat rejection/unit mass
(c) Thermal efficiency of the cycle
64BTU/lbm 05.402R, 1623.5
395.74 1600 8.263
7.93
409.13 1650 7.556
7.932144.32/18.
144.32 BTU/lbm, 92.04 R, 54017E,-A Table
ncompressio isentropic:21
22
22
11122
111
==
=======
−
h T
h T
h T v
/rv/vvvv
vuT
r
rrr
r
cf. Cengel & Boles
65%1.59
63.388
87/15811
BTU/lbm 87.15804.9291.250
processrejection heat olumeconstant v :4-1
BTU/lbm 19.2506212.11 17E-A Table
6212.11848.1/)18.1(2.18848.1/848.1//
expansion isentropic:43
BTU/lbm 53.38805.40258.790
18.1 BTU/lbm, 58.907 R, 000317E,-A Table
1848.05.1623
3000
additionheat isobaric:32
th
14
44
32343344
23
333
2
3
2
3
2
22
3
33
=−=−=
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−
=−=−====
===⇒=
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in
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r
rrrr
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r
q
q
uuq
uv
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v
v
T
vp
T
vp
η
66
% 59.1 efficiency thermal(c)
0.4089 ratiorejection heat (b)
0.1848 ratio off-cut (a)
:Summary
21
===
===
inout
c
r/vv
TURBOCHARGING
73
COMPRESSOR
INTERCOOLERCARBURETOR
INTAKE MANIFOLD
6 =INLET VALVE7 = EXHAUST VALVE8 = EXHAUST MANIFOLD
TURBINE
AIR FLOW PASSES THROUGH THE COMPRESSOR, INTERCOOLER, CARBURETOR, INTAKE MANIFOLD, AND INLET VALVEEXHAUST FLOWS THROUGH EXHAUST VALVE EXHAUST MANIFOLD, WHICH DRIVES THE TURBINE WHICH POWERS THE COMPRESSOR
11 = WASTEGATE LINKAGE CONTROLLED BY A BOOST PRESSURE REGULATOR
ENGINE INLET PRESSURES (AKA BOOST) OF ~ 100 kPa ABOVE ATM. PRESSURES ARE TYPICAL
CF. HEYWOOD
TURBOCHARGING
• TURBOCHARGING IS A FORM OF SUPERCHARGING IN WHICH A COMPRESSOR IS DRIVEN BY AN EXHAUST GAS TURBINE, THUS SUPPLYING PRESSURIZED AIR TO THE ENGINE
• BY PRESSURIZING THE AIR AT INLET TO THE ENGINE, THE MASS FLOW RATE OF AIR INCREASES AND THUS AN INCREASE IN THE FUEL FLOW RATE OCCURS, LEADING TO AN INCREASE IN THE POWER OUTPUT, AND AN IMPROVEMENT IN THE EFFICIENCY
74
TURBOCHARGING
• THE OVERALL IMPROVEMENT IN EFFICIENCY ULTIMATELY DEPENDS ON THE MATCHING OF THE SUPERCHARGER TO THE ENGINE
• SUPERCHARGING DOES NOT HAVE A SIGNIFICANT EFFECT ON EXHAUST EMISSIONS
• AXIAL AND RADIAL COMPRESSORS CAN BE DRIVEN BY A TURBINE, THUS FORMING A TURBOCHARGER
76
TURBOCHARGING• THE THERMODYNAMIC ADVANTAGE OF
TURBOCHARGERS OVER SUPERCHARGERS STEMS FROM THEIR USE OF THE EXHAUST GAS ENERGY DURING BLOW-DOWN AS SHOWN BELOW
77
CF. Stone
TURBOCHARGING
• THE CHARACTERISTICS OF TURBOCHARGERS ARE FUNDAMENTALLY DIFFERENT FROM THOSE OF RECIPROCATING I.C.E.’S, THUS LEADING TO MATCHING PROBLEMS WHEN THE TWO ARE COMBINED
• SUPERCHARGERS HAVE A MECHANICAL DRIVE, AND THE COMPRESSOR EFFICIENCIES CAUSE THE OVERALL ECONOMY TO DECREASE, BUT THE FLOW CHARACTERISTICS ARE BETTER MATCHED
78
TURBOCHARGING
• CHARACTERISTICS– RADIAL FLOW COMPRESSOR AND TURBINE– NO STATOR BLADES FOR SMALL
TURBOCHARGERS• OPERATION SATISFACTORY OVER A WIDER FLOW
RANGE, BUT WITH A LOWER PEAK EFFICIENCY
• WITH STATOR BLADES PRESENT, THE FLOW ANGLE FROM THE ROTOR TO THE STATOR HAS TO MATCH THE BLADE ANGLE, WHICH MEANS FOR EVERY VOLUMETRIC FLOW RATE, THERE IS ONLY ONE ROTOR SPEED AT WHICH THERE IS PROPER MATCHING (SEE NEXT CHART)
80
TURBOCHARGING
• RADIAL FLOW COMPRESSOR WORK WELL WITH OR WITHOUT STATOR BLADES, DUE TO THECONSERVATION OF THE MOMENT OF ANGULAR MOMENTUM IN THE DIFFUSER AND THE INCREASE OF THE FLOW AREA AS THE RADIUS INCREASES
82
THERMODYNAMICS OF TURBOCHARGING
• GOVERNING THERMODYNAMIC RELATIONSHIPS
( )( )
12
12
42
12
1212
4343
:ies EfficiencIsentropicnery Turbomachi
heat specific mean
:gasesperfect -semiFor
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rbinescharger tuflow turbo axialfor
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43
43
43
43
<η<
<η<
−−=
−−=η
<η<
t
t
sst
c
TT
TT
hh
hh
THERMODYNAMICS OF TURBOCHARGING
• IN A TURBOCHARGER, THE COMPRESSOR IS DRIVEN SOLELY BY THE TURBINE, CONSEQUENTLY, A MECHANICAL EFFICIENCY CAN BE DEFINED AS FOLLOWS
( )( )433434
121212
TTcm
TTcm
w
w
,p
,p
t
cmech −
−==η
&
&
86
THERMODYNAMICS OF TURBOCHARGING
• AS IN GAS TURBINES, THE PRESSURE RATIOS ACROSS THE COMPRESSOR AND TURBINE ARE PROFOUND AND CAN BE MODELED VIA THE ISENTROPIC RELATIONSHIPS
87
vp
s
s
c/c
p
p
T
T
p
p
T
T
=γ
=
=
γ−γ
γ−γ
1
4
3
4
3
1
1
2
1
2
THERMODYNAMICS OF TURBOCHARGING
• IN CONSTANT PRESSURE TURBOCHARGING, IT IS BEST TO HAVE THE INLET PRESSURE > THE EXHAUST PRESSURE (P2/P3 > 1) SO THAT GOOD SCAVENGING WILL OCCUR– SCAVENING = THE OPERATION OF CLEARING
THE CYLINDER OR BURNED GASES AND FILLING IT WITH FRESH MIXTURE (OR AIR) THIS COMIBINED INTAKE AND EXHAUST PROCESS IS CALLED SCAVENGING
• THIS PLACES A LIMIT ON THE OVERALL TURBOCHARGER EFFICIENCY η= η mechηt ηcFOR DIFFERENT ENGINE EXHAUST TEMPERATURES, T3 AS SHOWN ON THE NEXT CHART
88
TURBOCHARGING
• THE FLOW CHARACTERISTICS OF AN AXIAL VS. RADIAL FLOW COMPRESSOR ARE GIVEN ON THE NEXT CHART– IT CAN BE SEEN THAT THE RADIAL FLOW
COMPRESSOR HAS A WIDER RANGE OF OPERATION DUE AS STATED BEFORE WHEN CONSIDERING MATCHING ISSUES
• THE SURGE LINE MARKS THE REGION OF UNSTABLE OPERATION, WHERE FLOW REVERSAL, ETC. CAN OCCUR
• THE POSITION OF THE SURGE LINE IS INFLUENCED BY THE INSTALLATION OF THE ENGINE
90
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• COMPRESSION IGNITION ENGINE –NEAR THE END OF THE COMPRESSION STROKE, LIQUID FUEL IS INJECTED AS ONE OR MORE JETS– THE INJECTOR RECEIVES FUEL AT VERY HIGH
PRESSURES IN ORDER TO PRODUCE RAPID INJECTION WITH HIGH VELOCITY JETS OR SMALL CROSS-SECTIONAL AREA
– THE FUEL JETS ENTRAIN AND BREAK UP INTO DROPLETS, LEADING TO RAPID MIXING
92
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
– RAPID MIXING IS ESSENTIAL FOR THE COMBUSTION TO OCCUR SUFFICIENTLY FAST
– SOMETIMES THE FUEL JET IMPINGES ON THE WALL TO VAPORIZE AND BREAK UP THE JET
– THERE IS A LARGE VARIATION IN FUEL/AIR MIXTURES ON BOTH THE LARGE AND SMALL SCALES WITHIN THE COMBUSTION CHAMBER
93
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• PLATE 2 OF STONE PP. 78-79 SHOWS COMBUSTION IN A HIGH-SPEED DIRECT INJECTION DIESEL ENGINE
94
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• PRESSURE DIAGRAM FO COMPRESSION IGNITION ENGINE
96CF. Stone
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• RECALL, TURBOCHARGING IS USED TO INCREASE THE ENGINE OUTPUT BY INCREASING THE DENSITY OF THE AIR DRAWN INTO THE ENGINE
• THE PRESSURE RISE ACROSS THE COMPRESSOR INCREASES THE DENSITY, BUT THE TEMPERATURE RISE REDUCES THE DENSITY
97
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• THE LOWER THE ISENTROPIC EFFICIENCY OF THE COMPRESSOR, THE GREATER THE TEMPERATURE RISE FOR A GIVEN PRESSURE RATIO
• NOW, WE DEVELOP A THERMODYNAMIC MODEL…
98
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
99
η
−
+=
η=−−
=
γ−γ
γ−γ
c
cs
s
p
p
TT
TT
TT
p
pTT
1
1
Tfor solve
into
Plug
1
1
2
12
2
12
12
1
1
212
11
1
2
1
2
1
2
1
1
gas ideal anfor
−
γ−γ
η
−
+=ρρ
ρ=
c
p
p
p
p
RTp
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• THE EFFECT OF COMPRESSOR EFFICIENCY ON CHARGE DENSITY IF SHOWN BELOW
100
CF. Stone
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• THE EFFECT OF FULL COOLING (ISOTHERMAL COMPRESSION) IS SHOWN ON THE PREVIOUS CHART
• IT IS SEEN THAT THE TEMPERATURE RISE IN FTHE COMPRESSOR SUBSTANTIALLY DECREASES THE DENSITY RATIO, PARTICULARLY AT HIGH PRESSURE RATIOS
101
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• ENSURING THAT THE COMPRESSOR OPERATES IN THE EFFICIENT PART OF THE REGIME, MINIMIZES THE WORK INPUT AND THE TEMPERATURE RISE
• HIGHER ENGINE INLET TEMPERATURES RAISE THE TEMPERATURE THROUGHOUT THE CYCLE, AND WHILE THIS REDUCES IGNITION DELAY, IT INCREASES THE THERMAL LOADING ON THE ENGINE
102
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• THE ADVANTAGES OF CHARGE COOLING LEAD TO THE USE OF INTERCOOLERS, WHICH HAVE THE EFFECTIVENESS METRIC
103
( )( ) ( )
( ) ( )
η−ε−+=
ε+ε−
η+=
ε+ε−=
=−−==ε
γ−γ
γ−γ
c
/
c
/
p/pTT
p/pTT
TTT
T
TT
TT
111
11
1
gasesperfect for cooler,-inter thefromExit
mperatureambient teat or water)(air medium cooling
ferheat trans possible max.
ferheat trans actual
112
13
112
13
123
1
12
32
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• NEGLECTING THE PRESSURE DROP IN THE INTER-COOLER
• EFFECT OF CHARGE COOLING ON THE DENSITY RATIO IS SHOWN ON THE NEXT CHART FOR EFF. 70%, AND AMB. TEMP. = 20 C…
104
( ) ( ) 1112
1
3
1
3 111
−γ−γ
η−ε−+=
ρρ
c
/p/p
p
p
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• INTERCOOLING INCREASES THE AIR FLOW RATE AND WEAKENS THE A/F RATIO FOR A FIXED FUELING RATE– THE TEMPERATURES WILL BE REDUCED
THROUGHOUT THE CYCLE, INCLUDING THE EXHAUST STAGE
– THE TURBINE OUTPUT WILL THEN BE REDUCED, UNLESS IT IS REMATCHED
– THE REDUCED HEAT TRANSFER AND CHANGES IN COMBUSTION LEAD TO AN INCREASE IN BMEP AND A REDUCTION IN SPECIFIC FUEL CONSUMPTION ON THE ORDER OF 6% FOR P2/P1= 2.25 & ε =0.7
106
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• LOW HEAT LOSS ENGINES IMPROVE THE PERFORMANCE OF DIESEL ENGINES– FIRST, THERE IS IMPROVEMENT IN EXPANSION WORK AND
HIGHER EXHAUST TEMPERATURE, WHICH LEADS TO GAINS WHEN THE ENGINE IS TURBOCHARGED
– THE REDUCED COOLING REQUIREMENTS ALLOW FOR A SMALLER CAPACITY COOLING SYSTEM
– THE ASSOCIATED REDUCTION IN POWER CONSUMED BY THE COOLING SYSTEM IS A SIGNIFICANT PART OF THE LOAD
– REDUCING THE HEAT TRANSFER FROM THE COMBUSTION CHAMBER LEADS TO REDUCED IGNITION DELAY, AND HENCE REDUCED DIESEL KNOCK (SPONTANEOUS IGNITION OF UNBURNT GASES LEADS TO RAPID PRESSURE RISE WHICH IS AUDIBLE, CAUSED BY RESONANCES OF THE COMBUSTION CHAMBER WALLS), BUT THE HIGHER COMBUSTION TEMPEATURES WILL LEAD TO AN INCREASE IN NOX EMISSIONS
107
TURBOCHARGING THE COMPRESSION IGNITION ENGINE
• STUDY OF AUGMENTED HEAT TRANSFER ON DIESEL ENGINE
108
cf. Stone
115115
SPECIFIC IMPULSE
• A metric used to compare propulsion systems is the specific impulse, defined as
• Example, say a Nitrogen rocket nozzle has an exit velocity of 550 m/s (Ma = 550/340 = 1.62), then:
slbm
lbf
skg
N
g
V
m
FI
c
exitsp /
~/
~rate flow mass
thrust ===&
(kg/s)
N550
s-Nm-kg
1
m/s 550
2
==spI
116116
SPECIFIC IMPULSE
second ingper thrust N of kg 0.18 consume We
kg/s 18.0
(kg/s)N
550
N 100
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117117
SPECIFIC IMPULSE
consumed) is fuel of (lots lbm/s 161
(lbm/s)lbf
124.2
lbf 20,000
thenused, is thruster lbf 20,000 aSay
(lbm/sec)
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ityexit velocan h rocket witHydrogen aConsider
:Example
2
=
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VI
118118
THE GAS TURBINE FOR JET PROPULSION
• The turbojet engine used extensively for aircraft propulsion is a simple modification of the open gas-turbine cycle:
cf. Reynolds & Perkins
119119
THE GAS TURBINE FOR JET PROPULSION
• Hot combustion gases are expanded in the turbine only far enough to generate work to drive the compressor• The remaining energy is then converted into high velocity K.E. by expanding it in a nozzle downstream of the turbine• The jet thrust results from the difference in momentum of the entering air flow to the compressor and the high-velocity exhaust gases
)( 05 VVg
mF
c
−=&
120120
TURBOJET WITH AFTERBURNER
• Afterburners are used to provide greater nozzle exhaust velocities and substantially increase the engine thrust• The afterburner is essentially a reheat device:
cf. Reynolds & Perkins
121121
TURBOJET WITH AFTERBURNER
• Typically, specific impulses ~ 600 N/(kg/s) hold for modern jet engines• Afterburners increase this number somewhat, however, fuel consumption is greatly increased• Thus, afterburners are “cut in” only for short periods of time for high thrust
122122
THE RAMJET
• The ramjet is a device employed for aircraft propulsion at very high speeds• Uses the momentum of the on-rushing air for compression, thus eliminating any moving parts• Operates on the same principle as the turbojet, except for the use of a flow passage which converts K.E. of incoming flow into pressure
cf. Reynolds & Perkins
123123
THE RAMJET
• The incoming flow velocity is reduced in the inlet nozzle in process 0-1• Fuel is burned at constant pressure in process 1-2, thus increasing the temperature•Finally, the flow is expanded to a high velocity, V3 in the outlet nozzle•Due to the energy which has been added to the flow, the velocity at exit is greater than that at inlet
124124
THE RAMJET
• The net thrust is given as the change in momentum flux
•The ramjet is only feasible for high inlet-flow velocities, otherwise it would not be possible to achieve a high enough compression ratio in the inlet to counteract the losses incurred due to friction and engine irreversibilities
)( 03 VVg
mF
c
−=&
125125
THE CHEMICAL ROCKET ENGINE
• A rocket is a simple device whereby a fuel and an oxidizer are supplied to a combustion chamber (a.k.a. burner) at high pressure
127127
THE CHEMICAL ROCKET ENGINE
• The high-pressure products of combustion are then expanded through a nozzle to high velocities, thus producing thrust
• Specific Impulse ranges from 1500 ~ 3000 N/(kg/s)
4Vg
mF
c
&=
128128
THE NUCLEAR HYDROGEN ROCKET ENGINE
• Highest specific impulse is obtained for the lightest possible particles• Specific impulse ~ 9000 N/(kg/s)
cf. Reynolds & Perkins
130130
THE ION ENGINE
• Easily ionized element (e.g. cesium or xenon) is evaporated and then brought into contact with a high-temperature surface (grid)• Electrons are removed at the surface, and the cesium ions and electrons are accelerated by appropriate electric fields and then mixed to form a neutral plasma beam which is shot from the end of the engine• Specific impulses ~ 100,000 N/(kg/s) are possible• XIPS (Xenon Ion Propulsion System) Deep Space 1 NASA
131131
SPECIFIC IMPULSE
Propulsion Device Specific Impulse N/(kg/s)
[lbf/(lbm/s)]
Pros/Cons
Turbojet 600 [5886] Inoperable in space flight
Chemical Rocket 1500~3000[14,715~29,340]
Consume large amounts of expensive fuel
Hydrogen Nuclear 9000 [88,290] Important in space flights, risks of nuclear power source
Ion Propulsion 100,000[981,000]
Small flow rate, small thrust, used for accelerating a vehicleonce into deep space
Note: 9.81 N/(kg/s) = 1 lbf/(lbm/s)
134
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