Chapter 3 Thermochemistry of Fuel Air Mixtures 3-1 Thermochemistry 3 2 Ideal Gas Model 3-2 Ideal Gas Model 3-3 Composition of Air and Fuels 3 4 C b ti St i hi t 3-4 Combustion Stoichiometry 3-5 The1 st Law of Thermodynamics and Combustion 3-6 Thermal conversion efficiency 3-7 Chemically Reacting Gas Mixtures 1
73
Embed
Chapter 3 3... · Chapter 3 Thermochemistry of Fuel Air Mixtures 3-1 Thermochemistry 3-2 Ideal Gas ModelIdeal Gas Model ... 3-5 The1st Law of Thermodynamics and Combustion
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Chapter 3pThermochemistry of Fuel Air Mixtures3-1 Thermochemistry3 2 Ideal Gas Model3-2 Ideal Gas Model3-3 Composition of Air and Fuels3 4 C b ti St i hi t3-4 Combustion Stoichiometry3-5 The1st Law of Thermodynamics and Combustion3-6 Thermal conversion efficiency3-7 Chemically Reacting Gas Mixtures
1
• The composition and thermodynamic properties of the pre- and postcombustion working fluids inthe pre and postcombustion working fluids in engines
The energy changes associated with the• The energy changes associated with the combustion processes
2
An “ideal” gas exhibits certain theoretical propertiesAn ideal gas exhibits certain theoretical properties. Specifically, an ideal gas …Does not condense into a liquid when cooledDoes not condense into a liquid when cooled.Shows perfectly straight lines when its V vs. T & P vs. T relationships are plotted on a graphT relationships are plotted on a graph.In reality, there are no gases that fit this definition perfectly We assume that gases are ideal to simplifyperfectly. We assume that gases are ideal to simplify our calculations.
If temperature is not too low or pressure not to high, the ideal gasmodel can be applied.
Gas species (working fluids) in engines (e.g., O2, N2, fuel vapor CO water vapor etc ) may be treated as idealvapor, CO2, water vapor, etc.) may be treated as ideal gases.
R~ TRnTMRmmRTpV ~===
where m = mass of gasR = gas constant (J/kg⋅K)n = number of moles~ = Universal gas constant
8.3143 J/mol⋅K or 1,543.3 ft⋅lbf/lb-mole⋅°RR~
4
M = Molecular weight
•Air typically contains about 1-3 percent by mass of vapor
irair
Nor
mal
Ai
norm
al a
ion
from
%
var
iati
%
5
TABLE 3.1 Principle constituents of dry air
Gas ppm by volume
Molecular weight
Mole fractions
Molar ratio
O2 209,500 31.998 0.2095 12 ,
N2 780,900 28.012 0.7905 3.773
A 9 300 38 948A 9,300 38.948
CO2 300 44.0092
Air 1,000,000 28.962 1.0000 4.773
6
Atmospheric Nitrogen
2222
~~aNaNOOair MXMXM +=
iX~ is a mole fraction of gas i with respect to total number of mole of the mixture gas.
where
22
2 ~
~OOair
aN X
MXMM
−=
2aNX
209501998.312095.0962.28 ×−
=
Molecular weightof air
2095.01−
16.28=2aNM
7
2
TABLE 3-2 Molecular weights
8
The fuels most commonly used in IC engines are blends of many
different hydrocarbon compounds obtained by refining petroleum or
crude oil. These fuels are predominantly carbon and hydrogen
though diesel fuels can contain up to about 1 percent sulfur. Other
f l f i t t l h l f l ( t l d li fi dfuels of interest are alcohols, gaseous fuels (natural gas and liquefied
petroleum gas), and single hydrogen compounds (e.g., methane,
propane, isooctane).
9
Substances contained in a crude oilSubstances contained in a crude oil have different boiling points, the substances in crude oil can be separated using fractional distillation. The crude oil is evaporated and its vapours are allowed to condense at different temperatures in the fractionatingtemperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms.
10
- Paraffins
11
12
13
- Olefins
14
- Alcohol What is the component of alcohol that differsWhat is the component of alcohol that differs from other fuels ?
15
Relationships between composition of reactants (fuel and air) and composition of products
For example, overall chemical equation for the complete combustion of one mole of propane C3H8 :
OHCOOHC 22283 435 +=+ OHCOOHC 22283 435 ++
16
What mass of O2 is required for the complete combustion of 8.58 g of C4H10?
2C4H10 + 13O2 --> 8CO2 + 10H2O
combustion of 8.58 g of C4H10?
4 10 2 2 2
mole1476.0/ l12358g58.8
=Moles of C4H10 :g/mol123.584 10
2 moles of C4H10 require 13 moles of O2
Molecular weight of C4H10 2 moles of C4H10 require 13 moles of O2
0.1476 mol x (13/2) = 0.9595 mole O2
4 10
0.9595 mol x 31.998 g/mol = 30.7 g O2Ans
17
Stoichiometric Mixture of products contain CO2, H2O and N2
Hydrocarbon fuel of composition 84.1 percent by mass C and 15.9 percent by mass H has a molecular weight of 114.15. Determine the number of moles of air required for stoichiometric combustion and the number of moles of products produced per mole ofnumber of moles of products produced per mole of fuel. Calculate (A/F)S and the molecular weight of the reactants and products.
•Assuming fuel chemical formulation is of the form CaHb, we have the relation
Based on 1 kg of fuel:Based on 1 kg of fuel:3.18428.172715.114
=+
14.1614.151 =+
15.11415.11415.114+
24
From the above equation, 1 mol of fuel requires 59.66 mol of air to get 65 16 mol of products
14.15114.15)/( ==sFA Ans
65.16 mol of products.
1
0661.01415
1)/( ==sAF Ans14.15
RM MMolecular weight of reactants ( ) and molecular weight of products ( ) areRM PMMolecular weight of reactants ( ) and molecular weight of products ( ) are
RM 1 1 (1 114.15 59.66 28.96)i in M= = × + ×∑Rtot
( )60.66i in ∑
36.30= Ans
PMtot
1 1 (8 44.01 9 18.12 47.16 28.16)64.16i in M
n= = × + × + ×∑
2571.28= Ans
3.5 THE FIRST LAW OF THERMODYNAMICS AND COMBUSTION
3.5.1 Energy and Enthalpy Balances
Reactants
ProductsCombustion
System changing from reactants to products
26
The first law of thermodynamics: relating changes in internal energy (enthalpy) to heat and work transferinternal energy (enthalpy) to heat and work transfer interaction
RPPRPR UUWQ −=− −− PRU −or
+ work transfer to the system, - work transfer from the system
C iCombustion process► Constant volume
initial and final temperature = T ′
∫ ==−
P
PR pdVW )0(27
∫R
TVRPPR UUUQ ′Δ=′−′= )( TVRPPRQ − ,)(
)(0 RPRP UUUU ′<′<′−′ )(0 RPRP UUUU <<
TVU ′Δ− )( T ′= heat of reaction at constant volume at temperature TV ,)( p
If all the reactant and product species are ideal gasesp p g
TVTp UH ′′ Δ−Δ ,, )()( TnnR RP ′′−′= )(~
One of the products H2O, can be in the gaseous or liquid phase. The internal energy (or enthalpy) of the products will depend on the relative proportions of water in gaseous or liquid phase.
u ′ i t l t T & f th d tOfgHu2 = internal energy at T & p of the products
The relationships applied for enthalpy
2 2 2 2, , , ,( ) ( )p T H Oliq p T H Ovap H O fgH OH H m h′ ′ ′Δ − Δ =
33
Products
Reactants ReactantsProducts
VaporVapor fuel
Liquid
Vapor Liquid Fuel
(a) (b)
Schematic plots of internal energy of reactants and products as a function of temperature. (a) Effect of water in products as either vapor or liquid.
(b) Effect of fuel in reactants as either vapor or liquid.
34
( ) p q
3.5.2 Enthalpies of Formation οfh~Δ
The enthalpy of formation of a chemical compoundf
is the enthalpy increase associated with the reaction ofis the enthalpy increase associated with the reaction of forming one mole of the given compound from its element in the thermodynamics standard [298.15 K (25°C), 1 atm]in the thermodynamics standard [298.15 K (25 C), 1 atm]
from )()()( RPTVTP VVpUH −=Δ−Δ ′′ )()()( ,, RPTVTP VVpU
(already discussed previously) ( y p y)
)()()(00 RPTpTV VVpHU −−Δ=Δ )()()(00 ,, RPTpTV p
The case of ideal gasThe case of ideal gas
0)(~)()(00
TnnRHU RPTpTV −−Δ=Δ36
0,, )()()(00 RPTpTV
TABLE 3.2 Standard enthalpies of formation (25°C, 1 atm)Species State
MJ/kmol
O2 Gas 0
οfh~Δ
N2 Gas 0
H2 Gas 0
C Gas 0
CO2 Gas -393.52
H O Gas -241 83H2O Gas 241.83
H2O Liquid -285.84
CO Gas -110.54
CH4 Gas -74.87
C3H8 Gas -103.85
CH3OH Gas -201.17
CH3OH Liquid -238.58
C H G 208 45
37
C8H18 Gas -208.45
C8H18 Liquid -249.35
3.5.3 Heating Values HVQ3.5.3 Heating Values
Heating value of a fuel is a magnitude of
HVQ
heat of reaction that is measured directly at a standard temperature (25°C or 77 °F) for complete combustion of
i f funit mass of fuel.
Reaction at constant pressure
)( HQ Δ−=0,)( TpHV HQ
PΔ=
Reaction at constant volumeReaction at constant volume
0,( )VH V V TQ U= − Δ
38
Higher heating value
OH formed in products is condensed to liquid phase
(HHV)
OH 2 formed in products is condensed to liquid phase
L h ti l (LHV)Lower heating value
OH 2 formed in products is in vapor phase
(LHV)
OH 2 formed in products is in vapor phase
39
HHV and LHV are related by
(constant pressure)
HHV and LHV are related by
OHhm
mQQ fg
OHLHVHHV pp 2)( 2+=
(constant pressure)
m f
(constant volume)
OHum
mQQ fg
f
OHLHVHHV vv 2)( 2+=
f
OH
mm
2 : ratio of mass of H2O produced to mass of fuel burnedfm
40
Heating values of fuels are measure in calorimeter
For gaseous fuels, using continuous-flow atmosphere pressure
g
calorimeter.
An entering fuel is saturated with water vapor and mixedAn entering fuel is saturated with water vapor and mixed with sufficient saturated air for complete combustion at reference temperature. p
For liquid and solid fuels, fuel is burned with oxygen under q ygpressure at constant volume in a bomb calorimeter.
41
Used for solid/liquid f lfuelsBurning under pressure at constant volumeObtain HHV
42
Example 3.2Liquid kerosene fuel of the heating value (d t i d i b b l i t ) f 43 2 MJ/k d(determined in a bomb calorimeter) of 43.2 MJ/kg and average molar CH / ratio of 2 is mixed with the,
t i hi t i i i t t 298 15 K C l l t thstoichiometric air requirement at 298.15 K. Calculate the enthalpy of the reactant mixture relative to the datum of zero enthalpy for C O N Hd t 298 15 Kzero enthalpy for C 2O 2N 2H, and at 298.15 K.,
In bomb, const volume {V 0
298.15
HHV V, TQ ( U)= − Δ = 43.2 MJ/kg
The combustion equation per mole of C can be written
222222 66.5).77.3(23 NOHCONOCH ++=++
43
6605)773(3 NOHCONOCH
kmolkmolkmol 66716071 ⎫⎫⎫
222222 660.5).77.3(2
NOHCONOCH ++=++
productskgkmolairkg
kmolfuelkgkmol
4.22166.7
4.207160.7
141
⎭⎬⎫
→⎭⎬⎫
+⎭⎬⎫
οVU )(Δ−The heating value given is at constant volume, = 43.2 MJ/kg
can be obtained, noting that the fuel is in the liquid phase:
οH )(Δ οTnnRU )(~)( +Δ=
οpH )(Δ
pH )(Δ ο TnnRU RpTV)()(
,−+Δ=
15.298)1607667(1031438243 3 ××+= −
14)160.766.7(103143.82.43 ×−×+−=
fuelkgMJ /1.4309.02.43 −=+−=
44
fg
The enthalpy of the products per kilogram of mixture is found from the enthalpies of formation (with H2O vapor):the enthalpies of formation (with H2O vapor):
οPH ∑ Δ=
productsifi hn ο
,~
products
4221)83.241(1)52.393(1 −+−
=οPH
4.221
2.87 /MJ kg= −
( )PH οΔ οοRPTp HHH −=Δ
0,)(οοο HHH )(Δ= pPR HHH )(Δ−=
⎟⎟⎞
⎜⎜⎛−−−=
fuelkgMJH R14.1.4387.2ο
AnsEnthalpy of the reactants per kilogram of mixture
⎟⎟⎠
⎜⎜⎝ productkgfuelkgR 4.221
..7.
145.0−= MJ/kg product
45
Anskilogram of mixture
6605)773(3 NOHCONOCH
kmolkmolkmol 66716071 ⎫⎫⎫
222222 660.5).77.3(2
NOHCONOCH ++=++
productskgkmolairkg
kmolfuelkgkmol
4.22166.7
4.207160.7
141
⎭⎬⎫
→⎭⎬⎫
+⎭⎬⎫
οVU )(Δ−The heating value given is at constant volume, = 43.2 MJ/kg
can be obtained, noting that the fuel is in the liquid phase:
The enthalpy of the products per kilogram of mixture is found from the enthalpies of formation (with H2O vapor):the enthalpies of formation (with H2O vapor):
οPH ∑ Δ=
productsifi hn ο
,~
products
4221)83.241(1)52.393(1 −+−
=οPH
4.221
2.87 /MJ kg= −
( )PH οΔ οοRPTp HHH −=Δ
0,)(οοο HHH )(Δ= pPR HHH )(Δ−=
724.287.2 +−=οRH
Ans
Enthalpy of the reactants per kilogram of mixture
R
146.0−= MJ/kg product
47
Ans
3.5.4 Adiabatic Combustion Process
-> adiabatic constant volume processdi b ti t tadiabatic constant pressure process
-> Constant volume -> UWQ Δ=Δ−Δ
∫P
∫−ΔR
pdVQ UΔ=adiabatic
{0 0
( )P RQ p V VΔ − −14243 UΔ=
adiabatic
UΔ 0=
048RP UU − 0=
Reactants Productsor Products
or
Adiabatic combustion process: constant-volume on U-Tdiagram and constant-pressure on H-T diagram
Consider a mass m which passes through the controlConsider a mass m which passes through the control volume, the net chemical energy release due to combustion iscombustion is
[ ] ⎟⎟⎞
⎜⎜⎛
Δ−Δ=− ∑∑ ~~)()( οοifiifiAPAR hnhnmTHTH[ ] ⎟⎠
⎜⎝
∑∑productsi,
,reactantsi,
,)()( ifiifiAPAR
where οifh ,
~Δ enthalpy of formation of speciesif , py p
51
The combustion efficiency cη c
The fraction of fuel energy supplied to the control volume around the engine which is released by combustion
THTH )()(
HVf
APARc Qm
THTH )()( −=η
energy releasedHVf Q
where HVf Qm is the amount of energy supplied
due to combustion
where HVf Qm is the amount of energy suppliedto the control volume around the engine which can be released by combustionreleased by combustion
52
Variation of engine combustion efficiency with fuel/air equivalence ratio
53
equivalence ratio
3.6 Thermal conversion efficiency tη
Efficiency which relates the actual work per cycle to amo nt of f el chemical energ released in comb stion process
cycleperworkActual
amount of fuel chemical energy released in combustion process.
releaseenergychemicalFuelyp
=tη
ccct Qm
WHW
THTHW
ηΔη =−=
−=
)()()( HVfcTAPAR QmHTHTHA
ηΔ )()()(
h fTherefore tcf ηηη =Fuel conversion efficiency
( b f !!)54
(seen before!!)
Example 3.3 The brake fuel conversion efficiency is 0.3. Themechanical efficiency is 0 8 The combustion efficiency is 0 95 Themechanical efficiency is 0.8. The combustion efficiency is 0.95. Theheat losses to coolant and oil are 60 kW. The fuel chemical energyentering the engine per unit time, is 190 kW. What percentage of thisg g p , p genergy become ?
a) brake work b) friction work c) heat lossesa) brake work b) friction work c) heat losses
d) Exhaust chemical energy e) Exhaust sensible energy
kW4.11190)94.01( =×−=due to combustion inefficiency
%6100190
4.11=×=
e) Exhaust sensible energy = 190 - (a + b + c + d)
4.116025.1457190 −−−−= 4.116025.1457190
kW35.47=
57%9.24100190
35.47=×= Ans
3.7 CHEMICALLY REACTING GAS MIXTURES
Depending on problem and portion of engine cycle, chemical reaction may:chemical reaction may:
1) to be very slow -> negligible effect on mixture i i (f )composition (frozen)
2) to be rapid > the composition remains in chemical2) to be rapid -> the composition remains in chemical equilibrium
3) rate-controlling -> determine how composition changes with timechanges with time
58
3.7.1 Chemical EquilibriumThe chemical reaction by which individual species in the burned gases react together, produce and remove species g g , p pat equal rates.
221 COOCO =+ 222
COOCO +
Consider a system undergoing a constant-pressure, constant temperature process.In the absence of work transfer, the first law gives
HQ Δ=Δ1st Law 0≤Δ−Δ STH2nd Law STQ Δ≤Δ Δ
43421G
0)( ≤ΔGor
59
0)( , ≤Δ TpG
Thus reactions can occur if Gibbs free energy GG <Thus reactions can occur if Gibbs free energy RP GG <
0)( , =Δ TpGAt equilibrium, ,pq ,
or RP GG =
Consider general reaction whose stoichiometry is given by
...... ++=++ mmllbbaa MMMM υυυυ
The mixture composition in equilibrium can be determined b ilib i t t Kby equilibrium constant, PK
GKlnοΔ
−=60TR
K P ~ln −=
++=++ MMMM υυυυ
The equilibrium constant for a specific reaction is obtained via
...... ++=++ mmllbbaa MMMM υυυυ
q pthe relation
∑=i
iPireactionP KK )(log)(log 1010 ν
reactionPK )( = equilibrium constant of reaction
iPK )( = equilibrium constant of formation of species iν = stoichiometric coefficientsiν = stoichiometric coefficients
Positive (+) for product species
61Negative (-) for reactant species
Ex 2221 COOCO =+ reaction r1 at 2,500 K2
r1reaction10 )(log pK ∑=i
iPi K )(log10ν
22)(log
21)(log1)(log1 101010 OPCOPCOP KKK −−=
PK can also be obtained from partial pressure
i
i
iP p
pK
ν
∏ ⎟⎟⎠
⎞⎜⎜⎝
⎛=
0
...321 ννν
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛pp
pp
pp iii->
i p ⎠⎝ 0 000 ⎠⎝⎠⎝⎠⎝ ppp
Π = multiplicationΠ multiplication
ip = partial pressure of species i
62
( ) ( ) ii i ip pK ν ν νχ χ∑= =∏ ∏% %
0 0
( ) ( ) ii i ip i i
i i
Kp p
χ χ= =∏ ∏where iχ
~= mole fractionw e e iχ mole fraction
p0 = standard state pressure
If ∏∑ ==i
iPi
iiK νχν ~,0
gmolecK ( )gmole
3cmAn equilibrium constant, , based on concentrations
[ ]∏= iciMK ν[ ]∏
iic
Relation between CP KK & CP KK &∑= i iTRKK CPν)~( R~; -> universal gas const
63Hence if Cpi i KK ==∑ ,0ν
Example 3.4 In fuel-rich combustion product mixtures, equilibrium between the species CO H O CO and H isequilibrium between the species CO2, H2O, CO and H2 is often assumed to determine the burned gas composition. For φ 1 2 f C H i b ti d t d t iFor φ = 1.2, for C8H18-air combustion products, determinethe mole fractions of product species at 1,700 K
Analysis:
Th i l i h i f ll d “ hThe reaction relating these species often called “the water gas reaction”.
OHCOHCO +=+ OHCOHCO 222 +=+
64
Question
What is the chemical equation for a stoichiometric combustion of C H ?
22222188 N16.47OH9CO8)N773.3O(5.12HC ++→++
combustion of C8H18 ?
Remember that these gases are the products for stoichiometric combustion !!
φ
for stoichiometric combustion !!
The combustion of C8H18 with air for φ = 1.2 can be written as
Combustion products are in chemical equilibrium.Combustion products are in chemical equilibrium.
OHCOHCO 222 +=+
5100log =K 2363=K
This chemical reaction is called “water gas reaction”. At 1,700 K,
The equilibrium relation then gives
510.0log10 =PK 236.3=PK
(5)0 0
( ) ( ) ii i ip i i
i i
p pKp p
ν ν νχ χ∑= =∏ ∏% %
66
0 0i ip p
What is Kp for this reaction?
OHCOHCO 222 +=+
p
?=∑i
iνIn this case,i
01111 =++−−=∑i
iν
0p ∑ ∏ ∏
i
0
( ) i i ip i
i
pKp
ν νχ∑= ∏ % ∏=i
iPiK νχ~Thus
67
OHCOHCO 222 +=+ 222
1111222
~~~~~OHCOHCOiP
iK χχχχχν −−==∏
3039)7733(5.12 NdHcCOObHaCONOHC ++++→++
222i∏
222222188 30.39)773.3(2.1
NdHcCOObHaCONOHC ++++→++
a b c d
PCO n
a=
2
~χP
OH nb
=2
~χP
CO nc
=χ~P
H nd
=2
~χ
⎟⎞
⎜⎛⎟⎞
⎜⎛ bc
30.39++++= dcbanp
Therefore⎟⎞
⎜⎛⎟⎞
⎜⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
= PPp da
nb
nc
K
68
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
PP nd
na
⎟⎞
⎜⎛⎟⎞
⎜⎛ bc
⎞⎛⎞⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
= PPp d
nb
nc
Kd
cb=236.3 (6)
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
PP
p
nd
na ad
From the equations (2), (3), (4) and (6) we can solve for1 b 69 2 86 d 1 31a = 5.14, b = 7.69, c = 2.86, d = 1.31
3039++++ db
30.3931.186.269.714.5 ++++=
30.39++++= dcbanp
30.56=
69
Mole fractions of products become
091.030.56
14.5~2
===P
CO naχ Ans30.56Pn
137.0305669.7~
2===OH n
bχ Ans30.56Pn
051.0305686.2~ ===CO
cχ Ans30.56PCO n
023.031.1~ ===Hdχ Ans023.0
30.562P
H nχ
3039~
s
698.030.5630.39~
2==Nχ Ans
70
For a chemical reaction that is not in equilibrium, the processes are controlled by rates at which the actual reactions occurby rates at which the actual reactions occur.
For a chemical reaction
ddccbbaa MMMM νννν +=+
The reaction rate in forward (+) direction:The reaction rate in forward (+) direction:
[ ] [ ] [ ] [ ] baba
ca MMk
dtMd
MdtdR υυ+
+++ ==−=
dtdt
The reaction rate in backward ( - ) direction:
[ ] [ ] [ ] [ ] dcdc
ac MMk
dtMd
MdtdR νν−
−−− ==−=
71
k+ and k- are the rate constants in forward and reverse directions
The net rate of production of products or removal of reactants:The net rate of production of products or removal of reactants: