1 isat 413 - module iv: combustion and power generation topic 2:chemical reactions and the first...
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1
ISAT 413 - Module IV:Combustion and Power Generation
Topic 2: Chemical Reactions and The First & Second Laws of Thermodynamics
Fuels and Combustion
Theoretical and Actual Combustion Processes
Enthalpy of Formation and Enthalpy of Combustion
First-Law Analysis of Reacting Systems
Adiabatic Flame Temperature
Entropy Change of reacting Systems
Second-Law Analysis of Reacting Systems
2
Chemical Reactions
• We need to consider the chemical internal energy (which is the energy associated with the destruction and formation of chemical bonds between the atoms) when dealing with reacting systems.
• Any material that can be burned to release energy is called a fuel, and a chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combustion.
3
Most liquid hydrocarbon fuels (CnHm) are obtained from crude oil distillation.Most liquid hydrocarbon fuels (CnHm) are obtained from crude oil distillation.
CRUDEOIL
Gasoline
Kerosene
Diesel fuel
Fuel oil
Fuels and Combustion Fuels and Combustion
The most volatile hydrocarbons vaporize first, forming what we know as gasoline. The less volatile fuels obtained during distillation are kerosene, diesel fuel, and fuel oil.
The most volatile hydrocarbons vaporize first, forming what we know as gasoline. The less volatile fuels obtained during distillation are kerosene, diesel fuel, and fuel oil.
4
Each kmol of O2 in Air is Accompanied by 3.76 kmol of N2
Each kmol of O2 in Air is Accompanied by 3.76 kmol of N2
• 1 kmol 02 + 3.76 kmol N2 = 4.76 kmol air
The oxidizer most often used in combustion processes is air. The dry air can be approximated as 21 % oxygen and 79% nitrogen (0.9% argon, and small amount of CO, He, Neon, and H2) by mole numbers. Therefore,
5
Steady-Flow Combustion ProcessSteady-Flow Combustion Process
Reactionchamber
In a steady-flow combustion process, the components that exist before the reaction are called reactants and the components that exist after the reaction are called products. Chemical equations are balanced on the basis of the conservation of mass principle, which states that the total mass of each element is conserved during a chemical reaction.
For example, C + O2 CO2 , where C and O2 are the reactants, and CO2 is the product.
6
Air-Fuel RatioAir-Fuel Ratio
Combustionchamber
AF =17
The ratio of the mass of air to the mass of fuel during a combustion process is called the air-fuel ratio AF:
fuel. of mass the is
and air, of mass the is where
fuel
air
iifuel
air
fuel
air
Mnm
nMm
m
m
F
AAF
For example,
7
Example IV-2.1Example IV-2.1
One kmol of octane (C8H18) is burned with air that contains 20 kmol of O2. Assuming the products contain only CO2, H2O, O2, and N2, determine the mole number of each gas in the product and the air-fuel ratio for this combustion process.
fuel kg
air kg
then, , and , , ,
elments, the of each to
balance mass the applyingby determined are and , , ,
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..
nMnM
nM
m
mAF
.w.zyx
wzyx
wNzOOyHxCON.OHC
:Solution
HC
air
fuel
air
8
Completion of the Combustion ProcessCompletion of the Combustion Process
Combustionchamber
• The combustion process is complete if all the combustible components in the fuel are burned to completion. That is, a combustion process is complete if all the carbon in the fuel burns to CO2, all the hydrogen burns to H2O , and all the sulfur (if any) burns to SO2.
• Insufficient oxygen causes incomplete combustion, unburned fuel, C, H2, CO, or OH would be in the products.
• At ordinary combustion temperatures, nitrogen behaves as an inert gas and does not react with other chemical elements.
9
Theoretical and Actual Combustion Theoretical and Actual Combustion
• The complete combustion process with no free oxygen in the products is called stoichiometric, or theoretical combustion. For example, the theoretical combustion of methane is
• The amount of air in excess of the stoichiometric amount is called excess air, or percent excess air.
• Amounts of air less than the stoichiometric amount are called deficiency of air, or percent deficiency of air.
10
Stoichiometric AirStoichiometric Air
The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. The theoretical air is also referred to as the chemically correct amount of air or 100 percent theoretical air. The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel.
Excess AirExcess Air
The air in excess of the stoichiometric amount is called the excess air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air.
11
Example IV-2.2Example IV-2.2
Ethane (C2H6) is burned with 20 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 100 kPa, determine the air-fuel ratio for this combustion process.
fuel kg
air kg
then ,
be to balance the from determined is tcoefficien The
air. for tcoefficientric stoichiome the is where
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2
2
22222262
...
nMnM
nM
m
mAF
N.O.OHCON.O.HC
.aa..a.:O
Oa
a
Na..Oa.OHCON.Oa.HC
:Solution
HC
air
fuel
air
ththth
th
th
ththth
12
Example IV-2.3Example IV-2.3
A certain natural gas has the following volumetric analysis: 72% CH4, 9% H2, 14% N2, 2% O2, and 3% CO2. This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 20oC, 1 atm, and 80% relative humidity. Assuming complete combustion and a total pressure of 1 atm, determine the dew-point temperature of the products.
ly.respective , and , , , on balances mass from
and , be to
determined are equation above the in tscoefficien unknown The
fuel, of gConsiderin
22
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1
NOHC
.z.a,., y.x
zNOyHxCO
N.OaCO.O.N.H.CH.
kmol
:Solution
th
th
13
C.TT
kPa.P.
.
.
P
n
n
P
P
N.OH.CO.OH.
N.O.CO.O.N.H.CH.
OHkmol.n
.
.
.n
n
P
P
n
n
kPa.kPa..PP
...a.
a
okPa.@satdp
prod,vprod,v
prod
prod,v
prod
prod,v
air,v
air,v
air,v
total
air,v
total
air,v
Co@satair,v
th
th
960
88200598
6611
325101
648566117501310
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1310
325101
8711
976
87113392800
9764651764764
8820
2222
2222224
2
20
Thus,
Also,
have weequation,
the of sides both to of amount this addingBy .
behavior, gas-ideal Assuming
is air the in moisture the of pressure partial The
air.dry of kmol is that
4.76 saccompanie that moisture of amount the determine weNext
14
Forms of EnergyForms of Energy
The microscopic form of energy of a substance consists of sensible, latent, chemical, and nuclear energies.
Sensible and latent energies are associated with a change of state (temperature for sensible and phase for latent), chemical energy associates with the molecular structure, and nuclear energy associates with the atomic structure.
15
Chemical Bonds in the Combustion ProcessChemical Bonds in the Combustion Process
When the existing chemical bonds are destroyed and new ones are formed during a combustion process, usually a large amount of sensible energy is released.
The chosen reference state is 25oC and 1 atm, which is known as the standard reference state. Property values at the standard reference state are indicated by a superscript o such as (ho and uo). For N2 at 500 K,
kJ/kmol,hh oK 5912866958114500
16
Enthalpy of Combustion Enthalpy of CombustionThe difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction is called the enthalpy of reaction hR. For combustion processes, the enthalpy of reaction is usually referred to as the enthalpy of combustion hc, which represents the amount of heat released during a steady-flow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure. For example,
kmol/kJ,HHh reactprodC 520393
-393,520 kJ/kmol is the enthalpy of combustion for C at the standard reference state. The enthalpy of combustion of a particular fuel will be different at different temperatures and pressures.
17
Enthalpy of FormationEnthalpy of Formation
The enthalpy of a substance at a specified state due to its chemical composition is called the enthalpy of formation hf. The enthalpy of formation of all stable elements is assigned a value of zero at the standard reference state of 25oC and 1 atm. For example, the enthalpy of formation of CO2 at the standard reference state is
kmol/kJ,h oCO,f 520393
2
The negative sign is due to the fact that the enthalpy of 1 kmol of CO2 at 25oC and 1 atm is 393,520 kJ less than the enthalpy of 1 kmol of C and 1 kmol of O2 at the same state. In other words, 393,520 kJ of chemical energy released (leaving the system as heat) when C and O2 combine to form 1 kmol of CO2.
18
Heating ValuesHeating Values
The heating value of a fuel is defined as the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel. Heating value is called the higher heating value (HHV) when the H2O in the products is in the liquid form, and it is called the lower heating value (LHV) when the H2O in the products is in the vapor form. The two heating values are related by
where n is the number of moles of H2O in the products and hfg is the enthalpy of vaporization of water at 25oC.
fuel kJ/kmol OHfgnhLHVHHV
2
19
Example IV-2.4Example IV-2.4Determine the enthalpy of combustion of gaseous octane (C8H18) at 25oC and 1 atm, using enthalpy-of-formation data from Table A-26. Assume the water in the products is in the liquid form.
188
188
22o
HCkJ/kmol
becomes HC of combustion
ofenthalpy the Then zero. is formation ofenthalpy their thus
and elements, stable are O and N Also,atm. 1 and C25 of state
reference standard the at are products the and reactants the Both
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,,
,,,
hnhnhnhnhn
HHh
Na.liqOHCON.OaHC
:Solution
HCofliqOH
ofCO
of
or,fr
op,fp
reacprodc
thth
20
First-Law Analysis of Reacting Systems First-Law Analysis of Reacting Systems
The enthalpy of a chemical compound at a specified state is the sum of the enthalpy of the compound at 25°C, 1 atm (hf°), and the sensible enthalpy of the compound relative to 25°C, 1 atm.
21
Steady-flow SystemsSteady-flow Systems
Taking heat transfer to the system and work done by the system to be positive quantities, the conservation of energy relation for chemically reacting steady-flow systems can be expressed per unit mole of fuel as
Where the superscript o represents properties at the standard reference state of 25oC and 1 atm.
r
oofrp
oofp hhhNhhhNWQ
22
Closed SystemsClosed Systems
For a closed system, the conservation of energy relation becomes
The terms are negligible for solids and liquids and can be replaced by RuT for gases that behave as ideal gases.
r
oofrp
oofp vPhhhNvPhhhNWQ
oof
oof
systemoutin
uvphhuuu
vphuUEE
Thus,
and , that Note
vp
23
Example IV-2.5Example IV-2.5
Liquid propane (C3H8) enters a combustion chamber at 25oC at a rate of 0.05 kg/min where it is mixed and burned with 50% percent excess air that enters the combustion chamber at 7oC. An analysis of the combustion gasses reveals that all the hydrogen in the fuel burns to H2O but only 90 percent of the carbon burns to CO2, with the remaining 10 percent forming CO. If the exit temperature of the combustion gases is 1500 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber.
.negligible are energies potential andKinetic 3.
gases. ideal are gases combustion the and Air2.
exist. conditions operatingSteady 1. :sAssumption
:Solution
24
roo
frpoo
fpout
fuelair
fuel
air
th
thth
hhhNhhhNQ
b
.
..mAFm
.kmol/kgkmolkmol/kgkmol
kmol/kgkmol..
m
mAF
a
N.O.gOHCO.CO.N.O.lHC
CO
C
a
Na.gOHCON.OalHC
is process combustionflow -steady this for transfer heat The
air/min kg
fuel/min kgfuel air/kg kg Thus,
fuel air/kg kg
is process combustion this for ratio fuel-air The )(
products, the in
forming of 10% and air excess 50% withncommbustio actual For
523 :balance O2
181
0505325
532524123
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2222283
25
kW.kmol
kmol
kJ,
min
kg.QmQ
,,.
,.,,
,,.,,.
..
.hh,HCkmol
hhhNhhhNQ
outout
o
po
KKofpr
oKK
ofrout
896HC of kg44
1882363050 Thus,
HC of kJ/kmol8823638669073470228
86822924906529904999578202414
8669517475301103093640787152039372
86698141076357
868281500579101181
83
83
29828083
2981500298280
*49,29247,07357,99971,07847,517
*86828669990493648669
*81508141
***
-118,91000
-241,820-393,520-110,530
C3H8 (l)
O2
N2
H2O (g)
CO2
CO
h 1500 K
(kJ/kmol)
h 298 K
(kJ/kmol)
h 280 K
(kJ/kmol)
h of
(kJ/kmol)Substance
26
Adiabatic Flame Temperature Adiabatic Flame Temperature
In the absence of any heat loss to the surroundings (Q = 0), the temperature of the products will reach a maximum, which is called the adiabatic flame temperature of the reaction. The adiabatic flame temperature of a steady-flow combustion process is determined from Hprod = Hreact or
r
oofrp
oofp hhhNhhhN
Combustionchamber
27
Theoretical Adiabatic Flame TemperatureTheoretical Adiabatic Flame Temperature
The maximum temperature encountered in a combustion chamber is lower than the theoretical adiabatic flame temperature
The adiabatic flame temperature of a fuel is not unique. Its value depends on (1) the state of the reaction, (2) the degree of completion of the reaction, and (3) the amount of air used.
28
Example IV-2.6Example IV-2.6
Liquid octane (C8H18) enters the combustion chamber of a gas turbine steadily at 1 atm and 25oC, and it is burned with air that enters the combustion chamber at the same state. Disregarding any changes in kinetic and potential energies, determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 400 percent theoretical air, and c) incomplete combustion (some CO in the products) with 90 percent theoretical air.
gases. ideal are gases combustion the and Air4.
0.PEkE ns.interactio workno are There 3.
adiabatic. is chamber combustion The 2.
process. combustionflow -Steady 1. :sAssumption
:Solution
29
hNhNhhhN
NOHCON.O.lHC
a
HCof
or,frp
oofp
188
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is case this in etemperatur flameadiabatic The
is air of amount ltheoretica
the withprocess combustion the for equation balanced The ).(
Substance
kJ,,hhh
HCkmol/kJ,HCkmolh
h,h,COkmol
NOHCO
N
OHCO
08164654798
95024918669047
9904820241993645203938
222
1881882
222
yieldswhich
(Please refer to the ideal gas Tables A─18 ~ A─27 for enthalpy of N2, O2, CO2, CO, H2, H2O, and enthalpy of formation of fuels)
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C8H18 (l)
O2
N2
H2O (g)
CO2
h 298 K
(kJ/kmol)
h of
(kJ/kmol)
30
have wepairs, two these using ioninterpolatBy 4.
than greater is since K, 2350Try 3.
etemperatur this at and K, 2650 to close is K which 2400
etemperatur a guess we, are moles the ofmajority the that Noting 2.
.CO forK 1800 and , forK 2100 , forK 2650
to scorrespond which Take 1.
. determine to approach error-and-trial
a use to have will weTherefore, gases. ideal for since ,
product, the of etemperatur the-unknown oneonly have eactually w
But unknowns. three withequation one have wethat appears It
2
K.T
,,,,,hhh
kJ,, kJ,,
,,,,,hhh
N
OHN
kmol/kJ,)/(,,
T
h(T)hT
prod
NOHCO
NOHCO
prod
prod
52394
65452654967747846100909112284798
08164658286605
82866053207947508103951212584798
2208847980816465
222
222
2
22
31
KT
a
NO.OHCON.OlHC
b
prod 962
1885379876350 222222188
be to determined is case this in
etemperatur flameadiabatic the ),( in used procedure the followingBy
is air ltheoretica 400% with
process combustion complete the for equation balanced The
KT
a
N.OHCO.CO.N.O.lHC
c
prod 2236
342952557632511 22222188
be to determined is case this in
etemperatur flameadiabatic the ),( in used procedure the followingBy
is air ltheoretica 90% with
process combustion incomplete the for equation balanced The
air. of amount ltheoretica the withoccurs
combustion complete whenachieved is etemperatur flameadiabatic
maximum the Also,air. excess using or combustion incomplete
of result a as deceases etemperatur flameadiabatic the thatNotice
32
Entropy Change of Reacting Systems Entropy Change of Reacting Systems
systemgenoutin SSSS
The entropy balance for any system (including reacting systems) undergoing any process can be expressed as
33
Entropy ChangesEntropy Changes
Taking the positive direction of heat transfer to be to the system, the entropy balance relation can be expressed for a closed system or steady-flow combustion chamber as
kJ/K 0 reactprodgenk
k SSST
Q
For an adiabatic process the entropy balance relation reduces to
0 reactprodadiabatic,gen SSS
34
The third law of thermodynamics The third law of thermodynamics
• The entropy relations for combustion processes involve the entropies of the components, not entropy changes, which was the case for non-reacting system. The search for a common base for the entropy of all substances led to the establishment of the third law of thermodynamics.
• The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero. The third law provides a common base for the entropy of all substances, and the entropy values relative to this base are called the absolute entropy, so, the values are listed in Tables A-18 through A-25.
35
Absolute Entropy of an Ideal GasAbsolute Entropy of an Ideal Gas
kmol.K
kJ
ouo
o
P
PlnRP,TsP,TS
The ideal-gas tables list the absolute entropy values over a wide range of temperatures but at a fixed pressure of Po = 1 atm. Absolute entropy values at other pressures P for any temperature T are determined from
36
Absolute Entropy for Ideal-Gas MixtureAbsolute Entropy for Ideal-Gas Mixture
• For component i of an ideal-gas mixture, the absolute entropy can be written as
where Pi is the partial pressure, yi is the mole fraction of the component, Po = 1 atm, and Pm is the total pressure of the mixture in atmospheres.
kmol.K
kJ
o
miuo
oiii P
PylnRP,TsP,TS
37
Second-Law Analysis of Reaction Systems Second-Law Analysis of Reaction Systems
The difference between the availability of the reactants and of the products during a chemical reaction is the reversible work associated with the reaction.
• The exergy destruction or irreversibility and the reversible work associated with a chemical reaction are determined from
kJ genoreactrevdestroyed STWWXI
38
Reversible WorkReversible Work
• When both the reactants and the products are at the temperature of the surroundings T0, the reversible work can be expressed in terms of the Gibbs functions as
ooToo
poo
fproo
frrev
gsThsTh
gggNgggNW
where,
kJ
kJ po
oofpro
oofrrev sThhhNsThhhNW
The reversible represents the maximum work that can be done during a process. In absence of any changes in kinetic and potential energies, the reversible work relation for a steady-flow combustion process is
39
Operation of a Hydrogen - Oxygen Fuel CellOperation of a Hydrogen - Oxygen Fuel Cell
The second law of thermodynamics suggests that there should be a better way of converting the chemical energy to work. The energy conversion devices that work on controlling the irreversibility are called fuel cells.
40
• The operation of a hydrogen-oxygen fuel cell is illustrated in the Figure on the previous slide.
• Hydrogen is ionized at the surface of the anode, and hydrogen ions flow through the electrolyte to the cathode. There is a potential difference between the anode and cathode, and free electrons flow from the anode to the cathode through an external circuit (such as a generator). Hydrogen ions combine with oxygen and the free electrons at the surface of the cathode, forming water.
• In steady operation, hydrogen and oxygen continuously enter the fuel cell as reactants, and water leaves as the product.
• Fuel cells are not heat engines, and thus their efficiencies are not limited by the Carnot efficiency.
41
Example IV-2.7 (adiabatic)Example IV-2.7 (adiabatic)
Methane (CH4) gas enters a steady-flow adiabatic combustion chamber at 25oC and 1 atm. It is burned with 50% excess air that also enters at 25oC and 1 atm. Assuming complete combustion, determine (a) the temperature of the products, (b) the entropy generation, and c) the reversible work and exergy destruction. Assume that To = 298 K and the products leave the combustion chamber at 1 atm pressure.
complete. is Combustion 5.
gases. ideal are gases combustion the and Air4.
0.PEkE ns.interactio workno are There 3.
adiabatic. is chamber combustion The 2.
process. combustionflow -Steady 1. :sAssumption
:Solution
42
hNhNhhhN
HH
N.OOHCON.OgCH
a
CHof
or,frp
oofp
reactprod
4
2222224 281127633
to reduces which, from determined
is etemperatur flameadiabatic the ,conditionsflow -steady Under
is air excess 50% with
process combustion complete the for equation balanced The ).(
kJ,h.hhh
CHkmol/kJ,CHkmolhh.
h,h,COkmol
NOOHCO
ON
OHCO
95093728112
850741868201866902811
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4422
222
yieldswhich
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CH4 (g)
O2
N2
H2O (g)
CO2
h 298 K
(kJ/kmol)
h of
(kJ/kmol)Substance
43
K.kmol/kJ.RP/PylnRP,TsNP,TsNS
iyPyP
sNsNSSSSSS
b
T
uomiuoiiiiii
itotalii
rrppreactprodgensurrsysgen
prod
314348
where,
and , component of fraction mole the is where
to equal is whichpressure, partial the at calculated be to are entropies the but
atm, 1 of pressure total a at are gases product and air both 26,- ATable On
from determined is process this during generationentropy The ).(
K 1789.0
be to found is products the of etemperatur the error,-and-trialBy
4029666923023713989 CHK.kmol/kJ...SSS reactprodgen Thus,
Ni yi soi(T, 1 atm) Ruln(yiPm/Po) Nisi
CH4 O2 N2
1 3 11.28
1.00 0.21 0.79
186.16 205.04 191.61
* 12.98
1.96
186.16 654.06
2183.47 sreact = 3023.69 CO2 H2O O2 N2
1 2 1 11.28
0.0654 0.1309 0.0654 0.7382
302.517 258.957 264.471 247.977
22.674 16.905 22.674 2.524
325.19 551.72 287.15
2825.65 sprod = 3989.71
44
wasted.is potential workentire the Instead,
not. is but process, this during done be could workof kJ 287,874 is, That
CHkJ/kmol287,874
:identical are destroyedexergy
and workreversible the Therefore, work.actual no involves process This
burned. methane of kmol each for process
combustion this during wastedis potential workof kJ 287,874 is, That
from determined is
process this withassociatedility irreversib or ndestructioexergy The ).(
4
rev
genodestroyed
W
CHkmol
kJ,CH)K.kmol/(kJ.K
STX
c
44 87428702966298
45
Example IV-2.8 (isothermal)Example IV-2.8 (isothermal)
2222224 281127633 N.OOHCON.OgCH
:Solution
same the remains
equation combustion the Thus them. from heat gtransferinby gssurroundin
the of state the to brought are products combustion the that except
example, previous the in discussed as process combustion same the is This
Methane (CH4) gas enters a steady-flow combustion chamber at 25oC and 1 atm. It is burned with 50% excess air, which also enters at 25oC and 1 atm. After combustion, the products are allowed to cool to 25oC. Assuming complete combustion, determine (a) the heat transfer per kmol of CH4, (b) the entropy generation, and c) the reversible work and exergy destruction. Assume that To = 298 K and the products leave the combustion chamber at 1 atm pressure.
46
Thus
balanceenergy flow -steady the
from determined is process combustion
flow -steady this during transfer Heat
atm. 1 and C25 at removed be
will whichform, liquid the in be willformed the of kmol 1.57 Therefore,
from determined is products the in remains that
vapor waterof amount The condense. willvapor waterthe of part C, At
o
rofrp
ofpout
vv
v
total
Co@v
gas
v
o
hN hNQ
OH
kmol.NkPa.
kPa.
N.
N
P
P
N
N
2
25430
325101
1693
2813
25
422
22
22
44
406871830285571
820241430
5203931
850741
CHkmol/kJ,lOHkmol/kJ,lOHkmol.
gOHkmol/kJ,gOHkmol.
COkmol/kJ,COkmol
CHkmol/kJ,CHkmolQout
Substance h of
(kJ/kmol)
CH4 (g)
H2O (l)
H2O (g)
CO2
-74,850-285,830-241,820-393,520
47
where
chamber combustion the of gssurroundin immediate
the includes that system extended an on applied balanceentropy an
from determined is process this during generationentropy total The ).(
4842745
298
406871693023352845
CHK.kmol/kJ.
K
kmol/kJ,K.kmol/kJ..
T
QSSS
b
surr
outreactprodgen
Ni yi soi(T, 1 atm) -Ruln(yi)Pm Nisi
CH4 O2 N2
1 3 11.28
1.00 0.21 0.79
186.16 205.04 191.61
* 12.98 1.96
186.16 654.06
2183.47 sreact = 3023.69 CO2 H2O (l) H2O O2 N2
1 1.57 0.43 1 11.28
0.0729 1.0000 0.0314 0.0729 0.8228
213.80 69.92
188.83 205.04 191.61
21.77 *
28.77 21.77 1.62
235.57 109.77 93.57
226.81 2179.63
sprod = 2845.35
48
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