1 ISAT 413 - Module IV: Combustion and Power Generation Topic 2:Chemical Reactions and The First & Second Laws of Thermodynamics Fuels and Combustion

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 , , ,

22429128

2976420

2755798

76320

2

222222188

..

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


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

31923122

2976424

7915703276324

532051221

76321203276321

2

22222262

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


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

222

2222224

64854651531750

763030020140090720

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

7634651030020140090720

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

1805125

450208153028595203938

76398763

18822

22222188

,,

,,,

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


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

2976457

2286524307276357

76343763

22222283

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


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.


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

22222188 4798763512

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)

*8682866999049364

-249,95000

-241,820-393,520

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.


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

9904820241293645203931

2222

4422

222

yieldswhich

*8682866999049364

-74,85000

-241,820-393,520

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

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Documents

1 ISAT 413 - Module IV: Combustion and Power Generation Topic 2:Chemical Reactions and The First & Second Laws of Thermodynamics Fuels and Combustion