at09 10 combustion

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1

Department of Engineering Physics, Faculty of Engineering, Gadjah Mada University

(Study Programs of Engineering Physics & Nuclear Engineering)J l. Grafika 2, Yogyakarta 55281,(+62 274) 580882, http://www.tf.ugm.ac.id/

Department of Engineering Physics, Faculty of Engineering, Gadjah Mada University

(Study Programs of Engineering Physics & Nuclear Engineering)J l. Grafika 2, Yogyakarta 55281,(+62 274) 580882, http://www.tf.ugm.ac.id/

Combustion

10 - 00

UGM

DepartmentofEngineeringPhysics,

FacultyofEngineering

StudyProgramsofEngineeringPhysics&NuclearEngineering

Chemical Equations

2

Mass ConservationMass Conservation

ReactantsReactants ProductsProducts HeatHeat++

Total MassTotal Massof Reactantsof Reactants

Total MassTotal Massof Productsof Products==

Energy Conservation (1Energy Conservation (1stst Law)Law)

EEsyssys== EEstatestate++ EEchemchem

00chapter 10
http://www.tf.ugm.ac.id/http://www.tf.ugm.ac.id/http://www.tf.ugm.ac.id/http://www.tf.ugm.ac.id/


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DepartmentofEngineeringPhys

ics,



Reference States

To determine the heat of reaction, QP, a referencestate (0) is needed from which to compute

internal energy and enthalpy of the reactants and

products.

chapter 10 3

pRchemP HEQ

UGM




The standard reference state

chapter 10 4

TheThe standard reference statestandard reference state is 25is 25oo CC

and 1 atm pressure, where theand 1 atm pressure, where the

enthalpy isenthalpy is assumedassumed to be zero for allto be zero for all

of theof the elementselements involved.involved.

),( TPh

298)(KT

1

)(atmP 0oh


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Enthalpy of Formation

The enthalpy of formation of a chemicalcompound (x) is its enthalpy at 25o C and 1 atm

pressure with respect to the

standard reference state.

5

h xo

= Enthalpy of Formation, kJ/kmole

chapter 10

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Examples Carbon, C

Butane, C4H10

Carbon Dioxide

A negative enthalpy of formation for a compoundindicates that heat is released during the formation ofthat compound from its stable elements.

A positive value indicates heat is absorbed.

6

hCo

0

150,126104

o

HCh

520,3932

o

COh

chapter 10


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Enthalpy of Reaction

Definition:

o the difference between the enthalpy of the products at aspecified state and the enthalpy of the reactants at the samestate for a complete reaction

7

0

,

0

, rfrpfp

reactprodR

hNhN

HHH

chapter 10

UGM




Enthalpy of Combustion

Definition:

o the difference between the enthalpy of the products at aspecified state and the enthalpy of the reactants at the samestate for a complete reaction

8

0

,

0

, rfrpfp

reactprodC

hNhN

HHH

chapter 10


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Example

9

Given:

Octane(C8H18). Water in the products is inthe liquid form

Find:

enthalpy of combustion of octane at 25 Cand 1 atm

chapter 10

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Equation

10

The combustion equation

C8H18+ath(O2+3.76N2)

8CO2

+ 9 H2

O + 3.76 ath

N2

18822)()()(

,,

HC

o

fOH

o

fCO

o

f

o

rfr

o

pfp

reactprodc

hNhNhN

hNhN

HHh

022 ,, o Nf

o

Of hhBecauseBecause

ThusThus

chapter 10


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Heat Released

11

188

188

,,

/47891

/000,471,5

)/249950)(1(

)/285830)(9()/393520)(8(

)()()(18822

HkgCkJ

HkmolCkJ

kmolkJkmol

kmolkJkmolkmolkJkmol

hNhNhN

hNhN

HHh

HC

o

fOH

o

fCO

o

f

o

rfr

o

pfp

reactprodc

Heat is releasedHeat is released

chapter 10

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Example

12

Find:

the enthalpy of combustion of hydrogen at

25 C and 1 atm, assuming water to be in the

liquid and the vapor form

Answers: -285,830kJ/kmolH2 (l)

-241,820kJ/kmolH2 (v)

chapter 10


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Heating value

Definition:

o the amount of heat released when a fuel is burnedcompletely in a steady-flow process and the products arereturned to the state of the reactants.

13

chueHeatingVal

The absolute value of the enthalpy of combustionThe absolute value of the enthalpy of combustion

chapter 10

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Carbon-based Fuels

14

Higher Heating Value(HHV)Higher Heating Value(HHV)::

Water in the products is in theWater in the products is in the

liquid state.liquid state.

Lower Heating Value(LHV)Lower Heating Value(LHV)::Water in theWater in the productsproducts is in theis in the

gaseous state.gaseous state.

)/()(2

kgfuelkJmhLHVHHVOHfg

Depends on temperature, is 44000kJ/Depends on temperature, is 44000kJ/kmolkmol HH22O at 25O at 25 CC

Mass of water per un it mass of fuel

chapter 10


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Bomb Calorimeter

15

KalorqV =C (T1 - T0)

Proses isochorisqV =U

Entalpi:Hm =Um +RT ngas

chapter 10

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Example

16

Find:

the heating values of hydrogen, octane, and themixture of hydrogen/octane (50%/50%) at 25 C and1 atm

chapter 10


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Steady-Flow Systems

17

outin EE

Reactants

Products

Heat Transfer

chapter 10

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Steady-Flow Systems

18

poofpoutoutroofrinin hhhnWQhhhnWQ )()(

Molar flow rateMolar flow rate

Energy balance per mole of fuelEnergy balance per mole of fuel

poofpoutoutroofrinin hhhNWQhhhNWQ )()(

Number of molesNumber of moles

roofrpoofpreactprod hhhNhhhNHHWQ )()(

roofrpoofpout hhhNhhhNQ )()(For a typical steadyFor a typical steady--flow combustion processflow combustion process

chapter 10


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Steady-Flow Combustion

The heat output during a combustion process issimply the difference between the energy of thereactants entering and the energy of the productsleaving the combustion chamber.

19chapter 10

Combustion chamber

CC33HH88(l)(l)

2525 C, 0.05kg/minC, 0.05kg/min

AIRAIR

77 CC

1500K1500K

HH22OO

COCO22COCO

OO22

NN22

Q=?Q=?

50 % excess air, 90% C in CO50 % excess air, 90% C in CO22, 10% C in CO, 10% C in CO

FindFind: the mass flow rate of air and the rate of heat transfer from the: the mass flow rate of air and the rate of heat transfer from the

combustion chambercombustion chamber

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Solution

20

Theoretical combustion equationTheoretical combustion equation

CC33HH88(l)+(l)+aathth(O(O22+3.76N+3.76N22))3CO3CO22+4H+4H22O+3.76aO+3.76aththNN22

OO22 balance:balance: aathth=3+2=5=3+2=5

CC33HH88(l)+5(O(l)+5(O22+3.76N+3.76N22))3CO3CO22+4H+4H22O+18.8NO+18.8N22

chapter 10


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Equation

21

Actual combustion equation

C3H8(l)+7.5(O2+3.76N2)2.7CO2+0.3CO+4H2O+2.65O2+28.2N2

kgfuelkgair

kmolkgkmolkmolkgkmol

kmolkgkmol

m

mAF

fuel

air

/53.25

)/2)(4()/12)(3(

)/29)(76.45.7(

The airThe ai r--fuelfuel ratioratio

min/18.1

min)/05.0)(/53.23(

kgair

kgfuelkgfuelkgairmAFm fuelair

chapter 10

UGM




Energy Balance

22

SubstanceSubstance

CC33HH

88(l)(l)

--118,910118,910 -- hhfuel,298Kfuel,298K --

OO22 00 81508150 86828682 49,29249,292

NN22 00 81408140 86698669 47,07347,073

HH22O(g)O(g) --241,820241,820 -- 99049904 57,99957,999

COCO22 --393,520393,520 -- 93649364 71,07871,078

COCO --110,530110,530 -- 86698669 47,51747,517

o

fh Kh280o

Kh298 Kh1500

chapter 10

poofproofrout hhhNhhhNQ )()(


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Heat

23

Qout = (1kmol C3H8)[(-118,910+hfuel,298K-hfuel,298K)kJ/kmolC3H8

+(7.5kmolO2)[(0+8150-8682)kJ/kmol O2]

+(28.2kmol N2)[(0+8141-8669)kJ/kmol N2]

-(2.7kmol CO2)[(-393,520+71,078-9364)kJ/kmol CO2]

-(0.3kmol CO)[(-110,530+47517-8669)kJ/kmol CO]

-(4kmol H2O)[(-241,820+57999-9904)kJ/kmol H2O]

-(2.65kmol O2)[(0+49292-8682)kJ/kmol O2]

-(28.2kmol N2)[(0+47073-8669)kJ/kmol N2]

= 363880 kJ/kmol C3H8

= 8270 kJ/kg C3H8

kWkgkJkgQmQ outout 89.6)/8270min)(/05.0(

chapter 10

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Closed Systems

24

Reactants

Products

Heat Transfer

systemoutin EEE

chapter 10


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Energy Balance

25

reactprodoutinoutin UUWWQQ )()(

Internal energyInternal energy

)(

)(

)(

TRhhhN

vPhhhN

PVhhhN

PVHU

u

oo

f

oo

f

oo

f

puoofpruoofrout TRhhhNTRhhhNQ )()(

For a typical closed combustion processFor a typical closed combustion process

chapter 10

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Closed Combustion

26

Find: the final pressure and the heattransfer during combustion

COCO22HH22OO

OO22

1000K1000KPP22

0.5 kmol CH0.5 kmol CH441.5 kmol O1.5 kmol O22

25 C25 C

1 atm1 atm

BeforeBefore AfterAfter

chapter 10


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Equation

27

The combustion equation

0.5CH4+1.5O2 0.5CO2 +H2O + 0.5O2

produprodprod

reactureactreact

TRNVP

TRNVP

atmK

K

kmol

kmolatm

T

T

N

NPP

react

prod

react

prod

reactprod

36.3298

1000

2

2)1(

chapter 10

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Table

28

SubstanceSubstance

CHCH44 --74,85074,850 hh --

OO22 00 86828682 31,38931,389

COCO22 --393,520393,520 93649364 42,76942,769

HH22O(g)O(g) --241,820241,820 99049904 35,88235,882

o

fho

Kh298 Kh1000

chapter 10


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Heat

29

Qout = (0.5kmol CH4)[(-74,850+hfuel,298K-hfuel,298K)kJ/kmolCH4]

+(1.5kmolO2)[(0+8682-8682-8.314*298)kJ/kmol O2]

-(0.5kmol CO2)[(-393,520+42769-9364-8.314*1000)kJ/kmol CO2]

-(1kmol H2O)[(-241,820+35882-9904-8.314*1000)kJ/kmol H2O]

-(0.5kmol O2)[(0+31389-8682-8.314*1000)kJ/kmol O2]

= 358794 kJ/kmol CH4

= 22425 kJ/kg CH4

chapter 10

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Combustion Processes

30

roofrpoofpout hhhNhhhNQ )()(

For a typical steadyFor a typical steady--flow combustion processflow combustion process



chapter 10


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Adiabatic flame temperature

Assume:

o Heat transfer from the system is zero, i.e., adiabaticcombustion.

Thus:

o All the heat of reaction goes into the products.

o The temperature in this case is called the adiabatic flametemperature.

31chapter 10

UGM




Adiabatic flame temperature

32

roofrpoofp hhhNhThhN )())(( *


puoofpruoofr TRhThhNTRhhhN ))(()( **


T*T* -- adiabatic flame temperatureadiabatic flame temperature

chapter 10


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Adiabatic Flame Temperature

Not unique

Depends on the state of the reactants, the degree ofcompletion of the reaction (products), the amount ofair used.

The actual maximum temperature encountered in acombustion chamber is lower than the theoreticaladiabatic flame temperature, due to incompletecombustion (bad mixing), heat loss, dissociation of

products.

33chapter 10

UGM




Determining T*

T* is obtained by assuming a value to satisfy theenergy balance by trial and error.

Note that the specific heat is not a constant, and thusa table of values would be needed.

34chapter 10


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Problem

35

Find: the adiabatic temperature for (a) complete combustionwith 100% theoretical air, (b) complete combustion

with 400% theoretical air, (c) incomplete combustion(with CO) with 90% theoretical air

CombustionCombustion

chamberchamber

CC88HH1818(l)(l)

2525 C, 1 atmC, 1 atm

AirAir2525 C, 1 atmC, 1 atm

TTpp

1 atm1 atm

COCO22HH22OO

NN22OO22

chapter 10

UGM




Balance

36

(a) The combustion equation

C8H18(l)+12.5(O2+3.76N2)

8CO2+9H2O+47N2

roofrpoofp hhhNhThhN )())(( *Energy balance for steadyEnergy balance for steady--flow systemflow system

chapter 10


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Table

37

SubstanceSubstance

CC88HH1818(l)(l) --249,950249,950 hh --

OO22 00 86828682 --

NN22 00 86698669 ??

COCO22 --393,520393,520 93649364 ??

HH22O(g)O(g) --241,820241,820 99049904 ??

o

fho

Kh298 *Th

chapter 10

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Temperature

38

(8kmol CO2)[(-393520+hCO2,T*-9364)kJ/kmolCO2]

+(9kmol H2O)[(-241,820+hH2O,T*-9904)kJ/kmol H2O]

+(47kmol N2)[(0+hN2,T*-8669)kJ/kmol N2]

= (1kmol C8H18)[-249950kJ/kmol C8H18]

which yields

8hCO2,T*+9hH2O,T*+47hN2,T*=5646081kJ

Assume the value of T* until

Left=right=5646081kJ

T*=2394.5 KT*=2394.5 K

chapter 10


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39

(b) The combustion equation

C8H18(l)+50(O2+3.76N2->

8CO2+9H2O+37.5O2+188N2

Repeat the procedure in (a)

T*=962 KT*=962 K

chapter 10

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40

(c) The combustion equation

C8H18(l)+11.25(O2+3.76N2)

5.5CO2+2.5CO+9H2O+42.3N2

Repeat the procedure in (a)

T*=2236 KT*=2236 K

chapter 10


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First Law

41

roofrpoofpout hhhNhhhNQ )()(




chapter 10

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Absolute entropy

42

The entropy value relative to a state ofThe entropy value relative to a state ofabsolute zero, 0 K or 0 R, is called theabsolute zero, 0 K or 0 R, is called the

absolute entropyabsolute entropy..

The entropy of a pure crystalline substanceThe entropy of a pure crystalline substanceat absolute zero temperature is zero.at absolute zero temperature is zero.

T0 273.15

25oC

298.15o K

0o C

chapter 10


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The standard reference state

43

Standard reference stateStandard reference state

at p = 1 atmat p = 1 atm

orefref sPTs ),(Molar basis:Molar basis:

Mass basis:Mass basis:o

refref sPTs ),(

T0 273.15

25oC

298.15 K

0o C

T = 298.15 K and pref = 1 atm.

chapter 10

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Entropy at any other state

44

When the absolute entropy is known at theWhen the absolute entropy is known at the

standard state, the entropy at any given statestandard state, the entropy at any given state

can be found by adding the entropy changecan be found by adding the entropy change

between the standard state and the givenbetween the standard state and the given

state.state.

T0 273.15

25oC

298.15 K

0o C

),( refrefo

PTs ),( refo

PTs

chapter 10


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Evaluating entropy at T and pEvaluating entropy at T and prefref

45

From the T-dS equation, evaluate entropy relative to

the reference pressure and temperature via the

integrated result on a molar basis.

refref PTTrefo

ref SPTsPTs ,),(),(

T0 273.15

25oC

298.15 K

0o C

),( refrefo

PTs ),( refo

PTs

chapter 10

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Absolute entropy at T and p

46

so(T) Absolute entropy per mole at

T and p = 1 atm.

),(),(),(),( refref pTspTspTspTs

To evaluate the second term, recall

p

dpR

T

TCds

dpp

V

T

dpCds

VdPdTCsTd

p

P

p

dpT

VdT

T

Cds

p

dpp

RdT

T

Cds

p chapter 10


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Absolute entropy at T and p

47

The standard reference state is 298.15 K and 1 atm(537 R and 1 atm). Absolute entropy is determined

relative to this reference state.

ref

o

p

p

p

RTspTs

p

dpR

T

Tcsd

ln)(),(

Absolu te entropy at T and

p ref = 1 atm.

dpp

RdT

T

Cds

p

chapter 10

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Entropy of each component

48

Mixture at T and P

n1+n2+n3+...nN = nTotal

ref

ioi

ref

ioii

P

PyRTs

P

pRTspTs

ln)(

ln)(),(

For each component in the mixture:For each component in the mixture:

chapter 10


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Entropy balance

49

Reaction chamberReaction chamber

Heat Transfer

1 2

j jj

genreactprodT

Q

SSS

SSreactreact SSprodprod

chapter 10

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Exergy destroyedExergy destroyed

50

)(kJSTX genodestroyed

Absolute temperature of the surroundingsAbsolute temperature of the surroundings

chapter 10


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Reversible Work

51

p

o

T

o

fpr

o

T

o

fr

p

oo

fpr

oo

frrev

gggNgggN

sThhhNsThhhNW

00

00

The maximum work that can be producedThe maximum work that can be produced

during a processduring a process

Gibbs function of formationGibbs function of formationIf TIf Tprodprod=T=Treactreact=T=T00=25=25 CC

ofpofrrev gNgNW

chapter 10

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Second-law analysis of adiabatic combustion

52

Find: (a) the temperature of the products (b) theentropy generation (c) the reversible work (d)exergy destructed

AdiabaticAdiabatic

combustioncombustion

chamberchamber

CHCH44

2525 C, 1atmC, 1atm

AIRAIR

2525 C, 1atmC, 1atm

COCO22HH22OO

OO22

NN22

50% excess air, complete combustion50% excess air, complete combustion

chapter 10


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53

(a) Combustion equation with 50% excess air

CH4(g)+3(O2+3.76N2)->

CO2+2H2O+O2+11.28N2

The adiabatic flame temperature is determined

chapter 10

roo

frpoo

fp hhhNhhhN )()(

UGM




54

SubstanceSubstance kJ/kmolkJ/kmol kJ/kmolkJ/kmol

CHCH44(g)(g) --74,85074,850 --

OO22 00 86828682

NN22 00 86698669

HH22O(g)O(g) --241,820241,820 99049904

COCO22 --393,520393,520 93649364

o

fh Kh298

chapter 10


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55

)/74850)(1(

]/)86820)[(1(

]/)86690)[(28.11(

]/)9904241820)[(2(

]/)9364393520)[(1(

44

222

222

222

222

kmolCHkJkmolCH

kmolOkJhkmolO

kmolNkJhkmolN

OkmolHkJhOkmolH

kmolCOkJhkmolCO

O

N

OH

CO

kJhhhh ONOHCO 93795028.112 2222

KTprod 1789

chapter 10

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56

j j

j

genreactprodT

QSSS

i

rmiu

o

ii

i

pmiu

o

ii

rrppreactprodgen

PyRPTsN

PyRPTsN

sNsNSSS

]}ln),([{

]}ln),([{

0

0

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57

NNii yyii

CHCH44 11 11 186.16186.16 -- 186.16186.16

OO22 33 0.210.21 205.04205.04 12.9812.98 654.06654.06

NN22 11.2811.28 0.790.79 191.61191.61 1.961.96 2183.472183.47

SSreactreact=3023.69=3023.69

COCO22 11 0.06540.0654 302.517302.517 22.67422.674 325.19325.19

HH22OO 22 0.13090.1309 258.957258.957 16.90516.905 551.72551.72

OO22

11 0.06540.0654 264.471264.471 22.67422.674 287.15287.15

NN22 11.2811.28 0.73820.7382 247.977247.977 2.5242.524 2825.652825.65

SSprodprod=3989.71=3989.71

),( 0PTso

i miu PyR ln iisN

chapter 10

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58

KkmolkJ

KkmolkJ

SSS

CH

CH

reactprodgen

4

4

/02.966

/)69.302371.3989(

4

4

0

/874,287

/874,287

)/02.966)(298(

kmolCHkJXW

kmolCHkJ

kmolKkJK

STX

destroyedrev

gendestroyed

chapter 10

at09 10 combustion

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