8 Combustion Bejan

8/2/2019 8 Combustion Bejan

1/24

Thermodynamicsof

Steady Flow Combustion


2/24

SteadyFlowCombus2on

Combus2onstoichiometry Applica2on of the 1st law and nd law tocombus2onprocesses

Combined1standndlaw ndlawefficiency


3/24

Combus2onStoichiometry

Combus'onburningorrapidoxida2onofHCfuels GenericformofHCfuelsCaHb CommonOxidantAtmosphericair 0.1O+0.79N (for1moleofair) O+3.76N (for1moleofO)

CompleteCombus,on

CaHb + Air

CO+HO IncompleteCombus,on

CaHb + AirCO+CO+HO

3


4/24


StoichiometricRela2on

Wheretheistheminimumno.ofmolesofOper

moleofCaHb effec2ngthecompletecombus2onofCaHb

Defining4

CH

+ O

2+3.76 N

2( )CO2 + H2O+N2wheretheunknownstoichiometriccoefficients are= =

2=+

4= 3.76 +

4"#$ %

&'

4

+

Actual air flow rate1

Theorectical air flow rate = >


5/24


Whereisdefinedastheexcessair

IfIncompletecombus,on IfpureOisused:

5

( )

( )

2 2

2 2 2 2

3.76

4

3.76 12 4 4

C H O N

CO H O N O

+ + +

+ + + + +

( )1

1<

Thereac2onissaidtobewith400%theore2calairorwith

300%excessair


6/24

1stLawAnalysis

Open,Steadyflowsystem

Let,themolalflowrateoffuel

6

Reactants ProductsHr HP

Q

W

ControlVolumeCombus2onChamber

0 = Q W+ nrihri

i=1

m

npihpii=1

n

nr1=

nfuel

0 =

Q

nfuel

Q

W

nfuel

W

+ rih

ri

i=1

m

Hr

pihpi

i=1

n

Hp

0 =QW+HrHp


7/24

1stLawAnalysis

Case:W=0andtempofproductsandreactantsare

T0,P0

quan22esevaluatedatreferenceToandPo.

Q


8/24

1stLawAnalysis

Enthalpyofforma'on

Combus2onofCHwith(-1)excessair

Q:Enthalpyofcombus2onofCH.

8

( )

( )

, 0 Elementary substances 0

, For Compounds

o

o o f

o

o o f

h T P h

h T P h

= =

=

Hr= h

f CH

o

+ 0O

2

+ 0

N2

H

p=h

f CO2

o+

2h

f H2O

o+0+0

Heat transfer per mole of fuelQ =hf CO

2

o+

2h

f H2O

o

hf C

H

o


9/24

1stLawAnalysis

Hea2ngvalueoffuel=|Q|=HrHp

IfHydrogenappearsasHO(vapor)LHV IfHydrogenappearsasHO(liquid)HHV

9


10/24

1stLawAnalysis

Case:Reactants&ProductsareatdifferentToandPo

LetTandPrepresentthetemperatureandpressureof the one par,cular cons,tuent in the stream of

reactantsorproductsandletbeitsenthalpy:

10

0

r p

ri pi o o

Q W H H

but h and h are not at T and P

= +

( ),h T P

h T,P( )= h To ,Po( )+ h T,P( ) h To ,Po( )

h

h T,P( ) = hfo+h where h is the enthalpy change


11/24

1stLawAnalysis

At low pressures, cons2tuents obey the ideal gasmodel:

Inconclusion,thepermoleoffuelenthalpyis:

Fig 7.7 in the textbook shows that isprac2callylinearintemp.

11

( ) ( ) 0oh h T only h T = =

( ) ( )1 1

m no o

r ri f p pi f ri pii i

H h h H h h = =

= + = +

h T( )


12/24

1stLawAnalysis

Case: W = 0 and Q = 0 (Adiaba2c). Adiaba2c Flame

Temperature

heretheproductsreachextremelyhightemperatures.

System external to combus2on chamber receivesjoulespermoleoffuelLHV

SystemisinsulatedCannotdisposeLHV1

r pH H=

Hea2ngValue

Combus2on Products,To,Po

Products

Taf,Po

Reactants

To,Po

Adiaba2candzeroworkcontrolsurface


13/24

1stLawAnalysis

LHVheats themixtureof products to theAdiaba,cFlameTemperature(Taf)

TafisobtainedfromHr(To,Po)=Hp(Taf,Po) Tafistheore2callyhighestvaluebecause:

nocombus2onchamberisperfectlyinsulated atTafoneoftheproductscanundergochemicaldissocia2on

E.g.COCO+O

13


14/24

14


15/24

1stLawAnalysis

Dissocia2onendsTnewproducts=Toriginalproducts

Tafd


16/24

ndLawAnalysis

Representsentropyinterac2ons

16

ControlVolume(C.C)Reactants

Sr

Products

SPTi

Qi

/Ti


17/24

ndLawAnalysis

Thepermoleoffuelstatement

C.C.actsasproducerofentropy Adiaba2ccombus2on: Inthe Incombus2onprocesses

17

Sgen

=

Qi

Tii

Sr + Sp 0

0gen p r

S S S=

0lim 0 ( )T s s s T = =

( , )s s T P=


18/24

ndLawAnalysis

AtP=Po(atm)

isduetopressurechangeatconstanttemperature

Inmanycasestemperaturesareveryhighandpar2alpressurePsarelowtojus2fytheidealgasmodel

18

( ) ( ), oos T P s T=

s T,P( ) = s o T( )+ s T,P( ) s T,Po( )s


19/24

ndLawAnalysis

Assumingallcons2tuentsareidealgases:

If the cons2tuents appear in liquid form, ICsubstancemodelisused

19

1

1

ln

ln

m

r ri o

i o ri

n

p pi o

i o pi

PS s R

P

PS s R

P

=

=

=

=


20/24

MaximumPowerOutput

How chemical reac2ons are used to producemechanicalpower?

1stLaw ndLaw

0

W

Products

Hp,Sp,Bp

Reactants

Hr,Sr,Br Combus2onchamber

andpowerplant

Steadyflowapparatus

TemperaturereservoirToQo

To

0o r p

Q W H H =

+

0o

gen r p

o

QS S S

T= +


21/24

MaximumPowerOutput

Combined1standndlaw

Workoutputpermoleoffuel


22/24

MaximumPowerOutput

SpecialCase:inletandoutletmixturesareatTo,Po

whereisthepar2almolalGibbsfreeenergy

Worklost(ToSgen)isdueto: combus2onprocesses powerplantbetweencombus2onchamberandambient

( ) ( ), ,

1 1

, ,

1 1

m n

rev ri o pi oo ri o pii i

m n

rev ri o ri pi o pi

i i

W h T s h T s

W

= =

= =

=

=

( )oo

h T s

rev r p

rev

W G G

W W

=


23/24

MaximumPowerOutput

Case:PowerplantreplacedbyCarnotengine

Irreversibility(WrevW)isduetoC.C.alone

3

W

QL

Carnotcycle

Powerplant

Sameas-Qo

QH=HrHp

Tf

ReservoirTo

Combus2onChamber

(To,Po)Reactants

Hr,Br

(To,Po)Products

Hp,Bp


24/24

MaximumPowerOutput

Effec2veflametemperatureoftheC.C.

ndlawefficiency

4

Wexergy

= H

rH

p( )heattransfer

1T

o

Tf

"

#$$

%

&''

1r p o

II

rev r p f

H H TW

W B B T

= =

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8 Combustion Bejan