Thermochemistry_2

8/12/2019 Thermochemistry_2

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Mole

Formally defined as the amount of substance that contains as many elementary

entities as there are in exactly 0.012 kg of Carbon 12.

Abbreviation of mole is mol 1kmol=103mol.

But how many elementary entities are there?

6.022141991023entities molecules or atoms;

This number is called Avogadro number (NAV)

Alternate definition

A mole is an Avogadro number of units of a substance

Avogadro, who?

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Ideal Gas Mixtures (contd.)

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Relations between mole fractions and mass fractions :

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Mixture molecular weight:

imixii

mixiii

MWMWYX

MWMWXY

/

/

i

ii

mix

iiimix

MWYMW

MWXMW

)/(

1


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Pressure of Ideal Gas Mixtures

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Partial Pressureof ithspecies :

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PXPii

i

iPP

Mixture Pressure:

Daltons Model


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H, U, cpandcvof Ideal Gas Mixtures

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i

ii

i

ii ThmThNTH )()()(

Mixture Enthalpy:

Mixture Specific Enthalpy:

)()( ThXThi

ii

)()( ThYThi ii

U, cpandcvexhibit similar behaviors


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Standard Enthalpy / Heat of Formation

is defined as the heat evolved when 1 mole of thesubstance is formed from its elements in their respective

standard states. (istands for the ithcompound)

Standard-state temperature:T0 = 25C (298.15K)

Standard-state pressure:

P0= 1 atm (101,325Pa)

Elements at T0and P0are assigned

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)( 00, Th if

0)( 00

, Th if


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Standard Values

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Compound / Element (kJ/mol)

CH4(g) 74.81

CO(g) 110.53

CO2(g) 393.51

C2H4(g) +52.26

H2O(g) 241.82

H2O(l) 285.8NO(g) +90.25

)( 00

, Th if

)( 00

, Th if


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Absolute Enthalpy,

at any state (T,P) is defined as the sum ofand the sensible enthalpy change between the standard state

(T0,P0) and the given state (T,P). (istands for the ithcompound)

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)(Thi

)()()( 0,00

, TThThTh isenifi

)(Thi

)( 00, Th if


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Ideal GasTables (Turns P. 622)

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Tables (Turns P. 646)

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Example

Express at any state (T,P0) in terms of andthe enthalpy change of all elements involved in the formation

reaction.

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j

jjf ThThTh )()()( 0'0

)(Th)(0 Thf

)(' compoundMj

jj T

T0Reac Prod


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Example

A gas stream at 1 atm contains a mixture of CO, CO2,and N2in which the CO mole fraction is 0.10 and the CO2mole

fraction is 0.20. The gas stream temperature is 1200K.

Determine the absolute enthalpy of the mixture on bothmole-basis (kJ/kmol) and mass-basis (kJ/kg).

Also determine the mass fractions of the three components.

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S of Ideal Gas Mixtures

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Mixture Entropy:

Mixture Specific Entropy:

),(),(

),(),(

ii

i

i

ii

i

i

PTsXPTs

PTsYPTs

i

iii

i

iii PTsmPTsNPTS ),(),(),(


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Pure Species Entropy,

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0

0 ln)(),(P

PRTsPTs iuii

),( PTsi

0

0 ln)(),(P

PRTsPTs iii

T

dTcTsTs

T

T iPii

0,0

00 )()(Where,


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Stoichiometry

Stoichiometric mixture:Amount of oxidizer required to completely

burn/oxidize a given amount of fuel (Oxst) (no dissociation)

Fuel Lean mixture:Ox > Oxst

Fuel Rich mixture:

Ox < Oxst

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Hydrocarbon Stoichiometry (contd.)

Assume air is composed of 79% N2

and 21% O2by volume

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22222 76.3)2/()76.3( aNOHyxCONOaHC yx

4/yxa


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Air-Fuel Ratio

Stoichiometric air-fuel ratio :

17

)/(

1

76.4)/( kgkg

MW

MWa

m

mFA

fuel

air

stfuel

airst

Fuel (A/F)st

CH4+ air 17.11H2+ O2 8.0

C(s) + air 11.4


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Equivalence Ratio

18

st

st

AF

AF

FA

FA

)/(

)/(

)/(

)/(

Mixture

Stoichiometric 1

Lean 1


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Example

A natural gas fired boiler operates with an O2concentration of 3% by mole (wet basis) in the flue gases.

Determine the operating air-fuel ratio and the equivalence

ratio. Treat the natural gas as methane.

Recalculate using dry basis of measurement of O2.

If the inlet fuel flow rate is 20 kg/s, find the flue gas flow

rate.

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Enthalpy or Heat of Reaction,

Steady State Steady Flow Reactor with complete combustion

Reactants and products at T0and P0

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RH

i

ii

i

iiRPR hhHHH '"

i

ii

i

ii MM "'

(KJ)


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Forms of Enthalpy of Reaction

Per unit mole of fuel : (KJ/mol of fuel)

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Rh

Per unit mass of fuel : (KJ/kg of fuel)fuelRR MWhh /

Per unit mass of mixture : (KJ/kg of

mixture)1)/(

FA

h

m

mh R

mix

fuel

R


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Enthalpy of Reaction for CH4

Per unit mole of fuel : (KJ/mol of fuel)

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405,802 Rh

Per unit mass of fuel : (KJ/kg of fuel)016,50 Rh

Per unit mass of mixture : (KJ/kg of

mixture)8.2761 Rh

kJHR 405,802


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Heat of Combustion,

Upper / Higher Heating Value (HHV): Heat of combustioncalculated assuming all water in product stream has been

condensed to liquid

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RC HH

CH

Lower Heating Value (LHV) : Heat of combustion calculated

assuming all water in product stream is in gaseous form


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Tables (Turns P. 649)

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Example

Express the heat of reaction at any state (T,P0) interms of the standard enthalpy of formation and the specific

heats at constant pressure of the reactants and products.

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i

T

T ipifi

i

T

T ipifiR dTcThdTcThTh

00,0

0

,

'

,0

0

,

" )()()(

)(ThR

T

T0Reac Prod

iii

iii MM

"'


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Example

Determine the HHV and the LHV at 298 K of n-decane (gas)(C10H22) (MW = 142.284 kg/kmol)

per kmol of fuel.

per kg of fuel.

Recalculate for n-decane (liquid). (hfg= 359 kJ/kgfuelat 298 K)

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Adiabatic Flame Temperature,

Adiabatic Flame Temperature for P= Constant : Temperature ofthe products after complete adiabatic combustion of a fuel-air

mixture at constant pressure

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adT

Reac to

Prod

),(),( PTHPTH adprodireac

Gas turbine combustorDiesel engine

Furnace

Also called the Adiabatic Frozen Flame Temperature


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Adiabatic Flame Temperature,

Adiabatic Flame Temperature for V= Constant : Temperature ofthe products after complete adiabatic combustion of a fuel-air

mixture at constant volume

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adT

Reac to

Prod

),(),( fadprodiireac PTUPTU

Bomb calorimeterGasoline engine

0)(),(),( adprodireacufadprodiireac TNTNRPTHPTH


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Example

Estimate the constant pressure adiabatic flametemperature for the combustion of a stoichiometric methane-

air mixture. The pressure is 1 atm and the initial reactant

temperature is 298 K. Use the following assumptions

Complete combustionUse constant specific heats at 1200 K for evaluating

the product enthalpies

Recalculate using variable specific heats

Recalculate using tables from Appendix A

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http://www.youtube.com/watch?v=tvvuUILhHUo

http://www.youtube.com/watch?v=65KIexy4New
http://www.youtube.com/watch?v=tvvuUILhHUohttp://www.youtube.com/watch?v=65KIexy4Newhttp://www.youtube.com/watch?v=65KIexy4Newhttp://www.youtube.com/watch?v=tvvuUILhHUo


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Example

Calculate the constant pressure adiabatic flame temperatureof water vapor based on the reaction of gaseous H2and O2.

Recalculate the same starting with liquid H2at -255C and

liquid O2at -225C (cryogenic combustion). Boiling point ofliquid O2is 182.96C and that of liquid H2is 252.87 C.

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Dissociation of Species

Unknowns :

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2222224 2222)76.3( OnNnOHnCOnNOaCH ONOHCO

No dissociation (Low flame temperature) :

adONOHCO Tnnnn ,,,, 2222

OnNnOHnNOnCOnOnNnOHnCOnNOaCH

ONOHNOCO

ONOHCO 2222224 2222)76.3(

Dissociation (High Flame temperature):


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Equilibrium Reactions

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OHHOH 2221

OHHOH 2

2221 OCOCO

222 ONNO

HH 22

OO 22

NN 22

GRI-Mech 3.0: Methane-air

reaction mechanism with53 species and 325 reactions.

http://www.me.berkeley.edu/gri_mech/
http://www.me.berkeley.edu/gri_mech/http://www.me.berkeley.edu/gri_mech/


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Condition for Equilibrium

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Since most combustion systems are in equilibrium at a

particular pressure and temperature, the general criterion is :

0, PTdG

where,

),()(, PTTSTHPTG


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Free Energy

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0

0 ln,P

PTNRTGPTG u

0

0 ln,

P

PTRTgPTg u

elementsj

jjf TgTgTg )()()( 0'00


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Derivation of (Equilibrium Constant)

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PK

dDcCbBaA

Consider the following simplified reaction at P0and T

Gof the mixture of A,B,C, and D:

ii

uiiPPTRTgNPTG 0

0 ln,


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Derivation of (contd.)

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PK

Differentiating :

i

iuii

i

iuii

P

PTRTgdN

P

PTRTgdNPTdG

0

0

0

0

ln

ln,

At equilibrium: 0, PTdG


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Derivation of (contd.)

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PK

i

ii

u

i

ii dNP

PTRdNTg 0ln

0

0

Since dNiis proportional to the stoichiometric coefficients:

Pu KTRTG ln)(0

TR

TGK

u

P

)(exp

0


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bB

a

A

dD

cC

P

PPPP

PPPPK

00

00

PK

Alternatively for a general reaction,

i

isi

i

isi MM "

,

'

,

Where:

i

iP

sisi

PPK',

",0


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)(0 TG

G0(T)is defined as the Standard state Gibbs Energy Change,

)()(

)(

0Reac0 odPr

0000

00

TGTG

TgdTgcTgbTga

dNTgTG

DCBA

iii


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4141

Alternatively for a general reaction,

i

ifsi

i

ifsi TgTgTG )()()( 0

,

'

,

0

,

"

,0

Where is defined as the Gibbs Free Energy of Formationof

the ithcompound, listed in Appendix A.

)(0, Tg if

)(0 TG

i

isi

i

isi MM "

,

'

,


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Observations about

P in KPsignifies that the equilibrium constant is written interms of partial pressures.

Other quantities that may be used to define Kare

concentration, CiNumber of moles, NiMole fraction,Xi

For a positive value of G0, KPis a fraction, thus reactants would

be favored at equilibrium

For a negative value of G0, KPis greater than unity, thus

products would be favored at equilibrium

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PK


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Example

Consider the dissociation of CO2as a function of Pand T.

State the necessary relations required to derive thecomposition of the mixture that results from subjecting

originally pure CO2to T1and P1.

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)(2

1)()( 22 gOgCOgCO


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Example

Consider the combustion of methane under fuel-rich conditions.

Assume that the following water-gas equilibrium reaction occurs

within the product species.

Namely, a portion of the combustion product undergoes a further

reaction. The equilibrium mixture may be assumed to consist of

Write the necessary mathematical relationships required for

solving the equilibrium composition and the adiabatic flame

temperature. 45

OHCOOCH 224 223

222 HCOOHCO

222)2()1( xHxCOOHxCOx


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Propane-Air Combustion (Turns P. 46)

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CO2, CO,

H2O, H2, H,OH, O2, O,

NO, N2, and

N


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NASA CEA Code

Equilibrium compositions are important in various systems

Gas turbines

Aircraft combustors

Rocket motors

Shock tubes

Automobile engines

Nozzles and diffusers

Gun propulsion systems 48

Chemical Equilibrium with

Applications (CEA) :

Developed at NASA Lewis(Glenn) Research Centre by

Gordon, McBride, Zeleznik,

and Svehla

http://www.grc.nasa.gov/WWW/C

EAWeb/ceaguiDownload-win.htm
http://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htmhttp://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htmhttp://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htmhttp://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htmhttp://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htmhttp://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htm


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NASA CEA Code Problems Handled

Equilibrium compositions of assigned thermodynamic

states

Theoretical rocket performance

ChapmanJouguet detonations

Shock-tube parameter calculations for both incident andreflected shocks

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NASA CEA Code Problems Handled

Equilibrium compositions of assigned thermodynamic states Temperature and Pressure

Enthalpy and Pressure - Constant Pressure Combustion

Entropy and Pressure

Temperature and Volume or Density

Internal Energy and Volume - Constant VolumeCombustion

Entropy and Volume

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CEA Input

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CEA Input (contd.)

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CEA Input (contd.)

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CEA Input (contd.)

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CEA Output

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5656

CEA Output for = 0.6


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CEA Output for = 0.6 (Contd.)

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CEA Output for = 1.05

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Thermochemistry_2

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