Stoichiometry, Energy Balances, Heat Transfer, …geoff/air_poll/lectures_handouts/combustion...1 Stoichiometry, Energy Balances, Heat Transfer, Chemical Equilibrium, and Adiabatic

1

Stoichiometry, Energy Balances, Heat Transfer, Chemical Equilibrium, and Adiabatic Flame

Temperatures

G f SilGeof Silcox

[email protected]

(801)581-8820

University of Utah

Chemical EngineeringChemical Engineering

Salt Lake City, Utah

May 2007

Stoichiometry

2

STOICHIOMETRY

• Outlinecombustion reactions– combustion reactions

– combustion with air

– equivalence ratio, stoichiometric ratio

– percent theoretical air, percent excess air

– percent excess oxygen in stack gases

– dry and wet basis for concentrationsdry and wet basis for concentrations

– actual and theoretical concentrations

– the effects of mixing and in leakage

STOICHIOMETRY

• Combustion reactions

2 2CH O CO H O In general4 2 2 2

2 2

2 2 2

2 2

1

21

2

CH O CO H O

CO O CO

H O H O

In general

2 2

2 24 2

nC nO nCO

m mmH O H O

or 2 2 24 2n m

m mC H n O nCO H O

3

STOICHIOMETRY

• Combustion with air

Assume air is 21% O2 and 79% N2, by volume.Then each mole O2 brings with it 3.76 mole N2.

2 23.764 4n m

m mC H n O n N

2 2 23.762 4

m mnCO H O n N

STOICHIOMETRY

• Equivalence ratio

• Stoichiometric ratio

mols fuelmols oxidant supplied by operator

mols fuelmols oxidant reqd for complete combustion

1SR

SR = 1/ = 1 means that conditions arestoichiometric or chemically balanced.

4

STOICHIOMETRY

• and SRF l i h ( d i ) 1 1SR d – Fuel rich (reducing):

– Fuel lean (oxidizing):

– Alternate forms w/ constant amount of fuel

1 1SR and 1 1SR and

mols oxidant reqd

l id t f d

mols oxidant fed

mols oxidant fedSR

mols oxidant reqd

STOICHIOMETRY

• Percent theoretical air• Percent theoretical air

• Percent excess air

( )100

%TA 100 SR

1 1%EA %TA 100 100 SR 1 100 1

5

STOICHIOMETRY

• Percent excess oxygen in stack gases– Defined as actual mole (volume) percent inDefined as actual mole (volume) percent in

stack. Not to be confused with %EA or %TA.

– Can be measured on a dry or wet basis.

• Dry or wet basis for stack measurement– Dry basis gives higher value because y g g

water vapor acts as a diluent and is 10 –15% of stack gases

STOICHIOMETRY• Actual and theoretical concentrations

0.18

0.2

sis effects of

imperfect mixing

0.06

0.08

0.1

0.12

0.14

0.16

s m

ole

fra

ctio

ns,

dry

ba

s

CO2 theoretical

O2 theoreticalCO theoretical

imperfect mixingeffects of airleakage

0

0.02

0.04

0.8 1 1.2 1.4 1.6 1.8 2

SR

Flu

e g

as

6

STOICHIOMETRY

• Controlling excess air

100 Th i i CO i t

50

100

ppm

CO

The rise in CO is steepas you approach SR = 1and it is insensitive toair leakage. Air flowcontrol by CO shouldlead to a close approachto SR = 1

1.0 1.1 1.2

Stoichiometric ratio

to SR = 1.

STOICHIOMETRY

• Spreadsheets for furnace material• Spreadsheets for furnace material balance calculations– FURN_MB.xls: balance for liquid and solid

fuels

– GAS_MB.xls: balance for gaseous fuels

• Author: Jost Wendt

7

ENERGY BALANCES

ENERGY BALANCES

• Outline– Conservation of mass and energyConservation of mass and energy– Example: drying salt with hot flue gas– Energy balances on systems that involve

reactions - the absolute enthalpy– Example: adiabatic combustion of H2 in air– Heats of combustion and higher and lower heating

values– Energy balance on a furnace – available heat and

the efficiency of combustion

8

ENERGY BALANCES

• Energy and mass are conserved

gz2

1ue 2 V

h = u + pv

Controlvolume

or system

Massentering (1)

Massleaving (2)

Q W

outinCV mm

dt

dm

2 2

, , 1 2

1 2

V V

2 2CV

in net in net

dEQ W m h gz m h gz

dt

ENERGY BALANCES

• Changes in kinetic and potential energy are usually small compared to changes in internal energy

, , 1 1 2 2CV

in net in net

dUQ W m h m h

dt Unsteady

, , 1 1 2 2 0in net in netQ W m h m h Steady

9

ENERGY BALANCES

• Drying salt with combustion gases

Flue gas at 1000F to be used to dry 40 tons/h saltthat is 3 wt. % water. The inlet temperature of thesalt is 40F. The dry salt exits the dryer at190-210F. What flow rate of flue gas do we need if we assume the flue gas leaves the dryer at 400F?

ENERGY BALANCES

• Drying salt with combustion gases

Dryer

1

2

3

4

5

6

1 ?m flue gas, 1000F

dry salt, 40F77,600 lb/h

H O(l) 40F

1 4m m flue gas, 400F


H O(g) 400FH2O(l), 40 F2400 lb/h

H2O(g), 400 F2400 lb/h

1 4m m 3 6m m

2 5m m

10

ENERGY BALANCES

• Drying salt with combustion gases– energy balance over adiabatic dryer

Dryer

1

2

3

4

5

6

1 ?m flue gas, 1000F


H2O(l), 40F

1 4m m flue gas, 400F


H2O(g), 400F

1 1 4 2 2 5 3 3 6 0m h h m h h m h h

2 5 2 3 6 31

1 4

m h h m h hm

h h

2400 lb/h 2400 lb/h

ENERGY BALANCES

• Drying salt with combustion gasesenergy balance over adiabatic dryer– energy balance over adiabatic dryer

2 5 2 3 2 ( ) 5 3 2 ( ) 6 5

11 4

salt H O l fg pH O g

pgas

m c T T m c T T h c T Tm

c T T

978 0.254fg pgas

Btu Btuh c

lb lb F

2 ( )

2 1

0.21 0.47

1.0 36,500

fg pgas

salt pH O g

H O

lb lb FBtu Btu

c clb F lb FBtu lb

c mlb F h

11

ENERGY BALANCES

• Energy balances on reacting systems –the absolute enthalpypy

( ) ( )T

oi f i ref p ih T h T c dT

The reference state for systems in whichreactions are occurring is conveniently chosen to be the elements at Tref = 25 C (298.15 K) and 1 bar.

, ,( ) ( )ref

i f i ref p i

T

h T h T c dT Absoluteenthalpy

Enthalpyof formationat 1 bar, 298 K

Sensibleenthalpychange

ENERGY BALANCES

• Heat capacity as a function of T (K)

( ) ( )T

oh T h T dT, ,( ) ( )ref

oi f i ref p i

T

h T h T c dT 2 3 4

1 2 3 4 5

2 3 43 5 62 4

pca a T a T a T a T

Ra a aa ah

a T T T T

1 2 3 4 5a T T T T

RT T

where T is in Kelvin and R is the universal gasconstant, R = 8.3145 kJ/(kmol K).

Spreadsheet: ther_coef.xls

12

ENERGY BALANCES

• Adiabatic combustion of H2 with air, no dissociation. What is exit temperature?p

Controlvolume

Inlet (i)0.42 kmol/s H2

0.21 kmol/s O2

0.79 kmol/s N2

300 K, 1 bar

Exit (e)0.42 kmol/s H2O0.79 kmol/s N2

T = ?, 1 bar

0inQ 0inW

, , 0in net in net i i e ei e

Q W n h n h

Spreadsheet: h2_flame.xls

ENERGY BALANCES

• Adiabatic combustion of H2

0i i e ei e

n h n h

, ,( ) ( )ref

To

i f i ref p iT

h T h T c dT We have one equation in one unknown T This isWe have one equation in one unknown, T. This isconveniently solved in Excel with the Goal Seek toolto give an exit temperature of 2530K. If we allow fordissociation of the water (at equilibrium), Te = 2390K.


13

ENERGY BALANCES

• Heats (enthalpies) of combustion and higher (gross) and lower (net) heating

l (HHV d LHV)values (HHV and LHV)

298rxn p fp r fr

prod react

h h h

4 2 2 22 2 ( )CH O CO H O liq

2984

298

74.600f

kJCH h

mol

4 2 2 2 ( )C O CO O q2982

2982

2982

2982

0

393.510

( ) 285.830

( ) 241.826

f

f

f

f

O h

CO h

H O l h

H O g h

4 2 2 22 2 ( )CH O CO H O gas

ENERGY BALANCES

• Heats of reaction and HHV and LHV

4 2 2 22 2 ( )CH O CO H O liq

298

4

( ) 393.51 2 285.83

74.6 0 890.57 /

rxnh l

kJ mol CH

4 2 2 22 2 ( )CH O CO H O gas

298

4

( )

890.57 /rxnHHV h l

kJ mol CH

Gross heating value

Net heating value4 2 2 2 ( )g

298

4

( ) 393.51 2 241.826

74.6 0 802.56 /

rxnh g

kJ mol CH

298

4

( )

802.56 /rxnLHV h g

kJ mol CH

Net heating value

14

ENERGY BALANCES

• Energy balance over a furnace

Controlvolume

Inlet (i) Exit (e)

inQ 0inW

Use absoluteenthalpies forenergy balancesinvolving reactions.

, 0in net i i e ei e

Q n h n h

, ,( ) ( )ref

To

i f i ref p i

T

h T h T c dT Tref = 298.15 K

ENERGY BALANCES

• Energy balance over a furnace –available heatavailable heat

, , , ,( ) ( )ei

ref ref

TTo o

avail i f i ref p i e f e ref p ei eT T

Q n h T c dT n h T c dT

Available heat = gross heat input Available heat = net heat inputg p– latent heat loss – unburned fuel loss – sensible flue gas heat loss

Available heat = net heat input – unburned fuel loss – sensible flue gas heat loss

15

ENERGY BALANCES

• Thermal efficiency of a combustion process– Definition in terms of HHV

– Definition in terms of LLV

avail

grossfuel

Q

m HHV

Definition in terms of LLV

avail

netfuel

Q

m LHV

HEAT TRANSFER

16

HEAT TRANSFER

• Outline• Outline– Conduction

– Convection

– Radiation

– Example: calculation of inside wall temperature

HEAT TRANSFER

• Review of basic heat transfer

T1

L

– Conduction (Fourier’s law)

1 2 ,K

kAQ T T W

L

M t i l k W/( K)

T1

T2

x

k = thermal conductivity, W/(m K) or W/(m C)

L Material k, W/(m K)

air 0.0263water 0.613copper 401

,K

dTQ kA W

dx

2, /dT

q k W mdx

17

HEAT TRANSFER

– Convection (Newton’s law of cooling)

Th h t t f ffi i t h i d fi d 2 ,C CQ h A T T W

Flow & fluid h, W/m2K

free conv, air 5-12f d i 10 300

T1

L

The heat transfer coefficient, hc, is definedby this equation. Newton’s “law” is not reallya law. It is a useful definition.

forced conv, air 10-300forced conv, water 300-12,000phase change 3,000-50,000(boiling)

T2

T

HEAT TRANSFER

– Radiation (Stefan-Boltzmann law) 4 4

2R surQ A T T

2 22 2 2

2

sur sur sur

R sur

A T T T T T T

h A T T

T1

T2

L

= 5.67x10-8 W/m2K4

T2

T

Tsur

Material , normal emissivity

polished Al 0.05water 0.95planed oak 0.90

18

HEAT TRANSFER

– Example. Given the outside wall temperature in a rotary kiln (T2), estimate the inside wall temperature (T1).the inside wall temperature (T1).

T1

C RK Q QQQq

A A A

At steady state

2 = 0.76h2C = 0.88 Btu/(h ft2 F)

T2

Steel shell, LS = 1“, k = 26 Btu/(h ft F)

Brick, LB = 6“, k = 0.98 Btu/(h ft F)

Deposit, LD = 0.262’, k = 0.9 Btu/(h ft F)

cL

ri = 7.75’T2 = 350 C = 1122 RT = Tsur = 15 C = 519 R

HEAT TRANSFER

– Solution to find T1

T2

T

qcond

qconv

Ti2Ti1T1

Ch

1

sur22

SBD

211TT

1TT

LLLTT

q

2qcond

qradTsur

i2i11

D

D

kL

S

S

kL

B

B

kL

Rh

1

r/rlnTT

length

Q baK

RCSBD hhkkk

where we have neglected the curvature of the wall.

Solving for T1 gives inside wall T = 2600 F = 1460 C.

k2r/rlnlength ab

why?

19

CHEMICAL EQUILIBRIUM


OutlineGibbs free energy and the criterion of gy

equilibriumGas-phase: water-gas shift reactionPresence of solid or liquid phase: steam-

carbon reactionEffect of inerts and pressure: steam-carbon

tireactionFuel-rich combustion of propane

20


Gibbs free energy and the criterion of equilibriumequilibriumG = U + pV – TS = H – TS

The criterion of equilibrium, for a closed system that is held at constant temperature and pressure, is that G has reached its minimum value:minimum value:

0 0dG

or dGdt


Gas-phase reaction: perfect gas mixtureConsider the general reactiong

The criterion of equilibrium means that the equilibrium constant, Kp, and the free energy change for the reaction, G, are related by

aA bB cC dD

y

0ln p TRT K G c dC D

p a bA B

p pK

p p and

21


Gas-phase reaction: water-gas shift

2 2 2CO H O CO H

T = 1000K, p = 1 bar. Initial CO and H2O, 1 mole each.

0 0 0 0 02 2 2

0

( ) ( ) ( ) ( )

395.865 0 ( 200.281 192.603) 2.981

T f f f f

T

G G CO G H G CO G H O

kJG

mol

0G = free energy of formation at 1 bar and 1000K

0 2981exp exp 1.431

8.314 1000

Tp

kJG kmol

KkJRT K

kmol K

fG = free energy of formation at 1 bar and 1000K.


Gas-phase reaction: water-gas shiftLet x moles of H2 form at equilibrium.

2 2

2

2 2 1.4311 1

tot totCO H

pCO H O

x xp pp p

Kx xp p p p

CO 1 – x

H2O 1 – x

CO

2 2 2CO H O CO H

2

2 2tot totp pCO2 x

H2 x

Total 2x = 0.5447

22


Immiscible solid or liquid phase: steam-carbon reaction

2 2( ) ( ) ( ) ( )C s H O g CO g H g

T = 1000K, p = 1 bar. Initially 1 mole H2O, excess C(s).

0 0 0 0 02 2

0

( ) ( ) ( ) ( )

200.281 0 (0 192.603) 7.678

T f f f f

T

G G CO G H G C G H O

kJG

0 7678exp exp 2.518

8.314 1000

Tp

kJG kmol

KkJRT K

kmol K

( )T mol


Immiscible solid or liquid phase: steam-carbon reactionLet x moles of H2 form at equilibrium

2 2( ) ( ) ( ) ( )C s H O g CO g H g

2

2

1 11

tot totCO H

pH O

x xp pp p x xK

xp p

CO x

H2O 1 – x2

2

2

1

2.5181

tot

tot

px

xp

x

H2 x

Total 1 + x

x = 0.8460Spreadsheet: minimizeG.xls

23

CHEMICAL EQUILIBRIUMDoes this value of x (0.8460) really

minimize G?Spreadsheet: minimizeG xls

-205000

-200000

-195000

-190000G

, kJ/

kmo

lSpreadsheet: minimizeG.xls

-215000

-210000

0 0.2 0.4 0.6 0.8 1

extent of reaction

G


Effect of pressure: steam-carbon reaction

2 2( ) ( ) ( ) ( )C s H O g CO g H g

1/ 2

1

/ 1x

K

At 1 bar and 1000 K, x = 0.8460. In general, for thisreaction,

/ 1tot pp K

At 10 bar and 1000 K, x = 0.4485.

Spreadsheet: minimizeG.xls

24


Effect of inert gases: steam-carbon reaction 2 2( ) ( ) ( ) ( )C s H O g CO g H g

Nitrogen will be present in gasifiers if air is used as an oxidizer. Assume that we start with 1 mole of steam, 1 mole of nitrogen, and

b S

CO x

H2O 1 – x

H2 xexcess carbon. Suppose we have x moles of H2 at equilibrium at 1000 K and 1 bar.

N2 1

Total 2 + x

Spreadsheet: minimizeG.xls


Effect of inert gases: steam-carbon reaction 2 2( ) ( ) ( ) ( )C s H O g CO g H g

2

2

2

2 212

( ) 2 0

tot totCO H

pH O

tot

x xp pp p x xK

xp px

p K x K x K

( ) 2 0

1 , 2.518, 0.8910

tot p p p

p

p K x K x K

p bar K x

Recall that with no N2, x = 0.8460

25


Fuel-rich combustion of propane in airReactants: O N C HReactants: O2, N2, C3H8

Products: N2, H2, CO, CO2, H2O, O2

1500K, 1 bar

ER = 1.20, SR = 0.833

1 kmol propane, 19.84 kmol air

Approach: minimize G subject to conservation of C, H, O. The Solver tool in Excel is convenient.

Spreadsheet: cxhy.xls


Fuel rich combustion of propane in airFuel-rich combustion of propane in airGram atoms C

Gram atoms H2x yC H CO COxn n n

2 22 2

x yC H H O Hyn n n Gram atoms O

2 2x y

2 2 22 * 0.21 2 2air CO CO H O On n n n n


26


Fuel-rich combustion of propane in airMinimization of GMinimization of G

0 ln

i i

i i i

G n

RT p

0 0fi ig = free energy of formation of i at 1 bar and T

ii tot

tot

np p

n

fi ig free energy of formation of i at 1 bar and T.

pi = partial pressure of ini = kmol i



Fuel-rich combustion Species mole frFuel-rich combustion of propane in air (1500K, 1 bar) –equilibrium concentrations (calculated using

Species mole fr.

N2 0.6913

H2 0.02909

CO 0.04444

CO2 0.08787( gSolver in Excel).

2

H2O 0.1473

O2 0.0000


27


SummaryAt equilibrium, for a system held at constant

T and p, the Gibb’s free energy will reach its minimum value.

The equilibrium constant is defined by

aA bB cC dD

0ln p TRT K G

aA bB cC dD c dC D

p a bA B

p pK

p p and


SummaryAt lo to moderate press res immiscibleAt low to moderate pressures, immiscible

liquids or solids do not appear in the equilibrium constant but must be included in the calculation of .

If c + d < a + b, an increase in pressure will increase the extent of reaction.

0TG

If c + d > a + b, an increase in pressure will decrease the extent of reaction.

aA bB cC dD

28


SummaryIf c + d > a + b, an increase in the partial pressure of

inert gases will increase the extent of reaction.

Complex equilibrium are conveniently solved with specialized software. One such program is GASEQ. It is available for download at http://www.gaseq.co.uk/.

The Solver in Excel is useful for simpler systems.

aA bB cC dD

ADIABATIC FLAME TEMPERATURE

29


OutlineGibbs free energy and the criterion ofGibbs free energy and the criterion of

equilibrium

Adiabatic flame temperature – no dissociation

Adiabatic flame temperature – dissociation

Calculation of g from heat capacity dataCalculation of gf from heat capacity data


Gibbs free energy and the criterion of equilibriumequilibriumG = U + pV – TS = H – TS

The criterion of equilibrium, for a closed system that is held at constant temperature and pressure, is that G has reached its minimum value:minimum value:

0 0dG

or dGdt

30


Combustion of H2 in air (no dissociation)

Controlvolume

0.42 kmol/s H2

0.21 kmol/s O2

0.79 kmol/s N2

300 K, 1 bar

0.42 kmol/s H2

0.79 kmol/s N2

T = ?, 1 bar

0inQ 0inW


Q W n h n h



Combustion of H2 in air (no dissociation)

0h h 0i i e ei e

n h n h

, ,( ) ( )ref

To

i f i ref p i

T

h T h T c dT Absoluteenthalpy

Enthalpyof formation

Sensibleenthalpy

We have one equation in one unknown, T. This isconveniently solved in Excel with the Goal Seek toolto give an exit temperature of 2530K.

enthalpy of formationat 1 bar, 298 K

enthalpychange


31


Adiabatic combustion of H2 in air (allow dissociation of products)

Controlvolume

0.42 kmol/s H2

0.21 kmol/s O2

0.79 kmol/s N2

300 K, 1 bar

? kmol/s H2, H2O, OH,…0.79 kmol/s N2

T = ?, 1 bar

0inQ 0inW


Q W n h n h

0i i e ei e

n h n h


Combustion of H2 in air (allow dissociation)dissociation)Reactants: O2, N2, H2

Products: N2, H2, H, OH, H2O, O2, O

ER = 1.0, SR = 1.0

Approach: minimize G subject to conservation of H, O while holding enthalpy constant. Enthalpy and p are specified rather than T and p.

32


Combustion of H2 in air (allow dissociation)element balances

Gram atoms H

Gram atoms O2 2 2

2 2 2H i H Oe H e OHe Hen n n n n

Gram atoms O

2 2 22 2O i H Oe O e OHe Oen n n n n

ADIABATIC FLAME TEMPERTURE

Combustion of H2 in air (allow dissociation)dissociation)Minimization of G

0 ln

i i

i i i

G n

RT p

0 0

ii tot

tot

np p

n

0 0fi ig = free energy of formation of i at 1 bar and T.

pi = partial pressure of ini = kmol i

33


Combustion of hydrogen in air

Species kmol

N 0 9hydrogen in air (enthalpy constant, p = 1 bar) – equilibrium composition calculated using GASEQ.

T d = 2390 K

N2 0.79

H2O 0.3964

O2 0.00687

H2 0.01794

OH 0 00913Tad 2390 K. OH 0.00913

H 0.00223

O 7.46x10-4

http://www.gaseq.co.uk/


Calculation of gf from ideal gas heat capacity datacapacity data

2 3 41 2 3 4 5

2 3 43 5 62 41 2 3 4 5

pca a T a T a T a T

Ra a aa ah

a T T T TRT T

3 5

2 3 43 541 2 7ln

2 3 4

a aasa T a T T T T a

R

Spreadsheet: therm_coef.xls

34


Calculation of gf from heat capacity data

Example: free energy of formation of CO

0

1

i iM

Example: free-energy of formation of CO.C(graphite) + 1/2O2 = CO

0 0 0

0 0 0

elements

fi i j jj

fi i j j

h h h

s s s

2

10 (1) ( 1)

2CO C O

0 0 0

elements

fi i j jj

fi fi fi

s s s

g h Ts

Spreadsheet: therm_coef.xls


Calculated values for t i hi t i

Fuel HHV LHV Tad (K)stoichiometric combustion with air; 300 K inlet, 1 bar, and allowing for dissociation. Enthalpy and pressure held

(kJ/kg) (kJ/kg)

H2 141,800 120,000 2390

CH4 55,528 50,016 2226

C3H8 50,368 46,357 2267and pressure held constant. n-

octane

C8H18

48,275 44,791 2275

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Stoichiometry, Energy Balances, Heat Transfer, …geoff/air_poll/lectures_handouts/combustion...1 Stoichiometry, Energy Balances, Heat Transfer, Chemical Equilibrium, and Adiabatic