Wet Basis => Dry Basis

2^1 = K{T) ycoyH 2 o

1. Wet Basis => Dry Basis

4.8

Combustion

Reactions 143

When

a fuel is burned, carbon in the fuel reacts to

form

either

C0

₂ or

CO, hydrogen

forms

H

20,

and

sulfurforms

S0

₂^.

At

temperaturesgreaterthan approximately 1800°C,

some

of the nitrogenin theairreacts to

form

nitricacid

(NO). A ^combustion

reactionin

which CO

is

formed from

a

hydrocarbon

is referred to as partial

combustion

orincomplete

combustion

of thehydrocarbon.

Examples:

C + o

₂ ^->

co

₂

C

₃

H

₈

+

5

0

2

—

3

C0

₂

Complete combustion

ofcarbon 4

H

₂

₀ Complete combustion

of

propane

C

₃

H

₈

+ \Oi

^-* 3

CO +

4

H

2

0

^Partial

combustion

_of

propane

SOLUTION

Basis:

100 mol Wet Gas

60.0

mol N

2

15.0

mol C0

2

10.0

mol Q

2

85.0

mol

drygas

60.0

0.706

mol N

2

85.0

mol

drygas

15.0

0.176

mol C0

2

85.0

_

mol

dry gas 10.0

0.118

mol 0

2

85.0 moldrygas

2.

Dry

Basis

Wet

Basis.

An

Orsatanalysis (atechniqueforstackanalysis) yieldsthe followingdrybasiscomposition:

N

2

65%

CO, 14%

CO 11%

0

2

10%

A

humidity

measurement

showsthatthe

mole

fractionof

H

2

0

inthe stackgasis0.0700. Calculate thestackgascomposition

on

awetbasis.

SOLUTION

^Basis:

¹⁰⁰

^lb-moles

Dry Gas

0.0700

lb^-

m

°

leH2 ° _<=>

_0.930 ^lb-

^m0kdrygas

lb-molewetgas lb-molewetgas

I

0.0700lb-mole

H

2Q/lb-mole wetgas

_ _0Q753

^lb-mole

H

2

0

0.930lb-mole dry gas/lb-molewetgas ^" lb-mole dry gas

Hence

thegasinthe

assumed

basiscontains

lb-moleH-,0

=

7.53lb-moles

H

2

0

=

65.0lb-moles

N

2

100 lb-moles dry gas 0.0753 lb-mole

H

2

0

lb-mole drygas 100 lb-moles dry gas 0.650lb-mole

N

2

lb-mole dry gas

(100)(0.140)lb-moles

C0

2

=

14.0lb-moles

C0

2

(100)(0.110)lb-moles

CO ⁼

11.0lb-moles

CO

(100)(0.100) lb-moles

0

2

=

10.0lb-moles

0

2

107.5lb-moleswetgas

The

molefractionsofeachstackgas

component may now

easilybecalculated:

7.53 lb-moles

H

2

Q

lb-mole

H

2

0

yn2o

= <mt

„ ,

— =

0.070

107.5 lb-moleswetgas ^' lb-molewetgas^'

"

4.8

Combustion

Reactions 145

TEST

^!•

^What

^isthe

approximate molar

composition ofair?

What

^istheapproximate

molar

ratio

YOURSELF

^of

^N

2 to

0

₂ ^inair?

(Answers,

_p.

657)

^2.

A

^gascontains1

^mol H

₂,1

mol 0

₂,

and

2

mol H

20.

What

isthe

molar

compositionofthis gas

on

a

wet

basis?

On

adrybasis?

3.

A

^fluegas contains⁵

mole% H

20. Calculate theratios (a)

kmol

flue gas/kmol

H

20.

(b)

kmol

dryfluegas/kmolflue gas.

(c)

kmol H

2

Q/kmol

dryfluegas.

4.8b Theoretical and Excess Air

If

two

reactants participateina reaction

and one

^isconsiderably

more

expensivethantheother, the usualpracticeis tofeedthelessexpensivereactantinexcess of the valuable one. Thishas theeffectof increasing theconversionofthevaluable reactantattheexpenseof thecostof the excess reactant

and

additional

pumping

costs.

The

extreme case of

an

inexpensive reactant is air,

which

is free.

Combustion

reactions are therefore invariablyrun with

more

airthan is

needed

tosupply

oxygen

in stoichiometric proportion to the fuel.

The

following terms are

commonly

used to describe the quantities of fuel

and

airfed to a reactor.

Theoretical

Oxygen: The moles

(batch)or

molar

^flowrate(continuous) of

O? needed

for

com-

plete

combustion

ofall the fuelfed to the reactor, assuming that all carbon in the fuelis

oxidizedto

C0

2

and

all the

hydrogen

is oxidizedto

H

20.

Theoretical Air:

The

quantity ofairthatcontains the theoreticaloxygen.

ExcessAir:

The amount by which

theairfed to the reactorexceedsthetheoreticalair.

(molesair)fed

-

(molesair)theoretical

Percent ExcessAir:

x 100%

(4.8-1)

(molesair) the0retical

If

you know

the fuelfeedrate

and

the stoichiometricequation(s) forcomplete

combustion

of the fuel,

you

can calculate the theoretical

0

₂

and

airfeed rates. Ifinaddition

you know

the actual feed rateofair,

you

cancalculate thepercentexcess air

from

Equation 4.8-1. Itisalso easyto calculate the airfeedrate

from

the theoretical air

and

a given value of the percentage excess: if

50%

excessairissupplied, forexample, then

(molesair)_fed

=

1.5 (molesair)_theoreticai

EXAMPLEt4i8-2'

Theoretical

and Excess Air

One

hundred mol/h of butane

(QHio)

and 5000 mol/h of air are fed into a combustion reactor.

Calculate thepercent excessair.

SOLUTION

^First, calculate the theoretical air from the feed rate of fuel and the stoichiometric equation for complete combustionof butane:

C

4

H

10

+ ^A2Q

2

—

4

C0

2

+

5

H

2

0

("Oitheoretical

—

¹⁰⁰

^mol C

4H,₀ 6.5

mol 0

2 required

h

mol

C4H10

650

mol 0

2

("air)theoretical

650mol

0

2 4.76

mol

air

h

mol 0

2

=

3094

mol

air

146

Chapter

4

Fundamentals

ofMaterialBalances

Hence

%

excessair ("air)fed (wa ir)theoretical

X 100%

⁵⁰⁰⁰

-

3094 3094

X 100% = 61.6%

(^air)theoretical

Ifinsteadyou had beengiven

61.6%

excessair,youcouldhavecalculated thefeedrate ofairas ("air)fed

=

l-616(/t_{air )theoreticai}

=

1.616(3094 mol/h)

=

5000mol/h.

Two

pointsofconfusionoftenarise inthe calculation oftheoretical

and

excessair,bothof

which

arecaused

by

ignoring thedefinitionsof these terms.

1.

The

theoretical airrequiredto

burn

a given quantity

of

fueldoes not

depend on how much

isactuallyburned.

The

fuel

may

notreactcompletely,

and

it

may

reactto

form

both

CO

and CO2,

butthe theoreticalair is stillthat

which would be

required to reactwithallof thefuel to

form C0

2 only.

2.

The

valueof^thepercent excessair

depends

only

on

the theoretical air

and

theairfeedrate,

and

not

on how much O2

^is

consumed

inthereactoror whether combustioniscompleteor partial.

(Answers,

_p.

₆₅₇₎

TEST ^Methane

^burnsinthe reactions

YOURSELF _{C H}

4

+

2

o

2 -*

C0

2

+

2

H

₂

0 CH

₄

+ §0

^{2 -»}

CO + 2H

2

0

One hundred

mol/hof

methane

isfedtoareactor.

1.

What

isthetheoretical

0

2 flow rate ifcompletecombustionoccursin thereactor?

2.

What

^is the theoretical

0

2 flow rate assuming that only

70%

of the

methane

reacts?

(Careful!)

3.

What

isthetheoreticalairflow rate?

4. If

100%

excessairissupplied,

what

istheflowrateofairentering the reactor?

5. Ifthe actual flowrateofairissuchthat300

mol 0

2/henters thereactor,

what

isthe percent excessair?

CREATIVITY EXERCISES

Equipment Encyclopedia heattransfer

boilers

1. Years agoit

was common

to operateboilerfurnaceswithairfedin

20%

^{excess or}more, while today

improved

boiler designs enable the use of

5-10%

excess air. Cite as

many

possiblenegativeconsequencesas

you

canthink offortheair-to-fuelfeedratiobeing(a) too

low and

(b) toohigh.

2.

The

costsofpetroleum

and

naturalgas

have

increased dramaticallysince the early1970s,

and

thereis

some

question abouttheircontinued long-termavailability.Listas

many

al- ternative energy sources as

you

can think of, being as creative as

you

can,

and

thengo

back and

suggest possible

drawbacks

toeach one.

4.8c Material Balances on Combustion Reactors

The

procedure for writing

and

solvingmaterialbalancesfora

combustion

reactoristhe

same

asthat for

any

otherreactivesystem.

Bear

in

mind

these points,however:

1.

When you draw and

labelthe flowchart,besure theoutletstream(thestack gas) includes (a) unreactedfuelunless

you

aretold thatallthefuelis

consumed,

(b)unreacted oxygen,

(c) water

and

carbon dioxide, aswell ascarbon

monoxide

ifthe

problem

statementsays

any

^ispresent,

and

(d) nitrogenifthefuelis

burned

withair

and

notpure oxygen.

4.8

Combustion

Reactions 147

2.

To

calculatethe

oxygen

feedrate

from

a specified percent excess

oxygen

orpercentexcess air (bothpercentages

have

the

same

value,so itdoesn'tmatter

which one

is stated),_first calculate the theoretical

O2 from

the fuel feed rate

and

the reaction stoichiometry for complete combustion, then calculate the

oxygen

feedrate

by

multiplying thetheoretical

oxygen by

₍₁

+

fractionalexcessoxygen).

3. Ifonly

one

reaction is involved, all three balance

methods

(molecular species balances, atomic species balances, extent of reaction) are equally convenient. Ifseveral reactions occursimultaneously,

however —

suchas

combustion

of afuelto

form

both

CO and C0

₂

—

atomicspecies balancesare usually

most

convenient.

EXAMPLE 4.&3\ Combustion _of Ethane

Ethaneisburnedwith

50%

excessair.

The

percentage conversionof theethaneis

90%;

oftheethane burned.

25%

reactstoform

CO

andthebalancereacts toform

C0

2^.Calculatethemolarcomposition of thestackgas onadrybasisandthe

mole

ratioofwatertodrystackgas.

SOLUTION

^Basis:

^{100 mol} C

2

H

6

Fed

100 molC₂H₆ ^njtmoiC2H_S) /i2(mol0₂) /i3(molN₂)

50%excessair n₄(molCO)

n6(molC0₂)

n₆(molH₂0) n₀(mol)

0.21 mol0₂/mol 0.79 molN₂/mol

C₂H₆+ -£-0

2

—

- 2C0₂+3H₂0 C₂H₆+

J-0²

—

- 2C0+3H₂0

Notes

1. Since noproduct stream

mole

fractions are

known,

subsequentcalculations are easierifindi- vidual

component amounts

ratherthana total

amount

and molefractionsare labeled.

2.

The

composition ofairistaken tobe approximately21

mole% 0

2^,79

mo!e% N

2^.

3. Ifthe ethane reacted completely, n\

would

be omitted. Sinceexcess airis supplied,

0

2 must appearintheproduct stream.

4. Inmaterialbalancecalculationson combustionprocessesitisreasonable toassumethatnitro-

genisinert

—

^thatis,toneglect the trace

amounts

of

NO, NO2,

and

N

2

0

4 (collectivelyreferred toas

NO*)

thatmightformintheburner.

On

the otherhand,in environmental impactstud- ies

NO* may

not automaticallybeneglected;traceamountsofnitrogenoxides

may

havelittle

impact

on

the nitrogen balance but

may

haveasignificantpolluting effectiftheyarereleased intothe atmosphere.

Degree-of-

Freedom

Analysis

7

unknowns

(n_0,«i,..

.,n₆₎

-

3 atomic balances(C,H,

O) -

1

N

2 balance

-

_{1 excess}airspecification (relatesn₀tothe quantity of fuel fed)

-

1 ethane conversionspecification

-

1

CO/C0

₂ ratiospecification

=

0 degrees offreedom

50%

ExcessAir

(n02 theoretical

=

¹⁰⁰

^mol C

2

H

6 3.50mol

0

2 1 mol

C

2

H

6

=

350mol

0

2

J|50% ^excess

air

0.21no

=

1.50(350

mol 0

2 )

=>

n₀

=

2500molairfed

90% Ethane

Conversion: (

=> 10%

unreacted)

«i

=

0.100(100

mol C

2

H

6 fed)

=

^I 10.0mol

C

2

H

6

0.900(100

mol C

2

H

6fed)

=

90.0

mol C

2

H

6react

25%

Conversionto

CO

(0.25

X

90.0)

mol C

2

H

6react toform

CO

2

mol CO

generated

1

mol C

₂

H

6react

45.0

mol CO

Nitrogen Balance: output

=

_input

«₃

=

0.79(2500mol)

Atomic Carbon

Balance: input

=

output

1975

mol N

₂

100

mol C

2

H

6

2molC _«,(molC

₂

H

6) 2 mol

C

^«4(mol

CO)

1

molC

1

mol C

2

H

6 ¹

mol C

2

H

6 1

mol CO

+

^«5(mol

C0

2)

ImolC

1

mol C0

2

N

n\

=

10mol n4

=

45mol n₅ ^-= 135

mol C0

2

A

tornic

Hydrogen

Balance: input

—

output

100

mol C

2

H

6

6molH

10

mol C

2

H

6

6molH

1

mol C

2

H

6 1

mol C

2

H

6

n₆(mol

H

2

0) 2molH

1

mol H

2

0

I

n₆

=

₂₇₀

mol H

₂

0

Atomic Oxygen

Balance: input

=

output 525

mol 0

2

2molO

1 mol

0

2

n₂(mol

0

2) 2

mol O

45

mol CO

+

¹

molO

1

mol 0

2 1 mol

CO

^

¹³⁵

mol C0

2

2molO

1

mol C0

2

n₂

=

232

mol 0

2

+

²⁷⁰

^mol H

2

0

1

mol O

1

mol H

2

0

4.8

Combustion

Reactions 149

The

analysis of the stackgasis

now

complete. Summarizing:

m ^{= 10molC}

²

H

6

«₂

=

232 mol

0

2

n₃

=

1974

mol N

2

n₄

=

45

mol CO

«₅

=

135mol

CQ

₂

+

«₆

=

2396

mol

dry gas 270mol

H

2

Q

2666

mol

total

Hence

the stackgascomposition

on

adrybasisis

y\

=

y2

=

y*

ys

=

10

mol C

2

H

6

2396

mol

dry gas 232

mol Q

2

2396

mol

dry gas 1974

mol N

2

2396

mol

dry gas 45

mol CO

2396

mol

dry gas 135

mol CQ

2

2396

mol

dry gas

=

0.00417

=

0.0970

mol C

2

H

6

mol mol

Oi

=

0.824

=

_0.019

=

0.0563

mol mol N

2

mol mol CO

mol mol C0

2

mol andthe moleratioofwatertodrystackgasis

270

mol H

2

Q

2396

mol

drystackgas

=

0.113

mol H

2

0

mol

drystackgas

EXAMPLE

4.8-4i

Ifa fuelof

unknown

compositionis burned,

you may

be ableto

deduce something

about

its composition

by

analyzing the

combustion

products

and

writingand solving atomicspecies balances.

The

procedureisillustrated in thenextexample.

Combustion of a Hydrocarbon Fuel of Unknown Composition

A

hydrocarbongasisburnedwithair.

The

dry-basisproduct gas compositionis1.5

mole% CO, 6.0%

C0

2^,

8.2% 0

2^,and84.3%

N

2^.Thereisnoatomicoxygeninthefuel.Calculatetheratioofhydrogen tocarbonⁱⁿ the fuelgasandspeculate

on

whatthefuelmightbe.

Then

calculatethepercentexcess airfedtothe reactor.

SOLUTION

^Basis:

^{100 mol}

^Product

^Gas

Sincethemolecularcompositionofthefuelis

unknown, we

labelitsatomicspeciescomposition.

We

alsorecognizethatsince the fuelisahydrocarbon, watermust be oneof thecombustionproducts.

n_c(mol C) n_H(mol H) n_a(molair)

0.21 mol0₂/mol 0.79 mol N₂/mol

100moldrygas

0.015molCO/moldrygas 0.060mol C02/moldrygas 0.082mol0₂/moldrygas 0.843 mol N₂/moldrygas

«w^(molH20)

150

Chapter

4

Fundamentals

of MaterialBalances

C + o

2

— _co

₂

2C + 0

2

—

*

2CO 4H + 0

2

— _2H

_z

_O

Degree-of-FreedomAnalysis

4

unknowns

(«H,He,_"a,«w)

-3

independentatomic balances(C,H,

O) -1 N

2balance

=

0degrees offreedom

A

solutionprocedurethatdoesnotrequire solvingsimultaneous equationsisasfollows:

iV₂

Balance

^0.79na

=

(100)(0.843)

mol N

2

=>

«_a

=

106.7molair

Atomic C Balance

n c

=

¹⁰⁰

^mol

^0.015

^mol CO

1

molC