How much oxygen was consumed? What is the fractional conversion of oxygen?

2^1 = K{T) ycoyH 2 o

2. How much oxygen was consumed? What is the fractional conversion of oxygen?

3.

Write

the extent of reaction equation (4.6-3) for methane, oxygen,

and CO2. Use

each equationtodeterminethe extent ofreaction,_£,substituting inlet

and

outletvalues

from

theflowchart.

4.7 Balances

on

Reactive Processes 135

4.

How many

independent molecularspeciesbalances canbewritten?

How many

indepen- dent atomicspeciesbalances can

be

written?

!

5. Write the followingbalances

and

verify thatthey areallsatisfied.

The

solution of thefirst

one

isgivenasan example.

(a) Methane. (I

= O +

C. 100

mol CH

4 in

=

40

mol CH

₄ out

+

60

mol CH

4

consumed)

(b)

Atomic oxygen

(O).

(c) Molecular

oxygen (0

2^).

(d) Water.

(e)

Atomic

hydrogen.

4.7f

Product Separation and Recycle

Two

definitionsof reactantconversionare

used

intheanalysisofchemicalreactorswith product separation

and

recycleof

unconsumed

reactants:

Overall Conversion:

™

^{ntinpUttQprOCeSS}

^~

reactant 0Ut

P

^ut

from P

^rocess

(4.7.x) reactant inputtoprocess

Single-Pass Conversion: reactant input to reactor

-

reactant

o

utput

from

reactor reactant inputtoreactor

"

As

usual,thecorresponding percentage conversionsareobtained bymultiplying these quanti- ties

by 100%.

For example,consider the following labeled flowchartfora simplechemical process based

on

the reaction

A

^{-» B:}

75 mol A/min 100molA/min

REACTOR ^{25mol A/min} 75mol B/min

PRODUCT SEPARATION

UNIT

75mol B/min

25mol A/min

TEST

YOURSELF

(An^*ei^p. 657)

EXAMPLE4jf-2\

The

overallconversionof

A

^is

from Equation

4.7-1:

(75

mol

A/min)j_n

-

₍₀

_mol/min)

_out

(75

moi

A/min)i_n

The

single-passconversionis

from Equation

4.7-2:

(100

mol

A/min)j_n

-

₍₂₅

mol A/min)

out

X 100% = 100%

X 100% = 75%

(100

mol A/min)

in

This

example

providesanotherillustrationof the object ofrecycle.

We

have achieved

com-

plete use of the reactant for

which we

are

paying—the

fresh

feed— even though

only

75%

of the reactant entering the reactoris

consumed

before emerging.

The

reasonthe overall conver- sionis

100%

^isthatperfect separation

was assumed: any A

^thatdoesnot reactgetssent

back

to thereactor. Ifa less-than-perfect separation

were

achieved

and some A

^leftwith theproduct stream,the overallconversion

would

be lessthan

100%,

althoughit

would

alwaysbe greater thanthesingle-passconversion.

What

are the overall

and

single-pass conversions for the process

shown

in Figure ^4.5-1

on

p. 110?

Dehydrogenation _of Propane

Propane

isdehydrogenatedtoformpropyleneinacatalytic reactor:

CsHg —

_*

C

3

H

6

+ H

2

The

processisto be designed fora

95%

overallconversion of propane.

The

reactionproducts are separated into two streams: the first, which contains

H

2, C3H6, and 0.555% of the propane that leaves the reactor, is taken off as product; the second stream, which contains the balance ofthe unreactedpropane and

5%

of thepropylene inthe firststream,isrecycled to the reactor.Calculate the composition of the product, the ratio (moles recycled)/(mole fresh feed), and the single-pass conversion.

SOLUTION

^Basis:

^{100 mol}

^FreshFeed

Fresh feed

j_

100molC₃H₈ 1i~

_i/ijtmolC₃H

8⁾¹ n₂(molC₃H₆) |

REACTOR

Recycle

n₃(molC₃H_g)i

n₄(molC₃H₆)|

n₅(molH₂)

/!9(molC₃H₈)

n₁₀{molC₃H₆)(5%ofn₇)

SEPARATOR

' Product

n₆(molC₃H_a) (0.555%ofn₃)

n₇(molC₃H₅) /i8(molH₂)

In terms ofthe labeled variables,thequantities tobecalculated are themole fractionsofproduct stream

components

[n₆/⁽ⁿ⁶

+

n7

+

n_{s )],}... , the recycleratio [(n₉

+

n₁₀)/100 mol],andthesingle- pass conversion

[100% X (m -

^«3)/m].

We

musttherefore calculaterti, «3,and throughn_lQ.

As

usual,

we

begin withthe degree-of-freedomanalysis todeterminewhetherthe problemisproperly specified(i.e.,whetheritispossibletosolveit).

Degree-of-Freedom

Analysis

When

analyzing subsystems in whichreactions occur (the overallsystem and the reactor),

we

will countatomicbalances; fornonreactivesubsystems(therecyclemixingpointandtheseparationunit)

we

^willcountmolecularspecies balances.

• Overallsystem (the outerdashed

box on

the flowchart). 3

unknown

variables (n₆, n₇, n_{s )}

-

₂ independentatomic balances(C and

H) —

₁additional relation

(95%

^overallpropaneconversion)

==>

0degrees offreedom.

We

willthereforebeable todetermine«₆,«₇,andn$byanalyzingthe overallsystem. Let uscountthese three variablesas

known

at thispoint.

• Recycle-freshfeed mixing^point. 4

unknown

variables (n_9,n

w

^,nx,n2 )

—

_2balances(C

3

H

g,CiH$) 2 degrees offreedom. Since

we do

not have

enough

equationsto solve for the

unknowns

associated withthissubsystem,

we

proceedtothe nextone.

• Reactor. 5

unknown

variables(n\throughn$)

-2

atomic balances(C and

H)

3degreesof freedom.

No

helphere.Letusconsider theremainingunit.

• Separator. 5

unknown

variables (n₃, n4,n₅, n₉, «io) («6 through ng are

known from

the overall systemanalysis)

-

₃balances (C3H8, C3H6,

H

2 )

-

₂additional relations(«6

-

0.00555«3, n

w ⁼

0.05n7 )

=>

0 degrees of freedom.

We

can therefore determinethe fivegivenvariables associatedwiththeseparatorand then return toanalyzeeither themixingpointorthe reactor;ineithercase

we

canwritetwo atomicbalancesto solve forthetwo remaining

unknowns

_(«i andn_{2 )»}thereby completingthe solution. (Infact,notall

of thesystemvariables are requiredbytheproblemstatementso

we

willbeable tostopwell short of the fullanalysis.)

The

calculations follow, beginning withthe analysis of the overallsystem.

95%

^Overall

^Propane

Conversion

(=> 5%

unconverted) n₆

=

0.05(100mol)

=

5

mol C

3

H

8

We

are leftwith two overall system atomic balances to write.

An H

^{balance involves} both re-

maining

unknowns

(n₇ andng)but a

C

^balanceinvolvesonlyn₇;

we

^{thereforebegin with} the latter balance.

4.7 Balances

on

Reactive Processes 137 Overall

C

Balance

(100

mol C

3

H

8)(3

mol

C/mol

C

₃

H

8 )

=

_[«6(mol

C

3

H

8)](3mol C/mol

C

3

H

8)

rtf,

=

5mol

+

^[;i7(moI

C

3

H

6)](3mol C/mol

QH

6)

n₇

=

95

mol C

3

H

6

Overall

H

Balance(Fillin units.)

(100)(8)

=

n₆₍₈₎

+

«₇₍₆₎

+

n₈₍₂₎

The

product_thereforecontains

5

mol C

₃

H

8

95

mol C

3

H

6

=

95

mol H

2

n6

=

5mol. _«7

=

95mol

n₈

=

95

mol H

2

2.6

mole% C

3

H

8

48.7

mole% C

3

H

6

48.7

mole%H,

GivenRelations

Among

Separator Variables n6

=

5mol

n₆

=

0.00555«₃

> =

₉₀₀mol

C

₃

H

8

n10

=

0.0500n₇

«7

=

95 mol

«io

=

4.75mol

C

3

H

6

Propane

Balance

About

Separation Unit

n3

=

900 mol, n6

=

5mol

n3

=

n₆

+

n_g n_g

=

895mol

C

3

H

8

We

could continue writingbalances aboutthe separation unit todetermine the values ofnAand n₅ butthere is

no

reasonto

do

so,since these values werenot requestedinthe problemstatement.

The

only valuestilltobe determinedisthatofn

u

^whichcan becalculatedfrom apropane balance aboutthemixingpoint.

Propane

Balance

About Mixing

_Point 100

mol +

«₉

= m

n9

=

895mol

m ⁼

^995mol

C

3

H

8

We now

haveallthevariablevalues

we

need.

The

desiredquantities are (n₉

+

n

w

⁾

^mol

^{recycle "9}

⁼

^{895 mol. n10}

⁼

^4.75mol

Recycleratio

=

100

mol

fresh feed ^9.00

mol

recycle molfreshfeed

„. ,

m ^- m

^n\

⁼

^{995 mol, ni}

⁼

^900mol

Single-passconversion

= — _{X 100%}

_>

"1

9.6%

Consider

what

is

happening

intheprocessjust analyzed.

Only

about

10%

of the

propane

entering the reactoris converted topropylene in a single pass;however, over

99%

of the un-

consumed propane

inthe reactoreffluentisrecoveredinthe separationunit

and

recycled

back

tothereactor,

where

itgetsanother chanceto react.

The

netresultisthat

95%

of the

propane

entering the processisconverted

and 5%

leaves withthe finalproduct.

In general, high overallconversions can be achievedin

two

ways:(a)design the reactorto yield a highsingle-passconversion, or(b) designthe reactor_{to yield}alowsingle-passconver- sion(e.g.,

10%,

^asinthepreceding example),

and

followitwith_aseparationunit torecover

and

recycle

unconsumed

reactant. Ifthesecond

scheme

is used, the reactor

must

handle a larger throughput, but it takes a

much

larger reaction

volume

to achieve a

95%

conversion than a

10%

conversioninasingle pass.

The

lowersingle-passconversion consequentlyleads to ade- crease in the costofthereactor.

On

the otherhand, the savings

may

be offset by the cost of

Recycle 40molC₂H₄/s

20mol0₂/s

452molN₂/s

Purge stream

50molC₂H₄/s

25mol0₂/s

565mol N_{2 /s}

10 molC₂H₄/s

5 mol0₂/s

113mol N₂/s

Fresh feed 60molC₂H_4/s 30mol0₂/s

113 molN₂/s

100mol C₂H₄/s

50mol0₂/s

565molN₂/s

REACTOR

50molC₂H_4/s 25mol0₂/s

565molN₂/s

50molC₂H₄0/s

ABSORBER

Solvent

Product 50molC₂H₄0/s Solvent

Figure4.7-2 Process withrecycleandpurge.

theseparation processunit

and

the

pump,

pipes,

and

fittingsinthe recycleline.

The

finaldesign

would

be based

on

^adetailed

economic

analysisof thealternatives.

4.7g Purging

A problem may

arise inprocesses thatinvolverecycling.

Suppose

a materialthatenters with the fresh feed or is

produced

in a reaction remains entirely in a recycle stream, rather than beingcarriedoutinaprocess product.Ifnothing

were done

aboutthissituation,thesubstance

would

continuouslyenter the process

and would have no way

ofleaving; it

would

therefore steadilyaccumulate,

making

theattainmentofsteadystateimpossible.

To

preventthisbuildup, aportion of the recyclestream

must

be

withdrawn

^asapurge streamtoridtheprocess of the substanceinquestion.

The

flowchart

shown

in Figure 4.7-2 forthe production of ethylene oxide

from

ethylene illustratesthissituation.

The

reactionis2

C

₂

H

4

+ O2

^-* 2

C

2

H

40.

A

mixtureof ethylene

and

airconstitutesthefreshfeedtotheprocess.

The

effluent

from

the reactor passestoan absorber

and

iscontactedwith aliquid solvent.All oftheethyleneoxideisabsorbedintothe^solvent.

The

gasstreamleaving the absorber,whichcontainsnitrogen

and

unreactedethylene

and

oxygen,

isrecycled to thereactor.

If there

were no

nitrogen (or any other inert

and

insoluble substance) inthe feed, there

would

be

no need

for a purgestream.

The

recycle

would

contain only ethylene

and

oxygen;

the freshfeed

would

containjust

enough

of these substances to

make up

for the

amount

lost inthe reaction,

and

thesystem

would

be atsteady^state.

However,

thereis nitrogen. Itenters thesystematarateof113mol/s

and

leaves thesystematthe

same

rateinthepurge stream.If thesystem

were

not purged,nitrogen

would

accumulateatthisrate untilsomething

—

^probably

unpleasant

—

^{occurredtoshut}

down

theprocess.

Material balance calculations

on

systems involving recycle

and

purge follow the proce- dures giveninprevioussections.

When

labelingthe flowchart,notethatthe purge stream

and

the recyclestream before

and

afterthepurgetakeoffall

have

the

same

composition.

TEST

YOURSELF

(Answers,

_p.

657)

A

^reactionwith stoichiometry

A

^-*

B

takes place inaprocess withthefollowingflowchart:

REACTOR SEPARATOR

60molA ^'1 200molA 150molA 50mol B

10 molA 50mol B

140molA

4.7 Balances

on

Reactive Processes 139

1.

What

istheoverallconversionof

A

^{for this} process?

What

isthe single-passconversion?

2.

The

separationunit

and

recycle

pump and

piping are expensive.

Why

noteliminate

them and

selltheeffluent

from

thereactor asis?Alternatively,

why

not

keep

the separatorbut discard the

bottom

streaminsteadof recyclingit?

3.

Suppose

a trace

amount

(say,

0.1%)

of

an

inert material

C

iscontainedin the freshfeed

and

allofitstays inthe

bottom

effluentstream

from

the separation unit (andso isrecy- cled).

Why would

the process eventuallyshut

down? What would you

haveto

do

togetit

to

work?

4.

Why

not design the reactorto

produce

10

mol A and

50

mol B from

60

mol A

^inasingle

pass,therebyeliminating the

need

fortheseparation

and

recycle?

E^lflPEEt^^ Recycle and Purge

in theSynthesis

of Methanol

Methanol

isproducedinthe reactionofcarbondioxideandhydrogen:

C0

2

+

3

H

2

— CH

3

OH + H

2

0

The

fresh feedto theprocess contains hydrogen, carbon dioxide,and0.400

mole%

inerts (I).

The

reactoreffluentpassestoacondenserthat

removes

essentiallyallof themethanol and water formed and

none

ofthereactantsorinerts.

The

lattersubstances arerecycled to the reactor.

To

avoid buildup of theinerts in thesystem,apurge streamiswithdrawn fromtherecycle.

The

feed to the reactor (not the fresh feed to the process) contains 28.0

mole% C0

2,

70.0

mole% H

2, and 2.00

mole%

inerts.

The

single-pass conversion of hydrogen is 60.0%. Cal- culatethemolarflowrates and molarcompositions ofthe fresh feed, thetotalfeed to the reactor, therecyclestream, andthepurgestream_{for a}methanolproductionrateof155

kmol CH

3

OH/h.

SOL UTION

Basis:

100 mol Combined Feed

totheReactor Equipment

Encyclopedia reactor, condenser

«r(mol)

jr5c(molC0₂/mol)

*5H (molH^mol)

(1-*5c-*5h)^(moll/mol)

np(mol)

x

x

^(mo\COymol)

jr5H (molH^mol)

(1-x

x

^-jt5h⁾(moll/mol) n₅(mol)

.*5c(molC0₂/mol)

jr5H (mol Hj/mol) (1-.T5c-*5h)^(moll/mol)

n₀{mo\) 100mol

xoclmolCO^mol) (0.996-jtqc)(molH^mol) 0.00400moll/mol

0.280molCO^mol 0.700 molH^mol 0.020moll/mol

REACTOR

rtjfmolC0₂)

n2(molH₂)

2.0molI

n3(molCH3OH) n_{4 (mol}H₂0)

CONDENSER

n₃(molCH3OH) n₄(molH₂0)

As

ageneralrule,the

combined

feedtothe reactorisa convenient streamtouseasabasisofcalcu- lation for recycle problems

when

the streamcompositionisknown.

We

willtherefore temporarily ignorethe specifiedmethanolproductionrate,balance theflowchart for theassumedbasis,andthen scale theprocesstotherequired_{extent. In}terms of the labeledvariables,theproblemstatementwill

be solvedbydetermining n_0, x_oc,n₃,x_5C,x_5H, n_p,

and

n_{ for theassumedbasis, thenscalingup n0,

100

mol

(fedtoreactor),n_p,andnt bythe factor(155

kmol CH

3OH/h)/n₃^.

Degree-of -Freedom

Analysis

In the analysis that follows,

we

will count molecular species balances for all systems.

(We

could equallywell useatomicspecies balances or the extent ofreaction.) Note that the reaction occurs

within the overall system and the reactor subsystem and so must be included in the degree-of- freedomanalysesforboth^systems.

• Overall system. 7

unknowns

(no,.*oc>«3>''4>*p.*5Ci-*5H)

+

1 reaction

-

5 independentbalances

(C0

₂

,H

2

,I,CH

3

OH,H

2

0) ==>

3 degreesoffreedom. Since

we

do nothave

enough

equations tosolve forthe

number

of

unknowns

intheoverallsystem,

we

checksubsystemstoseeifonecan befoundwithzerodegreesoffreedom.

• Recycle-freshfeed mixing point. 5

unknowns

(no,*oc,«r,-*5C,-*5H)

~

3 independent balances (CO2,

H

2^, 1)

==>

2degrees offreedom.

• Reactor. 4

unknowns

(n

u

n₂,n₃,n4)

+

1 reaction

-

4independentbalances

(C0

2,

H

2,

CH3OH,

H

2

0)-

1single-passconversion 0 degrees of freedom.

We

willthereforebeabletodetermine

«i,«₂,"3,andn₄and proceed fromthere.

Noticethat

we

only subtracted four balancesandnot

one

foreachof thefivespecies.

The

reason

isthat

when we

labeledthe outletflow ofIas2.0mol,

we

implicitlyusedthebalanceonI(input

=

output) and so can no longercount it inthe degree-of-freedom analysis.

We

^will use the

same

reasoninginthe analysis of thecondenser.

• Condenser. 3

unknowns

(h5,*5c,*5h)

-

3 independent balances

(C0

2,

H

2^,1)

=>

0 degreesof freedom.

We may now presume

thatn₅,xsc,and*sh^are known.

Inthisanalysis

we presumed

that

we knew

n\,n₂,«3,andn₄fromthe reactoranalysis,andsince

we

used themethanol and waterbalances

when we

labeled the bottom productstream

we

only countedthreeavailablebalancesinthe degree-of-freedomanalysis.

• Purge-recyclesplitting point. 2

unknowns

(n_T,n_p)

-

₁ independent balance 1 degree of freedom. Since the labeled

component mole

fractions are the

same

inall three streams in this subsystem, balances

on

allthreespeciesreducetothe

same

equation(tryitand^see).

• Recycle-freshfeed mixing point (revisited). 3

unknowns

(/io,*oc,«r)

~

₃ independentbalances

=>

0 degrees of freedom.

We

can

now

determine (no,xqc,andn_{r ).}

• Purge-recyclesplittingpoint (revisited). 1

unknown

(n

p⁾

-

₁independentbalance

=>

0degrees offreedom.

The

final

unknown

variablecan

now

becalculated.

The

solutionprocedure^willthereforebetowritebalanceson the reactor,thenthecondenser,then the freshfeed-recyclemixingpoint,andfinallythepurge-recyclesplittingpoint.

The

flowchart

may

thenbescaledup bytherequired

amount

toobtainamethanolproductionrate of155kmol/h.

The

calculations follow.

Reactor Analysis

We

^willusemolecularbalances. Recallthatthestoichiometricreactionis

C0

2

+ 3H

2

—

*

CH3OH + H

2

0

60%

Single-Pass

Hz

Conversion: (

40%

isunconvertedand emergesatthe reactor outlet)

m ⁼

^0.40(70.0

^{mol H}

2fed)

=

28.0

mol H

2

Hi

Balance: consumption

=

input

-

output

Cons

_H2

=

(70.0

-

_28.0)

_mol H

2

=

42.0

mol H

2

consumed COz

Balance: output

=

input

-

consumption

hi

=

28.0

mol C0

2

-

^42.0

^mol H

2

consumed

1

mol C0

2

consumed

3

mol H

2

consumed

=

14.0

mol C0

2

CH3OH

Balance: output

=

generation

"3

42.0

mol H

2

consumed

1

mol CH3OH

^generated

3

mol H

2

consumed

H2O

^Balance: output

=

generation

«4

=

^42.0

^mol H

2

consumed

1

mol H

2

0

^generated

3

mol H

2

consumed

=

14.0

mol CH3OH

=

14.0

mol H.O

4.7 Balances

on

ReactiveProcesses 141 CondenserAnalysis

Total

Mole

Balance: input

=

output

«1

+

«2

+

"3

+

w₄

+

^2-0

mol =

n₃

+

«4

+

^«5

J|n²

=

28.0mol,rti

=

n₃

=

h4

=

14.0mol n₅

=44.0 mol

COt

^{Balance: input}

=

output

=

14.0mol, n₅

=

44.0mol

jc5C

=

0.3182

mol C0

2/mol

Hz

^{Balance: input}

=

output

"2

=

_«5*5H

j|«²

=

28.0mol, n₅

-

44.0mol x_5H

=

0.6364

mol C0

2/mol

x\

=

1

-

_xsc

-

x₅h

=

0.04545

mol

I/mol /revA Feed-Recycle

Mixing

PointAnalysis

Total

Mole

Balance: input

=

output n₀

+

n_r

=

100

mol

IBalance: ^input

=

output

n₀(0.00400)

+

nr(0.04545)

=

2.0

mol

I

Solvingthesetwoequations simultaneouslyyields

no

=

61.4

mol

fresh feed, nr

=

38.6molrecycle

COi

^{Balance: input}

=

output

«o-*oc

+

nTxsc

=

28.0

mol C0

2

jjn0

=

61.4mol, nr = 38.6mol, x_5C

=

_0.3182mol

C0

2/mol xqc

=

0.256

mol C0

2/mol

*oh

=

(1

_

*oc

~

_xqi)

=

0.740

mol H

2/mol Recycle-PurgeSplittingPointAnalysis

Total

Mole

Balance: input

=

output n_s

=

nT

+

«

p

jj^n₅

=

44.0mol, n_r

=

38.6mol np

=

5.4

mol

purge

FlowchartScaling

Forthe assumedbasisof100

mol

feed to the reactor,the productionrate ofmethanolis /13

=

14.0

mol CH3OH. To

scale theprocesstoamethanolproduction_{rate of}155

kmol CH

₃

OH/h, we

multiply eachtotaland

component

molarflowratebythefactor

155

kmol CH

3

OH/h

^\ 11.1kmol/h 14.0

mol CH3OH

^/

mol

142

Chapter

4

Fundamentals

of MaterialBalances

The mole

fractionsremain unchanged by^thescaling.

The

results follow.

Variable BasisValue ScaledValue Fresh feed 61.4

mol

25.6

mole% C0

2

74.0

mole% H

2

0.400

mole%

I

681kmol/h 25.6

mole% C0

2

74.0

mole% H

2

0.400

mole%

I

Feedtoreactor 100

mol

28.0

mole% C0

2

70.0

mole% H

2

2.0

mole %

I

1110kmol/h 28.0

mole% C0

2

70.0

mole% H

2

2.0

mole%

I

Recycle 38.6

mol

31.8

mole% C0

2

63.6

mole%H

2

4.6

mole%

^I

428kmol/h 31.8

mole% C0

2

63.6

mole% H

2

4.6

mole%

I

Purge 5.4

mol

31.8

mole% C0

2

63.6

mole% H

2

4.6

mole%

I

59.9kmol/h 31.8

mole% C0

2

63.6

mole% H

2

4.6

mole%

I

4.8

COMBUSTION REACTIONS

Combustion —

^the rapid reaction of a fuelwith

oxygen—

is perhaps

more

important than any otherclassofindustrial chemical reactions, despite the factthat

combustion

products

(C0

₂,

H

20,

and

possibly

CO and SO2)

are

worth much

less than the fuels

burned

to obtain them.

The

significance of these reactions liesin the

tremendous

quantities ofenergy theyrelease energythat^isusedto boilwaterto

produce

steam,whichisthenusedtodrivethe turbinesthat generate

most

oftheworld's electricalpower.

The

job of designing

power

_generation

equipment

usually fallsto mechanical engineers, but theanalysis of

combustion

reactions

and

reactors

and

the

abatement and

controlofenvi- ronmentalpollutioncaused

by combustion

productslike

CO, C0

₂^,

and S0

₂ are

problems

with which chemical engineersare heavily involved. In Chapter 14, forexample,

we

present acase study involving thegeneration ofelectricity

from

the

combustion

of coal

and removal

of

S0

2 (apollutant)

from combustion

products.

In the sections that follow,

we

introduce terminology

commonly

used in the analysis of

combustion

reactors

and

discuss material balance calculations for such reactors.

Methods

of determiningthe energy that can be obtained

from combustion

reactions are given in

Chap-

ter9.

4.8a Combustion Chemistry

Most

of thefuelusedin

power

plant

combustion

furnaces iseithercoal (carbon,

some

hydro- gen

and

sulfur,

and

variousnoncombustiblematerials), fuel oil(mostlyhighmolecular weight hydrocarbons,

some

sulfur),gaseousfuel(suchasnaturalgas,

which

isprimarilymethane),or liquefiedpetroleumgas,

which

isusually

propane

and/or butane.

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