Balances on Reactive Systems - جامعة نزوى€¦ · · 2016-02-294/2/2012 1 Balances on...

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Balances on Reactive SystemsMaterial balance no longer takes the form g

INPUT = OUTPUT

Must account for the disappearance of reactants and appearance of products through stoichiometry.

Stoichiometric EquationsThe stoichiometric equation of a chemical reaction is a statement qof the relative amounts of reactants and products that participate in the reaction.

2 SO2 + O2 → 2 SO3

A stoichiometric equation is valid only if the number of atoms of q yeach atomic species is balanced.

2 S → 2 S4 O + 2 O → 6 O

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Stoichiometric Equations

The stoichiometric equation of a chemical reaction is a qstatement of the relative amounts of reactants and products that participate in the reaction.

2 SO2 + O2 → 2 SO3

A stoichiometric ratio of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients:stoichiometric coefficients:

2 mol SO3 generated / 1 mol O2 consumed2 mol SO3 generated / 2 mol SO2 consumed

Stoichiometric Equations

C4H8 + 6 O2 → 4 CO2 + 4 H2OIs this stoichiometric equation balanced?What is the stoichiometric coefficient of CO2?What is the stoichiometric ratio of H2O to O2?How many lb‐mol O2 react to form 400 lb‐mol CO2?

22

2 Olbmol600COlb l4O lbmol 6

COlbmol 400 =×

100 lbmol/min C4H8 is fed and 50% reacts. At what rate is water formed?

22

2 COlbmol4

minOH lbmol

200HC lbmol 1OH lbmol 4

50.0min

HClbmol 100 2

84

284 =××

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Limiting and Excess ReactantsTwo reactants are said to be in stoichiometric proportion if the p pratio (moles A present/moles B present) equals the stoichiometric ratio from the balanced reaction equation.

2 SO2 + O2 → 2 SO3

the feed ratio that would represent stoichiometric proportion is nSO2/nO2 = 2:1f f d h d hIf reactants are fed in stoichometric proportion, and the reaction proceeds to completion, all reactants are consumed.

Limiting and Excess ReactantsStoichiometric Proportion – Reactants are present in a ratio p pequivalent to the ratio of the stoichiometric coefficients.

A + 2B→ 2C

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Limiting and Excess ReactantsLimiting reactant – A reactant is limiting if it is present g g pin less than stoichiometric proportion relative to every other reactant.

A + 2B→ 2C

Excess reactant – All other reactants besides the limiting reactant.

Limiting and Excess Reactants

fractional excess (fXS) – ratio of the excess to the stoichiometricf ( fXS)proportion.

A + 2B→ 2C

( ) ( )( )

25.0445n

nnf

stoichA

stoichAfeedAXS

=−

=

−=

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fractional conversion (f) – ratio of the amount of a fractional conversion (f) ratio of the amount of a reactant reacted, to the amount fed.

A + 2B→ 2C ( )( )fedA

reactedA

nn

f =

0.050fA == 0.0

80fB ==


fractional conversion (f) – ratio of the amount of a reactant f ( f)reacted, to the amount fed.

A + 2B→ 2C

( )( )fedA

reactedA

nn

f =

2.051fA == 25.0

82fB ==

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A + 2B→ 2C

( )( )fedA

reactedA

nn

f =

4.052fA == 5.0

84fB ==



A + 2B→ 2C( )

( )fedA

reactedA

nn

f =

6.053fA == 75.0

86fB ==

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Limiting and Excess Reactantsfractional conversion (f) – ratio of the amount of a reactant f ( f)reacted, to the amount fed.

A + 2B→ 2C( )

( )fedA

reactedA

nn

f =

8.054fA == 0.1

88fB ==

Extent of Reaction

extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .p g

ν – stoichiometric coefficients: νA = ‐1, νB = ‐2, νC = 2

A + 2B→ 2C ni = ni0 + ν iξ

0=ξ

nA = nA0 − ξ nB = nB0 − 2ξ nC = nC0 + 2ξ

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Extent of Reaction

extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .


A + 2B→ 2C ni = ni0 + ν iξ ξ = 0

nB = 8 − 2ξ = 8 nC = 0 + 2ξ = 0 nA = 5− ξ = 5

Extent of Reaction



A + 2B→ 2C ni = ni0 + ν iξ ξ = 1

nB = 8 − 2ξ = 6 nC = 0 + 2ξ = 2 nA = 5− ξ = 4

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Extent of Reaction



A + 2B→ 2C ni = ni0 + ν iξ ξ = 2

nB = 8 − 2ξ = 4 nC = 0 + 2ξ = 4 nA = 5− ξ = 3

Extent of Reaction

extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction .p g


A + 2B→ 2C ni = ni0 + ν iξ ξ = 4

nB = 8 − 2ξ = 0 nC = 0 + 2ξ = 8 nA = 5− ξ = 1

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Assume an equimolar reactant feed of 100 kmol:What is the limiting reactant?

OHCOHC 42242 22 →+

g

ethylene

A reactant is limiting if it is present in less than stoichiometricproportion relative to every other reactant.

12

nn

stoichO

HC

2

42 =⎟⎟⎠

⎞⎜⎜⎝

⎛

11

nn

feedO

HC

stoich

2

42

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⎠⎝

Assume an equimolar reactant feed of 100 kmol:

OHCOHC 42242 22 →+

qWhat is the percentage excess of each reactant?

( ) ( )( )

%10000150100

nnn

fstoichO

stoichOfeedOO,XS

2

22

2

−

−=

%10000.150

===

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Assume an equimolar reactant feed of 100 kmol:If the reaction proceeds to completion:

2C2H4 + O2 → 2C2H4O

If the reaction proceeds to completion: (a) how much of the excess

reactant will be left?

(b) H h C H O

( )50

21000

nn424242 HC

oHCHC

=ξξ−+=

ξν+= ( )50n

501100n

nn

2

2

222

O

O

OoOO

=

−+=

ξν+=

nn OHCo

OHCOHC 424242ξν+=

(b) How much C2H4O will be formed?

(c) What is the extent of reaction?

( )100n

5020n

OHC

OHC

OHCOHCOHC

42

42

424242

=

+=

Assume an equimolar reactant feed of 100 kmol:If the reaction proceeds to a point where the fractional

2C2H4 + O2 → 2C2H4O

p pconversion of the limiting reactant is 50%, how much of each reactant and product is present at the end? What is ξ?

( )( ) 5.0nn

ffedHC

reactedHC

42

42 == ( )25

210050100

nn424242 HC

oHCHC

=ξξ−+=−

ξν+=

nn o ξν+( )2520n

nn

OHC

OHCo

OHCOHC

42

424242

+=

ξν+=

5.0nn

f HCoHC 4242 =

−=

( )75n

251100n

nn

2

2

222

O

O

OOO

=

−+=

ξν+=50n OHC 42

42

=5.0

nf o

HC 42

50n

5.0100n100

42

42

HC

HC

=

=−

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Assume an equimolar reactant feed of 100 kmol:

2C2H4 + O2 → 2C2H4O

Assume an equimolar reactant feed of 100 kmol:If the reaction proceeds to a point where 60 mol of O2 is left, what is the fractional conversion of C2H4? What is ξ?

( )20n

402100n

nn

42

42

424242

HC

HC

HCoHCHC

=

−+=

ξν+=

( )40

110060

nn222 O

oOO

ξξ−+=

ξν+=

( )( ) 8.0

10020100

nn

ffedHC

reactedHC

42

42 =−

==

40=ξ

Reaction StoichiometryAcrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:

C3H6 + NH3 + 3

2O3 → C3H3N + 3H2O

limitingdetermine limiting reactant

nNH3nC3H6( )0

= 0.120 ×100 0.100 ×100( )= 1.20

nNH3nC3H6( )stoich

= 1 1( )= 1

nO2nC3H6( )0

= 0.780 × 0.21×100 0.100 ×100( )= 1.64

nO2nC3H6( )stoich

= 1.5 1( )= 1.5

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Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:

f 0 20determine fractional excesses

C3H6 + NH3 + 3

2O3 → C3H3N + 3H2O

fXS = 0.20

fXS = 0.093

determine fractional excesses limiting

nNH3( )stoich= 10.0 mol C3H6

1 mol NH3

1 mol C3H6

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 10.0 mol NH3

fXS( )NH3=

NH3( )0− NH3( )stoich

NH3( )stoich

= 12.0−10.010.0

= 0.20

nO2( )stoich= 10.0 mol C3H6

1.5 mol O2

1 mol C3H6

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 15.0 mol O2

fXS( )O2=

O2( )0− O2( )stoich

O2( )stoich

= 16.4−15.015.0

= 0.093


fXS = 0.20limiting

C3H6 + NH3 + 3

2O3 → C3H3N + 3H2O

fXS = 0.093

fXS 0.20

use fractional conversion to determine amount of propylene that leaves the reactor

limiting

nC3H6

= 1− f( ) nC3H6( )0= 1− 0.30( )10.0 mol C3H6( )= 7.0 mol C3H6

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C3H6 + NH3 + 32

O3 → C3H3N + 3H2O fXS = 0.093

fXS = 0.20limiting

ni = ni0 + ν iξ

2

nC3H6= 7.0 mol C3H6

ξ = 3 mol

determine extent ofreaction by applying molebalance to propylene

nC3H6= nC3H6( )0

+ −1( )ξ7.0 mol = 10.0 mol − ξ

ξ = 3 mol


fXS = 0.093fXS = 0.20

OHNHCONHHC 233223

363 3+→++

( )( ) ( )( ) 93110012.0 =−+=NHn

nC3H6= 7.0 mol C3H6

ξ = 3 mol ni = ni0 + ν iξapply mole balance to all remaining species

( )( ) ( )( )( )( )( ) ( )( )

( ) ( )( )( )( )( ) ( )( )( ) ( )( ) 90.330

6.613010078.079.0

30.310

9.11310078.021.0

93110012.0

2

2

33

2

3

23

=+=

=+=

=++=

=−+=

+

OH

N

NHC

O

NH

n

n

n

n

n

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Fuel for motor vehicles other than gasoline are being eyedbecause they generate lower levels of pollutants thandoes gasoline Compressed propane has been suggesteddoes gasoline. Compressed propane has been suggestedas a source of economic power for vehicles. Suppose thatin a test 22 kg of C3H8 is burned with 400 kg of air toproduce 44 kg of CO2 and 12 kg of CO.a. What was the percent excess air?b. The components in the product stream.

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Component Amount(kg)

Molecular weight

Input(kg-mol)

Output(kg-mol)

C H 22 44 0 50 ?

C3H8 + 5O2 3CO2 + 4H2O

C3H8 22 44 0.50 ?Air 400 29 13.80 ?O2 0.21*13.80

=2.90?

N2 0.79*13.80=10.9

?

CO2 44 44 - 1

30

CO 12 28 - 0.43

H2O. 18 - ?

Total 29.87 ?

Percent excess air = (O2 in – O2 used)/(O2 theory)= (2.90 – 5*0.5)/(5*0.5)=16%

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Component Input(kg-mol)

Consumed [1]

Consumed [2]

Generated [1]

Generated [2]

Total Output(kg-mol)

C3H8 + 5O2 3CO2 + 4H2O [1]

C3H8 + 7/2O2 3CO + 4H2O [2]

(kg-mol) [ ](kg-mol)

[ ](kg-mol)

[ ](kg-mol)

[ ](kg-mol)

(kg-mol)

C3H8 0.50 0.33 0.43/3=0.14 - - 0.03

O2 0.21*13.80=2.90

5*0.33= 1.65 3.5*0.14= 0.49

- - 0.76

N2 0.79*13.80=10.9

- - - - 10.9

CO2 - - - 1 - 1

31

CO - - - - 0.43 0.43

H2O. - 4*0.33= 1.32

4*0.14= 0.56

1.88

Total 29.87 15

Example

A gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned inA gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned in an engine with 200% excess air. If 80% of the ethane goes to CO2, 10% goes to CO and the remaining unburned, determine the amount of the excess air per 100 moles of the feed gas.

C2H6 + 7/2O2 2CO2 + 3H2O [1]C2H6 + 5/2O2 2CO + 3H2O [2]

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Component Input Gas Stream Output

C2H6 + 7/2O2 2CO2 + 3H2OBasis=100 mole feed gas

[1]

(kg-mol)

C2H6 80

O2 20 + 0.21*Gair= 799.7

Air Gair= 4.76*780 = 3,712.8

N2 0.79*Gair= 2,933.1

CO

Percent excess air (based on air):

(O2 in – O2 used)/(O2 theory)=2O2 theory (Eq 1) = 3.5*80=280O2 used (air) =O2 theory (Eq 1)-20=260 moles(O2 in (air)-260)/(260)=2O2in (air)=260+(2*260)=780molesN2 in = 3.76*780=2,932.8

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CO2

CO

H2O.

Total

N 3.76 780 ,93 .8Air stream=2,932.8 + 780=4.76*780=3,712.8

Component Input(mol)

Consumed [1]

Consumed [2]

Generated [1]

Generated [2]

Total Output(mol)

C2H6 + 7/2O2 2CO2 + 3H2O [1]C2H6 + 5/2O2 2CO + 3H2O [2]

(mol) [ ](mol)

[ ](mol)

[ ](mol)

[ ](mol)

(mol)

C2H6 80 64 8 - - 8

O2 799.7 224 20 - - 204

N2 2,933.1 - - - - 2933.1

CO2 - - - 128 - 128

CO - - - 16 16

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H2O. - - - 192 24 216

Total 3812.8 3,505.1

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Chemical EquilibriumGivenGiven

a set of reactive species, andreaction conditions

Determine1. the final (equilibrium) composition of the reaction

mixture2. how long the system takes to reach a specified state

short of equilibrium

Chemical EquilibriumIrreversible reactionIrreversible reaction

reaction proceeds only in a single direction A → Bconcentration of the limiting reactant eventually approaches zero (time duration can vary widely)

Equilibrium composition of an irreversible reaction is that which corresponds to complete conversion.

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Chemical EquilibriumReversible reactionReversible reaction

reaction proceeds in both directions A ↔ Bnet rate (forward – backward) eventually approaches zero (again, time can vary widely)

Equilibrium composition of a reversible reaction is that which corresponds to the equilibrium conversion.

Equilibrium Composition

An equilibrium reaction proceeds ( ) HCO 22

yy

yyTK =An equilibrium reaction proceeds

to an extent at temperature T based on the equilibrium constant, K(T).

where yi is the mole fraction of species i

OHCO 2yy

ξν+= i0ii nn

total

ii n

ny =

Water‐gas shift reaction:Assume 1 mole CO and 2 mole H2OK(1105 K) = 1.00

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+

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( ) HCO 22yy

TK =( ) ( ) ( )11nn 0COCO ξ−=ξ−+= ( )OHCO 2

yyTK( ) ( ) ( )

( ) ( ) ( )( ) ( )

( ) ( )3nnnnn

1nn

1nn

21nn

11nn

222

22

22

22

HCOOHCOtotal

0HH

0COCO

0OHOH

0COCO

=+++=

ξ=ξ++=

ξ=ξ++=

ξ−=ξ−+=

ξξ+

ξν+= i0ii nntotal

ii n

ny =


222

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+


( ) HCO 22

yy

yyTK =

1n ξ−=( )( )

( )( )==

ξξξξ

1213330= OHCO 2

yy

ξν iii nn += 0

total

ii n

ny =

3n

n

n

2n

1n

total

H

CO

OH

CO

2

2

2

=

ξ=

ξ=

ξ−=ξ−= ( )( )ξ−ξ− 21

( )( )

mol667.0

22

2122

2

=ξξ+ξ−ξ−=ξ

ξ−ξ−=ξ

667.0

667.0

333.1

333.0

====


( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+

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Equilibrium Composition( )( )

( )( ) 121

=−− ξξ

ξξ ( )OHCO

HCO 22

yy

yyTK =

3n

222.03/667.0y

222.03/667.0y

444.03/333.1y

111.03/333.0y

total

H

CO

OH

CO

2

2

2

=

==

==

====

( )( )21 ξξ

( )( )

mol667.0

22

2122

2

=ξξ+ξ−ξ−=ξ

ξ−ξ−=ξ

OHCO 2yy

ξν+= i0ii nntotal

ii n

ny =


( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+

limiting reactant is CO

ξν+nnat equilibrium111.03/333.0yCO ==

ξν+= i0ii nn ξ = 0.667mol

( )( )mol 333.0

667.011nCO

=−+=

at equilibrium,

fractional conversion at equilibrium

667.0f 00.1333.000.1 == −

3n

222.03/667.0y

222.03/667.0y

444.03/333.1y

total

H

CO

OH

2

2

2

=

==

==

==


( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+

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Multiple Reactions

C2H4 + 12

O2 → C2H4OC H 3O 2CO 2H O C2H4 + 3O2 → 2CO2 + 2H2O

( ) ∑ ξν+=j

jij0ii nnfor j reactions of i species,mole balance becomes

( ) ( ) ( ) 21HCHC 11nn ξ−+ξ−+= ( ) ( ) ( )( ) ( ) ( )

( ) ( ) 10OHCOHC

2121

0OO

210HCHC

1nn

3nn

11nn

4242

22

4242

ξ++=

ξ−+ξ−+=

ξ+ξ+( ) ( )( ) ( ) 20OHOH

20COCO

2nn

2nn

22

22

ξ++=

ξ++=

Multiple Reactions

C2H4 + 12

O2 → C2H4OC2H4 + 3O2 → 2CO2 + 2H2O 2 4 2 2 2

reactionssidenowithconversion100%atformed molesformed product desired moles

yield =

( ) ∑ ξν+=j

jij0ii nnfor j reactions of i species,mole balance becomes

formed product undesired molesformed product desired moles

yselectivit =

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Multiple Reactions

100 moles A fed to a batch reactorproduct composition: 10 mol A, 160 B, 10 C

What is1. fA?2. YB?3. SB/C?

CA

B2A

→→

10100−4. ξ1, ξ2 9.0

10010100

fA ==

Multiple Reactions


What is1. fA?2. YB?3. SB/C?

160

CA

B2A

→→

4. ξ1, ξ2( )( ) 889.0

10100160

Y12B =

−=

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What is1. fA?2. YB?3. SB/C? 16

10160

/ ==CBS

CA

B2A

→→

4. ξ1, ξ2 10

100 moles A fed to a batch reactorproduct composition: 10 mol A 160 B 10 Cproduct composition: 10 mol A, 160 B, 10 C

What is1. fA?2. YB?

S ? nn 22A11AAA ξν+ξν+=nn ξν+

CA

B2A

→→

3. SB/C?4. ξ1, ξ2 1090

10010

nn

12

21

22A11AAoA

=ξ−=ξξ−ξ−=

ξν+ξν+

80

20160

nn

1

1

11BBoB

=ξξ+=

ξν+=

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Balances on Reactive Processes

Continuous, steady‐state dehydrogenation of ethane, y y gTotal mass balance still has INPUT = OUTPUT formMolecular balances contain consumption/generationAtomic balances (H and C) also have simple form

24262 HHCHC +→


Continuous, steady‐state dehydrogenation of ethaneFirst consider molecular balances:

Molecular H2 balance: generation = output generation H2 = 40 kmol H2/min

C2H6 balance: input = output + consumption 100 kmol C2H6/min = n1 + (C2H6 consumed)

C2H4 balance: generation = output(C H t d) (C2H4 generated) = n2

24262 HHCHC +→

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Continuous, steady‐state dehydrogenation of ethaneAt i C b l i t t t HHCHC +→Atomic C balance: input = output

Atomic H balance: input = output

( ) ( ) ( )426262 HC mol 1

C mol 22HC mol 1

C mol 21HC mol 1

C mol 262 nnHC mol 100 && +=

( ) ( ) ( ) ( )Hmol4Hmol6Hmol2Hmol 6 40mol100 nnHC && ++=

24262 HHCHC +→

( ) ( ) ( ) ( )4262262 mol 12 mol 11 mol 1 mol 162 40 mol 100 HCHCHHC nnHC ++=

Independent Equations

To understand the number of independent species balances in areacting system requires an understanding of independent algebraicequations.Algebraic equations are independent if you cannot obtain any of themby adding/subtracting multiples of the others.

2][12y6x3

[1] 4y2x

=+=+

[3] 4y2x =+2][ 12y6x3 =+

[5] 6zy4

[4] 2zx2

=+=−

3×[1] = [2] 2×[3] – [4] = [5]

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Independent Equations

To understand the number of independent species balancesp pin a reacting system requires an understanding ofindependent algebraic equations.Algebraic equations are independent if you cannot obtainany of them by adding/subtracting multiples of the others.

2][ 12y6x3

[1] 4y2x

=+=+ ( )

( )1212

12y6y612

12y6y243

==+−=+−

Independent Species

If two molecular or atomic species are in the same ratio topeach other where ever they appear in a process and thisratio is incorporated in the flowchart labeling, balances onthose species will not be independent equations.

31 nn && =

31

31

n76.3n76.3 && =

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Independent Chemical Reactions

When using molecular species balances or extents ofg preaction to analyze a reactive system, the degree of freedomanalysis must account for the number of independentchemical reactions among the species entering and leavingthe system.

Independent Chemical Reactions

Chemical reactions are independent if the stoichiometricpequation of any one of them cannot be obtained by addingand subtracting multiples of the stoichiometric equationsof the others.

[2]CB

[1] 2BA

→→

[3] 2CA

[2] CB

→→

2×[2] + [1] = [3]

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Solving Reactive Systems

There are 3 possible methods for solving balances around a reactivesystem:

1. Molecular species balances require more complex calculations thanthe other methods and should be used only for simple (singlereaction) systems.

2. Atomic species balances generally lead to the most straightforwardsolution procedure, especially when more than one reaction isinvolved

3. Extents of reaction are convenient for chemical equilibriumproblems.

Molecular Species Balances

To use molecular species balances to analyze a reactive system, the balances must contain generation and/or consumption terms.

The degree‐of‐freedom analysis is as follows:# unknown labeled variables

+ # independent chemical reactions‐ # independent molecular species balances

# h i l i k i bl‐ # other equations relating unknown variables# of degrees of freedom

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2 unknown labeled variables+ 1 independent chemical reactions

at steady state24262 HHCHC +→

‐ 3 independent molecular species balances‐ 0 other equations relating unknown variables0 degrees of freedom


at steady state24262 HHCHC +→

H2 Balance: generation = output

2 402

HkmolgenH =

C2H6 Balance: input = output + consumption

( )( )1 1

min

1min

2

62262 40 1000 HkmolHCkmolHkmolHCkmol n += & ( )( )

min

162

2

60 HCkmoln =&

C2H4 Balance: generation = output

( )( )min

HC kmol2

H kmol 1HC kmol 1

minH kmol

2

42

2

422

40n

40n

=

=

&

&

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Atomic Species Balance

24262 HHCHC +→

All atomic species balances take the form INPUT = OUTPUT

Degree‐of‐freedom analysis, ndf =# unknown labeled variables

‐ # independent atomic species balances p p‐ # molecular balances on independent nonreactive species‐ # other equations relating unknown variables


All atomic species balances take the form INPUT = OUTPUT

Degree‐of‐freedom analysis, ndf = 0 =2 unknown labeled variables

‐ 2 independent atomic species balances ‐ 0 molecular balances on independent nonreactive species

‐ 0 other equations relating unknown variables

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HHCHC +→

C Balance: input = output

( )( ) ( ) ( )21

1 2

2 1 2

1 1 2

min

mol 100

100426262

42

nnk

nn HCkmolCkmol

HCkmolCkmol

HCkmolCkmolHCkmol

&&

&&

+=

+=

24262 HHCHC +→

H Balance: input = output

( )( ) ( )( )( ) ( )

21

HC kmol 1H kmol 4

2HC kmol 1H kmol 6

1

H kmol 1H kmol 2

minH kmol

HC kmol 1H kmol 6

minHC kmol

n4n6+molk 80molk 600

nn

40 100

4262

2

2

62

42

&&

&&

+=

++

=

Atomic Species BalanceHHCHC +→

Solve simultaneously

n4n6+molk80molk600:H

nnmolk 100 :C 21

+=+=

&&

&&

24262 HHCHC +→

min/HC kmol 40n

min/HC kmol 60n

n4n6+molk 80molk 600 :H

422

621

21

==

+=

&

&

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Extent of Reaction

The 3rd method by which to determine molar flows in a reactive system is using expressions for each species flow rate in terms of extents of reaction (ξ).

Degree‐of‐freedom analysis for such an approach:ndf = # of unknown labeled variables

( ) ∑ ξν+=j

jij0ii nn

+ # independent reactions‐ # independent nonreactive species‐ # other relationships or specifications

Incomplete Combustion of CH4

Methane is burned with air in a continuous steady‐state reactor to yield a mixture of carbon monoxide, carbon dioxide, and water.

The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO per mole CO.

CH4 + 32O2 → CO+2 H2O

CH4 +2 O2 → CO2 +2 H2O

contains 8 mol CO2 per mole CO.

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CH4 + 32O2 → CO+2 H2O

fCH4 = 0.9

The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.

ndf = 5 unknowns + 2 independent reactions

CH4 +2 O2 → CO2 +2 H2O

p‐ 5 expressions for ξ (CH4, O2, CO, CO2, H2O)‐ 1 nonreactive species balance (N2)‐ 1 specified methane conversion

=0

CH4 + 32O2 → CO+2 H2O

CH4 +2 O2 → CO2 +2 H2O

fCH4 = 0.9

N2 balance: nonreactive species, INPUT = OUTPUT

CH4 conversion specification:

( ) 2molN mol

N N mol 8.72mol 100728.0n 2

2==

( )( )( )CHmol ll( )( )( ) 4molCHmol

CH CHmol780.0mol1000780.0900.01n 4

4=−=

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OHCOOCHOHCOOCH

2224

2223

4

2 2 2

+→+

+→+

Extent of reaction mole balances:

( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( ) 210OHOH

20COCO

10COCO

210CHCH

22nn

1nn

1nn

11nn

22

22

44

ξ++ξ++=

ξ++=

ξ++=

ξ−+ξ−+= 0.78 =7.80− ξ1 − ξ2nCO = ξ1nCO2

=8nCO = ξ2nH2O =2ξ1 +2ξ2( )

( ) ( ) ( ) 2123

0OO

0

2nn22

22

ξ−+ξ−+= nO2

=19.4− 32

ξ1 −2ξ2

Product Separation and Recycle

Two definitions of reactant conversion are used in the analysis of chemical reactors with product separation and recycle of unconsumed reactants.

⎟⎟⎞

⎜⎜⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛=

reactorfromoutput‐reactortoinputreactant

pass glesin

process to inputprocess from output ‐ process to input

reactantconversion

overall

⎟⎟⎠

⎜⎜⎝

=reactor to input

reactantconversion

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%75%100A/min mol 100

A/min mol 25 ‐ A/min mol 100conversion

pass glesin

%100%100A/min mol 75

0 ‐ A/min mol 75conversion

overall

=×⎟⎠⎞

⎜⎝⎛=

=×⎟⎠⎞

⎜⎝⎛=

Catalytic Propane Dehydrogenation

C3H8 → C3H6 +H2

95% overallconversion

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Catalytic Propane Dehydrogenation95% overallconversion

C3H8 → C3H6 +H2

Overall Process

n = 3 unknowns (n n n )ndf = 3 unknowns (n6, n7, n8) – 2 independent atomic balances (C and H) – 1 relation (overall conversion)= 0

consider n6, n7, n8 known for further DOF analyses

Catalytic Propane Dehydrogenation95% overallconversionndf = 2

C3H8 → C3H6 +H2

Mixingpoint

n = 4 unknowns (n n n n )ndf = 4 unknowns (n9, n10, n1, n2) – 2 balances (C3H8 and C3H6) = 2

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ndf = 3

C3H8 → C3H6 +H2

reactor

n = 5 unknowns (n n n n n )ndf = 5 unknowns (n3, n4, n5, n1, n2) – 2 balances (C and H) = 3


ndf = 3 ndf = 0

C3H8 → C3H6 +H2

separator

n = 5 unknowns (n n n n n )ndf = 5 unknowns (n3, n4, n5, n9, n10) – 3 balances (C3H8, C3H6, and H2)– 2 relations (reactant and product recovery fractions)= 0

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C3H8 → C3H6 +H2

overall

( )( ) HCl5l1009501

conversionrelationship

( )( ) 836 HCmol5mol10095.01n =−=


836 HCmol 5n =

C3H8 → C3H6 +H2

overallC atomic balance

( )( ) ( )( ) ( )637

HC mol 1C mol3

7HC mol 1C mol3

83HC mol 1C mol3

HC mol 95n

nHC mol 5mol 100638383

=

+=

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836 HCmol 5n =

C3H8 → C3H6 +H2

overall

H atomic balance

637 HCmol 95n =

( )( ) ( )( )( )( ) ( )

263

8383

H mol 1H mol 2

8HC mol 1H mol 6

63

HC mol 1H mol8

83HC mol 1H mol 8

nHC mol 95

HC mol 5mol 100

++

=

28 H mol 95n =


836 HCmol 5n =

C3H8 → C3H6 +H2

separator

given relations

637 HCmol 95n =

28 H mol 95n =

( )( ) 6310710

83336

HC mol 75.4nn0500.0n

HCmol900nn00555.0n

=→==→=

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833 HCmol900n = 836 HCmol 5n =

C3H8 → C3H6 +H2

separator

propane balance n10 = 4.75 mol C3H6

637 HCmol 95n =

28 H mol 95n =

n3 =n6 +n9 →n9 = 895 mol C3H8



C3H8 → C3H6 +H2

mixerpropane balance

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

637 HCmol 95n =

28 H mol 95n =

100+n9 =n1 →n1 = 995 mol C3H8

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n1 = 995 mol C3H8


C3H8 → C3H6 +H2

mixer

propylene balance n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

637 HCmol 95n =

28 H mol 95n =

n10 =n2 →n2 = 4.75 mol C3H6


n1 = 995 mol C 3H8 833 HCmol900n = 836 HCmol 5n =

C3H8 → C3H6 +H2

reactor

C atomic balance n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n1 995 mol C 3H8

n2 = 4.75 mol C3H6

( )( ) ( )( )Cmol3Cmol3 HCl754HCl959 +

637 HCmol 95n =

28 H mol 95n =

( )( ) ( )( )( )( ) ( )( )

634

HC mol 1C mol 3

4HC mol 1C mol3

83

HC mol 1Cmol3

63HC mol 1Cmol3

83

HC mol 75.99n

nHC mol 009

HCmol.754HC mol 959

6383

6383

=

+=

+

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C3H8 → C3H6 +H2

reactor

H atomic balance n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n1 995 mol C 3H8

n2 = 4.75 mol C3H6634 HCmol75.99n = 637 HCmol 95n =

28 H mol 95n =

( )( ) ( )( )( )( ) ( )( ) ( )

25

H mol 1H mol 2

5HC mol 1H mol 6

63HC mol 1H mol 8

83

HC mol 1H mol 6

63HC mol 1H mol 8

83

H mol 95n

nHC mol 75.99HC mol 009

HC mol .754HC mol 959

26383

6383

=

++=

+



C3H8 → C3H6 +H2

single‐passconversion

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n1 995 mol C 3H8

n2 = 4.75 mol C3H6634 HCmol75.99n =

25 Hmol95n =637 HCmol 95n =

28 H mol 95n =

( ) ( )( ) %55.9%100

HC mol 959HCmol009HCmol959

f83

8383passsingle =×

−=−

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836 HCmol 5n =833 HCmol900n =n1 = 995 mol C 3H8

fsingle−pass =9.55%

C3H8 → C3H6 +H2

Recycle ratio

637 HCmol 95n =

28 H mol 95n =

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n1 995 mol C 3H8

n2 = 4.75 mol C3H6634 HCmol75.99n =

25 Hmol95n =

( )( ) ( )

( ) feed fresh molrecycle mol109 0.9

mol 001mol .754mol 958

feed mol 001nn

R =+

=+

=

PurgingNecessary with recycle to prevent accumulation of a species that is both present in the fresh feed and is recycled rather than separated with the

dproduct.

mixed fresh feedand recycle is aconvenient

basis selection fsingle−pass = 60%

CO2 +3H2 → CH3OH+H2O

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Methanol Synthesis

ndf = 7 unknowns (n0, x0C, np, x5C, x5H, n3, n4) + 1 rxnp 5 5 3

‐ 5 independent species balances = 3

fsingle−pass = 60%


Methanol Synthesisndf = 4 unknowns (n1, n2, n3, n4) + 1 rxn

– 4 independent species balances 4 p p– 1 single pass conversion = 0



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ndf = 3 unknowns (n5, x5C, X5H)– 3 independent species balances

= 0



ndf = 3 unknowns (n0, x0C, nr)– 3 independent species balances

= 0



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ndf = 1 unknowns (np)– 1 independent species balance

investigatemole balances andtheir solution in the textp p

= 0


the text


Combustion Reactions

Combustion ‐ rapid reaction of a fuel with oxygen.p ygValuable class of reactions due to the tremendous amount of heat liberated, subsequently used to produce steam used to drive turbines which generates most of the world’s electrical power.Common fuels used in power plants:

coalcoalfuel oil (high MW hydrocarbons)gaseous fuel (natural gas)liquified petroleum gas (propane and/or butane)

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Combustion Chemistry

When a fuel is burnedC forms CO2 (complete) or CO (partial combustion)H forms H2OS forms SO2

N forms NO2 (above 1800°C)Air is used as the source of oxygen. DRY air analysis:

78.03 mol% N2ll f20.99 mol% O2

0.94 mol% Ar0.03 mol% CO2

0.01 mol% H2, He, Ne, Kr, Xe

usually safe to assume:79 mol% N221 mol% O2


Stack (flue) gas – product gas that leaves a furnace.(f ) g p gComposition analysis:

wet basis – water is included in mole fractionsdry basis – does not include water in mole fractions

Stack gas contains (mol) on a wet basis:60.0% N2, 15.0% CO2, 10.0% O2, 15.0% H2ODry basis analysis:Dry basis analysis:

60/(60+15+10) = 0.706 mol N2/mol15/(60+15+10) = 0.176 mol CO2/mol10/(60+15+10) = 0.118 mol O2/mol

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Stack gas contains (mol) on a dry basis:65% N2, 14% CO2, 10% O2, 11% CO

assume 100 mole dry gas basis7.53 mole H2O65 mole N2

14 mole CO2

xH2O = 7.53107.5

= 0.0700

xN2= 65

107.5= 0.605

xCO2= 14

107 5= 0.130

10 mole O2

11 mole COtotal = 107.5 mole

CO2 107.5

xO2= 10

107.5= 0.0930

xCO = 11107.5

= 0.102

Theoretical and Excess AirThe less expensive reactant is commonly fed in excess of stoichiometricratio relative to the more valuable reactant thereby increasingratio relative to the more valuable reactant, thereby increasingconversion of the more expensive reactant at the expense of increaseduse of excess reactant.In a combustion reaction, the less expensive reactant is oxygen,obtained from the air. Conseqently, air is fed in excess to the fuel.Theoretical oxygen is the exact amount of O2 needed to completelycombust the fuel to CO2 and H2O.Theoretical air is that amount of air that contains the amount ofTheoretical air is that amount of air that contains the amount oftheoretical oxygen.

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Theoretical and Excess Air

Excess air is the amount by which the air fed to the Excess air is the amount by which the air fed to the reactor exceeds the theoretical air.

% excess air =

moles air fed ‐ moles air theoretical

moles air theoretical×100%

Theoretical and Excess Air

C4H10 + 13

2O2 → 4CO2 +5H2O

nC4H10 = 100 mol/hr; nair = 5000 mol/hr

( )hrO mol

650HC molO mol 5.6

hrHC mol 100

n 2

104

2104ltheoreticaO2

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

( ) airmol3094

airmol.764Omol506n 2

⎟⎟⎞

⎜⎜⎛

⎟⎞

⎜⎛( )

hr3094

O molhrn

2

2ltheoreticaair =⎟⎟

⎠⎜⎜⎝

⎟⎠⎞

⎜⎝⎛=

% excess air =

5000−30943094

×100% =61.6%

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Combustion Reactors

Procedure for writing/solving material balances for a g/ gcombustion reactor1. When you draw and label the flowchart, be sure the outlet

stream (the stack gas) includesa. unreacted fuel (unless the fuel is completely consumed)b. unreacted oxygenc. water and carbon dioxide (and CO if combustion is

i l )incomplete)d. nitrogen (if air is used as the oxygen source)

2. Calculate the O2 feed rate from the specifed percent excess oxygen or air

3. If multiple reactions, use atomic balances

Combustion of Ethane

OHCOOHCOHCOOHC

5

22227

62

3232

+→+

+→+fC2H6 = 0.9OHCOOHC 222

562 32 +→+ C2H6 9

25% of the ethane burned forms CO

degree‐of‐freedomanalysis

ndf = 7 unknowns‐ 3 atomic balances 1 nitrogen balance‐ 1 nitrogen balance

‐ 1 excess air specification‐ 1 ethane conversion specification‐ 1 CO/CO2 ratio specification

= 0

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fC2H6 = 0.9excess air specification

OHCOOHCOHCOOHC

2225

62

22227

62

3232

+→+

+→+


( ) 262

262ltheoreticaO O mol 350

HC molO mol 5.3HC mol 100

n2

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

( )( )air mol 2500n

Omol 3505.1n21.0

0

20

==

( )( ) 62621 HCmol0.10HCmol10090.01n =−=

Ethane conversion specification

C H 7 O 2CO 3H O

fC2H6 = 0.9CO/CO2 ratio specification n0 = 2500 mol air

n1 =10.0 mol C2H6

C2H6 + 72O2 → 2CO2 +3H2O

C2H6 + 52O2 → 2CO+3H2O


( )( )( ) OC mol 0.45react HC mol 1gen CO mol 2

HC mol 1009.025.0n62

624 =⎟⎟⎠

⎞⎜⎜⎝

⎛=

Nitrogen balance

( )( ) 23 Nmol1975iramol250079.0n ==

( )( ) ( ) ( ) ( )25

CO mol 1C mol 1

5CO mol 1C mol 1

4HC mol 1C mol 2

1HC mol 1C mol 2

62

CO mol 135n

nnnHC mol 10026262

=

++=

Atomic C balance

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C2H6 + 72O2 → 2CO2 +3H2O

fC2H6 = 0.9

n1 =10.0 mol C2H6

n3 =1975 mol N2

C2H6 + 5

2O2 → 2CO+3H2O

atomic H balance

( )( ) ( )( ) ( )OH mol 270n

nHC mol 10HC mol 100

26

OH mol 1H mol 2

6HC mol 1H mol 6

62HC mol 1H mol 6

62 26262

=

+=

n0 = 2500 mol air

n4 =45.0 mol CO

n5 =135 mol CO2

( )( ) ( ) ( )( )( )( ) ( )( )

22

OH mol 1O mol 1

2CO mol 1O mol 2

2

CO mol 1O mol 1

O mol 1O mol 2

2O mol 1O mol 2

2

O mol 232n

OH mol 702CO mol 135

CO mol 54nO mol 255

22

22

=

++

+=

atomic O balance

C2H6 + 72O2 → 2CO2 +3H2O

C2H6 + 52O2 → 2CO+3H2O

fC2H6 = 0.9

n1 =10.0 mol C2H6

n3 =1975 mol N2

C2H6 9

25% of the ethane burned forms COatomic O balance

( )( ) ( ) ( )( )( )( ) ( )( )OHmol1

O mol 12COmol1

O mol 22

CO mol 1O mol 1

O mol 1O mol 2

2O mol 1O mol 2

2

OHmol702COmol135

CO mol 54nO mol 25522

++

+=

n0 = 2500 mol air

n4 = 45.0 mol CO

n5 =135 mol CO2

n6 = 270 mol H2O

( )( ) ( )( )22

OHmol12COmol12

O mol 232n22

=

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C2H6 + 72O2 → 2CO2 +3H2O

C2H6 + 52O2 → 2CO+3H2O

fC2H6 = 0.9

n1 =10.0 mol C2H6

n3 =1975 mol N2

n2 = 232 mol O2

C2H6 9


stack gas composition (dry basis)

sum=10+232+1974+45+135= 2396

n0 = 2500 mol air

n4 = 45.0 mol CO

n5 =135 mol CO2

n6 = 270 mol H2O

y1 =10 2396= 0.00417 mol C2H6 mol

y2 = 232 2396= 0.0970 mol O2 mol

y3 =1974 2396=0.824 mol N2 mol

y4 = 45 2396=0.019 mol CO mol

y5 =135 2396=0.0563 mol CO2 mol

270 mol H2O

2396 mol dry stack gas

=0.113mol H2O

mol dry stack gas

Balances on Reactive Systems - جامعة نزوى€¦ · · 2016-02-294/2/2012 1 Balances on...

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Transcript of Balances on Reactive Systems - جامعة نزوى€¦ · · 2016-02-294/2/2012 1 Balances on...