reactive system mb

8/13/2019 Reactive System MB

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Balances on Reactive Systems

Material balance no lon er takes the form

INPUT = OUTPUT

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

Stoichiometric Equations

The stoichiometric e uation of a chemical reaction is a statement of 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 each atomic species is balanced.

2 S → 2 S

4 O + 2 O → 6 O


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The stoichiometric e uation of a chemical reaction is a statement 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

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


C4H8 + 6 O2→ 4 CO2 + 4 H2O

Is 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 Olbmol600Olbmol6

COlbmol400 =×

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

2COmo4

min

OHlbmol 200

HClbmol1

OHlbmol450.0

min

HClbmol 100 2

84

284 =××


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Limiting and Excess Reactants

Two reactants are said to be in stoichiometric ro ortion if the ratio (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 = 2:1 I reactants are e in stoic ometric proportion, an t e reaction

proceeds to completion, all reactants are consumed.


Stoichiometric Pro ortion – Reactants are resent in a ratio equivalent to the ratio of the stoichiometric coefficients.

A + 2B→ 2C


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Limiting and

Excess

Reactants

Limitin reactant – A reactant is limitin if it is resent in less than stoichiometric proportion relative to every other reactant.

A + 2B→ 2C

Excess reactant – All other reactants besides the limiting reactant.


ractional excess ( ) – ratio of the excess to the stoichiometricproportion.

A + 2B→ 2C

( ) ( )( )

25.04

45

n

nnf

stoichA

stoichAfeedA

XS

=−

=

−=


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Limiting and

Excess

Reactants

ractional conversion – ratio of the amount of a reactant reacted, to the amount fed.

A + 2B→ 2C ( )( )

fedA

reactedA

n

nf =

0.05

0f A == 0.0

8

0f

B ==


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

A + 2B→ 2C

( )( )

fedA

reactedA

n

nf =

2.05

1f

A == 25.0

8

2f

B ==


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Limiting and

Excess

Reactants


A + 2B→ 2C

( )( )

fedA

reactedA

n

nf =

4.05

2f

A == 5.0

8

4f

B ==

Limiting and

Excess

Reactants


A + 2B→ 2C( )

( )fedA

reactedA

n

nf =

6.05

3f

A == 75.08

6f

B ==


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A + 2B→ 2C

( )( )

fedA

reactedA

n

nf =

8.05

4f

A == 0.1

8

8f

B ==

Extent of

Reaction

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

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

A + 2B→ 2C n i = ni0 + νiξ

0=ξ

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


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Extent of

Reaction



A + 2B→ 2C n i = ni0 + νiξ ξ = 0

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

Extent of

Reaction



A + 2B→ 2C n i = ni0 + νiξ ξ =1

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


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Extent of

Reaction



A + 2B→ 2C n i = ni0 + νiξ 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 ro ress of a chemical reaction .


A + 2B→ 2C n i = ni0 + νiξ 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 limitin reactant?

O H C O H C 42242 22 →+

ethylene

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

1

2

n

n

O

HC

2

42 =⎟⎟ ⎞

⎜⎜⎛

1

1

n

n

feedO

HC

2

42

=⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

Assume an e uimolar reactant feed of 100 kmol:

O H C O H C 42242 22 →+

What is the percentage excess of each reactant?

) )( )

50100

n

nnf

stoichO

stoichOfeedO

O,XS

2

22

2

−

−=

.

50

===


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2C2H4 + O2 → 2C2H4O

e reac on procee s o comp e on

(a) how much of the excess reactant will be left?

( )50

21000

nn424242 HC

o

HCHC

=ξ

ξ−+=

ξν+= ( )

50n

501100n

nn

2

2

222

O

O

O

o

OO

=

−+=

ξν+=

nnOHC

o

OHCOHC ξν+=

ow muc 2 4

will be formed?

(c) What is the extent of reaction?

( )

100n

5020n

OHC

OHC

42

42

=

+=


If the reaction proceeds to a point where the fractional

2C2H4 + O2 → 2C2H4O

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

( ) 5.0

n

nf

fedHC

reactedHC

42

42 == ( )25

210050100

nn424242 HC

o

HCHC

=ξ

ξ−+=−

ξν+=

o ( )2520n

nn

OHC

OHC

o

OHCOHC

42

424242

+=

ξν+=

5.0nn

f HC

o

HC 4242 =−

=

( )75n

251100n

2

2

222

O

O

OOO

=

−+=

=

50n OHC 42 =n

o

HC42

50n

5.0100

n100

42

42

HC

HC

=

=−


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Assume an e uimolar reactant feed of 100 kmol:

2C2H4 + O2 → 2C2H4O

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

HC

o

HCHC

=

−+=

ξν+=

( )110060

nn222 O

o

OO

ξ−+=

ξν+=

( ) 8.0

100

20100

n

nf

fedHC

reactedHC

42

42 =−==

=

Reaction Stoichiometry

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

C3H6 + NH3 +

3

2O3 → C3H3 N + 3H2O

limitingdetermine limiting reactant

n NH3 nC3H6( )0 = 0.120 ×100 0.100 ×100( )= 1.20n NH3 nC3H6( )stoich = 1 1( )= 1nO2 nC3H6( )0 = 0.780 × 0.21×100 0.100 ×100( )= 1.64nO2 nC3H6( )stoich = 1.5 1( )= 1.5


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C3H6 + NH3 +

3

2O3 → C3H3 N + 3H2O

XS = .

f XS = 0.093

limiting

n NH3

( )stoich

= 10.0 mol C3H61 mol NH3

1 mol C3H6

⎛

⎝

⎜ ⎞

⎠

⎟ = 10.0 mol NH3

f XS( ) NH3

= NH3( )0 − NH3( )stoich

NH3( )stoich= 12.0−10.0

10.0= 0.20

nO2( )stoich = 10.0 mol C3H61.5 mol O2

1 mol C3H6

⎛⎝⎜ ⎞

⎠⎟ = 15.0 mol O2

f XS( )O2

=O2( )0 − O2( )stoich

O2( )stoich= 16.4−15.0

15.0= 0.093


= 0.20

C3H6 + NH3 +

3

2O3 → C3H3 N + 3H2O

f XS = 0.093

.

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

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


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Acrylonitrile produced by reaction of ammonia, propylene, and O2 at

30% conversion of limiting reactant:

C3H6 + NH3 +3 O3 → C3H3 N + 3H2O

f XS = 0.093 f XS = 0.20limiting

ni = ni0 + νiξ

nC3H6 = 7.0 mol C3H6

ξ = 3 mol

determine extent of reaction by applying molebalance to propylene

nC3H6 = nC3H6( )0 + −1( )ξ7.0 mol =10.0 mol − ξ

ξ = 3 mol


f XS = 0.093f XS = 0.20

O H N H C O NH H C 233223

363 3+→++

93110012.0 =−+=n

nC3H6 = 7.0 mol C3H6

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

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

( )( )( ) ( )( )

( ) ( )( ) 90.330

6.613010078.079.0

30.310

9.11310078.021.0

.

2

2

33

2

3

23

=+=

=+=

=++==−+=

O H

N

N H C

O

n

n

n

n


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Fuel for motor vehicles other than gasoline are being eyedbecause they generate lower levels of pollutants than

.as 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.

29

Component Amount

(kg)

Molecular

weight

Input

(kg-mol)

Output

(kg-mol)

C3H8 + 5O2 3CO2 + 4H2O

3 8 .

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

-

Consumed

1

Consumed

2

Generated

1

Generated

2

Total Output

-

C3H8 + 5O2 3CO2 + 4H2O [1]C3H8 + 7/2O2 3CO + 4H2O [2]

-

(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

2 6 2an engine with 200% excess air. If 80% of the ethane goes to CO2,10% goes to CO and the remaining unburned, determine the amountof the excess air per 100 moles of the feed gas.

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

32


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Component Input Gas Stream Output

C2H6 + 7/2O2 2CO2 + 3H2O

Basis=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

Percent excess air (based on air):

(O2 in – O2 used)/(O2 theory)=2

O2 theory (Eq 1) = 3.5*80=280

O2 used (air) =O2 theory (Eq 1)-20=260 moles

(O2 in (air)-260)/(260)=2

O2in (air)=260+(2*260)=780moles

N2 in = 3.76*780=2 932.8

33

2

CO

H2O.

Total

Air stream=2,932.8 + 780=4.76*780=3,712.8

Component Input Consumed

1

Consumed

2

Generated

1

Generated

2

Total Output

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

(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

34

H2O. - - - 192 24 216

Total 3812.8 3,505.1


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Chemical Equilibrium

Given

a set of reactive species, and

reaction conditions

Determine

1. the final (equilibrium) composition of the reaction mixture

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

short of equilibrium

Chemical Equilibrium Irreversible reaction

reaction proceeds only in a single direction A → B

concentration 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 Equilibrium

Reversible reaction

reaction proceeds in both directions A ↔ B

net 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 e uilibrium reaction roceeds ( ) HCO 22

yyTK =

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

where y i is the mole fraction of species i

OHCO 2

ξν+= i0ii nn

total

ii

n

ny =

Water‐gas shift reaction: Assume 1 mole CO and 2 mole H2O

K(1105 K) = 1.00

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+


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HCO 22

yyTK =11nn −=−+=

OHCO 2yy

( ) ( ) ( )( ) ( )

( ) ( )3nnnnn

1nn

1nn

21nn

222

22

22

22

HCOOHCOtotal

0HH

0COCO

0OHOH

0

=+++=

ξ=ξ++=

ξ=ξ++=

ξ−=ξ−+=

ξν+= i0ii nntotal

ii

n

ny =


K(1105 K) = 1.00

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+


( ) HCO 22yy

TK =−

( )( )==

ξξ1

OHCO 2

ξ ν iii nn += 0

total

ii

n

ny =

3n

n

n

2n

total

H

CO

OH

CO

2

2

2

=

ξ=

ξ=

ξ−=

− −−

( )( )

mol667.0

22

21

22

2

=ξ

ξ+ξ−ξ−=ξ

ξ−ξ−=ξ

667.0

667.0

333.1

.

=

=

=


K(1105 K) = 1.00

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+


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( )( )1

21=

−−ξ ξ

( ) HCO 22yy

TK =

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

=

==

==

==

==

( )( )

mol667.0

22

21

22

2

=ξ

ξ+ξ−ξ−=ξ

ξ−ξ−=ξ

2

ξν+= i0ii nntotal

ii

n

ny =


K(1105 K) = 1.00

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+

limiting reactant is CO

111.03/333.0yCO === i0ii

ξ = 0.667mol( )( )

mol333.0

667.011nCO

=

−+=

,

fractional conversion at equilibrium

667.0f 00.1

333.000.1 == −3n

222.03/667.0y

222.03/667.0y

444.03/333.1y

total

H

CO

OH

2

2

2

=

==

==

==


K(1105 K) = 1.00

( ) ( ) ( ) ( )g2g2g2g HCOOHCO +⇔+


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Multiple

ReactionsC2H4 +

1

2O2 → C2H4O

2 4 + 2 → 2 + 2

( ) ∑ ξν+= j

jij0ii nn

for j reactions of i species,mole balance becomes

−−=

( ) ( ) ( )( ) ( ) 10OHCOHC212

1

0OO

0

1nn3nn

4242

22

4242

ξ++=ξ−+ξ−+= ( )( ) ( ) 20OHOH

20COCO

2nn2nn

22

22

ξ++=ξ++=

Multiple Reactions

C2H4 +1

2O2 → C2H4O

C H + 3O → 2CO + 2H O

reactionssidenowithconversion100%atformedmoles

formedproductdesiredmolesyield =

( ) ∑ ξν+= j

jij0ii nnfor j reactions of i species,

mole balance becomes

formedproductundesiredmoles

formedproductdesiredmolesyselectivit =


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Multiple Reactions

100 moles A fed to a batch reactor

product composition: 10 mol A, 160 B, 10 C

What is

1. f A ?

2. Y B?

3. SB/C?

CA

B2A

→

→

−4. ξ1, ξ2 9.0

100f A ==

Multiple Reactions



What is

1. f A ?

2. Y B?

3. SB/C?

CA

B2A

→

→

4. ξ1, ξ2( )( )

889.010100

160Y

12B =−=


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What is

1. f A ?

2. Y B?

3. SB/C? 16160

/ ==C BS

CA

B2A

→

→

4. ξ1, ξ2


pro uc compos on: 10 mo , 1 0 , 10

What is

1. f A ?

2. Y B?=

CA

B2A

→

→

3. B/C4. ξ1, ξ2 1090

10010

12

21

22A11AAoA

=ξ−=ξξ−ξ−=

80

20160

1

1

11BBoB

=ξξ+=

=


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Balances on

Reactive

Processes

Continuous stead ‐state deh dro enation of ethane

Total mass balance still has INPUT = OUTPUT form

Molecular balances contain consumption/generation

Atomic balances (H and C) also have simple form

24262 HHCHC +→

Balances on

Reactive

Processes

Continuous, steady ‐state dehydrogenation of ethane

First 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

2 4 genera e = n2

24262 HHCHC +→


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Balances on

Reactive

Processes

Continuous, steady ‐state dehydrogenation of ethane

om c a ance: npu = ou pu

Atomic H balance: input = output

426262 HCmol1Cmol2

2HCmol1Cmol2

1HCmol1Cmol2

62 nnHCmol100 && +=

Hmol4Hmol6Hmol2Hmol6 &&

24262

4262262 mol12mol11mol1mol162 H C H C H H C

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.

[1] 4y2x =+[3] 4y2x =+

[5] 6zy4

[4] 2zx2

=+

=−

3×[1] = [2] 2×[3] – [4] = [5]


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Independent Equations

To understand the number of inde endent s ecies balancesin a reacting system requires an understanding of independent 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 s ecies are in the same ratio toeach other where ever they appear in a process and thisratio is incorporated in the flowchart labeling, balances onthose species will not be independent equations.

nn && =

31 n76.3n76.3 &&

=


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Independent Chemical

Reactions

When usin molecular s ecies balances or extents of reaction 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 inde endent if the stoichiometricequation of any one of them cannot be obtained by addingand subtracting multiples of the stoichiometric equationsof the others.

[1] 2BA →

[3]

2CA

→ 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 equilibrium

problems.

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

‐ # ot er equat ons re at ng un nown var a es

# of degrees of freedom


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Molecular Species

Balances

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

Molecular Species

Balances

at steady state

24262 H H C H C +→

H2 Balance: generation = output

2402 H kmolgen H =

C2H6 Balance: input = output + consumption

) )1

1

min

1min

62262 401000 H kmol

H C kmol H kmol H C kmoln += &

min

1 6260 H C kmol

n =&C2H4 Balance: generation = output

( )( )

min

HCkmol

2

Hkmol1

HCkmol1

min

Hkmol

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

‐ # molecular balances on independent nonreactive species

‐ # other equations relating unknown variables

Atomic Species Balance

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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Atomic Species

Balance

C Balance: input = output

( )( ) ( ) ( )

21

1 2

21 2

11 2

min

mol100

100426262

42

nnk

nn H C kmolC kmol

H C kmolC kmol

H C kmolC kmol H C kmol

&&

&&

+=

+=

24262

H Balance: input = output

( )( ) ( )( )

( ) ( )21

HCkmol1Hkmol4

2HCkmol1Hkmol6

1

Hkmol1Hkmol2

min

Hkmol

HCkmol1Hkmol6

min

HCkmol

n4n6+molk80molk600

nn

40100

4262

2

2

62

42

&&

&&

+=

++

=

Atomic Species Balance

Solve simultaneously

nnmolk100 :C 21

=

+=&&

&&

24262

min/HCkmol40n

min/HCkmol60n

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

CH4 + 3

2O2 → CO+2 H2O

CH4 +2 O2 → CO2 +2 H2O

2 .


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CH4 + 3

2O2 → CO+2 H2O

f CH4 = 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 inde endent reactions

CH4 +2 O2 → CO2 +2 H2O

‐ 5 expressions for ξ (CH4, O2, CO, CO2, H2O)

‐ 1 nonreactive species balance (N2)‐ 1 specified methane conversion

=0

CH4 + 3

2O2 → CO+2 H2O

CH4 +2 O2 → CO2 +2 H2O

f CH4 = 0.9

N2 balance: nonreactive species, INPUT = OUTPUT

CH4 conversion specification:( ) 2mol

Nmol

N Nmol8.72mol100728.0n 2

2==

4molCH CHmol780.0mol1000780.0900.01n 4

4=−=


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O H COOCH

O H COOCH

2224

2223

4

22

2

+→+

+→+

Extent of reaction mole balances:( ) ( )

( ) ( )

( ) ( )( ) ( ) ( ) 21OHOH

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

( ) ( ) ( ) 21230OO 2nn 22 ξ−+ξ−+= nO2 =19.4 −

3

2ξ1 −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‐reactortoinputpassglesin

processtoinput

processfromoutput‐processtoinputreactant

conversion

overall

⎠⎝

=reactortoinput

reac anconversion


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%75%100A/minmol100

A/minmol25‐A/minmol100

conversion

passglesin

%100%100A/minmol75

0‐A/minmol75

conversion

overall

=×⎟ ⎠

⎞⎜⎝

⎛ =

=×⎟ ⎠

⎞⎜⎝

⎛ =

Catalytic Propane Dehydrogenation

C3H8 → C3H6 +H2 95% overallconversion


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Catalytic Propane

Dehydrogenation

95% overall

conversion

C3H8 → C3H6 +H2

Overall

Process

=df 6, 7, 8

– 2 independent atomic balances (C and H)

– 1 relation

(overall

conversion)

= 0

consider n6, n7, n8 known for further DOF analyses

Catalytic Propane Dehydrogenation95% overall

conversionndf = 2

C3H8 → C3H6 +H2

Mixing

point

=df 9, 10, 1, 2

– 2 balances

(C3H8 and

C3H6)

= 2


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Catalytic Propane

Dehydrogenation

95% overall

conversionndf = 2 ndf = 3

C3H8 → C3H6 +H2

reactor

=df 3, 4, 5, 1, 2 – 2 balances (C and H)

= 3


conversionndf = 2 ndf = 3 ndf = 0

C3H8 → C3H6 +H2

separator

=df 3, 4, 5, 9, 10

– 3 balances

(C3H8,

C3H6,

and

H2)

– 2 relations (reactant and product recovery fractions)= 0


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Catalytic Propane

Dehydrogenation

95% overall

conversion

C3H8 → C3H6 +H2

overall

conversion

relationship

836 momo.n =−=


conversion

836 HCmol5n =

C3H8 → C3H6 +H2

overall

C atomic balance

( ) ( )637

HCmol1

Cmol3

7HCmol1

Cmol3

83HCmol1

Cmol3

HCmol95nnHCmol5mol100 638383

=+=


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Catalytic Propane

Dehydrogenation

95% overall

conversion

836 HCmol5n =

C3H8 → C3H6 +H2

overall

H atomic balance

637 HCmol95n =

( ) ( )

( )( ) ( )263

8383

Hmol1Hmol2

8HCmol1Hmol6

63

HCmol1Hmol8

83HCmol1Hmol8

nHCmol95

HCmol5mol100

++

=

28 Hmol95n =


conversion

836 HCmol5n =

C3H8 → C3H6 +H2

separator

given relations

637 HCmol95n =

28 Hmol95n =

( )( ) 6310710

83336

HCmol75.4nn0500.0nHCmol900nn00555.0n

=→==→=


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Catalytic Propane

Dehydrogenation

95% overall

conversion

833 HCmol900n = 836 HCmol5n =

C3H8 → C3H6 +H2

separator

propane balance

n10 = 4.75 mol C3H6

637 HCmol95n =

28 Hmol95n =

n3 = n6 +n9 → n9 = 895 mol C3H8


conversion


C3H8 → C3H6 +H2

mixer

propane balance

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

637 HCmol95n =

28 Hmol95n =

100+n9 = n1 → n1 = 995 mol C3H8


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Catalytic Propane

Dehydrogenation

95% overall

conversion

n1 = 995 mol C3H8


C3H8 → C3H6 +H2

mixer

propylene balance

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

637 HCmol95n =

28 Hmol95n =

n10 = n2 → n2 = 4.75 mol C3H6


conversion

n = 995 mol C H 833 HCmol900n = 836 HCmol5n =

C3H8 → C3H6 +H2

reactor

C atomic balance

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n2 = 4.75 mol C 3H6637 HCmol95n =

28 Hmol95n =

( )( ) ( )( )634

HCmol1Cmol3

4HCmol1Cmol3

83

HCmol163HCmol183

HCmol75.99n

nHCmol009

mo.mo

6383

6383

=

+=

+


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Catalytic Propane

Dehydrogenation

95% overall

conversion


C3H8 → C3H6 +H2

reactor

H atomic balance

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n2 = 4.75 mol C 3H6 634 HCmol75.99n = 637 HCmol95n =

28 Hmol95n =

( ) ( )

( )( ) ( )( ) ( )25

Hmol1Hmol2

5HCmol1Hmol6

63HCmol1Hmol8

83

HCmol1Hmol6

63HCmol1Hmol8

83

Hmol95n

nHCmol75.99HCmol009

HCmol.754HCmol959

26383

6383

=

++=

+


conversion


C3H8 → C3H6 +H2

single‐pass

conversion

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n2 = 4.75 mol C 3H6 634 HCmol75.99n =

25 Hmol95n =

637 HCmol95n =

28 Hmol95n =

( ) ( )( )

%55.9%100HCmol959

HCmol009HCmol959f 83

8383passsingle =×−=−


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Catalytic Propane

Dehydrogenation

95% overall

conversion

836 HCmol5n =833 HCmol900n =n = 995 mol C H

f single−pass = 9.55%

C3H8 → C3H6 +H2

Recycle ratio

637 HCmol95n =

28 Hmol95n =

n10 = 4.75 mol C3H6

n9 = 895 mol C3H8

n2 = 4.75 mol C 3H6 634 HCmol75.99n =

25 Hmol95n =

( )

( ) ( )

( ) feedfreshmolrecyclemol109

0.9mol001

mol.754mol958

feedmol001

nn

R =

+

=

+

=

Purging Necessary with recycle to prevent accumulation of a species that is both

present in the fresh feed and is recycled rather than separated with the product.

mixed fresh feed

and recycle is a

convenient

basis selection

f single−pass = 60%

CO2 + 3H2 → CH3OH+H2O


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Methanol Synthesis

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

‐ 5 independent species balances = 3



Methanol Synthesis

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

– inde endent s ecies balances

– 1 single pass conversion = 0




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ndf = 3 unknowns (n5, x5C, X 5H)

– 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 inde endent s ecies balance

investigate

mole balances and

their solution in

= 0



Combustion Reactions

Combustion ‐ ra id reaction of a fuel with ox en.

Valuable 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:

fuel 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 burned

C forms CO2 (complete) or CO (partial combustion)

H forms H2O

S forms SO2 N forms NO2 (above 1800°C)

Air is used as the source of oxygen. DRY air analysis: 78.03 mol% N2 20.99 mol% O2

0.94 mol% Ar 0.03 mol% CO2 0.01 mol% H2, He, Ne, Kr, Xe

usua y sa e to assume:

79 mol% N221 mol% O2


Stack lue) as – roduct as that leaves a furnace.

Composition analysis: wet basis – water is included in mole fractions

dry 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% H2O

60/(60+15+10) = 0.706 mol N2/mol 15/(60+15+10) = 0.176 mol CO2/mol

10/(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 basis

7.53 mole H2O

65 mole N2 14 mole CO2

xH2O = 7.53

107.5= 0.0700

xN2 = 65

107.5= 0.605

xCO = 14 = 0.130

10 mole O2

11 mole CO total = 107.5 mole

.

xO2 = 10

107.5= 0.0930

xCO = 11

107.5= 0.102

Theoretical and

Excess

Air

The less expensive reactant is commonly fed in excess of stoichiometric,

conversion 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 completely combust the fuel to CO2 and H2O.

theoretical oxygen.


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Theoretical and

Excess

Air

Excess air is the amount b 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

( )hr

Omol650

HCmol

Omol5.6

hr

HCmol100n 2

104

2104

ltheoreticaO2=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟ ⎠ ⎞

⎜⎝ ⎛ =

airmolairmol.764Omol506 2 ⎞⎛ ⎞⎛

hrOmolhr

n

2

ltheoreticaair =

⎠⎝ ⎠⎝

=

% excess air = 5000−3094

3094×100% = 61.6%


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Combustion Reactors

Procedure for writin /solvin material balances for a combustion reactor

1. When you draw and label the flowchart, be sure the outlet stream (the stack gas) includes

a. unreacted fuel (unless the fuel is completely consumed)

b. unreacted oxygen

c. water and carbon dioxide (and CO if combustion is ncomp ete

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

O H COO H C

5

22227

62 32 +→+f = 0.9

22262

25% of the ethane burned forms CO

degree‐of ‐freedom

analysis

ndf = 7 unknowns

‐ 3 atomic balances

‐

‐ 1

excess

air

specification

‐ 1 ethane conversion specification

‐ 1 CO/CO2 ratio specification

= 0


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f C2H6 = 0.9excess air specification

O H COO H C

O H COO H C

2225

62

22227

62

32

32

+→+

+→+


( ) 262

262

ltheoreticaO Omol350

HCmol

Omol5.3HCmol100n

2=⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ =

airmol2500n

Omol3505.1n21.0

0

20

=

=

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

Ethane conversion specification

f C2H6 = 0.9CO/CO2 ratio specification

n0 = 2500 mol air

n1 =10.0 mol C2H6

C2H6 + 2 O2 → 2CO2 +3H2O

C2H6 + 5

2O2 → 2CO+3H2O


( )( )( ) OCmol0.45reactHCmol1

genCOmol2HCmol1009.025.0n

62

624 =⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

Nitrogen balance( )( ) 23 Nmol1975iramol250079.0n ==

( )( ) ( ) ( ) ( )

25

COmol1Cmol1

5COmol1Cmol1

4HCmol1Cmol2

1HCmol1Cmol2

62

COmol135n

nnnHCmol10026262

=

++=

Atomic C balance


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53

C2H6 + 7

2O2 → 2CO2 +3H2O

f C2H6 = 0.9

n1 =10.0 mol C2H6

n3 = 1975 mol N2

C2H6 + 5

2O2 → 2CO+3H2O

atomic H balance( ) ) ( ) ) )

OHmol270n

nHCmol10HCmol100

26

OHmol1Hmol2

6HCmol1Hmol6

62HCmol1Hmol6

62 26262

=

+=

n0 = 2500 mol air

n4 = 45.0 mol CO

n5 =135 mol CO2

( ) ( )( )

( )( ) ( )( )

22

OHmol1Omol1

2COmol1Omol2

2

COmol1Omol1

Omol1Omol2

2Omol1Omol2

2

Omol232n

OHmol702COmol135

COmol54nOmol255

22

22

=

++

+=

atomic O balance

C2H6 + 72 O2 → 2CO2 + 3H2O

C2H6 + 5

2O2 → 2CO+3H2O

f = 0.9

n1 = 10.0 mol C2H6

n3 = 1975 mol N2

25% of the ethane burned forms COatomic O balance

( ) ( )( )Omol1Omol2

COmol1Omol1

Omol1Omol2

2Omol1Omol2

2

OHmol702COmol135

COmol54nOmol25522

++

+=

n0 = 2500 mol air

n4 = 45.0 mol CO

n5 = 135 mol CO2

n6 = 270 mol H2O

22 Omol232n

22

=


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C2H6 + 7

2

O2 → 2CO2 + 3H2O

C2H6 + 5

2O2 → 2CO+3H2O

f C2H6 = 0.9

n1 = 10.0 mol C2H6

n3 = 1975 mol N2

n2 = 232 mol O2


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

y 4 = 45 2396 = 0.019 mol CO mol

y 5 =135 2396 = 0.0563 mol CO2 mol

270 mol H2O

2396 mol

dry

stack

gas

= 0.113 mol H2O

mol dry stack gas

reactive system mb

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