Reactive Mass Balance

7/28/2019 Reactive Mass Balance

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CHEMICAL REACTION

STOICHIOMETRY

CHAPTER 3- Material balance for

SIMPLE reactive system

BALANCES REACTIVE

PROCESSES

COMBUSTION REACTIONS

SEPARATION & RECYCLE



1) Stoichiometry



CHEMICAL REACTIONSTOICHIOMETRY

! Stoichiometry – theory of the proportions in which chemical speciescombine with one another.

*Note: Refer Felder pp.116.

! The Stoichiometric Equation of chemical reaction – statement of therelative number of molecules or moles of reactants and products thatparticipate in the reaction – must be balanced (no. of atoms of eachatomic species must be the same on both sides of the equation).Example:

2 SO2 + O2 → 2 SO3






! The Stoichiometric ratio two molecular species participating in areaction – ratio of their stoichiometric coefficients in the balancedequation.

1) Stoichiometry






! The limiting reactant – reactant that would run out if a reactionproceeded to completion, other reactants are termed excess reactant.


! A reactant is limiting if it is present in less than its stoichiometric proportionrelative to every other reactants.

! Fractional excess of reactant is the ratio of the excess to thestoichiometric requirement.

*Note: Refer Felder pp.120 for Example 4.6 – 1.

2) Limiting and Excess Reactants, Fractional Conversion and Extent

of Reaction






! Fractional conversion of a reactant is the ratio:

2) Limiting and Excess Reactants, Fractional Conversion and Extent of Reaction

( ) ( )

( )

of Anal excess x Fractio Aexcess of Percentage

n

nn Aexcess of Fractional

stoich A

stoich A feed A

100=

−

=

sion of Anal conver x Fraction of Aconversio Percentage

moles feed

ted moles reac n of Aconversio Fractional

100=

=






! In chemical processes, some reactants can usually combine in morethan one way, and the product once formed may react to yieldsomething less desirable.

*Note: Refer Felder pp.123 and Example 4.6 – 3 pp 124.

! For example, ethylene can be produced by the dehydrogenation ofethane:

C2H6 → C2H4 + H2

3) Multiple Reactions Yield and Selectivity






! Once hydrogen is produced, it can react with ethane to producemethane:

C2H6 + H2 → 2 CH4

• Moreover, ethylene can react with ethane to form propylene andmethane:

C2H4 + C2H6 → C3H6 + CH4

3) Multiple Reactions Yield and Selectivity






! Yield :

mpletelyreacted co

t had taniting reacd theactions anno side re

re weremed if thee been for would havmoles that

d duct formeesired promoles of d Yield

lim

=

! Selectivity :

med roduct for ndesired pmoles of u

d duct formeesired promoles of d ySelectivit =



MOLECULAR SPECIES BALANCE



ATOMIC SPECIES BALANCE

EXTENT OF REACTION



! Example

Methane is burned with air in a continuous steady state combustion reactorto yield a mixture of carbon monoxide, carbon dioxide and water. Thereaction taking place are:

CH4 + 3/2 O2 → CO + 2 H2O

CH4 + 2 O2 → CO2 + 2 H2O

The feed to the reactor contains 7.80 mole% CH4 , 19.4 % O2 , and 72.8% N2.The percentage conversion of methane is 90.0%, and the gas leaving thereactor contains 8 mol CO2/mol CO.

Calculate the molar composition of the product stream.

*Note: Refer felder pp.131





HO W ? ? ?

Analyze the information….

! Feed: 0.078 mol CH4/mol, 0.194 mol O2/mol, 0.728 mol N2/mol

! Percentage conversion of CH4: 90.0% @ 0.90

! Product gas: 8 mol CO2/mol CO

CHAPTER 3- Material balance for SIMPLE reactive system



! Basis of calculation: 100 mol of feed

! Process flow chart:

Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O





! Write system equation andoutline a solution procedure

Additional information – fractionalconversion CH 4 = 0.90

( )

4

4

4

4

4

4

CH moles

CH unreactedof Moles

mol7.020.901000.078reactedCH Moles

1000.078reactedCH moles

fedCH molesreactedCH moles

7800

027807

900

1

1

.n

..n

. f

=

−=

=××=

=

×

==

Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

Additional information – N 2 in = N 2 out = 0.728 x 100 = 72.8 mol N 2

n3 = 72.8 mol N 2





CH 4 Balance – From first reaction:- Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

CH4 + 3/2 O2 → CO + 2 H2O

! 1 mol CH4 → 1 mol CO

! x mol CH4 → x mol CO = n4 → x = n4 → Eq.(1)

CH 4 Balance – From second reaction:-

CH4 + 2 O2 → CO2 + 2 H2O

! 1 mol CH4 → 1 mol CO2 ! (7.02 – x) mol CH4 → (7.02 – x) mol CO2 = 8n4

→ 8n4 = 7.02 – x→ Eq.(2)

! Insert Eq. 1 into 2 COmol

9

780

9

027

0270278

4

444

.

.

n

.nn.n

==

=→−=





O2 Balance – From first reaction:- Processunit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

CH4 + 3/2 O2 → CO + 2 H2O

! 3/2 mol O2 → 1 mol CO

! 0.78 mol CO → 0.78 x 3/2 mol O2 = 1.17 mol O2 consumed

O2 Balance – From second reaction:-

CH4 + 2 O2 → CO2 + 2 H2O

! 2 mol O2→ 1 mol CO2 ! 6.24 mol CO2→ 6.24 x 2 mol O2 = 12.48 mol O2 consumed

mol

12.48-1.17-0.194x100consumed-

22

222

755 O.n

O Fed On

=

==





H 2O Balance – From first reaction:- Processunit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

CH4 + 3/2 O2 → CO + 2 H2O

! 2 mol H2 O→ 1 mol CO

! 0.78 mol CO → 0.78 x 2 mol H2 O = 1.56 mol H2 O produced

H 2O Balance – From second reaction:-

CH4 + 2 O2 → CO2 + 2 H2O

! 2 mol H2 O → 1 mol CO2 ! 6.24 mol CO2→ 6.24 x 2 mol H2 O = 12.48 mol H2 O produced

O H .n

O H n

mol

12.481.56 produced

22

25

0414=

+==∑





Processunit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

Component Mole Composition (%)

O2 5.75 5.72

CO 0.78 0.78

CO2 6.24 6.22

CH4 0.78 0.78

H2O 14.04 13.99

N2 72.8 72.51

Total 100.39 100%

A NS W E R





! Example


CH4 + 3/2 O2 → CO + 2 H2O

CH4 + 2 O2 → CO2 + 2 H2O








HO W ? ? ?











Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O



C f



! Write system equation andoutline a solution procedure


( )

4

4

4

4

4

4

CH moles





7800

027807

900

1

1

.n

..n

. f

=

−=

=××=

=

×

==

Processunit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O


n3 = 72.8 mol N 2



CHAPTER 3 M i l b l f



! Atom C: 1 unk. (n4)

! Atom H: 1 unk. (n5)

! Atom O: 3 unk. (n2, n4, n5)

Analyze the atomic balance for:

mol0.78

9

mol7.02 0.780-7.89 80.7807.8

CO mol1

Cmol1COmol8

COmol1

Cmol1COmol

CH mol1

Cmol1CH mol0.780

CH mol1

Cmol1CH mol.8

2

2

4

4

4

4

==→=→++=

×+×+×=×

4444

447

nnnn

nn

Solve the atomic balance on C…



Processunit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

0.780 mol CH4

n2 mol O2

72.80 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

mol14.042mol28.08 3.12-31.22 23.1231.2

OH mol1

H mol2OH mol

CH mol1

H mol4CH mol0.780

CH mol1

H mol4CH mol.8

2

2

4

4

4

4

==→=→+=

×+×=×

555

57

nnn

n

Solve the atomic balance on H…

CHAPTER 3 M t i l b l f



mol5.752

mol11.5 11.52 14.042.480.78238.8

H mol1Omol1H mol14.04

CO mol1Omol2COmol6.24

COmol1

Omol1COmol0.78

Omol1

Omol2Omol

Omol1

Omol2Omol9.4

2

2

2

2

2

2

2

2

==→=→+++=

×+×+

×+×=×

222

2

1

1

nnn

OO

n

Solve the atomic balance on O…



Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

0.780 mol CH4

n2 mol O2

72.80 mol N2

0.78 mol CO

6.24 mol CO2

14.04 mol H2O





O2 5.75 5.72

CO 0.78 0.78

CO2 6.24 6.22

CH4 0.78 0.78

H2O 14.04 13.99

N2 72.8 72.51

Total 100.39 100%

A NS W E R



Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

0.780 mol CH4

5.75 mol O2

72.80 mol N2

0.78 mol CO

6.24 mol CO2

14.04 mol H2O




! Example


CH4 + 3/2 O2 → CO + 2 H2O

CH4 + 2 O2 → CO2 + 2 H2O









HO W ? ? ?












Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1

mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O






!Write system equation andoutline a solution procedure


( )

4

4

4

4

4

4

CH moles





7800

027807

900

1

1

.n

..n

. f

=

−=

=××=

=

×

==

Processunit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

n1 mol CH4

n2 mol O2

n3 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O


n3 = 72.8 mol N 2






From first reaction:-

Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

0.78 mol CH4

n2

mol O2

72.8 mol N2

n4 mol CO

8n4 mol CO2

n5 mol H2O

CH4

+ 3/2 O2

→ CO + 2 H2

O

From second reaction:-

CH4 + 2 O2 → CO2 + 2 H2O



∑+= jiji0i ξvnn:equationFrom

1ξ

2ξ

( ) ( )[ ]

( )

( ) ( )

( ) ( ) 3Eq. 0

2Eq. 2--19.419.4

1Eq. mol7.02 0.780--7.80

mol0.780 n,calculatioFrom

--7.807.80

→=→=×++==

→=⎥⎦

⎤⎢⎣

⎡×−+⎟

⎠

⎞⎜⎝

⎛×−+==

→=+→=

=

=×−+×−+==

14114

21212

2121

1

21211

1

2

32

2

3

11

2

4

ξ ξ ξ

ξ ξ ξ ξ

ξ ξ ξ ξ

ξ ξ ξ ξ

nnn

nn

n

nn

CO

O

CH

CHAPTER 3 Material balance for





( ) ( )

( )

( ) ( )

( ) ( )

( ) ( )( ) ( )[ ] ( )

( ) ( )

( ) ( ) mol14.046.2420.7822ξ2ξnn

mol5.756.242-0.782

3 -19.42ξ-ξ

2

3 -19.4nn

mol6.240.7888nξn

mol0.78ξnnmol0.78n mol7.029n mol7.028nn mol7.02ξξ

(1)Equationinto(4)and(3)EqSubstitute

5Eq. 2ξ2ξn 2ξ2ξξ2ξ20nn

4Eq. ξ8n ξξ108nn

3Eq. ξn ξξ10nn

2Eq. 2ξ-ξ2

3 -19.4ξ2ξ

2

3 19.4nn

1Eq. mol7.02ξξ 0.780ξ-ξ-7.80mol0.780n n,calculatioFrom

ξ-ξ-7.80ξ1ξ17.80nn

215OH

212O

42CO

14CO

444421

21521215OH

24224CO

14114CO

21212O

2121

1

21211CH

2

2

2

2

2

2

4

=+=+==

====

=×===

===

=→==+→=+

→+=→+=×++×++==

→=→=×++==

→=→=×++==

→=⎥⎦

⎤⎢⎣

⎡×−+⎟

⎠

⎞⎜⎝

⎛×−+==

→=+→=

=

=×−+×−+==





O2 5.75 5.72

CO 0.78 0.78

CO2 6.24 6.22

CH4 0.78 0.78

H2O 14.04 13.99

N2 72.8 72.51

Total 100.39 100%

A NS W E R



Process

unit

100 mol

0.078 mol CH4/mol

0.194 mol O2/mol

0.728 mol N2/mol

0.780 mol CH4

5.75 mol O2

72.80 mol N2

0.78 mol CO

6.24 mol CO2

14.04 mol H2O




1) Overall Conversion and Single Pass Conversion



! Overall Conversion :

SEPARATION & RECYCLE

processthetoinput reactant

processthe fromoutput reactant processthetoinput reactant ConversionOverall

−

=

! Single – Pass Conversion :

reactor thetoinput reactant

processthe fromoutput reactant reactor thetoinput reactant Conversion Pass-Single

−

=






! Example (Felder pp.135)

Dehydrogenation of propane

Propane is dehydrogenated to form propylene in a catalytic reactor:

C3 H8 → C3 H6 + H2 The process is to be designed for a 95% overall conversion of propane. The

reaction products are separated into two streams:

1st : contains H2 , C3H6 and 0.555% of the propane that leaves the reactor,

is taken off as product

2nd : contains balance of the unreacted propane and 5% of the propylene in the first stream.

Calculate the composition of the product, the ratio (moles recycled)/(molefresh feed) and the single pass conversion.






HO W ? ? ?

Basis of calculation.. reactor theto fed H C mol 100 3 8

! 95% overall conversion of propane

Analyze the information..

! Sketch and insert the known and unknown values….






Reactor Separator 100 mol C3H8

n1 mol C3H8

n2 mol C3H6

n3 mol C3H8

n4 mol C3H6

n5 mol H2

5.5x10-3 n3 mol C3H8

0.95 n4 mol C3H6

n5 mol H2

0.99445 n3 mol C3H8

0.05 n4 mol C3H6






! Overall conversion of propane = 0.95

83

3

3

90.900

100

1055.5100950

H C mol n

n .

3=

×−

=

−

! Mol balance at fresh feed – recycle on propane:

( )

mol n

n

nn

9.995

90.90099445.0100

99445.0100

1

1

13

=

=+

=+






! Mol balance at fresh feed – recycle on propylene:

2405.0 nn =

! Atomic Balance at Reactor:

C: n4

H : n5 , n4

! Atomic Balance on C:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )64

44

4341

100

339.900305.039.995

33305.03

H C mol n

nn

nnnn

3=

+=+

+=+

62

24

510005.0

05.0

H C mol n

nn

3

=×=

=






!Atomic Balance on H:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

25

5

54321

95

2610089.9006589.995

26868

H mol n

n

nnnnn

=

++=+

++=+

! Product gas outlet:

( )

( )mol n H

mol n H C

H C mol n H C 3

95

9510095.095.0

59.9001055.51055.5

52

462

8

3

3

3

83

==

===

=×=×=

−−




1) Combustion Chemistry



! Combustion – the rapid reaction of a fuel with oxygen.


! When a fuel is burned, carbon in the fuel reacts to form either CO2 orCO, hydrogen forms H2O, and sulfur form SO2.

! Partial or incomplete combustion – a combustion reaction in whichCO is formed from a hydrocarbon.







! Example:

C + O2 → CO2 Complete combustion of carbon

C3 H8 + 5O2 → 3CO2 + 4H2 O Complete combustion of propane

C3 H8 + 3.5O2 → 3CO + 4H2 O Incomplete combustion of propane


! Composition on a wet basis – mole fraction of a gas that containswater.

! Composition on a dry basis (Orsat Analysis)– mole fraction of thesame gas that without the water.

! Stack gas or flue gas – product gas that leaves a combustion furnace.




2) Theoretical and Excess Air



! Theoretical Oxygen – the moles (batch) or molar flow rate(continuous) of O2 needed for complete combustion of all the fuel fedto the reactor, assuming that all carbon in the fuel is oxidized to CO2 and all hydrogen is oxidized to H2O.

! Theoretical Air – the quantity of air that contains theoretical oxygen.

! Percent Excess Air:



( ) ( )

( )100%

airof moles

airof molesairof moles AirExcess%

ltheoretica

ltheoretica fed×

−

=





A stack gas contains 60.0 mole% N2 , 15.0% CO2 , 10.0% O2 , and the balanceH2O. Calculate the molar composition of the gas on a dry basis.



HO W ? ? ?

Basis of calculation.. gaswetmol100

Comp.Mol (wet

basis)

O2 10

CO2 15

N2 60

H2O 15

Total 100 11.8%%O

10085

10%O

2

2

=

×=

Mol (Dry

basis)

% Mol

(Dry basis)

10 11.8

15 17.7

60 70.5

85 1.00





An Orsat analysis (a technique for stack analysis) yields the following dry basis composition: 65.0 mole% N2 , 14.0% CO2 , 11% CO and 10.0% O2.

A humidity measurement shows that the mole fraction of H2O in the stackgas is 0.0700. Calculate the stack gas component on a wet basis.



HO W ? ? ?

Basis of calculation.. basisdrymol100






Comp.Mol (Dry

basis)

O2 10

CO2 14

N2 65

CO 11

H2O 0

Total 100

Mol (Wetbasis)

% Mol(Dry basis)

10

14

65

11

x 0.0700

100 + x 1.00

mol7.527x 70.93x

70.070x-x x0.07x7

x100

x0.070 nn y

total

OH

OH 2

2

=→=

=→=+

+

=→=






Comp.Mol (Dry

basis)

O2 10

CO2 14

N2 65

CO 11

H2O 0

Total 100

0930.==

107.527

10

y 2O

Mol (Wetbasis)

Mol fraction(Wet basis)

10 0.093

14 0.130

65 0.605

11 0.102

7.527 0.070

107.527 1.00





100 mol/h of butane (C4H10) and 5000 mol/h of air are fed into a combustionreactor. Calculate the percentage of excess air.



HO W ? ? ?

Find the theoretical air needed

OH 5CO4O2

13 H C

222104+→+

22104

2104

Omol650Omol

2

13100 H Cmol100

Omol2

13 H Cmol1

=×→

→






( )

mol3095

0.21

mol650n

mol650nn0.21

mol650nl,theoreticaO

airltheoretica

Oairltheoretica

O2

2

2

==

==

=

1.6%airexcessof Percentage

100%3095

35 airexcessof Percentage

100%n

nn airexcessof Percentage

airltheoretica

airltheoreticaairexcess

6

095000

=

×

−

=

×

−

=





Ethane is burned with 50% excess air. The percentage conversion of theethane is 90%; of the ethane burned, 25% react to form CO and the balancereact to form CO2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas.



OH 3CO2O2

5 H C

OH 3CO2O2

7 H C

2262

22262

+→+

+→+






HO W ? ? ?


! Feed: 50% excess air

! Percentage conversion of C2H6: 90.0% @ 0.90 – 25% form CO, balanceform CO2




! Basis of calculation: 100 mol of C2H6 fed


Processunit

100 mol C2H6 n1 mol C2H6

n2 mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol






! Write system equation and

outline a solution procedure

Additional information – fractionalconversion C2H 6 = 0.90

( )

621

162

62

62

H Cmol10n

mol100.1100nH Cunreactedof Moles

mol900.90100reactedH C Moles

mol100 fedH C Moles

0.90 f

=

=×=

=×=

=

=

C 3 ate a ba a ce o


Process

unit


n2

mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Find the amount of air fed into the process unit:

! Calculate the theoretical O2 needed

! Calculate the theoretical air needed

! From % excess air, calculate the amount of excess air fed to the process unit





Process

unit


n2

mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol

( )

mol16670.21

Omol350

n

Omol350n0.21

2airltheoretica

2airltheoretica

==

=

Calculate the theoretical O2 needed

22262

262

OlTheoreticaOmol350Omol2

7 100 H Cmol100

Omol2

7 H Cmol1

:combustioncompleteFor

==×→

→

Calculate the theoretical air needed





Process

unit


n2

mol O2

n3 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

n0 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Calculate the amount of excess air fed to the process unit

2500moln

0.501667

1667n 0.50

n

nn

100%n

nnair%excess

0

0

airltheoretica

airltheoretica0

airltheoretica

airltheoretica0

=

=

−

→=

−

×

−

=

N 2 in = N 2 out = 0.79 x 2500 = 1975 mol N 2

n3 = 1975 mol N 2




SIMPLE reactive systemProcess

unit

100 mol C2H6 10 mol C2H6

n2 mol O2

1975 mol N2

n4 mol CO

n5 mol CO2

n6 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol0.21 mol O2/mol

Given 75% ethane reacted form CO2

COmol135n

formedCOmol135267.5 H Cmol67.5

COmol2 H Cmol1

:1reactionFrom

mol67.5mol900.75CO formtoreactH C Amount

25

262

262

262

=

=×→

→

=×=

Given 25% ethane reacted form CO

COmol45n

formedCOmol45222.5 Cmol22.5

COmol2 Cmol1

:2reactionFrommol22.5mol900.25CO formtoreactC

4

2

2

2

=

=×→

→

=×

=

6

6

6

H

H

H Amount





Process

unit


n2 mol O2

1975 mol N2

45 mol CO

135 mol CO2

n6 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol

0.21 mol O2/mol

! Atom C: 0 unk.

! Atom H: 1 unk. (n6)

! Atom O: 2 unk. (n2, n6)


mol2702

mol540 n 60-6002n 2n60600

OH mol1

H mol2OH moln

H Cmol1

H mol6H Cmol10

H Cmol1

H mol6H Cmol100

666

2

26

62

62

62

62

==→=→+=

×+×=×






Process

unit


n2 mol O2

1975 mol N2

45 mol CO

135 mol CO2

n6 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol

0.21 mol O2/mol

mol232.52

mol465 n 4652n 452n1050

OH mol1

Omol1OH mol

molCO1

Omol2COmol1

COmol1

Omol1COmol45

Omol1

Omol2Omoln

Omol1

Omol2Omol25000.21

222

2

2

2

2

2

2

2

2

==→=→+++=

×+×

+×+×=××

270270

27035

2





Component Mole Composition (y)

O2 232.5 0.097

CO 45 0.019

CO2 135 0.056

C2H6 10 0.004

N2 1975 0.824

Total 2397.5 100%

A NS W E R


Process

unit


232.5 mol O2

1975 mol N2

45 mol CO

135 mol CO2

270 mol H2O

50% excess air

2500 mol air

0.79 mol N2/mol

0.21 mol O2/mol

Moles ratio of water todry stack gas:

gasstackdrymolOH mol0.113

gasstackdrymol2397.5

OH mol270

2

2

=

=





A hydrocarbon gas is burned with air. The dry basis product gas compositionis 1.5 mol%CO, 6.0 mol% CO2 , 8.2 % O2 , and 84.3% N2. There is no atomicoxygen in the fuel. Calculate the ratio of hydrogen to carbon in the fuel gasand speculate on what the fuel might be. Then calculate the percent excess airfed to the reactor.


HO W ? ? ?


! Dry basis: 0.015 mol CO/mol, 0.06 mol CO2/mol,0.082 mol O2/mol, 0.843 mol N2/mol




! Basis of calculation: 100 mol of dry basis product


Process

unit

n1 mol C

n2 mol H1.5 mol CO

8.2 mol O2

84.3 mol N2

6.0 mol CO2

n4 mol H2O

n3 mol air 0.79 mol N2/mol

0.21 mol O2/mol


N 2 in = N 2 out = 84.3 mol N 2

Calculate the air fed to the unit

( )

mol106.70

0.79

mol84.3n

mol84.3n0.79

3

3

==

=





Processunit

n1 mol C

n2 mol H1.5 mol CO

8.2 mol O2

84.3 mol N2

6.0 mol CO2

n4 mol H2O

106.70 mol air

0.79 mol N2/mol

0.21 mol O2/mol

! Atom C: 1 unk. (n1)

! Atom H: 2 unk. (n2, n4)

! Atom O: 1 unk. (n4)


mol7.5n 61.5n

COmol1

Cmol1COmol6

COmol1

Cmol1COmol1.5Cmoln

11

2

21

=→+=

×+×=

Solve the atomic balance on C…





Processunit

n1 mol C

n2 mol H1.5 mol CO

8.2 mol O2

84.3 mol N2

6.0 mol CO2

n4 mol H2O

106.70 mol air

0.79 mol N2/mol

0.21 mol O2/mol

mol14.9n n1.51216.444.814

OH mol1

Omol1OH moln

COmol1

Omol1COmol1.5

COmol1

Omol2COmol6

Omol1

Omol2Omol8.2

Omol1

Omol2Omol106.700.21

44

2

24

2

2

2

2

2

2

=→+++=

×+×

+×+×=××



mol9.8n

H mol1

H mol2H mol1H moln

2

2

22

2

94

=

×=

OO.





Calculate the H/C in the fuel

)(CH methane CH/molmol4CH/molmol3.977.5

29.8

n

n4

C

H →≈==

Percentage of excess air

2actionReOH O2

1 2H

1Reaction CO OC

22

22

→+

→+

mol14.95n

H 2mol

Omol0.5H mol29.8

C1mol

Omol1Cmol7.5n

nnn

ltheoreticaO

22

ltheoreticaO

2ltheoreticaO1ltheoreticaOltheoreticaO

2

2

222

=

×+×=

+=





( )

airmol71.190.21

Omol14.95n

Omol14.95n0.21

2

airltheoretica

2airltheoretica

==

=

Calculate the theoretical air needed

Calculate the amount of excess air fed to the process unit

49.9%100%71.19

71.19106.7air%excess

100%n

nnair%excess

airltheoretica

airltheoretica0

=×

−

=

×

−

=




! Exercise

A researcher burned n-Pentane (C5H12) with excess air in a continuouscombustion chamber.

The analysis on the product gas and report shows that product gas contains0.304 mole% pentane, 5.9 mole% oxygen, 10.2 mole% carbon dioxide and the balance nitrogen on a dry basis. Based on 100 mol of dry product gas,

i) Draw and label a flowchart for this process

(2 marks)

ii) Calculate the mol of n-Pentane and air fed in the combustion chamber,the percentage of excess air and the fractional conversion of n-Pentanein this process.

(8 marks)


O H COO H C 222125

658 +→+




! Exercise

Ethylene (C2H4) has been commercially used for production of ethanol(C2H5OH) by hydration process:

However, some of the product is converted to diethyl ether ((C2H5)2O) in theside reaction:

The feed to the reactor contains ethylene (C2H4), steam (H2O) and inert gas(G). A sample of the reactor effluent gas is analyzed and found to contain 43.3mole% ethylene, 2.5 mole% ethanol, 0.14% ether, 9.3% inert gas andthe balance water. Based on 100 mol of effluent gas,

i) Draw and label a flowchart for this process (3 marks)

ii) Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol and theselectivity of ethanol production relative to diethyl ether production.

(9 marks)


OH H C O H H C 52242

→+

( ) ( ) O H O H C OH H C 2252522 +→

CHAPTER 4- ENERGY balance for non



! A common practice is to arbitrarily designate a reference state for a

substance at which U or H is declared to equal zero, and then tabulateU and/or H for the substance relative to the reference state. *Note: ReferFelder pp. 339 and 359

! In Chapter 7 (Felder), U and H are state properties of a species; theirvalues depend only on the state of the species – primarily on itstemperature and state of aggregation (solid, liquid or gas) and, to alesser extent, on its pressure (and for mixtures of some species, on itsmole fraction in the mixture).

! When a species passes from one state to another, both ΔU and ΔH forthe process are independent of the path taken from the first state to

reactive system

Reference State

Reactive Mass Balance

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