Chapter 3 Mass Balance

Balance on Reactive Processes System: Part B

Balance of Reactive Processes Balance on reactive process can be solved based

on three method:

1. Atomic Species Balance

2. Extent of Reaction

3. Molecular Species Balance

Degree-of-Freedom Analysis

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Independent EquationIndependent Species Independent Reaction

Independent Equation Algebraic equation are independent if we cannot obtain any

one of them by adding and subtracting multiples of any of the others

x + 2y = 4 [1]3x + 6y = 12 [2]

Only one independent equation because [2]= 3 x [1]

x + 2y = 4 [1]2x – z= 2 [2]4y + z= 6 [3]

Although 3 equation, but only two independent equation exist because [3]=2x[1] –[2]

If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

Similarly

If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

Independent Species

Independent Molecular Species

If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

n3 mol O2

3.76 n3 mol N2

n4 mol CCl4(v)

ProcessUnit

n1 mol O2

3.76 n1 mol N2

n2 mol CCl4(l) n5 mol CCl4(l)

Since N2 and O2 have a same ratio wherever they appear on the flowchart (3.76 mol N2/ mol O2), only ONE independent balance can be obtained.

Let’s make a molecular balance on both species to prove it

Balance on O2:

n1=n3 [1]

Balance on N2:

3.76 n1=3.76n3

n1=n3 [2]

Eq. [1] and [2] are SAME. Only ONE INDEPENDENT EQUATION OBTAINED although two species are involved.

Independent Molecular Species

Independent Atomic Species If two ATOMIC species are in the SAME RATIO to each other

wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is obtained)

n3 mol O2

3.76 n3 mol N2

n4 mol CCl4(v)

ProcessUnit

n1 mol O2

3.76 n1 mol N2

n2 mol CCl4(l) n5 mol CCl4(l)

Independent Atomic Species Atomic N and O are always in same proportion to each other in the

process (3.76:1), similar for atom C and Cl which always have the same ratio too (1:4).

Although FOUR atomic species exist, only TWO independent equation can be obtained for this case.

Prove:Balance on atomic O: 2n1=2n3

n1=n3 [1]

Balance on atomic N: 2(3.76)n1=2(3.76)n3

n1=n3 [2]

Balance on atomic C: n2=n4+n5 [3]

Balance on atomic Cl: 4n2=4n4 +4n5

n2=n4+n5 [4]

Eq. [1]=[2] and [3]=[4], only TWO independent equation obtained

Independent Reaction Used when we are using either molecular species balance

or extent of reaction method to analyze a balance on reactive process

Chemical reaction are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others

A ------> 2B [1]B ------> C [2]A ------> 2C [3]

Only TWO independent eq. can obtained although three equation exist since [3]=[1] + 2[2].

Balance of Reactive Processes Balance on reactive process can be solved based

on three method:

1.Atomic Species Balance

2.Extent of Reaction

3.Molecular Species Balance

[arranged according to the easiest method(1) to more difficult method(2), but not always true]

Atomic Species Balance

No. of unknowns variables

- No. of independent atomic species balance- No. of molecular balance on indep. nonreactive species- No. of other equation relating the variable

=============================No. of degree of freedom

=============================

Extent of Reaction

No. of unknowns variables+ No. of independent chemical reaction- No. of independent reactive species- No. of independent nonreactive species- No. of other equation relating the variable=============================

No. of degree of freedom=============================

Molecular Species Balance

No. of unknowns variables+ No. of independent chemical reaction- No. of independent molecular species balance- No. of other equation relating the variable=============================

No. of degree of freedom=============================

Application of Method C2H6 -------> C2H4 + H2

40 kmol H2/min

n1 kmol C2H6/min

n2 kmol C2H4/min

Reactor100 kmol C2H6/min

Method 1: Atomic Species Balance

All atomic balance is INPUT=OUTPUT Degree-of-freedom analysis

2 unknowns variables (n1, n2)

- 2 independent atomic species balance (C, H)- 0 molecular balance on indep. nonreactive species- 0 other equation relating the variable

=============================0 No. of degree of freedom

=============================

Method 1: Atomic Species Balance

Balance on atomic C (input= output)

200=2n1 + 2n2

100=n1 + n2 [1]

Balance on atomic H (input = output)100(6)=40(2) + 6n1+4n2

520 = 6n1 + 4n2 [2]

Solve simultaneous equation, n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

100 kmol C2H6 2 kmol C

=

n1 kmol C2H6 2 kmol C

+ n2(2)1 kmol C2H6 1 kmol C2H6

Method 2: Extent of Reaction Degree-of-freedom analysis

2 unknowns variables (n1,n2)+ 1 independent chemical reaction- 3 independent reactive species (C2H6, C2H4, H2)

- 0 independent nonreactive species- 0 other equation relating the variable

=============================0 No. of degree of freedom

=============================

Method 2: Extent of Reaction Write extent of reaction for each species

C2H6 : n1 = 100-ξ

C2H4 : n2= ξ

H2 : 40= ξ

Solve for n1 and n2 (ξ =40)

n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

Method 3: Molecular Species Balance

Degree-of-freedom analysis

2 unknowns variables (n1, n2)

+1 independent chemical reaction- 3 independent molecular species balance (C2H6, C2H4, H2)

- 0 other equation relating the variable=============================

0 No. of degree of freedom=============================

Method 3: Molecular Species Balance

H2 balance (Gen=Output):

H2 Gen= 40 kmol H2/min

C2H6 Balance (input=output + cons.):

100 kmol C2H6/min = n1kmol C2H6/min +

40 kmol H2 gen X (1 kmol C2H4 gen/1 kmol H2 gen)n1= 60 kmol C2H6/min

C2H4 balance (Gen.=Ouput):

40 kmol H2 gen x (1 kmol C2H4 gen./ 1 kmol H2 gen) = n2

n2= 40 kmol C2H4/min

CLASS DISCUSSION

EXAMPLE 4.7-1

Product Separation & Recycle

Overall ConversionReactant input to Process – reactant output from Process

Reactant input to Process

Single Pass ConversionReactant input to Reactor – reactant output from Reactor

Reactant input to Reactor

75 mol B/min100 mol A/min75 mol A/minReactor

Product Separation

Unit

25 mol A/min75 mol B/min

25 mol A/min

Purging To prevent any inert or insoluble substance build up and

accumulate in the system Purge stream and recycle stream before and after the

purge have a same composition.

ProductFresh FeedReactor

Product Separation

Unit

Recycle Purge

CLASS DISCUSSION

EXAMPLE 4.7-2

CLASS DISCUSSION

EXAMPLE 4.7-3

ANY QUESTION?