1. Chapter 3 Mass Balance Balance on Reactive Processes System: Part B

2. Balance of Reactive Processes
Balance on reactive process can be solved based on three method:
Atomic Species Balance
Extent of Reaction
Molecular Species Balance

3. Degree-of-Freedom Analysis?
Independent Equation
Independent Species
Independent Reaction

4. 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]
2x – z= 2 [2]
4y + z= 6 [3]
Although 3 equation, but only two independent equation exist because [3]=2x[1] –[2]

5. Independent 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)
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)

6. 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)
Process
Unit
n1 mol O2
3.76 n1 mol N2
n2 mol CCl4(l)
n5 mol CCl4(l)

7. Independent Molecular Species
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.

8. 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)
Process
Unit
n1 mol O2
3.76 n1 mol N2
n2 mol CCl4(l)
n5 mol CCl4(l)

9. 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

10. 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].

11. Balance of Reactive Processes
Balance on reactive process can be solved based on three method:
Atomic Species Balance
Extent of Reaction
Molecular Species Balance
[arranged according to the easiest method(1) to more difficult method(2), but not always true]

12. 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

13. =============================
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

14. 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

15. Application of Method C2H6 -------> C2H4 + H2 Reactor
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min

16. 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

17. 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 +
n2(2)
1 kmol C2H6

18. 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

19. 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

20. 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

21. 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

22. CLASS
DISCUSSION
EXAMPLE 4.7-1

23. Product Separation & Recycle
75 mol B/min
100 mol A/min
75 mol A/min
Reactor
Product Separation Unit
25 mol A/min
Overall Conversion
Reactant input to Process – reactant output from Process
Reactant input to Process
Single Pass Conversion
Reactant input to Reactor – reactant output from Reactor
Reactant input to Reactor

24. Product Separation Unit
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.
Product
Fresh Feed
Reactor
Product Separation Unit
Recycle
Purge

25. CLASS
DISCUSSION
EXAMPLE 4.7-2

26. CLASS
DISCUSSION
EXAMPLE 4.7-3

27. ANY QUESTION?