2. The values of the flows are constant
Recycle, Bypass and Purge Streams
Definition of each term
Why do we use recycle streams ? Examples: recycled cans, newspapers, exit catalyst, and the reflux stream in a distillation column
Systems are operating in the steady state
The values of the flows are constant
The steps in the analysis and solution
are exactly the same as before.
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

3. For reaction systems:
We use either atomic balances or the stoichiometric equations.
We carry component balances for components not involved in the reaction
For component balances around the reactor, we use the steady state general MB equation :
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

4. Input through 
boundary
Output through +
boundary
Generation within system
– Consumption within the system
= 0
With the help of the single-pass (or the once-through) fractional conversion of a single component.
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

5. Note : normally this component is the limiting reactant
The single-pass (or the once-through) fractional conversion of a single component A =
Mass (or moles) of A fed into the reactor - mass (or moles) of A exiting the reactor
mass (or moles) of A fed into the reactor
Note : normally this component is the limiting reactant
Sometimes we are also given another conversion, called the overall fractional conversion of a reactant B , which is =
Mass (or moles) of B in fresh feed  mass (or moles)of B in output of the overall process Mass (or moles) of B in fresh feed
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

6. Reactor FF PF GP P
The single-pass (or the once-through) fractional conversion of a component
the overall fractional conversion of a component
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

7. PF Process Feed GP Gross Product Fresh Feed Net Product
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

8. We have three types of junction points: 1- A Splitter where the stream leaving the reactor splits into 2 or more streams. The composition in all these streams is same. 2- A Separator where the stream leaving the reactor enters into a separation process (such as distillation, extraction, etc..). The 2 or more streams leaving the separator have different compositions 3- A mixing point where 2 streams or more join together.
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

9. Splitter A 20% B 50% C 30% A 20% B 50% C 30% A 20% B 50% C 30%
A Splitter where the stream leaving the reactor splits into 2 or more streams. The composition in all these streams is same.
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

10. A 20% B 50% C 30% A 10% B 75% C 15% Separator A 70% B 28% C 2 %
A Separator where the stream leaving the reactor enters into a separation process (such as distillation, extraction, etc..). The 2 or more streams leaving the separator have different compositions
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

11. Mixing pt
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

12. Example
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

13. FF = 100 Kg xG xF xwater FIND the once-through fr conversion
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

14. Let the once-through fractional conversion of G = f
Basis: FF = 100 kg
Let the once-through fractional conversion of G = f
1. System is : the OVERALL system
Carry Total mass balance:
Carry water component balance: (why?)
2. System is the MIXING point
Carry balances around the MIXING point
3. System is the reactor + splitter (why?)
Carry balances around the reactor + splitter
Solve for the unknowns
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

15. 1- SYSTEM : Overall system a- Total mass balance: IN = OUT 100 kg = P
b- Water Balance (water is not reacting)
= P . xwater = xwater
THUS, xwater = 0.6
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

16. P/R = 8.33 THUS R = 100/8.33 = 12 Kg Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

17. G 40% W 60% F xF G xG F 4% G xG,PF W xw,PF FF R PF MIXING Pt Balance G
2- System is the MIXING point
Carry balances around the MIXING point
G 40%
W 60%
F xF
G xG
F 4%
G xG,PF
W xw,PF FF R PF
F= 100 kg
R= 12 kg
PF= 112 kg
MIXING Pt
Balance G F W
* xG
12 * xF
* 0.6
MIXING Pt
= 112 * xG,PF
= 112 * 0.04
= 112 * xW,PF
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

18. Balance G F W
* xG
12 * xF
* 0.6
= 112 * xG,PF
= 112 * 0.04
= 112 * xW,PF
XF = 0.373
XG = =
XG,PF = (40+12 * )/112 = 0.360
XW,PF = 1- ( ) =0.6
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

19. Splitter P PF Reactor R G xG F xF W xwater F 4% G xG,PF W xw,PF G xG F
3. System is the reactor + splitter (why?)
Carry balances around the reactor + splitter P PF xG xF
xwater G F W
Reactor
Splitter
F 4%
G xG,PF
W xw,PF R xG xF
xwater G F W
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

20. Splitter 100 112 kg Reactor 12 0.0267 0.373 0.600 G F W F 4% G 0.36
G Balance
Input –
Output +
Generation –
Consumption – =0
112* (100* *0.0267) (112*0.36 ) * f = 0
Note: G consumed = G entered into reactor * once-through fractional conversion
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

21. 112* (100* *0.0267)
(112*0.36 )
f =
= 0.93
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

22. New Example (take home)
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

23. Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

24. Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

25. System is : the OVERALL system Carry Total mass balance. (why?)
Basis: FF = 100 lb
Unknowns: R, P, PF, S,PF W,PF S,R IS,R W,R
Equations:  i,PF =1  i,R =1, 3 component balances for the mixing point, 3 component balances for the reactor , 3 component balances for the separator, component balances for the overall system.
System is : the OVERALL system
Carry Total mass balance. (why?)
2. System is the MIXING point
Carry balances around the MIXING point
3. System is the reactor + separator (why?)
Carry balances for S, water, G and F around
the reactor + separator
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

26. Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

27. New Example
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

28. R
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

29. Basis: FF = 28 moles (10moles N2 + 18 moles H2)
Limiting Reactant is H2 (Why?)
System is : the OVERALL system
Carry atomic balances
N bal x 10 = 1 x nNH3,P + 2 x nN2,P
H bal x 18 = 3 x nNH3,P
Therefore, nNH3P = 12 moles and nN2,P = 4 moles
Thus R = 4 x P = 4 x 16 = 64 moles
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

30. 2. System is the reactor + separator (why?)
Carry balances around the reactor + separator
H2 balance:
In  Out + Generation  Consumption = 0
(18+nH2R)  nH2R + 0  (18+nH2R) x 0.4 = 0
Solve to get nH2R = 27 moles ,
Then , nN2R = = 37 moles
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

31. To check our answers, carry N2 balance:
In  Out + Generation  Consumption = 0
(10+nN2R)  (nN2R+ nN2P)  [(18+nH2R) x 0.4]x1/3 = 0
(10+37) - (37+4) [(18+27) x 0.4]x1/3 = 0
Note : Consumption term =
[(18+nH2R) x 0.4] moles H2 x 1 mole N2
3 moles H2
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

32. 3. System is the SEPARATOR
Carry component balances, then you will get the composition of GP.
N2 37
H2 27
IN MOLES R GP P
Separator
N2 x1
H2 x2
NH3 x3 N
NH3 12
x1 =
x2 = 27 +0
x3 = 12
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

33. 4. System is the MIXING point
Carry balances around the MIXING point , then you will get the composition of PF
IN MOLES R N H
N2 x4
H2 x5
= x4
=x5 FF N H PF
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

34. New Example
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

35. A Purge : a stream bled off to remove an accumulation of inerts or unwanted material that might otherwise build up in the recycle stream R
CO + 2H2  CH3OH
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

36. the fractional conversion of CO and the stoichiometric equation
Given: the once-through fractional conversion for CO = 18% Basis: FF = 100 mol
CARRY the following balances:
- For the overall system
Carry atomic balances
- Carry component balances around the mixing pt
- Carry balances around the reactor using
the fractional conversion of CO and the
stoichiometric equation
 for CO  for H  for CH3OH
- carry component balances around the separator
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

37. 1- For the overall system Carry atomic balances
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

38. CH4 Bal 0.2 = P (0.032) THUS P= 6.25 moles
Not reacting
H2 x
CO y
CH4 3.2%
CH4 Bal = P (0.032) THUS P= 6.25 moles
H * * 4 = P(2x+0.032*4) + E *4
C * * 1 = P(y *1) + E *1
O * = P(y) E *1
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

39. GET E GET x GET y SOLVE the equations Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

40. 2. System is the reactor + separator
Carry balances around the reactor + separator
CO balance:
In  Out + Generation  Consumption =  
(32.5 +R y )  (R+P) * y  (32.5+ R y) x 0.18 = 0
Thus get R
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

41. 3. System is the reactor + separator (why?)
Carry balances around the reactor + separator
H2 balance:
In  Out + Generation  Consumption =  
(67.3+Rx)  (R+P) * x +0  [(32.5+Ry) x 0.18] * 2 =0 1
THUS GET R
[(32.5+Ry) x 0.18] moles CO * 2 moles H2 = 0
1 mole CO
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004

42. 4. System is the MIXING point
Carry balances around the MIXING point , then you will get the composition of PF
IN MOLES R
H2 x
CO y
CH %
H2 x2
CO x3
CH4 x4
x2 = x R
x3 = y R
x4 = R FF H CO CH PF
Dr. Faisal Iskanderani
ChE 201 Spring 2003/2004