1. Conversion and reactor sizing
Objective: sizing a reactor once the relationship between reaction rate ra and concentration is known

2. Conversion: definition
Given the general reaction:
A (limiting reactant) is the base for the calculation
For irreversible reactions, the maximum value for X is for complete conversion
For reversible reactions, the maximum value for X is at equilibrium
Batch
Flow
               
             

3. Conversion (X) Quantification of how a reaction has progressed
Batch Reactors
Continuous (or Flow) Reactors

4. Mole Balance for Batch Reactor
Two different cases:
1. Constant Volume
2. Constant Pressure
If reactor is perfectly mixed

5. Design equations in terms of conversion: batch reactor
Mole balance for a batch reactor is
Definition of conversion
General equation
Batch time

6. Flux reactors
Conversion grows with raising residence time  with volume
Molar flow rate is used instead of mole number: FA = FAo (1-X)
Inlet Molar flow rate is the product of inlet concentration CAo times the volumetric inlet flow rate vo FAo = CAo vo
CAo is:
Liquids: molarity (mol/l)
Gas: calculated from EOS (Ideal gas law): CAo= PAo /RTo = yAo Po/RTo

7. Example of calculation of CA0
A gas mixture 50% A and 50% inert at 10 atm enters the reactor with constant flow rate of 6 dm3/s at 300°F (422.2 K)
Estimate the inlet concentration CA0 and volumetric flow rate FA0. Use dm3 atm/mol K as gas constant.

8. Solution
A gas mixture 50% A and 50% inert at 10 atm enters the reactor with constant flow rate of 6 dm3/s at 300°F (422.2 K)
Estimate the inlet concentration CA0 and volumetric flow rate FA0. Use dm3 atm/mol K as gas constant.
Solution
For an ideal gas

9. Design equation: CSTR (back mix)
Mass balance in terms of conversion
Introduction of ra
Volume of CSTR for obtaining a given conversion

10. Design equation: PFR Similar equation for Packed Bed Reactor
Mole balance
In terms of conversion
Differentail form of the design equation
Integral Equation
Similar equation for Packed Bed Reactor

11. Design equations in terms of conversion

12. Design equations: application
Given -rA as a function of conversion -rA=f(X),…
… it is possible to size any (ideal) reactor
Using the Levenspiel plots …
…. In which Fao / (-ra) or 1 / (-ra) is plotted as a function of X
In the Levenspiel plots Fao / (-ra) vs. X, the volume of a CSTR and that of a PFR are represented by the following areas:

13. Levenspiel Plots
PFR
FAO
-rA X
FAO
-rA X
CSTR
FAO
-rA
X=X

14. Example
The isoptherma reaction A  B + C takes place in a CSTR. Calculate the volume of the CSTR and that of the PFR for consuming 80% of A with the following conditions:
Inlet Volumetric flow rate is constant at 6 litri/sec
Pressure 10 atm
Initail concentration yA0 = 0.5
Temperature K
Ideal gas law is valid
Data reaction rate and conversion X .1 .2 .3 .4 .5 .6 .7 .8
.85
-rA
0.0053
0.0052
0.005
0.0045
0.0040
0.0033
0.0025
0.0018
0.001

15. Solution: CSTR FAO -rA X Report data of r vs. X in the Levenspiel plot
X=X
Report data of r vs. X in the Levenspiel plot
Calcualtion of FA0
Value of rA at X= 0.8 is , therefore 1/rA = 800
Substituting in the design equation:
Area of the rectangle is 800 * 0.8 = 640 : this should be multiplied by FA0 and gives the same result as above

16. Solution: PFR FAO -rA X Report data of r vs. X in the Levenspiel plot
Calculation of FA0
Applying Simpson method for integration:
Gives the value of V= 225 dm3
Direct measure of the area = 260 dm3 ; this should be multiplied by FA0 and gives the same result as above

17. Value of conversion along the reactorX V
250 1
Along the reactor:
Concentration of reactant drops in the reactor
Reaction rate drops too
Conversion raises
Calculations are made at different conversion (likewise for X=0.8) , giving the table:
V (dm3)
33.4
71.6
126
225 X
0.2
0.4
0.6
0.8
-rA (mol/dm3 s)
0.0053
0.005
0.004
0.0025

18. Comparison between CSTR e PFR
For same conversion, Is the CSTR volume always higher than PFR volume ?
It seems so, but is it ALWAYS true??
FAO
-rA X
X=X
FAO
-rA X

19. For same conversion, Is the CSTR volume always higher than PFR volume ?
FAO
-rA X
VPFR
VCSTR < VPFR
VCSTR

20. For same conversion, Is the CSTR volume always higher than PFR volume ?
FAO
-rA X
VPFR
VCSTR
VCSTR = VPFR

21. Levenspiel plot for bacteria growth

22. Reactors in series

23. PFR in series x3 FAO -rA x2 x1 Vsingle = V1+V2+V3 FAO X=0 FA1 X=X1 FA2
Let us compare two scenarios
single reactor achieves X3
3 reactors in series achieve X3
How are the total volume of 3 reactors in series related to single reactor ??
FAO
-rA x2 x1
Vsingle = V1+V2+V3

24. CSTR in series FAO -rA V1 + V2 + V3 < Vsingle Vsingle V3 X3
X=X1
FA2
X=X2
FA3; X=X3
FAO
-rA
V1 + V2 + V3 < Vsingle
Vsingle V3 X3
Can we model PFR as a series of “n” equal volume CSTRs?? X2 X1 V2 V1

25. Reactors in series: numerical example
Same data of the previous example. Determine:
The total volume of 2 CSTR in series for a total conversion of 80% with the first reactor has a conversion of X=0.4.
The total volume of 2 PFR in series for a total conversion of 80% with the first reactor has a conversion of X=0.4.
Solution CSTR
FA0= mol/s
V1=86.7; V2=  V= V1+V2 = 364 l
Only one CSTR V= 555 l (see previous example)
Solution PFR
V1=71.6; V2= 153  V=V1+V2= 225 l (=to previous example)

26. Order of the sequence for intermediate X=.5
Case A
PFR equation
Integration from 0 to 0.5 gives V1= 97 l
CSTR
V2= (0.8 – 0.5) (800) = 208 l
Vtot= V1+V2= 305 l
Case B
CSTR equation V1= (0.5 – 0) (303) = l
PFR V2 = l
Vtot= V1+V2= 262 l
FAO
FA1 X1 X2
FAO
FA1 X1 X2

27. Modelling of Hippopotamus Digestive System

28. Blood perfusion
Stomach
VG = 2.4 l
Gastrointestinal
tG = 2.67 min
Liver
Alcohol
VL = 2.4 l
tL = 2.4 min
Central
VC = 15.3 l
tC = 0.9 min
Muscle & Fat
VM = 22.0 l
tM = 27 min
It is the process in which the blood is moved to the muscles and fat (using capillary vases).
Iteractions are given by arrows.
VG, VL, VC, e VM are the volumes of tissues-water representing gastrointestinal, liver and muscles.
VS is the volume of the stomach.

29. Some more definitions
Space time (or Mean residence time) () is the Time required to process 1 reactor volume of fluid at inlet conditions
Space velocity:
Other conditions for flow rate v0 (in terms of velocity)
LHSV - Liquid Hourly Space Velocity
GHSV - Gas Hourly Space Velocity
Actual Residence Time: The time actually spent by fluid inside the reactor.
Plug flow

30. More definitions Relative rates of reaction aA + bB  cC + dD
How is (-rA) related to (-rB), (rC) and (rD) ?

31. Exapmple
The exothermic reaction A  B + C was carried out adiabatically and the following data recorded:
The entering molar flow rate of A was 300 mol/min
What PFR volume is necessary to achieve 40% conversion
What conversion can be achieved if the PFR in part(a) is followed by a 2.4 L CSTR?

32. Solution h1 w Design Equation Method-1: Area of Trapezoid
Area under curve
= w x [h1+h2]/2 h2 w
= 0.4 x [ ]/2
= 0.4 x 0.06 = L·min/mol
V = x 300 L = 7.2 L

33. Solution h Design Equation Method-2: Simpson’s Rule Area under curve
= [h/3]x [f(XO) + 4f(X1) +f(X2)]
where, h = [X2-XO]/2 & X1=XO+h
Area = [0.4/3] x [ x ]
= [0.2/3] x [0.36] = L·min/mol
V = x 300 L = 7.2 L

34. Solution VCSTR = 2.4 L Part(b) Conversion in CSTR in series with PFR
FAO
X=0
FA1
X=0.4
FA2 = ?
X=X2=?
VCSTR = 2.4 L
Design Equation
Step-1: Evaluate if X2<0.6
Area under flat portion of curve
= [ ] x 0.02 = 0.004
V = 300 x = 1.2 L
Therefore, we know that X2>0.6

35. Solution Part(b) Conversion in CSTR in series with PFR
Step-2: for X>0.6
find relationship between (1/rA) and X
1/(-rA) = f(X)
for X>0.6; f(X) = (X - 0.6)
= 0.3 X -0.16
Area under curve = f(X2).[X2-X1]
V = Area under curve x FAO =2.4 L
= [0.3 X ] [X2 -0.4] 300 L
Solving for X2, we get X2 =0.6425