1. Lecture 8
Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place.

2. Today’s lecture Block 1: Mole Balances on PFRs and PBR
Must Use the Differential Form
Block 2: Rate Laws
Block 3: Stoichiometry
Pressure Drop:
Liquid Phase Reactions:
Pressure Drop does not affect the concentrations in liquid phase rx.
Gas Phase Reactions:
Epsilon not Equal to Zero
d(P)/d(W)=..
Polymath will combine with d(X)/f(W)=..for you
Epsilon = 0 and Isothermal
P=f(W)
Combine then Separate Variables (X,W) and Integrate

3. Reactor Mole Balances in terms of conversion
Differential
Algebraic
Integral
Batch X t
CSTR
PFR
PBR X W

4. Concentration Flow System:
Gas Phase Flow System:

5. Pressure Drop in Packed Bed Reactors
Note: Pressure drop does NOT affect liquid phase reactions
Sample Question:
Analyze the following second order gas phase reaction that occurs isothermally in a PBR:
AB
Mole Balance:
Must use the differential form of the mole balance to separate variables:
Second order in A and irreversible:
Rate Law:

6. Pressure Drop in Packed Bed Reactors
Stoichiometry:
Isothermal, T=T0
Combine:
Need to find (P/P0) as a function of W (or V if you have a PFR)

7. Pressure Drop in Packed Bed Reactors
Ergun Equation:
Constant mass flow:

8. Pressure Drop in Packed Bed Reactors
Variable Density
Let

9. Pressure Drop in Packed Bed Reactors
Catalyst Weight
Where
Let

10. Pressure Drop in Packed Bed Reactors
We will use this form for single reactions:
Isothermal case

11. Pressure Drop in Packed Bed Reactors
and or
The two expressions are coupled ordinary differential equations. We can only solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution.
Polymath will combine the mole balance, rate law and stoichiometry.

12. PBR
AB
1) Mole Balance:
2) Rate Law:

13. PBR

14. W P 1

15. CA 2 W

16. -rA 3 W

17. X 4 W

19. 5 W
1.0

20. Example 1: Gas Phase Reaction in PBR for δ = 0
Gas Phase Reaction in PBR with δ = 0 (Polymath Solution) A + B  2C Repeat the previous one with equil molar feed of A and B and kA = 1.5dm9/mol2/kg/min α = kg-1 Find X at 100 kg

21. Example 1: Gas Phase Reaction in PBR for δ = 0
A + B  2C
Case 1:
Case 2:

22. Example 1: Gas Phase Reaction in PBR for δ = 0
1) Mole Balance:
2) Rate Law: 3) 4)

23. Example 1: Gas Phase Reaction in PBR for δ = 05)

24. Example 1: Gas Phase Reaction in PBR for δ = 0

25. Example A + B → 2C

26. Example A + B → 2C

27. Example 2: Gas Phase Reaction in PBR for δ ≠ 0
Polymath Solution A + 2B  C is carried out in a packed bed reactor in which there is pressure drop.The fed is stoichiometric in A and B. Plot the conversion and pressure ratio y = P/P0 as a function of catalyst weight upto 100 kg. Additional Information kA = 6dm9/mol2/kg/min α = 0.02 kg-1

28. Example 2: Gas Phase Reaction in PBR for δ ≠ 0
A + 2B  C
1) Mole Balance:
2) Rate Law:
3) Stoichiometry: Gas, Isothermal

29. Example 2: Gas Phase Reaction in PBR for δ ≠ 04) 5) 6) 7)
Initial values: W=0, X=0, y=1  W=100
Combine with Polymath.
If δ≠0, polymath must be used to solve.

30. Example 2: Gas Phase Reaction in PBR for δ ≠ 0

31. Example 2: Gas Phase Reaction in PBR for δ ≠ 0

32. T = T0

33. Engineering Analysis

34. Engineering Analysis

35. Engineering Analysis

36. Pressure Change – Molar Flow Rate
Use for heat effects, multiple rxns
Isothermal: T = T0

37. Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance

38. End of Lecture 8