1. Steady-state Nonisothermal reactor Design Part I
CSTRs with Heat Exchange.
PFRs with Heat Exchange.

2. Non Isothermal Process
Most reactions are not carried out isothermally
Focus is then on heat effects in chemical reactors
Basic design equation such as rate laws, stoichiometric relationships used for isothermal reactor design are valid for non-isothermal reactors
The major difference is method of evaluating the temperature effect
e.g. length of PFR, or heat is removed from a CSTR

3. Non Isothermal Process
So what additional information is needed to design non-isothermal reactors ?
Example: For the following reaction
A → B
Calculate the reactor volume necessary for 70% conversion
Note: The reaction is exothermic, reactor is operated adiabatically. So temperature increases with conversion down the length of reactor

4. Non Isothermal Process
Solution:
Mole Balance (design equation)
Rate Law (Arrhenius equation)
k = f(T)

5. Non Isothermal Process
Stoichiometry (liquid phase), v = vo
Combining
and cancelling the entering concentration CA0=0

6. Non Isothermal Process
The final equation
We require a relationship which relates X and T, T and V to solve the above equation
The energy balance will provide the relationship

7. Non Isothermal Process
Energy Balance
we need appropriate energy balance to relate temperature and conversion or reaction rate
For adiabatic reaction we have
So we have all equations needed to solve the conversion and temperature profile

8. The Energy Balances
We know that for a closed system, the energy balance is
For open (continuous) flow system where mass crosses the system boundaries

9. The Energy Balances
The energy balance of the case of only one species entering and leaving becomes

10. The Energy Balances
The unsteady-state energy balance for an open well-mixed system that has n species, entering and leaving the system at their respective molar flow rates Fi (moles of i per time) and their respective energy Ei (joules per mole of i)

11. The Energy Balances Evaluating work term
W is separated into flow work (W) or other work (Ws)
Ws is shaft work and produce such stirrer in CSTR or turbine in PFR
Flow work is the work necessary to get the mass into and out of the system
J/s

12. The Energy Balances
Combining the energy balance term derived earlier and work term we get
We know that
In chemical reactor these energies are often negligible as compared to enthalpy, heat transfer and work term
J/mol

13. The Energy Balances
Enthalpy in case of into and out of the system can be expressed as the sum of the net internal energy carried into (or out of) the system by mass flow plus the flow work
Combining we get
We can also write

14. The Energy Balances
Adiabatic (Q=0) CSTR, PFR, Batch or PBR. The relationship between conversion, XEB and temperature for Ws = 0, constant CPi and ∆CP = 0
For an exothermic reaction

15. CSTRs with Heat Exchange. Evaluating the Heat Exchange Term, Q

16. Evaluating the Heat Exchange Term, Q
Rate of heat transfer from Exchanger

17. Evaluating the Heat Exchange Term, Q
Energy transferred between the reactor and the coolant
Assuming the temperature inside the CSTR, T, is spatially uniform
Page 555

18. Evaluating the Heat Exchange Term, Q

19. Evaluating the Heat Exchange Term, Q
Sub into the energy balance

20. CSTR sizing Algorithm Example AB 1. Design Eqn 2. Rate Law
3. Stoichiometry
4. Combining
X specified
Calculate V & T
V specified
Calculate X & T
Calculate T
Calculate k and KC
Calculate V=FA0X/kCA0(1-X)
Calculate XMB
Calculate XEB
Plot XEB vs T
Plot XMB vs T
Evaluate V from the graph
Page 557

21. CSTR Algorithm

22. CSTR Algorithm XMB Conversion from mole balance
XEB Conversion from energy balance

23. Example: Adiabatic Liquid Phase in A CSTR
Given T Find X and V Also given k, E, CPA=CPB, ΔHRx  , CAo, and vo

24. Example: Adiabatic Liquid Phase in A CSTR

25. Example: Adiabatic Liquid Phase in A CSTR

26. Sketch XEB versus T

27. Mole Balance to Determine XMB=f(T)
rearranging...

28. At the intersection of XMB vs. T and XEB vs. T
both the mole balance and the energy balances are satisfied.
The steady-state conditions (XSS,TSS) occur here.
Examples 8-8, Example 8-9 (Page 563)
Both the mole and energy balances are satisfied when XMB=XEB. The steady state temperature and conversion are TSS and XSS respectively for an entering temperature TO

29. Energy Balance on a PFR
Page 496 , pdf page 528
a = heat exchange area per unit volume of reactor; for a tubular reactor, a = 4/D

30. Energy Balance on a PFR Energy Balance ΔQ+∑Fi0Hi0 -∑FiHi=
          
Catalyst weight is related to reactor volume by:
Differentiate with respect to W

31. Energy Balance on a PFR Conversion as Variable or
Flow Rate as Variable

32. Energy Balance on a PFR

33. Energy Balance on a PBR
If we want to include pressure drop:

34. Balance on Heat Exchanger Coolant
If Ta is not constant, then we must add an additional energy balance on the coolant fluid:
with Ta = Tao at W = 0

35. Example 8-4
When we checked the vapor pressure at the exit to the adiabatic reactor in Example 8-3 where the temperature is 360K, we found the vapor pressure to be 1.5MPa for isobutane, which is greater than the rupture pressure of

42. P8-2 (d)
How would the answers change if the reactor were in a counter current exchanger where the coolant temperature was not constant along the length of the reactor

46. P8-5 (3rd edn)
The elementary irreversible organic liquid phase reaction A + B  C is carried out adiabatically in a flow reactor. An equal molar feed in A and B enters at 27C and volumetric flow is 2 dm3/s.
Additional information
HAo (273) = -20kcal/mol
HBo (273)= -15kcal/mol
HCo (273)= -41kcal/mol
CA0= 0.1kmol/m3
CPA=CPB=15cal/mol.K CPC=30cal/mol.K
k = 0.01 / dm3/mol.s at 300K
E= cal/mol

47. (a) Calculate the PFR and CSTR volumes necessary to achieve 85% conversion
A + B  C

48. (a) Calculate the PFR and CSTR volumes necessary to achieve 85% conversion

51. PFR X T k
-rA
FA0/-rA
300
0.01 1
2000

52. PFR X T k
-rA
FA0/-rA
300
0.01 1
2000
0.2
340
0.072
4.6
435
0.4
380
0.34
12.3
163
0.6
420
1.21
19.3
103
0.8
460
3.42
13.7
146
0.85
470
4.31
9.71
206

54. (b) What is the maximum inlet temperature one could have so that the boiling point of the liquid (550K) would not be exceeded even for complete conversion?

55. (b) What is the maximum inlet temperature one could have so that the boiling point of the liquid (550K) would not be exceeded even for complete conversion?

56. (b) What is the maximum inlet temperature one could have so that the boiling point of the liquid (550K) would not be exceeded even for complete conversion?

57. (c) Plot the conversion and temperature as a function of PFR volume
(c) Plot the conversion and temperature as a function of PFR volume. Use Polymath

58. (d) Calculate the conversion that can achieved in one 500dm3 CSTR volume and in two 250dm3 CSTR in series Use Polymath for CSTR