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

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

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

5. Evaluating the Heat Exchange Term, Q

6. Evaluating the Heat Exchange Term, Q

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

8. CSTR Algorithm

9. CSTR Algorithm

10. CSTR Algorithm

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

14. Sketch XEB versus T

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

16. 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.
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.

17. Energy Balance on a PFR
a = heat exchange area per unit volume of reactor; for a tubular reactor, a = 4/D

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

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

20. Energy Balance on a PFR

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

22. 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

23. 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

30. 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

34. 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

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

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

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

40. 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

42. (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?

43. (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?

44. (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?

45. (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

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