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Sec. 6.3 Algorithm for Solution to Complex Reactions 309
3. Stoichiometry u = uo (ECi-7.8)
FH = vOCH (E6-7.9)
FM = VOCM (E6-7.10)
Fx =v& (E6-7.11)
FT = VOCT = (FMO - FM) - Fx (E6-7.12)
FMe = UOCMe = vO (CHO - cH ) (E6-7.13)
4. Combining and letting z = V/u, (space-time) yields:
c~0-c~ =(ki(C:’CM + k2C;’Cx)~ (E6-7.14)
CM, - CM = (k, ci’cc,) T (E6-’7.15)
CX =(klc, cM-k~c~’Cx)‘T
112
(E6’7.16)
Next, we put these equations in a form such that they can be readily solved using
POIXMATH.
0
(E6-’7.17)
f(c~) = ~H-~HO+(~~C:~CM+~~C~’C~)T
=
f(cM) = o= c~-cp,,jo+k,C;~CMT (E6-’7.18)
f(cx) = 0 = (k,CFCM-k,C[’Cx)z- c, (E6-’1.19)
The POLYMATH program and solution are shown in Table E6-7.1. The problem
was solved for different values of z and the results are plotted in Figure E6-7.1.
For a space-time of z = 0.5, the exiting concentrations are CH = 0.0089,
CM = 0.0029, and Cx = 0.0033. The overall conversion is
Hydrogen: X, = - - -
FHO - FH CHO - CH - 0.021 - 0.0089 - o.58
-
=
FHO cHO 0.02 1
Mesitylene: XM = - - -
FMO- FM = CMO - CM 0.0105 - 0.0029 - o.72
-
FMO 0.0 105
TABLE E6-7.1. POLYMATH how AND SOLUTION
~~~
Equations Initial Values
fC~ch~=ch-.0rM+~55.2W~mWchWW.5+30.2X~xXch~.5~~tau 0.006
f <cml=cm-. OIO59CS5. PXcmXchW. 5)Xtau 0.0033
fC1:x>=~55.2Wcm#~M~.5-30.2XoxXchXX.S~Xtau-cx 0.005
I tacJ=O. 5