CWRU ECHE 364 Chemical Reaction Processes
Lecture 22
Chapter 6.1 – 6.4
CYCLYHEXANE
BENZENE + HYDROGEN
"
1
Example: WGS (water-gas shift) with Pd membrane reactor
• CO + H2O ↔ CO2 + H2
•A+B↔C+D
What is the material balance with rate expression for CO2 (component C) in terms of
FA
molar flow rates? Assume that the reaction is elementary and reversible, and that k
kA ")
=
-
describes the rate constant for the forward reaction.
rc -TA
=
E r k((( ((/k)
=
-TA
=
=
-
Now assume that hydrogen can diffuse out the sides of the membrane, while CO, H2O
and CO2 cannot. What is the rate of reaction of hydrogen (component D) in terms of
(
=
molar flow rates? Assume kM, KC, and k are positive numbers.
B rA
-
rc
=
F*
-rB
E
=
=
k(CA) c(/x) kmCp
-
=
2
Example continued: Fogler P6-7
WGS (water-gas shift) with Pd membrane reactor
CO + H2O ↔ CO2 + H2
• The volumetric feed is 10 dm3/min at 10 atm, and the equimolar feed of CO and
water vapor with CT0=0.4 mol/dm3. Bulk density, �=1,000 kg/m3
• The equilibrium constant is Keq = 1.44 and the forward rate constant k=1.37
dm6/mol kg-cat·min
• mass transfer coefficient, kH2 = 0.1 dm3/kg-cat·min
Plot the concentrations and molar flow rates of each of the reacting species down the
length of the membrane reactor.
(a) What is the necessary catalyst weight to achieve 85% conversion of
CO? (b) Compare with a PFR
(c) For the same reactor volume, what would be the conversion if the feed rate
were doubled?
equations:
&Es:
# A -rA r r k((( CCp/k) kmCD
=
=
=
-
-
Ru
↳E -rB -FA rCA
=
=
=
=
=
GTo
Er
=
=
rA
=-
r
CB Cto FB/FT
=
Cc (To Fc/FT
=
= R Ct
-
r-
Cp
=
constants:
k 1.37
=
kc
10 1.44
Vo =
=
km 0.1
=
FTo =
4
CT0
=0.4
FD/FT
a
FD
+
FT FA + F + Fc 3
=
partal
to
Polymath online
/solution
#@Chart_Size = 400; 250
#@Report_Show_Intermediate_dat
a_points = yes
#@DEQ_Intermediate_data_points
= 50 #@Report_show_Settings = yes
d(F_A)/d(W)=r
d(F_B)/d(W)=r
to see the evaluated
E
add this linevalues.
need to
to
your
0.3
d(F_C)/d(W)=-r
d(F_D)/d(W)=-r-R_H
r=-k*(C_A*C_B-C_C*C_D/K_C) R_H=k_m*C_D
C_A=C_T0*F_A/F_T
C_B=C_T0*F_B/F_T
C_C=C_T0*F_C/F_T
C_D=C_T0*F_D/F_T
F_T=F_A+F_B+F_C+F_D
k=1.37
K_C=1.44
k_m=0.1
v_0=10
C_T0=0.4
F_T0=4
X 85% FA
=
W430
=>
Or
value
from plots.
approx
* See
the
plots
of molar
flow rates (and reaction
rated
inthe nextpage.
F_A(0)=2
F_B(0)=2
F_C(0)=0
F_D(0)=0
W(0)=0
W(f)=500
at
solution
=
For this
we
part;
will
remove fam
Ryettan
imiskey---Yx"
from
Co CC
s
GACB
the mole balance for dio/dw.
gal 1.44
solve
⑥ If Fr&F's
f-54
doubled - X
0.4597
=
v
Solve the problem again including the following text in order to plot for variable r
specifically (reaction rate; already defined in the program).
#@Chart_Y_Curves = r
8
Other general examples (we will solve if time allows; otherwise work on your own):
Example: The positive-order reaction R→P takes place in two CSTRs that are not
identical but are at the same temperature. The flow rate into both reactors is 0.5 mol/s.
Which reactor has the higher reaction rate per volume?
Example: Many proteins unfold and fold reversibly in a unimolecular reaction. The
ርሲ
elementary rate law for this folding ౡ౫
and unfolding mechanism is : U
ౡ ሱሮ
N
where U is the unfolded protein, and N is the folded protein. In a typical protein drug
product, the total concentration of protein (folded and unfolded combined) is 0.1mM.
a) If the free energy of folding is - 40 kJ/mol at 295 K, what is the concentration of
unfolded protein in the product at equilibrium?
b) If the rate constant for folding kf =1000 s-1, what is the rate constant ku for
unfolding?
4