6-1
Solutions for Chapter 6 – Multiple Reactions
P6-1 Individualized solution
P6-2 (a) Example 6-2
For PFR,
0.3
0.35
0.4
CA
0.5
0.6
0.7
POLYMATH Results
No Title 02-16-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
tau 0 0 783 783
Ca 0.4 0.0166165 0.4 0.0166165
6-2
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] Cao = 0.4
(2) Pressure increased by a factor of 100.
P6-2 (b) Example 6-3
(a) CSTR: intense agitation is needed, good temperature control.
(b) PFR: High conversion attainable, temperature control is hard – non-exothermic reactions,
selectivity not an issue
6-3
P6-2 (c) Example 6-4
(1) For k1 = k2, we get
( )
‘
1
exp
!
kCC AOA “=
and
( )
‘
111
‘exp
!
!
kCkCk
d
dC
AOB
B“=+
(2) For a CSTR:
A
AA
A
AA
Ck
CC
r
CC
1
0
‘
0!
=
!
!
=
“
#
01 AAA CCCk =+
!
“
BBA CCkCk =
!
“
!21
##
( ) ( )( )
111 1
0
2
1
2
1
+
!
+
!
!
=
+
!
!
=k
C
k
k
k
Ck
CA
A
B
““
“
“
“
( )( )
[ ] [ ]
( )( )
[ ]2
12
2121011201
11
211
0
+
!
+
!
!
++
!
“+
!
+
!
==
!kk
kkkkCkkkCk
d
dC AA
B
##
####
#
6-4
( )( )
11 12
01
+
!
+
!
!
=kk
Ck
CA
B
““
“
POLYMATH Results
No Title 02-14-2006, Rev5.1.233
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 100 100
T 300 300 400 400
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 1
Explicit equations as entered by the user
[1] Cao = 5
6-5
P6-2 (d) Example 6-5
KC = Ke = 0.25
!
“
#
$
%
&‘=
C
ON
NONN K
CC
Ckr 22
22
2
22
P6-2 (e) Example 6-6
For liquid phase,
joj CvF =
P6-2 (f) Example 6-7
For equal molar feed in hydrogen and mesitylene.
CHO = yHOCTO = (0.5)(0.032)lbmol/ft3 =0.016 lbmol/ft3
Ex 6-7
This question
XH
0.50
0.99
CH
0.0105
0.00016
6-6
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
tau 0 0 0.43 0.43
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(CH)/d(tau) = r1H+r2H
Explicit equations as entered by the user
[1] k1 = 55.2
Increasing θH decreases τopt and
~
S
X/T.
P6-2 (g) Example 6-8
Using equation from example 6-8:
Polymath code:
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
CH 4.783E-05 –4.889E-11 1.0E-04
6-7
NLES Report (safenewt)
Nonlinear equations
[1] f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0
Explicit equations
A plot using different values of
!
is given.
For
!
=0.5, the exit concentration are
CH = 4.8 ×10-5 lbmol/ft3 CM =0.0134 lbmol/ft3
CX =0.00232 lbmol/ft3
The yield of xylene from mesitylene based on molar
flow rates exiting the CSTR for
!
=0.5:
Ex 6-8
This Question
CH
0.0089
4.8 x 10-5
P6-2 (h) Example 6-9
(1)
SD/U Original Problem
SD/U P6-2 h
Membrane Reactor
2.58
1.01
P6-2 (i) Example 6-10
Original Case – Example 6-10 P6-2 i
The reaction does not go as far to completion when the changes are made. The exiting concentration of D,
E, and F are lower, and A, B, and C are higher.
See Polymath program P6-2-i.pol.
P6-2 (j)
At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As
the two components are created, the reactant concentration drops and equilibrium forces the production to
6-9
P6-2 (k)
Base case Equal-molar feed
P6-2 (l) Individualized solution
P6-3 Solution is in the decoding algorithm given with the modules ( ICM problem )
P6-4 (a)
Assume that all the bites will deliver the standard volume of venom. This means that the initial
concentration increases by 5e-9 M for every bite.
P6-4 (b)
The victim was bitten by a harmless snake and antivenom was injected. This means that the initial
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 0.5 0.5
fsv 0 0 0 0
fs 1 0.6655661 1 0.6655661
Cv 0 0 0 0
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(fsv)/d(t) = kv * fs * Cv – ksv * fsv * Ca
[2] d(fs)/d(t) = -kv*fs*Cv – ka * fs * Ca + kia * fsa + g
Explicit equations as entered by the user
[1] kv = 2e8
[2] ksv = 6e8
[3] ka = 2e8
6-11
P6-4 (c)
The latest time after being bitten that antivenom can successfully be administerd is 27.49 minutes. See the
cobra web module on the CDROM/website for a more detailed solution to this problem
P6-4 (d) Individualized Solution
P6-5 (a)
Plot of CA, CD and CU as a function of time (t):
See Polymath program P6-5-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 15 15
Ca 1 0.0801802 1 0.0801802
ODE Report (RKF45)
Differential equations as entered by the user
6-12
Explicit equations as entered by the user
[1] k1 = 1.0
P6-5 (b)
Conc. Of U is maximum at t = 0.31 min.(CA = 0.53)
P6-5 (c)
Equilibrium concentrations:
P6-5 (d)
See Polymath program P6-5-d.pol.
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
Ca 0.0862762 –3.844E-14 1
NLES Report (safenewt)
Nonlinear equations
Explicit equations
[1] Ca0 = 1
P6-6 (a)
XA !
2/1
1AX Ckr =
min004.0
2/1
3
1!
#
$
&
=dm
mol
k
See Polymath program P6-6-a.pol.
1) SB/X =
2/1
1
2
2/1
1
2
A
A
A
X
BC
k
k
Ck
Ck
r
r==
CAexit
0.295
0.133
0.0862
3) SB/XY =
2
3
2/1
1
2
AA
A
YX
B
CkCk
Ck
rr
r
+
=
+
P6-6 (b)
Volume of first reactor can be found as follows
We have to maximize SB/XY
From the graph above, maximum value of SBXY = 10 occurs at CA
* = 0.040 mol/dm3
3
2
1
AAA
P6-6 (c)
Effluent concentrations:
P6-6 (d)
6-15
Conversion of A in the first reactor:
74.0
00 =!=“XXCCC AAA
P6-6 (e)
A CSTR followed by a PFR should be used.
Required conversion = 0.99
P6-6 (f)
If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower
temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of B,
decrease as temperature drops. So we have to compromise between high selectivity and production. To do
Now we use a mole balance on species A
0A A
A
F F
V
r
!
=!
6-16
Using these equations we can make a Polymath program and by varying the temperature, we can find a
maximum value for CB at T = 306 K. At this temperature the selectivity is only 5.9. This may result in too
much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal temperature
will depend on the price of B and the cost of removing X and Y, but without actual data, we can only state
for certain that the optimal temperature will be equal to or less than 306 K.
See Polymath program P6-6-f.pol.
POLYMATH Results
NLE Solution
Variable Value f(x) Ini Guess
Ca 0.0170239 3.663E-10 0.05
T 306
NLE Report (safenewt)
Nonlinear equations
[1] f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0
Explicit equations
[1] T = 306
P6-6 (g)
Concentration is proportional to pressure in a gas-phase system. Therefore:
P6-7
US legal limit: 0.8 g/l
6-17
Two tall martinis = 80 g of ethanol
Body fluid = 40 L
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 10 10
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
P6-7 (a)
In the US the legal limit it 0.8 g/L.
This occurs at t = 6.3 hours..
P6-7 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs
P6-7 (d)
For this situation we will use the original Polymath code and change the initial concentration of A to 1 g/L.
6-18
See Polymath program P6-7-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0.5 0.5 10 10
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] k1 = 10
P6-7 (e)
The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g of
ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the rate
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 11 11
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
6-19
Russia: t = 5.2 hours
P6-7 (f)
60 g of ethanol immediately CA = 1.5 g/L
CB = 0.8 g/L at 0.0785 hours or 4.71 minutes.
P6-7 (g)
A heavy person will have more body fluid and so the initial concentration of CA would be lower. This
P6-8 (a)
See Polymath program P6-8-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 4 4
Ca 6.25 0.3111692 6.25 0.3111692
ODE Report (RKF45)
Explicit equations as entered by the user
6-20
P6-8 (b)
P6-8 (c)
If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first six
hours. Doing so will result in build up of the drug in the bloodstream that can cause harmful effects.