4-41
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
w 0 0 100 100
X 0 0 0.5707526 0.5707526
Q 0.5625 0.5625 0.5625 0.5625
Cao 0.207 0.207 0.207 0.207
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(w) = -ra/Fao
[2] d(y)/d(w) = -alpha/2/y
Explicit equations as entered by the user
[1] Dp = .0075
P4-20 (d) Individualized solution
P4-20 (e) Individualized solution
P4-20 (f) Individualized solution
P4-20 (g) Individualized solution
P4-20 (h) Individualized solution
P4-21 (a)
Assume constant volume batch reactor
Mole balance:
0A A
dX
C r
dt =!
Rate law and stoichiometry:
( )
01
AAA
r kC kC X!= = !
4-42
( )
01
A A
C C X=!
but since volume and molecular weight are constant the equation can be written as:
( )
0
6500 1 0.108
A
IU m=!
07287
A
m IU=
07287 6500
% *100 *100 12.1%
6500
A A
A
C C
OU
C
!!
= = =
P4-21 (b)
10,000,000 lbs/yr = 4.58 * 109 g/yr of cereal
Serving size = 30g
P4-21 (c)
If the nutrients are too expensive, it could be more economical to store the cereal at lower temperatures
where nutrients degrade more slowly, therefore lowering the amount of overuse. The cost of this storage
P4-21 (d)
( ) 1
40 0.0048k C weeks!
=
o
6 months = 26 weeks
( ) ( )
0
0
0
1ln 1
1
X
A
A
dX
t C X
kC X k
!
= = !
!
“
( )
1
1
26 ln 1
0.0048
weeks X
weeks!
!
=!
0.12X=
( )
01
A A
C C X=!
but since volume and molecular weight are constant the equation can be written as:
( )
0
6500 1 0.12
A
IU m=!
07386
A
m IU=
07386 6500
% *100 *100 13.6%
6500
A A
A
C C
OU
C
!!
= = =
P4-22 No solution necessary
4-43
P4-23B
Suppose the volumetric flow rate could be increased to as much as 6,000 dm3/h and the total time to fail,
heat, empty and clean is 4.5 hours. What is the maximum number of moles of ethylene glycol (CH2OH)2
you can make in one 24 hour period? The feed rate of ethylene cholorhydrin will be adjusted so that the
volume of fluid at the end of the reaction time will be 2500 dm3. Now suppose CO2 leaves the reactor as
fast as it is formed.
A+B“C+D+CO2
Mole Balance
dNA
dt =r
AV
dNB
dt =FB0+r
BV
dNC
dt =r
CV
ND=NC
Overall Mass Balance
Accumulation = In – Out
dm
dt =
“#
0$˙
m
CO2
m=
“
V Assume constant density
The rate of formation of CO2 is equal to the rate of formation of ethylene glycol (C).
˙
m
CO2=r
CV•MW
dV
dt =
0#r
CVMW
$
Rate Law and Relative Rates
“r
A=kCACB
r
B=r
A
r
C=“r
A
4-44
Stoichiometry
CA=NA
V
CB=NB
V
4-45
For two runs per day
Clean time = 2 x 4.5 = 9 h
Reaction time = 24 – 9 = 15 h
Runs/
day
Down
Time
Reaction
Time
Time/
Batch
FB0
mol
h
“0
dm 3h
V
dm 3
X
NC per Batch
Moles
Total
moles
2
9h
15h
7.5h
209.5
mol
h
139.6
2500
0.958
1077.7x2 =
2155
3
13.5
10.5
3.5
447
298
24993
0.88
993x3 =
2979
We will use four batches per day to make the most ethylene glycol (i.e., C)
4-46
P4-24
3 2 5 3 2 5
NaOH CH COOC H CH COO Na C H OH
!+
+“”# +
A B C D+!!” +
Mole balance:
( )
AAO
O
ACC
V
v
r
dt
dC !+=
( )
B
O
BC
V
v
r
dt
dC !+=
( )
C
OC C
V
v
r
dt
dC !+!=
To produce 200 moles of D, 200 moles of A and 200 moles of B are needed. Because the concentration of
A must be kept low, it makes sense to add A slowly to a large amount of B. Therefore, we will start with
pure B in the reactor. To get 200 moles of B, we need to fill the reactor with at least 800 dm3 of pure B.
Assume it will take 6 hours to fill, heat, etc. the reactor. That leaves 18 hours to carry out the reaction. We
Now vary the initial amount of B in the reactor, the flow rate of A, and the temperature to find a solution
that satisfies all the constraints. The program below shows one possible solution.
4-47
See Polymath program P4-24.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 6.5E+04 6.5E+04
Ca 0 0 0.1688083 0.1688083
Cb 0.25 0.0068364 0.25 0.0068364
Cc 0 0 0.0151725 0.0142903
Cao 0.2 0.2 0.2 0.2
T 308 308 308 308
k 1.224E-04 1.224E-04 1.224E-04 1.224E-04
ra 0 0 1.397E-06 1.412E-07
V 1200 1200 1.42E+04 1.42E+04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -ra+(vo/V)*(Cao-Ca)
[2] d(Cb)/d(t) = -ra-(vo/V)*Cb
[3] d(Cc)/d(t) = ra-(vo/V)*Cc
Explicit equations as entered by the user
[1] ko = 5.2e-5
[2] Fao = .04
[3] Cao = .2
[4] Vo = 1200
[5] vo = Fao/Cao
P4-25 (a)
A ↔ B + 2C
To plot the flow rates down the reactor we need the differential mole balance for the three species, noting
that BOTH A and B diffuse through the membrane
A
A A
dF r R
dV =!
B
B B
dF r R
dV =!
C
C
dF r
dV =
4-48
Next we express the rate law:
First-order reversible reaction
& ‘
Transport out the sides of the reactor:
RA = kACA =
0A T A
T
k C F
F
RB = kBCB =
0B T B
k C F
Stoichiometery:
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
v 0 0 20 20
Fa 100 57.210025 100 57.210025
Ft 100 100 122.2435 121.06199
Co 1 1 1 1
Ka 1 1 1 1
Ra 1 0.472568 1 0.472568
Rb 0 0 2.9904791 0.6396478
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra – Ra
4-49
Explicit equations as entered by the user
[1] Kc = 0.01
[2] Ft = Fa+ Fb+ Fc
[3] Co = 1
[4] K = 10
P4-25 (b)
The setup is the same as in part (a) except there is no transport out the sides of the reactor.
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
v 0 0 20 20
Fa 100 84.652698 100 84.652698
Ft 100 100 130.6946 130.6946
Fao 100 100 100 100
X 0 0 0.153473 0.153473
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra
[2] d(Fb)/d(v) = –ra
Explicit equations as entered by the user
[1] Kc = 0.01
[2] Ft = Fa+ Fb+ Fc
[3] Co = 1
4-50
P4-25 (c) Conversion would be greater if C were diffusing out.
P4-25 (d) Individualized solution
P4-26
CO + H2O ↔ CO2 + H2
A + B ↔ C + D
Assuming catalyst distributed uniformly over the whole volume
Mole balance:
r
dW
dFA=
r
dW
dFB=
r
dW
dFC!=
2
H
DRr
dW
dF !!=
Rate law:
C D
A B C D A B
C C
r r r r r k C C K
! “
= = = #=#=# #
$ %
Stoichiometry:
T
A
TOA F
F
CC =
T
B
TOB F
F
CC =
T
C
TOC F
F
CC =
D
TOD F
F
CC =
4-51
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 100 100
Fa 2 0.7750721 2 0.7750721
Fd 0 0 0.7429617 0.5536716
Keq 1.44 1.44 1.44 1.44
Ft 4 3.3287437 4 3.3287437
Cto 0.4 0.4 0.4 0.4
Cb 0.2 0.0931369 0.2 0.0931369
Cc 0 0 0.147194 0.147194
r -0.0548 -0.0548 –0.002567 -0.002567
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(W) = r
[2] d(Fb)/d(W) = r
[3] d(Fc)/d(W) = -r
Explicit equations as entered by the user
[1] Keq = 1.44
[2] Ft = Fa+Fb+Fc+Fd
[3] Cto = 0.4
[4] Ca = Cto*Fa/Ft
For 85% conversion, W = weight of catalyst = 430 kg
In a PFR no hydrogen escapes and the equilibrium conversion is reached.
solve this for X,
If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of .459
4-52
P4-27 Individualized solution
P4-28 (a)
P4-29 Individualized solution
P4-30 (a)
First order gas phase reaction,
C6H5CH(CH3)2 C6H6 + C3H6
4-54
But alpha is also a function of superficial mass velocity (G). If the entering mass flow rate is held constant,
then increasing pipe diameter (or cross-sectional area) will result in lower superficial mass velocity. The
relationship is the following for turbulent flow:
G“1
AC
and
#
“G2
therefore,
“
#1
2
.
P4-30 (c) Individualized Solution
P4-31 (a)
ε = 0.33(1-3) = -0.666 PAO = 0.333*10 FAO = 13.33 K = 0.05
Mole balance: dX/dW = -ra/Fao
4-55
See Polymath program P4-31-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
W 0 0 100 100
X 0 0 0.995887 0.995887
K 0.05 0.05 0.05 0.05
Pao 3.33 3.33 3.33 3.33
Pb 6.66 0.0813464 6.66 0.0813464
ra -0.333 -0.333 –0.0040673 -0.0040673
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ra/Fao
Explicit equations as entered by the user
[1] K = 0.05
[2] Pao = 0.333*10
P4-31 (b)
For α = 0.027 kg-1,
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
w 0 0 30 30
X 0 0 0.4711039 0.4711039
Pao 3.33 3.33 3.33 3.33
Pb 6.66 0.9734003 6.66 0.9734003
ra -0.333 -0.333 –0.04867 -0.04867
alfa 0.027 0.027 0.027 0.027
4-56
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(X)/d(w) = -ra/Fao
Explicit equations as entered by the user
[1] K = 0.05
[2] Pao = 0.333*10
[3] Pa = Pao*(1 – X)*y/(1 – 0.666*X)
P4-31 (c)
1) For laminar flow:
Diameter of pipe = D and diameter of particle = DP
Now D1/Do = 3/2 so G1 = 4/9Go
α = (constant)(G/DP2)(1/AC)
2) For turbulent flow:
β ∞ G2/DP
α = (constant)(G2/DP)(1/AC)
P4-32 (a)
At equilibrium, r = 0 =>
C
DC
BA K
CC
CC =
V = V0 + vot
4-58
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra – Ca*vo/V
[2] d(Cb)/d(t) = ra – vo/V*(Cb– 10.93)
[3] d(Cc)/d(t) = -ra – vo*Cc/V
Explicit equations as entered by the user
[1] Kc = 1.08
[2] k = 0.00009
Polymath solution
P4-32 (c)
Change the value of vo and CAO in the Polymath
P4-32 (d)
As ethanol evaporates as fast as it forms: CD = 0
Now using part (b) remaining equations,
4-59
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 6000 6000
Ca 7.72 0.0519348 7.72 0.0519348
Cb 10.93 6.9932872 10.93 7.8939348
ra -0.0075942 -0.0075942 –3.69E-05 -3.69E-05
Vo 200 200 200 200
V 200 200 500 500
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra – Ca*vo/V
[2] d(Cb)/d(t) = ra – vo/V*(Cb– 10.93)
Explicit equations as entered by the user
[1] k = 0.00009
[2] ra = -k*Ca*Cb
P4-32 (f) Individualized solution
P4-33 (a)
Mole balance on reactor 1:
1
0 0 1 1
A
A A A A
dN
C v C v r V
dt
! ! =
with
0 0
1
2
A
v v=
01
011
2
AA
A A
CdN
v C v r V
dt
! ! =
Liquid phase reaction so V and v are constant.
4-60
01 1
1
2
AA A
A
CC dC
r
dt
! !
“ ” =
Mole balance on reactor 2:
2
1 0 2 0 2
A
A A A
dN
C v C v r V
dt
! ! =
Mole balance for reactor 3 is similar to reactor 2:
3
2 0 3 0 3
A
A A A
dN
C v C v r V
dt
! ! =
Rate law:
Ai Ai Bi
r kC C!=
Stoichiometry
For parts a, b, and c CAi = CBi
so that
2
Ai Ai
r kC!=
Combine:
2
01 1
1
2
AA A
A
CC dC
kC
dt
! !
” “ =
2
1 2 2
2
A A A
A
C C dC
kC
dt
! !
” “ =
2
3 3
2
3
A A
A
A
C dC
CkC
dt
! !
” “ =
See Polymath program P4-33.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 100 100
Ca1 0 0 0.8284264 0.8284264
Ca2 0 0 0.7043757 0.7043757
Cao 2 2 2 2