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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
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
P4-20 (d) Individualized solution
P4-20 (e) Individualized solution
P4-21 (a)
Assume constant volume batch reactor
Mole balance:
0A A
dX
C r
dt =!
4-42
P4-21 (b)
10,000,000 lbs/yr = 4.58 * 109 g/yr of cereal
P4-21 (c)
If the nutrients are too expensive, it could be more economical to store the cereal at lower temperatures
P4-21 (d)
( ) 1
40 0.0048k C weeks!
=
o
6 months = 26 weeks
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,
Mole Balance
dNA
dt =r
AV
Overall Mass Balance
Accumulation = In – Out
The rate of formation of CO2 is equal to the rate of formation of ethylene glycol (C).
˙
m
CO2=r
CV•MW
Rate Law and Relative Rates
4-44
Stoichiometry
V
For two runs per day
Clean time = 2 x 4.5 = 9 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
4-46
P4-24
3 2 5 3 2 5
NaOH CH COOC H CH COO Na C H OH
!+
+""# +
A B C D+!!" +
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
will need to add 1000 dm3 of A to get 200 moles in the reactor. We need to check to make sure the reactor
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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -ra+(vo/V)*(Cao-Ca)
Explicit equations as entered by the user
[1] ko = 5.2e-5
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
4-48
Next we express the rate law:
First-order reversible reaction
T
Stoichiometery:
-rA = rB =1/2 rC
Combine and solve in Polymath code:
See Polymath program P4-25-a.pol.
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
ODE Report (RKF45)
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
P4-25 (b)
The setup is the same as in part (a) except there is no transport out the sides of the reactor.
See Polymath program P4-25-b.pol.
POLYMATH Results
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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra
Explicit equations as entered by the user
[1] Kc = 0.01
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 !!=
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
Fb 2 0.7750721 2 0.7750721
Fc 0 0 1.2249279 1.2249279
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(W) = r
Explicit equations as entered by the user
[1] Keq = 1.44
[2] Ft = Fa+Fb+Fc+Fd
For 85% conversion, W = weight of catalyst = 430 kg
In a PFR no hydrogen escapes and the equilibrium conversion is reached.
P4-27 Individualized solution
P4-28 (a)
P4-28 (b)
P4-29 Individualized solution
4-53
P4-30 (a)
First order gas phase reaction,
C6H5CH(CH3)2 C6H6 + C3H6
1, 1 1
AO
y
! "
= = #=
X = 0.064 and y = 0.123,
Solving (1) and (2) by trial and error on polymath we get,
0.000948
!
=
(kg of catalyst)-1
a = 0.000101 dm-3
Now solve for a fluidized bed with 8000 kg of catalyst.
P4-30 (b)
For a PBR:
( )
( )
1
1
a X
dX y
dW X
"
=+
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
so decreasing the particle diameter has a larger effect on alpha and will increase pressure drop resulting in a
lower conversion.
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
Rate law: rA = -KPB
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
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 code with pressure drop equation:
See Polymath program P4-31-b.pol.
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
4-56
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(X)/d(w) = -ra/Fao
[2] d(y)/d(w) = -alfa*(1 - esp*X)/(2*y)
Explicit equations as entered by the user
P4-31 (c)
1) For laminar flow:
Diameter of pipe = D and diameter of particle = DP
2) For turbulent flow:
β ∞ G2/DP
4-57
P4-32 (a)
At equilibrium, r = 0 =>
C
DC
BA K
CC
CC =
V = V0 + vot
!
( ) ( )
C
AO
AO
O
O
BOAO K
XC
XCt
V
v
CXC
2
1=
!
!
"
#
$
$
%
&''
!
( ) !
"
#
$
%
&+
'
=X
XK
X
Cv
CV
t
CBOO
AOO
1
2
Now solving in polymath,
See Polymath program P4-32-a.pol.
P4-32 (b)
See Polymath program P4-32-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 1.5E+04 1.5E+04
Ca 7.72 0.2074331 7.72 0.2074331
X 0 0 0.9731304 0.9731304
4-58
ODE Report (RKF45)
Differential equations as entered by the user
Explicit equations as entered by the user
[1] Kc = 1.08
Polymath solution
P4-32 (c)
P4-32 (d)
As ethanol evaporates as fast as it forms: CD = 0
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
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)
Explicit equations as entered by the user
[1] k = 0.00009
P4-32 (e) Individualized solution
P4-32 (f) Individualized solution
P4-33 (a)
Mole balance on reactor 1:
01 1
1
2
AA A
A
CC dC
r
dt
! !
" " =
Mole balance on reactor 2:
Mole balance for reactor 3 is similar to reactor 2:
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:
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
