4-1
Solutions for Chapter 4 – Isothermal Reactor Design
P4-1 Individualized solution.
P4-2 (a)
P4-2 (c) Example 4-2
For 50% conversion, X = 0.5 and k = 0.311min-1
P4-2 (d) Example 4-3
For P = 60atm,
P4-2 (e) Example 4-4
New Dp = 3D0/4
Because the flow is turbulent
4-2
P4-2 (f) Example 4-5
For without pressure drop, conversion will remain same as example X = 0.82.
With Pressure drop,
For turbulent flow:
“
#1
Dp
and
$
#1
P0
%
$
2=
$
1
P01Dp1
P02Dp2
(2) Optimum diameter would be larger
Now 1-αW < 0, too much pressure drop due to higher superficial velocity.
4-3
P4-2 (g) Example 4-6
For turbulent flow
“
#1
Dp
and
“
#1
P0
Therefore there is no change.
P4-2 (h) Exmple 4-7
For pressure doubled and temperature decrease
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
V 0 0 1.0E-04 1.0E-04
Fa 2.26E-04 1.363E-05 2.26E-04 1.363E-05
Fb 0 0 2.124E-04 2.124E-04
Fc 0 0 1.062E-04 1.062E-04
E 2.4E+04 2.4E+04 2.4E+04 2.4E+04
ODE Report (RKF45)
Differential equations as entered by the user
4-4
Explicit equations as entered by the user
[1] E = 24000
[2] T = 688
P4-2 (i) Example 4-8 Individualized solution.
P4-2 (j) Example 4-9
Using trial and error, we get maximum feed rate of B = 0.0251dm3/s to keep concentration of B
0.01mol/dm3.
See Polymath program P4-2-j.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 500 500
ca 0.05 0.0063485 0.05 0.0063485
cb 0 0 0.009981 0.009981
4-5
Differential equations as entered by the user
[1] d(ca)/d(t) = -k*ca*cb-v00*ca/v
Explicit equations as entered by the user
If the concentration of A is tripled the
maximum feed rate becomes
0.064 dm3/s
P4-2 (k through r) Individualized solution.
P4-3 Solution is in the decoding algorithm given with the modules.
P4-4
We have to find the time required to cook spaghetti in Cuzco, Peru.
Location
Elevation (km)
Pressure (mm Hg)
Boiling Point (°C)
Time (min)
Ann Arbor
0.21
739
99.2
15
Assume reaction is zero order with respect to spaghetti conversion:
E
A
RT
A
dC
r k Ae
dt
!
!= = = !
so that
4-6
Now, plot the natural log of the cooking time versus 1/Tb and get a linear relationship. Extrapolation to Tb =
88.3°C = 361.45 K yields t = 21 minutes.
P4-5 (a)
2 2
(1 )
AO
A
kC X
r
dX
dV F F
!
!
= =
P4-5 (b)
P4-5 (c)
P4-5 (d)
1) CSTR and PFR are connected in series:
4-8
3 2
(0.07 / / min) (1 )(1 )
10 / min
AO
dm mol C X X
dX dV
mole
! !
=
2) when CSTR and PFR are connected in parallel,
P4-5 (e)
To process the same amount of species A, the batch reactor must handle
3
5 60 min 24
2 14400
min
dm h mol
Mhr day day
! “ ! “
! ” =
# $ # $
# $
% &% &
% &
If the reactants are in the same concentrations as in the flow reactors, then
Now we find the time required to reach 90% conversion. Assume the reaction temperature is 300K.
( )
2
0
0 0
1
A
A
A A
kC X V
r V
dX
dt N N
!
!
= =
0
2
01
A
R
A
NX
t
VkC X
=!
, and since
0
0
A
A
NC
V=
Assume that it takes three hours to fill, empty, and heat to the reaction temperature.
tf = 3 hours
P4-5 (f)
The points of the problem are:
1) To note the significant differences in processing times at different temperatures (i.e.
P4-6 (a)
P4-6 (b)
To increase conversion, use PFR, higher temperature, or use better catalyst.
P4-6 (c)
P4-6 (d)
4-11
P4-6 (e)
P4-6 (f) Individualized solution
4-12
P4-7 (a)
Elementary gas phase reaction.
P4-7 (b)
P4-7 (c)
For α = 0.001dm-3
4-13
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
v 0 0 500 500
x 0 0 0.656431 0.656431
Differential equations as entered by the user
[1] d(x)/d(v) = –r/fo
[2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y)
Explicit equations as entered by the user
P4-7 (d) Individualized solution
P4-7 (e)
A ↔ B + 2C
4-14
Using these equations in Polymath we get the volume to be 290 dm3.
P4-7 (f)
4-15
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
X 0 0 0.47 0.47
V 0 0 290.23883 290.23883
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(V)/d(X) = Fao/(-ra)
Explicit equations as entered by the user
[1] Kc = .025
PFR with pressure drop: Alter the Polymath equations from part (c).
See Polymath program P4-7-f-pressure.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
v 0 0 500 500
x 0 0 0.5077714 0.5077714
4-16
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(x)/d(v) = –r/fo
[2] d(y)/d(v) = -alfa*(1+esp*x)/(2*y)
Explicit equations as entered by the user
P4-7 (g)
Membrane reactor: A → B + 2C
CA = COFA/FT CB = COFB/FT CC = COFC/FT
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
v 0 0 1040 1040
Fa 2.5 1.3231889 2.5 1.3231889
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra
[2] d(Fb)/d(v) = –ra – kc*Co*Fb/Ft
[3] d(Fc)/d(v) = -2*ra
Explicit equations as entered by the user
[1] Kc = 0.025
4-17
P4-8 (a)
The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal.
P4-8 (b)
A CSTR is been created at the bend due to backmixing, so the effective arrangement is a PFR is in series
with a CSTR.
A → B
k = 5 min-1 vo = 5 dm3/min.
Xexpected = 0.632 Xactual = 0.586
4-19
Also,
1
C
PV
V
V V
+ =
………………………………………..3
Solving 1, 2 and 3 by using polymath,
See Polymath program P4-8-b.pol.
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
X1 0.350949 3.148E-10 0
NLES Report (safenewt)
Nonlinear equations
Explicit equations
[1] V = 1
P4-8 (c)
CAO = 2 mol/dm3
A
!
B
Assuming 1st order reaction,
For CSTR,
AO
A
C X
r
!
=“
Now assuming 2nd order reaction,
For CSTR, Now, assuming 2nd order reaction,
For CSTR,
AO
A
C X
r
!
=“
4-20
P4-8 (d)
A graph between conversion and particle size is as follows: Originally we are at point A in graph, when
P4-9
A
!
B
To = 300K KCO (300K)= 3.0 V = 1000gal = 3785.4 dm3
Mole balance:
A
AO
r
XF
V
!
=