2-1
Solutions for Chapter 2 – Conversion and Reactor
Sizing
Synopsis
General: The overall goal of these problems is to help the student realize that if they
have –rA=f(X) they can “design” or size a large number of reaction systems. It sets the
stage for the algorithm developed in Chapter 4.
P2-1. This problem will keep students thinking about writing down what they learned
every chapter.
P2-2. This “forces” the students to determine their learning style so they can better use
the resources in the text and on the CDROM and the web.
P2-3. ICMs have been found to motivate the students learning.
P2-4. Introduces one of the new concepts of the 4th edition whereby the students
“play” with the example problems before going on to other solutions.
P2-5. This is a reasonably challenging problem that reinforces Levenspiels plots.
P2-6. Novel application of Levenspiel plots from an article by Professor Alice Gast at
Massachusetts Institute of Technology in CEE.
P2-7. Straight forward problem alternative to problems 8, 9, and 12.
P2-9. The answer gives ridiculously large reactor volume. The point is to encourage
the student to question their numerical answers.
P2-10. Helps the students get a feel of real reactor sizes.
P2-11. Great motivating problem. Students remember this problem long after the
course is over.
P2-12. Alternative problem to P2-7 and P2-9.
CDP2-A Similar to 2-9
2-2
CDP2-C Similar to problems 2-8, 2-9, 2-12.
CDP2-D Similar to problems 2-8, 2-9, 2-12.
Assigned
Alternates
Difficulty
Time (min)
P2-1
O
15
P2-2
A
30
P2-4
O
75
P2-6
S
M
60
P2-8
S
FSF
45
P2-10
S
SF
15
P2-12
AA
7,8,9
SF
60
CDP2-A
O
9,B,C,D
FSF
5
CDP2-B
O
9,B,C,D
FSF
30
CDP2-C
O
9,B,C,D
FSF
30
Assigned
= Always assigned, AA = Always assign one from the group of alternates,
O = Often, I = Infrequently, S = Seldom, G = Graduate level
Alternates
In problems that have a dot in conjunction with AA means that one of the
problems, either the problem with a dot or any one of the alternates are always
assigned.
Time
Difficulty
SF = Straight forward reinforcement of principles (plug and chug)
FSF = Fairly straight forward (requires some manipulation of equations or an
intermediate calculation).
*Note the letter problems are found on the CD-ROM. For example A ≡ CDP1-A.
2-3
Summary Table Ch-2
Straight forward
1,2,3,4,10
Fairly straight forward
7,9,12
More difficult
5,6,8
P2-1 Individualized solution.
P2-2 Individualized solution.
P2-3 Solution is in the decoding algorithm given with the modules.
P2-4 (a) Example 2-1 through 2-3
If flow rate FAO is cut in half.
v1 = v/2 , F1= FAO/2 and CAO will remain same.
Therefore, volume of CSTR in example 2-3,
If the flow rate is doubled,
F2 = 2FAO and CAO will remain same,
Volume of CSTR in example 2-3,
V2 = F2X/-rA = 12.8 m3
P2-4 (b) Example 2-5
2-4
New Table: Divide each term
A
A
r
F
!
0
in Table 2-3 by 2.
X
0
0.1
0.2
0.4
0.6
0.7
0.8
[FAO/-rA](m3)
0.445
0.545
0.665
1.025
1.77
2.53
4
Reactor 1 Reactor 2
V1 = 0.82m3 V2 = 3.2 m3
V = (FAO/-rA)X
( )
182.0
1
0X
r
F
X
A
A
!
!
“
#
$
$
%
&
‘
=
( )
22.3
2
0X
r
F
X
A
A
!
!
“
#
$
$
%
&
‘
=
By trial and error we get:
X1 = 0.546 and X2 = 0.8
Overall conversion X0verall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673
P2-4 (c) Example 2-6
Now, FAO = 0.4/2 = 0.2 mol/s,
0201
202101
0
AA
AA
FF
XFXF
X
+
+
=
X
0
0.1
0.2
0.4
0.6
0.7
0.8
[FAO/-rA](m3)
0.445
0.545
0.665
1.025
1.77
2.53
4
V1 = 0.551m3 V2 = 1.614 m3
1
0
X
AO
A
dX
V F
r
=!
“
Plot FA0/-rA versus conversion. Estimate outlet conversions by computing the integral of the
plotted function.
2-5
P2-4 (d) Example 2-7
(1)
For PFR,
0.2
2
0
3
0.222
AO
A
F
V dX
r
m
! “
=# $
%
& ‘
=
(
For first CSTR,
X2 = 0.6,
3
1.32
AO
Fm
rA =
!
, V1 =
3
2 1
( ) .528
AO
F X X m
rA
!=
!
(2)
First CSTR remains unchanged
For PFR:
0.5
0.2
AO
A
F
V dX
r
! “
=# $
%
& ‘
(
,
Using the Levenspiel Plot
2-6
For CSTR,
(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the
smaller CSTR.
Conversion
Original Reactor Volumes
Worst Arrangement
X1 = 0.20
V1 = 0.188 (CSTR)
V1 = 0.23 (PFR)
X2 = 0.60
V2 = 0.38 (PFR)
V2 = 0.53 (CSTR)
For PFR,
X1 = 0.2
1
1
X
AO
F
V dX
r
! “
=# $
%
(
Using trapezoidal rule,
XO = 0.1, X1 = 0.1
( ) ( ) ( )
[ ]
1
1 1
3
3
0.2 1.28 0.98
2
0.23
O
O
X X
V f X f X
rA
m
m
!
= +
” #
$ %
!
= +
=
For CSTR,
For X2 = 0.6,
3
1.32
AO
Fm
rA =
!
, V2 =
( )
2 1
AO
FX X
r!
!
= 1.32(0.6 – 0.2) = 0.53 m3
P2-4 (e) Example 2-8
hrs5=
!
v0 = 1dm3/min = 60dm3/hr CA = 2.5 mol/dm3 X = 0.8
2-7
(1)
hrdmmol
hrdmmol
XC
rA
A
3
3
0
/4.0
/
5
8.05.2
=
!
==“
#
(2) V = 300dm3
P2-5
X
0
0.1
0.2
0.4
0.6
0.7
0.8
V = 1.0 m3
P2-5 (a) Two CSTRs in series
For first CSTR,
V = (FAo/-rAX1) X1
=> X1 = 0.44
For second CSTR,
P2-5 (b)
Two PFRs in series
1 2
1
X X
Ao Ao
F F
V dX dX
r r
! “ ! “
= +
# $ # $
% %
( (
By extrapolating and solving, we get
X1 = 0.50 X2 = 0.74
P2-5 (c)
Two CSTRs in parallel with the feed, FAO,
divided equally between two reactors.
P2-5 (d)
Two PFRs in parallel with the feed equally divided between the two reactors.
FANEW/-rAX1 = 0.5FAO/-rAX1
2-8
P2-5 (e)
A CSTR and a PFR are in parallel with flow equally divided
Since the flow is divided equally between the two reactors, the overall conversion is the average
P2-5 (f)
A PFR followed by a CSTR,
XPFR = 0.50 (using part(b))
V = (FAo/-rA-XCSTR) (XCSTR – XPFR)
P2-5 (g)
A CSTR followed by a PFR,
XCSTR = 0.44 (using part(a))
P2-5 (h)
A 1 m3 PFR followed by two 0.5 m3 CSTRs,
For PFR,
P2-6 (a) Individualized Solution
P2-6 (b)
1) In order to find the age of the baby hippo, we need to know the volume of the stomach.
The metabolic rate, -rA, is the same for mother and baby, so if the baby hippo eats one half of
2-9
Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of
the baby’s stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.
4.5 2.25
2
years
Age years= =
2) If Vmax and mao are both one half of the mother’s then
0
2 2
1
2mother
A
Ao
AM AM
m
m
r r
! “
# $
% &
=
‘ ‘
2-10
2 2
2
1
2
1
2
Ao
Ao Ao
AM AM mother
baby AM
mother
m
m m
r r
r
! “
! ” # $ ! “
= =
# $ # $ # $
# $
% %
& ‘
# $
& ‘ %
& ‘
2
Ao
AM
m
r!
will be identical for both the baby and mother.
Assuming that like the stomach the intestine volume is proportional to age then the volume of the
intestine would be 0.75 m3 and the final conversion would be 0.40
P2-6 (c)
Vstomach = 0.2 m3
From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:
For the intestine:
The Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178
Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot
survive.
2-11
P2-6 (d)
PFR CSTR
PFR:
Outlet conversion of PFR = 0.111
We must solve
Since the hippo gets a conversion over 30% it will survive.
P2-7
Exothermic reaction: A B + C
X
r(mol/dm3.min)
1/-
r(dm3.min/mol)
0
1
1
0.20
1.67
0.6
0.40
5
0.2
0.50
5
0.2
0.80
1.25
0.8
P2-7 (a)
To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as
2-12
Mole balance:
( )( )
( )
min./5
4.0min/300
F
V 3
A0
dmmol
mol
r
X
A
CSTR =
!
=
=>
=>VCSTR = 24 dm3
PFR:
Mole balance:
!“
=
X
A
APFR r
dX
FV
0
0
= 300(area under the curve)
VPFR = 72 dm3
P2-7 (b)
For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any
conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is
constant over this conversion range.
! ! “
=
“
=
“
=
6.
4.
6.
4.
0
6.
4.
00 X
r
F
dX
r
F
dX
r
F
V
A
A
A
A
A
A
PFR
P2-7 (c)
VCSTR = 105 dm3
Mole balance:
A
CSTR r
X
!
=A0
F
V
moldm
dm
X
min/35.0
105
3
3
==
2-13
P2-7 (d)
From part (a) we know that X1 = 0.40.
Use trial and error to find X2.
2
Rearranging, we get
008.0
40.0
0
2
2
==
!
!
A
X
AF
V
r
X
At X2 = 0.64,
008.0
40.0
2=
!
!
r
X
P2-7 (e)
From part (a), we know that X1 = 0.40. Use trial and error to find X2.
Mole balance:
!! “
=
“
==
22
030072
X
X
APFR r
dX
r
dX
FV
P2-7 (f)
See Polymath program P2-7-f.pol.
P2-8 (a)
S
S
r
XF
V
!
=0
FS0 = 1000 g/hr
P2-8 (b)
At a conversion of 80%,
hrdm
3
8.0
1=
2-15
P2-8 (d)
To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to
have a CSTR with a volume to achieve a conversion of about 45%, or the conversion that
corresponds to the minimum value of 1/-rs. Next is a PFR with the necessary volume to achieve
P2-8 (e)
SM
CS
sCK
CkC
r
+
=!
and
[ ] 001.01.0 0+!=SSC CCC
Let us first consider when CS is small.
CS0 is a constant and if we group together the constants and simplify
CskCk
K
rS
M
s2
2
1
1
+
=!
which is consistent with the shape of the graph when X is large (if CS is
small X is large and as CS grows X decreases).
Now consider when CS is large (X is small)
As CS gets larger CC approaches 0:
[ ] 001.01.0 0+!=SSC CCC
and
0SS CC !
2-16
And
CCS
S
skCCkC
C
r
11 ==!
P9-2
Irreversible gas phase reaction
P2-9 (a)
PFR volume necessary to achieve 50% conversion
Mole Balance
1)(
Volume = Geometric area under the curve of
(FA0/-rA) vs X)
( )
5.01000005.0400000
2
1!+
“
#
$
%
&
‘!!=V
V = 150000 m3
One of the points of this problem is for the students to
recognize 150,000 m3 is ridiculously reactor volume.
P2-9 (b)
CSTR Volume to achieve 50% conversion
Mole Balance
)(
0
A
A
r
XF
V
!
=
1000005.0 !=V
V = 50000m3
recognize 150,000 m3 is ridiculously reactor volume.
2-17
P2-9 (c)
Volume of second CSTR added in series to achieve 80%
conversion
)(
)( 120
2
A
A
r
XXF
V
!
!
=
)5.08.0(500000
2!“=V
V2 = 150000m3
P2-9 (d)
Volume of PFR added in series to first CSTR to achieve
80% conversion
P2-9 (e)
For CSTR,
V = 60000 m3 (CSTR)
Ridiculous volume
Mole Balance
r
XF
V
A
A
)(
0
!
=
For PFR,
V = 60000 m3 (PFR)
Mole balance
!“
=
X
A
Ar
dX
FV
0
0)(
dXX
X
)100000800000(60000
!+“=