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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.
P2-5. 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.
P2-11. Great motivating problem. Students remember this problem long after the
course is over.
2-2
CDP2-C Similar to problems 2-8, 2-9, 2-12.
CDP2-D Similar to problems 2-8, 2-9, 2-12.
Summary
Assigned
Alternates
Difficulty
Time (min)
P2-1
O
15
P2-2
A
30
P2-9
AA
7,9,12
SF
45
P2-10
S
SF
15
P2-11
AA
SF
1
P2-12
AA
7,8,9
SF
60
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
Time
Approximate time in minutes it would take a B/B+ student to solve the problem.
Difficulty
2-3
Summary Table Ch-2
Straight forward
1,2,3,4,10
Fairly straight forward
7,9,12
P2-1 Individualized solution.
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.
P2-4 (b) Example 2-5
New Table: Divide each term
A
A
r
F
!
0
in Table 2-3 by 2.
Reactor 1 Reactor 2
P2-4 (c) Example 2-6
Now, FAO = 0.4/2 = 0.2 mol/s,
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
X1 = 0.603 for V1 = 0.551m3 X2 = 0.89 for V2 =
1.614m3
P2-4 (d) Example 2-7
(1)
For PFR,
0.2
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:
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
X3 = 0.65
V3 = 0.10 (CSTR)
V3 = 0.10 (CSTR)
For PFR,
X1 = 0.2
For CSTR,
For X2 = 0.6,
3
1.32
AO
Fm
, V2 =
( )
2 1
AO
FX X
= 1.32(0.6 – 0.2) = 0.53 m3
rA =
!
P2-4 (e) Example 2-8
hrs5=
!
(1)
hrdmmol
XC
rA
A
3
0
/
5
8.05.2
!
=="
#
= 0.5 mol/dm3
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,
P2-5 (b)
Two PFRs in series
P2-5 (c)
Two CSTRs in parallel with the feed, FAO,
P2-5 (d)
Two PFRs in parallel with the feed equally divided between the two reactors.
2-8
P2-5 (e)
A CSTR and a PFR are in parallel with flow equally divided
P2-5 (f)
A PFR followed by a CSTR,
P2-5 (g)
A CSTR followed by a PFR,
P2-5 (h)
A 1 m3 PFR followed by two 0.5 m3 CSTRs,
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.
2-9
2
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
1
2
Ao
Ao Ao
m
m m
! "
! " # $ ! "
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:
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
CSTR:
We must solve
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
P2-7 (a)
To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as
given below.
CSTR:
2-12
Mole balance:
( )( )
( )
min./5
4.0min/300
F
V 3
A0
dmmol
mol
r
X
A
CSTR =
!
=
=>
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
P2-7 (c)
VCSTR = 105 dm3
Mole balance:
A
CSTR r
X
!
=A0
F
V
2-13
P2-7 (d)
From part (a) we know that X1 = 0.40.
Use trial and error to find X2.
Conversion = 0.64
P2-7 (e)
P2-7 (f)
See Polymath program P2-7-f.pol.
P2-8 (a)
S
S
r
XF
V
!
=0
P2-8 (b)
At a conversion of 80%,
g
hrdm
rS
3
8.0
1=
!
P2-8 (c)
!"
=
X
S
SPFR r
dX
FV
0
0
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
P2-8 (e)
SM
CS
sCK
CkC
r
+
=!
and
[ ] 001.01.0 0+!=SSC CCC
then
CskCk
CK
rS
SM
s2
2
1
1
+
+
=!
since CS < KM
P9-2
Irreversible gas phase reaction
2A + B 2C
See Polymath program P2-9.pol.
P2-9 (a)
PFR volume necessary to achieve 50% conversion
Mole Balance
P2-9 (b)
CSTR Volume to achieve 50% conversion
Mole Balance
P2-9 (c)
Volume of second CSTR added in series to achieve 80%
conversion
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)
For PFR,
V = 60000 m3 (PFR)
Mole balance
