2-18
P2-9 (f)
Real rates would not give that shape. The reactor volumes are absurdly large.
P2-10
Problem 2-10 involves estimating the volume of three reactors from a picture. The door on the
side of the building was used as a reference. It was assumed to be 8 ft high.
The following estimates were made:
CSTR
h = 56ft d = 9 ft
P2-11 No solution necessary.
P2-12 (a)
The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR
2-19
The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the
PBR. This figure shows that the smallest amount of catalyst is used when the CSTR is upstream
P2-12 (b)
Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by
determining the area of the shaded region in the figure below.
P2-12 (c)
The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating
the area of the shaded rectangle shown in the figure below.
2-20
P2-12 (d)
The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the
area of the shaded region in the figure below.
P2-12 (e)
The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from
2-21
P2-12 (f)
P2-12 (g)
For different (-rA) vs. (X) curves, reactors should be arranged so that the smallest amount of
catalyst is needed to give the maximum conversion. One useful heuristic is that for curves with a
negative slope, it is generally better to use a CSTR. Similarly, when a curve has a positive slope,
CDP2-A (a)
Over what range of conversions are the plug-flow reactor and CSTR volumes identical?
We first plot the inverse of the reaction rate versus conversion.
Mole balance equations for a CSTR and a PFR:
CSTR:
A
XF
V
=0
PFR:
!“
=
X
dX
V
Until the conversion (X) reaches 0.5, the reaction rate is independent of
conversion and the reactor volumes will be identical.
2-22
i.e.
CSTR
A
A
PFR V
XF
dX
F
dX
V=
=
“
=
“!
=0
5.0
0
5.0
CDP2-A (b)
What conversion will be achieved in a CSTR that has a volume of 90 L?
For now, we will assume that conversion (X) will be less that 0.5.
CSTR mole balance:
A
A
A
A
r
XCv
r
XF
V
!
=
!
=000
13
3
8
3
3
3
00
103
.
1032005
09.0 !
“=
“““
=
!
=
mol
sm
m
mol
s
m
m
r
Cv
V
X
A
A
CDP2-A (c)
This problem will be divided into two parts, as seen below:
The PFR volume required in reaching X=0.5 (reaction rate is independent of conversion).
2-23
Finally, we add V2 to V1 and get:
Vtot = V1 + V2 = 2.3 x1011 m3
CDP2-A (d)
What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a
CSTR to raise the conversion to 90 %
We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new
value into our CSTR mole balance equation:
311
000 104.1 m
r
XCv
r
XF
V
A
A
A
A
CSTR !=
“
#
=
“
#
=
CDP2-A (e)
If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to
the reactor, what length of time is necessary to achieve 40% conversion?
Since there is no flow into or out of the system, mole balance can be written as:
Mole Balance:
dN
Vr A
A=
2-24
Combine:
dX
NVr AA 0
=
From the stoichiometry of the reaction we know that V = Vo(1+eX) and e is 1. We insert
this into our mole balance equation and solve for time (t):
dt
dX
X
N
V
r
A
A=+!)1(
0
0
!+“
=
!
X
A
A
t
Xr
dX
Cdt
0
0
0)1(
After integration, we have:
)1ln(
1
0XC
r
tA
A
+
!
=
Inserting the values for our variables:
t = 2.02 x 1010 s
That is 640 years.
CDP2-A (f)
Plot the rate of reaction and conversion as a function of PFR volume.
The following graph plots the reaction rate (-rA) versus the PFR volume:
2-25
CDP2-A (g)
Critique the answers to this problem.
The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain
the desired conversion, it would require a reactor of geological proportions (a CSTR or PFR
approximately the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a
CDP2-B Individualized solution
CDP2-C (a)
For an intermediate conversion of 0.3, Figure below shows that a PFR yields the smallest volume,
since for the PFR we use the area under the curve. A minimum volume is also achieved by
following the PFR with a CSTR. In this case the area considered would be the rectangle bounded
2-26
CDP2-C (b)
CDP2-C (c)
CDP2-C (d)
2-27
2-28
CDP2-C (e)
CDP2-D
2-29
CDP2-D (a)
CDP2-D (b)
CDP2-D (c)
CDP2-D (d)
2-30
CDP2-D (e)
CDP2-D (g)
CDP2-D (h)
2-31
CDP2-E
CDP2-F (a)
Find the conversion for the CSTR and PFR connected in series.
X
-rA
1/(-rA)
0
0.2
5
0.1
0.0167
59.9
0.7
0.00286
349.65
2-32
CDP2-F (b)
2-33
CDP2-F (c)
CDP2-F (d)
2-34
CDP2-F (e)
CDP2-F (f)
CDP2-F (g) Individualized solution