Solutions.
Problem 1.
a) For first-order reaction kinetics
10000 L
Da
kτ
0.3L / sec =0.58
XA =
=
=
1 + Da 1 + kτ 1 + 2.5 × 10-3 min −1× 10000 L
0.3L / sec
2.5 × 10-3 min −1×
b) Consider the non-steady state design equation for CSTR, where we can have
FA0 − FA + rAV =
dN A
dt
Since in liquid phase with constant density, we have
C A0 − C A - kC Aτ = τ
dC A
dt
or equivalently
dCA 1 + kτ
C
CA = A0
+
dt
τ
τ
with the initial conditions CA,t=0=(1-0.58)×0.12 mol/L=0.05 mol/L, new τ=10000L/(0.7×0.3*60
L/min)=793.7 min
Therefore, integrate this equation we can have
C A (t ) =
C A0
1 + kτ
1 + kτ
{1 − exp[−
t ]} + C A, t = 0 exp(−
t)
1 + kτ
τ
τ
Therefore the conversion at time t is
X A (t ) = 1 −
C
1 + kτ
1
1 + kτ
t ]} − A, t = 0 exp[−
t]
{1 − exp[−
τ
τ
C A0
1 + kτ
After 60 min of changing flow rate,
X A (60 min) = 1 −
C
1 + kτ
1
1 + kτ
t ]} − A, t = 0 exp[−
t ] = 0.598
{1 − exp[−
τ
τ
C A0
1 + kτ
The new steady state fractional conversion is
10000 L
2.5 × 10-3 min −1×
Da
kτ
0.3L / sec× 0.7 =0.665
XA =
=
=
1 + Da 1 + kτ 1 + 2.5 × 10-3 min −1× 10000 L
0.3L / sec× 0.7
c) The steady state productivity (moles/time) of B is FB, and the ratio of that in b) vs. that in a) is
B
FB, b 0.3L/sec × 0.7 × 0.665
=
= 0.803
FB, a
0.3L/sec × 0.58
Therefore, the productivity will be decreased by lowering the flow rate even though a higher
conversion is to be achieved.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Problem 2.
We know for first order reaction, conversion XA has the following relation with rate constant k,
CSTR reactor volume V and volumetric flow rate v
V
v
XA =
V
1+ k
v
k
When we have a dead volume (denoted as subscript 1) which does not interact with the input and
output streams, we can deduce this amount of volume from the overall reactor volume.
V1
v
X A1 =
V
1+ k 1
v
k
We
are
told
V1=600L-400L=200L,
XA1=0.75,
therefore
we
can
calculate
X A1
k
==
=0.015.
v
V1 (1 - X A1 )
Then for the well-stirred reactor, V=600L
V
v =0.9
XA =
V
1+ k
v
k
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
Problem 3.
For a PFR reactor, the design equation is
dX celluose
-r
= S celluose
dz
Fcelluose, in
Now S=10cm2, length=2m, Fcelluose, in=vin[cellulose]in, in order to know v, we have to convert 60
grams of slurry into total moles of cellulose and water per second, that is
n total, in =
=
60 gram/sec × 20% 60 gram/sec × (1 − 20%)
+
=
MWcellulose
MWwater
60 gram/sec × 20% 60 gram/sec × (1 − 20%)
= 2.683 mol/sec
+
720 gram/mol
18 gram/mol
Using ideal gas law to get the volumetric flow rate:
v in =
n total, in RT 2.683 mol/sec × 8.314 J/mol/K × 600K
=
= 6.61 liter/sec
20 atm
P
Also
20%
y
P
20 atm
720 gram/mol
[cellulose]in = cellulose,0 =
= 2.52 mol/m3
20
%
1
−
20
%
RT
(
+
) 8.314J/(mol K) × 600K
720gram/mol 18gram/mol
where ycellulose,0 is the molar fraction of cellulose in the inlet.
In this problem, since the gas phase reaction creates more molecules, the volumetric flow rate is
not a constant.
- rcelluose = k
[cellulose]
1 + a[CO]
where the concentrations should be expressed (constant pressure and temperature are assumed)
[cellulose] = [cellulose]in
(1 - X)
(1 + εX)
[CO] = [H 2 ] = [cellulose]in
24X
(1 + εX)
and
Here ε=(24+24-1)ycellulose,0=47*0.0062=0.292.
Now the final molar flow rate of H2 is
vfinal[H2]=24vin[cellulose]inXf
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].