CHE 420
CHEMICAL REACTION ENGINEERING
Depnag, Chrysler Kane
Fernandez, Ramyll
Laroco, Chester John
Macalino, Ryan
Minong, Hugh
Perez, Angelika
Ticbayan, Krisha Ann
Turalva, Vanessa
March 25, 2019
1:30-4:00 MF
Submitted To: Joseph Marie Kitma, BSCHE, MSME
1. A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in a series. The
concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit
stream of the second reactor. The reaction is second-order with respect to A and V2/V1 =2.
GIVEN:
FA0=1mol/L
V2/V1 =2
CA1 =0.5 mol/L
n=2
SOLUTION:
REQUIRED: CA2
 C A1  C A2 


2
V2 k 2  C A 2 


V1 k 1  C A0  C A1 


 C 2 
A1


 0. 5  C A 2 


 C 2 
A2

2 
 1  0.5 

2 
 0. 5 
CA1= 0.25mol/L
2. Water containing a short-lived radioactive species flows continuously through a well-mixed
holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now
operates, the activity of the exit stream is 117 of the feed stream. This is not bad, but we'd like to
lower it still more. One of our office secretaries suggests that we insert a baffle down the middle
of the tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this
would help? If not, tell why; if so, calculate the expected activity of the exit stream compared to
the entering stream.
It should help, as dividing the volume of one CSTR into two CSTR’s operating in series
will result in a higher overall conversion for any positive order reaction
Radioactive decay typically follows a first order kientics:
- dN = λdt
Therefore, the activity of the stream should be lower if a baffle is used to split the
holding tank in two.
3. An aqueous reactant stream (4 mol A/liter) passes through a mixed flow reactor followed by a
plug flow reactor. Find the concentration at the exit of the plug flow reactor if in the mixed flow
reactor C, = 1 mol/liter. The reaction is second-order with respect to A, and the volume of the
plug flow unit is three times that of the mixed flow unit.
REQUIRED: CA2
SOLUTION:
For second order reaction
For mixed flow reactor:
For plug Flow reactor:
And
Therefore,
4. Reactant A (A -> R, C,, = 26 mol/m3)p asses in steady flow through four equal-size mixed
flow reactors in series (Ttotal = 2 min). When steady state is achieved the concentration of A is
found to be 11, 5, 2, 1 mol/m3 in the four units. For this reaction, what must be τPlug so as to
reduce CA from CA0 = 26 to CAf = 1 mol/m3?
REQUIRED: τPlug
SOLUTION:
For 1st CSTR:
2nd CSTR:
3rd CSTR:
4th CSTR:
FOR PFR:
Xa= (26-1)/26 = 0.9615
5. Originally we had planned to lower the activity of a gas stream containing radioactive Xe-138
(half-life = 14 min) by having it pass through two holdup tanks in series, both well mixed and of
such size that the mean residence time of gas is 2 weeks in each tank. It has been suggested that
we replace the two tanks with a long tube (assume plug flow). What must be the size of this tube
compared to the two original stirred tanks, and
what should be the mean residence time of gas in this tube for the same extent of radioactive
decay?
GIVEN:t1/2= 14 min
ti= 2 weeks= 20160 min
REQUIRED:V of PFR
Ʈ, mean residence time
SOLUTION:
Radioactive decay follows first order kinetics:
k=
Ʈ=
Ʈ =4.65 hr (instead of 2 weeks in MFR)
6. At 100°C pure gaseous A reacts away with stoichiometry 2A + R + S in a constant volume
batch reactor as follows:
t, sec
0
20
40
60
80
100
120
140
160
PA, atm 1.00
0.96
0.80
0.56
0.32
0.18
0.08
0.04
0.02
What size of a plug flow reactor operating at 100°C and 1 atm can treat 100 moles A/hr in a feed
consisting of 20% inerts to obtain 95% conversion of A?
GIVEN:
Batch reactor:
T= 100°C
Plug flow reactor:
T= 100°C
P= 1 atm
FAO= 100 mol/hr
XA= 0.95
With 20% inerts
REQUIRED: V of PFR
SOLUTION:
20% inerts
t
0
20
40
60
80
100
120
140
160
1
0.96
0.80
0.56
0.32
0.18
0.08
0.04
0.02
1
0.96
0.80
0.56
0.32
0.18
0.08
0.04
0.02
ln(p/p0)
0
0.040822
0.223144
0.579818
1.139434
1.714798
2.525729
3.218876
3.912023
Graph is approximately a straight line hence we can consider it as a first order system
For plug flow reactor:
7.) Aqueous feed containing reactant A (CAO= 2 mol/liter) enters a plug flow reactor (10 liter)
which has a provision for recycling a portion of the flowing stream. The reaction kinetics and
stoichiometry are,
A
R
-rA= 1CACR mol/liter·min
and we wish to get 96% conversion. Should we use the recycle stream? If so, at what value
should we set the recycle flow rate so as to obtain the highest production rate, and what
volumetric feed rate can we process to this conversion in the reactor?
8. From steady-state kinetics runs in a mixed flow reactor, we obtain the following data on the reaction A
R.
Ʈ, sec
60
35
11
20
11
CAO, mmol/liter
50
100
100
200
200
CA, mmol/liter
20
40
60
80
100
Find the space time needed to treat a feed of CAO= 100 mmol/liter to 80% conversion
(a) in a plug flow reactor
(b) in a mixed flow reactor
GIVEN:
CAO= 100 mmol/liter= 0.1 mol/ liter
XA= 0.8
CA= CAO (1-XA) = 0.1 (1-0.8) = 0.02 mol/liter
REQUIRED:
a.) Ʈ for PFR
b.) Ʈ for MFR
SOLUTION:
a.) The holding time required for plug flow reactor for 80% conversion is
From the given data, calculate 1/Ʈ, sec
60
35
11
20
11
CAO, mmol/liter
50
100
100
200
200
CA, mmol/liter
20
40
60
80
100
1/-rA
2
0.58
0.275
0.166
0.11
Plotting the graph between 1/-rA and CA, the area under this curve from concentration of
0.02 to 0.1 will give holding time in plug flow reactor:
We calculate the area using Excel tools and we ger 18.9 which is holding time for plug flow
reactor.
a.) Ʈ= 18.9 sec
b.) The holding time for mixed flow reactor for 80% conversion is
Ʈ=
It is the area of a rectangle:
With 1/-rA from 2 sec/mmol-liter to 0.11 sec/mmol-liter
Or 2000 sec/mol/liter to 110 sec/mol/liter
And with CAO 0.1 mol/liter to CA 0.02 mol/liter, we get
Ʈ = (2000-110)sec/mol/liter x (0.1-0.02)mol/liter
b.) Ʈ = 151.2 sec
9. A first order liquid-phase reaction, 92% conversion, is taking place in a mixed flow reactor. It
has been suggested that a fraction of the product stream, with no additional treatment, be
recycled. If the feed rate remains unchanged, in what way would this affect conversion?
GIVEN:
Liquid-phase reaction, therefore ƐA= 0
XA= 0.92
REQUIRED: How is the conversion affected?
SOLUTION:
The purpose of recycling is to increase back mixing in the reactor but here we are already
using a mixed flow reactor then it does not alter the conversion by recycling stream, and the
reaction here is a first order which gives the same conversion in the plug and mixed flow reactor
for same volume reactor.
10. At present the elementary liquid-phase reaction A+B
R+ S takes place in a plug flow
reactor using equimolar quantities of A and B. Conversion is 96%, CAO=CBO= 1 mol/liter. If a
mixed flow reactor ten times as large as the plug flow reactor were hooked up in series with the
existing unit, which unit should come first and by what fraction could the production be
increased for that set-up?
GIVEN: A+B
R+ S
Rate expression of reaction:
-rA= kCACB
CAO= CBO = 1 mol/liter
SOLUTION:
Performance equation for plug flow reactor is
ƮP= CAO
XA1= 0.96
Volume of CSTR is ten times of PFR thus,
ƮM= 10ƮP
Performance equation for CSTR is
CA1= CAO (1-XA1) =1(1-0.96) = 0.04
0.04 x10 x 24 =
XA2= 0.727
CAf= CA1(1-XA1) =0.04 (1-0.727) =0.011
Final conversion, XA = 0.989
Fraction of production increase = (0.989-0.96/0.96)x 100% = 3.02%
The plug flow reactor comes first before the mixed flow reactor and with this set-up
production is increased by 3.02%.
11. Using a color indicator which shows when the concentration of A falls below 0.1 mol/liter,
the following scheme is devised to explore the kinetics of the decomposition of A. A feed of 0.6
mol A/liter is introduced into the first of the two mixed flow reactors in series, each having a
volume of 400 cm3. The color change occurs in the first reactor for a steady-state feed rate of 10
cm3/min, and in the second reactor for a steady-state feed rate of 50 cm3/min. Find the equation
for the decomposition of A from this information.
GIVEN: CA1= 0.1 mol/L
V of each MFR= 400 cm3
FA1= 0.6 mol/L
1. (vo/V)(Ca_(N-1)-Ca_N)=kaCa^n
2.
For case #2 Ca_1 = 0.2 mol/L
3. For case #2 perform performance
equation again but over the first CSTR.
Which yields 0.05=ka(0.2)^n
Need to find ka and n
Use performance equation of CSTR
Substitute values for the two different flow
rate cases
#1 vo=0.01 C_a1=0.1 (perform over first
CSTR)
#2 vo=0.05 c_a2=0.1 (perform over second
CSTR)
For case #2 C_a1 is unknown but can be
found by dividing case #1 by #2
Substitution of Ca_1 back into performance
equation for case #2 yields 0.0125=ka(0.1)^n
2 equations and 2 unknowns (ka and n) can
noq be solved for to give n=2 and ka=1.25.
RESULT
-Ra=1.25Ca^2
12 At present 2/3 for an elementary second order liquid phase reactions 2A 2R, when
operating in an isothermal plug flow reactor with a recycle ratio of unity. What will be the
conversion if the recycle stream is shut off?
Solution:
13. We wish to explore various reactor setups for the transformation of A into R. The feed
contains 99% A, 1% R; the desired product is to consist of 10% A, 90% R. The transformation
takes place by means of the elementary reaction
A+R→R+R
with rate constant k = 1 liter/mol . min. The concentration of active materials is
=
1 mol/L
throughout. What reactor holding time will yield a product in which C, = 0.9 moll liter (a) in a
plug flow reactor, (b) in a mixed flow reactor, and (c) in a minimum-size setup without recycle?
GIVEN: A+R→R+R
=
1 mol/L
k = 1 mol/L
Feed contains 99% A and 1% B due to elementary reaction rate
=
REQUIRED: (a)
(b)
(c)
+
SOLUTION: (a) Holding time for PFR,
=
We get the area under the curve from x axis point 0.1 to 0.9 is 6.8. So holding time is
=
(b) Now for CSTR,
and (0.1, 11.11)
= 6.8 min
is area of rectangle formed by points (0.1,0), (0.9,0), (0.9, 11.11)
= (11.11)(0.9-0.1)= 8.8 min
(c) For the best system are under the curve should we use the minimum, it will happen
when we operate PFR followed by CSTR. Holding time in this case is 4.2 min, based
from the figure shown below.
14. Reactant A decomposes with stoichiometry A-R and with rate dependent only on C,. The following
data on this aqueous decomposition are obtained in a mixed flow reactor:
τ, sec
14
25
29
30
29
27
24
200
190
180
170
160
150
140
100
90
80
70
60
50
40
19
15
12
20
130
120
110
101
30
20
10
1
Determine which setup, plug flow, mixed flow, or any two-reactor combination gives
minimum τ for 90% conversion of a feed consisting of
. Also find this T
minimum. If a two reactor scheme is found to be optimum, give C, between stages and T
for each stage.
GIVEN: AA→R
= 100 mol/L
REQUIRED: Which gives minimum τ?
SOLUTION:
=
(1= 100(1-0.9) =10 mol/L
Now we have checked in which reactor system required holding time. The best way to
find out holding time is plot curve between
and
.
From the given data of mixed flow reactor, we calculate value of
.
From CSTR mole balance we get rate expression is
Here we assume ɛ=0 and plot
, the area under this curve from
concentration 10 to 100 will give holding time in PFR
We calculated area under the curve by using excel tools, we get area under the curve from
x axis point from 10 to 100 is 22.64. Thus holding time for PFR is 22.64 min.
Now for CSTR holding time will area of rectangle by points 910,0), 9100,0), (100,0.14),
and (10,0.14)
Ʈm = 90.14 x 90 = 12.6 min
15. For an irreversible first-order liquid-phase reaction (CAO = 10 mol/liter) conversion
is 90% in a plug flow reactor. If two-thirds of the stream leaving the reactor is recycled to
the reactor entrance, and if the throughput to the whole reactor-recycle system is kept
unchanged, what does this do to the concentration of reactant leaving the system?
GIVEN: R= 2
= 10 mol/L
REQUIRED:
SOLUTION:
=
(1-
) = 10(1-0.9) = 1 mol/L
Performance equation without recycle is
kτ = (R=1)ln(
2.3 = (2+1)ln(
16) At room temperature the second-order irreversible liquid-phase reaction proceeds as
follows:
A batch reactor takes 18 min to fill and empty. What percent conversion and reaction time
should we use so as to maximize the daily output of product R?
GIVEN:
At room temperature the second-order irreversible liquid-phase reaction proceeds as
follows:
REQUIRED: percent conversion and reaction time
SOLUTION:
Performance equation in batch reactor for second order is
Now for maximum conversion reaction time required will be infinite which is not
possible. Conversion at given time fill to empty of reactor is
=