Lab PFR
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Transcript of Lab PFR
ABSTRACT
From this experiment, our objectives are to carry out the saponification reaction
between NaOH and Et(Ac) in plug flow reactor, to determined the reaction rate constant and
the rate of reaction of the saponification process. First of all, the equipment is set up before
started the experiment. After collecting the data, the value of reaction rate constant and rate
of reaction is calculated. The reaction rate constant we get for 600ml/min flowrate is
17.25L/mol.min, for the 500mL/min reaction rate constant is 17.44L/mol.min, for the
400mL/min reaction rate constant is 15.93L/mol.min, for the 300mL/min reaction rate
constant is 18.32L/mol.min, for the 200mL/min reaction rate constant is 25.58L/mol.min and
for the 100mL/min reaction rate constant is 34.10L/mol.min. Besides that, we are also able to
determine the rate of reaction for this process. The rate of reaction we get for flowrate of
600ml/min is 0.0373mol/L.min, for the 500mL/min the rate of reaction is 0.0304mol/L.min,
for the 400mL/min the rate of reaction is 0.0237mol/L.min, for the 300mL/min the rate of
reaction is 0.0155mol/L.min, for the 200mL/min the rate of reaction is 0.0068mol/L.min and
for the 100mL/min the rate of reaction is 0.0016mol/L.min. Then, a graph of conversion
factor against residence time is plotted. From the graph we can see that the conversion factor
is directly proportional to the residence time. As the residence time increases, the conversion
factor also increases.
1
INTRODUCTION
In addition to the Continue Stir Tank Reactor (CSTR) and batch reactors, another type of
reactor commonly used in industry is the tubular flow reactor. It consists of a cylindrical pipe
and is normally operated at steady state, as is the CSTR [1]. Tubular flow reactors are usually
used for gas phase-reactions. A schematic of industrial tubular reactors are shown in figure
below:
Figure 1: Tubular reactor schematic. Longitudal flow reactor.Excerpeted by special permission from Chem. Eng., 63(10),
211(Oct.1956). Copyright 1956 by McGraw-Hill, Inc., New York
In the tubular reactor, the reactants are continually consumed as they flow down the length of
the reactor. In the ideal tubular reactor, which is called the “plug flow” reactor, specific
assumptions are made about the extent of mixing:
1. no mixing in the axial direction, i.e., the direction of flow
2. complete mixing in the radial direction
3. a uniform velocity profile across the radius.
Plug flow reactor is an ideal tubular reactor with laminar flow behaviour. The reactants pass
through the tube; the reactants are converted progressively along the length of reactor. The
reactants are continuously consumed and the product is formed as the flow preceded the
length of the reactor. There is no radial variation in concentration. Consequently, the reaction
rate, which is function of concentration for all but zero-order reactions, will also vary axially.
Plug flow-no radial variations in velocity,concentrations, temperature, or reaction rate
2
Figure 2: Plug flow reactor
The validity of the assumptions will depend on the geometry of the reactor and the flow
conditions. Deviations, which are frequent but not always important, are of two kinds:
1. mixing in longitudinal direction due to vortices and turbulence
2. incomplete mixing in radial direction in laminar flow conditions
Flow in tubular reactor can be laminar, as with viscous fluids in small-diameter tubes, and
greatly deviate from ideal plug-flow behaviour, or turbulent, as with gases. Turbulent flow
generally is preferred to laminar flow, because mixing and heat transfer are improved. For
slow reactions and especially in small laboratory and pilot-plant reactors, establishing
turbulent flow can result in conveniently long reactors or may require unacceptable high feed
rates.
For most chemical reactions, it is impossible for the reaction to proceed to 100% completion.
The rate of reaction decreases as the percent completion increases until the point where the
system reaches dynamic equilibrium (no net reaction, or change in chemical species occurs).
The equilibrium point for most systems is less than 100% complete. For this reason a
separation process, such as distillation, often follows a chemical reactor in order to separate
any remaining reagents or by products from the desired product. These reagents may
sometimes be reused at the beginning of the process, such as in the Haber process.
Tubular flow reactors are usually used for this application which are:
1. Large scale reactions
2. Fast reactions
3. Homogeneous or heterogeneous reactions
4. Continuous production
5. High temperature reactions
3
OBJECTIVES
The objectives of this experiment are:
1. To carry out the saponification reaction between NaOH and Et(Ac) in tubular flow
reactor.
2. To determine the reaction rate constant.
3. To determine the effect of residence time on the conversion in the tubular flow
reactor.
THEORY
In a plug flow reactor, the feed enters at one end of a cylindrical tube and the product stream
leaves at the other end. The long tube and the lack of provision for stirring prevent complete
mixing of the fluid in the tube. Hence the properties of the flowing stream will vary from one
point to another, namely in both radial and axial directions.
The rate expression can be shown to be
-rA = k [A] [B]
Where if [A] is equal to [B], this simplify to
-rA = k [A]2
In the general case the order of the reaction η is not known and is shown by
-rA = k [A]η
If the inlet concentration, [A] is known, k can be determined. The reaction:
NaOH + CH3COOC2H5 → CH3COONa + C2H5OH
Sodium Hydroxide + Ethyl Acetate → Sodium Acetate +
Ethyl Alcohol
can be considered equal-molar and first order with respect to both sodium hydroxide and
ethyl acetate i.e. second order overall, within the limits of the concentration (0-0.1M) and
temperature (20-40oC) studied.
4
The steady state conditions will vary depending on concentration of reagents, flowrate,
volume of reactor and temperature of reaction. For this experiment, we will let the flowrate as
the vary component.
For a time element Δt and a volume element ΔV, the mass balance for species ‘i’ at the
reactor is given by the following equation:
QA CA │v Δt- QA CA│v+Δv Δt - rAΔVΔt = 0
where QA : molar feed rate of reactant A to the reactor, mol/sec
CA : concentration of reactantA
rA : rate of disappearance of reactant A, mol/lt•sec
The conversion, X, is defined as:
X = (initial concentration - final concentration) / (initial concentration)
Since the system is at steady state, the accumulation term in Equation (3.1) is zero.
Equation (3.1) can be written as:
-QA ΔCA - rAΔV = 0
Dividing by ΔV and taking limit as ΔV → 0
dCA/dV = -rA/QA
This is the relationship between concentration and size of reactor for the plug flow reactor.
Here rate is a variable, but varies with longitudinal position (volume in the reactor, rather
than with time). Integrating,
-dV/ QA = dCA/rA
At the entrance: V = 0
CA = CA0
At the exit: V = VR (total reactor volume)
CA = CA (exist conversion)
−V RQA
=∫C AO
C A dCAr A
5
The new expression can be shown to be
-rA = k CA2(1-X)2
-rA = FAO dX/dV = voCAO dX/dV
V TFR=vokCAO
∫0
XdX
(1−X )2 =vokC AO
( X1−X )
For constant Tubular Flow Reactor volume, flow rate and initial concentrations, the reaction
rate constant is:
k=v0
V TFRCAO ( X1−X )
Where k = reaction constant rate
vo = total inlet flow rate of solutions (ml/min)
CAO = inlet concentration of reactant NaOH in the reactor (M)
VTFR = volume for reactor (ml)
X = conversion
For a tubular reactor, the mass balance for a reactant a is represented by
∫0
X dC A
−r A= VV o
Arranging for initial concentrations of CA and CB and integrate the equation
∫0
X dC A
kCA2 =τ
1k∫0
X dC A
C A2 =τ
1k [−1CA ]0
X
=τ
Solving the integration will give you this equation
kτ= X1−X
6
Where, τ is the residence time.
7
APPARATUS
1. Heater
2. Water bath
3. Coiled reaction tube
4. Agitator motor
5. Variable-area flow meter
6. Switch box
7. Steel tube support
8. Collecting tank
Figure 3: Plug flow reactor
8
PROCEDURES
SECTION A: SET UP THE EQUIPMENT
1. All valves were ensured closed at initial except for valves V4, V8 and V17.
2. The following solutions were prepared:
a. 20 liter of sodium hydroxide, NaOH (0.1M)
b. 20 liter of ethyl acetate, Et(Ac)(0.1M)
c. 1 liter of hydrochloric acid, HCl(0.25M), for quenching.
3. The both feed tank B1 and tank B2 was filled with the NaOH solution and with the
Et(Ac) solution.
4. The water jacket B4 and pre-heater B5 were filled with clean water.
5. The power was turned on at the control panel.
6. Valves V2, V4, V6, V8, V9 and V11 were opened.
7. Both pumps P1 and P2 were switched on. P1 and P2 were adjusted to obtain flow of
approximately 300ml/min at both flow meters Fl-01 and Fl-02. Make sure both flow rates
are the same.
8. Both solutions were allowed to flow through the reactor R1 and overflow into the waste
tank B3.
9. Valves V13 and V18 were opened on pump P3 to circulate the water through pre-heater
B5. The stirrer motor M1 was switched on and the speed was set to about 200 rpm to
ensure homogeneous water jacket temperature.
SECTION B: START THE EXPERIMENT
1. Procedures for section A were performed.
2. Valves V9 and V11 were opened.
3. Both NaOH and Et(Ac) solutions was allowed to enter the plug reactor R1 and empty into
the waste tank B3.
4. Both pumps P1 and P2 were switched on. P1 and P2 were adjusted to obtain flow of
approximately 300ml/min at both flow meters Fl-01 and Fl-02. Make sure both flow rates
are the same. The flow rates were record.
5. The inlet (Ql-01) and outlet (Ql-02) conductivity values were monitored until they do not
change over time. This is to ensure that the reactor has reached steady state.
9
6. Both inlet and outlet steady state conductivity values were recorded. The concentration of
NaOH exiting the reactor and extent of conversion were found from the calibration curve.
7. Sampling valve V15 was opened and a 50ml sample was collected. A back titration
procedure was carried out manually and the concentration NaOH in the reactor and the
extent of conversion were determined.
8. Steps 4 until 7 was repeated for different values of feed flow rates of NaOH and Et(Ac).
Make sure that both flow rates are the same.
10
RESULTS
Experiment 1
Conversion Solution Mixture Concentration Conductivity(%) 0.1M NaOH (mL) 0.1M Na(Ac) (mL) Water (mL) Of NaOH (M) (mS/cm)
0 100 0 100 0.0500 6.94
25 75 25 100 0.0375 5.79
50 50 50 100 0.0250 4.64
75 25 75 100 0.0125 3.57
100 0 100 100 0.0000 2.58
0 20 40 60 80 100 1200
1
2
3
4
5
6
7
8
f(x) = − 0.04376 x + 6.892R² = 0.9989850190142
Conductivity vs Conversion
Conductivity (mS/cm)Linear (Conductivity (mS/cm))
Conversion (%)
Cond
uctiv
ity (m
S/cm
)
11
Experiment 2
Reactor volume, VTFR = 0.4 L
Concentration of NaOH in feed tank, CAO = 0.1 M
Concentration Et(Ac) in feed tank, CBO = 0.1 M
No
Flow
rate of
NaOH
(ml/min)
Flow
rate of
Et(Ac)
(ml/min
)
Total
flow rate
of
solutions
(ml/min)
Residenc
e time, τ
(min)
Inlet
conductivity
(mS/cm)
Outlet
conductivity
(mS/cm)
Conversio
n X (%)
Reaction
rate
constant
(L/mol.min)
Rate of
reaction,
(mol/L.min)
1 300 300 600 0.667 7.6 5.8 99.9 1485 1.485 x 10-3
2 250 250 500 7.6 5.7
3 200 200 400 7.0 5.6
4 150 150 300 6.8 5.0
5 100 100 200 6.4 4.6
6 50 50 100 5.8 4.2
12
Residence time, τ (min)X
1−X
0.67 1.13
0.80 1.38
1.00 1.56
1.33 2.45
2.00 5.25
4.00 13.29
0 20 40 60 80 100 1200
1
2
3
4
5
6
7
8
f(x) = − 0.04376 x + 6.892R² = 0.9989850190142
Conductivity vs Conversion
Conductivity (mS/cm)Linear (Conductivity (mS/cm))
Conversion (%)
Cond
uctiv
ity (m
S/cm
)
13
0 2 4 6 8 10 120
2
4
6
8
10
12
f(x) = NaN x + NaNR² = 0 Graph X/(1-X) against residence time
Graph X/(1-X) against residence timeLinear (Graph X/(1-X) against residence time)
residence time
X/(1
-X)
Graph 1: Graph X/(1-X) against residence time
14
SAMPLE CALCULATION Experiment 1: Slope = -0.043x Y-intercept = 6.892 Y = -0.043x + 6.892 Experiment 2: To find conversion of NaOH in the reactor
For volume of titrated NaOH, V1 = 0.0185L 1. Conc. Of NaOH entering the reactor, CNaOH,0,
Where concentration of NaOH in the feed vessel, CNaOH, f = 0.1 M CNaOH,0 = (CNaOH,f) / 2 = 0.1 M / 2 = 0.05 M 2. Volume of unreacted quenching HCl, V2,
Where concentration of NaOH used for titration, CNaOH,s = 0.1 M; Concentration of HCl in standars solution, CHCl,s = 0.25 M; Volume of titrated NaOH, V1 = 0.0185L.
V2 = (CNaOH,s / CHCl,s ) x V1
= (0.1M / 0.25 M) x 0.0185 L = 7.40 x 10-3 L
15
3. Volume of HCl reacted with NaOH in sample, V3 Where volume of HCl for quenching, VHCl,s = 0.01 L
V3 = VHCl,s – V2 = 0.01 L – 7.40 x 10-3 L = 2.6 x 10-3 4. Moles of HCl reacted with NaOH in sample, n1
n1 = CHCl,s x V3 = 0.25 M x (2.6 x 10-3 L) = 6.5 x 10-4 mol 5. Moles of unreacted NaOH in sample n2 = n1 = 6.5 x 10-4 mol
6. Conc. Of unreacted NaOH in the reactor, CNaOH Where: volume of sample, Vs = 0.05 L
CNaOH = n2 / Vs = (6.5 x 10-4 mol) / 0.05 L = 3.25 x 10-5M 7. Conversion of NaOH in the reactor, X
X = (1 – (CNaOH / CNaOH,0 )) x 100% = (1– (3.25 x 10-5M / 0.05 M)) x 100% = 99.9 %
16
To find residence time Where reactor volume, VTFR = 0.4 L and total feed flowrates, V0 = 0.593 L/min Residence time, = VTFR / V0 = 0.4 / 0.6 = 0.667 min To find reaction rate constant K = V0 (X) VPTRC (1-X) = 0.6 . 0.99 0.4 x 0.1 1- 0.99 = 1485 L/mol.min To find rate of reaction -rA = kCA02 ( 1 – X )2
= 1485 ( 0.1 ) 2 x ( 1 – 0.99 )2 = 1.485 x 10-3 mol / L.min
17
SAMPLE OF CALCULATION
Reactor volume, υo = 0.4L Conversion, X = 53.49%
Flowrate of NaOH = 300mL/min Inlet conductivity = 9.3mS/cm
Flowrate of Et(Ac) = 300mL/min Outlet conductivity = 7.5mS/cm
Concentration of NaOH in feed tank = 0.1M
Concentration of Et(Ac) in feed tank = 0.1M
Residence time,
= V 0
= (0.4)/(0.6)
18
= 0.67min
Reaction rate constant
k=v0
V TFRCAO ( X1−X )
k = 0.6/ (0.4 x 0.1) [0.5349/(1 – 0.5349)]
= 17.25 L/mol.min
Rate of reaction
-rA = kC2AO(1-X)2
= (17.25) (0.1)2(1 – 0.5349)
= 0.0373 mol/L.min
19
DISCUSSIONS
By doing this experiment, we are able to carry out the saponification reaction between NaOH
and Et(Ac) in tubular flow reactor. At the end of the experiment, we are also able to
determine the reaction rate constant by using the formula and to determine the effect of
residence time on the conversion in the tubular flow reactor.
The experiment is started by running up the equipment in order to start the saponification
process. From Figure 3, the coiled reaction tube is where the saponification process to occur.
The saponification process can be done in two ways whether variation in temperature or
variation in contact time. In this experiment, we will let the flowrate of both solutions as the
varying components because the flowrate of both solutions is controlled by the temperature
of the reactor. At the end of the experiment, the saponification process is successfully done.
After that, we are needed to determine the reaction rate constant and the rate of the reaction
for the saponification process depends on the vary flowrate of both solution sodium
hydroxide and ethyl acetate.
NaOH + CH3COOC2H5 → CH3COONa + C2H5OH
Sodium Hydroxide + Ethyl Acetate → Sodium Acetate +
Ethyl Alcohol
The overall reaction order for the saponification process is second ordered, the reaction rate
constant can be determined by applied the equations below, where
-rA = k CA2(1-X)2
-rA = FAO dX/dV = voCAO dX/dV
V TFR=vokCAO
∫0
XdX
(1−X )2 =vokC AO
( X1−X )
For constant plug flow reactor volume, flow rate and initial concentrations, the reaction rate
constant is:
k=v0
V TFRCAO ( X1−X )
20
The reaction rate constant we get for flowrate of 600ml/min is 17.25L/mol.min, for the
500mL/min reaction rate constant is 17.44L/mol.min, for the 400mL/min reaction rate
constant is 15.93L/mol.min, for the 300mL/min reaction rate constant is 18.32L/mol.min, for
the 200mL/min reaction rate constant is 25.58L/mol.min and for the 100mL/min reaction rate
constant is 34.10L/mol.min. From the reaction rate constant we determined, we can see that
the value is increase as the flowrate is decrease. Only for 400mL/min we can see that the
value of reaction rate constant is decrease that is from 17.44L/mol.min to 15.93L/mol.min.
This is maybe due to the same value of the flowrate and the volume of the tank reactor. When
the volume and the flowrate is the same, this will give effect to the value of reaction rate
constant as the both are equipped in the formula in calculate the reaction rate constant.
The rate of reaction also can be determined after we had done find the reaction rate constant.
The rate of reaction we get for 600ml/min flowrate is 0.0373mol/L.min, for the 500mL/min
the rate of reaction is 0.0304mol/L.min, for the 400mL/min the rate of reaction is
0.0237mol/L.min, for the 300mL/min the rate of reaction is 0.0155mol/L.min, for the
200mL/min the rate of reaction is 0.0068mol/L.min and for the 100mL/min the rate of
reaction is 0.0016mol/L.min. After all value of rate of reactions has been calculated, a graph
of conversion factor against residence time is plotted. From the graph that had been plotted,
we can say that the conversion factor is directly proportional to the residence time. Where,
when the residence time increases, the conversion factor also increases.
Although we had also done the titration in this experiment, but the data collected in the
titration is quite useless because we doesn’t use the data for any calculation. It is just like
doing it without any reasons. But we are still thankful because we can improve our titration
skill by doing this experiment.
21
CONCLUSION
From this experiment, we are able to carry out the saponification reaction between NaOH and
Et(Ac) in tubular flow reactor. We are also able to determine the reaction rate constant for the
saponification process. The reaction rate constant we get for flowrate of 600ml/min is
17.25L/mol.min, for the 500mL/min reaction rate constant is 17.44L/mol.min, for the
400mL/min reaction rate constant is 15.93L/mol.min, for the 300mL/min reaction rate
constant is 18.32L/mol.min, for the 200mL/min reaction rate constant is 25.58L/mol.min and
for the 100mL/min reaction rate constant is 34.10L/mol.min. Besides that, we are also able to
determine the rate of reaction for this process. The rate of reaction we get for 600ml/min
flowrate is 0.0373mol/L.min, for the 500mL/min the rate of reaction is 0.0304mol/L.min, for
the 400mL/min the rate of reaction is 0.0237mol/L.min, for the 300mL/min the rate of
reaction is 0.0155mol/L.min, for the 200mL/min the rate of reaction is 0.0068mol/L.min and
for the 100mL/min the rate of reaction is 0.0016mol/L.min. A graph of conversion factor
against residence time is plotted. From the graph that had been plotted, we can say that the
conversion factor is directly proportional to the residence time. Where, when the residence
time increases, the conversion factor also increases. This experiment was a success.
22
RECOMMENDATIONS
There are several recommendations that can be taken in order to get more accurate result that are:
1. Before carry out the experiment, please consult with technician on how to run the
equipment so that you can save your time and energy while doing the experiment.
2. It is recommended that this experiment should be repeated at various other
temperatures to investigate the relationship between the reaction rate constant and the
rate of reaction.
3. It is further recommended that the experiment be repeated using dissimilar flow rates
for the NaOH solution and ethyl acetate solutions to investigate the effect that this will
have upon the saponification process.
4. For obtained more accurate results, run several trials on tubular flow reactor so we can
take the average value from each different molar rates.
5. Be careful when doing the titration because we only want the last drop of NaOH that
will convert the solution to light pale purple colour. The excess of drop of NaOH will
give effect on the result in the calculations.
23
REFERENCES
1. Fogler, H.S (2006). Elements of Chemical Reaction Engineering, 4 th Edition, New
Jersey:Prentice Hall
2. Perry, R. H., D. Green, ‘Perry’s Chemical Engineer’s Handbook’, 6 th edition,
McGraw-Hill, 1988.
3. Instruction Manual Turbular Flow Reactor, Jan2006,
http://eleceng.dit.ie/gavin/DT275/CET%20MKII%20manual%20issue%2016.pdf at
8.00pm at 14 Feb 2011
4. http://en.wikipedia.org/wiki/Plug_flow_reactor_model at 8.30pm on 14 Feb 20115. http://www.che.boun.edu.tr/courses/che302/Chapter%2010.pdf at 9.45pm on 14 Feb
2011
24
Figure 4
25
Figure 5: Structure of reactor in plug flow reactor
Figure 6: Example graph conversion against residence time
26