Full
-
Upload
eja-rotikeju -
Category
Documents
-
view
31 -
download
4
Transcript of Full
Continuous Stirred Tank Reactor 40L CHE574
ABSTRACT/SUMMARY
On 29th March 2012, our group consists of four members have conducted an experiment on
Continuous Stirrer Tank Reactor (CSTR) 40 L for the purpose of completing the syllabus of
the Chemical Engineering Laboratory subject. In this experiment, we wanted to observe and
know more details about Continuous Stirred Tank Reactor (CSTR) 40 L especially about its
operation and to carry out the saponification reaction between NaOH and Et (Ac) in CSTR.
This experiment was carried out to determine the effect of the residence time onto the reaction
extent of conversion and also to determine the reaction rate constant. Before the experiment
started, we ensured the general start up procedure was carried out properly so that the the
apparatus will run without any disturbance. After the methodology of start up process was
done, we began the experiment by opening valves V5 and V10 to obtain the highest possible
flow rate into the reactor. For this experiment, the feed flow rates were adjusted in increasing
order. For each flow rate, 50 mL sample was taken to be used in the back titration procedure.
In back titration, the samples were titrated with NaOH for saponification reaction. The
amounts of NaOH titrated were recorded. A few calculations had been conducted to
determine the conversion, residence time and concentration of NaOH. Two graphs were
plotted where the slope of conductivity v's conversion graph was defined to be -0.1233. While
for the second graph, the descending order of the 50th minutes to 33.33th minutes showed that
the objectives were successfully achieved. More details can be viewed in the result.
1
Continuous Stirred Tank Reactor 40L CHE574
INTRODUCTION
In a continuous-flow stirred-tank reactor (CSTR), reactants and products are continuously
added and withdrawn. In practice, mechanical or hydraulic agitation is required to achieve
uniform composition and temperature, a choice strongly influenced by process considerations.
The CSTR is the idealized opposite of the well-stirred batch and tubular plug-flow reactors.
Analysis of selected combinations of these reactor types can be useful in quantitatively
evaluating more complex gas-, liquid-, and solid-flow behaviors.
Because the compositions of mixtures leaving a CSTR are those within the reactor, the
reaction driving forces, usually the reactant concentrations, are necessarily low. Therefore,
except for reaction orders zero- and negative, a CSTR requires the largest volume of the
reactor types to obtain desired conversions. However, the low driving force makes possible
better control of rapid exothermic and endothermic reactions. When high conversions of
reactants are needed, several CSTRs in series can be used. Equally good results can be
obtained by dividing a single vessel into compartments while minimizing back-mixing and
short-circuiting. The larger the number of CSTR stages, the closer the performance
approaches that of a tubular plug-flow reactor.
The continuous stirred tank reactor (model BP 143) unit is suitable for student experiments on
continuous chemical reactions. The unit consists of jacketed reaction fitted in the agitator and
condenser. The unit comes complete with vessels for raw materials and product, feed pumps,
and thermostat.
2
Continuous Stirred Tank Reactor 40L CHE574
OBJECTIVE
To verify the conductivity values by manual determination on experiment sample
To carry out saponification reaction between NaOH and Et(Ac) in a CSTR
To determine the effect of residence time onto the reaction extent of conversion
To determine the reaction rate constant
3
Continuous Stirred Tank Reactor 40L CHE574
THEORY
Rate of Equation and Rate Law
The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a
particular reaction is intuitively defined as how fast or slow a reaction takes place. For
example, the oxidation of iron under the atmosphere is a slow reaction that can take many
years, but the combustion of butane in a fire is a reaction that takes place in fractions of a
second.
Consider a typical chemical reaction:
aA + bB → pP + qQ (eq 1.1)
The lowercase letters (a, b, p, and q) represent stoichiometric coefficients, while the capital
letters represent the reactants (A and B) and the products (P and Q).
According to IUPAC's Gold Book definition the reaction rate r for a chemical reaction
occurring in a closed system under constant-volume conditions, without a build-up of reaction
intermediates, is defined as:
where [X] denotes the concentration(Molarity, mol/L) of the substance X. (NOTE: Rate of a
reaction is always positive. '-' sign is present in the reactant involving terms because the
reactant concentration is decreasing.) The IUPAC recommends that the unit of time should
always be the second. In such a case the rate of reaction differs from the rate of increase of
concentration of a product P by a constant factor (the reciprocal of its stoichiometric number)
and for a reactant A by minus the reciprocal of the stoichiometric number. Reaction rate
usually has the units of mol L−1 s−1. It is important to bear in mind that the previous definition
4
Continuous Stirred Tank Reactor 40L CHE574
is only valid for a single reaction, in a closed system of constant volume. This most usually
implicit assumption must be stated explicitly, otherwise the definition is incorrect: If water is
added to a pot containing salty water, the concentration of salt decreases, although there is no
chemical reaction.
For any system in general the full mass balance must be taken into account:
IN - OUT + GENERATION -CONSUMPTION= ACCUMULATION
When applied to the severe case stated previously this equation reduces to:
For a chemical reaction n A + m B → C + D, the rate equation or rate law is a mathematical
expression used in chemical kinetics to link the rate of a reaction to the concentration of each
reactant. It is of the kind:
In this equation k(T) is the reaction rate coefficient or rate constant, although it is not really a
constant, because it includes all the parameters that affect reaction rate, except for
concentration, which is explicitly taken into account. Of all the parameters described before,
temperature is normally the most important one.
The exponents n and m are called reaction orders and depend on the reaction mechanism.
Stoichiometry, molecularity (the actual number of molecules colliding), and reaction
order coincide necessarily only in elementary reactions, that is, those reactions that take place
5
Continuous Stirred Tank Reactor 40L CHE574
in just one step. The reaction equation for elementary reactions coincides with the process
taking place at the atomic level, i.e. n molecules of type A colliding with m molecules of type
B (n plus m is the molecularity).
For gases, the rate law can also be expressed in pressure units using, e.g., the ideal gas law.
By combining the rate law with a mass balance for the system in which the reaction occurs, an
expression for the rate of change in concentration can be derived. For a closed system with
constant volume, such an expression can look like
Conversion
Using the equation shown in Equation 1.1 and taking species A as the basis of calculation, the
reaction expression can be divided through by the stoichiometric coefficient of species A, in
order to arrange the reaction expression in the form:
aA + baB →
pa P +
qaQ
The expression has now put every quantity on a ‘per mole of A basis’.
A convenient way to quantity how far the reaction has progressed, or how many moles of
products are formed for every mole of A consumed; is to define a parameter called
conversion. The conversion XA is the number of moles of A that have reacted per mole of A
fed to the system.
X A=molesof A reacted
molesof A fed
To perform a mole balance on any system, the system boundaries must first be specified. The
volume enclosed by these boundaries is referred to as the system volume. We shall perform a
6
Continuous Stirred Tank Reactor 40L CHE574
mole balance on species j in a system volume, where species j represents the particular
chemical species of interest, such as water or NaOH.
Mole balance on species j in a system volume, V
A mole balance on species j at any instant in time, t, yields the following equation:
where Nj represents the number of moles of species j in the system at time t. If all the system
variables (e.g., temperature, catalytic activity, and concentration of the chemical species) are
7
Continuous Stirred Tank Reactor 40L CHE574
spatially uniform throughout the system volume, the rate of generation of species j, Gj, is just
the product of the reaction volume, V, and the rate of formation of species j, rj.
G j=r j ∙V
molestime
= molestime ∙ volume
∙ volume
Now suppose that the rate of formation of species j for the reaction varies with position in the
system volume. That is, it has a value rj1 at location 1, which is surrounded by a small volume,
ΔV1, within which the rate is uniform: similarly, the reaction rate has a value rj2 at location 2
and an associated volume, ΔV2, and so on.
Dividing up the system volume, V
The rate of generation, ΔGj1, in terms of rj1 and subvolume ΔV1, is
ΔGj1 = rj1 ΔV1
Similar expressions can be written for ΔGj2 and the other system subvolumes, ΔVi. The total
rate of generation within the system volume is the sum of all the rates of generation in each of
the subvolumes. If the total system volume is divided into M subvolumes, the total rate of
generation is
8
Continuous Stirred Tank Reactor 40L CHE574
G j=∑i=1
M
∆ G ji=∑i=1
M
r ji ∆ V i
By taking the appropriate limits (i.e., let M → ∞ and ΔV → 0) and making use of the
definition of an integral, we can rewrite the foregoing equation in the form
G j=∫0
V
r j dV
From this equation we see that rj will be an indirect function of position, since the properties
of the reacting materials and reaction conditions (e.g., concentration, temperature) can have
different values at different locations in the reactor volume.
Replacing Gj,
F j 0−F j+G j=dN j
dt
by its integral form to yield a form of the general mole balance equation for any chemical
species j that is entering, leaving, reacting, and/or accumulating within any system volume V.
F j 0−F j+∫0
V
r j dV=dN j
dt
This is a basic equation for chemical reaction engineering.
From this general mole balance equation, we can develop the design equations for the various
types of industrial reactors: batch, semibatch, and continuous-flow. Upon evaluation of these
equations, we can determine the time (batch) or reactor volume (continuous-flow) necessary
to convert a specified amount of the reactants into products.
9
Continuous Stirred Tank Reactor 40L CHE574
Continuous Stirred Tank Reactors (CSTR)
CSTR runs at steady state with continuous flow of reactants and products; the feed assumes a
uniform composition throughout the reactor, exit stream has the same composition as in the
tank
Kinds of Phases
Present
Usage Advantages Disadvantages
1. Liquid phase
2. Gas-liquid rxns
1. When agitation is
required
2. Series
1. Continuous operation
2. Good temperature
1. Lowest
conversion per
10
Continuous Stirred Tank Reactor 40L CHE574
3. Solid-liquid rxns configurations for
different
concentration
streams
control
3. Easily adapts to two
phase runs
4. Good control
5. Simplicity of
construction
6 Low operating (labor)
cost
7. Easy to clean
unit volume
2. By-passing and
channeling
possible with poor
agitation
General Mole Balance Equation
F A 0−F A+∫0
V
r A dV=dN A
dt
Assumptions
Steady state, therefore dN A
dt=0
11
Continuous Stirred Tank Reactor 40L CHE574
Well-mixed therefore r A is the same throughout the reactor
∫0
V
r A dV=r A∫0
V
dV=r A V
Rearranging the generation,
V=F A 0−F A
−r A
In terms of conversion
X=F A 0−FA
F A 0
V=F A 0 X−r A
APPARATUS
12
Continuous Stirred Tank Reactor 40L CHE574
Continuous stirred tank reaction (Model: BP143)
0.1M sodium hydroxide, NaOH
0.1M sodium acetate, Et(Ac)
0.25 hydrochloric acid, HCl
De-ionized water,H2O
Burette
Retort stand
Conical flask
pH indicator
Measuring cylinder
PROCEDURE
13
Continuous Stirred Tank Reactor 40L CHE574
General Start-up Procedure
1. The following solutions were prepared:
i. 40 L of sodium hydroxide, NaOH (0.1M)
ii. 40 L of ethyl acetate, Et (Ac) (0.1M)
iii. 1 L of hydrochloric acid, HCl (0.25M), for quenching
2. All valves were initially closed.
3. The feed vessels were charged as follows:
i. The charge port caps for vessels B1 and B2 were opened.
ii. The NaOH solution was carefully poured into vessel B1, and the Et (Ac)
solution was poured into vessel B2.
iii. The charge port caps for both vessels were closed.
4. The power for the control panel was turned on.
5. Sufficient water in thermostat T1 was checked. Refill as necessary.
6. The overflow tube was adjusted to give a working volume of 10 L in the reactor R1.
7. Valves V2, V3, V7, V8 and V11 were opened.
8. The unit was ready for experiment.
General shutdown procedure.
1. The cooling water valve V13 was kept open to allow the cooling water to continue
flowing.
2. Pumps P1 and pump P2 were switched off. Stirrer M1 was switched off.
3. The thermostat T1 was switched off. The liquid in the reaction vessel R1 was let to
cool down to room temperature.
4. Cooling water valve V13 was closed.
5. Valves V2, V3, V7 and V8 were closed. Valves V4, V9 and V12 were opened to drain
any liquid from the unit.
6. The power for control panel was turned off.
14
Continuous Stirred Tank Reactor 40L CHE574
Preparation of Calibration Curve for Conversion vs. Conductivity
1. The following solutions were prepared:
i. 1 L of sodium hydroxide, NaOH (0.10M)
ii. 1 L of sodium acetate, Et (Ac) (0.10M)
iii. 1 L of deionised water, H20
2. The conductivity and NaOH concentration for each value were determined by mixing
the following solutions into 100 mL of deionised water:
i. 0% conversion : 100 mL NaOH
ii. 25% conversion : 75 mL NaOH + 25 mL Et (Ac)
iii
.
50% conversion : 50 mL NaOH + 50 mL Et (Ac)
iv
.
75% conversion : 25 mL NaOH + 75 mL Et (Ac)
v. 100% conversion : 100 mL Et (Ac)
Back Titration Procedures for Manual Conversion Determination
1. A burette was filled up with 0.1 M NaOH solution.
2. 10 mL of 0.25 M HCl was measured in a flask.
3. A 50 mL sample was obtained from the experiment and immediate the sample was
added to the HCl in the flask to quench the saponification reaction.
4. A few drops of pH indicator were added into the mixture.
5. The mixture was titrated with NaOH solution from the burette until the mixture was
neutralized. The amount of NaOH titrated was recorded.
15
Continuous Stirred Tank Reactor 40L CHE574
16
Continuous Stirred Tank Reactor 40L CHE574
17
Continuous Stirred Tank Reactor 40L CHE574
0 20 40 60 80 100 1200
2
4
6
8
10
12
1412.22
7.9
3.52
1.2760.116
f(x) = − 0.123328 x + 11.1728
Conductivity vs Concentration
Conversion of NaOH (%)
cond
uctiv
ity (m
s/cm
)
Graph 1:Conductivity vs Concentration
18
Continuous Stirred Tank Reactor 40L CHE574
20 30 40 50 60 70 80 90 100 1100
20
40
60
80
100
120
79.286
98.892.4
82
Conversion vs Residence Time
residence time (min)
conv
ersio
n, X
(%)
Graph 2:conversion vs residence time
19
Continuous Stirred Tank Reactor 40L CHE574
CALCULATION
1) F0=0.1 L/min
Known quantities
Volume of sample,V s = 50 mLConcentration of NaOH in the feed vessel,CNaOH f
= 0.1 mol /LVolume of HCl for quenching,V HCls
= 10 mLConcentration of HCl in standard solution,CHCl s
= 0.25 mol /LVolume of titrated NaOH,V 1 = 30.2 mLConcentration of NaOH used for titration,CNaOH s
= 0.1 mol /L
i. Concentration of NaOH entering the reactor , CNaOH 0
CNaOH 0= 1
2CNaOH f
= 12(0.1)
= 0.05 mol /L
ii. Volume of unreacted quenching HCl , V 2
V 2 = CNaOH s
C HCls
× V 1
= 0.10.25
×30.2
= 12.08 mL
iii. Volume of HCl reacted with NaOH in sample , V 3
V 3 = V HCls−V 2
= 10−12.08= −2.08
20
Continuous Stirred Tank Reactor 40L CHE574
iv. Moles of HCl reacted with NaOH in sample , n1
n1 = (CHCl s×V 3 ) /1000
= 0.25 ×2.08 /1000= 0.00052 mol
v. Moles of unreacted NaOH in sample , n2
n2 = n1
= 0.00052 mol
vi. Concentration of unreacted NaOH in the reactor , CNaOH
CNaOH = n2/V s×1000= 0.00052/50 ×1000= 0.0104 mol / L
vii. Conversion of NaOH in the reactor , X
X = (1−CNaOH
CNaOH 0)×100 %
= (1−0.01040.05 )×100 %
= 79.2 %
viii. Residence time , τ
τ = V CSTR
F0
= 100.10
= 100 min
ix. Reaction rate constant , k
k = (CA 0−C A )
τ C A2
21
Continuous Stirred Tank Reactor 40L CHE574
= (0.05−0.0104 )100 ×0.01042
= 3.66 M−1 min−1
x. Rate of reaction , −r A
−r A = kC A2
= 3.66 ×0.01042
= 3.96 ×10−4 mol /L . min
2) F0=0.15 L/min
Known quantities
Volume of sample,V s = 50 mLConcentration of NaOH in the feed vessel,CNaOH f
= 0.1 mol /LVolume of HCl for quenching,V HCls
= 10 mLConcentration of HCl in standard solution,CHCl s
= 0.25 mol /LVolume of titrated NaOH,V 1 = 28.5 mLConcentration of NaOH used for titration,CNaOH s
= 0.1 mol /L
i. Concentration of NaOH entering the reactor , CNaOH 0
CNaOH 0= 1
2CNaOH f
= 12(0.1)
= 0.05 mol /L
ii. Volume of unreacted quenching HCl , V 2
V 2 = CNaOH s
C HCls
× V 1
= 0.10.25
× 28.5
22
Continuous Stirred Tank Reactor 40L CHE574
= 11.40mL
iii. Volume of HCl reacted with NaOH in sample , V 3
V 3 = V HCls−V 2
= 10−11.40= −1.40 mL
iv. Moles of HCl reacted with NaOH in sample , n1
n1 = (CHCl s×V 3 ) /1000
= 0.25 ×1.40 /1000= 0.00035 mol
v. Moles of unreacted NaOH in sample , n2
n2 = n1
= 0.00035 mol
vi. Concentration of unreacted NaOH in the reactor , CNaOH
CNaOH = n2/V s×1000= 0.00035/50 ×1000= 0.007 mol / L
vii. Conversion of NaOH in the reactor , X
X = (1−CNaOH
CNaOH 0)×100 %
= (1−0.0070.05 )× 100 %
= 86 %
viii. Residence time , τ
τ = V CSTR
F0
23
Continuous Stirred Tank Reactor 40L CHE574
= 100.15
= 66.67 min
ix. Reaction rate constant , k
k = (CA 0−C A )
τ C A2
= (0.05−0.007 )100 ×0.0072
= 8.78 M−1 min−1
x. Rate of reaction , −r A
−r A = kC A2
= 8.78 ×0.0072
= 4.30 × 10−4 mol / L. min
3) F0=0.20 L/min
Known quantities
Volume of sample,V s = 50 mLConcentration of NaOH in the feed vessel,CNaOH f
= 0.1 mol /LVolume of HCl for quenching,V HCls
= 10 mLConcentration of HCl in standard solution,CHCl s
= 0.25 mol /LVolume of titrated NaOH,V 1 = 25.3 mLConcentration of NaOH used for titration,CNaOH s
= 0.1 mol /L
i. Concentration of NaOH entering the reactor , CNaOH 0
CNaOH 0= 1
2CNaOH f
= 12(0.1)
24
Continuous Stirred Tank Reactor 40L CHE574
= 0.05 mol /L
ii. Volume of unreacted quenching HCl , V 2
V 2 = CNaOH s
C HCls
× V 1
= 0.10.25
× 25.3
= 10.12 mL
iii. Volume of HCl reacted with NaOH in sample , V 3
V 3 = V HCls−V 2
= 10−10.12= −0.12 mL
iv. Moles of HCl reacted with NaOH in sample , n1
n1 = (CHCl s×V 3 ) /1000
= 0.25 ×0.12 /1000= 0.00003 mol
v. Moles of unreacted NaOH in sample , n2
n2 = n1
= 0.00003 mol
vi. Concentration of unreacted NaOH in the reactor , CNaOH
CNaOH = n2/V s×1000= 0.00003/50 ×1000= 0.0006 mol / L
vii. Conversion of NaOH in the reactor , X
X = (1−CNaOH
CNaOH 0)×100 %
25
Continuous Stirred Tank Reactor 40L CHE574
= (1−0.00060.05 )× 100 %
= 98.8 %
viii. Residence time , τ
τ = V CSTR
F0
= 100.20
= 50 min
ix. Reaction rate constant , k
k = (CA 0−C A )
τ C A2
= (0.05−0.0006 )50 ×0.00062
= 2744 M−1 min−1
x. Rate of reaction , −r A
−r A = kC A2
= 2744× 0.00062
= 9.88 ×10−4 mol /L . min
4) F0=0.25 L/min
Known quantities
Volume of sample,V s = 50 mLConcentration of NaOH in the feed vessel,CNaOH f
= 0.1 mol /LVolume of HCl for quenching,V HCls
= 10 mLConcentration of HCl in standard solution,CHCl s
= 0.25 mol /LVolume of titrated NaOH,V 1 = 23.1 mL
26
Continuous Stirred Tank Reactor 40L CHE574
Concentration of NaOH used for titration,CNaOH s= 0.1 mol /L
i. Concentration of NaOH entering the reactor , CNaOH 0
CNaOH 0= 1
2CNaOH f
= 12(0.1)
= 0.05 mol /L
ii. Volume of unreacted quenching HCl , V 2
V 2 = CNaOH s
C HCls
× V 1
= 0.10.25
× 23.1
= 9.24 mL
iii. Volume of HCl reacted with NaOH in sample , V 3
V 3 = V HCls−V 2
= 10−9.24= 0.76 mL
iv. Moles of HCl reacted with NaOH in sample , n1
n1 = (CHCl s×V 3 ) /1000
= 0.25 ×0.76 /1000= 0.00019 mol
v. Moles of unreacted NaOH in sample , n2
n2 = n1
= 0.00019 mol
vi. Concentration of unreacted NaOH in the reactor , CNaOH
CNaOH = n2/V s×1000
27
Continuous Stirred Tank Reactor 40L CHE574
= 0.00019/50 ×1000= 0.0038 mol /L
vii. Conversion of NaOH in the reactor , X
X = (1−CNaOH
CNaOH 0)×100 %
= (1−0.00380.05 )× 100%
= 92.4 %
viii. Residence time , τ
τ = V CSTR
F0
= 100.25
= 40 min
ix. Reaction rate constant , k
k = (CA 0−C A )
τ C A2
= (0.05−0.0038 )40 × 0.00382
= 79.99 M−1min−1
x. Rate of reaction , −r A
−r A = kC A2
= 79.99 ×0.00382
= 1.16×10−3mol / L.min
5) F0=0.30 L/min
28
Continuous Stirred Tank Reactor 40L CHE574
Known quantities
Volume of sample,V s = 50 mLConcentration of NaOH in the feed vessel,CNaOH f
= 0.1 mol /LVolume of HCl for quenching,V HCls
= 10 mLConcentration of HCl in standard solution,CHCl s
= 0.25 mol /LVolume of titrated NaOH,V 1 = 20.5 mLConcentration of NaOH used for titration,CNaOH s
= 0.1 mol /L
i. Concentration of NaOH entering the reactor , CNaOH 0
CNaOH 0= 1
2CNaOH f
= 12(0.1)
= 0.05 mol /L
ii. Volume of unreacted quenching HCl , V 2
V 2 = CNaOH s
C HCls
× V 1
= 0.10.25
× 20.5
= 8.2 mL
iii. Volume of HCl reacted with NaOH in sample , V 3
V 3 = V HCls−V 2
= 10−8.2= 1.8 mL
iv. Moles of HCl reacted with NaOH in sample , n1
n1 = (CHCl s×V 3 ) /1000
= 0.25 ×1.8 /1000= 0.00045 mol
v. Moles of unreacted NaOH in sample , n2
n2 = n1
= 0.00045 mol
29
Continuous Stirred Tank Reactor 40L CHE574
vi. Concentration of unreacted NaOH in the reactor , CNaOH
CNaOH = n2/V s×1000= 0.00045/50 ×1000= 0.009 mol /L
vii. Conversion of NaOH in the reactor , X
X = (1−CNaOH
CNaOH 0)×100 %
= (1−0.0090.05 )× 100 %
= 82 %
viii. Residence time , τ
τ = V CSTR
F0
= 100.30
= 33.33 min
ix. Reaction rate constant , k
k = (CA 0−C A )
τ C A2
= (0.05−0.009 )33.33× 0.0092
= 15.19 M−1min−1
x. Rate of reaction , −r A
−r A = kC A2
= 15.19 ×0.0092
= 1.23 ×10−3 mol / L.min
30
Continuous Stirred Tank Reactor 40L CHE574
DISCUSSION
This experiment was conducted to achieve three adjectives; to carry out saponification
reaction between NaOH and Et (Ac) in a CSTR, to determine the effect of the residence time
onto the reaction extent of conversion and to determine the reaction rate constant. We plotted
two graphs from this expeiment which are the graph of conductivity vs conversion and the
graph of conversion vs residence time.
For the first graph, we can see that the conductivity inversely proportional to the conversion.
This shows that there is a negative slope obtained from the line which is -1.233. This slope
can be seen through the trend line which is in black color. The y axis intercept is at x=0,
where the value for conductivity is 11.173. for the second graph, we can see there is an
ascending order from minutes 33 to 50. But it begins to decrease to the 100th minute. The
ascending trend may due to several errors during the experiment was conducted. The
decreaseing trend shows that our experiment was succesfully conducted.
Saponification is the process to make a soap. Saponification is a continuous reaction. In this
experiment, the reaction of saponification is quench with hydrochloride acid to stop the
reaction. The reaction rapidly react in increasing of temperature. Back titration is done to
investigate if the reaction is stop.
31
Continuous Stirred Tank Reactor 40L CHE574
CONCLUSION
Based on the objectives of the experiment to determine the effect of the residence time, we
found that conversion inverse proportionally with the residence time. This means, increasing
in resident time decreases the conversion. The objectives were succesfuly achieved.
32
Continuous Stirred Tank Reactor 40L CHE574
RECOMMENDATION
The samples need to immediately quench with hydrochloride acid (HCl) to stop the
reaction. The reaction still occurred as long as no quenching is done. The
recommendation is store the HCl near to the sample so it can immediately quench.
Immediately stop the titration after the colour turns light pink. The long titration may
cause error to the calculation which the flow rate of NaOH could be more than the
initial.
During titration, the indicator must be put to HCl then followed by sample.
33
Continuous Stirred Tank Reactor 40L CHE574
REFERENCES
http://www.solution.com.my/pdf/BP143(A4).pdf
http://www.engin.umich.edu/~cre/asyLearn/bits/cstr/
http://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor
34
Continuous Stirred Tank Reactor 40L CHE574
APPENDIX
35