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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.

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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added and withdrawn.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.

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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

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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

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http://en.wikipedia.org/wiki/Oxidation

http://en.wikipedia.org/wiki/Chemical_reaction

http://en.wikipedia.org/wiki/Product_(chemistry)

http://en.wikipedia.org/wiki/Reactant


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

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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

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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

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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

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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.

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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

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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

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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

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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


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.

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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.

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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

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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

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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

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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

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= (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





= 0.1 mol /L


CNaOH 0= 1

2CNaOH f

= 12(0.1)

= 0.05 mol /L


V 2 = CNaOH s

C HCls

× V 1

= 0.10.25

× 28.5

22


= 11.40mL


V 3 = V HCls−V 2

= 10−11.40= −1.40 mL


n1 = (CHCl s×V 3 ) /1000

= 0.25 ×1.40 /1000= 0.00035 mol


n2 = n1

= 0.00035 mol


CNaOH = n2/V s×1000= 0.00035/50 ×1000= 0.007 mol / L


X = (1−CNaOH

CNaOH 0)×100 %

= (1−0.0070.05 )× 100 %

= 86 %


τ = V CSTR

F0

23


= 100.15

= 66.67 min


k = (CA 0−C A )

τ C A2

= (0.05−0.007 )100 ×0.0072

= 8.78 M−1 min−1


−r A = kC A2

= 8.78 ×0.0072

= 4.30 × 10−4 mol / L. min

3) F0=0.20 L/min

Known quantities





= 0.1 mol /L


CNaOH 0= 1

2CNaOH f

= 12(0.1)

24


= 0.05 mol /L


V 2 = CNaOH s

C HCls

× V 1

= 0.10.25

× 25.3

= 10.12 mL


V 3 = V HCls−V 2

= 10−10.12= −0.12 mL


n1 = (CHCl s×V 3 ) /1000

= 0.25 ×0.12 /1000= 0.00003 mol


n2 = n1

= 0.00003 mol


CNaOH = n2/V s×1000= 0.00003/50 ×1000= 0.0006 mol / L


X = (1−CNaOH

CNaOH 0)×100 %

25


= (1−0.00060.05 )× 100 %

= 98.8 %


τ = V CSTR

F0

= 100.20

= 50 min


k = (CA 0−C A )

τ C A2

= (0.05−0.0006 )50 ×0.00062

= 2744 M−1 min−1


−r A = kC A2

= 2744× 0.00062

= 9.88 ×10−4 mol /L . min

4) F0=0.25 L/min

Known quantities




= 0.25 mol /LVolume of titrated NaOH,V 1 = 23.1 mL

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Concentration of NaOH used for titration,CNaOH s= 0.1 mol /L


CNaOH 0= 1

2CNaOH f

= 12(0.1)

= 0.05 mol /L


V 2 = CNaOH s

C HCls

× V 1

= 0.10.25

× 23.1

= 9.24 mL


V 3 = V HCls−V 2

= 10−9.24= 0.76 mL


n1 = (CHCl s×V 3 ) /1000

= 0.25 ×0.76 /1000= 0.00019 mol


n2 = n1

= 0.00019 mol


CNaOH = n2/V s×1000

27


= 0.00019/50 ×1000= 0.0038 mol /L


X = (1−CNaOH

CNaOH 0)×100 %

= (1−0.00380.05 )× 100%

= 92.4 %


τ = V CSTR

F0

= 100.25

= 40 min


k = (CA 0−C A )

τ C A2

= (0.05−0.0038 )40 × 0.00382

= 79.99 M−1min−1


−r A = kC A2

= 79.99 ×0.00382

= 1.16×10−3mol / L.min

5) F0=0.30 L/min

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Known quantities





= 0.1 mol /L


CNaOH 0= 1

2CNaOH f

= 12(0.1)

= 0.05 mol /L


V 2 = CNaOH s

C HCls

× V 1

= 0.10.25

× 20.5

= 8.2 mL


V 3 = V HCls−V 2

= 10−8.2= 1.8 mL


n1 = (CHCl s×V 3 ) /1000

= 0.25 ×1.8 /1000= 0.00045 mol


n2 = n1

= 0.00045 mol

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CNaOH = n2/V s×1000= 0.00045/50 ×1000= 0.009 mol /L


X = (1−CNaOH

CNaOH 0)×100 %

= (1−0.0090.05 )× 100 %

= 82 %


τ = V CSTR

F0

= 100.30

= 33.33 min


k = (CA 0−C A )

τ C A2

= (0.05−0.009 )33.33× 0.0092

= 15.19 M−1min−1


−r A = kC A2

= 15.19 ×0.0092

= 1.23 ×10−3 mol / L.min

30


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.

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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


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


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

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http://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor

http://www.engin.umich.edu/~cre/asyLearn/bits/cstr/

http://www.solution.com.my/pdf/BP143(A4).pdf


APPENDIX

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