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.

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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

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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

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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

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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 %

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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

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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)

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= 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

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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.

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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

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Figure 4

25

Figure 5: Structure of reactor in plug flow reactor

Figure 6: Example graph conversion against residence time

26