08-16-2017, 10:19 PM
attachment=5153]
Continuous stirrer tank reactors in series.
Dpeartment of Chemical Engineering
NWFP UET PESHAWAR
Pakistan
Muhammad Azeem
Muhammad Shakir
Qazi Muhammad Ali
Sami Ullah
Mehfooz Ur Rehman
Jehangir Khan
Group Members
Muhammad Azeem
Muhammad Shakir
Qazi Muhammad Ali
Sami Ullah
Mehfooz Ur Rehman
Jehangir Khan
Supervisor:
Engr. Ihsan ullah
Mini Project
Continuous stirrer tank reactors in series.
Summary of previous work.
1. Introduction
2. Commissioning
3. Calibration of pumps
Results of calibration of pumps.
Time constant calculation for different flow rates.
For liquid level system time constant is given by
T = V/q
Where V is the operating volume of tank
And q is flow rate.
Determining the operating volume of one tank.
Fill the first two tanks until the second tank overflows.
Open the drain valve and collect all liquid in a graduated beaker.
Divide this volume by 2 to get the operating volume of single tank.
Results
Total volume = 1350ml
thus V1 = 1350ml/2 = 675ml
Now T can be found by our adjusted flow rate.
Experimental work to be performed.
Section 1:
Dynamics of stirred tank.
Section 2:
Chemical reaction in stirred tank reactors in series
Dynamics
The way how the system responds when any type of changes in its input are introduced is known as its dynamic behavior.
Experiments performed on CSTRs
Section 1:
Effect of step input change.
Response of tank concentration to an impulse Change.
Influence of flow rate on a three tank system following a step change in input concentration.
Response to step change in input concentration of system comprising on one stirred vessel and dead time module.
Time constant
The value of Y(t) reaches 63.2 percent of its ultimate value when the time elapsed is equal to one time constant T. When the time elapsed is 2T,3T, and 4T, the percent response is 86.5, 95, and 98, respectively. From these facts, one can consider the response essentially completed in three to four time constants.
Experiment no 1:
Effect of a step input change.
Procedure:
1. Make up 5 litres of 0.1M potassium chloride solution.
Formula of potassium chloride = KCl
Molecular weight of potassium ,K= 39gm
Molecular weight of Chlorine, Cl= 35.5gm
Molecular weight of KCl = (39+35.5)gm
Molecular weight of KCl = 74.5gm
For .1M KCl solution
as KCl is in solid form
.1M KCl solution = Molarity *Molecular weight of KCl
= .1*74.5gm = 7.45gm (for 1litre solution)
For 5 litre solution;
5 litres of 0.1M KCl solution = 5lit * 7.45gm = 37.25gm KCl
Procedure
1. Dissolve 37.25gm of KCl in 5litre water to make.1M KCl solution.
2. Fill one of the reagent feed vessel with this solution.
3. Half fill the other feed vessel with demineralised or deionised water.
4. Pour this water into each of the reactors until they are full to standpipe overflow levels. This is to save time as it would take several minutes to fill the reactors using the feed pumps.
5. Start the water feed pump and set to 6.0 on speed adjust dial.
6. Start the stirrers and set to 8.0 on speed adjust dial.
7. Start taking conductivity reading after every 30sec for period of 45min
8. After few minutes, when water is flowing from the tank, through each reactor in turn and eventually out to drain. Stop the water feed pump and immediately start the KCl pump with the speed set to 6.0 on the speed adjust dial.
9. When experiment is complete, use Microsoft excel to plot graphs of conductivity vs. time and explain their shape.
Effect of step input change in tank 1,2 and 3 respectively by using experimental data.
Finding time constant from experimental data.
First method:
Potentiometer speed = 6
Corresponding flow rate = 54.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/54.1(ml/min)
T = 12.5min
Continued ..
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .56 + 7.3 = 7.84mS
Corresponding time from experimental data:
T1 = 15min
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .5 + 7.3 = 7.8mS
Corresponding time from experimental data:
T2 = 32min
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .45 + 7.3 = 7.75mS
Corresponding time from experimental data:
T3 = 49min
Experiment no 2:
Response of tank concentration to an impulse Change.
Procedure:
1. Make up 5 litres of 0.1M potassium chloride solution.
2. So we have to dissolve 37.25gm of KCl in 5litre water to 5 litres of .1M KCl solution.
3. Fill one of the reagent feed vessel with this solution.
4. Half fill the other feed vessel with demineralised or deionised water.
5. Pour this water into each of the reactors until they are full to standpipe overflow levels. This is to save time as it would take several minutes to fill the reactors using the feed pumps.
6. Start the water feed pump and set to 8.0 on speed adjust dial.
7. Set 5.0 as the flow rate on the salt feed pump.
8. Start the stirrers and set to 8.0 on speed adjust dial.
9. Start taking conductivity reading after every 30sec for period of 60min.
10. An impulse change is applied by switching off the water feed pump and instantly switching on the salt solution feed pump.
11. Allow the salt solution to be pumped into the first reactor for 15 minutes then instantly switch back to water again.
12. When experiment is complete, use Microsoft excel to plot graphs of conductivity vs. time and explain their shape.
Effect of impulse change in tank 1,2 and 3 respectively by using experimental data.
Experiment no 3:
Influence of flow rate on a three tank system following a step change in input concentration.
Procedure:
1. Make up 5 litres of 0.1M potassium chloride solution.
2. So we have to dissolve 37.25gm of KCl in 5litre water to make 0.1M KCl solution.
3. Fill one of the reagent feed vessel with this solution.
4. Half fill the other feed vessel with demineralised or deionised water.
5. Pour this water into each of the reactors until they are full to standpipe overflow levels. This is to save time as it would take several minutes to fill the reactors using the feed pumps.
6. Start the water feed pump and set to 4.0 on speed adjust dial.
7. Start the stirrers and set to 8.0 on speed adjust dial.
8. Start taking conductivity reading after every 30sec for period of 45min
9. After few minutes, when water is flowing from the tank, through each reactor in turn and eventually out to drain. Stop the water feed pump and immediately start the KCl pump with the speed set to 6.0 on the speed adjust dial.
10. Repeat the experiment twice with a KCl flow rate set 6.0 and 8.0 on the speed adjust dial.
11. When experiment is complete, use Microsoft excel to plot graphs of conductivity vs. time and explain their shape.
12. Determine time constant and verify that they are inversely proportional to respective flow rates.
Effect of step input change in tank 1,2 and 3 respectively when KCl feed pump set on 4.0 on speed adjust dial.
Finding time constant from experimental data.
First method:
Potentiometer speed = 4
Corresponding flow rate = 35.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/35.1(ml/min)
T =19.2min
Finding time constant from experimental data.
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .58 + 7.3 = 7.88mS
Corresponding time from experimental data:
T1 = 39.5min
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .56 + 7.3 = 7.86mS
This response is not achieved in the process period.
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .54 + 7.3 = 7.84mS
This response is not achieved in the process period.
Effect of step input change in tank 1,2 and 3 respectively when KCl feed pump set on 6.0 on speed adjust dial.
Finding time constant from experimental data.
First method:
Potentiometer speed = 6
Corresponding flow rate = 54.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/54.1(ml/min)
T = 12.5min
Continued .
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .56 + 7.3 = 7.84mS
Corresponding time from experimental data:
T1 = 15min
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .5 + 7.3 = 7.8mS
Corresponding time from experimental data:
T2 = 32min
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .45 + 7.3 = 7.75mS
Corresponding time from experimental data:
T3 = 49min
Effect of step input change in tank 1,2 and 3 respectively when KCl feed pump set on 8.0 on speed adjust dial.
Finding time constant from experimental data.
First method:
Potentiometer speed = 8
Corresponding flow rate = 74.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/74.1(ml/min)
T = 9.1min
Continued .
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .58 + 7.3 = 7.88mS
Corresponding time from experimental data:
T1 = 13.5min
[/b]
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .56 + 7.3 = 7.86mS
Corresponding time from experimental data:
T2 = 27.5min
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .54 + 7.3 = 7.84mS
Corresponding time from experimental data:
T3 = 39.5min
Relation between flow rate and time constant.
Experiment no 4:
Response to step change in input concentration of system comprising on one stirred vessel and dead time module.
Procedure:
1. Prepare 5 litres of 0.1M and 0.01M KCl.
2. Fill tank no. 3 and the dead time coil up to the overflow level with 0.1M KCl solution.
3. Connect the feed pump discharge to the tank no. 3 inlet fitting.
4. Start the pump from the 0.1M feed tank and adjust the flow to position 8.0 on the speed adjust dial.
5. Switch on the agitators and adjust the speed the speed to position 5.0 on the speed adjust dial.
6. Adjust the overflow height to give an operating level in tank no. 3 almost at the standpipe overflow height.
7. Start taking readings of conductivity after 30sec over a 45 minutes period.
8. Stop the 0.1M feed pump and immediately start the 0.01M feed pump adjusted to same flow rate.
9. Convert conductivity readings to concentration.
10. Plot curves relating concentration level (conductivity) and time for tank no. 3 and dead time coil.
Response to step change in input concentration of system comprising on one stirred vessel and dead time module using experimental data.
Finding time constant from experimental data.
First method:
Potentiometer speed = 8
Corresponding flow rate = 74.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/74.1(ml/min)
T = 9.1min
Continued .
Second method:
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 2.1(.01M KCl)
Calculating T for tank 3.
Tank 3:
K(t) = -.632 * 2.1
kt = ks -1.33
= 11.42 - 1.33 = 10.09mS
Corresponding time from experimental data:
T3 = 11min
Finding time constant from experimental data for dead time coil.
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for dead time coil.
Dead time coil:
K(t) = -.632 * 2.1
kt = ks -1.33
= 11.42 - 1.33 = 10.09mS
Corresponding time from experimental data:
T4 = 15min
Response of tank 3
C3=Co(Ku-K3/Ku-Ko) (a)
Response of dead time coil
C4=(Cb-Ca)*(1-exp(t-T4)/T3)+Ca (b)
Research paper
Process improvement approach to the saponification reaction by using statistical experimental design.
Author:
Nihal Bursali a, Suna Ertunc b, Bulent Akayb,
a Yenimahalle Municipality Precidency, 06170 Yenimahalle, Ankara, Turkey
Work:
The saponification reaction yields ethyl acetate and ethyl alcohol which are very useful products.
Therefore, it is quite demanding to determine best operating conditions for the maximum production. Present study is made for this purpose for the batch saponification reaction. The factors examined are:
Temperature.2.Agitation rate. 3. Initial NaOH concentration. 4. And initial CH3COOC2H5 concentration. Selected process response is conversion of NaOH.
Statistical design of experiments method has been used to plan the experiments to fulfill the purpose of optimization. This is divided into three phases:
1-Screening 2-Comparing 3-and influence of significant parameters.
At very first, the most effective parameters for process response are detected. The screening is planned by using the full-two-level factorial experimental design. This made convenience for evaluating the number of experiments to be performed in a mathematical way. For the four parameters, it gives 2^4=16 experiments. Randomization of experiments is made by using Gnumeric programme of Linux operation system. Temperature and agitation rate were declared to have no effects on the process response. Sum of squares of factors by using Yates Alogrithm was used to get the magnitude of effected parameters. Lastly, Variance analysis is made for comparison. Then, response surface method (RSM) is applied to get optimal conditions by using the quadratic second order polynomial model. In the response surface method design, Face-centered central composite technique is used to identify the second order polynomial model. This helped to illustrate the dependence of response on significant factors.
The results yielded by the constructed models in this study were checked for adequacy with the help of an efficient model validation test method known as Graphical residual analysis .
The maximum conversion of NaOH for the optimum conditions that are
Conc. Of NaOH = 0.01M and Conc. Of CH3COOC2H5 = 0.1M
is 98%.
Section 2:
Chemical reaction in stirred tank reactors in series
Sponification reaction:
NaOH +CH3COOC2H5 = CH3COONa + C2H5OH
Section2:
Chemical reaction in stirred tank reactors in series.
Experiments to optimize the saponification reaction:
Demonstration of the progress of a second order chemical reaction through three continuous stirred tank reactors in series and optimization of the operating parameters for maximum conversion .
Experiment no 5:
demonstration of the progress of a second order chemical reaction through three continuous stirred tank reactors in series.
Procedure:
1. Make up 5 litres batches of 0.05M sodium hydroxide and 0.05M ethyl acetate.
2. Fill the reagent feed vessels with reactants and refit the lids carefully.
3. Switch on both feed pumps and agitator motor with specific settings.
4. Start recording the conductivity with respect to time
5. The conversion is evaluated using the steady state concentration of sodium hydroxide.
6. These conductivity measurements at the drain through the dead time coil are now to be translated into degree of conversion of the constituents.
Conclusions and future suggestions
Continuous stirred tank reactors in series have been studied experimentally in detail for their dynamic behavior. By introducing the step and impulse input changes in concentrations and flow rates of reagents, the system is confirmed to be first order. The step and impulse response curves are found to be the sigmoidal and exponential respectively. Furthermore, saponification reaction is carried out and the effects of parameters are analyzed using the factorial design of experiments method. Best operating parameters for saponification reaction in CSTRs in series recommended are:
NaOH Feed Rate = 45 ml/min,
CH3COOC2H5 Feed Rate = 85 ml/min
and Agitators Speed = 9.
This combination of factors yielded 96.4 % Sodium Hydroxide conversion.
The reagents concentrations and system temperature effects have been neglected during the analysis for optimization, but these may affect the process response considerably. Taking these into account for optimization of process parameters, one may get more beneficial values of conversion.
First two and the third reactors may be connected in parallel, but obviously will have a low performance than the existing ones in series.
Future work
Finding rate constant for sponification reaction.
Summarizing of whole work.
Report preparation.
Reference:
armfield.org.uk
PROCESS SYSTEMS ANALYSIS AND CONTROL , Donald R. Coughanowr , 2nd edition.
Equation (a) and (b) , taken from manual of CSTRs in series.
Continuous stirrer tank reactors in series.
Dpeartment of Chemical Engineering
NWFP UET PESHAWAR
Pakistan
Muhammad Azeem
Muhammad Shakir
Qazi Muhammad Ali
Sami Ullah
Mehfooz Ur Rehman
Jehangir Khan
Group Members
Muhammad Azeem
Muhammad Shakir
Qazi Muhammad Ali
Sami Ullah
Mehfooz Ur Rehman
Jehangir Khan
Supervisor:
Engr. Ihsan ullah
Mini Project
Continuous stirrer tank reactors in series.
Summary of previous work.
1. Introduction
2. Commissioning
3. Calibration of pumps
Results of calibration of pumps.
Time constant calculation for different flow rates.
For liquid level system time constant is given by
T = V/q
Where V is the operating volume of tank
And q is flow rate.
Determining the operating volume of one tank.
Fill the first two tanks until the second tank overflows.
Open the drain valve and collect all liquid in a graduated beaker.
Divide this volume by 2 to get the operating volume of single tank.
Results
Total volume = 1350ml
thus V1 = 1350ml/2 = 675ml
Now T can be found by our adjusted flow rate.
Experimental work to be performed.
Section 1:
Dynamics of stirred tank.
Section 2:
Chemical reaction in stirred tank reactors in series
Dynamics
The way how the system responds when any type of changes in its input are introduced is known as its dynamic behavior.
Experiments performed on CSTRs
Section 1:
Effect of step input change.
Response of tank concentration to an impulse Change.
Influence of flow rate on a three tank system following a step change in input concentration.
Response to step change in input concentration of system comprising on one stirred vessel and dead time module.
Time constant
The value of Y(t) reaches 63.2 percent of its ultimate value when the time elapsed is equal to one time constant T. When the time elapsed is 2T,3T, and 4T, the percent response is 86.5, 95, and 98, respectively. From these facts, one can consider the response essentially completed in three to four time constants.
Experiment no 1:
Effect of a step input change.
Procedure:
1. Make up 5 litres of 0.1M potassium chloride solution.
Formula of potassium chloride = KCl
Molecular weight of potassium ,K= 39gm
Molecular weight of Chlorine, Cl= 35.5gm
Molecular weight of KCl = (39+35.5)gm
Molecular weight of KCl = 74.5gm
For .1M KCl solution
as KCl is in solid form
.1M KCl solution = Molarity *Molecular weight of KCl
= .1*74.5gm = 7.45gm (for 1litre solution)
For 5 litre solution;
5 litres of 0.1M KCl solution = 5lit * 7.45gm = 37.25gm KCl
Procedure
1. Dissolve 37.25gm of KCl in 5litre water to make.1M KCl solution.
2. Fill one of the reagent feed vessel with this solution.
3. Half fill the other feed vessel with demineralised or deionised water.
4. Pour this water into each of the reactors until they are full to standpipe overflow levels. This is to save time as it would take several minutes to fill the reactors using the feed pumps.
5. Start the water feed pump and set to 6.0 on speed adjust dial.
6. Start the stirrers and set to 8.0 on speed adjust dial.
7. Start taking conductivity reading after every 30sec for period of 45min
8. After few minutes, when water is flowing from the tank, through each reactor in turn and eventually out to drain. Stop the water feed pump and immediately start the KCl pump with the speed set to 6.0 on the speed adjust dial.
9. When experiment is complete, use Microsoft excel to plot graphs of conductivity vs. time and explain their shape.
Effect of step input change in tank 1,2 and 3 respectively by using experimental data.
Finding time constant from experimental data.
First method:
Potentiometer speed = 6
Corresponding flow rate = 54.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/54.1(ml/min)
T = 12.5min
Continued ..
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .56 + 7.3 = 7.84mS
Corresponding time from experimental data:
T1 = 15min
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .5 + 7.3 = 7.8mS
Corresponding time from experimental data:
T2 = 32min
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .45 + 7.3 = 7.75mS
Corresponding time from experimental data:
T3 = 49min
Experiment no 2:
Response of tank concentration to an impulse Change.
Procedure:
1. Make up 5 litres of 0.1M potassium chloride solution.
2. So we have to dissolve 37.25gm of KCl in 5litre water to 5 litres of .1M KCl solution.
3. Fill one of the reagent feed vessel with this solution.
4. Half fill the other feed vessel with demineralised or deionised water.
5. Pour this water into each of the reactors until they are full to standpipe overflow levels. This is to save time as it would take several minutes to fill the reactors using the feed pumps.
6. Start the water feed pump and set to 8.0 on speed adjust dial.
7. Set 5.0 as the flow rate on the salt feed pump.
8. Start the stirrers and set to 8.0 on speed adjust dial.
9. Start taking conductivity reading after every 30sec for period of 60min.
10. An impulse change is applied by switching off the water feed pump and instantly switching on the salt solution feed pump.
11. Allow the salt solution to be pumped into the first reactor for 15 minutes then instantly switch back to water again.
12. When experiment is complete, use Microsoft excel to plot graphs of conductivity vs. time and explain their shape.
Effect of impulse change in tank 1,2 and 3 respectively by using experimental data.
Experiment no 3:
Influence of flow rate on a three tank system following a step change in input concentration.
Procedure:
1. Make up 5 litres of 0.1M potassium chloride solution.
2. So we have to dissolve 37.25gm of KCl in 5litre water to make 0.1M KCl solution.
3. Fill one of the reagent feed vessel with this solution.
4. Half fill the other feed vessel with demineralised or deionised water.
5. Pour this water into each of the reactors until they are full to standpipe overflow levels. This is to save time as it would take several minutes to fill the reactors using the feed pumps.
6. Start the water feed pump and set to 4.0 on speed adjust dial.
7. Start the stirrers and set to 8.0 on speed adjust dial.
8. Start taking conductivity reading after every 30sec for period of 45min
9. After few minutes, when water is flowing from the tank, through each reactor in turn and eventually out to drain. Stop the water feed pump and immediately start the KCl pump with the speed set to 6.0 on the speed adjust dial.
10. Repeat the experiment twice with a KCl flow rate set 6.0 and 8.0 on the speed adjust dial.
11. When experiment is complete, use Microsoft excel to plot graphs of conductivity vs. time and explain their shape.
12. Determine time constant and verify that they are inversely proportional to respective flow rates.
Effect of step input change in tank 1,2 and 3 respectively when KCl feed pump set on 4.0 on speed adjust dial.
Finding time constant from experimental data.
First method:
Potentiometer speed = 4
Corresponding flow rate = 35.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/35.1(ml/min)
T =19.2min
Finding time constant from experimental data.
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .58 + 7.3 = 7.88mS
Corresponding time from experimental data:
T1 = 39.5min
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .56 + 7.3 = 7.86mS
This response is not achieved in the process period.
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .54 + 7.3 = 7.84mS
This response is not achieved in the process period.
Effect of step input change in tank 1,2 and 3 respectively when KCl feed pump set on 6.0 on speed adjust dial.
Finding time constant from experimental data.
First method:
Potentiometer speed = 6
Corresponding flow rate = 54.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/54.1(ml/min)
T = 12.5min
Continued .
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .56 + 7.3 = 7.84mS
Corresponding time from experimental data:
T1 = 15min
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .5 + 7.3 = 7.8mS
Corresponding time from experimental data:
T2 = 32min
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .45 + 7.3 = 7.75mS
Corresponding time from experimental data:
T3 = 49min
Effect of step input change in tank 1,2 and 3 respectively when KCl feed pump set on 8.0 on speed adjust dial.
Finding time constant from experimental data.
First method:
Potentiometer speed = 8
Corresponding flow rate = 74.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/74.1(ml/min)
T = 9.1min
Continued .
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for each tank.
Tank 1:
K(t) = .632 * 11.52
kt = ks +7.3
= .58 + 7.3 = 7.88mS
Corresponding time from experimental data:
T1 = 13.5min
[/b]
Tank 2:
K(t) = .632 * 11.52
k2 = ks +7.3
= .56 + 7.3 = 7.86mS
Corresponding time from experimental data:
T2 = 27.5min
Tank 3:
K(t) = .632 * 11.52
k3 = ks +7.3
= .54 + 7.3 = 7.84mS
Corresponding time from experimental data:
T3 = 39.5min
Relation between flow rate and time constant.
Experiment no 4:
Response to step change in input concentration of system comprising on one stirred vessel and dead time module.
Procedure:
1. Prepare 5 litres of 0.1M and 0.01M KCl.
2. Fill tank no. 3 and the dead time coil up to the overflow level with 0.1M KCl solution.
3. Connect the feed pump discharge to the tank no. 3 inlet fitting.
4. Start the pump from the 0.1M feed tank and adjust the flow to position 8.0 on the speed adjust dial.
5. Switch on the agitators and adjust the speed the speed to position 5.0 on the speed adjust dial.
6. Adjust the overflow height to give an operating level in tank no. 3 almost at the standpipe overflow height.
7. Start taking readings of conductivity after 30sec over a 45 minutes period.
8. Stop the 0.1M feed pump and immediately start the 0.01M feed pump adjusted to same flow rate.
9. Convert conductivity readings to concentration.
10. Plot curves relating concentration level (conductivity) and time for tank no. 3 and dead time coil.
Response to step change in input concentration of system comprising on one stirred vessel and dead time module using experimental data.
Finding time constant from experimental data.
First method:
Potentiometer speed = 8
Corresponding flow rate = 74.1 ml/min
Volume of one tank = 675 ml
since,
Time constant = T = V/q
Time constant = T = 675ml/74.1(ml/min)
T = 9.1min
Continued .
Second method:
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 2.1(.01M KCl)
Calculating T for tank 3.
Tank 3:
K(t) = -.632 * 2.1
kt = ks -1.33
= 11.42 - 1.33 = 10.09mS
Corresponding time from experimental data:
T3 = 11min
Finding time constant from experimental data for dead time coil.
Graphical method:
In step response curve the time constant is achieved when ordinate of graph
Y(t)/A = .632
Where,
Y(t) = K (conductivity)
And,
A = ultimate value of K = 11.5(.1M KCl)
Calculating T for dead time coil.
Dead time coil:
K(t) = -.632 * 2.1
kt = ks -1.33
= 11.42 - 1.33 = 10.09mS
Corresponding time from experimental data:
T4 = 15min
Response of tank 3
C3=Co(Ku-K3/Ku-Ko) (a)
Response of dead time coil
C4=(Cb-Ca)*(1-exp(t-T4)/T3)+Ca (b)
Research paper
Process improvement approach to the saponification reaction by using statistical experimental design.
Author:
Nihal Bursali a, Suna Ertunc b, Bulent Akayb,
a Yenimahalle Municipality Precidency, 06170 Yenimahalle, Ankara, Turkey
Work:
The saponification reaction yields ethyl acetate and ethyl alcohol which are very useful products.
Therefore, it is quite demanding to determine best operating conditions for the maximum production. Present study is made for this purpose for the batch saponification reaction. The factors examined are:
Temperature.2.Agitation rate. 3. Initial NaOH concentration. 4. And initial CH3COOC2H5 concentration. Selected process response is conversion of NaOH.
Statistical design of experiments method has been used to plan the experiments to fulfill the purpose of optimization. This is divided into three phases:
1-Screening 2-Comparing 3-and influence of significant parameters.
At very first, the most effective parameters for process response are detected. The screening is planned by using the full-two-level factorial experimental design. This made convenience for evaluating the number of experiments to be performed in a mathematical way. For the four parameters, it gives 2^4=16 experiments. Randomization of experiments is made by using Gnumeric programme of Linux operation system. Temperature and agitation rate were declared to have no effects on the process response. Sum of squares of factors by using Yates Alogrithm was used to get the magnitude of effected parameters. Lastly, Variance analysis is made for comparison. Then, response surface method (RSM) is applied to get optimal conditions by using the quadratic second order polynomial model. In the response surface method design, Face-centered central composite technique is used to identify the second order polynomial model. This helped to illustrate the dependence of response on significant factors.
The results yielded by the constructed models in this study were checked for adequacy with the help of an efficient model validation test method known as Graphical residual analysis .
The maximum conversion of NaOH for the optimum conditions that are
Conc. Of NaOH = 0.01M and Conc. Of CH3COOC2H5 = 0.1M
is 98%.
Section 2:
Chemical reaction in stirred tank reactors in series
Sponification reaction:
NaOH +CH3COOC2H5 = CH3COONa + C2H5OH
Section2:
Chemical reaction in stirred tank reactors in series.
Experiments to optimize the saponification reaction:
Demonstration of the progress of a second order chemical reaction through three continuous stirred tank reactors in series and optimization of the operating parameters for maximum conversion .
Experiment no 5:
demonstration of the progress of a second order chemical reaction through three continuous stirred tank reactors in series.
Procedure:
1. Make up 5 litres batches of 0.05M sodium hydroxide and 0.05M ethyl acetate.
2. Fill the reagent feed vessels with reactants and refit the lids carefully.
3. Switch on both feed pumps and agitator motor with specific settings.
4. Start recording the conductivity with respect to time
5. The conversion is evaluated using the steady state concentration of sodium hydroxide.
6. These conductivity measurements at the drain through the dead time coil are now to be translated into degree of conversion of the constituents.
Conclusions and future suggestions
Continuous stirred tank reactors in series have been studied experimentally in detail for their dynamic behavior. By introducing the step and impulse input changes in concentrations and flow rates of reagents, the system is confirmed to be first order. The step and impulse response curves are found to be the sigmoidal and exponential respectively. Furthermore, saponification reaction is carried out and the effects of parameters are analyzed using the factorial design of experiments method. Best operating parameters for saponification reaction in CSTRs in series recommended are:
NaOH Feed Rate = 45 ml/min,
CH3COOC2H5 Feed Rate = 85 ml/min
and Agitators Speed = 9.
This combination of factors yielded 96.4 % Sodium Hydroxide conversion.
The reagents concentrations and system temperature effects have been neglected during the analysis for optimization, but these may affect the process response considerably. Taking these into account for optimization of process parameters, one may get more beneficial values of conversion.
First two and the third reactors may be connected in parallel, but obviously will have a low performance than the existing ones in series.
Future work
Finding rate constant for sponification reaction.
Summarizing of whole work.
Report preparation.
Reference:
armfield.org.uk
PROCESS SYSTEMS ANALYSIS AND CONTROL , Donald R. Coughanowr , 2nd edition.
Equation (a) and (b) , taken from manual of CSTRs in series.