LabManual Final

Kuwait UniversityCollege of Engineering and Petroleum

Mass Transfer OperationsLaboratory Manual

ChE 443

Edited by: Khaled A. MahdiChemical Engineering Department

2007

A

This is the second edition of the mass transfer operations laboratory manual. Themanual is the product of previous works by talented engineers in the chemicalengineering department. I have to mention the typing of the previous collectionsof experiments by eng. Hala Al-Fulaij, and Dr. Amal Al-Kilani. As of this newmanual, I appreciate the comments from eng. Intesar Al-Gunaim and the help ofeng Mubarak Ghanim. I enjoyed the practical comments by the technician in themass lab, Mr. Essam Abdul Ali. The figures of this manual were scanned by eng.Khaled Dhamiyar. To them I give my full appreciation and gratuities.

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CONTENTS

1 VAPOR-LIQUID EQUILIBRIA 5

2 PACKED AND BUBBLE CAP COLUMN CONTINUOUS DIS-TILLATION 12

3 Solid-Liquid Extraction (Leaching) 20

4 SPRAY DRYING 27

5 Fixed and Fluidized bed 32

6 DOUBLE EFFECT EVAPORATOR 39

7 TUBULAR REACTOR 44

8 Writing a Formal Technical Report

9 Sample Report

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L O F

1.1 Schematic diagram of Ellis Still. . . . . . . . . . . . . . . . . . . . 112.1 Schematic diagram of a distillation column. . . . . . . . . . . . . . 172.2 Refractive Index versus Mol% of Methanol in Water. . . . . . . . 182.3 Flow diagram of combined bubble-cap and packed distillation column. 193.1 Block diagram of solid-liquid extraction and solvent recovery. . . . 253.2 Flow diagram of leaching unit. . . . . . . . . . . . . . . . . . . . . 264.1 Spray Drying Equipment. . . . . . . . . . . . . . . . . . . . . . . 316.1 Schematic diagram of multiple effect evaporator. . . . . . . . . . . 426.2 Double effect evaporator expiremntal setup. . . . . . . . . . . . . 437.1 Differential mass balance of tubular reactor. . . . . . . . . . . . . 477.2 Schematic diagram of tubular reactor. . . . . . . . . . . . . . . . 48

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1. VAPOR-LIQUID EQUILIBRIA

1.1. Background

The study of vapor-liquid equilibria is important due to its applications in sep-aration processes involving volatile mixtures. For example, one major applica-tion of vapor-liquid equilibrium analysis is the calculation of dew and bubblepoints. Chemical thermodynamics provides us with tools to understand vapor-liquid equilibrium in different solutions. Generally, there are two types of solu-tions, ideal solutions such as Benzene-Toluene solutions and non-ideal solutionssuch as tetrachloride-methanol and acetone-chloroform solutions. The reason fornon-ideality is mainly due to the influence of molecular interactions whereas idealsolutions show very weak molecular interactions that can be neglected. Simplethermodynamics laws can describe ideal solutions, for instance, Roault’s law andHenry’s law. On the other hand, more advance models are needed to describethe thermodynamics of real systems. Examples of these models are Margulesequations and Van Laar equations.

1.2. Objectives

• To understand vapor-liquid equilibria of solutions.• To determine the activity coefficients of each components.• To evaluate the parameters of different thermodynamics models such as VanLaar, and Margules.

1.3. Theory

Under certain pressure and temperature, a binary solution of two volatile compo-nents can exist in 2 phases, liquid and vapor. Each component distributes its selfbetween the two phases. According to the phase rule, the compositions at eachphase are predetermined if the pressure and temperature are both fixed. When aliquid mixture of two components of different volatility is heated in a closed vessel,

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vaporization of both components will occur until the pressure of the vapor attainsa steady state corresponding to the bubble point of the mixture. If the tempera-ture of this mixture is maintained constant, vaporization and condensation of thetwo components will proceed until equilibrium between the vapor and the liquid isattained. This equilibrium depends on temperature and the total pressure of thevessel. The pressure exerted by each component will depend on the compositionof the mixture and the saturated vapor pressure of each component. Accordingto Roult’s law, in an ideal mixture of components 1 and 2, the partial pressureexerted by each components is proportional to its mole fraction, where the pro-portionality constant is the vapor pressure of that component. Mathematicallyexpresses as,

P1 = x1P∗1 , (1.1)

P2 = x2P∗2 , (1.2)

Where: Pi is partial pressure of component i at a fixed temperature; P ∗i is thevapor pressure of component i at a given temperature; xi is the liquid mole fractionof component i at equilibrium. The total pressure P according to Dalton’s law ofpartial pressures is given by

P = P1 + P2,

= x1P∗1 + x2P

∗2 , (1.3)

since the total mass is conserved, x1+x2 = 1. The total pressure can be expressedas

P = x1P∗1 + (1− x1)P ∗2 . (1.4)

In the vapor phase, the partial pressure of a component is given by, Pi = yiP ,where yi is the vapor mole fraction of component i. Using equations 1.1 and 1.2,

y1P = x1P∗1 , (1.5)

y2P = x2P∗2 . (1.6)

In practice, most mixtures are not ideal; therefore, equations 1.5 and 1.6 shouldbe modified. This can be done by introducing the activity coefficients γ in theliquid phase and the fugacity coefficient in the vapor phase φ.

y1φ1P = γ1x1P∗1 , (1.7)

y2φ2P = γ2x2P∗2 . (1.8)

For simplicity we will assume that the vapor phase is ideal hence the fugacitycoefficients are 1. The activity coefficient of a component in a mixture is a function

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of temperature, pressure and composition. Thus, it has to be determined fromexperiments.. Often, it is calculated from the vapor-liquid equilibria data atspecified conditions.

1.4. Procedure

Apparatus

The apparatus, shown in figure 1.1, consists of a boiling chamber A in which themixture under test is vigorously bullated to produce a vapor liquid mixture thatpasses through the glass spiral tube B at a high velocity. The rising vapor entrainsdrops of the liquid, which equilibrates with the vapor, and the mixtures emergefrom the tube at equilibrium, striking the thermocouple pocket C that containsa thermocouple to measure the boiling temperature. The vapor then rises gentlythrough the central tube E while the liquid drops fall back into chamber F. Thevapor emerging from the tube E passes through the annulus G which is heatedelectrically in order to prevent condensation and therefore all the vapors pass tocondenser H where they are totally condensed (the branch at H is fitted witha vapor trap). The liquid formed accumulates in the receiver J from which asample of the vapor phase is obtained. When the receiver J is full of condensate itoverflows and passes through tube P to mix with the liquid flowing down tube Kfrom chamber F. This mixture enters heater A to be re-vaporized and discharged asa mist into spiral B where it is equilibrated on emerging against the thermocouplepocket C.The contents of the still are recirculated in this manner for about 1.5 hours to

ensure that the equilibrium is attained. When the equilibrium is attained betweenthe liquid and the vapor phases, a sample of the vapor phase is taken from tapM and a sample of the liquid phase is taken from tap N. This concludes theexperiment with this specific mixture and the contents of the still are thereforedrained through drain cock Q. A fresh mixture of different composition is thencharged and the experiment is repeated.

Experimental steps

Pleas make sure that THE SAFETY SHIELD MUST ALWAYS BE INPLACE BEFORE THE EXPERIMENTCOMMENCES. Follow the stepsof the experiment carefully,

1. Prepare 350 ml of the mixture (provided by your lab instructores).

2. Remove access panel from base of safety case.

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3. Transfer the prepared mixture into equilibrium still until the mixture coversthe coil B and is above the thermocouple C.

4. Replace the vapor trap H and turn on the cooling water ensuring that thereis a rapid flow of water through the condenser. Do not drain the receiver J.

5. Ensure that the thermocouple pocket R and C are filled with sufficient oilto immerse the tips of the thermocouples and vent S is closed.

6. Replace access panel.

7. Switch on the still heater to boil the liquid, ensure that the jacket is switchedoff.

8. Once boiling starts, slowly adjust the jacket heater control until the ther-mocouple in the jacket indicate 2 - 5 C higher than the still’s temperature.

9. Adjust the still heater control so that only 20 drops per minute fall fromthe base of the condenser into the distillate receiver J.

10. Continue the boiling process for at least 1.5 hours. Monitor the still andjacket temperatures every 15 minutes and adjust the still and jacket heatercontrols as required.

11. After 45 minute, take samples (one from the still and one from the distillatereceiver) then measure their refractive indices to find their compositionsusing the calibration curve provided to you by your lab instructor.

12. After 1.5 hours, switch off the equipment then take another set of samplesas in step 11.

1.5. Practice Questions

(1) A pressurized pot always cook the food faster than a regular (non-pressurized)pot, explain using phase equilibrium thermodynamics.

(2) Does the system you studied in this experiment exhibit negative or positivedeviation from ideality? Use figures 3 and 4 to support your answer.

(3) How do you make sure that the solution, in your experiment, reach steadystate (equilibrium)?

(4) Name four non-ideal binary systems and specify if the system is positivelyor negatively deviated from ideality.

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(5) How do your results change if the fugacity of the vapor is considered?

(6) ”According to the phase rule, the compositions of a two-phase binary solutionare predetermined if the pressure and temperature are both fixed.” Explainthis statement.

(7) A Water/Ethanol solution is in equilibrium at 760 mmHg and 80oC. Theequilibrium data of the solution is shown in the table.

Material xi yi Ai Bi CiWater 0.320 8.16 1623 229Ethanol 0.506 7.97 1668 228

Use Antoine equation to calculate the vapor pressures (in mmHg) and Tem-perature (oC) in of the components.

logP ∗i = Ai −Bi

T + Ci

Calculate: (1) The K-values of Water and Ethanol?. (2) The activity coeffi-cients, γ of Water and Ethanol?

1.6. Report

You are expected to suggest answers to the practice questions in your report.Some of your answers can be used in the introduction and/or discussion. In yourreports, the following calculations MUST be reported and discussed:

• Find the composition from the refractive index using the calibration curve.

• Find the activity coefficient experimentally using equations 1.7 and 1.8.• Using the experimentally determined activity coefficients, evaluate the pa-rameters A and B in the 3-suffix Margules equation. If you are given theexcess free energy of the model,

gE = x1x2[A+B(x1 − x2)] (1.9)

Note: you have to derive the activity coefficients from the excess energybefore you determine the constants.

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• Determine the Van-Laar Equation parameters A and B using the activitycoefficients equations

ln γ1 =A

(1 +Ax1/Bx2)2(1.10)

ln γ2 =B

1 + (Bx2/Ax1)2(1.11)

• Discuss the use of your equilibrium data to predict the equilibrium at an-other set of conditions using 2-suffix Margules and Van Laar equations ??to 1.11.

References

H. C. Smith and J. M. Van ness. Introduction to Chemical Engineering Thermo-dynamics. McGraw-Hill, 4th edition, 1987.

J. M. Prausnitz. Molecular Thermodynamics of Fluid Phase Equilibria. Prentice-Hall, 1969.

B. G. Kyle. Chemical and Process Thermodynamics. Prentice Hall, 2nd edi-tion, 1992.

K. S. Pitzer. Thermodynamics. McGraw-Hill, 3rd edition, 1995.

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Figure 1.1: Schematic diagram of Ellis Still.

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2. PACKED AND BUBBLE CAP COLUMNCONTINUOUS DISTILLATION

2.1. Background

Distillation is one of the major operations in the chemical and petroleum indus-tries, especially oil refineries. The main purpose of a distillation column is toseparate two or more volatile components; an example of a distillation column isshown in figure 2.1. This experiment examines two types of distillation columns,bubble-cap and packed. Other types, not examined here, include ballast traysand perforated (Sieve) plates. Bubble-cap columns are classical in the chemicalindustries, they are known as early as 1800’s. Packed columns were introducedlater. The use of one type over the other is governed by the objectives of theprocess.The distillation column operates as follows, the feed, in a liquid state, is in-

troduced continuously at a midpoint in the column. The midpoint divides thecolumn into enriching and stripping sections. By mean of a reboiler located inthe bottom of the column, the liquid feed is vaporized. The vapor ascends withinthe column and contacts with the descending liquid. The plate or the packing inthe column allows for a thorough contact of the vapor and liquid. The reminderof the bottom liquid is withdrawn as bottoms. Vapor reaching the top of thecolumn is condensed to liquid in the overhead condenser. A portion of the con-densed liquid is returned to the column, known as a reflux liquid. The remainderof the condensate becomes the distillate product. The reboiler vapor and refluxliquid lead to higher-purity products and provides a much greater recovery of thefeed. Vapor and liquid phases on a given tray approach thermal equilibrium toan extent dependent upon the efficiency of the contacting trays.

2.2. Objectives

• To examine and operate bubble cap and packed bed distillation columnsunder total and finite reflux.

• To determine and compare the separation efficiencies of bubble cap and

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packed bed columns.

2.3. Theory

The number of theoretical stages (NTS) can be calculated graphically by means ofa McCabe-Theile diagram, or numerically by step wise solution of the componentmass balances for stage compositions for the enriching and stripping sections.Upon the calculation of NTS, we can determine the overall efficiency of the columnE0 simply from the following expression

E0 =NTS

NA∗ 100, (2.1)

where NA is the actual number of stages (trays or plates) to achieve a givenseparation.For packed column distillation, two simple concepts are used to analyze the

column: height equivalent to a theoretical stage (HETS) and the height of thetransfer unit (HTU), which is related to the mass transfer coefficient. HETS isbased on the number of theoretical stages (NTS) calculated above. If we knowthe height of the packing Z, HETS is calculated from the following equation:

HETS =Z

NTS(2.2)

where Z is the height of the packing in the column. This method is attractivebecause it provides a direct link to well-known heuristics in the design of packedcolumn. For example, an HETS of 50 cm, a 10 meter packed column will beequivalent to 20 theoretical stages. However, this approach is not theoreticallysound. The other approach uses a differential mass balance of the packed column.The mass balance gives,

G dy = KOGa(y∗ − y) dz. (2.3)

Upon integration, one obtains the following familiar expression for packed columnoperations;

Z =G

KOGa

y2

y1

dy

y∗ − y ,

= Gy2

y1

dy

y∗ − y , (2.4)

where Z is the height of the packing and G is a mass transfer parameter. Theintegral maybe evaluated graphically or numerically. Graphically, by calculatingthe area under the plot y vs. (y∗− y)−1, where y is the mole fraction at this stageand y∗ is the ”equilibrium” mole fraction at this stage.

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2.4. Procedure

Main equipment is shown on figure 2.2. Beside we need a stopwatch, samplingbottles, beaker and thermometer. The outline of the experiment steps are givennext.

Start-up (Atmospheric Operation)

1. Ensure that the concentration of feed in the reboiler vessel is correct andthat the liquid level is at the vessel equator (25L of a solution).- Complete valve check list.- Set-up for a bubble cap column (By closing V13 and V12 ).

2. Set column to total reflux condition (Check reflux control valve RCV.1 isclosed).- Start-up cooling water circuit:- Open slowly, FCV.2 to prevent a pressure surge to cooling system.- Don’t allow the water pressure to exceed 2.7 bar maximum as indicatedby PI.2

3. Start-up heating circuit:- Open slowly PCV.4, to prevent a sudden surge of steam condensate to thereboiler.- Do not close V.11 until a steady flow of steam issues from condensatedrain.- Ensure that the steam-trap drain hose is safely positioned.

4. Set the reboiler temperature, to approximately 90 C.- Keep the return line of the product distillate to the reboiler vessel open(open V.2 , close V.3 ).

Sampling and Readings

1. Allow equilibrium to be established and record the values of temperature,flow rate, etc. as given in the log sheet.

2. Collect a sample from the distillate by opening the sampling valve; thendetermine the refractive index of the sample taken

3. Adjust the reflex control valve, RCV.1 to obtain the desired reflex ratio,which is the ratio of rotameter (RI.1A and RI.1B) scale readings, (i.e. thescale reading RI.1A / scale reading RI.1B) = Reflux Ratio, to set reflux

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ratio, slowly open reflux control valve RCV.1 to obtain a distillate productflow through RI.1A. Monitor the ratio RI.1A : RI.1B.

4. Record the value of temperature indicators, flow meters, reflux indicators,etc. as given in the log sheet.

5. Collect a sample from distillate as you did before, then determine it’s re-fractive index. Read the composition from the chart in figure 2.3.

6. Measure the flow rate of distillate:- Close V2, and start timing.- When the distillate approach the top graduation , stop timing.- Open V2.

7. Measure the flow rate of steam condensate.

Set-up for the packed column

1. Ensure that the cooling water is passing through a packed-column overheadcondenser.- Set the reflux flow valve to the packed column.- Open V13, and close V12.

2. Repeat the steps in the sampling and readings section §2.4.


(1) What are the advantages and disadvantages of the bubble-cap and packedcolumns?

(2) What is the different between the overall and plate efficiency parameters?

(3) What are the measures you should consider if you have a cold feed?

(4) Name four different commercial packed bed column’s packings?

(6) Why a 100% separation of Water-Ethanol system is impossible to achieveusing only distillation? What is the name of the phenomenon associatedwith the Water-Ethanol system at higher ethanol liquid composition?

(7) Distillation columns exhibit different behavior at different vapor-liquid flowrates. List 4 different phenomena happened during distillation column op-eration at different vapor and liquid flows.

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(8) A saturated liquid solution of 55wt% ethanol and 45 wt% water is fed into adistillation column to give a distillate containing 87 wt% of ethanol and abottom containing 3 wt% ethanol. The feed rate is 500 kg/h. If the molaloverflow is variable, estimate the following:

(i) Minimum reflux, Rm and Minimum rectifying ∆Rm and Stripping ∆Rm oper-ating points

(ii) Minimum number of stages Nm.

(iii) The number of stages for a reflux equals 1.85 and the location of the feedplate

(vi) The Condenser duty, Qc, kW and Reboiler duty, QR, kW

(v) If you assume a constant molal overflow, estimate minimum reflux andnumber of minimum stages.

(vi) If you assume a constant molal overflow, determine the number of stagesfor RD = 1.75Rm, using the new minimum reflux found in part (v) forsuperheated feed (q = −1.0).

(vii) Write down the three operating lines mathematical equations.

(viii) Is a 100% wt Ethanol is possible to be obtained ? Explain your an-swer.Suggest a solution to this problem

The enthalpy and equilibrium charts are provided

2.6. Report

You are expected to suggest answers to the practice questions in your report.Your answers may be included in the discussion. In your report, the followingcalculations MUST be reported and discussed:

• Calculate the number of theoretical stages (NTS) in the column using theMcCabe-Thiele method.

• Determine the height equivalent to a theoretical stage (HETS) in the packedcolumn.

• Evaluate the mass transfer coefficient G .

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• Estimate the overall plate efficiency (E0) and Murphree plate efficiency(EM).

• Calculate the heat loss from the distillation column.

References

J. M. Coulson, J. F. Rihardson, J. R. Backhurst, and J. H. Harker. Particle Tech-nology and Separation Processes, volume 2 of Chemical Engineering. Pergamon,Oxford, 4th edition, 1991.

C. J. Geankoplis. Transport Processes and Unit Operations. McGraw Hill, NewYork, 3rd edition, 1993.

W. L. McCabe, J. C. Smith, and P. Harriott. Unit Operations of Chemical Engi-neering. McGraw-Hill, 5th edition, 1993.

P. C. Wankat. Equilibrium Staged Separations. Elsevier, New York, 1988.

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Figure 2.1: Schematic diagram of a distillation column.

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Figure 2.2: Refractive Index versus Mol% of Methanol in Water.

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Figure 2.3: Flow diagram of combined bubble-cap and packed distillation column.20

3. S -L E (L )

3.1. Background

Leaching is one of many solid-fluid separation processes that are carried out inthe chemical, mineral and related industries. In fact, it is one of the oldest unitoperations in the chemical industries. The fundamental principle behind leachingis the removal of a soluble material from an insoluble, permeable solid phase. Thesoluble fraction, solid or liquid, may exist mechanically in the pore structure ofthe insoluble material or chemically combined with that material. This solublematerial is removed through dissolution in a dissolving solvent. The most familiarexample of leaching is the extraction of tea and coffee, and most importantlymineral recovery.The rate of leaching (extraction) is affected by many physical and chemical

phenomena. Mass transfer and equilibrium phenomena obviously play major roles.The process of leaching consists of the following steps: (i) The solvent diffuses intothe solid phase; (ii) The diffused solvent dissolves the solutes (i.e. transfer thesolute to the liquid phase). Leaching is always followed by solvent recovery, whichinvolves another mass transfer operation; such as filtration. Block diagram ofleaching and solvent recovery is shown in figure 3.1.

3.2. Objectives

• To examine leaching operation.• To demonstrate solid extraction using upward and/or downward flow of thesolvent.

3.3. Theory

The model equations for the system described in this experiment are obtained fromsimple unsteady solute mass balances around the leaching and reboiler vessels;

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assuming no solute exists in the stream leaving the leaching vessel.

−CLQ = VLdCLdt, and (3.1)

CLQ = VBdCBdt, (3.2)

where Q is the solvent flow rate, V is the fluid volume and C is the solid con-centration; the subscripts L and B refer to the leaching and boiler vessels. K.Solving these equations simultaneously should lead to

CL = CL0 exp(−Q/VL t), (3.3)

CB =CL0VLVB

[1− exp(−Q/VL t)] , (3.4)

from equation 3.4 we can easily evaluate the rate constant Q/VL if we plot

− ln 1− CBVBCL0VL

vs. time. This rate constant can be viewed as an empirical masstransfer coefficient.

3.4. Procedure

Start-up

1. Weigh out the appropriate quantity of solids into the fibre bag (typically 4kg)

2. Carefully remove the solvent feed line and the vent line fitted to the extrac-tion vessel cover.

3. Undo the vessel-cover retaining clips and remove the cover to the groundlevel.

4. Place the fibre bag containing the solids into the extraction vessel, ensurethat the top of the bag is closed for upward flow (a suitable weight shouldbe placed on top of the bag to keep it closed).

5. Replace the vessel cover and hand-tighten the securing clips.

6. Replace the vent and solvent feed lines; Ensure that the solvent feed linebelow the distillate cooler is firmly attached to the bottom of the extractionvessel.

7. Close V.5.

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8. Open the cooling water line through the condenser and distillate cooler byopening and closing respective valves.

9. Set the steam on to the reboiler, ensure that the steam pressure nearly0.0 bar. Please, consult your technician before opening the steam to thereboiler.

Experiment

1. Fifteen minutes after solvent has begun to pass through the product cooler,measure the flow rate, as follows:- Set the equipment for return of solvent to reboiler section. Keep valve V.5in open position.- Close valve V.5 and when the solvent reaches the bottom mark on FI.1begin timing (the interval at which the solvent reaches the top mark of FI.1).- Adjust the heat input to give a flow rate of approximately 500 ml/min.- Repeat the measurement of flow rate till it comes to the steady state.

2. When a steady valve for FI.1 has been determined, record its value and alsorecord the reboiler temperature (TI.1). Record as well the steam pressureto the reboiler.

3. Carefully take a sample from the reboiler, by :- Opening sampling valve V.2 and remove sample.- Close V.2 when sample is removed.

4. Close valves V.5 and V.7 and begin solvent transfer to the extraction vessel.

5. Take sample of first effluent, as follows:- Close stop valve V.8.- Carefully open sampling valves V.3 and remove sample.- Close sampling valve V.3.- Open stop valve V.8.

6. Record the collection of this sample as t = 0 .

7. Record the temperature indicated by TI.2 (solvent temperature in extractionchamber).

8. Samples of reboiler contents and effluent should be taken every 2 minutesfor the first 10 minutes and then every 5 minutes until 60 minutes have beenreached.

9. Record the final temperature indicated by TI.1 and TI.2.

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10. Open valve V.7.

11. Shut down as instructed by the technician.

12. Record the final volume in the reboiler to check the loss of solvent.

13. Determine the refractive index for each of the sample drawn.


(1) Is the leaching process described in this experiment, batch, semi-batch orcontinuous? Justify your answer.

(2) How do you expect leaching rate to vary with solid pore size, i.e. the area ofcontact?

(3) What are some of the properties you would look for in a good solvent?

(4) What is the opposite of leaching?

(5) Give one example for the use of leaching process in Kuwait.

(6) Explain briefly why the effluent refractive index decreases during the courseof the experiment but the refractive index of the reboiler contents increase.

(7) A continuous countercurrent leaching process is used to extract a rose wateroil from a rose petals using ether as solvent. 558 kg/hr of feed stream (thatcontains 38 kg oil, 95 kg ether and the rest is inert). The feed stream istreated with 340 kg/hr of FRESH ether. The leached solid contains 5 kg ofrose oil and the flower petals (inert solid) per solution ratio is 1.67.

(1) Calculate the amounts of the stream leaving the process.

(2) Calculate the number of stages using McCabe-Theile method using the( y vs. x) chart.

(3)Calculate the number of stages required for the leaching of rose oil usingthe (N vs. x, y)

3.6. Report

You are expected to suggest answers to the practice questions in your report. Youranswers can be included in the introduction and/or discussion. In your reports,the following calculations MUST be reported and discussed:

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• Plot the graphs of: (a) refractive index of reboiler contents vs. time and (b)refractive index of effluent vs time.

• Use the calibration graph to convert the data for the refractive index of efflu-ent to effluent composition (kg solute / kg solvent). (Alternatively estimatethe composition by calculation).

• Estimate the amount of solute extracted during the course of the experimentusing the effluent composition vs. time plot. Compare it to the change inweight of solid change before and after the extraction process.

• Using the final refractive index and corrected volume for the reboiler con-tents, calculate the amount of solute extracted.

• Determine the empirical mass transfer coefficient using equation 3.4.• Derive and solve equations 3.1 and 3.2.

References



W. L. McCabe, J. C. Smith, and P . Harriott. Unit Operations of ChemicalEngineering. McGraw-Hill, 5th edition, 1993.

P. C. Wankat. Equilibrium Staged Separations. Elsevier, New York, 1988.

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Figure 3.1: Block diagram of solid-liquid extraction and solvent recovery.

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Figure 3.2: Flow diagram of leaching unit.

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4. SPRAY DRYING

4.1. Background

Drying generally refers to the removal of a liquid from a solid by evaporation, inother words, dryers produce a dry solid from a wet feed. Drying equipments areclassified based on the method of heat transfer to the wet material and the typeof the material itself. In general, there are direct and indirect dryers. One typeof direct drying is spray drying, the subject of this experiment. Spray drying issuitable when the feed is in liquid state, slurry, or paste capable of being dispersedin a liquidlike spray. The main application of spray dryers is for ordinary dryingof aqueous solutions and slurries, especially in food industry.. In practice, thereare three fundamental steps in spry drying process: (1) liquid atomization, (2) hotgas-droplet contact, and (3) liquid evaporating from droplets. The evaporationrate of the liquid from the droplet critically depends on the size of the droplet.The droplets formed by most industrial atomizers are 2-500μm in size. The largestdroplet rarely exceeds 500μm. For efficient drying, the hot gas should be thouroulymixed with the droplets to be dried.

4.2. Theory

There is no single theory of drying that explains the drying process regardlessof the materials or the dryer types. covers all materials and dryer types. Solidsdrying goes through two main time zones: constant rate period, and falling rateperiod and. During the constat rate period, moisture vaporize into the streamand the controlling factor is the transfer coefficient for diffusion across the gasfilm. In the falling rate periods, the surface is no longer wetted and the rate ofdrying falls. The movement of water is proportional to a concentration gradientbetween the moisture content and the water content in the drying gas.In spry drying, a large interfacial area is produced by mean of atomizer. Con-

sequently, a high rate of evaporation is obtained. The amount of drying a droplargely depends upon the evaporation rate and the contact time, which dependson the velocity of the fall and the length of path through the dryer. When thedrop is first ejected from the atomizer, its initial velocity relative to the surround-

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ing gas is high hence high rate of evaporation is achieved.. However, the dropdecelerates to its terminal velocity, therefore, the large portion of mass transferoccurs during the free-fall period. For a pure liquid drop, Marshall suggested thefollowing relation for evaporation time

t ∝ (d20 − d2t )∆T

(4.1)

Since most drops contains solids in suspension of solution, as in our case, it isfound that evaporation initially proceeds in accordance to the above equation forpure liquid. Once the solids deposition starts, a solid film is rapidly formed whichincrease the resistance of water transfer.

4.3. Objectives

• To be acquainted with spray drying operation.• To determine the efficiency of the drying equipment

4.4. Procedure

Please make sure before starting that YOU HAVE YOUR SAFETY GLASSESAND HEAT RESISTANT GLOVES ON. Then proceed with the experimentalprocedure safely.

1. Switch the system on with the blower and heater under a specified condition.

2. Prepare about 1 liter of the solution for drying.

3. Determine the flow rate of the feed pump for a particular speed using mea-suring cylinder and stopwatch.

4. Record surrounding air temperatures (wet bulb and dry bulb temperatures).

5. Switch on the air compressor to the atomizer.

6. Wait until the inlet temperature reaches the set point and check both theinlet and outlet temperatures to be constants.

7. Insert distilled water through the feed pump till the inlet and outlet tem-peratures are constants.

8. Introduce the solution to the atomizer and observe the phase of atomization.Start the stopwatch and record the temperatures every 5 minutes.

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9. After 20 minutes, stop the pump and weigh the collected powder.

10. Repeat the experiment for another two flow rates of the feed.

11. Once you finish your experiments, introduce again distilled water beforeshutting down the equipment.

12. Wait for cooling down, then switch off the pump, heater, blower and com-pressor. At the end, switch off the main.

13. Clean the parts with water.


(1) What are the differences between the wet bulb and dry bulb temperatures?How are they measured?

(2) If an average person of 70 kg was totally dried, for instance an Egyptianmummy!, estimate the final weight. Hint, estimate the water percentage inthe human body.

(3) Why dried food doesn’t rot faster than raw food?, explain in scientific terms.

(4) What is the opposite of drying operations in the chemical processes?

(5) What is the role of the atomizer in the spray dryer?

(6) Air at 75oF and 70% humidity is cooled to 50oF at a constant pressure of1 atm, The heat capacity of water is 4.186 J/g.oC. Use the psychometricchart to calculate

• (a) The fraction of water that condenses• (b) The rate at which heat must be removed to deliver 1000 ft3/min ofhumid air at the final condition.

4.6. Report

You are expected to suggest answers to the practice questions in your report.Your answers can be included in the discussion. In your reports, the followingcalculations MUST be reported and discussed:

• Determine the overall evaporative efficiency and the collection efficiency.

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• Calculate overall mass losses.• Do rigorous overall material and energy balances.

References

D. Perry and R. H. Green. Perry’s Chemical Engineers’ Handbook. McGraw-Hill,6th edition, 1984.



K. Masters. Spray Drying Handbook. 4th edition, 1985.

R. E. Treybal. Mass-Transfer Operations. McGraw Hill, New York, 3rd edition,1980.

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Figure 4.1: Spray Drying Equipment.

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5. F F

5.1. Background

The concept of fluidization is best visualized by reference to a simple experiment..Consider a bed of particles supported by a perforated place in a vertical tube.Further, consider that fluid is forced upward through the bed of particles andthat the particles have a density greater than that of the fluid. A pressure dropis obviously necessary to cause the fluid to flow through the bed and when theupward pressure force on the bed becomes equal to the net weight of the bedacting ”incipient fluidization”. However, unless the particles agglomerate, thebed does not lift as a unit but rather expands to give an increased porosity. Ifthe fluid flow rate is continuously increased the bed continues to expand until inthe limit the upward fluid velocity becomes equal to the terminal falling velocityof the particles and the particles are carried out of the tube. Between the limitsof incipient fluidization and particle carry-over, the bed is usually reasonably welldefined and, because of the appreciable circulation of particles within the bed, thebed is said to be ”fluidized”.Obviously, the fluidized bed provides an excellent means to bring fluid and

solid into intimate contact. Applications fall into two categories: chemical reac-tions (e.g., the catalytic cracking of hydrocarbons) and physical processes (e.g.,the heating and drying of particulate solids). Despite the advantages of fluidizedbeds outline above, fluidized systems do have certain disadvantages which mayprevent the use of fluidized beds for some applications. For example, it is diffi-cult to fluidize a bed of particles which agglomerate and it is sometimes difficultto maintain uniform fluidization in large diameter beds where channelling oftenresults, in the absence of baffles. Furthermore, particle attrition may be an over-riding disadvantage.

5.2. Objectives

• Determine the pressure drop through a fixed and fluidized bed.• Verify Carman-Kozeny equation.

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• Observe the onset of fluidization..

5.3. Theory

To determine the pressure drop through a fixed bed we can show that the Carman-Kozeny can be expressed in the form

∆P

L

DP

ρ (Vsm)2

3

1− = 1501−Re

+ 1.75, (5.1)

where

Vsm : Superficial velocity, m/s

Dp : Size of the particles (μm)

L : Height of the bed (m)

Re : Renolds Number

: The bed voidage,

= 1− mass of particlesparticles density * Total bed volume

If the flow rate is measured in liters/second, and Vsm is the superficial velocitym/s, then

Vsm =Q ∗ 10−3A

where A is the cross-sectional area.The pressure drop at fluidization can be predicted by using the equation,

∆P = L(1− )g ρs − ρf

where

ρs : The density of solid

ρf :The density of the fluid

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Another important process variable in fluidized bed is the minimum fluidizingvelocity. It is frequently used in fluid-bed calculations and in quantifying oneof the particle properties.. This parameter is best estimated using Baeyens andGelard correlation

Umf =0.0076(ρs − ρf)

0.934d1.8

μ0.87ρ0.006f

where

d : The size of the particle, m

μf :The viscosity of the fluid

5.4. Procedure

Study the attached schematic diagram, figure 7.2 for the equipment setup. Followthe procedure in the laboratory and prepare your own data sheet. If you haveany problem, please check with your lab instructor. You will be asked to performthe experiment for two types of fluid: water and air. To perform the experiment,check that the manometer reads zero, if it does not, adjust the level accordingly.Switch on water pump. Adjust water flow rate in increments of 0.1 l/min. readthe height of the bed, manometer reading, and the state of the bed. In case ofair, adjust the air flow control valve to give a flow rate of 2.0 l/min. Increase flowrate by 1.0 l/min. Report your data in the data sheet.


(1) In your car, there is a fluidized bed tubular reactor for the oxidation of NOto NO2, what is this equipment is called? and what is its main purpose?

(2) What do you expect to happen if the flow rate under a fluidized bed in verylarge? What if the flow is really small?

(3) In gas reactor, catalytic fluidized bed are very often been used, explain.

(4) What factors influence the pressure drop in a fluidized bed?

(5) Describe a Fixed-Bed adsorption column.

(6) Explain the advantages of fluidization in chemical processes.

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(7) A waste stream of alcohol vapor in air from a process was adsorbed byactivated carbon particles in a packed bed having a diameter of 4 cm andlength of 14 cm containing 79.2 g of carbon. The inlet gas stream havinga concentration (c0) of 1500 ppm and a density of 0.00225 g/cm3 enteredthe bed at a flow rate of 800 cm3/s. The break-point concentration is set atc/c0 = 0.01 and the breakthrough curve is given.

(a) Determine the length of the unused bed.

(b) Determine the saturation loading capacity of the carbon.?

5.6. Report

• Draw a graph of bed pressure drop against air flow rate from the experi-mental data obtained.

• Compare the predicted fluidization point with the actual fluidization pointobtained from the graph

• Give reasons for any differences between the two values.• Calculate the minimum fluidizing Velocity

References

D. Perry and R. H. Green. Perry’s Chemical Engineers’ Handbook. McGraw-Hill,6th edition, 1984.



36

Figure of the fluidised bed expirement.

37

Schematic drawing of the fluidised bed equipment.

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Water and Air properies.

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6. DOUBLE EFFECT EVAPORATOR

6.1. Background

Evaporation process is one of the major operations to concentrate aqueous solu-tions. It is based on the removal of water from a solution by boiling the solution ina vessel, which is called evaporator. If the solution contains dissolved solids, theresulting evaporated solution may become saturated. There are three major typesof solutions: (1) Those that can be heated to high temperature without decom-position as opposed to one decomposes at low temperature; (2) Those that yielddifferent crystal sizes of solid; (3) Those that boils at the same boiling temperatureof water as opposed to a higher boiling temperature.Evaporation occurs when heat is added to the solution to vaporize the solvent

(more volatile component). The supplied heat is to provide the latent heat ofvaporization. Evaporation can be carried out in one step (one effect) or in morestages (multi-effect). The use of more than a single evaporator is related to theeconomy of heat utilization. For example, in a single effect evaporator, largequantities of heat are wasted because the latent heat in the vapor leaving theboiling liquor is simply lost to the cooling water in the condenser. It would bemore efficient if the hot vapor leaving the first evaporator to be utilized to heatliquid in a second evaporator. In a double effect system, vapor from the firsteffect loses its heat to the liquor in the second effect. The major variable usedindustrially to control the efficient operation of a double effect evaporator, is theheat input to the first effect.

6.2. Objectives

• Investigate the effect of variation in steam pressure on the capacity of adouble effect evaporator.

• Calculate the economy of a double effect evaporator operating at vacuumpressure.

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6.3. Theory

Evaporator capacity is determined from the material balance equation. It is to benoted that the capacity is proportional to the rate of heat transfer, Q, throughthe heating surface of the evaporator

Q

A= U ∆T,

where U is the overall heat transfer coefficient, W/m2K, A is the heat transfersurface, m2, and ∆T is the overall temperature driving force, C.The evaporator economy may be determined by heat balance assuming that

heat losses are negligible and heat the condensate leaves the heat exchanger atsteam temperature. For the double effect evaporator in figure 6.1. The steameconomy is given by:

E =Mtot

M0,

where Mtot is the total evaporation flow rate and M0 is the input steam flow rateinto the first evaporator. For example in figure 6.1, the steam economy is

E =D1 +D2 +D3

D0.

6.4. Procedure

Follow the procedure in the laboratory and prepare the data sheet.


(1) What is the major possible application of multi-effect evaporator in Kuwait?

(2) Name two products from your grocery store that use evaporation process?Hint: one product is diary product, name it, the other is used as sweet.

(3) What are the differences between evaporation and distillation?

(4) If you can build an infinite number of evaporators, what is the efficiency ofthe process? Why is it impossible to have this efficiency?

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6.6. Report

• Briefly describe the experiment in your own words.• Do energy balance around first and second evaporators.• Determine the capacity of each effect and the overall capacity at each steampressure.

• Calculate the heat removed in the condenser Qc for each steam pressure andillustrate results on a graph.

• Calculate the economy of each effect and the overall value for double effectoperation. Which of the systems has the best economy and why?

• Construct a heat balance over the two effects and for each steam supplypressure calculate the heat loss QL for each effect. Draw a graph of steamsupply pressure against heat loss for each of the two effects. Explain howQL should vary with steam pressure and compare this with your graph.

• Calculate the overall heat transfer coefficient U for each effect, given thatthe heat transfer area of the boiler is 0.5 m2.

References



D. Perry and R. H. Green. Perry’s Chemical Engineers’ Handbook. McGraw-Hill, 6th edition, 1984.

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Figure 6.1: Schematic diagram of multiple effect evaporator.

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Figure 6.2: Double effect evaporator expiremntal setup.

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7. TUBULAR REACTOR

7.1. Background

There three major types of reactors in the chemical industries: Batch, continuousstirred tank reactor (CSTR), and plug flow reaction (PFR) also known as tubularreactor. It consists of cylindrical pipe and is normally operated at steady state.For the purpose of the material presented here, we consider systems in whichthe flow is highly turbulent and the flow field can be modeled as plug flow, i.e.there is no radial distribution of velocity. In the tubular reactor, the reactantsare consumed as they flow down the pipe. In practice, tubular reactors can bea single continuous tube or several tubes in parallel. The use of tubular reactoris favorable when the back mixing of reaction mixture is not important. Theheat transfer to or from the reactor may be accomplished by means of a jacketor shell-and-tube. Also the reactor can be packed with catalyst pellets or inertsolids.

7.2. Objectives

• Recognize plug flow reactor in practice.• Determine the variation of conversion with respect to the reactor’s residencetime.

7.3. Theory

For a tubular plug flow reactor, shown in figure 7.1, mass balance for a particularreactant A is

CAi

CA0

dCArA

=V

Q= τ , (7.1)

where CAi and CA0 (mole/litre) are the inlet and outlet concentrations of A re-spectively, and rA is the reaction rate of A in moles/seconds, V is the reactorvolume and Q is the total volumetric flow rate. For the reaction between ethylacetate and caustic soda, it is equi-molar and is second order in nature, thus

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rA = k CACB, (7.2)

where A is NaOH and B is CH3COOC2H5 and k is the reaction rate constant. Itis algebraically simpler if it is arranged for the initial concentrations in the mixedfeed stream of A and B to be equal,

FACAFFA + FB

=FBCBFFA + FB

= CAi = CBi (7.3)

whereCAF andCBF are concentrations in the feed tanks of NaOH and CH3COOC2H5respectively, and FA and FB are the individual feed flow rates (ml/min). The con-version is defined as

X =CAi − CA0CAi

(7.4)

Then equation can be solved to become

X

1−X = k τCAi (7.5)

Thus the conversion factor X/(1 − X) should be directly proportional to theresidence time, τ , provided k is kept constant (i.e. at constant temperature) forthis reaction

7.4. Procedure

Study the attached schematic diagram, figure 7.2 for the equipment setup. Followthe procedure in the laboratory and prepare your own data sheet, please checkwith your lab instructor.


(1) In your car, there is a very important tubular reactor for the oxidation of NOto NO2, what is this equipment is called? and what is its main purpose?

(2) Why is it essential that the flow in the tubular reactor to be turbulent?

(3) What is back mixing? and what are the reactors that have back mixing?

(4) What is the conversion if the reactor is CSTR? Show detailed calculations.

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7.6. Report

• Consider 0.4 litre as the reactor volume, calculate conversion and plot itagainst X/(1−X). Comment on the graph calculated.

• Consider that the reaction rate is second order, derive equation 7.5.• How do you prove that a reaction is first order?[?][?]

References

H.S. Fogler. Elements of Chemical Reaction Engineering. PTR Prentice Hall, 2ndedition, 1992.

K.B. Froment, G.F. ; Bischoff. Chemical Reactor Analysis and Design. Wiley,2nd edition, 1990.

D. Perry and R. H. Green. Perry’s Chemical Engineers’ Handbook. McGraw-Hill, 6th edition, 1984.

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Figure 7.1: Differential mass balance of tubular reactor.

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Figure 7.2: Schematic diagram of tubular reactor.

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8. W F T R

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9. S R

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