CROSS CURRENT LEACHING Aim: To study the principle governing leaching and compare the result obtained by conducting test on a mixture of Sodium Carbonate and sand. Apparatus Required:
500 ml Beaker
Funnel
Pipette
Burette
Glass Rod
Filter paper
Chemicals required:
Sodium Carbonate (20 g)
Sand (50 g)
Hcl (0.5 N)
Phenolphthalein Indicator
Principle: The process of leaching deals with removal of solvent constituent from solid by means of liquid solvent. It is generally necessary to recover or separate either by evaporation or any other means the component of new system i.e., solute and solvent which was extracted. It is important because it is generally necessary for product the most economical scheme for leaching process. Leaching differs very little from the washing of filtered solids as discussed and leaching equipment strongly resembles the washing section of various filters. In leaching the amount of solute material removed is often greater than in ordinary filtration washing and the properties of the solids may change considerably during the leaching operation. Coarse, hard or granular feed solids may disintegrate into pulp when their content of soluble material is removed. Procedure:
Cleaned and dried 500 ml beakers were taken and then numbered.
Weigh a quantity of 50 g of sand and 20 g of Na 2Co3.
Add 100 ml of distilled water and stirred well.
Filter the solution using the filter paper and the filtered off. 10 ml of the raffinate is taken and titrated against 0.5 N HCl Solution using Phenolphthalein as indicator and note the titre value.
Then add 100 ml of distilled water again to the titrate and the above procedure was repeated.
The same procedure was repeated for three stages.
Examples of leaching operation: o
Extraction of gold from its ores by using sodium cyanide solution.
o
Extraction of sugar from sugar beats using hot water
o
Extraction of tannin from tree bark using H 2O
o
Extraction of perfumes from flowers
o
Preparation of tee and coffee
Result: The stage efficiency using cross current leaching was found to be Stage I
=
Stage II
=
Stage III
=
50 g sand +
1
Extract
Extract
Extract
Raffinate
2
Raffinate
Raffinate
3
20 g Na2CO3
S.No
No. of Stage
Volume of
Volume of
extract Collected
extract taken for
(ml)
filtration against Hcl (ml)
Titre Value (ml)
100 ml
100 ml
100 ml
Water
Water
Water
DETERMINATION OF STAGE EFFICIENCY USING COUNTER CURRENT LEACHIN G AIM: AIM: To verify the properties of counter current leaching leaching of a mixture of oxalic acid and sand using water by a three stage process.
APPARATUS REQUIRED:
Conical flask-6 flask-6 no’s (500 ml)
Burette-50ml
Pipette-25ml,10ml
Beaker- 200ml (2 no’s)
Measuring jar-500ml
Conical flash-250ml(2 flash-250ml(2 no’s)
PROCEDURE: 1) 250g of sand, 2.5g of oxalic acid and 100 ml of water are taken in each of 6 conical flasks. 2) Another 100 ml of fresh water is added to the flask one, content are mixed well for 5 min and allowed to settle. 3) 100ml of leached solution is transferred from flask o ne to flask two, agitated well and allowed to settle. 4) 100ml of leached solution is transferred from flask 2 to flask 3, agitated well and allowed to settle.
5) 100ml of leached solution from flask 3 is discarded; flask 1 is removed after this trial. 6) Next 100ml of fresh water is once again added to flask 2. 7) Once again the above procedure is repeated with flask 2, 3, 4 with flask 2 removed after this trial. 8) Then with 3, 4,5 with flask 3 is removed and finally with 4,5,6. 9) From the last of trial 4, 5,6 leached solution from these are titrated against standard NaOH solution when phenolphthalein as indicator (colour changes from colourless to pale pink). 10) Readings are tabulated; calculation as done and graph are plotted.
TABULATION: FLASK NO
STAGE NO
TITRE VALUE
OXALIC ACID IN EXTRACT
% EXTRACT
CALCULATION: STANDARDISATION OF SODIUM HYDROXIDE (NaOH): Volume of oxalic acid, V 1 =
ml
Normality of oxalic acid, N 1 =
N
Volume of sodium hydroxide, V 2 =
ml
Normality of sodium hydroxide,N 2 =
N
V1N1 = V2N2 N2 = V1N1/V2 N2 =
N
STAGE 1: Volume of oxalic acid, V 1 =
ml
Normality of oxalic acid, N 1 = Volume of NaoH, V 2 = Normality of NaoH, N 2 =
N
ml N
V1N1 = V2N2 N1 =
N
Amount of oxalic acid present in extrac t = (volume * normality * equivalent weight)/1000 =
g
STAGE 2: Volume of oxalic acid, V 1 =
ml
Normality of oxalic acid, N 1 = Volume of NaoH, V 2 = Normality of NaoH, N 2 =
N
ml N
V1N1 = V2N2 N1 =
N
Amount of oxalic acid present in extrac t = (Volume* Normality * Equivalent weight)/1000 =
g
STAGE 3: Volume of oxalic acid, V 1 =
ml
Normality of oxalic acid, N 1 = Volume of NaoH, V 2 = Normality of NaoH, N 2 =
N
ml N
V1 N1 = V2N2 N2 =
N
Amount of oxalic acid present in in extract = (volume (volume * normality * equivalent equivalent weight)/1000 = Overall percentage extract = (stage 1 + stage 2 + stage 3)*100*(1/2.5) =
RESULT:
%
g
Overall percentage extracted by three stages process =
%
Amount of oxalic acid in
1) I stage =
%
2) II stage =
%
3) III stage =
%
LIQUID - LIQUID EXTRACTION Aim: To determine the extraction characteristics characteristics of the solvent at various volume. Apparatus required:
Beaker.
Burette.
Funnel.
Separating funnel column.
Chemicals required:
15% Acetic Acid solution.
Ethyl Acetate.
Sodium Hydroxide – Hydroxide – 0.5N.
Phenolphthalein Indicator.
Theory:
Liquid –liquid –liquid extraction, also known as solvent extraction and partitioning, is a method to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. It is an extraction of a substance from one liquid phase into another liquid phase. Liquid extraction is one of the main alternatives to consider. Close boiling mixture or substance cannot withstand the temperature of distillation. Even under a vacuum may often be separated from impurities by extraction which utilises chemical differences instead of vapour pressure differences.
When either distillation or extraction may be used the choice is usually distillation in spite of the fact that heating and cooling are needed. In extraction the solvent must be recovered for reuse usually by distillate and combined operations is more complicated and often more expensive then ordinary distillation without extraction. However, extraction does offer greater flexibility in choice of operating conditions, since the type and amount of solvent can be valid as well as the operating temperatures.
Procedure: 1. 50ml of 15% acetic acid is taken in four conical flasks. 2. Ethyl acetate solvent of different volume (say 25, 50, 75 & 100) in ml added to each conical flask. 3. After thorough mixing, ethyl acetate will extract the acetic acid and form two layers. 4. The two layers can be separated by using separating funnel and the raffinate was taken in the burette. 5. A standard solution of NaOH was titrated against the burette solution with the help of phenolphthalein indicator and calculations were made. Applications:
Long chain fatty acid can be separated from vegetable oils by high vacuum - distillation but more economically by extraction with liquid propane.
Liquid extraction which incurs no chemical consumption or by – product can be less costly.
Metal separation such as Uranium – Uranium – Vanadium, Tungsten – Tungsten – molybdenum and the fission products of atomic energy process are more economical by liquid extraction.
Result: The extraction characteristics characteristics of the solvent at various volumes can be determined.
S.No
Solvent amount
% of Extraction
Distillation co-efficient
Feed 5o ml Acetic Acid
Raffinate
1
Solvent
Extract
Ethyl Acetate
Feed 5o ml Raffinate Acetic Acid Solvent Ethyl Acetate
2 Extract
Feed 5o ml Acetic Acid
Raffinate
3
Solvent
Extract
Ethyl Acetate
Feed 5o ml Raffinate Acetic Acid
4
Solvent
Extract
Ethyl Acetate
Tabulation: S.No
15% Acetic Acid solution (ml)
Solvent solution (ml)
Titre Value (NaOH) (ml)
% of Extraction
Distribution Co-efficient
MOLECULAR DIFFUSIVITY AT CONSTANT PRESSURE AND DIFFERENT TEMPERATURE Aim: To determine the diffusivity of acetone air system at a known temperature. Apparatus Required:
Diffusivity measurement
Thermometer
Stop Clock
Formula Used: 1) Diffusivity,
DAB = NA*x (RTZ) x PBM / PT x (PA1-PA2)
Where, NA*- Molar flux of acetone = moles/ time x area R - Universal gas constant T - Temperature, K Z - Distance between the maximum and minimum concentration point PT - Total pressure PA1- Partial pressure of acetone at section 1 PA2- Partial pressure of acetone at section 2 2) PBM = PB2-PB1 / ln( PB2/PB1 ) Where, PBM – log mean pressure of B (mm of Hg) PB1 - pressure of area of section 1 PB2 – pressure of area of section 2 3) NA = V x ρ/ MA x T x A Where, V- Volume of acetone used ρ – Density of acetone and the temperature of water bath MA – molecular weight of acetone T- Time taken for diffusion
A – Area Procedure:
Acetone was added to the diffusivity cell surrounded by water and the heater blower was turned on.
When the temperature attained the constant value, initial distance of acetone level in the diffusivity cell from the top was noted down and the stop watch was started.
The final distance of acetone level from the top after the decrease of 1cm level from the initial value was noted down and the total time taken for the decrease in 1 cm in acetone level was recorded. The temperature was noted down. Readings were tabulated and diffusivity of acetone vapour in air was calculated.
Tabulation: TIME ( sec)
T-TUBE READING
Z1 (cm)
Z2 (cm)
MANOMETER READING
L.H.S
R.H.S
0
TEMPERATURE ( C)
Result: The diffusivity of acetone vapour in air DAB =
MOLECULAR DIFFUSIVITY AT CONSTANT TEMPERATURE AND DIFFERENT PRESSURE
Aim: To determine the diffusivity of acetone in air at constant temperature by Stephen’s tube method. Apparatus Required:
Stephen’s tube
Pipette
Rubber pipe
Stop watch
Manometer
Procedure:
Acetone is taken in Stephen’s tube and its level Z 1, is noted. A constant temperature bath is maintained with a magnetic stirrer and air is supplied at one end of the tube to ensure that there is i s no stagnant acetone at the tube outlet.
Stop watch is stored simultaneously and time required for that active level for drop by 2ml is noted. Again the level is noted (Z2).
Formula:
The experiment is repeated for different temperature.
The density of acetone is determined using specific gravity bottle.
2
1
1) DAB = PBM x RT x ( Z 2 –Z –Z1 ) ρ acetone / PT x (PA1 – PA2) x ς x MA 2) PBM = PB2 – PB1 / ln( PB2/PB1) Where, DAB = diffusivity of acetone in air (m2/s) PBM = log mean pressure of B (mm of Hg) ρ acetone = density of acetone (kg/m3) Z = Head of liquid l iquid column (mm) PT = Total pressure (mm of Hg) PB1 = Partial pressure of air at methyl isobutyl acetone surface PB2 = Partial pressure of air at other end PA1 = Vapour pressure of organic fluid PA2 = Vapour pressure of other fluid ς = Time required for decrease in level of acetone by 2mm (sec) MA = Molecular weight of acetone Tabulation: TIME ( sec)
T-TUBE READING
Z1 (cm)
Z2 (cm)
MANOMETER READING
L.H.S
R.H.S
0
TEMPERATURE ( C)
Result: The diffusivity of acetone in air pressure DAB =
PAN EVAPORATOR AIM:
To determine the rate of evaporation of varying heat.
To determine the mass transfer coefficient using Himus equation.
APPARATUA REQUIRED: Water, Scale, Thermometer, Pan, Stopwatch, Electric heater, 20% NaOH solution and Beaker THEORY: Evaporation is the process of concentrating the liquid solution by evaporating a part of solvent. It is the process where atoms and molecules in a liquid state or solid state enter the gaseous state. Evaporation of a solution is an external step in the operation of a crystalline unit. The crystallizing solution is super evaporated to make it super saturated solution. FORMULA: (I) Volume of water evaporated (V) 2
3
V = π/4(d π/4(d h) (m )
Where, d= diameter h=height of head level of solution decreased (m) (II) Mass of water evaporated (m) M = V x ρ (Kg) Where 3
V = Volume of water evaporated (m ) 3
ρ = Density of water (Kg/m ) (III) Rate of mass transfer (ω) ω = M/t (Kg/s) Where, t = time taken for evaporation (s) M = mass of water evaporated (Kg) (IV) Himus equation ω/A = K΄(Ps-Pw)
Where, 2
K΄ = Mass transfer coefficient (Kg/m s atm) Ps = Vapour pressure of water at operating condition (1 atm) Pw = partial pressure of water at operating condition (1 atm) PROCEDURE: th
Fill the pan up to 3/4 level with water and allow it to boil.
Note down the time taken for reduction of 0.2 cm of water level in the pan.
Measure the operating temperature of water using thermometer. Note down the time taken and temperature of water using thermometer.
The readings are tabulated.
FACTORS AFFECTING: Concentration, Focusing, Temperature of substance, Scale formation, Material of construction, Flow rate of water and Air current. PRECAUTIONS:
Check whether the pan evaporator is placed properly over the heater.
At any time empty pan should did not be placed on the heater.
INDUSTRIAL APPLICATIONS:
Evaporator is used to concentrate the organic solution.
Used for concentrating pharmaceutical pharmaceutical products.
Used in large scale industries (sugar industries).
TABULATION: -2
d = 18.3 x 10 m Sample
Temperature
Height
Time (min)
Rate of
Mass transfer
Evaporation
coefficient
-4
Initial Pure water
Final
10 (Kg/s)
2
Kg/m s atm
20% NaOH
40% NaOH
DIAGRAM:
RESULT: The experiment has been carried out to determine the rate of evaporation and mass transfer coefficient.
Rate of Evaporation:
Pure water
=
kg/s
20% NaOH solution
=
kg/s
40% NaOH solution
=
kg/s
Pure water
=
Kg/m s atm
20% NaOH solution
=
Kg/m s atm
40% NaOH solution
=
Kg/m s atm
Mass Transfer Coefficient: 2
2 2
SIMPLE DISTILLATION Aim: 1. To find the relative volatility of the given liquid mixture 2. To verify Rayleigh Equation analytically and also graphically. Theory: A distillation is the process of separation of liquid mixture of a material to obtain one or more desired components present. A bath of liquid changed to a kettle or steel fitted with some heating devices such as steam jacket .The change is boiled steady and the condensate (distillate) is collected in the reservoir. The apparatus is essentially a large unit or ordinary laboratory distillation will be the most volatile substance and as distillation proceeds, the vaporized product becomes leaner. The distillate can be collected in several repeated batches called out to give a series of distilled product of various positions .Thus the simple distillation procedure as a standard for composition. Materials Required:
Round bottomed flask
Thermometer
25ml Beaker
Condenser
Beaker
Measuring Jar
Glass Beads
Formula Used: (Graphically) 1) Rayleigh’s Equation (Graphically) xF
ln (F/W) = xW∫ dx/(y*-x) Where, F
-Total number of moles in feed samples (gmol)
W
-Total number of moles of residue (gmol)
y*
-Mole fraction of Vapor in equilibrium with a liquid
x
- Mole fraction of liquid
2) Rayleigh’s Equation (Analytical Method) ln (F/W) = ln { (xD-xW)/(xD-xF) } Where, xD
- Mole fraction of distillate.
xW
- Mole fraction of residue.
XF
- Mole fraction of feed.
3) Relative Volatility of Liquid mixture log (F xF/WxW) = α log {F (1- xF) / W(1- xW)} Where, α - Relative Volatility of the system. Procedure:
A Calibration chart of density of liquid mixture is mole fraction of ethanol was prepared.
100 ml of CH3CH2OH and 50 ml of H 2O is taken in a beaker and then the contents are transformed to round bottomed flask and heated.
Water is admitted to the condenser and heater is switched on.
The liquid mixture is gently until 2/3 of liquid mixture is vapourised by gradual increase of heat after regarded volume of reaction is obtained. The Heater is allowed to cool.
The volume of distillate and residue i s collected and measured.
Composition of residue and distillate is found out using calibration chart.
Rayleigh equation is verified by using using the area by plotting graph graph for (y*-x) vs x.
Industrial Applications:
To separate naphtha from crude oil
To distill Ethanol from sugar molasses
To separate Ammonia from aqueous solution of ammonia and water.
Result: The simple distillation experiment was carried out using ethanol-water system and Rayleigh’s equation was verified experimentally and theoretically. Experimental Value
=
Theoretical Value
=
The average relative volatility was found to be
α =
Tabulation: S.No
Vol. of
Vol. of
H20
CH3CH2OH
(ml)
(ml)
Wt. of H20 (ml)
Wt. of
Moles of
CH3CH2OH CH3CH2OH (ml)
(ml)
Moles of
Relative
H20
Density
(ml)
Model Graph 1
Model Graph 2
Moles of
Moles of
Mole
H20
CH3CH2OH
Fraction of
CH3CH2OH
x
1/y*-x
y
SOLID IN AIR DIFFUSION Aim: To determine the diffusivity co-efficient of the given solid in air and to determine the functional relationship among the dimensionless numbers. Introduction: Molecular Diffusion is one of the important parameter in mass transfer studies. This is movement molecule through a substance by virtue of thermal energy. The diffusivity or diffusion coefficient DAB of constituent A in a solution B, Which is a measure of diffusive mobility and is well defined by Fick’s first law. JA =DAB dCA/ dz
This experimental set up is designed and developed for a case of diffusion of (A) D Camphor through non diffusing (B) air.
For this we assume the following:
The air layer adjacent to Solid S olid is saturated with the solid.
The diffusion is steady and unidirectional.
The vapors of the solid at the end of the cell are completely driven away and only fresh air is present.
Procedure: 2
Switch ON the compressor and allow it reach sufficient pressure (4 kg/cm )
Note down the initial weight of the solid under study state.
Fix the solid to the clamp provided.
Open the air inlet valve and adjust the flow rate of air.
Start the stop watch.
Note down the final weight of the solid after two hours.
Switch OFF the compressor.
OBSERVATIONS AND CALCULATION: 1. Initial weight of the solid, m i (Z1) = _______Kg 2. Final weight of the solid, m f (Z2) = ________Kg 3
3. Air flow rate, Q = _______m /sec 4. Molecular weight of the solid, Camphor (C 10H16O) = 142 5. Total pressure = mm Hg 6. Room temperature TR
=
C
7. Density of organic liquid at T R = 999 Kg/m3 8. Diameter of diffusivity tube = 100 mm For diffusion of A through non-diffusing B, 9. NA = PT DAB ln[(1 – ln[(1 – Ya2)/(1 – )/(1 – Ya1)] RTZ 2
10. NA = Rate of diffusion/unit area, kmol/m -sec 11. PT = Total pressure, pressure, = mm hg = Bar = N/m2
3
12. R = 8.314 10 J/Kmol – J/Kmol – K 13. Z = Z1 – Z2 , Kg 14. PA1 =
Vapor pressure of the solid =
mm Hg th
Vapor pressure data for d-Camphor: Perry 6 Edition, Pg no. 3-52, Table 3-8
15. PA2
=
Partial pressure pressure at the top of the cell = 0 mm Hg at 27 Deg Celsius
16. Ya1 = mole fraction of organic liquid = Pa 1/PT 17. Ya2 = Pa2/PT = 0 The vapor pressure of organic liquid at ce rtain temperature can be obtained from the above graph. 18. NA = [Volume of Solid diffused in air density of solid at T C] Molecular weight of organic liquid
Cross sectional area of Tube X Time NA =
[Q [Q ] MA 2 ( d / 4) 20 60
(MA=Molecular wt. of organic liquid)
2
NA = ________________ Kg/m -sec
Diffusivity coefficient D AB = NA RTZ ln (1 – (1 – Ya1) PT
(1 – (1 – Ya2)
2
DAB = _____________ m /sec
Sl. No.
Flow rate
Initial
Final
Time
Level
Level
Sec
Z1 Cm
Z average
Z2 Cm
NA
DAB
Molar
Cm2/Sec
flux
1 2 3 4
Result:
The diffusivity co-efficient of the Naphthalene in air= KG = Kc =
The functional relationship was found from dimensionless numbers
Sh= Sc = Re =
SOXHLET EXTRACTION Aim: To determine the percentage extraction of solute from the given sample using different solvent. Apparatus required:
Soxhlet extraction.
Electrical heater.
Tea powder.
Solvent – Solvent – Benzene, Water.
Formula used: % Extraction =
×100
Theory: Leaching is defined as the separation of soluble material dissolved from its mixture with an inlet solid by means of liquid solvent. The leaching is used in varieties of industrial applications such as metallurgical sugar pharmaceutical etc. In the soxhlet extraction the solid from an open solvent precoated through the agitated bed of solids. The solids are dispersed into the solvent and later separated from it. Procedure:
The known amount of solid material was weighed and taken into the soxhlet apparatus.
The solvent is placed in the round bottom flask which is kept on the heater.
Then the entire apparatus setup is along with condenser and the solvent is heated.
The solvent vapour condensed in the condenser and it was sent down through the solid material.
The condensed solvent was extracted the soluble material from the mixture and it is collected in the round bottom flask.
The complete removal of soluble material from the solid material is taken into the drier.
Result: The percentage of extraction of the solute from the given samples by using soxhlet apparatus is found by,
% Extraction [ for Water]
=
% Extraction [ for Ethanol]
=
Keq
=
SOXHLET APPARATUS
STEAM DISTILLATION AIM: To determine the overall efficiency of steam distillation. To determine the theoretical temperature for the same operation. APPARATUS REQURIED:
Round bottom flask
Steam generator
Condenser
Heater
Measuring jar
PROCEDURE: 1) The feed is aniline (mixture of aniline and water) in the ratio 1:3 was feed to the column.
2) Then the heater was switched on to supply heat for steam generator. 3) The steam was passed into the separation vessel and packed up aniline and causes to condensate. 4) Upto 1/3
rd
of the mixture is obtained as distillate water is measured after separating it
into a conical flask. 5) The composition of distillate was calculated.
TABULATION: S.No
Temperature (K)
PA = PT – PT – PB (mm of Hg)
CALCULATION: Volume of aniline in distillation
=
Volume of water in distillate
=
Mass of aniline in distillate
=
Mass of waterr in the distillate
=
Volume of aniline in residue
=
Volume of water in the residue
=
Mass of aniline in residue
=
Mass of water in residue
=
Initial bioling point
=
Final boinling point
=
Vapor pressure of aniline PA
=
Vapour pressure of water P B=PT-PA = 760-
TABULATION: S.No
Temperature
Vapour pressure of Water (PB) mm of Hg
Temperature
Vapour pressure of Acetone mm of Hg
FORMULA USED: 1. WA= (PA-MA)/( PB – MB) =
kg of aniline/ kg of water
Where, MA- Molecular weight of aniline MB - Molecular weight of aniline 2. Actual amount of aniline /kg of steam used, WA’=
Kg of aniline / kg of steam
3. Efficiency = ( WA’/ WA)*100
RESULT:
1) Theoectical amount of aniline / kg of H2O in distillation =
kg of aniline/ kg of
steam 2) Actual amount of aniline/kg of steam used = 3) Vapouraisation efficiency =
kg of aniline/ kg of steam
%
4) Theoretical temperature of steam distillation=
0
C
STERLING JACKETED PAN EVAPORATOR
AIM:
To determine the mass transfer coefficient.
APPARATUA REQUIRED: Pan, Steel Pan, Stopwatch Thermometer, THEORY: Evaporation is an operation that is carried out in industry as a means of concentration of the weak liquor (or) solution by vaporising of solvent. The heat is utilised to (a) Increase the temperature of the solution to its boiling point and, (b) Supply the latent heat of vaporisation of solvent its process where the atoms and molecules in a liquid state or solid state in substance sufficient energy enter the gaseous state. FORMULA: (I) Mass of water evaporated (m) m = V x ρ (Kg) Where 3
V = Volume of water evaporated (m ) 3
ρ = Density of water (Kg/m ) (II) Rate of mass mass transfer or (Evaporation) (ω) ω = m / t (Kg/s) Where, t = time taken taken for evaporation evaporation of water (s) m = mass of water evaporated (Kg) (III) Himus equation ω/A = K΄(Ps-Pw) Where, 2
K΄ = Mass transfer coefficient (Kg/m s atm) Ps = Vapour pressure of water at operating condition (1 atm)
Pw = Partial pressure of water at operating condition (1 atm) PROCEDURE: th
Fill the pan up to 3/4 level with water and allow it to boil.
The steam is admitted in the jacket
Note down the time taken for reduction of 0.5 cm of water level in the pan.
Measure the operating temperature of water using thermometer. Note down the time taken and temperature of water using thermometer.
The readings are tabulated.
The thick liquor is then drained out from outlet.
Note down the temperature of steam inlet and steam outlet.
TABULATION:
S.No
Temperature
Temperature
Temperature of
Initial
Final
of Feed (˚C)
of Steam (˚C)
Condensate(˚C)
height(cm)
height(cm)
Time (S)
TABULATION:
S.No
Temperature (˚C)
Ps (atm)
Ps – Ps – Pw (atm)
W/A (Kg / m3)
K = W / A / Ps – Ps – Pw 2
(Kg/m s atm)
RESULT: The experiment has been carried out to determine the rate of evaporation and mass transfer coefficient by using sterling jacketed pan evaporator.
Mass Transfer Coefficient
Rate of evaporation
2
=
(Kg/m s atm)
=
TRAY DRIER Aim: To study the characteristics of a given sample (sand) in a tray drier and to find the vapour drying zone for the above sample and drying for constant rate period and falling period. Apparatus Required: Sand, tray drier, equipment, stopwatch. Formula used:
Rate of drying: 2
N=SS/A (-dx/dt) g evaporated/m s Where, Ss=dry weight of the sample (g) 2
A= area of the drying surface (m )
Time of drying for constant rate period: tC= SS/A(X1 –XC) s Where, tC= Rate of drying during constant rate period X1= Initial moisture content (g of moisture)/(g of dry solid) XC= Critical moisture content (g of moisture)/(g of dry solid)
Time of drying for falling rate period:
tF= SS/A ∫
Where,
∫ = Area under the curve
Total time of drying: t = tF + tC (s) Where, tF= Time of drying for falling rate period (s) tC= Time of drying for constant rate period (s)
Procedure:
Accurately weigh the sample and it is taken. It is made to paste by adding water in a tray.
It was placed inside the drier and the door was tightly closed after which the drier was balanced and heated to by switching on the heater and the temperature was made constant by switch on the blower.
Time was constant for every 6 minutes. The sample was taken from the drier and weighed.
The procedure was repeated and readings are noted.
Industrial Applications:
Preparation of organic compounds
They are used to dry heat sensitive materials
In pharmaceutical industry
Preparation of food materials.
Graph: Moisture content (X) Vs the rate of drying (N) Moisture content (X) Vs VN
Result: Drying characteristics of sample using tray drier is calculated: Time of falling rate period,
tF=
Time of constant rate period,
tC=
Total time for drying,
t=
Rate of drying
=
Critical moisture content
=
Tabulation: S.no
Time (min)
Weight
Initial weight of sand=
dt (s)
-
dx/dt*10
N
1/N 2
Moisture Content dx g of moisture/ g of solid (x)
m s/g
g of moisture/
g of moisture/
g of dry solid (w)
g of dry solid
2
g Evaporated/m s
VAPOUR-LIQUID EQUILBRIUM AIM: To determine the temperature (T), liquid composition (x) and vapor composition (y) at equilibrium and to draw the vapor-liquid equilibrium curve for the given system of liquids. l iquids. PROCEDURE: 1. Preparing Calibration Chart:
Mix known volumes of Water – Water – Ethanol (25:0), (20:5), (15:10), (10:15) and (5:20)ml.
Determine the densities or refractive index mixtures by using specific gravity bottle or apply refractometer.
Determine the density or RI of pure water and ethanol at room temperature.
Convert the volume fraction into mole fraction of ethanol (more volatile component).
A plot of density or RI (y-axis) Vs mole fraction of ethanol (x-axis) is the required calibration chart.
If density or RI is known mole fraction can be easily obtained from the calibration chart.
2. Experimental Procedure:
Take 250ml of ethanol and 200ml of water in the RB flask and switch “ON” ON” the heater and adjust for required temperature.
Allow water to flow through the condenser.
Once the reflux is obtained in the system, note the constant temperature. temperature.
Switch “OFF” OFF” the heater and determine the density of the distill ate and residue.
Repeat the procedure for different composition of feed.
Plot the equilibrium curve.
TABULATION I: Volume
Volume
Weight
Weight
Moles of
Moles of
Mole
Specific
of
of Water
of
of Water
Ethanol
Water
fraction
Gravity
Ethanol
Ethanol
of Ethanol
TABULATION II: Mixture – Mixture – I
Mixture – Mixture – II
Mixture-III
A:100ml
B:150ml
A:150ml
B:100ml
A:50ml
B:200ml
Distillate
Residue
Distillate
Residue
Distillate
Residue
x
y
x
Y
X
y
Weight of empty bottle
Weight of bottle + sample
Weight of sample
Volume of sample taken
Density
TABULATION III:
Volume of feed (ml)
Temperature
Density gm/cc
Mole fraction of A
Ethanol
Water
Distillate
Residue
Distillate
Residue
A
B
y
x
y
x
RESULT: The temperature of liquid composition and vapour composition at equilibrium stage was found. The vapour liquid equilibrium equil ibrium curve was drawn.
1. For 100ml ethanol and 150ml water Temperature
=
Vapour composition (y)
=
Liquid composition(x)
=
2. For 150ml ethanol and 100ml water Temperature
=
Vapour composition
=
Liquid composition
=
3. For 50ml ethanol and 200ml water Temperature
=
Vapour composition
=
Liquid composition
=
WETTED WALL COLUMN Aim: To determine the mass transfer coefficient by the evaporation of liquid into steam in the wetted wall column. Apparatus required:
Measuring flask
Conical flask
Toluene
Benzene
Stop watch
Formula: 2
1. NA = m/A x t
(kg/m s)
Where, 2
NA = Flux kg/m s A = Cross sectional area in m
2
t = Time taken for evaporation in s m = Mass of the sample evaporated 2. Cross sectional area = 3. KG = NA x 2/(PA2 – PA1) 4. Total pressure = mgh 0.83
5. NSH = 0.23(Nre) 6.
NRe = DVρμ
7. NSC = m/ρDAB 8. NSH = kAPBmRT
(NSC)
2
2
(m )
Tabulation: S.no. 1.
Data
Benzene
Sample taken from feed reservoir
2.
Sample collected
3.
Sample evaporated
4.
Time taken
data Mol. Wt.
For benzene 78.14
For air 28.5
CT
562.1 k
132 k
CV
259 m /kg mol
2
2
86 m /kg mol
Technical data: 3
Density of air = 1.2 kg/m
Length of column = 1.2 m Diameter of the column = 8.1 cm -5
2
Viscosity of air = 1.90X10 kg/m s Density of toluene = 862.27 kg/m
3
o
3
Density of benzene at 85 C = 796.8 kg/m Procedure:
Fill the reservoir with the measured quantity of volatile liquid. Switch on the blower.
Control the rate of air entering
Open the feed reservoir wall and note the time of liquid entering the wetted wall column.
Liquid entering runs in a thin film inside the vertical column while air enters from the bottom.
Measure the pressure difference.
The inlet air will vaporize some quantity of liquid and removing liquid is collected in the measuring jar.
Mass transfer co-efficient can be found from graph.
Result: 2
Mass transfer co-efficient of benzene (experimental)
=
(kg/m s)
Mass transfer co-efficient of benzene (theoretical)
=
(kg/m s)
2
