Schx4007 Mass Transfer Lab

SATHYABAMA UNIVERSITY

(Established Under Section 3 of the UGC Act 1956)

CHENNAI-600 119

DEPARTMENT OF CHEMICAL ENGINEERING

SCHX4007- MASS TRANSFER LABORATORY MANUAL

NAME : _________________________________

REG. NO : __________________________________

SEMESTER :

MASS TRANSFER LABORATORY

LIST OF EXPERIMENTS

Serial No TITLE OF THE EXPERIMENT PAGE NO

SOLID – GAS OPERATIONS

1. Tray Drier

2. Vacuum Drier

3. Rotary Drier

SOLID – LIQUID OPERATIONS

4. Leaching

5. Continuous counter current decantation (CCD)

6. Adsorption Studies

LIQUID – LIQUID OPERATIONS

7. Packed bed Extraction column

8. Rotating Disc Contactor (RDC)

9. Liquid – Liquid Diffusivity

GAS – LIQUID OPERATIONS

10. Simple Distillation

11. Steam Distillation

12. Packed Bed Distillation

13. Absorption studies

14. Wetted wall column

15. Diffusion Co-efficient of organic vapor in air

16. Surface Evaporation

TRAY DRYER Expt No: 1

Date:

AIM: To determine the drying characteristics of a given sample by drying in a Tray dryer.

Theory:

Drying is removing relatively small amounts of water from material. In drying water is usually removed

as a vapor by air. In evaporation large amounts of water is removed from material when the water reaches

its boiling point.

If the wet stock is kept in a compartment over a tray/plate and air is passed over it by a blower, the dryer

is known as atmospheric tray dryer. When a slab of wet stock is placed in contact with air, water

evaporates from its surfaces and diffuses as vapor through the air film that covers the surface. The

concentration of the moisture on the moisture on the surface decreases. A concentration difference is

setup, which makes them moisture in the interior to diffuse to the surface or it moves by capillary action.

The drying rate at this point is constant and independent of moisture content of the slab as there is

continuous supply of moisture to the surface, provided the air velocity, temperature, and humidity do not

change. During this constant-rate period the rate of drying is controlled by the rate of vapor diffusion

through the air film and is, therefore, depends upon the humidity across the film and the thickness of the

film.

When the moisture content become less than a certain value known as the critical moisture content, the

rate of diffusion of moisture from interior of the slab to the surface begins to fall hence the rate of drying

decreases. This is known as the falling rate, which continues till equilibrium moisture content is reached,

after further drying is not possible under the particular drying conditions.

PROCEDURE:

1. Take about 150 gm of the given sample (sand) in the tray, spread it uniformly.

2. Add about 30 ml of water to the sample.

3. Set the temperature at desired level (say 95 C), when the temperature is around 55 C switch on

the blower. Start the stop clock only when the temperature reaches the desired level.

4. Note down the initial weight of the sample.

5. Note down the time taken for the digital balance for every 2 gm loss in weight.

OBSERVATIONS AND CALCULATION:

Moisture content in dry basis (X) = weight of water remaining / weight of dry solid

dX = X2-X1 dt = t2-t1

Drying area, A = πd2/4= ____________m

2

TABULAR COLUMN:

Serial no. Weight

change,

dw (kg)

Time

taken,

(sec)

Moisture

content,

X

(kg

moisture /

kg dry

solid)

dX / dt,

Kg/sec

N,

kg/m2sec

1/ N

m2sec /

Kg

I. Graph of N vs X

Nc

Constant rate

N

Falling rate

X* XC

II. Graph of 1/N vs X

From Graph 2 Get the values of Xc,Nc,Xe

By Graphical method

Time of constant rate period, TC = Area under the curve from Xinitial to Xc / (A NC ) = ___________sec

Time of falling rate period, TF = Area under the curve from Xc to Xfinal /A NC = ___________sec

RESULT:

1. Equilibrium Concentration (Xe*) =

2. Critical Moisture Content (Xc) =

3. Time period of Drying = sec ( By graphical)

ROTARY DRYER

Expt No: 2

Date:

AIM:

To study the operation of rotary dryer and to determine the operating characteristics & efficiency of the

dryer.

Theory:

A single pass rotary dryer in which the feed is introduced through a hopper with a screw conveyor driven

by reduction geared motor is passed countercurrent hot air.

The air is obtained from a blower is measured by an orifice meter and heated with finned air heaters and

is regulated by a valve. The rotary dryer is driven by a reduction geared motor ,chain and sprocket.

Provision to measure the inlet, outlet Temperature and humidity are made. Also provision for visual

observation through the dryer is made.

PROCEDURE:

1. Take about 500 gm of the given sample (sand) in the tray, spread it uniformly.


3. Record the wet bulb and dry bulb temperature of atmospheric air.


the blower. Switch on the blower and heater and regulate the flow rate and temperature. Wait till

steady state is reached.

5. Switch on the motor so that the feed starts falling into the dryer and dryer is rotating.


7. Note down the weight of the feed collected for every 3 minutes.

8. Weigh the sample collected and find out difference in the weights (dw) at dt= 3 mins.

9. Note down the following data

a. Humidity and temperature of entering and exit air, temperature of the exit solid, moisture

content of entering and exit solid sample.


Dry bulb temperature = _____________C

Manometer reading Rm = _____________ m

Hot air temperature Th = ______________C

Room temperature TR = _______________C

Density of air air = PM/RT =_________ Kg/m3

(at room temp)

Density of water W at Tm =( Th + TR)/2 , = ____________Kg/m3

Drying area, A = πd2/4h= ____________m

2

Weight of dry solid WS = 0.5 Kg

Drying rate, = dw/(Adt) , where dw = change in weight = 0.002 Kg

Orifice diameter, Do = 0.028 m

Pipe diameter, DP = 0.05 m

Area of the orifice = Ao = πDo2/4

Co-efficient of discharge of orifice meter Cd = 0.62

= Do/Dpipe = 0.028/0.05 = 0.56

Chamber dimensions = 1500mm x 150 x150 x 3.14/4 = 0.026 m2

Cross section are of the duct = Ac = 150 x150 x 3.14/4 =

AC = 0.17 m2

Pressure drop across orifice meter is P = Rm g (W - air) = _____N/m2

Velocity of air at orifice = Vo

Vo = Cd (2P)1/2

= ___________m/sec

(1 -

4)

1/2 (air)

1/2

Mass of air, M = air Vo Ao = __________Kg/sec

Mass of air in the chamber, M = air Vair Ac

Velocity of air in the chamber, Vair = M/(air Ao) = ___________ m/sec

Mass velocity of air G = Vair Vair = __________Kg/(m2)(sec)

Humidity from Wet bulb and Dry bulb temperatures (from humidity chart)

H1 = _________________ Kg / Kg dry air

Heat transfer co-efficient, h = 24.2 G 0.37

= ___________W/m-C

Molal humidity, Hm1 = H1 (29/18) = ________Kgmoles / Kgmole dry air

Humidity at temperature Th (from chart) H2 = _____Kg/Kg dry air

Molal humidity, Hm2 = H2 (29/28) = ______ Kgmoles / Kgmole dry air

Mass transfer Co-efficient Kg = / (Hm2 – Hm1)

Kg = _____ Kg / (m2)(sec)(unit humidity difference)

TABULAR COLUMN:

Serial no. Moisture

content, X

Weight

change, dw

(gm)

Time

taken, sec

dw/dt, 10-

3, Kg/sec

, kg/m2

-sec

1/

1. Graph X vs T

Slope=dw/dt

Time, (sec)

2. Graph of N vs X

3. Graph of 1/N vs X

Drying time Calculation



RESULT:



3. Time period of Drying = T constant + T falling =

VACUUM DRYER

Expt No: 3

Date:

AIM:

To study the drying characteristics of the vacuum dryer

Theory:

Vacuum shelf dryers are extensively used for effective, efficient drying of Pharmaceuticals, Food and

Chemical Products. They are very useful in temperature sensitive, easily oxidized material, toxic material,

valuable solvents, dusty material and hygroscopic material.

PROCEDURE:

1. Take about 50 gm of the given sample (calcium carbonate) in the tray, spread it uniformly.


3. Set the vacuum to the desired value.


the blower. Start the stop clock only when the temperature reaches the desired level.


6. Note down the time taken for the digital balance for every 2 gm loss in weight.


Moisture content in dry basis (X) = weight of water remaining / weight of dry solid

dX = X2-X1 dt = t2-t1

Drying area, A = πd2/4= ____________m

2

TABULAR COLUMN:

Serial no. Weight

change,

dw (kg)

Time

taken,

sec

Moisture

content,

X

(kg

moisture/kg

dry solic)

dX / dt,

Kg/sec

N,

kg/m2sec

1/ N

m2sec /

Kg

1. Graph of N vs X

2. Graph of 1/N vs X

From Graph 1 Get the values of Xc,Nc,Xe

By Graphical method



RESULT:



3. Time period of Drying = T constant + T falling = (By graphical)

DIFFUSIVITY OF VAPOR IN AIR

Expt No: 4

Date:

AIM :

To determine the diffusion coefficient of organic vapor (ACETONE) in air.

THEORY :

Individual molecules in a gas move due to molecular diffusion by virtue of their thermal energy. Initially non-

uniform mixture eventually becomes uniform in concentration due to molecular diffusion. The driving force for

the diffusion is the concentration gradient. The fundamental equation for steady state molecular diffusion is the

Fick’s Law.

JA = -DAB dCA / dZ

The diffusion coefficient is the property of the system & depends on the temperature, pressure and nature of the

components.

EXPERIMENTAL PROCEDURE :

1. Fill the diffusion cell with acetone

2. Note the initial level of acetone

3. Start the blower and the stopwatch.

4. Note the time taken for level drop of every 2mm.

OBSERVATIONS :

Room Temperature, T = 30 0C= ____K

Normal boiling point of the given organic liquid (acetone) = 56.4 0C

Molecular weight of the given organic liquid (acetone) = 58.08 0C

Gas constant R = 8.314 KJ / Kg mol K

Total pressure P = 1.013 x 10-5

N / m2

OBSERVATION TABLE :

Sl.

No.

Zinitial

mm

Zi

mm

Z = Zi – Z1

mm

sec

DAB

m2/sec

1

2

3

CALCULATIONS :

Partial pressure of acetone P* (acetone) = 0.318 x 10

5 N / m

2 (by using Antoine equation)

(acetone) = Density of acetone = 0.792 g / cc = 792 Kg / m3

M(acetone) (Molecular weight of acetone) = 58.08

M(air) (Molecular weight of air) = 29

RT Z12 – Z

2i

DAB = ------------------------ ------------ --------------

P Z M(acetone)

P Ln --------------

P – P*(acetone)

Result

The diffusion coefficient of the organic vapor in air = _________ m2/sec

LIQUID - LIQUID DIFFUSION

Expt No: 5

Date:

AIM :

To determine, the diffusion coefficient of oxalic acid in water at different temperatures.

THEORY :

Diffusion is the movement of molecules under the influence of a physical stimulus, the most

common cause being the concentration gradient.

The theory of diffusion in liquids is not so advanced as the gas diffusion and also experimental data

are scarcely availability. The diffusion in liquids is generally 4 – 5 orders of magnitude smaller than

in gases. In liquids the diffusion occurs by random motion of the molecules but the average distance

traveled between collisions is less than the molecular diameters. Many researchers have given

relations that show the diffusivity dependence on viscosity and temperature. The formation of sharp

interface between adjacent phases in the system is done in this set up by the porous diaphragm.

EXPERIMENTAL SET UP:

The unit consists of a glass solvent section (outer bottle). The sample bottle consists of porous

diaphragm and a long stem which can be attached to the motor spindle by the clamp provided. A

Teflon stopper with a rotary joint is provided for free rotation of the bottle.

The solvent bottle is kept in a constant temperature bath, whose temperature can be regulated at

desired level. Concentrations are measured by the standard conductivity values.

EXPERIEMNTAL PROCEDURE :

A) To Determine the cell constant :

1. Prepare 0.1N standard Oxalic acid solution

2. Prepare 0.1N standard NaOH solution and standardize with standard oxalic acid

3. Fill the diffusion cell with known standard of NaOH solution and outer beaker contain

solvent (water)

4. Instantly immerse this bottle in the outer bottle containing water and fix it to the stirrer

motor shaft such that, the liquid level inside the inner bottle and outer bottle are same.

5. Start motor rotation and stop watch and observe for keeping the constant speed (about 30

rpm).

6. The experiment is carried for a known time (20 min) motor is stopped and the solute

bottle is instantly withdrawn.

7. Stop the stirrer and the solvent is taken for the analysis of NaOH concentration by

titrating with std.HCl

B) To determine the diffusivity :

1. Proceed the experiment as above with slight modification that the cell contain oxalic acid

and the outer beaker contains water

2. Conduct the experiment as with same rpm and same temperature as done above..

3. Then The above procedure can be repeated for different bath temperature.

Tabulation

1. Standardisation of NaOH

Indicator : Phenolpthalein

S.No Volume of oxalic acid (ml) Burette readings (ml) Volume of NaOH (ml

Initial Final

Volume of oxalic acid V1 = ml

Normality of oxalic acid N1 = N

Volume of NaOH V2 = ml

Normality of NaOH N2 = N

2. Standardisation of HCl

Indicator : Phenolpthalein

S.No Volume of NaOH (ml) Burette readings (ml) Volume of HCl (ml

Initial Final

Volume of NaOH V2 = ml

Normality of NaOH N2 = N

Volume of HCl V3 = ml

Normality of HCl N3 = N

Titration for Water diffused NaOH Vs HCl

S.No Volume of Water diffused NaOH (ml) Burette readings (ml) Volume of HCl (ml)

Initial Final

Volume of Water diffused NaOH = ml

Normality of Water diffused NaOH = ? N

Volume of HCl V3 = ml

Normality of HCl N3 = N

Titration for Water diffused Oxalic acid Vs NaOH

S.No Volume of Water diffused Oxalic

acid (ml)

Burette readings (ml) Volume of NaOH

(ml Initial Final

Volume of NaOH = ml

Normality of NaOH = N

Volume of Water diffused Oxalic acid = ml

Normality of Water diffused Oxalic acid = ? N

Calculation:

A) Initial conc. of NaOH in the inner cell (n1) = N

Initial conc. of NaOH in the outer cell (n2) = 0 N

Initial difference in conc. (n1-n2) = N

B) Final conc. of NaOH in the inner cell (n3) = N

Final conc. of NaOH in the outer cell (n4) = N

Final difference in conc. (n3-n4) = N

C) Cell Constant B = {1/DAB t} ln {initial conc difference / final conc difference}

DAB (for NaOH – Water system) = 1.5 x 10 -5

cm2/s

t is experiment time = ___ min = _________ sec

B =

Determination of Diffusivity

a. DAB = {1/Bt} ln [(na – nb) / (nc- nd)]

(na – nb) is the initial difference in oxalic acid concentration

(nc- nd) is the final difference in oxalic acid concentration

Result:

The diffusion coefficient of oxalic acid in water is = __________ m2/sec

Expt No: 5

Date:

SOLID LIQUID EXTRACTION (LEACHING)

LEACHING STUDIES

AIM:

To study the principles governing leaching and compare the results obtained by conducting the

batch leaching test and a mixture of sodium carbonate and sand.

Procedure:

STANDARDISATION OF 1N HCl:

Standard solution of sodium carbonate (1N) was prepared by dissolving 5.3g of sodium

carbonate in 100ml of water.

Wash 50ml burette and rinse with 1N HCl. Pipette out 20ml of std. sodium carbonate into

conical flask and titrate it with HCl in burette using methyl orange indicator. The end point is the

appearance of red colour.

Leaching Studies:

Take five clean and dry 500ml beakers and number them 1 to 5. To each beaker add 50g of sand

and 10g of sodium carbonate. Mix both the solids thoroughly. Add exactly 400ml, 200ml,

133ml, 100ml, and 80ml of distilled water in the beakers and note them. Stir the contents. Pipette

out 10ml, 8ml, 6ml, 4ml, 2ml of supernatant liquid from each beaker respectively and titrate it

with std. HCl using methyl orange indicator.

STANDARDISATION OF 1N HCl

Sl. No. Volume of Sodium carbonate (ml) Burette reading Volume of HCl, (ml) Indicator

Final Initial

Volume of sodium carbonate = V1 = ml

Normality of sodium carbonate = N1 = N

Volume of HCl = V2 = ml

Normality of HCl = N2 = N

Leaching

Bottle

No

Weight

of sand

taken (g)

Weight of

Sodium

carbonate

(g)

Volume

of water

added

(ml)

Volume of

supernatant

liquid (ml)

Volume of

Hcl

Consumed

Amount of

sodium

carbonate

recovered (g)

%

recovered

1 50 10 400 10

2 50 10 200 8

3 50 10 133 6

4 50 10 100 4

5 50 10 80 2

Amount of sodium carbonate present in supernatant liquid =

Volume of HCl required * N2 * equ. Weight of sodium carbonate * volume of water added

= -----------------------------------------------------------------------------------------------------------------

1000 * volume of supernatant liquid

Percentage of sodium carbonate recovered

Amount of sodium carbonate present * 100

= ----------------------------------------------------------

Amount of sodium carbonate added to beaker

RESULT:

The % of sodium carbonate extract was found to be =

LEACHING STUDIES - II

Expt No: 6

Date:

AIM:


batch leaching test and a mixture of sodium hydroxide and sand.

Procedure:


Standard solution of sodium carbonate (1N) was prepared by dissolving 4g of sodium hydroxide

in 100ml of water.

Wash 50ml burette and rinse with 1N HCl. Pipette out 20ml of std. sodium hydroxide into

conical flask and titrate it with HCl in burette using phenolpthalein indicator. The end point is

the disappearance of pink colour.

Leaching Studies:


and 10g of sodium hydroxide. Mix both the solids thoroughly. Add exactly 400ml, 200ml,



with std. HCl using phenolpthalein indicator.


Sl.

No.

Volume of Sodium hydroxide

(ml)

Burette

reading

Volume of HCl,

(ml)

Indicator

Final Initial

phenolpthalein

Volume of sodium hydroxide = V1 = ml

Normality of sodium hydroxide = N1 = N



Leaching

Bottle

No

Weight

of sand

taken (g)

Weight of

Sodium

hydroxide

(g)

Volume

of water

added

(ml)

Volume of

supernatant

liquid (ml)

Volume of

Hcl

Consumed

Amount of

sodium

hydroxide

recovered (g)

%

recovered

1 50 10 400 10

2 50 10 200 8

3 50 10 133 6

4 50 10 100 4

5 50 10 80 2


Volume of HCl required * N2 * equ. Weight of sodium hydroxide * volume of water added

= -----------------------------------------------------------------------------------------------------------------


Percentage of sodium hydroxide recovered

Amount of sodium hydroxide present * 100

= ----------------------------------------------------------

Amount of sodium hydroxide added to beaker

RESULT:

The % of sodium hydroxide extract was found to be =

Wetted Wall Column Expt No: 7

Date:

Aim:

To determine the mass transfer coefficient for the given system using the experimental setup

Introduction: Mass transfer coefficient is the rate of mass transfer per unit area per unit concentration

difference and is usually based on equal molal flows. Wetted wall column is used to determine the mass

transfer coefficient experimentally. This device can be used when the area of contact of the phases is

known and boundary-layer separation does not exist. Wetted wall column gives valuable information on

mass transfer to and from fluids in turbulent flow. A wetted wall tower is essentially a vertical tube with

means for admitting volatile pure liquid at the top and causing it to flow downward along the inside wall

of the tube, under the influence of gravity, and means for admitting gas to the inside of the tube, where it

flows through the tower in contact with the liquid. Generally the gas enters at the bottom countercurrent

to the liquid, but parallel flow can be used. Measurement of the rate of evaporation of the liquid into the

gas stream over the known surface permits calculation of the mass-transfer for the gas phase. In the

wetted wall column, the interfacial area A, except for some complications from ripple formation, is

known and form drag is absent.

Procedure:

1. The overhead tank and the wet bulb thermometer provided are filled with the liquid (preferably water

or NaOH) to be evaporated. The liquid is passed through the column at such a flow rate that it forms a

film along the walls of the wetted wall column.

2. The liquid flow rate is measured using a rotameter provided.

3. The blower is switched on and the flow rate of the air through the column is measured with the help of

the orifice provided. The coefficient of discharge of the orifice can be assumed to be 0.62

4. The wet bulb and dry bulb temperatures of the inlet and exit air are noted down with the help of the

thermometers provided.

5. The experiment is repeated for different air flow rates.

Observations:

Sl.No. Manometer Reading, m Wet bulb

temperature, C

Dry bulb

temperature,

C

Rotameter

reading, LPH LHS RHS

Initial Final Initial Final

Calculations:

Manometer reading Rm = LHS ~ RHS m

Density of the manometric fluid ρA = Kg/m3

Density of the air ρB = Kg/m3

Pressure difference ΔP = Rm g (ρA - ρB) Kg/m sec2

Expansion factor Y = 0.95

Diameter of the orifice do = 12.5mm = m

Area of the orifice = ao = πdo2/4 m

2

Mass flow rate of air m = 0.68 Y ao ( 2ΔP ρB)0.5

Kg/sec

Diameter of the column D = 25mm = m

Area of the column A = πd2/4 m

2

Mass velocity G = m/A Kg/sec m 2

Reynolds number NRe = DG/μ

Diffusion coefficient of the liquid DAB = m2/sec

Viscosity of the liquid μ = Kg/msec

Density of the liquid ρ = Kg/m 3

Schmidt Number NSc = μ/ρ DAB (no unit)

Total pressure of the system P = 1 atm

Wet bulb temperature of exit air = tw

Partial pressure of liquid at the inlet = pA1

Dry bulb temperature of exit air = td

Vapor pressure of liquid at this temperature (from Perry) = pA2 = atm

PBM = pA1 - pA2/ ln [(P - pA2)/(P - pA1)]

Mass transfer Coefficient = kg Kgmol/ sec m2

atm

kg = 0.023 (NRe)0.83

(NSC)0.44

πDAB / (d PBM)

Result

The mass transfer co-efficient for the given system = ------------ Kgmol/ sec m2

atm

ADSORPTION STUDIES

Expt: 8

Date:

AIM:

To determine the adsorption isotherms of acetic acid on activated charcoal.

THEORY:

At a given temperature, the mass of adsorbate adsorbed by a solid adsorbent for various concentrations is

given by the following equation:

x / m = k c1/n

The above equation is known as Freundlich adsorption isotherm. Here, x is the mass of the adsorbate

adsorbed by m g of the adsorbent at various concentrations of the adsorbate. For liquids, this

concentration is given in terms of g/mole; for gases, in terms of pressures.

The values of k and n are constants at a given temperature for the given solid adsorbent and adsorbate

PROCEDURE:

1. Transfer 1 g of charcoal to each of six reagent bottles Number the reagent bottles.

2. Add 60, 50, 40, 30, 20 & 10 ml of 1N oxalic acid and 0, 10, 20, 30, 40& 50 ml of distilled water by

means of burette to each of the bottles in that order.

3. Shake the bottles vigorously and leave for one hour in a thermostat at desired temperature with

intermittent shaking.

4. Filter the solution from each bottle using filter papers and collect the filtrate in a labeled conical flask.

5. Titrate 10 ml of each filtrate against std. NaOH (1N) using phenolphthalein as the indicator to obtain

the concentration of acetic acid.

6. Plot (i) a graph of log ( x/m) vs log c and (ii) a graph of x/m vs c.

Observations & Calculations:

Titration I: Standardization of oxalic acid:

Std. NaOH Vs oxalic acid

Indicator: Phenolphthalein

End point: Appearance of pale pink color

S.No Volume of

Oxalic acid , ml

Burette reading Volume of

NaOH, ml

Concordant

Value, ml Initial Final

VNaOH x NNaOH

Normality of acetic acid, Noxalic acid = ---------------------- = = N

VOxalic acid

Titration II: Estimation of Oxalic acid after adsorption:

Std. NaOH Vs Oxalic acid

Indicator: Phenolphthalein

End point: Appearance of pale pink color

Bottle No

S.No Volume of Oxalic acid , ml

Burette reading Volume of NaOH, ml

Concordant Value, ml

Initial Final

1

2

3

4

5

6

VNaOH x NNaOH

Normality of oxalic acid after adsorption = ----------------------

V oxalic acid

Nfinal = N

Bottle No

Concentration of oxalic acid, N

Moles of oxalic acid adsorbed, G mole

60 ml

Weight of Charcoal, g

x/m log (x/m) log c

Initial After adsorption

Ninitial Nfinal

1

2

3

4

5

6

Model Calculation:

Bottle No:

Voxalic.acid x Noxalic.acid x

Initial concentration of oxalic acid, Ninitial = -------------------------- = ---------------

Vtotal 60

= N

VNaOH x NNaOH

Strength of oxalic acid after adsorption, Nfinal = ------------------------

Voxalic acid

= N

Change in concentration of oxalic acid = Ninitial – Nfinal = c = N

Amount of acetic acid adsorbed, x = c x Eq.wt x 60 / 1000 = gmole / 60 ml

Weight of charcoal = m = 1 g

x/m = / 1 = gmole / 60 ml

log ( x/m) =

c =

log c =

Graph:

(i) log (x/m) vs log c

Slope of the plot = n =

Intercept = log k =

k =

RESULT:

An experiment was conducted to determine the adsorption isotherm of acetic acid. A linear plot over the

concentration range shows the validity of the Freundlich isotherm. The values of the constants were found

to be:

k = _________ n = _________ .

Packed Bed Distillation

Aim:

To determine the number of theoretical plate’s equivalent to packed column height (HETP) using

1. Fenske’s equation and

2. McCabe – Thiele method

Introduction:

Packed columns are often used for distillation, liquid - liquid extraction and humidification as well as for

gas absorption. A packed tower can be used as a fractionating column for either continuous or batch

distillation. The column height is usually based on the no of theoretical plates and the height equivalent to

a theoretical plate (HETP). In most cases the upper operating line is slightly steeper than the equilibrium

line and the reverse is true for the lower operating line. The pressure drop per equivalent theoretical plate

is generally less than that for a sieve or bubble cap tray, which is an important advantage for vacuum

operation. The smaller packing has lower values of HETP but they also have lower capacity and are not

likely to be used in large columns.

Procedure: I Preparing Calibration Chart

1. Mix known volumes of water-methanol (30:0. 25:5, 20:10, 15:15, 10:20, 5:25, 0:30 ml).

2. Determine the densities of the mixtures by using specific gravity bottle or

3. Determine the density of pure water and methanol at room temperature.

4. Convert the volume fractions into mole fractions of methanol (more volatile component).

5. A plot of density (Y-axis) versus mole fraction of methanol (X-axis) is the required calibration

chart.

If density or Refractive Index is known mole fraction can be obtained from the calibration chart.

Experiment part

1. Fill the SS feed tank with 250 ml of methanol and 350 ml of water.

2. Allow water to pass through the condenser.

3. Switch on the heater and heat the mixture.

4. Operate the packed bed at total reflux for about 30 minutes.

5. Collect the distillate in the distillate tank.

6. Switch of the heater and cool the apparatus.

7. Determine the density/refractive index of both residue and distillate.

Observations and Calculation:

Room temperature TR = ____________ C

Weight of empty sp. gravity bottle, W1 = ______gm

Weight of empty sp. gravity bottle + water, W2 = ________gm

Weight of sp.gr. bottle + feed, W3 = ___________gm

Density of water at TR =w = ___________gm/cc (From Perry)

Density of feed at TR =f = ___________gm/cc

f = [(W3 – W1)/(W2 – W1)]* w , gm/cc

Moles of feed, F = [(VA*m)/MA]Methanol + [(VB*w)/MB]water

MA & MB are mol. Wt. of methanol and water

Mole fraction of methanol, xF = [(VA*m)/MA]Methanol .

[(VA*m)/MA]Methanol + [(VB*w)/MB]water

Wt. of sp.gr. bottle + distillate W4 = ________gm

Density of distillate = d = [(W4 – W1)/(W2 – W1)] * w, gm/cc

Mole fraction of distillate, xD = ___________(from calibration chart)

Wt. of sp.gr. bottle + Residue W5 = ________gm

Density of residue = r = [(W5 – W1)/(W2 – W1)] * w, gm/cc

Mole fraction of residue, xR = ___________ (from calibration chart)

Wt. of sp. gravity bottle + methanol W6 = _____________gm

Density of methanol m = [(W6 – W1)/(W2 – W1)] * w, gm/cc

Determination of Number of Theoretical Plates:

(i) Fenske Equation

Nm+1 = ln [ xD (1 – xR)]

[(1 – xD) xR ]

ln αavg

Relative volatility of distillate αD = yD*/(1 – yD

*)

xD/(1 – xD)

Relative volatility of residue αR = yR*/(1 – yR

*)

xR/(1 – xR)

Average volatility = αavg = (αDαR)

Nm + 1 = ____________

Height of packed bed Z = _____________cm

HETP Nm = Z

HETP =Z / Nm = ______________cm

(ii) McCabe Thiele method

From graph of Y vs X, Nm +1 = __________

HETP = Z / Nm = ______________cm

Number of gas phase transfer units from graph (NTU)G = ____________

NTU)G HTU)G = Z [(HTU)G is height of gas phase transfer Units]

HTU)G = Z / NTU)G = ____________cm

Sl. No. Volume of

methanol, ml

Volume of

water, ml

Weight of

mixture, gm

Density of

mixture,

gm/cc

Mole

fraction of

methanol

1

2

3

4

5

6

7

30

25

20

15

10

5

0

0

5

10

15

20

25

30

II. Equilibrium data: (From Perry VI Ed, P 13-13, T 13-1)

No. Temperature,

C

Pa , mm

Hg

Pb , mm

Hg

X = PT-Pb

Pa-Pb

Y* =

PaX

PT

1 .

(Y* - X)

II. To find number of transfer units

1/y*– x

Area under

Curve = NTU)G

Result:

The HETP for the packed bed by Fenske Equation =

By Mc-cabe Thiele Method =

SIMPLE DISTILLATION

AIM: To verify ‘Rayleigh’s Equation’ by conducting simple distillation of given binary mixture

(Methanol+Water)

Introduction: Simple distillation is a process in which a binary mixture is distilled batchwise

without giving any reflux. It is also known as differential distillation because the composition of

the liquid in the distillation still changes with time. The liquid is heated slowly and the vapors are

withdrawn as rapidly as they form to a condenser, where the distillate is collected. The vapors

which form as soon as the vaporization begins will be richest in the more volatile component.

Eventually the vapors become leaner in more volatile component.

x1

ln(F/W) = dx / (y – x)

x2

F = Original moles charged.

W = moles left in the still

X1 = original composition & x2 = final composition of liquid

y = composition of the component in vapor phase

The above equation is known as Rayleigh’s equation.

The integration part of the equation is obtained graphically by plotting 1/(y – x) versus x .

Procedure: I Preparing Calibration Chart

1. Mix known volumes of water-methanol (30:0. 25:5, 20:10, 15:15, 10:20, 5:25, 0:30 ml).

2. Determine the densities or refractive index (RI) of the mixtures by using specific gravity

bottle or Abbey refractometer.

3. Determine the density or RI of pure water and methanol at room temperature.

4. Convert the volume fractions into mole fractions of methanol (more volatile component).

5. A plot of density or RI (Y-axis) versus mole fraction of methanol (X-axis) is the required

calibration chart.

If density or RI is known mole fraction can be obtained from the calibration chart.

II. Experiment Part

1. Fill the RB flask with known volumes of water and methanol (250 ml + 250 ml).

2. Allow running water to flow through the condenser.

3. Heat the still at a low heating rate till 2/3rd

of the liquid is distilled.

4. Cool the distillate and residue to room temperature and determine their densities or RI

and volume.

5. Using calibration chart note down the mole fractions of distillate and residue (xD and

xR).

6. Plot 1/(y* – x) versus x by using equilibrium data given in Perry (6th

Ed., 13-13;

Reference –13).

7. Determine the area under the curve between xR and xD which gives the RHS part of

the Rayleighs equation.

Observations and Calculation:

Weight of empty sp. gravity bottle, W1 = ______gm

Weight of empty sp. gravity bottle + water, W2 = ________gm

Weight of sp.gr. bottle + methanol, W3 = ___________gm

Room temperature, TR = ______C


Density of methanol at TR =m = ___________gm/cc (From Perry) or

m = [(W3 – W1)/(W2 – W1)]* w , gm/cc

Moles of feed, F = [(VA*m)/MA]Methanol + [(VB*w)/MB]water

MA & MB are mol. Wt. of methanol and water

Mole fraction of methanol, xF = [(VA*m)/MA]Methanol .

[(VA*m)/MA]Methanol + [(VB*w)/MB]water

Refractive index of methanol = m = _______

Refractive index of water = w = _________

Wt. of sp.gr. bottle + distillate W4 = ________gm

Volume of Distillate = Vod = __________cc

Density of distillate = d = [(W4 – W1)/(W2 – W1)] * w, gm/cc

Mole fraction of distillate, xD = ___________(from calibration chart)

Average molecular wt. of distillate, MD = xD * MA + (1 – xD) * MB

Moles of distillate, D = (Vod * D)/MD = ___________gm moles

Wt. of sp.gr. bottle + Residue W5 = ________gm

Volume of Residue = Vor = __________cc

Density of distillate = r = [(W5 – W1)/(W2 – W1)] * w, gm/cc

Mole fraction of residue, xR = ___________(from calibration chart)

Moles of Residue, W = F – D = _____________gm moles

RHS = dx/(y* – x) = __________ from graph of x vs 1/(y* – x)

LHS = Ln(F/W) = ____________

LHS RHS

Thus the Rayleigh’s equation is verified.

Tabular Column:

Calibration Chart:

Sl.n

o.

Volume of

Methano-l

VA ,(cc)

Volume of

water VB

,(cc)

Moles

A=(VAA)

MA

Moles

B=(VBw)

MB

Mole

fractio-n

of A

Densit

y ,

(gm/cc

)

Refract

ive

index,

1.

2.

3.

4.

5.

6.

10

8

6

4

2

0

0

2

4

6

8

10

Graph of Density vs Mole fraction

X

Equilibrium Data: (From Perry VI Edition; 13-13; T 13-1)

x y* 1/(y*-x)

0.0

0.02

0.04

0.06

0.08

0.1

0.15

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.95

1.0

0.0

0.134

0.230

0.304

0.365

0.418

0.517

0.579

0.665

0.729

0.779

0.825

0.870

0.915

0.958

0.979

1.0

Graph of 1/(y* – x) vs x

STEAM DISTILLATION

AIM: To verify the principle of immiscible liquid mixture boiling and finding the vaporization and

thermal efficiencies.

INTRODUCTION: In steam distillation a high-boiling liquid (A) is distilled along with steam (B),

which is immiscible with A. At atmospheric pressure high-boiling liquids cannot be purified by

distillation since the components of the liquid decompose at the high temperatures required. Steam

distillation is used to separate high-boiling component from small amount of non-volatile impurities.

When the two components A and B which are immiscible exert their own vapor pressure and when the

sum equals total pressure of the system the mixture begins to boil. By steam distillation as long as liquid

water is present component A vaporizes at a temperature well below its normal boiling point without

using a vacuum. The vapors of high-boiling liquid and water are condensed and the two immiscible

phases should be separated.

The mole fraction of either of the components in the vapor phase will be proportional to its vapor

pressure. Therefore, the amount of water distilled along with the given immiscible liquid (A) can be

calculated from the relationship,

WB = WA PBMB

PAMA

Where WA & WB are weights of component A and water in vapor phase, Kg

PA & PB vapor pressures of component A and water at distillation temp., atm

MA & MB molecular weights of A and water, Kgmole.

Steam distillation is used in food industry for removal of volatile impurities and flavors from edible fats

and oils.

PROCEDURE:

1) Fill the boiler with water till the drain cock level.

2) Switch on the steam boiler

3) Take 350ml of nitrobenzene in the still

4) Connect the condenser to the still and allow water to flow through the condenser

5) Open slightly the steam outlet valve [v1] on the boiler. Ensure that not much of steam comes out

of the condenser.

6) Maintain constant pressure of the steam throughout the experiment.

7) When the temperature inside the flask starts rising up and at a particular temperature it remains

constant, note down this value of as boiling point of mixture .

8) Collect the condensate in a conical flask.

9) After the operation is over collect the residue in a separating funnel

10) Transfer the distillate [receiving jar] into a separating funnel and allowed to settle in to two

phases.

11) Separate the two phases of both residue and distillate and measure their volumes and

note them down.

OBSERVATION AND CALCULATION:

Room temperature TR = _________ C

Density of water at TR (From Perry) W = _______________ gm/cc

Weight of sp. gravity bottle W1 = __________________ gm

Weight of sp. gravity bottle + water in distillate W2 = ______________ gm

Weight of sp. gravity bottle + organic liquid in distillate W3 = ____________ gm

Density of organic liquid in distillate, OD = [(W3 – W1) / (W2 – W1)] W = ____gm/cc

Weight of sp. gravity bottle + organic liquid in residue W4 = ____________ gm

Density of organic liquid in residue, OR = [(W4 – W1) / (W2 – W1)] W = ____gm/cc

Density of aqueous liquid is same in feed, distillate and residue.

W = WD = WR

Volume of organic liquid in distillate VOD = ______________ cc

Mass of organic liquid in distillate, mOD = VOD OD = ____________ gm

Volume of aqueous lager in distillate, VWD = _______________cc

Mass of aqueous layer in distillate, mWD = VWD W = ____________ gm

Moles of organic liquid in distillate, nA = mOD / MA = _____________ gm moles

Moles of aqueous liquid in distillate, nB= mWD / MB = _____________ gm moles

Actual mole ratio Nct = nA / nB = _____________

Hausbrand’s Chart:

Data from Perry (VI Edition; Pg 3-59 ; T 3-8)

Serial no.

Temperature,

C

Total

Pressure PT,

mm Hg

Vapor Pressure, mm Hg

PT – PB ,

mm Hg Nitro

benzene, PA

Water, PB

Plot temperature vs PA & PT – PB

PT – PB

PA & Nth

PT – PB

PA

T

Theoretical mole ration is NTH = PA / (PT – PB)

At point P, PA = PT - PB

PB = PT - PA

NTH = PA / PB

Vaporization efficiency is = (Nact / NTH ) 100 = ____________%

Volume of organic layer in residue, VOR = ________________ cc

Volume of aqueous layer in residue, VAR = ________________ cc

Mass of organic layer in residue, mOR = VOR OR = ___________ gm

Mass of organic layer in residue, mAR = VAR W = ___________ gm

OTHMER’S PLOT:

Log PA (Y-axis) vs Log PB (X-axis) on log-log graph

Log PA Log PB

Log PA

Slope = Log(PA/PB)

Log PB

Ratio of mass of organic liquid in residue to that in distillate is R = mOR/mOD = _______

Latent heat of water at TD is W = _________________ J/gm

Latent heat of water is OL =W[Log (PB/PA)] (MB/MA) _________________ J/gm

Theoretical steam required / gm of organic liquid is

A = (R+1) CP (TD – TR) + (W/W) + (100 – TD) + [(PB/PA)] (MB/MA)]

A= ____________________ gm steam / gm of organic liquid

Actual steam consumption B = (mWD + mAR)/mOD = _____________

Thermal Efficiency = A/B = _______________%

Result:

PACKED BED ABSORPTION

AIM: to demonstrate the principles involved in a packed bed absorption column and to obtain data

during absorption of CO2 in NaOH soln.

DESCRPTION :

The packed column is made out of a M75 glass tube and ½ “ packing material is placed on a grid

plate. The liquid enters at the top of the column by a liquid distributor. The gas mixture enters at

the bottom of the column. The flow rate of air, CO2 & liquid are measured from rotameters

{CO2 cylinder is partially opened (very slight open) and the flow is regulated by needle valves.

Any excessive open may damage the line and rotameter}.

PROCEDURE:

1. Prepare 20 lit. NaOH solution [say 7.5% by wt] and pour that into the feed tank

2. Set the air flow [say500 lph]& CO2 [say 15 lph] flow to a fixed position.

3. Allow liquid to flow from the nozzle at a particular rate [say 20 lph)

4. At steady state note down the flow rates and collect the samples of liquid

5. Determine the concentration of entering and exit liquor by titration method.

6. Repeat this experiment at different flow rates

VAPOR LIQUID EQUILIBRIUM SET UP

AIM: To obtain the equilibrium data for the given binary mixture.

PROCEDURE:

1) Place 400 ml[200ml of alcohol +200ml water] of the given mixture in the round bottom flask.

2) Assemble the set up and start heating the sample by heating mantle

3) Check that the distillate collected is totally refluxed.

4) Let the reflux be repeated for 3,4 times , note the temperature

5) Collect the distillate and residue samples

6) Check the distillate and residue’s composition either by refractive index or by density method.

7) Repeat this experiment with different concentrations

8) Plot T-x-y diagram with the obtained values

Packed Bed Extraction

Aim:

To determine

1. Number of transfer units

2. Number of theoretical stages required by KREMSER equation

3. Overall volumetric co-efficient

Theory: In continuous-contact equipment it is necessary to consider the rate of transfer and the

time of contact of the phases. Packed towers used in continuous countercurrent contacting of two

phases are vertical shells filled with a suitable material having a large surface area. liquid phase 1

enters the top of the column and is distributed over the surface of the packing either by nozzles

of distribution plates. Upon good initial distribution, the liquid will flow down through the

packing by tortuous routes, thereby exposing a large surface area for contacting with the rising

liquid phase 2. The liquid phase 2 will enter the bottom of the tower and rise upward through a

similarly tortuous path.

Procedure:

1. Take 20 lit of benzene in the benzene feed tank.

2. Take 20 lit of water and 200 ml of acetic acid mixture in the water feed tank.

3. Allow benzene to flow such that the air is removed and rotameter reading is constant.

4. Stop the benzene flow and allow water to flow at 10 LPH.

5. When the water level reaches to 1/4th

of the packed bed allow benzene to flow at a rate of

13-15 LPH.

6. When the packed bed is filled with liquid set the bottom flow at around 10 LPH.

7. Collect the extract layer at the bottom of the packed bed.

8. Titrate the extract against standardized 0.1 N NaOH and determine the concentration of

acetic acid.

9. Determine the density of the raffinate layer and read mole fraction of acetic acid from the

calibration chart.

Calculations and Observation:

HOE = Overall height transfer unit (Extract phase)

HOR = Overall height transfer unit (Raffinate phase)

NOR = Number of transfer units in the Raffinate phase

NOE = Number of transfer units in the Extract phase

KORa = FR

HOR (1 – X)*m

KORa = Overall volumetric mass transfer coefficient (Raffinate phase) in

Kg/m3 sec

FR = Mass velocity acetic acid free Benzene Kg/m2 – sec

FR = VR R (1 – XR1)

C/S area

Where VR = Flow rate of Raffinate, m3/s (rotameter reading)

XR1 = XR Kg acetic acid

1 + XR Kg raffinate

KOEa = FE

HOE (1 – Y)*m

FE = VE E (1 – YE1)

C/S area

Where VE = flow rate of Extract, m3/s

YE1 = YE Kg acetic acid

(1 + YE) Kg Extract

Equation of the Equilibrium line Y* = 10 X

*

Extract Solution:

NaOH vs Extract solution

Sl.No Volume of

Extract,

(ml)

0.5 NaOH solution in

ml

Titre

value, (ml)

Indicator

Phenolphthalein

Final Initial

Volume of NaOH, V1 = ___________ml

Normality of NaOH, N1 = ___________N

Volume of Extract solution, V2 = _________ml

Normality of Extract solution, N2 = V1N1/V2 = ________N

Concentration of acetic acid in the extract, CE = _______gm moles/lit

CE = _______Kg moles/m3

To find YE: (Kg acetic acid / Kg water)

YE = Concentration of acetic acid in the extract expressed as Kg acetic acid/Kg water

E = density of Extract, Kg/m3

Height of the packing =__________m

Diameter of the column =__________m

Density measurement:

Empty weight of specific gravity bottle, W1 =________gm

Empty weight of specific gravity bottle + water, W 2 =________gm

Density of water at room temp. (from Perry) , w =________gm/cc

Empty weight of specific gravity bottle + Feed solution, W3 =________gm

Empty weight of specific gravity bottle + Extract soln., W4 = ________gm

Empty weight of specific gravity bottle + Raffinate soln., W5 = ________gm

Density of feed solution, F = (W3 – W1) w =______gm/cc

(W2 – W1)

Density of Raffinate solution, R = (W4 – W1) w =______gm/cc

(W2 – W1)

Density of extract solution, E = (W5 – W1) w =______gm/cc

(W2 – W1)

Estimation of Concentration of solution:

Feed Solution:

NaOH Vs Feed solution

Sl.

No.

Volume of feed

solution, (ml)

0.5 N NaOH solution

in ml

Titre value,

(ml)

Indicator

Final Initial

Volume of NaOH, V1 =___________ml

Normality of NaOH, N1 =___________N

Volume of Feed solution, V2 =___________ml

Normality of Feed solution, N2 =V1N1/V2 = ______N

N2 = CF = gm moles/lit = _________Kg mol/m3

(1 – X)*m = (1 – X) – (1 – X*)

ln (1 – X)

(1 – X*)

(1 – Y)*m = (1 – Y) – (1 – Y*)

ln (1 – Y)

(1 – Y*)

n = number of transfer units by KREMSER equation

n = log [(XF – ys/m) (1 – 1/E) +1/E]

(XR – ys/m)

(1 – 1/E)

Where E = extraction factor = m FE / FR

m = slope of equilibrium line = 10

YE = 60CE

E – 60 CE

Raffinate solution:

NaOH Vs Raffinate solution (benzene + acetic acid)

Sl.

No.

Volume of feed

solution, (ml)

0.5 N NaOH

solution in ml

Titre value,

(ml)

Indicator

Final Initial

Phenolphthalein

Volume of NaOH, V1 =___________ml

Normality of NaOH, N1 =___________N

Volume of Raffinate solution, V2 =___________ml

Normality of Raffinate solution, N2 =V1N1/V2 = ______N

N2 = CR = gm moles/lit = _________Kg mol/m3

To find XR: (Kg acetic acid/Kg benzene)

XR = 60CR

E – 60 CR

HOE = Z / NOE HOR = Z / NOR

XR

NOR = dX/(X – X*)

XF

YE

NOE = dX/(X – X*)

Ys

CR = Concentration of acetic acid in the Raffinate, Kgmole / m3

R = Density of Raffinate, Kg/m3

XR = concentration of acetic acid in the Raffinate expressed as Kg acetic acid / Kg Benzene.

Graph: 1). To find the NOE by graph

From 1/(Y* - Y) vs Y between YS to YE

2). To find NOR by graph

From 1/(X – X*) vs X between XF to XR

Liquid – Liquid Extraction

Aim:

To determine the % recovery of Acetic acid from a solution of Benzene – Acetic acid

mixture using water as solvent by

1. Single stage

2. Three stage cross current extraction and to determine the stage efficiencies.

Introduction: Liquid extraction, sometimes called solvent extraction, is the separation of the

constituents of a liquid solution by contact with another insoluble liquid. When a solution of

benzene-acetic acid is agitated with water some of acetic acid will dissolve in water and only

little benzene mixes with the aqueous phase. Since at equilibrium the densities of the aqueous

and benzene layers are different, they form different layers on settling and can be decanted from

each other. The ratio of acetic acid to benzene in the aqueous layer is different from that in the

residual benzene solution, a certain degree of separation will have occurred.

The solution which is to be extracted is called feed, and the liquid with which the feed is

contacted is the solvent. The solvent rich product is called the extract, and residual liquid from

which solute has been removed is the raffinate.

Procedure: Calibration Chart for Benzene – Acetic acid

1. Prepare mixtures of 5:20, 10:10, 12.5:12.5, 15:10, 20:5 ml of benzene-acetic acid

mixtures.

2. Determine the refractive index or densities of each sample using Abbey refractometer or

specific gravity bottle.

3. Determine RI or density (from Perry, at room temp.) of pure benzene, acetic acid and

water.

4. Convert the volume ratios into mass fraction of acetic acid in benzene.

5. Plot RI vs mass fraction or density vs mass fraction. This is the required calibration chart.

Experiment Part: Single Stage

1. Take 100 ml of benzene in a conical flask and add 10 ml acetic acid and 150 ml

deionized water to it.

2. Stopper the flask, keep it in a flask shaker and shake for about 1 hr.

3. Separate the contents of flask using separating funnel.

4. Take about 30 ml of the organic layer and determine its refractive index, or

5. Determine the density of the layer using specific gravity bottle.

6. Read the mass fraction of acetic acid using the calibration chart.

7. Similarly determine the mass fraction of acetic acid in the aqueous layer or extract.

Multistage (Three stage) Cross current Operation:

1. Take 100 ml of benzene in a conical flask and add 10 ml of acetic acid and 50 ml of

deionized water.

2. Stopper the flask; agitate the contents using a flask shaker for about 20 minutes.

3. After 20 min. separate the two layers using separating funnel and add 50 ml of water to

the organic layer.

4. Agitate the contents for another 20 minutes.

5. Again separate the two layers and add another 50 ml of water to the organic layer.

Agitate for 20 minutes.

6. After third agitation separate the layers and determine the mass fraction of acetic acid in

extract and raffinate using calibration chart.

7. Both single and multistage can be carried out at a time by adding required quantities into

4 flasks, shaking one flask for 60 minutes continuously while removing the other 3 flasks

after 20 min.

Observations and Calculations:

Room temperature TR = ____________C

Weight of empty specific gravity bottle, W1 = _____________gm

Weight of specific gravity bottle + water, W2 = _________gm

Weight of specific gravity bottle + acetic acid,W3 = ____________gm

Weight of specific gravity bottle + benzene, W4 = ___________gm


Density of acetic acid, A = [(W3 – W1) / (W2 – W1)]* w = _______gm/cc

Density of benzene, B = [(W4 – W1) / (W2 – W1)]* w = __________gm/cc

Mass fraction of acetic acid in benzene xF = (V A)A

(V A)A (V B)A

V – volume of the compound

Weight ratio of acetic acid in benzene xF| = xF/(1 – xF

|)

Either density or refractive index can be used to plot calibration chart.

Weight of specific gravity bottle + feed solution = W5 = __________gm

Weight of specific gravity bottle + extract = W6 = ____________gm

Weight of specific gravity bottle + raffinate = W7 = ____________gm

Density of feed, f = [(W5 – W1) / (W2 – W1)]* w = __________gm/cc

Density of extract, E = [(W6 – W1) / (W2 – W1)]* w = __________gm/cc

Density of raffinate, R = [(W7 – W1) / (W2 – W1)]* w = __________gm/cc

Singlestage Calculations:

Time,

minutes

Volume of

AA, ml

Volume of

water

added, ml

Volume of

Benzene,

Ml

Raffinate

density,

gm/cc

Mole

fraction of

AA xR

(from

calibration

chart)

Weight

fraction of

AA, xR' =

xR/(1-xR)

60

10 150 100

%Acetic acid extracted theoretically = [(xF' - xR' )/ xF'] × 100

Multistage Calculations:

Time,

minutes

Volume of

AA, ml

Volume of

water

added, ml

Volume of

Benzene,

Ml

Raffinate

density,

gm/cc

Mole

fraction of

AA xR

Weight

fraction of

AA, xR' =

xR/(1-xR)

20 10 50 100

40 10 100 100

60 10 150 100

Titration against NaOH

Normality of NaOH = ___________N

Normality of Extract layer = (N×V)NaOH

VExtract

Mixture Titre value, ml Normality, N Mole fraction of

AA in extract, y

= N×60×10-3

Weight fraction

of AA in extract,

y' = y/(1-y)

1

2

3

Equilibrium Data:

Wt. ratio of AA in

Raffinate, x'

0

Wt. ratio of AA in

Extract, y'

0

Mass ratio of AA in feed, xF' =10 ml × ρC = mC/mA

100 ml ×ρA

Mass of AA extracted = (xF' - xR') × mA = ________gm

%AA extracted (experimental) = [(xF' - xR')/xF'] × 100

Weight of benzene in feed = A = mA ×(1 - 60×10-3

) gm

Weight of Water in feed = B = mB gm

Using equilibrium data Plot a graph of x' vs y'

From the graph read the value of xR|

%Acetic acid extracted theoretically = [(xF' - xR' )/ xF'] × 100

Stage Volume of solvent

added, ml

% Acetic acid

extracted (Expt)

% Acetic acid

extracted (The)

Single ------

Multi Stage – 1

2

3

LEACHING STUDIES

AIM:


batch leaching test and a mixture of sodium carbonate and sand.

Procedure:


Standard solution of sodium carbonate (1N) was prepared by dissolving 0.53g of sodium

carbonate in 100ml of water.

Wash 50ml burette and rinse with 1N HCl. Pipette out 20ml of std. sodium carbonate into

conical flask and titrate it with HCl in burette using methyl orange indicator. The end point is the

appearance of red colour.

Leaching Studies:


and 10g of sodium carbonate. Mix both the solids thoroughly. Add exactly 400ml, 200ml,



with std. HCl using methyl orange indicator.


Sl. No. Volume of Sodium carbonate (ml) Burette reading Volume of HCl, (ml) Indicator

Final Initial

Volume of sodium carbonate = V1 = ml

Normality of sodium carbonate = N1 = N



Leaching

Bottle

No

Weight

of sand

taken (g)

Weight of

Sodium

carbonate

(g)

Volume

of water

added

(ml)

Volume of

supernatant

liquid (ml)

Volume of

Hcl

Consumed

Amount of

sodium

carbonate

recovered (g)

%

recovered

1 50 10 400 10

2 50 10 200 8

3 50 10 133 6

4 50 10 100 4

5 50 10 80 2


Volume of HCl required * N2 * equ. Weight of sodium carbonate * volume of water added

= -----------------------------------------------------------------------------------------------------------------


Percentage of sodium carbonate recovered

Amount of sodium carbonate present * 100

= ----------------------------------------------------------

Amount of sodium carbonate added to beaker

RESULT:

The % of sodium carbonate extract was found to be =

Schx4007 Mass Transfer Lab

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