Final Mass Transfer - I All Practical Writeups
-
Upload
vkpaithankar -
Category
Documents
-
view
219 -
download
0
Transcript of Final Mass Transfer - I All Practical Writeups
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
1/36
DIFFUSION IN AIR
AIM: To determine the diffusion coefficient of Liquid Gas System (acetone airsystem).
APPARATUS:LiquidGas diffusion system setup, acetone, water (for water bath).
THEORY: When a concentration gradient exists within a system consisting of one or morecomponents, the components flow in such a direction, so as to reduce the concentration
gradient. This process is called Mass Transfer.
The mass transfer of components takes place by random motion of molecules and is known as
diffusion. In a liquid gas system, when the liquid evaporates into still gas, the vapour formedon the surface of the liquid is transferred into the bulk of gas by diffusion. The rate of
diffusion depends on the ease with which the molecules can travel through the gas. This factor
is called diffusivity.
Refer to the figure; here the liquid evaporates in a vertical glass tube in still air. By Stephens lawfor diffusion of vapour through a stationary gas, the mass transfer rate is given by the equation: -
1
2
12
lnAT
ATTA
CC
CC
yy
DCN
=
Where
D = diffusivity of vapour through air (m2/s)y2 y1 = the distance through which the vapour travels in the direction of diffusion (m).
CT = the total concentration of vapour and gas.
CA1, CA2 = concentrations of vapour at the surface of the liquid and at the end of the tube
respectively (kmol/m3).
Replacing (y2-y1) by x and writing the concentration in terms of vapour pressure we get
1
2lnB
BA
C
C
xRT
DPN =
=1
2lnB
B
P
P
xRT
DP
Where
CB1, CB2 = concentration of air the surface of the liquid and at the end of the tube respectively.(kmol/ m3).
PB1, PB2 = partial pressure of air at the above mentioned points, total pressure (kN/m2).
P = total pressure (kN/m2).
This rate of mass transfer will be equal to the rate of evaporation of liquid, which in turn can be
followed by the fall in level of liquid surface.
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
2/36
dt
dx
MN LA
=
Where,
L = density of liquid (kg /m3)M = molecular weight of liquid (kg/kmol)dx/dt = rate of fall of liquid level
dt
dx
MP
P
RT
P
x
D L
B
B
=1
2ln
Integrating between the limits x = x1 to x2 when t changes from 0 to t
dxxM
dtP
P
RT
DPx
x
L
t
=2
10
ln
2ln
12
1
2
2
1
2
xx
P
Pt
MP
RTD
B
B
=
All the other values in the above equation being known, D can be calculated.
EQUIPMENT DESCRIPTION:
The set up consists of: A diffusion column immersed in a water bath. A blower for circulation of air. Orificemeter and manometer for measuring the air flow rate. An electrical heater for heating the water in the water bath. A temperature sensor for measuring the temperature of the liquid. Digital temperature indicator.
PROCEDURE:
1. Fill acetone in the process column upto the desired level.
2. Check the water level in the water bath for heating acetone.3. Switch on the main switch of the control panel.
4. Close the flow regulating valve partially and start the blower.
5. Adjust the opening to get the desired flow rate of air.
6. Note the level of liquid in the tube at time t = 0 as x1 (m).7. Allow the liquid level to fall to another level (x2) and note the time required for it.
8. Repeat the experiment for different temperatures.
9. Switch off the heater and blower10. Switch off the main control
OBSERVATION TABLE: -
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
3/36
Sr.
No.
Temperature
C
Time
(sec)
Liquid level PB1
kN/m2
D
m2/s
Initial
x1(m)
Final
x2(m)
Using the formula below D is calculated
2ln
12
1
2
2
1
2
xx
P
Pt
MP
RTD
B
B
=
Where,P = atmospheric pressure (KN/m2)
PB2 = atmospheric pressure (KN/m2)PB1=difference in atmospheric pressure and vapour pressure of acetone at the operating
temperature (KN/m2)
t = time for which evaporation occurs (s)
RESULT: The calculated value of Diffusivity is
CONCLUSION:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
4/36
LIQUID-LIQUID DIFFUSION
AIM: To determine diffusion co-efficient for a liquid-liquid system (Acetic acid- water)
APPRATUS: Porous cylinder, beaker, conical flask, acetic acid, distilled water, burette, pipette.
THEORY: If in a liquid there exist a concentration gradient, the molecules of the liquid will
migrate in such a way so as to reduce concentration gradient. This phenomenon is called diffusion
depends on magnitude of concentration gradient. The basic law defining mass transfer rate is
known as Ficks law given by,
NA = DAB dCA/dy
Where,
NA = molar flux (Kmol/m2sec).
DAB = Diffusivity of compound A in B (m2/sec).
dCA/dy= Concentration gradient (Kmol/m3)/m.
In addition to diffusion movement of A occurs by bulk flow also. N=NA+NB gives the net flux &
total rate of transfer of A is given by,
dy
dCD
C
CNNN AAB
ABAA
+=
)(
This equation can be integrated if DAB & C are constant. In case of liquid, these vary considerably
with concentration. Integrated form of above equation is given by;
1
2
)/(
)/(ln
)(ABAA
ABAAAB
BA
AA
xNNN
xNNN
z
CD
NN
NN
++
+=
In case of liquid DAB & C both vary with concentration & therefore an average value of above
equation is written as,
1
2
)/(
)/(ln
)( ABAA
ABAA
avgBA
AA
xNNN
xNNN
MNN
NN
+
+
+=
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
5/36
For steady state diffusion of A through non-diffusing B, NB=0. In this particular case we consider
only diffusion of A in B
)1(
)1(ln
1
2
A
A
avg
ABA
x
x
Mz
DN
=
1
2lnB
B
avg
AB
x
x
Mz
D
=
PROCEDURE: -
1. Fill the porous pot with acetic-acid of known concentration (4N)
2. Place it in beaker.
3. Pour known quantity of distilled water in beaker.
4. Start the stop clock.
5. Note thickness of porous pot.
6. Acetic acid diffuses in to water.
7. After every 15 minutes take a sample from beaker & titrate it against 0.05 N NaOH
solution to find quantity of acetic acid in it, using phenolphthaliene indicator.
8. Using the formula given calculate DAB for each reading.
OBSERVATIONS:
Inner radius of porous pot (r1) =Outer radius of porous pot (r2) =
Thickness of porous pot (z) = (r2- r1) =
Initial conc. of acetic acid =
Length of porous pot involved in transfer =
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
6/36
OBSERVATION TABLE:
Sr.
No.
Time
(sec)
Burette
Reading
(ml)
Normality
of solution
in beaker
(N)
Strength of
solution in
beaker (g/lit)
Total amount
of acetic acid
in beaker
(gm)
Mass transfer
rate (NA)
(kmol/m2sec)
Diffusivit
DAB (m2/s
CALCULATIONS:
1. To calculate normality of solution in beaker using equationN1V1 = N2V2N1V1 = Solution in beaker
N2V2 = NaOH used for titration
1
221
V
VNN = =
2. To calculate strength of solution in beaker (which gives g/lit of acetic acid diffused)
= Normality (N1) x Equivalent weight of acetic acid
=
3. Total gm of acetic acid present in the beaker = (Strength of solution in beaker *total
volume of water in beaker).=
4. Initial conc. of acetic acid (A) = x N = _______N.= x * eq. wt. of A (gm/lit of solution)
=
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
7/36
= kg/lit of solution.
Initial mass fraction of acetic acid = (wt of acetic acid)/(wt.of acetic acid + wt. of water)=
Initial mass fraction of water = (wt.of water)/(wt.of acetic acid + wt. of water)
=
Density of solution inside pot (1) =(Mass fraction of acetic acid * density of acetic acid) + (mass fraction of water * density ofwater) kg/m3
=
5. Mole fraction of acetic acid (xA1) inside pot =(wt. of acetic acid inside the pot) / (molecular wt of AA)___
(wt of AA / mol wt of AA) + (wt of water / mol wt of water)
=
Mole fraction of water inside the pot (xB1) = 1 xA1 =6. Mole fraction of acetic acid outside pot (xA2 ) =
(wt of acetic acid diffused/mol.wt of acetic acid)__________(wt of AA / mol wt of AA) + (wt of water / mol wt of water)
=
Mole fraction of water xB2 = 1xA2 =
7. Molecular wt of solution inside pot (M1) = xiMi.
8. 1/M1 =
2/M2 =
(/M)Avg =
9.
1
2
12
lnB
B
BBBm
x
x
xxx
=
=
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
8/36
10. Area of mass transfer A = . DO L
11. NA = N /At kmol / m2 s =
12. Diffusivity =
RESULT: Value of diffusivity is
CONCLUSION:
.
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
9/36
WETTED WALL COLUMN
AIM: To find the Mass Transfer coefficient for air-water system using a wetted wall column.
THEORY: Wetted wall column is mainly used for experimental purpose, to estimate the value ofmass transfer coefficient for a gas-liquid system. In this equipment, liquid is allowed to flow down
the inside wall of the column, so that, the entire inner surface of the column is wetted. The gas is
allowed to enter at the bottom and mass transfer coefficient is evaluated. Ripples are avoided sothat the inner surface area of the column can be taken as mass transfer area.
The advantage of using a wetted wall column is that the mass transfer area is known.
Though, industrially it has limited application, it provides a very useful means for conductingexperiments in the lab. This provides a basis from which correlation can be developed for packed
towers. It also provides a means for determining the importance of various factors affecting the
mass transfer coefficient.
The general equation for mass transfer used for a wetted wall column is:
44.083.0Re023.0 ScP
P
D
dh Bm
V
D =
Where, hD = mass transfer coefficient = kG RT and
BmG
VG
PRTz
PDk =
ZG = gas film thickness (m)
Dv= diffusivity (m2
/s)P = total pressure (kN/m2)
d = diameter of the column
Re = Reynolds no. based on gas phase = du/Sc= Schmidt no. = / DV = gas phase viscosity = gas phase density.
EQUIPMENT DESCRIPTION: The wetted wall column set-up consists of the following:
A glass column with a small liquid reservoir at the top. An inlet for air at the bottom A rotameter for measuring the inlet airflow rate. A second liquid reservoir kept at such a height so as to allow flow of liquid to the first
reservoir by gravity.
A hygrometer to measure the humidity of inlet and outlet air.
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
10/36
PROCEDURE:
1. Allow the water from the reservoir at the top to flow down to the small reservoir in
such a way that the water overflows into the column, a thin layer covers the entire inner
surface of the column. Keep the flow rate minimum.2. Measure this flow rate of water.
3. Start the air supply and set it to the desired flow rate.
4. Measure the humidity of inlet air.5. Measure the humidity of outlet air.
6. Note the inlet air pressure.
7. Note the temperature of inlet and outlet air, and inlet and outlet water.
OBSERVATIONS:
1. Diameter of the column = m
2. Height of the column = m3. Inlet air pressure = kN/m2
OBSERVATION TABLE:
Sr.
No.
Flow rate
of water
(lpm)
Flow
rate of
air
(lpm)
Temp.
of
inlet
water
(oC)
Temp.
of inlet
air
(oC)
Temp.
of outlet
air
(oC)
Humidity
of inlet air
(RH1)
%RH
Humidity
of outlet
air (RH2)
%RH
Outlet
water
temp
(oC)
1
2
3
CALCULATIONS:
1. Inlet partial pressure of water vapour
100
11
1AS
A
PRHP
= (PAS1 = Vapour pressure of water at the inlet temperature of
air)
=
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
11/36
Inlet humidity of air
1
11
A
A
PPPH
= (P = total pressure)
=
2. The outlet air humidity in kmol vapour/kmol dry air
100
22
2
AS
A
PRHP
= (PAS2 = Vapour pressure of water at the outlet temperature of
air)
=
Outlet humidity of air
2
2
2
A
A
PP
PH
= (P = total pressure)
=
3. The dry air flow rate in kmol/s
Air flow rate in m3/s (moist) = air flow rate in lph x 10 -3
3600
=
Therefore, molar flow rate of air (moist)
= Volumetric flow rate of air x inlet air pressure
R x absolute temperature of inlet air
=
4. The rate of mass transfer:
NA = dry air flow rate (outlet molal humidity- inlet molal humidity)
Area for mass transfer
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
12/36
=
Area for mass transfer = dL =
d = diameter of the columnL = height of the column
5. The experimental value of mass transfer coefficient kG by
)( 12'
AAAG PPNk = (where, kG is in kmol/(m2s)(kN/m2)
=
6. The theoretical value of kG is
44.083.0Re023.0 ScP
P
D
dh Bm
V
D =
1
2
12
ln
)(
B
B
BBBm
P
P
PPP
=
=
PB1 = P - PA1, PB2 = P - PA2
Re = du / = where =PM/RT = obtained from literature
Sc = / DV =
Dv= taken from literature at temperature (T) for air water system.
hD =
Where, hD/RT = kG, theoretical mass transfer coefficient
kG =
RESULT:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
13/36
CONCLUSION:
PACKED BED ABSORPTION
Aim: To determine the overall mass transfer coefficient for gas, liquid system using packed bedabsorption.
Apparatus and Chemicals: - Packed bed absorption column assembly, NaOH pellets or flakes,hydrochloric acid, barium chloride and phenolphthalein indicator.
Theory: - Gas absorption is a process in which a gas mixture is contacted with a liquid for thepurpose of preferentially dissolving one or more components of the gas.
The absorption process may be a physical one or may be accompanied by a chemical
reaction. For absorption to take place it is necessary that the gas & liquid be brought in to intimate
contact with each other. For this purpose packed or plate towers are generally used, packed towers
being more common. In this system under consideration CO2 is being absorbed in NaOH solutionin a packed tower .The chemical reaction taking place is
CO2 + 2NaOH Na2CO3 + H2OAs the solution is concentrated it may be assumed that there will be negligible vapour
pressure of carbon dioxide over the solution, i.e. all the resistance to mass transfer lies in the gas
phase.The basic equation, which can be used to calculate the overall mass transfer coefficient, is
Gm (y1-y2) = KG a P (y-ye)ln Z
Where:
Gm = molar gas flow rate (kmol / m2
s)y = mole fraction of CO2KG a = overall mass transfer coefficient, kmol /m
3s (KN/m2)
P = operating pressure (kN/m2)
Z = ht. of packing (m)
Subscripts 1, 2 & e represents inlet, out late & equilibrium conditions.
Equipment descriptions: The set up consists of
A packed column (with Raschig rings) Feed tank (10 lit) to store NaOH solution fed to column.
Receiving tank (7 lit) for collecting the solution leaving the column. A compressor for air circulation. CO2 cylinder with regulator to supply CO2 to feed gas stream. Rotameters for measuring feed solution, air and CO2 flow rates
Procedure:
1. Fill the feed tank with 10 lit of 1N NaOH solution and close the tank cover.
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
14/36
2. Prepare 100 ml of BaCl2 solution 25% w/w (25%BaCl2 + 75% water, by wt.)
3. Adjust the air pressure regulator to 1kg/cm2 pressure, to maintain the flowrate of
NaOH.4. The flow rate of the gas (air + CO2) and liquid (NaOH) are adjusted to and
respectively by valves provided on the barometer
5. The CO2 flow rate should be such that the percentage in air-CO2 mixture is 8 to10%
6. The system is allowed to attain to steady state after fixing the liquid level. This
level is adjusted by changing the valve opening at the bottom.7. The time for steady state should be 3-4 times the mean residence time of the
liquid phase in the column. During this period the liquid level in the bottom section must be
maintained at marked height.
8. After steady state is achieved, liquid samples are collected at the outlet foranalysis
9. A known volume V of the sample is titrated against s tandard HCl using
phenolphthalein. End point is noted as T1, colour change:- pink to colourless
10. To the volume V of the sample 25% BaCl2 solution is added, to precipitateNa2CO3. Go on adding BaCl2 till precipitation is complete, add some excess amount. This
solution is then titrated against standard 1N HCl using phenolphthalein as indicator. End pointas T2, colour change: colourless to pink
11. Repeat the experiment for different flow rates.
Observations:-
Height of column (Z) = (m)Blank reading (T1) = (ml)
Normality of HCL = (N)
Operating pressure = (KN/m
2
)Cross sectional area of column = (m2)
Observation table:-
Sr.No.
Feedflow
(lph)
Air flow(lph)
CO2 flow(lph)
TitrationT1 (ml of HCl)
Titration T2after adding
BaCl2(ml of HCl)
KGa
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
15/36
Calculations:-
1. Concentration of inlet NaOH, (CNaOH)in T1 x concentration of HCLNormality of NaOH =
Volume of sample
X1 (N) =
X1 x Eq. Wt. of NaOH
(CNaOH)in = mol. Wt. Of NaOH
=
2. Concentration of out let NaOH,(CNaOH)out T2 x concentration of HCL
Normality of NaOH =
Volume of sample
X2 (N) =
X2 x Eq. Wt. of NaOH
(CNaOH)Out =mol. Wt. Of NaOH
=
3. Concentration of Na2CO3 in outlet stream
(CNa2CO3)Out =(1/2){(CNaOH)in (CNaOH)Out}
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
16/36
4. Rate of absorption of CO2
RCO2 = (1/2){(CNaOH)in (CNaOH)Out} x Liquid flow rate m3/s
=
Flow rate at inlet m3/s5. CO2 flow rate at inlet =
Molar volume m3/Kmol
=
Flow rate at inlet m3/s
6. Air flow rate at inlet =Molar volume m3/Kmol
=
Flow rate of CO2 (Kmol/s)in7. Mole fraction of CO2 entering y1 =
Flow rate of CO2 + Flow rate of air (Kmol/s)
=
Flow rate of CO2 (Kmol/s)outMole fraction of CO2 at outlet y2 =
Flow rate of CO2 + Flow rate of air (Kmol/s)
=
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
17/36
7.(y-ye)ln = The log mean average of the concentration differences at the end of tower.
2
1
21
ln
)(
y
y
yyyy lme
=
=
8. Overall gas transfer coefficient
Gm (y1-y2) = KGa P Z (y-ye)ln
Gm = molar flow rate at inlet = Kmol/s (unit cross sectional area of Column)
[Flow rate of CO2 + Flow rate of air (Kmol/s)]inGm =
Cross sectional area of column
=
P = Operating pressure = Atmospheric pressure (101.325 kN/m2)
ln
1
)(
)2(
e
mG
yyPz
yyGaK
=
=
Result: The value of KGa calculated =
Conclusion:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
18/36
ENHANCEMENT FACTOR
AIM: To determine the enhancement factor for absorption using CO2, H2O & NaOH System
APPARATUS: Packed bed absorption set up
CHEMICALS: CO2, NaOH solution (1N), distilled water, BaCl2 (25%W/W), HCl (1N),
phenolphthalein indicator
THEORY: Absorption is a process used for removal of unwanted elements from a gas mixture by
contacting it with a suitable solvent. If the gaseous component is easily soluble in water (e.g.
NH3) water can be used as solvent for removal of this component. But if solubility in water is
low (CO2) a solution with gaseous component react can be used as solvent. The reaction of
solution with the solvent leads to the consumption in the solution thus helping in maintaining
a low concentration gradient between the bulk of the gas & interface. Consequently increasing
the rate of absorption as compared to that with non-reactive solvent. The ratio of the two rates
obtained above is called as Enhancement factor
Rate of absorption with reactionEnhancement Factor = ---------------------------------------------
Rate of absorption without reaction
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
19/36
EQUIPMENT DESCRIPTION:
The set up consists of
1. A packed column (with Raschig rings)
2. Feed tank 10 lit. to store NaOH solution.
3. The receiving tank (1lit) for collection of the outlet solution.4. The compressor for air feed.
5. CO2 cylinder with regulator to supply CO2 to the feed gas stream
6. Rotameter to measure feed solution, air & CO2 flow rates
PROCEDURE:
1. Fill feed tank with 1N NaOH solution.2. Allow CO2 and air mixture to flow up the column at a fixed flow rate.
3. Allow NaOH solution to flow down the column at a fixed flow rate.
4. After steady state is reached collect the sample.5. Repeat the procedure for different flow rates of NaOH solution keeping CO2 and
airflow rates constant.6. Analyze the collected sample for Na2CO3 content to calculate rate of absorption of
CO2.7. Drain remaining NaOH from the tank and fill the tank with distilled water.
8. Wash entire system by flooding with water.
9. Next fill the tank with distilled water.10. Allow it to flow down the column at the same flow rate as those used for NaOH
solution.
11. Keeping the CO2 and air flow rates again at the same value, collect the samples forsame flow rates of H2O as flowrates of NaOH solution.
12. Analyze these samples for their CO2 content and calculate the rate of absorption of
CO2.13. The enhancement factor can now calculated by taking the ratio of the rate calculated in
steps from (1) to (6) to the rate calculated in remaining steps for respective NaOH and
water flow rates.
OBSERVATIONS:
Blank reading =
OBSERVATIONS TABLE:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
20/36
Sr.
No.
Feed Flow Rate
(LPH)
Air Flow
Rate(LPH)
CO2 Flow
Rate(LPH)
Titration Readings
NaOHT1 (ml)
WaterT2 (ml)
1
2
3
CALCULATIONS:
A) For absorption of CO2 in NaOH
1) Concentration of NaOH at inlet
Blank reading x 1
Normality = ---------------------------25
=
Normality * Equivalent weight
C(NaOH)inlet = -------------------------------------------------
Molecular weight
=
2. Concentration of NaOH at outlet
Normality =
Normality * Equivalent weightC(NaOH)out = ------------------------------------------------
Molecular weight
=
3. Concentration of Na2CO3 at outlet
C (Na2CO3) = (1/2) (C (NaOH)inlet C (NaOH)out)
=
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
21/36
4. Rate of absorption of CO2 1N NaOH
(R co2 ) 1 = C( Na2CO3) (Kmole/m3 ) * liquid flow rate (m3/sec)= Kmol/sec
=
B) For absorption of CO2 in water
1. Concentration of H2CO3:
Normality =
Normality * Eq. wt.C(H2CO3) = -------------------------------
Molecular weight
=
2. Rate of absorption of CO2 in water
C(H2CO3) * liquid flow rate (m3/sec)
(RCO2)2 = -------------------------------------------------Molecular weight of H2CO3
=
3. Enhancement Factor
(RCO2)1E = --------------------
(RCO2)2RESULT:
Sr.No. Rate of absorption of CO2 Enhancement Factor
EIn NaOH (RCO2)
(kmol / sec)
In water
(RCO2) 2
(kmol/sec)1
2
3
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
22/36
CONCLUSION:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
23/36
INDIVIDUAL AND OVERALL MASS TRANSFER COEFFICIENT
AIM: To determine individual and overall mass transfer coefficient for gas-liquid system.
APPARATUS AND CHEMICALS: Packed bed absorption column, sodium hydroxide solution
(1 N), Hydrochloric acid (1 N), BaCl2, CO2 gas cylinder, Air compressor, and Phenolphthalein
indicator.
THEORY: When gas is absorbed in solution, the solute component has to overcome interfacial
resistance before it can be absorbed in the bulk of liquid using two-film theory to represent theprocess. The transfer of solute component from bulk of gas to bulk of liquid is as shown in the
figure.
The solute particles from gas diffuse through gas film near interface and partial pressuredrops from PAG to PAi at interface. The component passes across the liquid film before it reaches
bulk of liquid phase. The rate of mass transfer through liquid and gas film becomes equal whensystem is at steady state and is,
NA = kG (PAG - PAi)
= kL (CAi CAL)
CAi = Concentration of solute at the interfacePAi = Partial pressure in gas phase at the interface
Relation between PAi and CAi is given byPAi = H CAi
Where
H = Henrys constantMeasurement of kG and kL requires concentrations at interface, which are not easily
measurable hence; overall mass transfer coefficients are measured. The rate of mass transfer is
given by
NA = KG (PAG - PA*)= KL (CA* CAL)
CA*= Concentration of solute in liquid phase which is in equilibrium with PAGPA*= Partial pressure in gas phase in equilibrium with CAL
Relation between Overall and Individual mass transfer coefficients is given as
1/KG = 1/kG + H/kL (1)
1/KL = 1/kL + 1/H kG (2)
When NaOH is contacted with mixture of air-CO2 overall KG involveskG and kLbut when
NaOH solution is contacted with CO2 alone resistance to transfer of solute component in the gasphase becomes negligible i.e. kGbecomes very large so the above equation reduces to
1/KG = H/kL (since 1/kG = 0)
and hence overall mass transfer coefficient gives values of kL. This value along with KG isused to find out kG. Thus individual and overall mass transfer coefficients can be calculated.
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
24/36
PROCEDURE:
1. Fill the tank with 1 N NaOH solution.
2. Allow it to flow down through column at desired rate.3. Allow mixture of air and CO2 to flow up column.
4. Let the system achieve steady state.
5. Collect the product sample at the outlet.6. Repeat steps from 1 to 5 for minimum two readings.
7. Stop airflow and pass only CO2.
8. Allow NaOH to flow down the column.9. Let the system achieve steady state.
10. Collect the product sample at the outlet.
11. Analyze samples collected in the step 5 and 10 for CO2 content by titrating it with 1N HCl.
OBSERVATIONS:
1) Height of the column = z = 0.5 m
2) Diameter of the column = d = 0.06 m
3) C/s of the column = m
2
4) Volume of the packed bed = m3
5) Blank reading = ml6) Normality of HCl = 1 N
7) Operating pressure = 101.325 KN/m2
OBSERVATION TABLES:
Part 1) For Air + CO2 mixture
Sr.
No.
Flow rate of NaOH
(LPH)
Air flow rate
(LPH)
CO2 flow rate
(LPH)
Titration Reading
(T1) (ml)
1
2
3
Part 2) For only CO2Sr.
No.
Flow rate of NaOH
(LPH)
Air flow rate
(LPH)
CO2 flow rate
(LPH)
Titration reading
(T2) (ml)
1 --
2--
3--
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
25/36
CALCULATIONS:
Part 1)
1. Normality (N1) = [ 1 x Blank reading] / 25
=
2. C(NaOH)in = [N1 x 40] / 40
=
3. C(NaOH)out = [T1 x 1 x 40] / [25 x 40]
=
4. C(Na2CO3) = 0.5 [C(NaOH)in - C(NaOH)out]
=
5. Rate of absorption of CO2 (RCO2)
RCO2per unit volume of packed bed = C(Na2CO3) x liquid flow rate (m3/sec)
= kmole/sec
6. Rate of CO2 at outlet = CO2 flow rate at inlet - RCO2
=
= kmol/sec
7. y1 = mole fraction of CO2 at inlet = (flowrate of CO2 in)/(flowrate of CO2 + air in)
=
8. y2 = mole fraction of CO2 at outlet = (rate of CO2 out)/( flowrate of CO2 + air out)
=
9. (y - ye)ln = log mean concentration difference = [y1 - y2] / ln(y1 /y2)
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
26/36
10. Gm = molar flow rate of air and CO2 mixture / c/s area of packed bed
=
11. KGa= Gm(y1 - y2) / P z (y - ye)=
Part ll) For Absorption of CO2 only, in NaOH
Rate of absorption of CO21. C(NaOH)out = [T1 x 1 x 40] / [25 x 40] (N = Normality of NaOH solution)
= kmole/m
3
2. R(CO2) = 0.5 [C(NaOH)out - C(NaOH)in] x liquid flow rate (m3/sec)
= kmole/sec
3. CAi = PAi / H [Where H = Henrys constant = 30 and PAi is total pressure]
4. CAL = 0
5. R(CO2) = kL (CAi - CAL)
=
6. 1/K Ga= 1/ kGa+ H/kLa
RESULT:
Sr. Flow rate of Overall mass Individual mass Individual mass transfer
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
27/36
no. NaOH (LPH) transfer coefficient
(KGa)
transfer coefficient
on liquid side (kLa)
coefficient on gas side
(kGa)
CONCLUSION:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
28/36
TRAY DRYER
AIM: To determine the drying rate characteristics using tray dryer
APPARATUS: Tray dryer assembly, sand and water.
THEORY: Drying refers to the final removal of moisture from a substance. It is
generally carried out after evaporation, filtration or crystallization. In most of the drying process
water is removed by vaporization.
The moisture content of a material is usually expressed as a percentage of weight of dry
material. It may be present in the material in 2 forms: -
Bound moisture: - Moisture present in any material in such a way that it exerts an equilibrium
vapor pressure less that of pure water at the same temperature. Eg: moisture in loose chemical
combination with cellulosic material, moisture absorbed, etc.
Unbound moisture: - Its the moisture contained by the substance, which exerts a vapor pressure
at equilibrium, equal to that of the pure liquid.
Rate-of-drying-curve: - When rate of drying is plotted against percentage moisture above
equilibrium value, curve, as shown in the fig. is obtained.
The curve can be divided into 2 parts:
1. Constant rate period: - In this region the rate of moisture removal / unit area remains
constant and it is assumed that drying takes place from a saturated surface of the material.2. Falling rate period: - In this region the rate of drying/unit area continuously decreases, till
the equilibrium moisture content is reached. In some cases this curve may have two
different slopes and this period is then divided into the first and second falling rate periods
Critical moisture content: - the moisture content of the material at the end of the constant rate
period is called critical moisture content.
Equilibrium moisture content: - the moisture content of the material at the end of the falling rate
period is called as the equilibrium moisture content. This does not change on further heating of the
material.
Free moisture: - Its the water in excess of the equilibrium moisture content.
EQUIPMENT DESCRIPTION:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
29/36
Tray dryer is a one pan dryer with a blower. The complete assembly consists of:-
1. A drying chamber with tray.
2. A blower for passage of air
3. An electrical heater placed after the blower
4. A pan placed in the path of air
5. A weighing balance for periodic weighing of the pan with the drying material.
PROCEDURE:
1. Take a measured quantity of material and add a measured quantity of water to it and place
it on the pan.
2. Note the total weight of the pan and wet material.
3. Start the blower and air heater.
4. As time elapses, drying proceeds and there is loss of water which leads to reduction in the
weight of the wet material.
5. Record this weight reduction against time by weighing the pan with material.
6. Continue heating till no further change in weight is observed even on continued heating.
7. Note the total time required for drying & the temperature of dryer.
OBSERVATIONS:
Dry weight of the material =
Weight of water added =
Dimensions of the tray (L x B) =
Area of the tray (m2) =
Temperature of the dryer =
OBSERVATION TABLE:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
30/36
S No. Weight of solid
(w) kg on tray
Time
(t)
secs
w
kg.
t
secs.
NA=w/A (t)
kg/s m2
X= kg moisture
kg dry solid
CALCULATIONS:
1. Calculate rate of drying:
NA = moisture evaporated kg
Area x time m2s
2. Calculate the mass ratio of solvent as:
X= Weight of solvent remaining =
Weight of dry solid
3. Plot a graph of NA Vs. X. Identify the const. Rate & falling rate periods & find out thevalues for critical & euilibrium moisture content.
4. Plot a graph of X Vs. t & calculate the value of slope = dx/dt. Using this value calculate the
rate of drying using the eqn:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
31/36
NA = Kg of dry solids dx kg
Area dt m2 s
=
RESULTS:
Equilibrium moisture content = kg/kg dry solid
Critical moisture content = kg/kg dry solid
Rate of drying :-
In constant rate period = kg/m2 s
In falling rate period = kg/m2 s
CONCLUSION:
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
32/36
COOLING TOWER
AIM: To study the characteristics of cooling tower.
APPARATUS: Cooling towers are equipment used for lowering the temperature of process water
so that it can be reused, by bringing it in contact with ambient air. Cooling of water takes placeboth by transfer of sensible heat and by evaporative cooling. As a result of this the necessary latent
heat for evaporation is provided by the sensible heat of water, resulting in the lowering of water
temperature.In a cooling tower, air and water are brought into intimate contact in a counter-current manner.
The warm process water flows down over a series of slots, which increase the interfacial area, and
promotes turbulence. As the air rises it gets heated and humidified, which the water is cooled.The air circulation in cooling tower can be effected in two ways-by natural draught and by
mechanical draught. Thus there are two types of cooling towers: -
1. Natural draught type, which depend on chimney effect for air circulation. The air andvapour present in the cooling tower have a higher temperature than atmospheric air. Hence,
they are less dense and rise upward through the tower, allowing fresh air to enter the towerat the bottom and thus setting up an air circulation pattern.
2. Mechanical draught type in which circulation of air is effected by either a forced draught inwhich a fan is placed at the bottom of the tower, or an induced draught in which a fan is
placed at the top of the tower removing moist air. In this type of tower the air velocity is
higher than in natural draught type.
The performance factor for cooling tower is given by the relation:
( )
+
+
+
=
= 4321
21
''*
1111
4
11
2
HHHH
TTdTHHL
KaVT
T
CP
Where,H = H* - H
H*= enthalpy of saturated air (kJ/kg)
H = enthalpy of air (kJ/kg)T1 = inlet temperature of water (K)
T2 = outlet temperature of water (K)
K = mass transfer coefficienta = interfacial area per unit volume (m2/m3)
Cp= specific heat of water = 4.18 kJ/kg K
V = active volume (m2
)L = liquid flow rate (kg/s)
H1 = value of H at T2+0.1 (T1-T2)
H2= value of H at T2+0.4 (T1-T2)H3 = value of H at T1-0.4(T1-T2)
H4 = value of H at T1+0.1 (T1-T2)
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
33/36
EQUIPMENT DESCRIPTION:
The setup consists of:
Cooling tower A tank at the bottom for water collection A second tank with heater to produce hot water
Pump for hot water circulation Rotameter for measuring hot water flow rate Thermometer for measuring inlet and outlet temperature of water Valve for controlling hot water flow Two sets of dry and wet bulb thermometer to measure dry and wet bulb temperature of
air at inlet and outlet of tower
Blower for supplying air to the tower at a fixed flow rate
PROCEDURE:
1. Fill water in hot water tank and ensure that the heater is completely immersed.2. Start the heater and heat the water upto the temperature of 40-430C
3. Start the hot water pump keeping the by-pass open and set the flow of hot water at the
desired rate.4. Start the blower for air circulation
5. Allow the system to stabilize
6. Note the inlet and outlet temperatures of water.7. Note the inlet dry and outlet dry and wet bulb temperatures of air.
8. Repeat the above for different flow rate of water.
OBSERVATIONS:
1. Height of tower = m2. Velocity of air inlet = m/s3. Velocity of air outlet = m/s
4. Cross- section area
of air inlet = m2
5. Relative Humidity of inlet Air = %
OBSERVATION TABLE: -
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
34/36
Sr.
No.
Hot
water
flow
rate
(lph)
Inlet
water
Temp.0C (T1)
Outlet
water
Temp.0C (T2)
Inlet
air
Temp.0C
Outlet
air
Temp.0C
Relative
Humidity
of Outlet
Air
%RH
Water flow
Rate (kg/s)
L
Gas
flow
rate
(kg/s)
(G)
1
2
3
CALCULATIONS:
1. Calculate the different temperatures using the formulae mentioned below, for each flowrate of water,
Sr. No. T2+(0.1)(T1-T2) T2+0.4 (T1-T2) T1-0.1 (T1-T2) T1-0.4 (T1-T2)
1.
2.
3.
4.
3. Calculate H* & H, the enthalpy of saturated air at these temperatures,using the humidity chart.
4. Calculate H, enthalpy of air, at all these temperatures using the operating line
equation
G (HG2-HG1) = L CL (L2- L1)
Where:
H = HG2 (kJ/kg)HG1=enthalpy of inlet air (kJ/kg) =G = gas flow rate (kg/s.m2) =
L = liquid flow rate (kg/s.m2) =
CL= specific heat of water (kJ/kg K) =L1=outlet temperature of water
L2 = various temperatures calculated above
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
35/36
Sr. No T2+(0.1)
(T1-T2)
T2+(0.1)
(T1-T2)
T2+0.4
(T1-T2)
T2+0.4
(T1-T2)
T1-0.1
(T1-T2)
T1-0.1
(T1-T2)
T1-0.4
(T1-T2)
T1-0.4
(T1-T2)
H* H H* H H* H H* H
1.
2.
3.
4.
5. Calculate H1, H2, H3, &H4 & their reciprocals.
Sr.
No.
H1 1/ H1 H2 1/ H2 H3 1/ H3 H4 1/ H4
1.
2.
3.
4.
6. Now calculate the performance factor using the formula given earlier.
RESULT:
The performance factors are calculated and their values are reported in the table below:
Sr. No. Performance Factor
-
7/31/2019 Final Mass Transfer - I All Practical Writeups
36/36
CONCLUSION: