Chemical Engineering & Fuels EXPERIMENT NO. – 01 11.11.2009
DATE:
NAME OF THE EXPERIMENT: Separation of the components of a binary mixture (methanol and water) by simple distillation.
THEORY: The term distillation is properly applied only to those operations where vaporization of a liquid mixture yields a vapor phase containing more than one constituent and it is desired to recover one or more of these constituents in a nearly pure state. The basis requirement for a separation of components by distillation is that the composition of the vapor be different from the composition of the liquid with which it is in equilibrium. If the vapor composition is the same as the liquid composition, distillation processes will not affect a separation. According to Raoult’s law, PA = PoA . XA
Where, PA = Partial pressure of component A in the vapor phase.
P B = P o B . XB
PB = Partial pressure of component B in the vapor phase. PoA = Vapor pressure of component A PoB = Vapor pressure of component B XA = Mole fraction of component A in liquid phase XB = Mole fraction of component B in liquid phase.
According to Dalton’s law, P = PA + PB
Where, P = Total pressure of vapor phase.
= PoA . XA + PoB . XB Let, A is low boiling component, therefore it is more volatile. So, Relative volatility,
PA . X B o A
α = P /P
o B
=
-----------
PB.XA
If Îą = 1, then separation by distillation will be impossible. The larger the value of Îą, the easier is the distillation. As A is more volatile component, the molar ratio of A to B in vapor phase will be greater than that in liquid phase.
XA (v) -------XB (v)
XA (L) > -------XB (L)
Simple distillation is approached by commercial batch distillation processes where the vapor generated by boiling the liquid is withdrawn from contact with liquid and condensed as fast as it is formed without appreciable condensation. A binary mixture of methanol and water can be distilled easily because of the difference of their boiling points. The boiling point of methanol is 64.7 0c whereas that of water is 1000C.
PROCEDURE: 1. Dry round bottom flask and conical flask are taken (if not dry, then these are dried using Acetone) and weighed. 2. Approximately 150 ml. of the sample is taken in the round bottom flask and wt. of it containing sample is taken. 3. The whole simple distillation unit is arranged. 4. Room temperature is measured and the distillation process is started by switching on the Mantle heater. 5. Temperature reading is taken when the first drop of distillate is found in the conical flask ( For Methanol – water mixture the temperature is approximately 65 0C ) 6. Temperature of the heating medium is controlled in such a way that the number of drops of distillate does not exceed more than 25 drops. 7. Distillation process is carried on up to the temperature when it just touches 100 0C mark i.e. the Mantle heater is switched off. 8. The distillation unit is let to be cooled for several minutes. 9. The wt. of both the round bottom flask containing residue and the conical flask containing distillate are taken. 10. Refractive index of the distillate is measured.
DATA AND CALCULATION: Data – 1:
Weight of round bottom flask
=101.226 gm
Weight of round bottom flask + Sample=239.030 gm Weight of sample = 137.804 gm. Weight of conical flask = 105.395 gm Weight of conical flask + condensate= 135.301 gm Weight of round bottom flask + residue = 203.247 gm Weight of residue in round bottom flask = 102.021 gm Weight of condensate in conical flask = 34.075 gm Room temperature = 280c
Data – 2:
First drop of distillate at 67.50c Constant distillation at 690c Distillation stopped at 1000c Data – 3:
Refractive index of distillate = 1.336
Calculation: Table: Material balance for simple distillation of methanol-water mixture. Input Material
CH3OH - H2O mixture Total
Weight (gm)
137.804
137.804
Output Wt./Wt. % 100
100
Material
Weight (gm)
Wt./Wt %
Distillate
34.075
24.73
Residue
102.021
74.03
Loss
1.708
1.24
137.804
100
EXPERIMENT NO. – 02 18.11.2009
DATE:
NAME OF THE EXPERIMENT: Separation of components from a binary mixture (methanol and water) by fractional distillation process.
THEORY: The term distillation is properly applied only to those operations where vaporization of a liquid mixture yields a vapor phase containing more than one constituent and it is desired to recover one or more of these constituents in a nearly pure state. The basis requirement for a separation of components by distillation is that the composition of the vapor be different from the composition of the liquid with which is in equilibrium. If the vapor composition is the same as the liquid composition, distillation processes will not affect a separation. According to Raoult’s law, PA = PoA . XA
Where, PA = Partial pressure of component A in the vapor phase.
P B = P o B . XB
PB = Partial pressure of component
B in the vapor phase.
PoA = Vapor pressure of component A PoB = Vapor pressure of component B XA = Mole fraction of component A in liquid phase XB = Mole fraction of component B in liquid phase. According to Dalton’s law, P = PA + PB
Where,
P = Total pressure of vapor phase.
= PoA . xA + PoB . xB Let, A is low boiling component, therefore it is more volatile. So, Relative volatility, o
α = P A/P
o
B
=
PA . xB -----------
PB.xA
If Îą = 1, then separation by distillation will be impossible. The larger the value of Îą, the easier is the distillation. As A is more volatile component, the molar ratio of A to B in vapor phase will be greater than that in liquid phase.
XA (v) -------XB (v)
XA (v) > -------XB (v)
Fractional distillation involves the returning of part of the condensate to the still under such conditions that this returning liquid is brought into intimate counter-current contact with the vapors on their way to condenser. This method is of such importance that it is given the special name of rectification. A binary mixture of methanol and water can be distilled easily because of the difference of their boiling points. The boiling point of methanol is 64.7 0c whereas that of water is 100 0c. The products of the fractional distillation are almost pure methanol and water.
PROCEDURE: 1.
Dry round bottom flask and conical flask are taken (if not dry, then these are dried
using Acetone) and weighed. 2.
Approximately 150 ml. of the sample is taken in the round bottom flask and wt. of it
containing sample is taken. 3.
The whole fractional distillation unit is arranged.
4.
Room temperature is measured and the distillation process is started by switching
on the Mantle heater. 5.
Temperature reading is taken when the first drop of distillate is found in the conical
flask. 6.
Temperature of the heating medium is controlled in such a way that the number of
drops of distillate does not exceed more than 25 drops. 7.
Distillation process is carried on up to the temperature (of the fractionating column)
when it just exceeds the constant temperature (for Methanol – water mixture the constant temperature is approximately 650C) i.e. the Mantle heater is switched off. 8.
The distillation unit is let be cooled for several minutes.
9.
The wt. of both the round bottom flask containing residue and the conical flask
containing distillate are taken. 10.
Refractive index of the distillate is measured.
DATA AND CALCULATION: Data – 1:
Weight of round bottom flask =110.761gm Weight of round bottom flask + Sample =251.264 gm Weight of conical flask =111.436 gm Weight of conical flask + Distillate =157.545 gm Weight of round bottom flask + Residue =201.571 gm Weight of sample = (251.264 – 110.761) = 140.503 gm Weight of distillate = (157.545 – 111.436) = 46.109 gm Weight of residue = (201.571 – 110.761) = 90.81 gm Room temperature = 27oc
Data – 2:
First drop of distillate at 64oc Constant distillation at 66oc Distillation stopped at 66.3oc Data – 3:
Refractive index of distillate = 1.3295
Calculation: Table: Material balance for fractional distillation of CH 3OH-H2O mixture: Input Material
Weight(gm)
CH3OH-H2O mixture
140.503
Total
140.503
EXPERIMENT NO. – 03 22.11.2009 NAME OF THE EXPERIMENT:
Output
Wt./Wt. % 100
100
Wt./Wt.
Material
Weight(gm)
Distillate
46.109
32.82
Residue
90.81
64.63
Loss
3.584
2.55
140.503
100
%
DATE:
Determination of viscosity index of lubricating oil from kinematic viscosity at 40 0C and 1000C.
THEORY: Viscosity: In case of a fluid a backward force acts against the direction of the flow when it flows due to friction at solid-fluid interface and also between the layers of fluid .This property of fluid is called viscosity. In a Newtonian fluid the share stress (Force per unit area of the sharing plane) is proportional to the share rate (Local velocity gradient). i.e. TV α
du/dy
Where, TV = Shear stress arising from viscous laminar flow = Fs/As
or ,
TV = μ . du/dy
du/dy = Shear rate μ
= Proportionality constant
The proportionality constant μ is called the co-efficient of viscosity or simply viscosity in a quantitative sense. In CGS system viscosity is expressed in gm/ cm-sec or poise (P). Kinematic viscosity: The ratio of the absolute viscosity to the density of a fluid is the kinematic viscosity for that fluid. γ = μ/ρ
Where, γ = Kinematic viscosity μ = Absolute viscosity ρ = Density of the fluid.
In CGS system kinematic viscosity is expressed in stokes (St). 1 cSt = 1 mm 2/ sec. The viscosity of liquid decreases significantly when the temperature is raised. Kinematic viscosity varies with temperature over a somewhat narrower range than absolute viscosity. Log μ = A + B/T Where, T = Kelvin temperature. A and B are constants. Higher the change of viscosity with temperature, lower the lubricating performance or suitability of a lubricant for the mechanical parts. Viscosity index: It is an arbitrary number indicating the effect of change of temperature on the kinematic viscosity of oil. A high viscosity index signifies relatively small change of kinematic viscosity with temperature. For oils of viscosity index of 0 to 100,
V.I. =
L–U x 100 Where, U = Kinematic viscosity at 40 0c of the sample. L–H L - U
L = Kinematic viscosity at 400c of an oil of 0 V.I.
--------------
L - H
H =Kinematic viscosity at 40 0c of an oil of 100 V.I
For oils of viscosity index of 100 and greater, (Antilog N) - 1 V.I. = ------------------ + 100 0.00715
Where, N = (log H- log U)/ log γ γ =Kinematic viscosity of the sample oil at 100 0c
Again, Kinematic viscosity, γ = K. T
Where, K = Constant for viscosity index measuring instrument. T = Time of fall of sample oil.
PROCEDURE: 1.
A large beaker is filled up with water.
2.
A viscosity index meter with sample oil is dipped into the beaker with the help of a
stand in such a way that the openings of that device are in the air. 3.
A thermometer is also dipped into the water and placed at a level between the
upper and lower marks of the viscosity index meter. 4.
Water is heated by using a Bunsen burner.
5.
The sample oil is sucked up over the upper mark of the device and time reading is
taken during the fall of oil from upper mark to lower mark. This is done 3 times at 40 0C and again 3 times at 1000C.
DATA AND CALCULATION: Kinematic viscosity of the sample oil at 40 0 C = 105.859 cSt i.e.
γ40 = U
= 105.859 cSt
Table: Required time for oil fall Temperature (oC) 40
Required time 7 min 35.11 sec 7 min 19.18 sec 7 min 45.02 sec
455.11 sec 439.18 sec 465.02 sec
453.10 sec
1 min 38.08 sec 1 min 48.09 sec 1 min 48. 28 sec
100
Now,
γ40 = =>
Again, Therefore,
K =
104.82 sec
K . T40 105.859 ----------cSt. = 453.10 0.2336 cSt. /sec
γ40 / T40 =
γ100 = Y =
98.08 sec 108.09 sec 108.28 sec
K. T100
= 0.2336 x 104.82 cSt. = 24.49 cSt.
γ 100 = 24.49 cSt.
From table – 3 (on the basis of Y), L = 709.54 cSt. And H = 311.1 cSt. Hence,
L - U -----------
V.I. = L -= H
709.54 - 105.859 ---------------------------x 100 = 151.51 709.54 - 311.1
As the viscosity index is more than 100, it should be recalculated. Now,
N = (logH – logU)/ logY log (311.1) – log (105.859) =
------------------------------x 100 log (24.49)
V.I. = =
=
0.3371
(Antilog N) - 1 ------------------+ 100 0.00715 Antilog(0.3371) - 1 -----------------------+ 100 0.00715
= 264.08
Results The viscosity index of the lubricating oil from the kinematic viscosity at 40°C and 100°C is 1264.08.