Simple and Differential Distillation Experiment Procedure | Simple distillation process | Differential Distillation process

AIM: Differential distillation, distillation process, distillation calculation, obtaining distillation data for the binary mixture.

OBJECTIVES: To compare experimentally (F/W) values to that calculated from Rayleigh’s equation.

PRINCIPLE/THEORY:

If an infinite number of successive flash vaporization of liquid is carried out with a number of decimally small portions of liquid each time, then the net result would be equivalent to a simple distillation of differential distillation. In practice, the method can be approximated to a batch of liquid containing charged to a kettle or still fitted with some of the heating devices such as a steam jacket. The charge is boiled slowly and the vapors are withdrawn rapidly as they form through a condenser where they are liquefied and the condensate is collected in the reservoir. The first portion of the distillate will be richest in a more volatile substance and as distillation proceeds, the vaporized product becomes lean. 

Rayleigh’s equation can explain this phenomenon by EQUATION

Modified Rayleigh’s equation
Log (FxF /Wxw) = a  log [F (1-xF) /W (1-xw )] 
a = ÖaW aF
Where F is the total number of moles of the feed, W is the number of moles of residue XF and XW are mole fractions of more volatile substances in feed, and residue, which relates the number of moles of ‘A’ remaining in the residue WXW to that of ‘B’ remaining W (1- Xw). These expressions are most likely to be valid for an ideal mixture, for which ‘a’ is most nearly constant.

EQUIPMENT USED:

  • Simple distillation column
  • Mantle heater
  • 250 ml beakers-2
  • 50 ml beaker-1
  • Specific gravity bottle (25 ml)
  • Thermometer

MATERIAL USED: Methanol –water.

EXPERIMENTAL PROCEDURE:

1. Measure exactly 300 ml using a measuring jar and transfer this 300 ml of feed into a beaker. In case of feed in the bottle is less than 300 ml make up with pure methanol.
2. Set aside 50 ml of the feed for measurement of specific gravity. Find specific gravity.
3. The balance of 250 ml is transferred into a round bottom flask of the setup.

4. Assemble the glass setup carefully. i.e. connect the condenser, thermal well, etc. Place a thermometer in the thermal well of the still.
5. Start the condenser water supply. Ensure a rich flow of water through the condenser.
6. Check all the connections and then slowly heat the feed by using a mantle heater.
7. When the vapor is produced, it will condense in the condenser and be collected in a 250 ml beaker.
8. This procedure is continued up to half the volume of feed is vaporized and condensed.
9. The volume of feed in the beaker is reduced to half the volume stopping the heating process while supplying water to the condenser for about 15 minutes. 

10. Make sure that no vapors are produced to shut down the water supply.
11. Cool the residue, the specific gravity, or R.I. for the residue and distillate are found.
12. The volume of the distillate and residue are also measured accurately using a measuring jar.
13. A calibration chart is prepared for (specific gravity or R . I.) to the mole fraction of methanol in the solution.

OBSERVATIONS:

The volume of feed = _____________ ml .
Room temperature = _____________ `C .
Distillate temperature =_______________`C.
Barometric pressure =728 mmHg .
Volume of distillate = _____________ml.
Volume of residue =______________ml.
Temperature of the first bubble starts at =_________`C.
Temperature at the end of distillation = ______________`C.
The specific gravity of the feed =_________
The specific gravity of the distillate =________
The specific gravity of the residue =_________

S.G.sample= (Wt of S.G. Bottle + Sample ) – (Wt of the empty S.G Bottle)/{(Wt of the S.G bottle + Distillate water )-(Wt of empty S.G Bottle )}


Temperature-Vapor pressure data:
Get the temperature–vapor pressure data for both methanol and water from Perry’s handbook.
Selection of upper and lower limits of temperatures.
First temperature—– Boiling point of more volatile component (here it is methanol ) corresponding to site barometric pressure (728 mmHg).
Last temperature is —-the boiling point of the less volatile component (here it is water ) corresponding to site barometric pressure (728 mmHg).
Any number of temperatures between the above two may be selected for filling up table 1

DATA ANALYSIS:

plot the graph between S.G Vs mole fraction X.
Get the temperature–vapor pressure data of water, and methanol from Perry’s handbook.
Then plot the graph between the vapor pressure of methanol Vs T and the Vapor pressure of water Vs Tin in the same graph.
Calculation X, Y* data for the methanol-water system at Pt =728 mmHg using Raoult’s law.( α are also computed).
NOTE: refer to example problem (1) in the distillation chapter of R.E Treybal (mass transfer book ).
Plot X Vs Y* (Take equal scale on both axes).
from graph 1, find X f, X d , Xw.
Calculate moles of feed, distillate & residue using the values of Xf, Xd and Xw obtained from step 6.
compute ln (F/W) ( i.e. experimental value ).
Plot the graph between 1/(Y*-X) Vs X. This graph was plotted using data available in step 4. The area under this curve between the limits Xf & Xw gives the RHS of Rayleigh‘s equation. This gives ln (F/W) (theoretical value).

Graphs to be plotted:
For a methanol-water system,
Specific gravity Vs mole fraction X.
Plot Y* Vs X.
Vapor pressure Vs Temperature of both methanol and water.
1/(Y*-X) Vs X (report area under the curve between the limits Xf and Xw ).
RESULTS:

ln(F/W) experiment =
ln(F/W) rayleigh’s equation =

DATA TABLES:

Preparation of calibration chart:

   S.NoA volume of methanol Volume of       waterSpecific  gravityMole fraction of methanol
     1        30              0
     2        25              5
     3        20            10
     4        15            15
     5          10            20
6         5            25
7          030

Table 1:

S.NoTemperaturePA(mmHg) (methanol)PB (mmHg)   (water)XA = (Pt-PA) /(PA- PB)                Y A= PA.XA/Pt

Table 2:

    S.No         X         Y*    1/(Y *-X)α =(PA/ PB)

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