DEPARTMENT OF CHEMICAL ENGINEERING INDIAN INSTITUTE OF TECHNOLOGY KANPUR

Che492-Unit Operations Laboratory 2 Experiment no. Continuous Distillation Instructors: Dr. Siddhartha Panda Dr. Deepak Kunzru

Name of TA: Prashant Kr. Gupta

Date of experiment: 12/08/2013Date of submission: 19/08/2013

Group No: 2

Roll no.Name10059Akshay Bansal10062Akshit Gupta10067Aman Jain

INDEX

S. NoTitlePage No.

1Introduction3

2Objective4

3Theory4

4Utulities required5

5Procedure5

6Observation & calculation5

7Inferences9

8Precautions9

9References9

INTRODUCTIONContinuous distillation, a form ofdistillation, is an on-going separation in which a mixture is continuously (without interruption) fed into the process and separated fractions are removed continuously as output streams. A distillation is theseparationor partial separation of aliquidfeed mixture into components or fractions by selectiveboiling(orevaporation) andcondensation. A distillation produces at least two output fractions. These fractions include at least onevolatiledistillate fraction, which has boiled and been separately captured as a vapor condensed to a liquid, and practically always a bottoms (or residuum) fraction, which is the least volatile residue that has not been separately captured as a condensed vapor.Each fraction may contain one or more components (types ofchemical compounds). When distillingcrude oilor a similar feedstock, each fraction contains many components of similar volatility and other properties. Although it is possible to run a small-scale or laboratory continuous distillation, most often continuous distillation is used in a large-scale industrial process.

OBJECTIVE

To study Continuous Distillation in a Sieve plate column

THEORY

The principle for continuous distillation is the same as for normal distillation: when a liquid mixture is heated so that it boils, the composition of the vapor above the liquid differs from the liquid composition. If this vapor is then separated andcondensedinto a liquid, it becomes richer in the lower boiling component(s) of the original mixture.This is what happens in a continuous distillation column. A mixture is heated up, and routed into the distillation column. On entering the column, the feed starts flowing down but part of it, the component(s) with lower boiling point(s), vaporizes and rises. However, as it rises, it cools and while part of it continues up as vapor, some of it (enriched in the lessvolatilecomponent) begins to descend again.The "lightest" products (those with the lowest boiling point or highest volatility) exit from the top of the columns and the "heaviest" products (the bottoms, those with the highest boiling point) exit from the bottom of the column. The overhead stream may be cooled and condensed using a water-cooled or air-cooledcondenser. Thebottoms reboilermay be a steam-heated or hot oil-heatedheat exchanger, or even a gas or oil-firedfurnace.

In a continuous distillation, the system is kept in asteady stateor approximate steady state. Steady state means that quantities related to the process do not change as time passes during operation. Such constant quantities include feed input rate, output stream rates, heating and cooling rates,refluxratio, andtemperatures, pressures, and compositions at every point (location). Unless the process is disturbed due to changes in feed, heating, ambient temperature, or condensing, steady state is normally maintained. This is also the main attraction of continuous distillation, apart from the minimum amount of (easily instrumentable) surveillance; if the feed rate and feed composition are kept constant, product rate andqualityare also constant. Even when a variation in conditions occurs, modernprocess controlmethods are commonly able to gradually return the continuous process to another steady state again.

TERMINOLOGYREFLUXReflux refers to the portion of the condensed overhead liquid product from a distillation tower that is returned to the upper part of the tower the downflowing reflux liquid provides cooling and partial condensation of the upflowing vapors, thereby increasing the efficacy of the distillation tower. The more reflux that is provided, the better is the tower's separation of the lower boiling from the higher boiling components of the feedReflux Ratio is given by L/D

Where,

L is the flow rate of the liquid sent back to the column D is the distillate flow rate.

Total Reflux

For, a column operating at Total Reflux (i.e. all of the distillate being sent back to the column)L=Vin both the enriching & stripping sections

y = x for all trays.Thus the operating line of McCabe Thiele diagram coincides with the y = x line.

Finite Reflux ratio (R)

For the column being operated at some other reflux ratio R, the quality is given by:q= Hg- Hf /(Hg-HL)

And the slope of this q-line is given by -> q/(q-1)

The enriching section operating line is given by-

It passes throughy = x = XD

The operating line of stripping section passes through the intersection of the enriching section operating line & the q line.

The equation for enriching section operating line is given by -

A column with a high reflux ratio may have fewer stages, but it refluxes a large amount of liquid, giving a wide column with a large holdup. Conversely, a column with a low reflux ratio must have a large number of stages, thus requiring a taller column.

EXPERIMENTAL APPARATUS

Chemical engineering schematic of Continuous Binary Fractional Distillation tower. A binary distillation separates a feed mixture stream into two fractions: one distillate and one bottoms fractions.UTILITIES REQUIRED: Distillation column made of glass having perforated plates

Ethanol-Water solution

pumps for feed and reflux

Refractometer for measuring the refractive index

Electricity Supply: 220 V AC

Floor Area Required: 2 m x 2 m.

PROCEDURE

We calibrated the rotameter with the given solution. We obtained the calibration curve for refractive index Vs. Alcohol percent. We noted the composition of feed in the boiler and initial conditions of setup such as temperature of feed and water flowing in and water flowing out etc,. as per in data sheet given. We started the reboiler heating, taking the initial reading of power meter and cooling water in the condenser also. We operated the system in total reflux and waited till steady state is obtained. Distillate and bottom samples are taken systematically and then we determined the refractive index and specific gravity . We noted the flow rate of distillate. Now we operated the system under a reflux ratio higher than minimum reflux ratio and obtained samples at steady state as same procedure explained above. We took readings of flow rate of, feed , distillate, product and reflux. The setup is shutdown and finally took the power reading at the end of the experiment. Quantity of sample collected from product stream is calculated and note down for mass balance calculations. We came next day to calculate the total amount of feed remain in re-boiler and its refractive index. The amount of secondary feed consumed is also noted.

OBSERVATIONS AND CALCULATION

Mole fraction of ethanolRefractive indexMole fraction of ethanolSpecific gravity

0.101.3430.4300.873

0.201.3530.6200.845

0.321.3580.7000.830

0.421.3590.7500.820

0.471.3590.8000.810

0.611.3600.9000.804

0.631.360

Finding the Value of q :Given Feed conditions, super cooled liquid we can calculate the value of q by following expression:q = 1 + Cpl(Tb Tf)/Hvap = Boiling point of the mixture = Freezing point of the mixture = Heat capacity of the mixture Hvap= Latent heat of vaporization of the mixture Given, refractive index = 1.356From the graph of Refractive index vs mole fraction plotted above, we find that at the mole fraction of ethanol in the feed is 0.2380.24

Mol. Wt. of Ethanol = 46Mol. Wt. of water = 18

Therefore, average mol. Wt. for the given composition = 0.24x46 + 0.76x18 = 24.72 g/mole

Calculating the weighted average of heat capacities of water and alcohol, we get = 111.46*0.24 + 75.6*0.76 = 84.21 J/mole K

= 351.37*.24 + 373*.76 = 367.81 K = 159*.24 + 273*.76 = 245.64 KHvap = 38.6*.24 + 40.68*.76 = 40.18 KJ/mole = 40180 J/molPutting these values in the equation obtains q = 1+84.21*(367.81-245.64)/40180 =1.256Slope of the q line is 1.256/(1.256-1) = 4.91 and Intercept of the y-axis is .24/(1-1.256)= .937

Calculation of R minimumRelative volatility of mixture at given feed conditions is given as = .76/.24 = 3.2 approximately. can be calculated by following expression ,where Zf = 0.24= 1/zf( - 1) = 1/(.24*(3.2-1)) = 1.894

ConditionWeight of sampleSp.gravityMolefraction

total reflux26.9320.8070.85

26.8840.8020.88

Partail Reflux 126.9120.8050.80

26.8970.8030.80

Partail Reflux 226.9140.8050.81

26.9340.8070.81

Used scientific calculator equation solving function to solve the best fitting curve in the fig.2ConditionRefractive indexMolefraction

total reflux1.3550.23

1.3450.11

Partail Reflux 11.3400.08

1.3460.12

Partail Reflux 21.3460.12

1.3470.13

Total Reflux:Here R is infinite so y intercept of operating line is 0 And operating Line is y = xObserved is 0.88 from table 4 given above.= 0.24 and after steady the mole fraction in the bottoms is = 0.11

Now using McCabe-Thiele method to calculate theoretical No. of stages No. of stage = 8 = 8*100/12 = 66.66%Partial Reflux 1Flow rate of reflux (R) = .152 ml/s = 9.12 ml/minFlow rate of product (D) = .0425 ml/s = 2.55 ml/min Reflux ratio = (R-D)/D = (9.12-2.55)/2.55 = 2.576Observed is .8By McCabe-Thiele method No. of stages = 9

= 9*100/12 = 75%Partial Reflux 2Flow rate of reflux (R) = .165ml/s = 9.9 ml/minFlow rate of product (D) = .0438ml/s = 2.628 ml/min Reflux ratio = (R-D)/D = (9.9-2.628)/2.628 = 2.767Observed is .81By McCabe-Thiele method No. of stages = 8

= 7*100/12 = 66.66%Material Balance (Mass in Mass out ) = Mass accumulating Mass coming in:Feed in = 1200mlDensity of feed = .24*.789 + .76*.9905 = .942g/mlMass coming in = 1200*.942 = 1130.568gMass coming out:Product taken out = 225mlAverage density of product = (.88*.789+.12*.9905 + .81*.789+.19*.9905 + .80*.789+.2*.9905)/3 = 0.823 gm/mlMass of product coming out = .823*225 = 185.23gMass of bottom = 970*(.13*.789+.87*.9905)= 935.4gMass accumulating:Mass accumulating = 1130.568-935.4 = 195.2g

Energy Balance: Latent heat of vaporization: - Ethanol:-38.6 K J/mol = 767.9 J/g, Water: - 40.68 KJ/mol = 2146 J/gHeat capacity: - Ethanol:-111.46 J/mol-K = 2.845 J/g-KWater:- 75.6 J/mol-K =4.184 J/g-K (Energy in Energy out) = (Energy accumulating + Energy loss) Energy coming in: Re-boiler energy input = (349.1-347.5)/ = 1.6 KWH = 5760 KJEnergy coming in by feed = 0 (at reference temp)Energy coming out:Bottom heat cap. = 0.13*2.845+0.87*4.184 =4.01 J/g-KEnergy coming out by product = 225*0.823*(.8*2.845+.2*4.184)*(76-28) = 27.67 KJEnergy out by Condenser = (flow rate (ml/s)*density (g/ml)*Cp(J/g-K)*T*time (s)) = (2.1*1000/60)*1*4.184*(31-28)*(180*60) = 4744 KJEnergy loss:Energy loss = 5760-4744 = 1016 KJ