Distillation I

Distillation I

Mass Transfer for 4th YearChemical Engineering Department

Faculty of EngineeringCairo University

THINGS YOU HAVE TO REMEMBER

• Phase Equilibria• Dalton’s Law• Raoult’s law• Antoine Equation

LET’S SAY THAT

For a binary system:A = More Volatile ComponentB = Less Volatile ComponentPo

A>PoB

a=Relative volatility (no more recovery)

oB

oA

PP

a

AND OF COURSE

xi= composition of component i in liquid phaseyi= composition of component i in vapour phasePi=yiPT Dalton’s lawPi=xiPo

i Raoult’s law

Antoine EquationWhere T in oK and Po in mmHgOr

TCBAPo

ln

TCBAo eP

Binary System Phase Diagram

T

x,y

Temperature-Composition Diagram

L

V

L+V

y

x

x-y Diagram

100% A0%B

0% A100%B

Liquidous Curve

Vapourous Curve

Equilibriu

m Curve

45 line (y

=x)

Binary System Phase Diagram

For a pure component evaporation occurs at a constant temperature, so it’s called “BOILING POINT”

For a mixture of two or more components evaporation occurs at increasing temperature so it’s called “BOILING RANGE” starting by Bubble point and ending at Dew point

T

x,y 100% A0%B

0% A100%B

Tbubble

Tdew

WHAT IS THIS??????

It is PTxy diagram for phase equilibrium

Effect of Pressure on Equilibrium

Some systems has what is so called AZEOTROPE like CS2-Acetone and Methanol-Water systems.An azeotropic mixture has a=1 so can’t be separated by distillation.This figure represents a “minimum azeotrope”

AZEOTROPIC MIXTURE

AZEOTROPIC MIXTURE

This figure represents a “maximum azeotrope” in the system acetone-chloroform

CALCULATING LIQUID VAPOUR EQUILIBRIA FOR BINARY SYSTEMS

RELATIONS USED

x-y diagram

T

y

x

1- Draw tie line2- Project x and y of that line on the x-y diagram and get a point on the equilibrium curve3- Repeat the last step several times4- Connect the points to get the equilibrium curve

1- I f we have TEMPERATURE-COMPOSITION DIAGRAM

2- If vapour pressures are known as a function of temperature:

TPA

o

PBo

BAT PPP

AB xx 1

oB

oA

oBx

PPPPT

A

oBB

oA xx PPP AT

oB

oB

oA

oBA

oA xxx PPPP1PP AAT

And by the general equilibrium relation for ideal binary systems:At equilibrium

interface Liquidphase Gas PP oAATA PxPy

The most used form of equilibrium relation for liquid-vapour systems(Either ideal or non-ideal, binary or multicomponent)

AAAT

A PP xKxyoA

3- Another form of equilibrium relation ( Only for binary systems if a is known):

AT

oA

A xPPy T

A

AoA P

xyP

oB

oA

PP

a

AA

AA

A

A

A

A

B

B

A

A

y-1xx-1y

x-1y-1

xy

xy

xy

a

AAAA x-1yy-1x a

A

AA x1-1

xαya

A NEW PHASE DIAGRAM

H= specific enthalpy of vapour.

h= specific enthalpy of liquid.

Liquidous Curve

Vapourous Curve

HBTU/Lb

x,y100% A

0%B0% A

100%B

ENTHALPY COMPOSITION DIAGRAM

H

h

Tie line is used for getting points on equilibrium curve as what was done in case of Temperature composition diagram.

This diagram is more difficult to draw, however it provides more information about enthalpy of streams.

H

y

x

THEORY OF DISTILLATION

For any system, once it’s in the wet region, it splits into two phases in equilibrium each of different compositions.The compositions of the two phases can be determined using “TIE LINE”

T

x,y 100% A0%B

0% A100%B

y

x

Distillation operations

Single Stage

Simple Differential Distillation

Steam Distillation

Flash vaporization Distillation

Multistage

Binary system

Multicomponent systems

Simple Differential DistillationSometimes called “ASTM distillation” and Used in labs

Distillation occurs so that the feed is slowly evaporated and then condensed “DISTILLATE (D)” and remains un-evaporated portion “RESIDUES (W)”.Now to get xw and ῩD

By Differential MB:Residues

(W)

Distillate(D)

f

w

x

x xydx

WF

*ln

Solved by graphical integrationORAlgebraically using the relation: (use when α given)

Then calculate ῩD using C.M.B equation

Simple Differential Distillation

f

w

x

x xydx

WF

*ln

F

W

FW

WF

xx

xxxx

WF

11ln

11ln

11ln

a

4- 100 kmoles of a mixture of A and B is fed to a simple

still. The feed contains 50 mole % A and a product

contains 5 mole % A is required. Calculate the

quantity of product obtained. The equilibrium data is

presented as follows:

Mole fraction of A in liquid

0 0.2 0.4 0.6 0.8 1

Mole fraction of A in vapor

0 0.35 0.58 0.75 0.9 1

Givens:F=100kmole xf=0.5xw=0.05

Here we will solve using graphical integration:From xw to xf

f

w

x

x xydx

WFln

x 1/(y-x)

0.05 26.67

0.2 6.67

0.4 5.56

0.5 6.06

kmolesWWF

xydx

WF

35.1

)3045.4exp(

3045.4)]]06.656.5(*1.0[)]56.567.6(*2.0[

)]67.2667.6(*)05.02.0[[(21ln

5.0

05.0

:polationinter by calculated are points additional The

5- A liquid containing 50% n-heptane and 50% n-octane is differentially distilled at 1 atmosphere to vaporize 60 mole% of feed. Find the composition of the distillate and the residue. Find the boiling range during distillation. (The average volatility α=2.15).

By trial and error : X w=0.3285From M.B: ῩD =0.614Boiling range =109.78-114.586oC

F

W

FW

WF

xx

xxxx

WF

11

ln11

ln1

1lna

Experimental method used for thermally sensible materials.For insoluble mixtures (hydrocarbon-water mixtures) each component exerts pressure equals the vapour pressure (x=1). If the summation of the vapour pressures equals the total pressure then the mixture will start boiling at a temperature lower than boiling point of the pure hydrocarbon.Steam is used to perform evaporation at a reduced temperature.

Steam Distillation

Steam functions:1- Heating the batch (material to be distilled) to the

bubble point.2- Giving the batch the latent heat needed to

vaporize.3- Carrying the vapours.

Steam Distillation

To calculate the necessary amount of steam needed we will need to know:1- Q1=heat required to heat the batch to Tb

2- Q2=Latent heat gained by the batch3- amount of steam carrying the vapours

Steam Distillation

Q1=mbatch*CP ) batch*(Tb-TF)Q2=mbatch*lbatch

Amount of steam carrying the vapours:Vapours leaving=water vapour+hydrocarbon vapourPT=Pwater+PHC

Steam Distillation

HC

w

HC

w

HCHC

ww

HC

w

MM

PP

MnMn

mm

Q1=mw1*l w mw1=Q1/l w

Q2=mw2*l w mw2=Q2/l w

Steam Distillation

batchHC

w

HC

ww3 m

MM

PPm

Problem 8:10 Kg batch of ethylaniline is to be steam distilled from small amount of non-volatile impurity. Saturated steam at 25 Psia is used. Initial temperature of ethylaniline is 40 oC and the distillation takes place at atmospheric pressure. a- At what temperature will the distillation proceeds? b- Determine the composition of the vapour phase. c- How much steam is used? Data: • Heat capacity of ethylaniline is 0.4 KCal/Kg.C• Heat capacity of steam is 0.35 KCal/Kg.C • Latent heat of vaporization of ethylaniline is 72 Kcal/Kg • Vapour pressures of water and ethylaniline are given in the table

below:

T (oC) 38.5 64.4 80.6 96 99.15 113.2Pw (mmHg) 51.1 199.7 363.9 657.6 737.2 1225PEA (mmHg) 1 5 10 20 22.8 40

T Pw PEA PT

38.5 51.1 1 52.1

64.4 199.7 5 204.7

80.6 363.9 10 373.9

96 657.6 20 677.6

99.15 737.2 22.8 760

113.2 1225 40 1265

Operation occurs at 1 atm=760 mmHgSO Temperature at which Ptotal=760 mmHg is 99.15oC

Composition of vapour phase:At 99.15oC:Pw=737.2 mmHgPEA=22.8 mmHgyw=737.2/760=0.97yEA=22.8/760=0.03

Amount of steam used:CP)EA= 0.4 KCal/Kg.C CP)Steam= 0.35 KCal/Kg.C lEA= 72 Kcal/Kg lSteam= 540 Kcal/Kg

Q1=10*0.4*(99.15-40)=236.5 Kcalm1=Q1/lSteam= 236.5/540=0.438 KgQ2=10*72=720 Kcalm2=Q2/lSteam= 720/540=1.333 Kg

mw=0.438+1.333+48.1=49.87 Kg

Kg 48.11012118

22.8737.2m

MM

PPm batch

HC

w

HC

ww3

Flash Vaporization Distillation

Flash vaporization or equilibrium distillation is a single stage operation where a liquid mixture is partially vaporized, the vapour allowed to come to equilibrium with the residual liquid and the resulting vapour and liquid phases are separated and removed.It’s easy but not efficient, and no packing or trays are needed.Generally used for easy separation (very high relative volatility) or as a primary separation step.

How to Reach Flashing Conditions

T

x,y 100% A0%B

0% A100%B

Changing Temperature Changing Pressure

T

x,y 100% A0%B

0% A100%B

How to Reach Flashing ConditionsChanging Temperature Changing Pressure

Fxf

Vy

Lx

Fxf

Vy

Lx

Calculating Tbubble and Tdew

Flashing occurs at a temperature between bubble and dew points. So calculating bubble and dew points is necessary to specify suitable flashing conditions.

T bubble > T flashing > T dew

Calculating Tbubble

Bubble point is the temperature at which JUST ONE bubble of gas evaporates and is in equilibrium with the liquid whose composition will not be affected.

T

xf100% A

0%B0% A

100%B

Tbubble

y

x

Calculating Tbubble

Then xi=xFi

And yi=KixFi

Where Ki=Poi/PT

TO GET Tbubble:1. Assume Tbubble

2. Calculate yi’s3. Calculate Syi. if equals 1

OK, else reassume Tbubble

4. Interpolate

T

xf100% A

0%B0% A

100%B

Tbubble

y

x

(Tbubble)1 Sy1

(Tbubble)TRUE Sy=1(Tbubble)2 Sy2

Sy

Tbubble

1

(Tbubble)TRUE

OR

21

1

21

1

ΣyΣy1Σy

bb

TRUEbb

TTTT

Calculating Tdew

Dew point is the temperature at which JUST ONE point of liquid condenses and is in equilibrium with the vapour whose composition will not be affected.

T

yf100% A

0%B0% A

100%B

Tdew

y

x

Calculating Tdew

Then yi=xFi

And xi=yFi/Ki

Where Ki=Poi/PT

TO GET Tdew:1. Assume Tdew

2. Calculate xi’s3. Calculate Sxi. if equals 1

OK, else reassume Tdew

4. Interpolate

T

xf100% A

0%B0% A

100%B

Tdew

y

x

(Tdew)1 Sx1

(Tdew)TRUE Sx=1(Tdew)2 Sx2

Sx

Tdew

1

(Tdew)TRUE

OR

21

1

21

1

ΣxΣx1Σx

dd

TRUEdd

TTTT

Flash Distillation Calculations

Overall Material Balance:F=L+VComponent Material Balance:F xFi=L xi+V yi

Equilibrium Relation:yi=ki xi

FxF

Vy

Lx


(L+V) xFi=L xi+V yi

(L+V) xFi=L xi+V ki xi

(L+V) xFi=(L+V ki) xi

It’s also solved by trial and error on liquid and vapour flow rates.

Fii

i xVkLVLx

FxF

Vy

Lx


For the easiness of trials:

(÷V)

FxF

Vy

Lx

ii

i xVkLVLx

if

ii x

kVL

1VL

x

i

i

i x1k V

L1V

Ly


Finally:

Now trials are done by changing ONLY L/V till Sx and Sy are BOTH equal one

Fii

i xkV

L1V

Lx

Fi

i

i x1k V

L1V

Ly


Once L/V is calculated, all needed variables can be calculated:F=L+V

V1F

VL

VF L

1

V

VLF

REMEMBER .....

The case we always deal with throughout our course is the ideal case, that is why we can say:

Ki=Poi/PT

HOWEVER .....In general, the systems are non-ideal. K values then can be calculated from the relation

pfKi

oii

i

Solving Flashing Problems IN EXAM

Due to time limitations:1- Assume L/V (say L/V=1)2- Calculate Sx and Sy.3- If Sx=1 AND Sy=1 L/V is correct4- If Sx ≠1 OR Sy≠1 assume another L/V5- Do the last step twice then interpolate

(L/V)1 Sx1

(L/V)operating Sx=1(L/V)2 Sx2

Sx

L/V

1

(L/V)operating

OR

21

1

21

1

ΣxΣx1Σx

VL

VL

VL

VL

operting

Solving Flashing Problems Practically

More iterations must be done to get more accurate results.Computer softwares such as Microsoft Excel can be used to do such iterations.Sometimes system is non-ideal, so other softwares like Hysys or Aspen can be used.We will now see how to do such calculations using Microsoft Excel.

Microsoft Excel

How about non-ideal systems

All calculation steps will not be changed, only the method of calculating “K” will change.K can be got thermodynamically using a proper equation of state (somehow difficult).K can be got from practical data, i.e. charts.For systems of oil fractionation it’s so common to use charts that are got from experiments instead of calculating them.

For More Information

For more information please check:1- Robert E. Treybal, “Mass Transfer Operations”, Third

ed., Ch. 9, PP. 3422- McCabe and Smith, “Unit Operations for chemical

engineers”, Fifth ed., Ch. 19, pp. 5883- Perry, R. H. and D. Green, eds., “Perry’s Chemical

Engineer’s Handbook”, Seventh ed., Section 13, McGraw-Hill, New York, 1997.

Documents

Distillation I