Distillation I
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Transcript of 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
Flash Distillation Calculations
(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
Flash Distillation Calculations
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
Flash Distillation Calculations
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
Flash Distillation Calculations
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.