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DISTILLATION

•  2 methods of distillation:

•  eg. Ethanol-water:

vapour phase = higher conc. of ethanol

liquid phase = higher conc. of water

•  separation of different components in a liquid solution

•  involved producing a vapour from a liquid by heating the liquid in a vessel

  rectification/fractional/distillation with reflux – part of the vapour is condensed & returned as

liquid back to the vessel

  all of the vapour is removed or is condensed as product

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DISTILLATION

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DISTILLATION

•  steam

•  fractional

•  extractive

Types of distillation:

•  simple

  immiscible solvent

  azeotropic

  vacuum

  molecular

  entrainer sublimation Distillation tower at an oil refinery.

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VAPOUR-LIQUID EQUILIBRIUM RELATIONS

•  ideal solution (substances very similar to each other)

RAOULT’S LAW

pA = partial pressure of component A in the vapour

•  vapour & liquid in intimate contact for a long time, equilibrium is attained

pA = PAxA or yA =(pA/P)=(PA/P)xA

where

PA = vapour pressure of pure A

xA = mole fraction of A in the liquid

yA = mole fraction of A in the gas

P= total pressure

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Vapour-liquid equilibrium for A-B mixture eg. benzene(A)-toluene (B)

BOILING-POINT DIAGRAMS & x-y PLOTS (VLE diagrams)

  heat a mixture of benzene-toluene at xA1 = 0.318( boil at 98oC), first vapour in equilibrium is yA1 = 0.532

  distance bet. equilibrium & 45o line = diff. bet. xA & yA ( 〉 diff., easier separation)

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VLE diagrams

  non-ideal systems which will present more difficult separation

  An azeotrope is a liquid mixture which when vaporised, produces the same composition as the liquid.

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RELATIVE VOLATILITY OF VAPOUR-LIQUID SYSTEMS

yA = mole fraction of A in the gas phase

Relative volatility, α - numerical measure of ease of separation

αAB - relative volatility of component A with respect to component B

where

xA = mole fraction of A in the liquid which is in equilibrium with yA phase

For an ideal system (obeys Raoult’s law):

€

αAB =

yAxAyBxB

=y

A/x

A

1−yA

/ 1−x

A

€

yA = αxA1+ α−1

xA

where

€

αAB = PAPB

= vapour pressure of pure Avapour pressure of pure B

when α 〉 1 , separation is possible

SINGLE-STAGE EQUILIBRIUM CONTACTOR

Binary distillation - components A & B

xA1 xA0

yA1 yA2

Total material balance: L0 + V2 = L1 + V1

Balance on A: L0xA0 + V2yA2 = L1xA1 + V1yA1

Unknown : x1 & y1 – solve simultaneously (graphically) between equilibrium line & overall material balance

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Constant molal overflow: V1 = V2

L0 = L1

Example 11.2-1

•  A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.4 benzene (A) and 0.6 toluene (B) and 100 kg mol total is brought into contact with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molar overflow. Calculate the amounts and compositions of the exit streams.

SINGLE-STAGE EQUILIBRIUM CONTACTOR

Example 11.2-1

Balance on A: L0xA0 + V2yA2 = L1xA1 + V1yA1

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Constant molal overflow: V1 = V2= 100 kmol

L0 = L1 = 110 kmol

xA1 xA0 =0.3

yA1 yA2 =0.4

=100 kmol

=110 kmol

110(0.3) + 100(0.4) = 110xA1 + 100yA1

lets xA1 = 0.2, yA1 = 0.51 lets xA1 = 0.4, yA1 = 0.29

lets xA1 = 0.3, yA1 = 0.4 At intersection, xA1 = 0.25, yA1 = 0.455

SIMPLE DISTILLATION METHODS

•  Single stage binary distillation

•  vapour & liquid then separated

Balance on A: FxF = Lx + Vy or FxF = (F-V)x + Vy Unknown : x & y – solve simultaneously (graphically) between equilibrium

line & overall material balance similar to eg. 11.2-1

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•  liquid mixture partially vapourized

•  vapour allowed to come to equilibrium with liquid

Separator F,xF

Heater

V,y

L,x

EQUILIBRIUM OR FLASH DISTILLATION

SIMPLE DISTILLATION METHODS

•  liquid mixture charged to a still (heated kettle)

•  first portion of vapour = richest in component A

unknown: x2

L1x1 = L2x2 + (L1-L2)yav

Average composition of total material distilled, yav:

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•  slowly boiled & vapourized part of the liquid •  vapour withdrawn rapidly to condenser

SIMPLE BATCH OR DIFFERENTIAL DISTILLATION

•  vapourized product gets learner in comp. A

V, y

L, x

L1 = original moles charge

x1 = original composition

L2 = moles left in the still

x2 = final composition of liquid

∫ −= 1

2

xx

2

1xy

dxLLln

Graphical solution:

area under the curve 1/(y-x) vs x plot = 2

1LLln

Example 11.3-2

•  A mixture of 100 mol containing 50 mol% n-pentane and 50mol% n-heptane is distilled under differential conditions at 101.3 kPa until 40 mol is distilled. What is the average composition of the total vapor distilled and the composition of the liquid left? The equilibrium data are as follows, where x and y are mole fractions of n-pentane:

The equilibrium data are as follows, where x and y are mole fractions of n-pentane:

x y x y x y

1.000 1.000 0.398 0.836 0.059 0.271

0.867 0.984 0.254 0.701 0 0

0.594 0.925 0.145 0.521

Example 11.3-2

Total balance : L1 = V + L2

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Equilibrium data:

100 = 40 + L2

L2 = 60 mol

L1 = 100 mol

x1 = 0.5 mol /mol

V = 40 mol

x2 = ? , yav = ?

x 1.0 0.867 0.594 0.398 0.254 0.145 0.059 0 y 1.0 0.984 0.925 0.836 0.701 0.521 0.271 0

∫ −= 1

2

xx

2

1xy

dxLLln

0.51xydx

60100ln 0.5

x2=∫ −=

By trial-&-error, x2 = 0.277

L1x1 = L2x2 + (L1-L2)yav

100(0.5) = 60(0.277)+ (100-60)yav

yav = 0.835

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McCABE-THIELE METHOD

•  assume equimolar overflow/constant molal overflow between

feed inlet & top tray

feed inlet & bottom tray

•  graphical method for determining the number of theoretical stages,N

•  binary mixture A-B

q line

•  Effect of feed condition (q line)

•  q=heat needed to vaporize 1 mol of feed at entering conditions per molar latent heat of vaporization of feed.

•  q-line equation:

McCabe-Thiele Method

•  Graphical method requires: •  i.Top Operating Line •  ii.Feed Operating Line (q-line) •  Iii.Bottom Operating Line

Feed Operating Line (q-line) •  q-line equation: •  Lm=Ln + qF •  Vn=Vm + (1-q)F •  Vny=Lnx+DxD (Top) ---(1) •  Vmy=Lmx – WxW (Bottom)-----(2) •  (2)-(1): •  (Vm-Vn)y=(Lm-Ln)x-(DxD-FxF) •  (Vm-Vn)y=(Lm-Ln)x – FxF

•  Lm-Ln=qF •  Vm-Vn=(q-1)F

q-line cont’

•  (q-1)Fy = qFx – FxF

•  y = [(q)/(q-1)]x – xF/(q-1)

q-line cont’

•  q value depends on feed conditions: •  If feed is liquid at boiling pt: q=1 •  If feed is saturated vapor:q=0 •  If feed is liquid below its boiling pt:q>1 •  If feed is a mix bet. Liq & vap: 0<q<1

Example 11.4-1 •  A liquid mixture of benzene-toluene is to be distilled in a

fractionating tower at 101.3 kPa pressure. The feed of 100 kg mol/h is liquid, containing 45 mol% benzene and 55 mol% toluene, and enters at 327.6 K(130oF). A distillate containing 95 mol% benzene and 5 mol% toluene and a bottoms containing 10 mol% benzene and 90 mol% toluene are to be obtained.The reflux ratio is 4:1. The average heat capacity of the feed is 159 kJ/kg mol.K (38 btu/lb mol.oF) and the average latent heat 32099 kJ/kg mol (13800 btu/lbmol). Equilibrium data for this system are given in Table 11.1-1.Calculate the kg moles per hour distillate, kg mole per hour bottoms, and the number of theoretical trays needed.

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Example 11.4-1 Binary mixture A-B (benzene-toluene) at 101.3kPa. Reflux ratio (R) = 4. Average heat capacity of feed = 159 kJ/kmol.K & average latent heat = 32099 kJ/kmol. Determine D kmol/h, W kmol/h & N theoretical trays needed.

D kmol/h

xD = 0.95

W kmol/h

xW = 0.1

F =100 kmol/h

xF = 0.45

TF = 327.6K

Total material balance: F =100 =D + W

Balance on benzene (A): FxF = DxD + WxW

100(0.45) = D(0.95) + W(0.1)

45 = (100-W)(0.95) + W(0.1)

D =100 - W

Substituting D = 100-W

45 = 95 +(W)(0.95) + W(0.1)

W = 58.8 kmol/h D = 100-58.8 = 41.2 kmol/h

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Example 11.4-1

2. Draw enriching operating line

R = 4 = L/D

D kmol/h

xD = 0.95

W kmol/h

xW = 0.1

F =100 kmol/h

xF = 0.45

TF = 327.6K

L

1Rxx1R

Ry D+

++

=

0.190.8x140.95x14

4y +=+

++

=

1. Plot equilibrium & 45o lines on x-y graph

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Example 11.4-1

3. Calculate q (fraction of feed that is liquid)

LV

FBPLV

LV

FVHH

)T(TcHHHHHHq

−−+−

=−−

=

32099327.6)(159)(T32099

q B −+=

Average heat capacity of feed =159 kJ/kmol.K

Average latent heat 32099 kJ/kmol

From Fig. 11.1-1, at xF = 0.45, TB = 93.5oC (366.7K)

1.19532099327.6)7(159)(366.32099

q =−+

=

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Example 11.4-1

q = 1 (liquid at its boiling point) , q = 0 (saturated vapour) , q 〉 1 (cold liquid feed)

q 〈 0 (superheated vapour) , 0 〈 q 〉 1 (mixture of liquid & vapour)

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Example 11.4-1

Slope = 6.12 = Δy/Δx = (0.45 – y)/(0.45-x)

q = 1.195

€

y= qq−1

x− xF

q−1

€

y= 1.1951.195−1x− 0.45

1.195−1= 6.12x−2.31

4. Draw q-line

0.05y0.45

0.40.45y-0.45

x0.45y-0.456.12 −=

−=

−=

Lets x = 0.40

y = 0.144 5. Draw stripping operating line

Connect xW(on 45o line) with the point of intersection of the q-line & the enriching operating line

q-line

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Example 11.4-1

Starting from xD, make steps bet. equilibrium line & enriching line to q-line

6. Stepping off from xD

7. Shift to stripping line after passing q-line

8. Feed location = tray on the shift

9. Ntheo. stages = number of steps

10. Ntheo. trays = theo. stages - reboiler

Feed tray = tray 5 from the top

Ntheo. trays = 8 – 1= 7 trays plus a reboiler

Ntheo. stages = 8 stages

Feed tray

12

3

4

5

6

7

8

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TOTAL REFLUX, R = ∞ •  minimum number of stages, Nmin

•  stepping off from xD to xW on the 45o line

•  operating lines coincide with 45o line •  infinite sizes of condenser, reboiler & tower diameter

•  or using Fenske equation (total condenser)

€

Nmin =log xD

1−xD

1−xWxW

log αav

αav = average value of relative volatility = (α1αW)½

α1= relative volatility of the overhead vapour

αW= relative volatility of the bottom liquid

Example 11.4-2

For the rectification in Example 11.4-1, where a benzene-toluene feed is being distilled to give a distillate composition of xD=0.95 and a bottoms composition of xW=0.10, calculate the following:

(a) Minimum reflux ratio Rmin

(b) Minimum number of theoretical plates at total reflux.

41.2 kmol/h

xD = 0.95

58.8 kmol/h

xW = 0.1

F =100 kmol/h

xF = 0.45

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Example 11.4-2

At R = ∞, Nmin = ?

Nmin = 5.8 stages or 4.8 trays plus a reboiler

Steps are drawn from xD to xW.

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MINIMUM REFLUX, Rmin

•  infinite number of stages/trays

•  Rmin at pinch point (x’,y’)

•  minimum vapour flow •  minimum condenser & reboiler

•  or when equilibrium line has an inflection, operating line tangent to the equilibrium line

1Rxx1R

Ry D+

++

=

Enriching op. line:

y-intercept: intercepty

1R0.95

1Rx

minmin

D −=+

=+

Minimum Reflux Ratio

•  The top operating line intercepts q-line at equilibrium line.

•  The line passes through the points (x’,y’) and (xD,xD):

OPERATING AND OPTIMUM REFLUX RATIO

Two limits of tower operation exist: At total reflux -minimum number of plates with infinite tower diameter -cost of tower, steam & cooling tower increases.

At minimum reflux -infinite number of tray -infinite cost of tower. So, actual operating reflux ratio lies between total reflux and

minimum reflux (Rmin).

Normally, Ractual=1.2 -1.5 of Rmin

EXAMPLE 11.4-2

•  For the rectification in Example 11.4-1, where a benzene-toluene feed is being distilled to give a distillate composition of xD=0.95 and a bottoms composition of xW=0.10, deretmine the following:

•  (a) Minimum reflux ratio Rmin. •  (b) Minimum number of theoretical plates at

total reflux (Nmin).

41.2 kmol/h

xD = 0.95

58.8 kmol/h

xW = 0.1

F =100 kmol/h

xF = 0.45

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Example 11.4-2

Given R = 4, Rmin = ?

Rmin = 1.21

The enriching op. line from xD is drawn through the intersection of the q-line & the equilibrium line to intersect the y-axis

1Rxx1R

Ry D+

++

=

Enriching op. line:

y-intercept: 0.43

1R0.95

1Rx

minmin

D =+

=+

SPECIAL CASE DISTILLATION

•  1.Stripping column distillation •  2.Enriching column distillation •  3.Rectification with direct steam injection •  4.Rectification tower with side streams •  5.Partial condensers

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STRIPPING-COLUMN DISTILLATION •  feed is saturated liquid at boiling point (q=1)

•  added to the top of the column •  overhead product is not returned back to the tower

•  operating line: 1m

Wm

1m

m

VWx

xVLy

++

−=

•  Ntheo. stages - starting from xW(on 45o line) draw a straight line to the intersection of yD with the q-line

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ENRICHING-COLUMN DISTILLATION •  feed is saturated vapour (q=0)

•  added to the bottom of the column •  overhead product is refluxed back to the tower

•  operating line: 1R

xx1R

Ry D

++

+=

•  N theo. stages - starting from xD(on 45o line) draw a straight line to the y-intercept, xD/(R+1)

F

D

W

xD xF

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DIRECT STEAM INJECTION

•  steam injected as small bubbles into liquid

•  stripping operating line: WxSWxS

Wy −=

•  heat provided by open steam injected directly at bottom of tower

•  draw a straight line from (xW,0) through WxW/(W-S) on the 45o line

•  use of open steam requires an extra fraction of a stage

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SIDE STREAM

•  side stream above feed inlet:

intermediate operating line: 1S

DO

1S

S

VDxOx

xVLy

++

++=

•  stream removed from sections of tower

liquid side stream: Ln = LS+O

VS+1 = Vn+1=V1=Ln+D

•  from the intersection of enriching op. line & xO, draw a straight line to y-intercept of intermediate op. line

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PARTIAL CONDENSERS

•  liquid condensate returned to tower as reflux

•  overhead product = vapour

•  one extra theoretical stage for partial condenser (both liquid & vapour in condenser is in equilibrium)

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TRAY EFFICIENCY

•  Overall tray efficiency, Eo 3 types of tray efficiency:

•  Murphree tray efficiency, EM

•  Point/local tray efficiency, EMP

trays actual of no.trays ideal of no.EO =

1nn

1nnMP -y'*y

-y'y'E+

+=

1nn

1nnM -y*y

y-yE+

+=

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TRAY EFFICIENCY

1 actual tray = triangle ‘abe’ on actual equil. line

1 ideal tray = triangle ‘acd’ on ideal equil. line

Eg. EM = 0.6 (60% efficiency) distance ‘ac’ = 10 cm

Graphical determination of actual trays given EM:

Reboiler = 1 stage (bet. ideal equil. & operating lines

distance ‘ab’ = 0.6(10 cm) = 6 cm

Get 4-5 points & draw actual equil. line thru’ each points

Step off actual trays between operating & actual equil. lines

Problem (Tray efficiency)

•  No 11.5-1 (pg 755): •  For the distillation of heptane and ethyl

benzene in problem 11.4-2, the Murphree tray efficiency is estimated as 0.55. Determine the actual number of trays needed by stepping off the trays using the tray efficiency of 0.55. Also, calculate the overall tray efficiency Eo.

Condenser Duty (qc) •  Enthalpy Balance Around the Condenser: •  V1(H1) =L(hD) + D(hD) + qc

•  qc=VHD – (LhD+DhD)

•  For total condenser: •  V1=VD, HD=H1 •  H1=the saturated vapor enthalpy (equation 11.6-2):

•  H1 can also be determined from the enthalpy-concentration diagram

Reboiler Duty (qR) •  Overall Enthalpy Balance :

•  Enthalpy in = Enthalpy out

•  qR +Fhf = qc + DhD + WhW

•  qR =qc + DhD + WhW – Fhf

•  hD and hf from equation (11.6-1) or from the enthalpy-concentration diagram.

Example •  Binary mixture A-B (benzene-toluene) is to be distilled in a

fractionation column 101.3kPa. The feed of 100 kgmol/h is liquid, containing 45 mol% benzene and 55 mol% toluene, and enters at 327.6 K. A distillate containing 95 mol% benzene and 5 mol% toluene and a bottoms containing 10 mol% benzene and 90 mol% toluene are to be obtained. A Reflux ratio (R) = 1.5Rm. Given that Rm=1.17. Determine the condenser duty and the reboiler duty required by the distillation column by assuming constant molar overflow. Physical property for benzene and toluene and enthalpy-concentration diagram are given in Table 11.6-1(pg 733) and Table 11.6-2 (pg 734), respectively.

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ENTHALPY-CONCENTRATION METHOD

•  no assumption of molal overflow rates

•  takes into account latent heats, heats of solution & sensible heats

•  Ponchon-Savarit method

•  graphical procedure combining enthalpy & material balances

•  provides information on condenser & reboiler duties

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•  calculation shown in eg. 11.6-1

•  data Table 11.6-2 pg. 734 (Geankoplis 4th Ed.)

Enthalpy-concentration for benzene-toluene

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ENTHALPY-CONCENTRATION METHOD

Drawing isotherms (tie lines) on the enthalpy-concentration diagram from

(a) temperature-concentration diagram (b) x-y diagram

41.2 kmol/h

xD = 0.95

58.8 kmol/h

xW = 0.1

F =100 kmol/h

xF = 0.45

TF = 327.6K

L

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Example 11.6-2

Given : R = 1.5 Rm = 1.5(1.17) = 1.755

1. Plot enthalpy-concentration diagram and x-y diagram on the same sheet of paper. Locate points D, W and F at xD, xW and xF, respectively.

hF = xFcpA(TF-T0) + (1-xF)cpB(TF-T0)

where cpA = cp of liquid benzene = 138.2 kJ/kmol.K

cpB = cp of liquid toluene = 167.5 kJ/kmol.K

T0 = Tref. = Tb.p. of liquid benzene = 80.1oC

hF= 0.45(138.2)(327.6-353.1) + (1-0.45)167.5(327.6-353.1) = -3938.1 kJ

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Example 11.6-2

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

W

DF

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Sat. vapour

Sat. liquid hD

H1

ΔR hD+QC/D

V1

1RHÄ

1DHh

Locating ΔR

2. Locate rectifying-section difference point,ΔR

D1

1R

D1

1C

D

hHHÄ

hH

HDQh

R =−

−+=

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

W

DF

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Example 11.6-2

1cmHÄ

1.755hHHÄ

hH

HDQh

R 1R

D1

1R

D1

1C

D===

−

−+=

1.755cm1cm1.755HÄ 1R ==

∆R ΔR =85x 103kJ/kmol

hW=5x 103kJ/kmol

H1=31x 103kJ/kmol V1

D1

1R

D1

1C

D

hHHÄ

hH

HDQh

R =−

−+=

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3. Locate stripping-section difference point, ΔS

4. Step off trays for rectifying section using ΔR. 5. Step off trays for stripping section using ΔS

D1

1C

D

hH

HDQh

R−

−+=

Example 11.6-2

6. Theoretical stages = numbers of tie lines

7. Theoretical trays = theoretical stages - reboiler

8. Feed tray = tie line that crosses the line SRFÄÄ

9. Condenser duty, QC = (ΔR –hD)D or

10. Reboiler duty, QR = (hW-ΔS)W

Theoretical trays = 11.9 – 1 = 10.9 trays

Feed trays = tray no. 7 from the top

QC = (85x103-0)41.2 = 3 460 800 kJ/h

QR = (hW-ΔS)W = (5 x 103 –[-64x103])58.8 = 4 057 200 kJ.h

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Locating ΔS

Draw a straight line from ΔR through F to intersect the vertical line at xW

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Example 11.6-2

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

W

DF

∆R

∆S -64 x103kJ/kmol

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Stepping off trays for the rectifying section

From V1 vertically to 45o line, horizontally across to equilibrium line and vertically back to saturated liquid line, to give L1. Connect L1 to V1 using a tie line. From L1 to ΔR intersecting the saturated vapour line to give V2.Repeat until a tie line crosses over the line ∆RF∆S.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

∆S

∆R

D

V1 W

F

V2

L1 L2

V3

L3

V4

L4

V5

L5

V6

L6

V7

L7

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Stepping off trays for the stripping section

Draw a line from ∆S through L7 up to the saturated vapour line to give V8. From V8 vertically to 45o line, horizontally to equilibrium line and vertically back to saturated liquid line to give L8. Connect L8 and V8 using a tie line..Repeat until a tie line touches W or exceed W.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

∆S

∆R

D

V1 W

F

V2

L1 L2

V3

L3

V4

L4

V5 V6 V7

L7 L5 L6

V8

L8

V9

L9

V10

L10

V11

L11

V12

L12

No. of theoretical stages = no. of tie lines = 11.9 stages

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MINIMUM REFLUX RATIO, Rmin

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

W

DF

Rmin occur when the line ∆RF∆S coincides with the tie line that passes through F

Get a tie line that passes through F and extend that line to intersect the vertical line xD to get ∆Rmin

∆Rmin

D1

1Rmin

D1

1min

CD

min hHHÄ

hH

HDQh

R =−

−+=

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MINIMUM STAGES, Nmin

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

W

DF

Nmin is obtained when the operating lines are vertical since ΔR and ΔS are at infinity.

Nmin = 5.9 stages

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PARTIAL CONDENSER

Partial condenser = 1 stage

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

∆S

∆R

VD W

F

V1

L0 L1

V2

L2

V3

L3

V4

L4

V5

L5

V6

L6

Distillate product, VD – vapour with the composition yD Condensed liquid/reflux, L0, have the composition xo which is in equilibrium with yD

yD

x0

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PARTIALLY VAPOURISED FEED,zF

-140.0

-120.0

-100.0

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 0.2 0.4 0.6 0.8 1.0

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

W

F with the composition zF lies on the tie line LF (composition = xF) and VF (composition = yF)

F

F

F

FFVFL

LV

=

F LF

VF

F LF

VF

By trial-and-error, locate the point F so as to satisfy the inverse lever rule: zF

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MULTICOMPONENT DISTILLATION

•  more than 2 components

•  shortcut calculation methods – an approximation

Equilibrium data •  Raoult’s law – for ideal mixture

•  hydrocarbon system:

AAA

A xPP

Ppy == B

BBB xP

PPpy == C

CCC xP

PPpy == D

DDD xP

PPpy ==

AAA xKy = BBB xKy = CCC xKy = DDD xKy =

where KA = vapour-liquid equilibrium constant or distribution coefficient

•  relative volatility, αi :

€

αA = KAKref.

€

αB = KBKref.

€

αC = KCKref.

€

αD = KDKref.

•  only allow separation between two components, heavy key and light key

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Example 11.7-2

F = 100 mol/h at boiling point at 405.3 kPa. xFA = 0.4, xFB = 0.25, xFC = 0.20 and xFD = 0.15 where components A = n-butane, B = n-pentane, C = n-hexane and D = n-heptane. 90% of B is recovered in the distillate and 90% of C in the bottoms. Calculate: (a) D and W moles/h (b) dew point of distillate and boiling point of bottoms (c) minimum stages for total reflux and distribution of other components in the distillate and bottoms. Solution:

1st trial: Assume all of component A will be in the distillate and all of component D will be in the bottom product

Components xF xFF yD yDD xW xWW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

40.0 25.0 20.0 15.0

F=100

0.62 0.349 0.031

0 ∑yD=1.00

40.0 22.5 2.0 0

D=64.5

0 0.070 0.507 0.423

∑xW=1.00

0 2.5 18.0 15.0

W=35.5

Assume all A go to distillate and all D go to the bottoms. yAD(D)=xAF(F)=0.4(100)=40 mol/h xDW(W)=xDF(F)=0.15(100)=15 mol/h N-pentane(B): Light Key Overall Balance: F =D+W Comp Balance: xBF(F)=0.25(100)=25 mol/h=yBD(D) + xBW(W) Since 90% of B is in distillate: yBD(D) = (0.90)(25) = 22.5, xBW(W)=2.5 N-hexane (C): Heavy Key xCF(F)=0.20(100) =20 mol/h Since 90% of C is in the bottoms: xCW(W)=0.90(20)=18 mol/h, yCDD=2.0 mol/h

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DEW POINT For a vapour mixture of A, B, C and D:

Liquid composition which is in equilibrium with the vapour mixture:

€

xi =

yi

αi

yi

αi

∑

€

xi =yiKi

∑ = yi

Kref.

yiαi

∑ =1

1. By trial-&-error, assume Td.p.

2. Get corresponding values of Ki and αi

3. Calculate yi/αi

4. From Kref. = ∑(yi/αi) , get the corresponding T

5. Compare latest T with assumed T. If differ, use latest T for next iteration by repeating steps 2-4

6. Once Td.p. is obtained, calculate liquid composition

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Example 11.7-2

Dew point: Assume Td.p. = 67oC (Kref. = KC = ∑(yi/αi) = 0.26)

Components xF xFF yD yDD xW xWW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

40.0 25.0 20.0 15.0

F=100

0.62 0.349 0.031

0 ∑yD=1.00

40.0 22.5 2.0 0

D=64.5

0 0.070 0.507 0.423

∑xW=1.00

0 2.5 18.0 15.0

W=35.5

Components yiD Ki αi yi /αi xi

A B(L) C(H)

D

0.62 0.349 0.031

0 ∑yiD=1.00

1.75 0.65 0.26 0.10

6.73 2.50 1.00

0.385

0.0921 0.1396 0.0310

0 ∑yi /αi=0.2627

0.351 0.531 0.118

0 ∑xi=1.00

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Example 11.7-2

1.75

0.65

0.26

0.10

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BOILING POINT

For a liquid mixture of A, B, C and D:

∑yi = ∑Kixi = K ref. ∑αixi = 1

Vapour composition which is in equilibrium with the liquid mixture:

€

yi =α

ix

i

αix

i

∑

1. By trial-&-error, assume Tb.p.

2. Get corresponding values of Ki and αi

3. Calculate αixi

4. From Kref. = 1/(αixi) , get the corresponding T

5. Compare latest T with assumed T. If differ, use latest T for next iteration by repeating steps 2-4

6. Once Tb.p. is obtained, calculate vapour composition

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Example 11.7-2

Boiling point: Assume Tb.p. = 132oC (Kref. = KC = 1/∑αixi = 1.144)

Components xF xFF yD yDD xW xWW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

40.0 25.0 20.0 15.0

F=100

0.62 0.349 0.031

0 ∑yD=1.00

40.0 22.5 2.0 0

D=64.5

0 0.070 0.507 0.423

∑xW=1.00

0 2.5 18.0 15.0

W=35.5

Components xiw Ki αi xi αi yi A

B(L) C(H)

D

0 0.070 0.507 0.423

∑xW=1.00

4.95 2.34 1.10 0.61

5.00 2.043 1.000 0.530

0 0.1430 0.5070 0.2242

∑xi αi=0.8742 1/∑αixi = 1.144

0 0.164 0.580 0.256 ∑xi=1.00

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Example 11.7-2

5.00

2.35

1.15

0.61

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MINIMUM STAGES,NMIN AT R = ∞ Minimum stages, Nmin, using Fenske equation:

where:

α=

av, log

WxWx

DxDx

log

NL

LW

HW

HD

LD

min

xLD = mole fraction of LK in distillate

xLW = mole fraction of LK in bottom product xHD = mole fraction of HK in distillate

xHW = mole fraction of HK in bottom product

αLD = relative volatility of LK at dew point. αLW = relative volatility of LK at boiling point.

αL,av =

€

αLDαLW

Distribution of other components:

€

xiDDxiWW = (αi,av

)Nm xHDDxHWW

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Example 11.7-2

stages ltheoretica 5.4042.258 log

0.0700.507

0.031(0.349)log

av, log

WxWx

DxDx

log

NL

LW

HW

HD

LD

min ==α

=

αL,av =

€

αLDαLW = (2.50)(2.043) = 2.258

Components yiD=xiD yDD αi xiw xWW αi A

B(L) C(H)

D

0.62 0.349 0.031

0 ∑yD=1.00

40.0 22.5 2.0 0

D=64.5

6.73 2.50 1.00 0.385

0 0.070 0.507 0.423

∑xW=1.00

0 2.5

18.0 15.0

W=35.5

4.348 2.043 1.000 0.530

Distribution of other components:

€

xiDDxiWW = (αi,av

)Nm xHDDxHWW

Distribution of component A & D:

€

xADDxAWW = (αA,av

)Nm xHDDxHWW = (αA,av

)5.404 (0.031)64.5(0.507)35.5 = (αA,av

)5.4040.1111

€

xDDDxDWW = (αD,av

)Nm xHDDxHWW = (αD,av

)5.404 (0.031)64.5(0.507)35.5 = (αD,av

)5.4040.1111

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Example 11.7-2

αA,av =

€

αADαAW = (6.73)(4.348) = 5.409

αD,av =

€

αDDαDW = 0.385(0.530) = 0.452

€

xADDxAWW = (αA,av

)Nm xHDDxHWW = (αA,av

)5.404 (0.031)64.5(0.507)35.5 = (5.409)5.4040.1111=1017

€

xDDDxDWW = (αD,av

)Nm xHDDxHWW = (αD,av

)5.404 (0.031)64.5(0.507)35.5 = (0.452)5.4040.1111= 0.001521

Material balance of component A & D:

Components yiD=xiD yDD αi xiw xWW αi A

B(L) C(H)

D

0.62 0.349 0.031

0 ∑yD=1.00

40.0 22.5 2.0 0

D=64.5

6.73 2.50 1.00 0.385

0 0.070 0.507 0.423

∑xW=1.00

0 2.5

18.0 15.0

W=35.5

4.348 2.043 1.000 0.530

40 Wx Dx AWAD =+ 15 Wx Dx DWDD =+

Solving : 0.039Wx39.961 Dx

AW

AD==

14.977Wx0.023 Dx

DW

DD==

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Example 11.7-2

Revised distillate and bottoms compositions:

Components xF xFF yD yDD xW xWW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

40.0 25.0 20.0 15.0

F=100

0.62 0.349 0.031

0 ∑yD=1.00

40.0 22.5 2.0 0

D=64.5

0 0.070 0.507 0.423

∑xW=1.00

0 2.5 18.0 15.0

W=35.5

Components xF xFF yD yDD xW xWW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

40.0 25.0 20.0 15.0

F=100

0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

39.961 22.5 2.0

0.023 D=64.484

0.0011 0.0704 0.5068 0.4217 ∑xW=1.00

0.039 2.500 18.00 14.977

W=35.516

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MINIMUM REFLUX RATIO,RMIN Underwood’s shortcut method for calculating Rmin:

where: xiD = mole fraction of component i in distillate taken at R = ∞

xiF = mole fraction of component i in the feed

αi = relative volatility of the top and the bottom of the tower

To determine Rmin: 1. By trial-and-error, assume θ ( αLK 〈 θ 〉 αHK )

€

1−q= αixiFαi−θ

∑

€

Rm +1= αixiDαi−θ

∑

2. Calculate 1-q for various θ

3. Use θ obtained to calculate Rmin

Assumes constant flows in both sections of tower

Uses constant average α (at Tave. = [Ttop + Tbottom ]/2)

Example 11.7-3

Using the conditions and results given in Example 11.7-2, calculate the following:

(a) Minimum reflux ratio using the Underwood method.

(b) Number of theoretical stages at an operating reflux ratio R of 1.5Rm using the Erbar-Maddox correlation.

(c) Location of feed tray using the method of Kirkbride.

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Example 11.7-3

Tb.p. = 132oC & Tdp = 67oC Tave. = (132 + 67)oC/2 = 99.5oC

Components xF xFF xiD xiDD xW xWW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

40.0 25.0 20.0 15.0

F=100

0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

39.961 22.5 2.0

0.023 D=64.484

0.0011 0.0704 0.5068 0.4217 ∑xW=1.00

0.039 2.500 18.00 14.977

W=35.516

Components xiF xiD Ki αi xiW A

B(L) C(H)

D

0.40 0.25 0.20 0.15

∑xiF=1.00

0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

3.12 1.38 0.60 0.28

5.20 2.30 1.00 0.467

0.0011 0.0704 0.5068 0.4217

∑xiW=1.00

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Example 11.7-3

3.12

1.38

o.60

0.28

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Example 11.7-3

Components xiF xiD Ki (99.5oC) αi (99.5oC) xiW

A B(L) C(H)

D

0.40 0.25 0.20 0.15

∑xiF=1.00

0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

3.12 1.38 0.60 0.28

5.20 2.30 1.00 0.467

0.0011 0.0704 0.5068 0.4217

∑xiW=1.00

€

1−q= αixiFαi−θ

∑ = 1−0=1= 5.2(0.4)5.20−θ

+ 2.3(0.25)

2.3−θ

+ 1.0(0.2)

1.0−θ

+ 0.467(0.15)

0.467−θ

1. By trial-and-error, assume θ ( αLK 〈 θ 〉 αHK )

θ (assumed)

∑(sum)

1.210 1.200

1.2096

0.5213 0.5200 0.5213

0.5275 0.5227 0.5273

-0.9524 -1.0000 -0.9542

-0.0942 -0.0955 -0.0943

+0.0022 -0.0528 +0.0001

€

0.467(0.15)0.467−θ

€

1.0(0.2)1.0−θ

€

2.3(0.25)2.3−θ

€

5.2(0.4)5.20−θ

2. Calculate 1-q for various θ

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Example 11.7-3

Components xiF xiD Ki (99.5oC) αi (99.5oC) xiW

A B(L) C(H)

D

0.40 0.25 0.20 0.15

∑xiF=1.00

0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

3.12 1.38 0.60 0.28

5.20 2.30 1.00 0.467

0.0011 0.0704 0.5068 0.4217

∑xiW=1.00

θ = 1.2096

€

Rm +1= αixiDαi−θ

∑ = 5.20(0.6197)5.20−1.2096+ 2.30(0.3489)

2.30−1.2096+1.00(0.0310)1.00−1.2096+ 0.467(0.15)

0.467−1.2096

3. Use θ obtained to calculate Rmin

€

Rm +1= αixiDαi−θ

∑ =1.395

Rmin = 0.395

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SHORTCUT METHOD NUMBER OF STAGES

Correlation of Erbar & Maddox:

Feed plate location using Kirkbride method:

where:

Ne = number of theoretical stages above the feed plate

NS = number of theoretical stages below the feed plate

=

2

HD

LW

LF

HF

xx

DW

xx

log 0.206sNeN

log

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Example 11.7-3

R = 1.5Rmin = 1.5(0.395) = 0.593

R/(R+1) = 0.593/(0.593+1) = 0.3723

Rmin/(Rmin+1) = 0.395/(0.395+1)

0.37

0.49

Nmin/N = 0.49 = 5.404/N

N = 11.0 theoretical stages or 10 theoretical trays plus a reboiler

Rmin/(Rmin+1) = 0.2832

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Example 11.7-3

N = 11.0 theoretical stages Ne + NS = 11.0 theoretical stages

=

2

HD

LW

LF

HF

xx

DW

xx

log 0.206sNeN

log

0.073440.03100.0704

64.48435.516

0.250.2 log 0.206

sNeN

log2

==

Components xiF (F=100 mol/h) xiD (D=64.484 mol/h) xiW (W=35.516 mol/h)

A B(L) C(H)

D

0.40 0.25 0.20 0.15

∑xiF=1.00

0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

0.0011 0.0704 0.5068 0.4217

∑xiW=1.00

1.184 sNeN= s1.184N eN = 1.184NS + NS = 11.0

NS = 5.0 Ne = 6.0 Feed tray = 6.0 trays from the top