Separation Process 1 1pef

DISTILLATION •  separation of different components in a liquid solution •  involved producing a vapour from a liquid by heating the liquid in a vessel •  eg. Ethanol-water: vapour phase = higher conc. of ethanol liquid phase = higher conc. of water •  2 methods of distillation:   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 Types of distillation: •  simple •  fractional •  steam   immiscible solvent   azeotropic •  extractive   vacuum   molecular Distillation tower at an oil refinery.

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  entrainer sublimation

VAPOUR-LIQUID EQUILIBRIUM RELATIONS •  vapour & liquid in intimate for a long time, equilibrium is attained RAOULT’S LAW •  ideal solution (substances very similar to each other) pA = PAxA

or

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

where pA = partial pressure of component A in the vapour PA = vapour pressure of pure A P= total pressure xA = mole fraction of A in the liquid yA = mole fraction of A in the gas FKK

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

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

  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) FKK

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. FKK

RELATIVE VOLATILITY OF VAPOUR-LIQUID SYSTEMS Relative volatility, α - numerical measure of ease of separation αAB - relative volatility of component A with respect to component B yA y A /x A xA αAB = y =     B 1− y  /1− x  A  A xB  where yA = mole fraction of A in the gas phase € xA = mole fraction of A in the liquid which is in equilibrium with yA phase For an ideal system (obeys Raoult’s law): αx yA =  A  1+ α −1x A where

PA vapour pressure of pure A = P vapour pressure of pure B € B

αAB =

when α 〉 1 , separation is possible FKK

€

SINGLE-STAGE EQUILIBRIUM OR Binary distillation - components A & B yA1

yA2

xA1

xA0

Constant molal overflow:

V1 = V2 L0 = L1

Total material balance: Balance on A:

L0 + V2 = L1 + V1 L0xA0 + V2yA2 = L1xA1 + V1yA1

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

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 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 ed 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 OR Example 11.2-1

=100 kmol yA2 =0.4

yA1 =110 kmol xA0 =0.3

Constant molal overflow:

xA1

V1 = V2= 100 kmol L0 = L1 = 110 kmol

Balance on A:

L0xA0 + V2yA2 = L1xA1 + V1yA1 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

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SIMPLE DISTILLATION METHODS EQUILIBRIUM OR FLASH DISTILLATION •  Single stage binary distillation V,y F,xF

Heater

Separator

L,x •  liquid mixture partially vapourized •  vapour allowed to come to equilibrium with liquid •  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 FKK

SIMPLE DISTILLATION METHODS SIMPLE BATCH OR DIFFERENTIAL DISTILLATION •  liquid mixture charged to a still (heated kettle) V, y •  slowly boiled & vapourized part of the liquid •  vapour withdrawn rapidly to condenser L, x •  first portion of vapour = richest in component A •  vapourized product gets learner in comp. A L1 = original moles charge x1 = original composition L2 = moles left in the still x2 = final composition of liquid

unknown: x2

L1 x dx = L 2 ∫x y − x Graphical solution: ln

2

L1 ln area under the curve 1/(y-x) vs x plot = L 2

Average composition of total material distilled, yav: L1x1 = L2x2 + (L1-L2)yav FKK

1

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 Equilibrium data:

L1 = 100 mol x1 = 0.5 mol /mol

x

1.0 0.867 0.594 0.398 0.254 0.145 0.059 0

V = 40 mol

y

1.0 0.984 0.925 0.836 0.701 0.521 0.271 0

x2 = ? , yav = ?

Total balance :

L1 = V + L2 100 = 40 + L2

L2 = 60 mol L ln 1 = ∫xx ydx −x L2 ln 100 = ∫x0.5 ydx − x = 0.51 60 1

2

2

L1x1 = L2x2 + (L1-L2)yav 100(0.5) = 60(0.277)+ (100-60)yav By trial-&-error, x2 = 0.277 FKK

yav = 0.835

McCABE-THIELE METHOD •  binary mixture A-B •  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

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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<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.

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. Total material balance: F =100 =D + W D kmol/h xD = 0.95

Balance on benzene (A):

F =100 kmol/h

FxF = DxD + WxW

xF = 0.45

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

TF = 327.6K

Substituting D = 100-W W kmol/h

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

xW = 0.1

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

W = 58.8 kmol/h FKK

D =100 - W

D = 100-58.8 = 41.2 kmol/h

Example 11.4-1 L F =100 kmol/h

D kmol/h xD = 0.95 1. Plot equilibrium & 45o lines on x-y graph

xF = 0.45 TF = 327.6K

W kmol/h xW = 0.1

R = 4 = L/D 2. Draw enriching operating line x y = R x+ D R +1 R +1 y = 4 x + 0.95 = 0.8x + 0.19 4 +1 4 +1

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Example 11.4-1 3. Calculate q (fraction of feed that is liquid) H V − H F  H V − H L  + (TB − TF ) q= = HV − HL HV − HL

Average heat capacity of feed =159 kJ/kmol.K Average latent heat 32099 kJ/kmol 32099 + (159)(TB − 327.6) q= 32099  

From Fig. 11.1-1, at xF = 0.45, TB = 93.5oC (366.7K) 32099 + (159)(366.7 − 327.6) q= = 1.195 32099  

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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 4. Draw q-line

q-line

q = 1.195   xF  q  y=  x− q−1  q−1   

y= 1.195 x− 0.45 = 6.12x− 2.31 1.195−1 € 1.195−1 €

Slope = 6.12 = Δy/Δx = (0.45 – y)/(0.45-x) Lets x = 0.40 0.45 − y 6.12 = 0.45 - y = 0.45 - y = 0.45 − x 0.45 − 0.4 0.05 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 FKK

Example 11.4-1 6. Stepping off from xD Starting from xD, make steps bet. equilibrium line & enriching line to q-line

2 3

Feed tray

4 5

7. Shift to stripping line after ing qline

6 7

8. Feed location = tray on the shift Feed tray = tray 5 from the top 9. Ntheo. stages = number of steps Ntheo. stages = 8 stages 10. Ntheo. trays = theo. stages - reboiler Ntheo. trays = 8 – 1= 7 trays plus a reboiler FKK

8

1

TOTAL REFLUX, R = ∞ •  minimum number of stages, Nmin •  operating lines coincide with 45o line •  infinite sizes of condenser, reboiler & tower diameter •  stepping off from xD to xW on the 45o line •  or using Fenske equation (total condenser)

xD 1− x W  log 1− xD x W  Nmin = log αav    

αav = average value of relative volatility =€ (α1αW)½ α1= relative volatility of the overhead vapour αW= relative volatility of the bottom liquid FKK

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.

Example 11.4-2

41.2 kmol/h

At R = ∞, Nmin = ? Steps are drawn from xD to xW.

xD = 0.95 F =100 kmol/ h xF = 0.45 58.8 kmol/h xW = 0.1

Nmin = 5.8 stages or 4.8 trays plus a reboiler FKK

MINIMUM REFLUX, Rmin •  infinite number of stages/trays •  minimum vapour flow •  minimum condenser & reboiler •  Rmin at pinch point (x’,y’) •  or when equilibrium line has an inflection, operating line tangent to the equilibrium line

Enriching op. line: x y = R x+ D R +1 R +1 y-intercept: xD 0.95 = = y − intercept R min + 1 R + 1 min FKK

Minimum Reflux Ratio •  The top operating line intercepts q-line at equilibrium line. •  The line es 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).

Example 11.4-2 41.2 kmol/h xD = 0.95 F =100 kmol/ h

Given R = 4, Rmin = ? The enriching op. line from xD is drawn through the intersection of the q-line & the equilibrium line to intersect the y-axis

xF = 0.45 58.8 kmol/h xW = 0.1 Enriching op. line: x y = R x+ D R +1 R +1 y-intercept: xD 0.95 = = 0.43 R min + 1 R + 1 min FKK

Rmin = 1.21

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

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 Wx W Lm y = x − •  operating line: Vm + 1 m V m +1

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

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

•  added to the bottom of the column •  overhead product is refluxed back to the tower xD R y= x+ •  operating line: R +1 R +1 •  N theo. stages - starting from xD(on 45o line) draw a straight line to the y-intercept, xD/(R+1)

F

xF FKK

xD

W

DIRECT STEAM INJECTION •  heat provided by open steam injected directly at bottom of tower •  steam injected as small bubbles into liquid W •  stripping operating line: y = W x − x W S S •  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 FKK

SIDE STREAM •  stream removed from sections of tower •  side stream above feed inlet:

Ox O + Dx D LS intermediate operating line: y = V x + V S +1

liquid side stream:

Ln = LS+O

S +1

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 •  overhead product = vapour •  liquid condensate returned to tower as reflux •  one extra theoretical stage for partial condenser (both liquid & vapour in condenser is in equilibrium)

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TRAY EFFICIENCY 3 types of tray efficiency: •  Overall tray efficiency, Eo •  Murphree tray efficiency, EM •  Point/local tray efficiency, EMP EO = no. of ideal trays no. of actual trays

EM =

yn - yn + 1 y *n -y n + 1

EMP =

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y'n -y'n + 1 y *n -y'n + 1

TRAY EFFICIENCY Graphical determination of actual trays given EM: 1 ideal tray = triangle ‘acd’ on ideal equil. line 1 actual tray = triangle ‘abe’ on actual equil. line Eg. EM = 0.6 (60% efficiency) distance ‘ac’ = 10 cm 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 Reboiler = 1 stage (bet. ideal equil. & operating lines FKK

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 enthalpyconcentration diagram are given in Table 11.6-1(pg 733) and Table 11.6-2 (pg 734), respectively.

ENTHALPY-CONCENTRATION METHOD •  Ponchon-Savarit method •  takes into latent heats, heats of solution & sensible heats •  no assumption of molal overflow rates •  graphical procedure combining enthalpy & material balances •  provides information on condenser & reboiler duties

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Enthalpy-concentration for benzene-toluene •  data Table 11.6-2 pg. 734 (Geankoplis 4th Ed.) •  calculation shown in eg. 11.6-1

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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 FKKKSA

Example 11.6-2 Given : R = 1.5 Rm = 1.5(1.17) = 1.755

L

41.2 kmol/ h

F =100 kmol/ h

xD = 0.95

xF = 0.45

58.8 kmol/h

TF = 327.6K

xW = 0.1

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 = xFA(TF-T0) + (1-xF)B(TF-T0) where A = of liquid benzene = 138.2 kJ/kmol.K B = 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 FKKKSA

Example 11.6-2 1

0.8

0.6

0.4

0.2

140.0 0 0

0.2

0.4

0.2

0.4

0.6

0.8

0.6

0.8

1

120.0

100.0

80.0

60.0

40.0

W

20.0

0.0

0.0 - 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

F

D

1.0

Locating ΔR 2. Locate rectifying-section difference point,ΔR QC  hD +  − H1 D  ÄH R= = R 1 H1 − hD Hh     

1 D

ΔR

Sat. vapour

Sat. liquid

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hD+QC/D ÄR H 1

V1

H1 h DH1

hD

Example 11.6-2 1

QC  hD + − H1 D  Ä H R= = R 1 H1 − h D Hh     

0.8

0.6

1 D

QC  hD +  − H1 ÄR H 1 D  ÄR H1 R= = = 1.755 = H1 − hD Hh 1cm     

0.4

1 D

ÄR H1 = 1.7551cm  = 1.755cm

0.2

140.0 0 0

0.2

0.4

0.6

0.8

1

120.0

100.0

ΔR =85x

103kJ/kmol

∆R

80.0

60.0

H1=31x 103kJ/kmol

V1

40.0

W

20.0

hW=5x 103kJ/kmol

0.0

- 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

0.0

0.2

0.4

F

0.6

0.8

D

1.0

Example 11.6-2 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 6. Theoretical stages = numbers of tie lines 7. Theoretical trays = theoretical stages - reboiler Theoretical trays = 11.9 – 1 = 10.9 trays 8. Feed tray = tie line that crosses the line ÄR FÄS Feed trays = tray no. 7 from the top 9. Condenser duty, QC = (ΔR –hD)D or  QC   hD +  − H1 3-0)41.2 = 3 460 800 kJ/h Q = (85x10 D   C  R= H1 − h D 10. Reboiler duty, QR = (hW-ΔS)W QR = (hW-ΔS)W = (5 x 103 –[-64x103])58.8 = 4 057 200 kJ.h FKKKSA

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 1

0.8

0.6

0.4

0.2

140.0 0 0

0.2

0.4

0.6

0.8

1

120.0

100.0

∆R

80.0

60.0

40.0

W

20.0

0.0

0.0

0.2

- 20.0

- 40.0

-64 x103kJ/kmol

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

∆S

0.4

F

0.6

0.8

D

1.0

Stepping off trays for the rectifying section 1

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.

0.8

0.6

0.4

0.2

140.0 0 0

0.2

0.6

0.8

1

100.0

∆R

80.0

60.0

V7 V6 V5 V V V V 1 4 3 2

40.0

W

20.0

0.0

0.0

0.2

- 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

0.4

120.0

∆S

0.4

L7 L6 L5 F

L4 0.6

L3

L2

0.8

L1 D

1.0

Stepping off trays for the stripping section 1

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.

0.8

0.6

0.4

0.2

140.0 0 0

0.2

0.4

0.8

1

100.0

∆R

80.0

60.0

V12

40.0

WL L 12 11 L10

20.0

0.0

0.0

0.2

- 20.0

V11 L9

V10

- 40.0

- 60.0

- 80.0

∆S

- 100.0

- 120.0

- 140.0

V9

L8 L7 L6 L5 F

0.4

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

0.6

120.0

V8 V7 V6 V5 V V V V 1 4 3 2 L4 0.6

L3

L2

0.8

L1 D

1.0

MINIMUM REFLUX RATIO, Rmin 1

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

0.8

0.6

0.4

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

0.2

140.0 0 0

0.2

0.4

0.6

0.8

1

120.0

100.0

QC  hD +  − H1 D min Ä H Rmin = = Rmin 1 H1 − hD Hh     

1 D

60.0

40.0

W

20.0

0.0

0.0 - 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

∆Rmin

80.0

0.2

0.4

F

0.6

0.8

D

1.0

MINIMUM STAGES, Nmin 1

0.8

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

0.6

0.4

Nmin = 5.9 stages

0.2

140.0 0 0

0.2

0.4

0.2

0.4

0.6

0.8

0.6

0.8

1

120.0

100.0

80.0

60.0

40.0

W

20.0

0.0

0.0 - 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

F

D

1.0

PARTIAL CONDENSER yD

1

Distillate product, VD – vapour with the composition yD Condensed liquid/reflux, L0, have the composition xo which is in equilibrium with yD 0.8

0.6

Partial condenser = 1 stage

0.4

0.2

140.0 0 0

0.2

0.4

0.6

0.8

x0

1

120.0

100.0

∆R

80.0

60.0

V6 V5 V4 V V V V D 3 2 1

40.0

W

20.0

0.0

0.0

0.2

- 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

∆S

0.4

L6 L5 L4 F

L3 0.6

L2

L1

0.8

L0 1.0

PARTIALLY VAPOURISED FEED,zF 1

0.8

F with the composition zF lies on the tie line LF (composition = xF) and VF (composition = yF) 0.6

0.4

By trial-and-error, locate the point F so as to satisfy the inverse lever rule: 0.2

140.0 0

0

VF FL F = LF FVF

0.2

zF

0.4

120.0

0.6

1

0.8

1.0

100.0

80.0

60.0

40.0

W

20.0

0.0

0.0 - 20.0

- 40.0

- 60.0

- 80.0

- 100.0

- 120.0

- 140.0

FKKKSA

0.8

0.2

LF LF

0.4

F F

VF VF 0.6

MULTICOMPONENT DISTILLATION •  more than 2 components •  shortcut calculation methods – an approximation •  only allow separation between two components, heavy key and light key

Equilibrium data •  Raoult’s law – for ideal mixture p P p P y B = B = B xB y A = A = A xA P P P P

yC =

pC PC = x P P C

yD =

pD PD = x P P D

•  hydrocarbon system: yC = K CxC

y B = K B xB

y A = K AxA

y D = K Dx D

where KA = vapour-liquid equilibrium constant or distribution coefficient •  relative volatility, αi :

αA =

KA K ref.

αB =

KB K ref.

αC =

KC K ref.

αD =

FKKKSA €

€

€

€

KD K ref.

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

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

FKKKSA

xW

xWW

0 0 0.070 2.5 0.507 18.0 0.423 15.0 ∑xW=1.00 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

DEW POINT For a vapour mixture of A, B, C and D:  yi   yi   yi   xi = ∑  =  =1 ∑α  K K  i   ref.   i  Liquid composition which is in equilibrium with the vapour mixture: yi €

xi =

α

i   ∑ i  α  i

y

    

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 FKKKSA

Example 11.7-2 Components

xF

xFF

yD

yDD

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

xW

xWW

0 0 0.070 2.5 0.507 18.0 0.423 15.0 ∑xW=1.00 W=35.5

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

yiD

Ki

αi

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

FKKKSA

yi /αi

xi

0.0921 0.351 0.1396 0.531 0.0310 0.118 0 0 ∑yi /αi=0.2627 ∑xi=1.00

Example 11.7-2

1.75

0.65

0.26

0.10

FKKKSA

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: αx yi =  i i  ∑ α x   i 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

FKKKSA

Example 11.7-2 Components

xF

xFF

yD

yDD

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

xW

xWW

0 0 0.070 2.5 0.507 18.0 0.423 15.0 ∑xW=1.00 W=35.5

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

FKKKSA

Example 11.7-2

5.00

2.35 1.15 0.61

FKKKSA

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

where:

     



x LD D  x HW W     

x HD D x LW W

   

log  α L ,av 

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 αL,av =

αLDαLW

αLD = relative volatility of LK at dew point. €

αLW = relative volatility of LK at boiling point. xiDD x D Distribution of other components: = (αi,av )Nm HD xiWW xHWW FKKKSA €

Example 11.7-2 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

αLDαLW = (2.50)(2.043) = 2.258

αL,av =    LD   € HD 



x D  x HW W 

(0.349)  0.507  log   x D x LW W   0.031  0.070  Nmin = = = 5.404 theoretical stages     log  α L ,av  log  2.258 xiDD x D Distribution of other components: = (αi,av )Nm HD xiWW xHWW Distribution of component A & D: x ADD x D = (αA,av )Nm HD = (αA,av )5.404 (0.031)64.5 = (αA,av )5.404 0.1111 x AWW xHWW € (0.507)35.5 log

€

   

   

xDDD x D = (αD,av )Nm HD = (αD,av )5.404 (0.031)64.5 = (αD,av )5.404 0.1111 xDWW xHWW (0.507)35.5 FKKKSA

€

   

Example 11.7-2 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

αA,av =

αADαAW = (6.73)(4.348) = 5.409

αD,av =

αDDαDW = 0.385(0.530) = 0.452

x ADD € x D = (αA,av )Nm HD = (αA,av )5.404 (0.031)64.5 = (5.409)5.404 0.1111=1017 x AWW xHWW (0.507)35.5 €

€

€

xDDD x D = (αD,av )Nm HD = (αD,av )5.404 (0.031)64.5 = (0.452)5.404 0.1111= 0.001521 xDWW xHWW (0.507)35.5 Material balance of component A & D: x ADD + x AW W = 40 x DDD + x DW W = 15 x ADD = 39.961 x DDD = 0.023 Solving : x AW W = 0.039 x DW W = 14.977 FKKKSA

Example 11.7-2 Components

xF

xFF

yD

yDD

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

xW

xWW

0 0 0.070 2.5 0.507 18.0 0.423 15.0 ∑xW=1.00 W=35.5

Revised distillate and bottoms compositions: Components

xF

xFF

yD

yDD

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

FKKKSA

xW

xWW

0.0011 0.039 0.0704 2.500 0.5068 18.00 0.4217 14.977 ∑xW=1.00 W=35.516

MINIMUM REFLUX RATIO,RMIN Underwood’s shortcut method for calculating Rmin: Assumes constant flows in both sections of tower Uses constant average α (at Tave. = [Ttop + Tbottom ]/2) αx 1−q= ∑ i iF α −θ i where:

x € α −θ i xiD = mole fraction of component i in distillate taken at R = ∞ R m + 1= ∑

α

i iD

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 ) 2. Calculate 1-q for various θ 3. Use θ obtained to calculate Rmin FKKKSA

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

Example 11.7-3 Components

xF

xFF

xiD

xiDD

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

Tb.p. = 132oC & Tdp = 67oC

xW

xWW

0.0011 0.039 0.0704 2.500 0.5068 18.00 0.4217 14.977 ∑xW=1.00 W=35.516

Tave. = (132 + 67)oC/2 = 99.5oC

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

FKKKSA

Example 11.7-3

3.12

1.38

o.60

0.28

FKKKSA

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

 x 5.2(0.4)   2.3(0.25)   1.0(0.2)   0.467(0.15)   1−q= ∑ = 1− 0=1=  + + +        α −θ  5.20− θ   2.3− θ   1.0− θ   0.467− θ  i 1. By trial-and-error, assume θ ( αLK 〈 θ 〉 αHK ) 2. Calculate 1-q for various θ         θ ∑(sum)  5.2(0.4)   2.3(0.25)   1.0(0.2)   0.467(0.15)        (assumed)  5.20−θ   2.3−θ   1.0− θ   0.467− θ 

α

€

1.210 1.200€ 1.2096 FKKKSA

i iF

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

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 3. Use θ obtained to calculate Rmin

R m + 1= ∑ €

x = 5.20(0.6197) + 2.30(0.3489) + 1.00(0.0310) + 0.467(0.15) 5.20−1.2096 2.30−1.2096 1.00−1.2096 0.467−1.2096 α −θ i αx R m + 1= ∑ i iD =1.395 α −θ i α

i iD

Rmin = 0.395 €

FKKKSA

SHORTCUT METHOD NUMBER OF STAGES Correlation of Erbar & Maddox:

Feed plate location using Kirkbride method: log

Ne Ns

= 0.206 log

     

x x

  HF    LF 

W x D x      

  LW    HD 

2     

where: Ne = number of theoretical stages above the feed plate NS = number of theoretical stages below the feed plate FKKKSA

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) Rmin/(Rmin+1) = 0.2832 Nmin/N = 0.49 = 5.404/N N = 11.0 theoretical stages

0.37

or 10 theoretical trays plus a reboiler

0.49 FKKKSA

Example 11.7-3 Components

xiF (F=100 mol/h)

A B(L) C(H) D

0.40 0.25 0.20 0.15 ∑xiF=1.00

xiD (D=64.484 mol/h) xiW (W=35.516 mol/h) 0.6197 0.3489 0.0310 0.0004 ∑yD=1.00

0.0011 0.0704 0.5068 0.4217 ∑xiW=1.00

N = 11.0 theoretical stages Ne + NS = 11.0 theoretical stages 2       x Wx  Ne LW  HF      log = 0.206 log       x Ns D x      HD    LF  2  Ne  0.2  35.516  0.0704   log = 0.206 log      = 0.07344     0.25  64.484  0.0310   Ns   Ne = 1.184 Ne = 1.184Ns 1.184NS + NS = 11.0 Ns NS = 5.0 Ne = 6.0 Feed tray = 6.0 trays from the top FKKKSA