DISTILLATION components in a liquid solution • separation of different components
• 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/d rectification/fractional/distillation istillation with reflux – part of the vapour vapour is condensed condensed & returned returned as liquid back to the vessel !
FKK
all of the vapour is removed or is condensed as product
DISTILLATION
FKK
DISTILLATION Types of distillation:
• simple • fractional • steam "
immiscible solvent
"
azeotropic
• extractive
Distillation tower at an oil refinery.
FKK
"
vacuum
"
molecular
"
entrainer sublimation
VAPOUR-LIQUID EQUILIBRIUM RELATIONS • vapour & liquid in intimate contact 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 98 oC), first vapour in equilibrium is yA1 = 0.532
"
distance bet. equilibrium & 45o line = diff. bet. x A & 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, AB
- numerical measure of ease of separation
- relative volatility of component A with respect to component B yA y A /x A xA " AB yB $ ' $ ' &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): " x yA = $ A ' 1+ &%" # 1)(x A where
"
AB
when FKK
=
PA PB
=
vapour pressure of pure A vapour pressure of pure B
1 , separation is possible
SINGLE-STAGE EQUILIBRIUM CONTACTOR 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 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
=100 kmol yA1
yA2 =0.4
=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) = 110x A1 + 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
FKK
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
1
2
area under the curve 1/(y-x) vs x plot =
Average composition of total material distilled, y av: L1x1 = L2x2 + (L1-L2)yav FKK
=
ln
L L
1 2
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 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
y
Example 11.3-2 Equilibrium data:
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
Total balance :
L1 = V + L2 100 = 40 + L2
L2 = 60 mol L ln 1 ! xx dx y"x L2 ln 100 ! x0.5 dx 0.51 y "x 60 =
1
2
=
=
2
L1x1 = L2x2 + (L1-L2)yav 100(0.5) = 60(0.277)+ (100-60)y av 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
FKK
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
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(130 oF). 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: D kmol/h xD = 0.95
D =100 - W 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
F =100 =D + 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
FKK
Example 11.4-1 3. Calculate q (fraction of feed that that is liquid) H ! HF q= V H V ! HL
'% H & V =
! H L $"# + cP (TB ! TF ) HV ! HL
Average heat capacity of feed =159 kJ/kmol.K Average latent heat 32099 kJ/kmol
! 327.6)
'% 32099$" + (159)(T B # q=&
32099
From Fig. 11.1-1, at xF = 0.45, TB = 93.5oC (366.7K) '% 32099$" + (159)(366.7 # q=&
32099
FKK
! 327.6)
=
1.195
Example 11.4-1 q = 1 (liquid (liquid at its boiling point) , q = 0 (saturated vapour) , q 1 (cold liquid liquid feed) q 0 (superheated vapour) , 0 q 1 (mixture of liquid & vapour)
FKK
Example 11.4-1 4. Draw q-line
q-line
q = 1.195 y y
=
=
# & % q ( % (x % q 1( $ '
"
"
xF q "1
1.195 x " 0.45 6.12x " 2.31 1.195 " 1 1.195 " 1 =
Slope = 6.12 = y/ x = (0.45 – y)/(0.45-x) Lets x = 0.40 6.12
=
0.45 - y 0.45 ! x
y = 0.144
=
0.45 - y 0.45 ! 0.4
=
0.45 ! y 0.05
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 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 stripping line after after passing qline
6 7
8. Feed location 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)
"
Nmin
# x D 1 x W &( % log% x W (' $1 x D
"
=
log )
av
av =
=( 1=
relative volatility of the overhead vapour
W=
FKK
1
average value of relative volatility " W)
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.
Example 11.4-2 At R = !, Nmin = ? 41.2 kmol/h
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, R min • infinite number of stages/trays • minimum vapour flow • minimum condenser & reboiler • R min 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 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).
Example 11.4-2 Given R = 4, R min = ?
41.2 kmol/h xD = 0.95 F =100 kmol/ h
The enriching op. line from x D 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: x R
0.95
D
min
+1
FKK
=
R min
+1
= 0.43
R min = 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 Lm W y x ! = • operating line: m Vm 1 V +
m +1
• Ntheo. stages - starting from xW(on 45o line) draw a straight line to the intersection of yD with the qline
FKK
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 R x + x D y = • 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 Wx! W x • stripping operating line: y W S S =
• draw a straight line from (x W,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 + Dx LS O D y x = + intermediate operating line: VS 1 V +
liquid side stream:
S +1
Ln = LS+O VS+1 = Vn+1=V1=Ln+D
• from the intersection of enriching op. line & x O, draw a straight line to y-intercept of intermediate op. line
FKK
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)
FKK
TRAY EFFICIENCY 3 types of tray efficiency:
• Overall tray efficiency, E o • Murphree tray efficiency, E M • Point/local tray efficiency, EMP
EO
=
no. of ideal trays no. of actual trays
EM =
yn - yn 1 y *n -y n 1 +
+
EMP =
y'n -y'n 1 y *n -y'n 1 +
+
FKK
TRAY EFFICIENCY Graphical determination of actual trays given E M: 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) = 1.5R m. Given that R m=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 account 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
FKK
Enthalpy-concentration for benzene-toluene • data Table 11.6-2 pg. 734 (Geankoplis 4 th Ed.) • calculation shown in eg. 11.6-1
FKKKSA
ENTHALPY-CONCENTRATION METHOD Drawing isotherms (tie lines) on the enthalpy-concentration diagram from
(a) temperature-concentration diagram FKKKSA
(b) x-y diagram
Example 11.6-2 Given : R = 1.5 R m = 1.5(1.17) = 1.755 F =100 kmol/ h xF = 0.45 TF = 327.6K
L
41.2 kmol/ h xD = 0.95 58.8 kmol/h 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 x D, 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 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
1.0
D
Locating
R
2. Locate rectifying-section difference point, ' % % % & =
R
R
QC $" !H D ""# 1 ÄR H1 = H1 ! hD Hh
hD +
1 D
R
Sat. vapour
hD+QC/D ÄR H 1
V1
H1 h D H1
Sat. liquid
FKKKSA
hD
Example 11.6-2 1
QC $" h D + " ! H1 D "# ÄR H1 R = H1 ! h D Hh ' % % % & =
0.8
0.6
1 D
QC $" h !H ÄR H1 D ""# 1 ÄR H1 R = = 1.755 = H1 ! hD Hh 1cm ' % % D+ % & =
0.4
1 D
0.2
ÄR H1 1.755&$%1cm !#" 1.755cm =
=
140.0 0 0
0.2
0.4
0.6
0.8
1
120.0
100.0
R =85x
3
10 kJ/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
1.0
D
Example 11.6-2 3. Locate stripping-section difference point,
S
4. Step off trays for rectifying section using
R .
5. Step off trays for stripping section using
S
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 Ä FÄ R S Feed trays = tray no. 7 from the top 9. Condenser duty, QC = ( R –hD)D or ' QC $" % %hD + " ! H1 QC = (85x103-0)41.2 = 3 460 800 kJ/h D % " & # R = H1 ! h D 10. Reboiler duty, QR = (hWQR = (hWFKKKSA
S)W
S)W
= (5 x 103 –[-64x103])58.8 = 4 057 200 kJ.h
Locating
S
Draw a straight line from R through F to intersect the vertical line at xW
FKKKSA
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
1.0
D
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 the R intersecting saturated vapour line to give V2.Repeat until a tie line crosses over the line #R F#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 V V V V V V 6 5 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 L L5 6 F
L4 0.6
L3
L2
L1
0.8
1.0
D
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
V11
V10
L9
-20.0
-40.0
-60.0
-80.0
#S
-100.0
-120.0
- 140.0
V9
L8 L7 L L5 6 F
0.4
No. of theoretical stages = no. of tie lines = 11.9 stages FKKKSA
0.6
120.0
V8 V7 V V V V V V 6 5 1 4 3 2 L4 0.6
L3
L2
L1
0.8
1.0
D
MINIMUM REFLUX RATIO, R min 1
0.8
R min occur when the line #R F#S coincides with the tie line that passes through F
0.6
0.4
Get a tie line that passes through F and extend that line to intersect the vertical line x D to get #Rmin
0.2
140.0 0 0
0.2
0.4
0.6
0.8
1
120.0
100.0
R min =
QC $" !H h D ""#min 1 ÄRmin H1 = H1 ! hD H1hD
' % % D+ % &
#Rmin
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
0.2
0.4
F
0.6
0.8
1.0
D
MINIMUM STAGES, Nmin 1
0.8
Nmin is obtained when the operating lines are vertical since R and S 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
1.0
D
PARTIAL CONDENSER yD
1
0.8
Distillate product, VD – vapour with the composition yD Condensed liquid/reflux, L0, have the composition x o which is in equilibrium with yD 0.6
0.4
Partial condenser = 1 stage 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 V V V V V V 5 4 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 L L4 5 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
0.2
zF
0.4
120.0
V
F
LF
FL =
0.6
0.8
1
100.0
F
FVF
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
0.2
LF LF
0.4
VF VF
F F 0.6
0.8
1.0
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 yA
pA P
=
=
PA x P A
yB
=
pB P
=
PB x P B
yC
=
pC P
=
PC x P C
yD
=
pD P
=
PD x P D
• hydrocarbon system: y A K AxA =
y B K BxB =
yC K CxC =
y D K DxD =
where K A = vapour-liquid equilibrium constant or distribution coefficient
• relative volatility, "
A
FKKKSA
=
K A K ref.
"
B
i : =
K B K ref.
"
C
=
K C K
ref.
"
D
=
K D K ref.
Example 11.7-2 F = 100 mol/h at boiling point at 405.3 kPa. x FA = 0.4, x FB = 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
x F
x F F
y D
y D D
x W
x W W
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 $xW=1.00
0 2.5 18.0 15.0 W=35.5
FKKKSA
Assume all A go to distillate and all D go to the bottoms. y AD(D)=xAF(F)=0.4(100)=40 mol/h xDW(W)=xDF(F)=0.15(100)=15 mol/h N-pentane(B): Light Key Overall Balance: F =D+W Comp Balance: xBF(F)=0.25(100)=25 mol/h=yBD(D) + xBW(W) Since 90% of B is in distillate: yBD(D) = (0.90)(25) = 22.5, xBW(W)=2.5 N-hexane (C): Heavy Key xCF(F)=0.20(100) =20 mol/h Since 90% of C is in the bottoms: xCW(W)=0.90(20)=18 mol/h, yCDD=2.0 mol/h
DEW POINT For a vapour mixture of A, B, C and D: " yi %' "$ yi %' "$ yi %' $ xi ($ ' $ 1 '($) ' K K # i & # ref. & # i & =
=
=
Liquid composition which is in equilibrium with the vapour mixture: yi xi
" =
i
# %y )% i %% " $ i
& ( ( (( '
1. By trial-&-error, assume Td.p. 2. Get corresponding values of K i and 3. Calculate yi/
i
i
4. From K ref. = $(yi/ i) , 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
x F
x F F
y D
y D D
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
x W
x W W
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 (K ref. = K C = $(yi/ i) = 0.26) Components
yi D
K i
A B(L) C(H) D
0.62 0.349 0.031 0 $yi D=1.00
1.75 0.65 0.26 0.10
FKKKSA
i
6.73 2.50 1.00 0.385
yi /
i
0.0921 0.1396 0.0310 0
x i
0.351 0.531 0.118 0 $ yi / i =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 = $K ixi = K ref. $ ixi = 1 Vapour composition which is in equilibrium with the liquid mixture: " x i i yi # & ) %" x ( $ i i' =
1. By trial-&-error, assume Tb.p. 2. Get corresponding values of K i and 3. Calculate
i
ixi
4. From K ref. = 1/( ixi) , 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
x F
x F F
y D
y D D
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
x W
x W W
0 0 0.070 2.5 0.507 18.0 0.423 15.0 $xW=1.00 W=35.5
Boiling point: Assume Tb.p. = 132oC (K ref. = K C = 1/$ ixi = 1.144) Components
x i w
K i
A B(L) C(H) D
0 0.070 0.507 0.423 $xW=1.00
4.95 2.34 1.10 0.61
FKKKSA
i
5.00 2.043 1.000 0.530
x i
i
0 0.1430 0.5070 0.2242 $ x i i =0.8742 1/$ ixi = 1.144
yi 0 0.164 0.580 0.256 $xi=1.00
Example 11.7-2
5.00
2.35
1.15 0.61
FKKKSA
MINIMUM STAGES,NMIN AT R = ! Minimum stages, Nmin, using Fenske equation:
Nmin where:
=
,& #) #& !' !$ x W D *$ x HW LD $ !' $ ! log*$ !' !$ * *$% x HD D !"$% x LW W !"' + ( log &$ 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 L,av =
"
"
LD
LW
LD =
relative volatility of LK at dew point.
LW =
relative volatility of LK at boiling point.
Distribution of other components: FKKKSA
x iDD x D (" i,av )Nm HD xiWW x HWW =
Example 11.7-2 Components
yi D=x iD
y D D
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
L,av =
Nmin
=
"
"
LW -' $* $' "( "% x W D +% x HW LD % "( % " + log % "( "% + +%& x HD D "#%& x LW W "#( , )
log ! ' % &
LD
=
,av $" L #
=
x i w
x W W
0 0.070 0.507 0.423 $xW=1.00
0 2.5 18.0 15.0 W=35.5
i
6.73 2.50 1.00 0.385
(2.50)(2.043) 2.258 =
-' (0.349) $' 0.507 $* "% "( log ++%% "% "( % "% " +,& 0.031 #& 0.070 #()
log
Distribution of other components:
=
'% 2.258$" & #
5.404 theoretical stages
xiDD x D (" i,av )Nm HD xiWW xHWW =
Distribution of component A & D: x ADD x D (" A,av )Nm HD (" A,av )5.404 (0.031)64.5 (" A,av )5.404 0.1111 x AWW xHWW (0.507)35.5 =
=
=
xDDD x D (" D,av )Nm HD (" D,av )5.404 (0.031)64.5 (" D,av )5.404 0.1111 xDWW x HWW (0.507)35.5 =
FKKKSA
=
=
i
4.348 2.043 1.000 0.530
Example 11.7-2 Components
yi D=x iD
y D D
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
A,av =
"
"
D,av =
"
"
AD
AW
DD
DW
=
=
x i w
x W W
0 0.070 0.507 0.423 $xW=1.00
0 2.5 18.0 15.0 W=35.5
i
6.73 2.50 1.00 0.385
i
4.348 2.043 1.000 0.530
(6.73)(4.348) 5.409 =
0.385(0.530) 0.452 =
x ADD x D (" A,av )Nm HD (" A,av )5.404 (0.031)64.5 (5.409)5.404 0.1111 1017 x AWW xHWW (0.507)35.5 =
=
=
=
xDDD x D (" D,av )Nm HD (" D,av )5.404 (0.031)64.5 (0.452)5.404 0.1111 0.001521 xDWW x HWW (0.507)35.5 =
=
=
=
Material balance of component A & D: x AD D + x AW W = 40
Solving :
x AD D
=
x AW W FKKKSA
x DD D + x DW W
39.961
x DD D
0.039
x DW W
=
=
0.023
=
14.977
= 15
Example 11.7-2 Components
x F
x F F
y D
y D D
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
x W
x W W
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
x F
x F F
y D
y D D
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
x W
x W W
0.0011 0.039 0.0704 2.500 0.5068 18.00 0.4217 14.977 $xW=1.00 W=35.516
MINIMUM REFLUX RATIO,R MIN Underwood’s shortcut method for calculating R min: Assumes constant flows in both sections of tower Uses constant average
(at Tave. = [Ttop + Tbottom ]/2) 1" q %
#
x
#
"$
i iF
=
i
R m + 1 = %
where:
" i " i
x
iD
#$
xiD = mole fraction of component i in distillate taken at R = ! xiF = mole fraction of component i in the feed i =
relative volatility of the top and the bottom of the tower
To determine R min: 1. By trial-and-error, assume (
LK
2. Calculate 1-q for various 3. Use FKKKSA
obtained to calculate R min
HK )
Example 11.7-3
Using the conditions and results given Example 11.7-2, calculate the following:
in
(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
x F
x F F
x i D
x i D D
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
x W W
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
x i F
x i D
K i
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
FKKKSA
x W
i
5.20 2.30 1.00 0.467
x i W 0.0011 0.0704 0.5068 0.4217 $xiW=1.00
Example 11.7-3
3.12
1.38
o.60
0.28
FKKKSA
Example 11.7-3 Components
x i F
x i D
K i (99.5oC)
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 $yD=1.00
3.12 1.38 0.60 0.28
1" q = %
#
x
i iF =
"$ i
#
1"
i
(99.5oC) 5.20 2.30 1.00 0.467
x i W 0.0011 0.0704 0.5068 0.4217 $xiW=1.00
& 5.2(0.4) )+ &( 2.3(0.25) )+ &( 1.0(0.2) )+ &( 0.467(0.15) )+ ( + + + 0 = 1= ( + ( + ( + ( + ' 5.20 $ * ' 2.3 $ * ' 1.0 $ * ' 0 467 $ *
"
"
1. By trial-and-error, assume (
LK
"
.
"
HK )
2. Calculate 1-q for various $ ' & 2.3(0.25) ) & ) % 2.3 # (
$ ' & 1.0(0.2) ) & ) % 1.0 # (
$ ' & 0.467(0.15) ) & ) % 0 467 # (
$(sum)
(assumed)
$ ' & 5.2(0.4) ) & ) % 5.20 # (
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
FKKKSA
"
"
"
.
"
Example 11.7-3 Components
x i F
x i D
K i (99.5oC)
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
i
(99.5oC) 5.20 2.30 1.00 0.467
x i W 0.0011 0.0704 0.5068 0.4217 $xiW=1.00
= 1.2096 3. Use
obtained to calculate R min
R 1 = % m
+
"
i
"
i
xiD
#$
=
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
R m + 1 = %
" i " i
x
iD =
#$
R min = 0.395 FKKKSA
1.395
SHORTCUT METHOD NUMBER OF STAGES Correlation of Erbar & Maddox:
Feed plate location using Kirkbride method: log
N N
e s
=
0.206 log
&, $* x $* HF $* $* x $%+ LF
) , 'W*x ' * LW ' * ' D *x ( + 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 = 1.5R min = 1.5(0.395) = 0.593 R/(R+1) = 0.593/(0.593+1) = 0.3723 R min/(R min+1) = 0.395/(0.395+1) R min/(R min+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
