Mass transfer and separation technology - Åbo …users.abo.fi/rzevenho/MOFST15-OH12.pdfRoNz 424302 Massöverföring och Separationsteknik Mass transfer and separation technology Massöverföring

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ik Mass transfer and separation technologyMassöverföring och separationsteknik (”MÖF-ST”) 404302, 7 sp

12. Continuous distillation

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering Laboratorytel. 3223 ; [email protected]

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12.1 General considerations

Åbo Akademi - kemiteknik Värme- och strömningsteknik - Biskopsgatan 8, 20500 Åbo

2

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mars 2015 Åbo Akademi - kemiteknik Värme- och strömningsteknik - Biskopsgatan 8, 20500 Åbo

Continuous distillation in a tray column

Pictures: T68

3

RoNzmars 2015 Åbo Akademi - kemiteknik Värme- och strömningsteknik - Biskopsgatan 8, 20500 Åbo

Crude oil distillation

Simple process ↑

Modern process →

Source: Moulijn et al., 2001

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Source: http://www.nesteoil.com (2007)

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Mass balance and equilibrium: absorption

yi = Kxi

Lxin

Lxout

Vyout

Vyinxoutxin

yout

yin

i

yi+1

yi

xi

xi-1

LV

mass balance: V·yi+L·xi= V∙yi+1 + L∙xi-1

→ working line: yi+1 – yi = (L/V)·(xi-1 – xi)

slope L/V

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Mass balance and equilibrium: desorption / stripping

yi = Kxi

Lxin

Lxout

Vyout

Vyinxinxout

yin

yout

i

yi+1

yi

xi

xi-1

LV

mass balance: V·yi+L·xi= V∙yi+1 + L∙xi-1

→ working line: yi+1 – yi = (L/V)·(xi-1 – xi)

slope L/V

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Continuous distillation (binary) /1

Separating a mixture of 2 components A and B, with A more volatile

Liquid for absorption section produced by condensing some top product

Gas for stripping section produced by boiling some bottom product

Top section:rectifying sectionabsorbing the less volatile component

Bottom section:stripping sectiondesorbing the more volatile component

Bottom product:less volatile component

Top product: more volatile component

Picture: WK92

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Continuous distillation /2

Roughly equimolarexchange: 1 mol A liquid → gas «-»1mol B gas → liquid

As a result: Gas and liquidstreams ~ constantin each section: L and V in top section, L´ and V´ in bottom section

Feed enters therewhere it is similar to the mixture inside

Top section:rectifying sectionabsorbing the less volatile component

Bottom section:stripping sectiondesorbing the more volatile component

Bottom product:less volatile component

Top product: more volatile component

Picture: WK92

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Continuous distillation /3

Some general considerations:– The equimolar exchange A ↔ B requires balanced energy

exchange as well: vaporisation heats of A and B should be roughly the same

– Pressure and the temperature ranges must be chosen:• Below the critical pressures (and temperatures) of the components• The components must be thermally stable (for hydrocarbon

fractions of crude oil: below 300 ~ 350°C)• The temperature at the top of the column should preferably be

> 40°C, allowing for heat transfer with surrounding air or water

– Equilibrium data is needed, and data on the feed (see later!) and the required products specifications

→ The necessary gas and liquid streams, the numberof stages and heat consumption can calculated.

!

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11/80

Separation process and entropy /1 Separation processes involve de-mixing of species or phases,

in thermodynamic sense resulting in a decrease of the total entropy of these

∑ Entropy of components or phases < Entropy of mix One way of acccomplishing a separation is to create more

phases; Gibbs’ phase rule gives for the degrees of freedom, f, for a system of p phases and n components: f = n + 2 – p. f↑ gives ΔS↑; f↓ gives ΔS↓.

A new phase can be created by 1) adding a new component(absorption/desorption, extraction), or 2) by adding or removingenergy (e.g. heat) (distillation, crystallisation)

Creating an ”un-equilibrium” gives rise to a separation (or a chemical reaction) if ΔG = ΔH-T·ΔS < 0, or ΔS >ΔH/T


12/80

Separation process and entropy /2

mmix

smix

Qin

Tin

Qout

Tout

mA

sA

mB

sB

BBAAmixmix

in

in

out

out

mixmix

in

in

BBAA

out

out

outBBAAinmixmix

BAmix

smsmsmTQ

TQ

smTQ

smsmTQ

QhmhmQhm

mmm

:Law 2nd Law, 1st balance,mass state-Steady

Qout

Tout

mmix

smix

mliquid

sliquid

mcrystals

scrystals

Distillation Crystallisation

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12.2 Selecting temperature & pressure; the x,y diagram

13Åbo Akademi - kemiteknik Värme- och strömningsteknik - Biskopsgatan 8, 20500 Åbo

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mars 2015 14Åbo Akademi - kemiteknik Värme- och strömningsteknik - Biskopsgatan 8, 20500 Åbo

Equilibriumdata for hydrocarbons

vapour pressures C1 ... C10

For two components”1” and ”2”

K = y/x = p°(T) / ptotal

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Example:C3 – iC4distillation /1

Give the ranges for temperature and pressure for a propane -isobutane distillation. Data:1) Equilibrium data →2) Components stable up

to 300°C3) Critical pressure 42 bar

for C3, 37 bar for iC4.

vapour pressures C1 ... C10

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Example:C3 – iC4distillation /2a

Answer:

pressures < critical pressures allowable for C3 – iC4 : p < 37 bar

37 bar boiling points 85 - 140°C: stability OK

Tboil iC4

at 37 bar

Tboil C3

at 37 bar

Raoults law: K = y/x = p°/ptotal

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Example:C3 – iC4distillation /2b

Answer:

Chose 40°C as the temperatur at the condensor.

At Tmin = 40°C, the vapour pressure of the more volatile component which is C3

= 14 bar → p ≥ 14 bar

Then Tboil ~ 81°C for iC4

Tboil C3

at 14 bar

p° C3

at 40°C

Tboil iC4

at 14 bar

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12.3 McCabe – Thiele procedureI: theoretical stages



Distillation: the x,y diagram /1

Mass balance top section for tray i: (x, y for volatile component)

with distillate product D (kmol/s)With Li+1 ~ Li = L, Vi+1 ~ Vi = V :

Li = RD· D and Vi = D + Li, gives

Dii xV

Dx

V

Ly

i+1i

V(kmol/s)

L(kmol/s)

RefluxRD·D

(kmol/s)

DD

iD

Di x

Rx

R

Ry

Dii xV

Dx

V

Ly

Diiii DxxLVy

Picture: after WK92

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Distillation: the x,y diagram /2

Mass balance bottom section for tray j: (x, y for volatile component)

with distillate product D (kmol/s)With L´j+1 ~ L´j = L´, V´j+1 ~ V´j = V´:

L´j = V´j + B and V´j = RB· B, gives

Dii xV

Dx

V

Ly

j+1j

V’(kmol/s)

L’(kmol/s)

RefluxRB·B

(kmol/s)

BB

jB

Bj x

Rx

R

Ry

Bjj x'V

Bx

'V

'Ly

jjBjj xLBxVy

Picture: after WK92

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Operation line top section

Source: T68, see also SH06

condenser

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R = RD

RB usuallynot used...


Operation line bottom section

Source: T68, see also SH06

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reboiler


McCabe -Thiele diagram /1

Working line topsection with slopeL/V = RD/(RD+1) < 1

Working linebottom section with slope L´/V´ = (RB+1)/RB >1

Products xB= yBand xD= yD on x=ydiagonal line

Next: the feed (xF,yF)Picture: WK92

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A third line in the x,ydiagram is the so-calledq-line through feedpoint (xF,yF) and crossing point (x*,y*)of operating lines for topand bottom section

For point x*, y* :

x* xF

y*

yF

slope- q / (1- q)

where ΔV = V - V’ is vapor entering with the feed, and ΔL = L’ - L is liquid entering with the feed

Fxq

xq

qy

1

1

1

**

BD

B

D

xBxDLxVy

xBxLVy

xDxLVy

**

*'' *

* *

Picture: after WK92

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McCabe -Thiele diagram: q-line /3

q = fraction of feed that gives liquid on the feeding tray: L´= L+q· F,

F· q = ΔL, F· (1-q) = ΔV

q is related to the energy neededto convert the feed completelyinto vapour. With enthalpies H for saturated liquid and saturatedgas it is found that

q >1q = 1

0 < q < 1q = 0q < 0

sat,Lsat,G

Fsat,G

HH

HHq

Picture: after T68

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Distillation: feed temperature

q = fraction of feed that gives liquid on the feeding trayq = 0 ~ saturated gas; q = 1 ~ saturated liquid

Picture: WK92

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Having determined the equilibrium curve; the operation lines for topand bottom section; the q-line, the number of equilibriumstages can be counted.

Note: the q-line is fixed by (xF,yF) and q. In practice the operaton line for the bottom section is usually determinedfrom the operation line for the topsection and the q-line !

McCabe -Thiele diagram: stages /4

Example: 4 stages bottom section +5.4 stages top section, total 10 stages

Picture: WK92

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Location of the feed tray

(a) feed on 7th tray from top; (b) feed on 4th trayfrom top; (c) feed on 5th tray (optimum)

above feed tray:count with operationline for top section;

below feed tray:continue with operation line for bottom section

Picture: after T68

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12.4 McCabe – Thiele procedureII. Minimum reflux, minimum number of stages




The crossing point of working lines and q-line is on the equilibrium line, if L/V = (L/V)min , or L/V = Rmin/(Rmin+1)

reflux ratio too small (not enough liquid for absorption)

This results in a problematic”pinch” point (x,y) = (x*,y*) where N = ∞ equilibriumstages are needed:

Rmin/(Rmin+1) = (xD-y*)/(xD-x*),

or: Rmin = (xD-y*)/(y*-x*)

The optimal reflux ratio R depends on costs, typicallyR = (1.1 ~ 1.3) · Rmin

(x*,y*)

Picture: after WK92

30


Minimum reflux Underwood

For feed at boiling point (q = 1), x* = xF and y* = α·xF /(1 + (α-1)·xF)

Using this in º) gives

Similarly for q = 0: y* = yF = xF, and

For feed at q = 0 or q =1, Underwood derived relations for Rmin, using Rmin = (xD-y*)/(y*-x*) º)

(x*,y*)

q=1

q=0

1y1

x1

y

x

1

1R

F

D

F

Dmin

F

D

F

Dmin x1

x1

x

x

1

1R

Picture: after WK92

31

If q ≠ 0 and q ≠ 1:see exam question

360 (May 2009)



The other extreme is foundfor ”total reflux” operation R = ∞, which physically means no feed or product streams

This requires a minimum number of equilibriumstages, N = Nmin.

Picture: WK92

32


Minimum number of stages Fenske

For constant α: y/(1-y) = α·x/(1-x)

Operation line (no products):

y1 = xD, y2 = x1, etc.

1st stage: [y1 /(1-y1)]·[(1-x1)/x1] = α

or [xD /(xD-1)]·[(1-x1) /x1] = α

2nd: [y2 / (1-y2)]·[(1-x2) /x2] = α, y2=x1

Combine: [xD / (xD-1)]·[(1-x2)/x2] = α2

Continue until bottom: [xD / (xD-1)]·[(1-xB)/xB] = αN+1

B

B

D

Dmin x

x1

x1

xlnln1N

x1x2

y2

y1

Picture: after WK92

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Example: C3 – iC4 distillation /4

For the C3 –iC4 distillation (see above), at p = 14 bar, with F = 500 mol/s, xF = 0.40, xD = 0.95, xB = 0.1 and q = 2/3:

– Produce an (x,y) equilibrium diagram using, for example, the nomogram given in §12.2.

– Determine the minimal (top section) reflux Rmin

– Construct the two operation lines in the (McCabe –Thiele) diagram for R = 1.2 × Rmin

– Determine the number of equilibrium stages in bottom (includes feed) and top section.

– Calculate the product streams (in mol/s or kg/s)

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Answer part 1: For the (x,y) equilibrium plot;

temperature varies from 40°C at the condensor to 81°C at the reboiler. p = 14 bar For temperatures in this range,

find KC3 and KiC4, and calculatex and y. For example: At 1.4 MPa for 60°C KC3 = 1.45 and KiC4 = 0.70. Thus 1) y/x = 1.45 and 2) (1-y)/(1-x) = 0.7. Fill in y = 1.45x in 2) gives(1-1.45x)=0.7·(1-x). This gives result x=0.40, y=0.58. et cetera


C3 –iC4 liquid – vaporequilibrium

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Example: C3 – iC4 distillation /5a

Method 1: for several T:x = (KiC4 – 1)/(KiC4 – K3) andy = KC3· x gives

(40°C,1,1), (45°C,0.8,0.9)

(50°C,0.67,0.82), (55°C,0.54,0.71),

(60°C,0.40,0.58), (65°C,0.31,0.48),(70°C,0.20,0.34), (75°C,0.11,0.20),(80°C,0.05,0.10) points

Method 2: averagetemperature ~ 60°C,α = Kc3 / KiC4 = 2.07plot y = α· x /(1 + (α-1)· x)

(at 40°C: α= 2.43, at 80°C: α=1.95)

curved line

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

y

relative volatility 1.9 ... 2.4

relative volatility 2.07

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Answer part 2: See McCabe-Thiele

diagram:Rmin = 2.45, R = 2.93

Top: 7.7 (8) stages, bottom 7.1 (8) stages

Mass balancesF = D + B, andF· xF = D· xD + B· xB

gives via 1-D/F = B/F and D/F = (xF-xB)/(xD-xB)

D = 0.353· F = 176 mol/s B = 0.647· F = 324 mol/s

C3 – iC4 separation

at 14 bar

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McCabe-Thiele diagram: /7 sidestreams

Between sidestream and feed:Vs = Ls + S´ + DVs· yn=Ls· xn+1 + S´· xs´ + D·xd

gives extra operating linewith slope Ls/Vs, throughpoint with x = y = (S´· xs’ + D· xD) / (S´ + D) Lines for top section and

sidestream intersect at x = xs

Stages counted as usualThe sidestream increases

the number of plates(a result of less liquid in topsection below sidestream)

Source: Coulson & Richardson vol. 2 (1983)

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Restrictions McCabe-Thiele method

Molar heats of vaporisation should not differ more than, say, ~10%

Heats of solution are negligible Relative volatilities should be 1.3 < α < 5 Reflux ratio’s should be R > 1.1· Rmin

Number of trays N < 25 preferably (but sometimes > 100)

Otherwise: operating lines are presumably not straight use a more exact method. for example, Ponchon - Savarit, based on the enthalpy-composition, or H,x chart- see §12.5 - see course 424304 Process Engineering Thermodynamics (Jan/Feb 2015, 2017, ....)


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12.5 Condensers, reboilers;Energy in-/output


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Task ö5.1

Methanol = 46 kg/kmol

For ethanol Δhvap,mol = Δhvap,m· M

= 841 kJ/kg . 46 kg/kmol = 38.67 MJ/kmol

For water Δhvap,mol = 2257 kJ/kg·18 kg/kmol = 40.67 MJ/kmol

Roughly the same, McCabe -Thieles method applicable.

At 90°C, ΔHvap,mol = 2283 kJ/kg ·18 kg/kmol = 41.15 MJ/kmol

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Distillation and energy /1Heat (enthalpy) balance top section, with enthalpy h and heat of vaporisation Δhv

V· hy = D· hD + L· hx + Qc and V = L + D

With hx = hº + cp· (T-Tº) and hy = hx + Δhv:

V· (cp·Ty + Δhv) = D· cp·TD + L· cp·Tx + Qc

since cp·T terms are relatively small:

V· Δhv ≈ Qc

Condensation of the less volatile component is compensated by vaporisation of equal amount of vaporisation of the more volatile component.

Similarly for bottom section:

V´· Δhv ≈ QB

Picture: WK9242/92


Distillation and energy /2

Example: see Figure

For cp = 150 J/(mol· K) for L and V; Δhv = 15 kJ/mol; and L = ½· V

→ Calculate ΔV/V.

Heat balance: hin = hout

(V+ΔV)· (cp·T0 + Δhv) + L· cp·T1 =

(L+ΔL)· cp· T0 + V· (cpT1 + Δhv)

L/V = ½ and ΔV = ΔL

ΔV· Δhv / V = ½· cp· (T1-T0) gives

ΔV / V = ½· cp· (T1-T0) / Δhv = 0.1

Picture: WK92

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Condensers

Selecting a condenser type depends very much on total pressure and costs:

– Total condensers (a) for < 15 bar (215 psi)

– Partial condenser (b) for 15 – 25 bar (215 – 365 psi)

(can give a vapour distillate product at lower pressures)– Mixed condensors (c) give vapour and liquid distillate products– At pressures > 25 bar (365 psi) a special condenser coolant is used

(i.e. a refrigerant) ! A partial condensor must be counted as an equilibrium stage !

Picture: SH06

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Reboilers

Reboilers provide ”boilup” vapour to the bottom section Liquid from the column is (partially) vaporised using an external heat

exchanger (with for example condensing steam), such as – (a) a kettle reboiler, giving partial vaporosation, or – (b),(c) a vertical syphon, giving total vaporisation– Note the difference in extracting bottom product from the unit!

Partial vaporisation gives a vapour that is somewhat richer in volatile component (making the bottom product a bit more heavy). ! A partial reboiler type (c) must be counted as an equilibrium

stage !

Pictures: SH06

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Determine the heat streams for the condensor QC, reboiler QB and feed pre-heat QF. Data: Δhv = 15 kJ/mol. Given: F = 500 mol/s, q = ⅔ Already calculated: R=2.93; D=176 mol/s, B=324 mol/s Gas to the condensor V = (R+1)· D = 694 mol/s

→ Heat streams:QC = - 694 mol/s·15 kJ/mol = -10.4 MWQF = (1-q)· F·Δhv = ⅓·500 mol/s·15 kJ/mol = 2.5 MWQB = 7.9 MW (from heat balance)

Q multiplied with heat quality factor (1 – Tsurroundings/T), Tsurroundings = 290 K gives the exergy (= useful part) of heat Ex(Q): Ex (Qc) = -10.4·(1- 290/313) = - 0.76 MW, Ex(QB) = 1.43 MW, Ex(QF assume 60°C) = 0.32 MW. Total Ex(Q) in = 1.77 MW, total Ex(Q) out = 0.76 MW. Lost 1.01/1.77 = 57 %

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Ponchon-Savarit method (intro)

For more detail: advanced courses like ÅA KT 424304, or literature.

h,x diagrams

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12.6 A few examples



F xF xD xB q D B V L L´ V´

Old 1 0.40 0.84 0.08 0.5 0.42 0.58 1.09 0.67 1.10 0.52

New 1 0.60 0.84 0.08 1

Using an existing distillation column for another feed.

7 equilibrium stages:- 3 above feed - 3 below feed- reboiler

Source: WK92

An existing distillation column /1

49


An existing distillation column /2

too low

too high

notpossible

These operational lines give a ”fit” but the feed is not optimal (not on crossing of q-line and equilibrium line). Stages 3 and 4 do not contribute much, another value for q is preferable.

50

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Task ö5.3 The question in Swedish


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Task ö5.3 Mass balance component A:

0.5· 5 mol/s = 0.9·1 mol/s + xA· 4 mol/s→ xA = 0.4 mol/s

yA for the vapor to condensor= 0.9 = xA for the reflux

2 points for working line:(0.40, 0.50) and (0.90, 0.90)

Equilibrium line: ö Fig. 3.2(assuming p = 1 atm.)

Count stages: ~ 2.

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Task ö5.4 /1

a. Total mass balance: ṅF= ṅW + ṅD(F = B + D)

For A: ṅF · xF = ṅW · xW+ ṅD· xD

2 mol/s = ṅW + ṅD

2 mol/s · 0.4 = ṅW · 0.1 + ṅD· 0.9gives ṅW = 1.25 mol/s, ṅD= 0.75 mol/s

b. ṅR = ṅD = ṅL = 0.75 mol/sTotal mass balance: ṅG= ṅL + ṅD= 1.5 mol/sFor A: 1.5 mol/s· yA = 0.75 mol/s· 0.9 + 0.75 mol/s· 0.4 gives yA = 0.65

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Task ö5.4 /2

c. Line through points (0.4, 0.65) and (0.9, 0.9)

d. For bottom section: (q=1)ṅL´ = ṅL + ṅF (L´ = L+ F)

ṅL´ = 0.75 mol/s + 2.0 mol/s

= 2.75 mol/sṅG´ = ṅG = 1.5 mol/s (V = V´)

Line through (0.1, 0.1) and (0.4, 0.65)

e. Count stages: ~ 4.5 → 54 stages + reboiler

q-line

Note: q = 1 is not explicitelymentioned in the question and no q-line used in ö - answer

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12.7 Stage efficiency(theoretical stages → real stages)



Murphree efficiency /1

In an ideal equilibrium stage, composition yn+1 > equilbrium (with liquid xn) composition yn. This causes masstransfer and y decreases to yn* which is at equilibrium with xn. Here, * means: if equilibrium is really reached!

Ifequilibriumis reached:

Δyn = Δyn* =

yn+1 – yn*

Picture: WK92

57/92



In real equipment a different Δys will be reached for a stage, typically Δys < Δy* (but also Δys > Δy* is possible!)

The ratio Δys / Δy* is referred to as the Murphree efficiency for the vapour phase, ηmv.

Similarly there is also

ηml = Δxs / Δx* for the

liquid phase but that is

scarcely used. Note that

factor) n(separatio

SL

KV

η

η

ml

mv

Picture: WK92

58/92



Using ηmv the number of practical(or ”real”) stages can be calculated from the number of ideal stages

real number ofstages (or trays)

apparent equilibrium

line

Pictures: WK92

59/92


Murphree efficiency /4 Tray columns

← Plate efficiencies ηmv

versus mass velocity for vapor for various plate spacings Zp and pressures ↓ Plate efficiencies ηmv

versus plate spacing Zpand versus pressure


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Overall efficiency /1

The total or overall efficiency Eo is defined as

= number of ideal stages / number of real stages

Eo and ηmv are related via separation factor S = KV/L:

practical

idealo N

N E

1S if ηE

1η if S ln

1)·η-(SE

S ln

1)·η-(S 1 ln E

mvo

mvmv

o

mvo

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Overall efficiency /2O’Connel’s (1946) relation for bubble-tray distillation overall efficiency

1 cP = 10-3 Pa.s

For bubble-tray distillation towers separating hydrocarbons

and similar mixtures (from Treybal, 1968)

Drickhamer & Bradford (’43) relation for efficiency of tray columns handling

hydrocarbons:

for feed fractions xF, i viscosity µL, i (centipoise = mPa.s) at average tower temperature

ii,Li,F μxlog..E

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12.8 Tray capacity, tray columndesign


Pictures: K80


Distillationcolumn trays x

Very popular: valve trays ↓; cheaper than bubble-caps ↑ and better than simple perforated plates

Bubble-cap or bubble tray

Valve tray

Perforated plate trayphot

os: C

ouls

on &

Ric

hard

son

vol.

2 (1

983)

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See

also

: http

://w

ww

.eur

oslo

tkds

s.co

m/m

tri/t

ower

-inte

rnal

s/di

still

atio

n-tr

ays.

aspx


Tray capacity /1 Minimum column diameter is

determined by gas flow that would otherwise entrain toomuch liquid upwards.

Free fall velocity of droplets with diameter d, density ρL, in gas with density ρG is approx.

vd ≈ (g·ρL·d /ρG )½

with d ~ 1 mm, ρL ~ 1000 kg/m3

and ρG ~ 1 kg/m3 this gives

vd ~ 3 m/s ~ vgas,max

Pictures: WK92 ↑, SH06 ↓

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Tray capacity /2

ExampleC3 – iC4 distillation,

top section: L = 150 mol/s, V = 300 mol/s, p = 10 bar, T = 30°C = 303 KFor C3: MG = ML

= 44 kg/kmol, ρL = 500 kg/m3

Find column diameter ?

ρG = M∙p/R∙T = 44×10-3·106 / (303· 8.314) = 17.5 kg/m3

For 1 mm droplets:vd ~ (9.8·500· 0.001/17.5)½

= 0.53 m/s = max. gas speedGas flow = V / ρG,mol

= R∙T∙V/p = 8.314·303·300/106= 0.75 m3/sGas flow = vmax· (π/4)·D2

min

Dmin = (4· 0.75/(π· 0.53))½ = = 1.33 m

66


Dimensioning tray columns /1 Important design features: column diameter D and

pressure drop per tray, Δps ( ~ 0.7 kPa / tray at 1 bar)

The trays must be capable of handling the (maximum) necessary flows of gas and liquid.

Perforated plate tray column general design features: 1) Downcomer, 2) Slit, 3) Weir, 4) Hole (typically 5 – 10 mm)

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Dimensioning tray columns /2 The load diagram is based on the requirement that the

pressure drops for liquid (over downcomer) and gas (over the tray) should be equal !

Load diagram for a perforated plate tray

vG, vL: velocitiesρG, ρL: densitiesQL, QG: volume flowsH: tray spacingg: gravityF: tray surface area

fraction of the holes

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½½

8707.0exp0421.0G

L

G

L

L

GG Q

Q

gHv


Dimensioning tray columns /3

Liquid: Δpdowncomer = ρL· g· H; Δpslit = - c1· ρL· vL² Gas: Δpholes = c2· ρG· vG² (with constants c1, c2)

Using vL/vG = QL/QG; QL = L· ML/ρL ; QG = V· MG/ρG

gives (ρG· vG²) / (ρL· g· H) = (c2 + c1· (ρL· QL²)/(ρG· QG²))-1

→ → load diagram (with c1 ≈ 2494,5; c2 ≈ 694,4) .Note: ratio QL/QG is strongly pressure dependent.

Gas Liquid

Δp Δp

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Design procedure for given tray spacing:– Determine QG, QL, ρG, ρL, and tray spacing H– Calculate parameter (QL/QG)· (ρL /ρG)½

– Read vG· (ρG / ρL· g· H)½ from load diagram

– Calculate the maximum gas velocity vG and to allow for some margin take vG × 0.85

– Column cross-sectional area A = QG/vG = (π/4)· D2

– For hole area surface fraction F, the pressure dropΔpholes = 1.1· ρG· ( vG/F )2, add pressure drop for ~ 5 cm liquid on the trays: ΔpLoT = ρL· 0.05· g gives an estimate of Δps per tray.

See literature for design of downcomers, weirs, etc.

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ExampleC3 – iC4 at 10 bar, MG= ML= 44 kg/kmol, gases ideal, ρL = 500 kg/m3, V = 100 mol/s, L = 50 mol/s

tray spacing H = 0.5 m

answer: ρG = MG· p/RT = 17.5 kg/m3, QG = V· MG/ ρG = 0.25 m³/s QL = L· ML/ ρL = 0.0044 m³/s, (QL/QG)· (ρL /ρG)½ = 0.093 Load diagram: vG· (ρG / ρL· g· H)½ = 0.037, vG = 0.44 m/s

0.85 vG = 0.375 m/s; QG/vG = 0.67 m² gives D = 0.92 m Load diagram F = 0.065 Δps = 1.1· ρG· (vG/F)2 + ρL· 0.05· g = 890 Pa / tray.

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12.9 Steam distillation



Using steam in distillation processes is beneficial if a volatile compound A must be separated from an aqueous solution.

Then a dilute solution of A in water is obtained at the bottom and insteadof reboiling it may be more beneficial to feed steam from the bottom. (If the costs for a few more trays compensate for reboiler and otheradditional costs.)

In the (x,y) diagram, with steam injection S (mol/s) theoperation line for the bottomsection becomesV’m = L’m+1 + S – B, with L’ ≈ B and S ≈ V’V’· ym = L’· xm+1 – B· xB

Steam distillation /1

BS

Lm+1Vm

x=yeq

x = B·xB/(B-S)x = xB

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Steam distillation /2

Using steam in distillation processes is also beneficial if substances with high boiling points must be distilled and decomposition or other chemical changes might occur during direct distillation.

For these cases pressures below 1 atm can be used (for example vacuum distillation), but a good alternative is to add steam (or water). This should not be soluble in the liquid(s) of the substances to be distilled. That allows for operating at atmospheric pressure.

The composition of the vapour produced, with A is the component recovered and B is steam: yA/yB = pA/pB = pA / (ptotal – pA)

If some liquid water is present, for a certain pressure ptotal the temperature and composition of vapour can be calculated using a diagram by Hausbrand (1926) → see next slide.

Liquid water 1 degree of freedom: p ↔ T; no liquid water 2 degrees of freedom: p and T can be selected

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Steam distillation /3Vapour pressure curves for steam distillation of several organicsubstances. For total pressures ptot= 101.3 kPa, 40 kPa and 9.3 kPa, the intersectionof the lines for organic substance and water gives the temperature; the molar ratio water / organic substance equals (ptotal - pA) / pA.

Coulson & Richardson vol. 2 (1983)

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(a continuation of ö5.4, xA = 0.4, R = 1, see above (53-55))

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Task ö5.5 /1

Component B = H2O Unchanged (with respect to task

ö5.4)

– ṅD = 0.75 mol/s

– zD = xD = 0.90– R = 1– ṅF = 2,0 mol/s

– xF = 0.40

ṅL,F = ṅR = R· ṅD = 0.75 mol/s (identical to ö5.4)

ṅG = ṅL,F + ṅD = (R+1)· ṅD

= 1.50 mol/s (identical to ö5.4)

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Task ö5.5 /2 yA1 = xR = xD = 0.9· D (= resultat ö5.4)

yA = 0.65 Operation line (= resultat ö5.4)

ṅL,a = ṅL,f + ṅF

gives = 2.75 mol/s (= ö5.4) Changed: ṅF + ṅÅ = ṅW + ṅD

→ ṅW = 2.75 mol/s xF· ṅF + ṅÅ· 0 =

ṅW· xW+ ṅD · xD

→ xW = (2.00×0.40 - 0.75×0.9)/2.75 = 0.04545

1-2 more stages ṅW > ṅG = ṅ’G diameter up

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Task ö5.6 /1

In principle, decanting may be possible, since A = water and B = aniline are immiscible.

50 g H2O = 2.775 mole, 50 g aniline = 0.413 mole Ptot = 1 bar = 105 Pa = pA + pB = p°A + p°B (at T = ?) According to the table, at T (= θ) = 99°C:

yB = pB/ptot = 3039 Pa / 105 Pa = 0.03 pA + pB = 101300 Pa

yA = pA/ptot = 98261 Pa / 105 Pa = 0.97This is the composition of the vapour as long as there is A and B in the still.

Most volatile is water, when the 50 g = 2.775 moles havevaporised then also (0.03/0.97)· 2.775 = 0.086 moles of aniline have vaporised.

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Task ö5.6 /2

When no more H2O is present in the still, the amount of aniline left equals 0.413 – 0.086 = 0.327 moles.

At 1 bar, aniline boils at 204°C, so the temperature has to rise until this temperature and then the aniline starts to boil producing a gas at pB = ptot = 1 bar and yB = pB/ptot= 1 (pure aniline).

time

tem

pera

ture

°C

99

204

boiling A + B

boiling B

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12.10 Multi-component distillation, Azeotropic distillation, Extractivedistillation (a brief introduction/overview)


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Multi-component distillation /1

Distillation column arrangements for a mixture of A, B and C

Separation of multi-component mixtures requires a series of columns, typically (n-1) for a mix with n components.

83/92



For more than two components A,B,C, ..... the vapourpressure of yA depends not only on xA but also on the relative amounts of the other components.

Relative volatilies (for chemically similar components) are ≈ constant over wide ranges of temperature and composition.

A trial-and-error method by Gilliland and Reed (1942) can be used, noting that a multicomponent mixture usually meansa separation of two key components.

Example: light key B, heavy key C.

84/92



For liquid mixtures xA, xB, xC, ... and relative volatilities αAB, αBC etc. the composition of the equilibrium vapour can be calculated: for example

Minimum reflux ratio RDm and minimum number of stages Nmin can be found similar to binary distillation, for key components A and B

i i

Fii

i i

DiiDm

average,AB

BA

B

DB

A

min γα

xα q1 with

γα

xα1R ;

ln

x

x

x

xln

)1N(

diagram by Gilliland(1940) relating

N, Nmin, RD and RDm

givesγ

85/92

5668,0

min

1175.0

1:(1975) Eduljee

D

DmD

R

RR

N

NN

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Azeotropic distillation /1

Source: A83

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Adding a new substance C (”entrainer”), increasing the relative volatility of key components A and B

Azeotropic distillation: creating a new azeotrope with one(or more) of the key components.

For example: adding benzene (B) to ethanol(E)/water (W) gives a ternary azeotrope at 65°C (1 bar) which is lower than the E/W azeotrope (78°C). This makes W more volatile → next slide:

1: Forming ternary azeotrope vapour + nearly pure E2: Ternary azeotrope is condensed, separated into two liquids3: a. B-rich fraction is reflux, b. W+E-rich fraction to two towers for B

and W recovery, respectively4: Azeotropic overheads from W+E distillation back to first tower

87/92


Source: CRBH83

1

2 3b3a

= near

azeotrope

A

44


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ikAzeotropes and activity coefficients

see also §3.4

The azeotropic mix composition gives the data needed for finding activity coefficients and Van Laar parameters:

For example, ethyl acate (1) – ethanol (2) has an azeotrope at 1 atm, 78,1°C, xEtOH = 0,462.

At 78,1°C, p°EtOH = 77,46 kPa; p°EtAc = 84,13 kPa.

At the azeotrope, x = y = 0,462 p = γi· pi° γ1 = p / p1° = 1/0,8413 = 1,204; γ2 = 1/0,7746 = 1,307

Van Laar coefficients: A*, B* etc.


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Extractive distillation /1

Adding a new substance C (”solvent”), increasing the relative volatility of key components A and B

Extractive distillation: a relatively non-volatilesubstance is added, present in the mixture at ratherhigh concentrations.

For example: adding phenol (P) to iso-octane (iO) / toluene (T) increases the relative volatility of iO.

(At 83 mol-% P the separation becomes easy).

→ next slide

see also L/L and S/L extraction: section 14

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Extractive distillation /2

Source: CRBH83

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Sources #12 A83 P.W. Atkins ”Physical chemistry” 2nd Edition, Oxford University Press (1983)

CRBH83 J.M. Coulson, J.F. Richardson , J.R. Blackhurst, J.H. Harker ”Chemical engineering”, vol. 2, 3rd ed. Pergamon Press (1983) Chapter 11

E75: H.E. Eduljee, ” Equations Replace Gilliland Plot ”Hydroc. Proc. 54(9) (1975) 120

K80 King, C.J. ”Separation processes” McGraw-Hill (1971, 2nd ed. 1980)

MMD01 Moulijn., J.A., Makkee, M., van Diepen, A. ”Chemical process technology” (Wiley 2001)

MSH93 W.L. McCabe, J.C. Smith. P. Harriott ”Unit operations of chemical engineering” 5th ed. McGraw-Hill 1993

SH06 J.D. Seader, E.J Henley ”Separation process principles” John Wiley, 2nd edition (2006) Chapter 4.2, 7.1-7.4, 7.6, 6.5

T68 R.E. Treybal ”Mass transfer operations” McGraw-Hill 2nd edition (1968)

WK92 J.A. Wesselingh, H.H. Kleizen ”Separation processes” (in Dutch: Scheidingsprocessen) Delft University Press (1992)

Z87 F. Zuiderweg ”Physical separation methods” (in Dutch: Fysische Scheidingsmethoden) TU Delft 1987 (vol. 1, vol. 2)

Ö96 G. Öhman ”Mass transfer” (in Swedish: Massöverföring) course compendium Åbo Akademi VT (1996) §5

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