mass transfer and separation technologyweb.abo.fi/~rzevenho/mofst15-oh13.pdf · mass transfer and...

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

13. Packed columns

Ron ZevenhovenÅbo Akademi University

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

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13.1 Principle of operation, Packings

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

2

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

Packed tower columns /1

Random packing materials

Pictures:T68, SH06

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Mass transfer from gas to liquid or vice versa where the liquid forms a (thin) film on the surface of packing material elements, creating a large contact surface ”a” (m2 / m3

apparatus) For relatively small

amounts of material transferred (say, < 2% of the streams) the process may be consideredisotherm (vaporisation and condensation have a heat effect!) and streams V and L may be consideredconstant. Picture: WK92

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Structured packing materials Support plate

”Mellapak”http://www.sulzerchemtech.com

Pictures: T68

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Structured packings

Pictures: SH06

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Structured packings or a random packed material can be used. Typicalsize of filling material 25 – 150 mm.

Mostly applied for absorption / desorption or distillation, with lowpressure drop as an advantage.

Compared to tray columns, more liquid with respect to gas flow. Typically countercurrent operation (gas upwards, liquid flows

downwards). Also co-current application as a gas-liquid reactor, and so-called trickle-beds where the packing acts as catalyst.

The use of packed tower columns is prefered over tray columns if 1) small units are needed (H < 10 m, D < 1 m); 2) corrosive fluids must be handled (using a ceramic or plastic packing); 3) pressure drop must be low; or 4) foaming can be a problem

Operation is limitated by ”flooding”, where the amount of gas flow prevents downflow of liquid. Typical operation at ~ 70% of floodinglimit. Important factors are liquid and gas flow per unit cross-section, and liquid viscosity.

7


Example: Raschig ring packing For Raschig rings with a 1” (inch)

diameter and height, calculate the contact surface ”a” , m2, per unit column volume, m3. Assume a cubic packing (see Figure). (1” = 2.54 cm)

Answer: The surface of inside + outside of the cylinders equals 2· π· d· l m2; in a cubic packing one cylinder occupies a volume equal to d²· l m³.

Thus, a = 2· π· d· l / (d²· l ) = 2· π / d with d = 0.0254 m gives a ~ 250 m2/m3 with respect to column volume.

The packing occupies fraction (1-ε) of the column volume(with voidage ε), thus the contact surface with respect to packing volume equals ap = a· (1-ε).

Picture: WK92

8

For Raschig rings: l = d

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Column packing characteristics

Source: http://www.cheresources.com/packcolzz.shtml

Generally, the column diameter to packing size ratio D/d should be > 30 for Raschig rings, > 15 for ceramic saddles, and > 10 for plastic rings or plastic saddles.

Fp = ap / ε3 unit. m-1

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NOTE: instead of ap/ε3 = a·(1-ε)/ε3 nowadays PACKING FACTOR Fp is used


Packed columns: flooding limits(random-dumped packings)

See also expression inÖhman (1996) Fig. 6.2 (extended to X = 100):

Y = - 1.68 – (10log X)/1.14 – ((10log X)/2.04)2

– ((10log X)/2.88)3

symbols: L and G: mass rate of liquid and gas (kg/s);densities ρG, ρL (kg/m3); SB is surface of packing per unit volume of bed (m2/m3) = S·(1-e) with packing voidage e = ε (m3/m3 ) and S is surface of packing per unit volumeof column (m2/m3); g is gravity; uG= gas velocity calculated over whole cross-section of the column; µL is liquid viscosity (Pa.s), µW is viscosity of water at 20°C (~ 0.001 Pa.s)

Coulson & Richardson vol. 2 (1983)

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Packed columns: pressure dropOften, column diameter is calculated from the gas velocity for a certain pressure drop!

Coulson & Richardson vol. 2 (1983)

GL

G

GLG

0.12

ρρ

ρ

G´

L´X

gρρρ

μFG´Y

Error: unit: 1/m (1/ft)

11/56


Packed columns: wetting rate

There is a minimum liquid rate for effective use of the packing surface area. A wetting rate can be defined as

Recommended minimum liquid wetting rates are LW = 2×10-5 m2/s for 25 -75 mm rings and grids with less than 50

mm spacings, and LW = 3.3×10-5 m2/s for larger packings.

)/()/(

)/(

)/(

)(/)/(smL

mma

smu

mma

mAsmW

p

L

p

columnL 23232

23

column the of volume unit per area surface Packing

column the of area sectional-cross unit per rate liquid Volumetric

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In a packed column no discrete, identifiable stages are seen

Similar to a plate column the goal is, however, not that equilbrium is reached at each stage

More importantly, mass transfer is accomplished that brings the phases closer to equilibrium

Pictures:↑ BSL60← K71

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13.2 Mass balance, Mass transfer

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


Task ö6.1 : the question (swedish)

70 mol/s luftyN+1 = 1.8%

y1 = 0.2 %vatten

Picture: after SH06

15

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Task ö6.1 /1

a) Assume that air = aceton-free air, A = aceton Mass flow gas in: ṅG, in = ṅair, in + ṅA, in

= 70.00 mol/s + 1.80 %· ṅG, in

→ ṅG, in = 70 / (1 - 0.018) = 71.2831 mol/sand ṅA, in = 1.2831 mol/s

Mass flow gas out: ṅG, out = ṅair, out + ṅA, out

= 70.00 mol/s + 0.2 % · ṅG,out

→ ṅG, out = 70 / (1 - 0.002) = 70.1403 mol/sand ṅA, out = 0.1403 mol/s

Transferred: ṅA = ṅA,in – ṅA,out = 1.2831 – 0.1403 = 1.1428 mol/s

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Task ö6.1 /2

b) With incoming gas stream 71.2831 · (1.8 % - 0.2 %)

= 1.1405 mol/s A With outgoing gas stream 70.1403 · (1.8 % - 0.2 %)

= 1.1222 mol/s A Transferred 70.00 · (1.8 % - 0.2 % ) = 1.1200 mol/s

Exact: ṅG, in · yA,in – ṅG,out · yA,ut = ṅA

ṅG, in · yA,in – (ṅG,in – ṅA) · yA,out = ṅA

gives ṅA = [ ṅG,in · (yA,in - yA,out)] / (1 – yA,out) ≈ ṅG,in · (yA,in - yA,out)

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Mass transfer in a packed column /1

Transport equation: transport across surface A

ṅ = koy· A· (y - y*) = kox·A· (x*- x) (mol/s) = k´oy· A· (cG - cG*) = k´ox· A· (cL*- cL) (mol/s)

defined like this: ṅ > 0 for G → L, ṅ < 0 for L → G

Equilibrium constant K: y* = K· x and y = K· x* at equilibrium: y = y*= K· x*= K· x = end of mass transfer process

K is related to kx and ky

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1. With K = β = Hc/ptotal (Henry), or

K = β = p°/ptotal (Raoult), or

K = β = γ∙p°/ptotal (Modified Raoult)

unit kox, koy, kx, ky = (m/s)·(mol/m3)

2. With volumetric distribution coefficient m: m∙cG = cL


Mass transfer in a packed column /2

11

oxxyoy k

β

k

β

kk

1111

oxxyoy k´m

k'mk' k´

unit kox´, koy´,k´x, k´y = m/s

massmolar

density massdensitymolar

M(kg/mol)

)(kg/mρ)(mol/mρ with

3x3

xmol,

mol,x

x'x ρ

k klinked by

19/56

Distribution coefficient β


Volumetric distribution coefficient m The volumetric distribution

coefficient, m, is usually°defined as cliquid = m· cgas

This is related to equilibruimconstant K = y/x usingcliq. = x· ρmol, liq. = x· ρliquid/Mliq.

cgas = y· ρmol, gas = y· (p/R·T)

with molar densities ρmol, (mass) densitiesρ, molar mass M, pressure p, temperatureT

This gives:

Volumetric distribution coefficient data for several gases in water at 20°C ceq

water = m·cgas

Table: BMH99See also #4 slides 6-10

° Not always, is sometimes definedas cgas = m.cliquid (see e.g. WK92)

RTpK

Mρ

mliquid

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Mass balance for a packed column Mass balance for a height section dℓ :

ṅ = V· dy = kG· (cG - cGi)· a· A· dℓ= kL· (cLi - cL)· a· A· dℓ = L· dx

with mass transfer coefficients kG and kL for gas and liquid side, cross-section area A and packingsurface per volume a.

With overall mass transfer coefficients koG or koL

(eliminating the phase interface concentrations), for the gas side, using koG :

V· dy = koG· (y* - y)· (p/R·T)· a· A· dℓ

yin∫ yout dy/(y* - y) = 0∫Z (1/V)· koG· ( p/R·T)· a·A· dℓ

and similar for the liquid phase……

Z

Picture: SH06

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mars 2015 22Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo

General approach height Z calculation

.kor kt coefficien transfer mass overall

eliminated be should and unknownusually c ionconcentrat

only; phase onefor balance mass theon based is This

Z @cc and 0 @ cc with

flow for volume also and

flow mass gives thisd sectionheight for

phase eachfor s)mol/(m )c(ckNflux :general

oxoy

interface

0interface

out in

int

2interface

daAk

cc

dc

dc daA)c(ck

dcnd

daANdSurfaceNnd

Zc

c

erface

out

in

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13.3 Transfer units: HTU, NTU(Height of a transfer unit, Number of transfer units)

Method 1, based on mass transfer resistances



Transfer units HTU, NTU /1

Absorption of a component from a gas into a liquid(or similarly, for desorption / stripping from a liquid to a gas)

Mass balance for transfer in a column heightsection with volume A· dz, with concentrations cG=cG(z) and cL=cL(z), volume streams ΦL, ΦG

with distribution constant m, cL/ m = cG* at equilbrium, and mass transfer surface “a” m2/m3

apparatus, and total height “Z”

koy (m/s) (or koG) is the overall mass transfer coefficient for the gas phase side

A similar mass balance can be set up for the liquidphase, using kox (or koL)

dzA)m

c(cakdc L

GoyGG dz

ΦL

ΦG

cross-sectionarea A m2

contact surfacea m2/m3

24



Integrating the left-hand part °, called the

Number of Transfer Units NTUoG gives

Zz

zoG

Zz

zG

oyZzc

zc LG

G

HTU

dzdz

Aak

mc

c

dcG

G 00

)(

)0()(

m

cccΔ

m

cccΔ

HTU

Z

cΔ

cΔln

cΔcΔ

cc

LZGZZ

LG

oGZZ

GZG

and

where

Integration defines the overall Height of a Transfer Unit HTU , here: HTUoG , subscript oG = overall, from gas phase

Column height Z = HTUoG × NTUoG.

Zz

° Operation line & equilibrium line must be straight lines!

25/56



Column height Z = HTUoG × NTUoG.

In a column section with volume HTUoG × A (m3), the mass transfer capacity koy· a· A· HTUoG (m3/s) is equal to ΦG (m3/s), i,e. HTUoG is the height of a section for which koy· a· A·HTUoG = ΦG

NTU equals the ratio between horizontal mass transfer and vertical transport by convection

note: 1) HTUoG = HTUoG(z), if koy varies with position; 2) axial dispersion is neglected

Zz

dz

ΦL

ΦG

cross-sectionarea A m2

contact surfacea m2/m3

26/56



The total mass balance gives (for counter-current plug flow)

lmGoy

Z

Zoy

oG

Z

ZG

oG

Z

ZGZGGZGGGmass

cZAak

cc

ccZAak

HTU

Z

cc

cc

HTU

Z

cc

cccccc

0

0

0

0

0

000

lnln

)ln

using(

Without thermodynamic limitations in the other phase (m=0 or =∞)the fraction transferred within one HTU is 1 - e-1 ≈ 63%.

Similarly, Z = HTUoG×NTUoG = HTUoL×NTUoL, but HTUoG ≠ HTUoL

with a logarithmic mean concentration drivingforce which can be used if the equilibrium line and operating line are straight (constant value for m), (see also #9: Φ = k·A·ΔYlm ) otherwise: plot 1/(cG - cL/m) versus cG and integrate: see next

Zz

27/56



Illustration for procedure of determining NTUoG by integrating the reciprocal of the driving force curve. Here: NTUoG= 6.2.

yAE = equilibrium value for yA Source: King 1971, 1980

yAE

yA

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Alternatively, HTUoG can be calculated from the individual values for the gas phase HTUG and for liquid phase HTUL. Combining these gives HTUoG or HTUoL, using the equilibrium data (K or m).

Similarly, the dimensionless ratio ΦL /(kL· a· A) = uL /(kL· a) is the height of a transfer unit HTUL (or HL) for the liquid phase.

The mass transfer capacity for the gas phase equals kG·a·A·Z (m3/s), with gas-side mass transfer coefficient kG (m/s).

The ratioΦG / (kG· a· A) = uG / (kG· a)

is the height of a transfer unitHTUG, (or HG) for the gas phase.

Z(m)

ΦG

(m3/s)

ΦL

(m3/s) crosssectionalarea A (m2)

packingwith contactsurface a(m2/m3)

uG

(m/s)

uL

(m/s)

29



Combining the gas-side and liquid-side HTUs:

with separation factor S. Similary, for the overall liquid-phase:

SHTUHTUmkAakAa

kmAakAakmkAakAaHTU

LGL

G

L

L

G

G

L

G

G

G

LG

G

oG

GoG

11

11111

SHTU

HTUS

HTUHTUHTU

oL

oG

GLoL

:note

Z(m)

ΦG(m3/s)

ΦL(m3/s) cross

sectionalarea A (m2)

packingwith contactsurface(m2/m3)

uG(m/s)

uL(m/s)

Z(m)

ΦG(m3/s)

ΦL(m3/s) cross



uG(m/s)

uL(m/s)

30/56

a



For the NTUs, for gas-side and liquid-side:

Usually the mass transfer is mainlylimited by the gas-side transfer,

→ koG is more constant than koL

→ more convenient to use NTUoG

and HTUoG than NTUoL and HTUoL

Column height

Z = HTUoG× NTUoG = HTUoL× NTUoL

:overall Liquid film Liquid

:overallGas filmGas

x*x

dxNTU

xx

dxNTU:

*yy

dyNTU

yy

dyNTU:

oLi

L

oGi

G

Z(m)

ΦG(m3/s)

ΦL(m3/s) cross



uG(m/s)

uL(m/s)

Z(m)

ΦG(m3/s)

ΦL(m3/s) cross



uG(m/s)

uL(m/s)

31

a



With straight operation lines and equibrium lines and liquid and gas molar streams V and L ~ constant:y = x· (L/V) + yout - xin· (L/V) for an absorbery = x· (L/V) + yin - xout· (L/V) for a strippery = K· x for the equilibrium

For an absorber, this gives the following for NTU (Colburn, 1939):

For a stripper (with S = KV/L) a similar expression can be derived for NTUoL or stick to NTUoG....

KV

LA S withabsorber, anfor

1

11ln

)1(*

SS

SxKyxKy

SS

xKL

KVyy

LKV

dy

yy

dyNTU

inout

inin

yin

youtinout

yin

yout

oG

x = (y-yout)·V/L - xin

y* = K∙x

32/56

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Example HTU, NTU Air containing 1.6 %-vol SO2 is scrubbed with clean water in an

absorber, which is a packed bed column with cross-sectional area A = 1.5 m2 and packing height Z = 3.5 m. Incoming gas and liquid flow rates are L = 62 mol/s and V = 2200 mol/s. For a concentration of 0.4 mol-% in the outgoing air and K = y/x = 40:

calculate NTUoG, HTUoG and koy· a (as mol/m3· s) for the SO2.

Answer:Separation factor S = A = L/KV = 0.89; with yin = 0.016, yout = 0.004, and xin = 0Colburn’s expression (see previous slide) gives NTUoG = 3.75Z = NTUoG × HTUoG = 3.5 m → HTUoG = 0.93 m

koy· a· A·HTUoG = ΦG, with unit for koy (m/s), V = ΦG· (p/RT)

koy· a· A·HTUoG = V , with unit for koy (m/s)· (mol/m3)

→ koy· a = 44 mol/m3.s

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Alternatively (notation Öhman 1996, §6.3-6.4) using– molar fractions y and x, describing equilibrium with y = β· x– notation for molar streams ṅ (mol/s), contact surface a (m2/m3) – transport coefficients kx and ky for each phase (mol/m3· m/s)– column height Z and cross sectional area Atv

– notation HG = HTUG; HL = HTUL; Z = HoG· NoG = HoL· NoL

lmAA

1A0A

G

GLL

lmAA

1A0A

L

LGG

xtv

LL

ytv

GG

x*x

xx

n

HnHZ

*yy

yy

n

HnHZ

akA

nH and

akA

nH

But note: using logarithmicmean concentrations so that NG = Δy/Δylm

and NL = Δx/Δxlm

requires that the equilibriumline and operating lines are straight !

ṅ = ṅG·(y0-y1) = Ky·a·Atv·Z·(y-y*)lm → Z = [ṅG /(Ky·a·Atv)]·[(y0-y1)/(y-y*)lm ]

→ Z = HoG·NoG .

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13.4 Theoretical stages (”plates”): HETP

Method 2, based on equilibrium stages



Height equivalent to a theoretical plate, HETP Important for mass transfer equipment is

the Height Equivalent to a Theoretical Plate (or equilibrium stage), HETP.

This is the height needed for equilibriumfor outgoing flows.

For liquid ”1” and gas ”2” this implies c1(x) = m· c2(x + HETP)

HETP and HTU are related viaconstant

balancemass

cΦcΦ vv

1

ln.......

21

2

S

S

cmc

dc

HTU

HETP HETPx

xoGS = 1: HETP = HTUoG

S > 1: HETP < HTUoGS < 1: HETP > HTUoG

Picture: BMH99

L V

36/56


HETP, HTU, NTU, Ntheoretical

(a) NTU = Ntheor. (b) NTU > Ntheor. (c) NTU < Ntheor.

Absorber S = A = 1 S = A < 1 S = A > 1

S = 1: HETP = HTUoG

S > 1: HETP < HTUoGS < 1: HETP > HTUoG

since

1

ln.......

..

21

2

NTUHTUNHETPN

NTU

S

S

cmc

dc

HTU

HETP

oGoGtheortheor

oG

HETPx

xoG

Picture: SH06

37/56

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HETP values for common column packings

CRBH83, dHB07

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For Raschig rings 25 < d < 50 mmHETP ≈ 18·d + 12·K·((V/L ) - 1)

Note: smaller packing elements give lowervalue for HETP but increase pressure drop.....


Definitions revisited.... HTU: height of mass transfer apparatus over which a

concentration is reduced by a factor 1/e, for c1 >> m· c2or for c1 << m· c2 (i.e. small thermodynamic limitation in other phase)(= also height needed to reach equilibrium between opposite streamsat equilibrium if separation factor S =1)

NTU: how many HTUs in apparatus height, is equal to residence time / time needed for transfer

HETP: height in transfer apparatus between oppositestreams at equilibrium

Column height Z= HTU × NTU; or Z = HETP × Ntheoretical stages

from countingstages in x,y diagram,or Kremser equation.

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Example: exam question 341 (7+3 p.)

I en motströmsprocess absorberas SO3 från en upgåendeluftström med 7 %-mol SO3 i koncentrerad H2SO4 vätske som rinner ner i en fyllkroppskolonn med Rashig-ring element som ger kontaktytan a = 100 m2/m3. Gasens hastighet (ytlig, dvs för tornet utan packning) är vG = 2 m/s. Partialtrycket p* för SO3 i jämvikt med koncentrerad H2SO4 är p* ≈ 0 bar. Massöverföringskoefficienten för gassidan är kG = 0.02 m/s medan massöverförings-resistansen för vätskesidan kan försummas. Också värmeeffekter kan försummas.

a. Om 98% av SO3 ska absorberas från luftströmmen, beräkna kolonnhöjden Z (i m).

b. Beräkna totala överföringshöjden HTUoG (= HoG) (i m), och antalet överföringsenheter NTUoG (= NoG) hänförd till gassidan. Beräkna även antalet teoretiska steg, Ntheory


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ikExample: exam question 341 answer


41/56

912.3m 1 HETP HETP- 1-or HETP,hfor 1- ln(1/e)

HETP.hfor /ecc(h) : limitation . thermodynno: .

912.3m. 1HTUvv

HTUb.

m. 3.912Z2

10002.0912.3)02.0ln(

02.0ln

:

removal) 98%(for 0.02 ; 1

~0*

K

111tcoefficien transfer Massa.

in

oGGG

G

00

02.0

,3

,3

3SO3

theoryG

G

theory

GGoG

oGG

G

G

in

in

Z

G

GZ

G

G

c

c G

GGGGG

LGoGGG

inSO

outSOGoG

SO

LGoG

Nv

HETPak

mHETPNHTUNTU

HTU

ZNTUNTU

kaka

Z

v

Zak

c

c

dzv

akdz

Aak

c

dcdzAcakdc

givesdzA)m

c(cakdcbalancemass

c

ckk

Km

p

p

kmkk

in

in

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13.5 Design calculations


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Dimensioning packed columns /1

ExampleA column for stripping aceton from water using an air stream at T = 20°C, 1 bar. x0 = 10-3 mol/mol in incoming water stream, xn = 10-4, yn+1 = 0 (aceton-free air). K = y*/x = 3 for the equilibriumkx = 3×10-5 m/s, ky = 4×10-2 m/s. Given vG = 1m/s.ρmol,L = 5.56×104 mol/m3, ρmol,G from ideal gas law.

Use a = 200 m2/m3.

How many (theoretical) equilibrium stages Nth if L/V = (L/V)max/1.2, using Kremser’s equation.

Calculate HTUoG and HETP and column height Z.

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answer:

y1eq.line = K· x0 = 0.003.

(L/V)max = (y1-yn+1)/(x0-xn) = (0.003–0)/(10-3-10-4) = 3.3.L/V = 3.3/1.2 = 2.75, y1

new= 2.475×10-3 ; S = K/(L/V)= 3 / 2.75 = 1.08; not transferred f = 0.1 Kremser Nth = 6.64

ρmol,L = 5.56×104 mol/m3, ρmol,G = p/RT = 41 mol/m3

equilibrium constant K = (ρmol,L / ρmol,G )/m

→ m = 452 → 1/koy = 1/(mkx)+ 1/ky → koy = 0.01 m/s

y = Kx

10-310-4

3×10-3

(L/V)max

(L/V)

V L

10-3

10-40

y1

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Mass transfer resistance:

1/100 = 1/0.0136 + 1/0.04.gas 25%, liq. 75%

See also #10 p. 20

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answer:

HTUoG = QG/(koy· a· koyA) = vG /(koy· a) = 1/(0.01·200) = 0.5 m

HETP =(ln S)/(S-1)· HTUoG

=(ln1.08)/(0.08)· 0.5 = 0.48 m Z = Ntheor· HETP

= 6.64· 0.48 m = 3.2 m

y = Kx

10-310-4

3×10-3

(L/V)

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Similar to tray column design, the most important features are column diameter and pressure drop.

Also here, a load diagram can be used.

ceramic

metal

Raschig ringsPacking factor

load diagram fora packed columnfor Raschig rings

Pictures: WK92

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The load diagram follows from a pressure drop analysis:Assume the packing to be a set of

vertical plates at a distance d, on which the liquid flows down as a layer with thickness δ (m), and velocity υL (m/s) υL is related to the superficial (= with respect to emptycolumn) liquid velocity vL (m/s) as vL = 2· υL· δ/d. Liquid volumefraction λ = 2· δ/d (m3/m3).

The gas moves upwards with superficial velocity vG (m/s).

Picture: WK92

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Forces as a result of flow in a packedcolumn (per m2 surface):1. Turbulent shear at the wall :

f1 = c1·ρL·υL²2. Gravity:

f2 = - ρL·δ·g3. Turbulent shear gas/liquid:

f3 = c2·ρL·(υL+vG)2 ≈ c2·ρG·vG2

With total force = 0 for stationary flow thisgives: c1·ρL·(½·vL·d/δ)² - ρL·δ·g + c2·ρG·vG

2 = 0 with λ = 2·δ/d ≈ c3·(vL

2/(g.d))⅓

gives final resultc´1·ρL·vL² - c´4 ρL·d·g + c´2·ρG·vG

2 = 0 (packing factor Fp is included in c´4 !)

Picture: WK92

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Design procedure:– Determine QG, QL, ρG, ρL, column height Z and diameter

d and packing factor Fp for the packing– Calculate parameter (QL/QG)· (ρL /ρG)½ and read

vG· (ρG · Fp / ρL· g)½ from load diagram

– Calculate the maximum gas velocity vG and to allow for some margin take vG × 0.70

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

– Calculate liquid superficial velocity vL = QL/A and volumefraction liquid λ =1.2· (vL

2/(g· d)⅓

– ”Dry” pressure drop Δpdry ≈ 0.8· ρG· vG2· Fp·Z

– ”Wet” pressure drop Δp ≈ Δpdry· (1-3.5·λ)-3


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Example (almost the same as above)

A column for stripping aceton from water using an air stream at T = 20°C, 1 bar. L = 0.4 mol/s, V = 0.25 mol/s, ML = 18 kg/kmol, MG = 29 kg/kmol, ρL= 1000 kg/m3, ρG = 1.2 kg/m3.

Column height Z = 2 m, packing material 10 mm ceramicRaschig rings.

Calculate column diameter and pressure drop

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answer: QL= L· ML/ρL = 7.2×10-6 m3/s

QG= V· MG/ρG = 6.0×10-3 m3/s. Fp = 1400 1/m and (QL/QG)· (ρL /ρG)½ = 0.034

Load diagram: vG· (ρG· Fp/ρL· g)½ = 0.37

→ vG = 0.90 m/s, 0.7· vG = 0.63 m/s. A = QG/VG = 0.0096 m2 → D = 0.11 m

vL = QL/A = 7.5×10-4 m/s, λ = 1.2· (vL

2/(g· d))⅓ = 0.021,Δpdry ≈ 0.8· ρG· vG

2· Fp· Z = 1075 Pa, Δp ≈ Δpdry· (1-3.5· λ)-3 = 1360 Pa

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Typical design procedure for packed columns Calculate the number of theoretical stages Nth. Select the type and size of the packing material Calculate the flooding limit from the two flows Calculate the column diameter for a gas velocity that

corresponds to 70% of the flooding limit. Calculate the pressure drop. Determine the values for HTUoG and HETP. Calculate the height of packing as HETP × Nth. Add 0.5 ... 1 m above and below packing.

for further details (inlet/outlet distributors etc.): see the literature

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mars 2015

13.6 Concentrated solutions; Distillation



Concentrated solutions; DistillationWhen concentrations are high and

significant amounts of mass is transferred, the operation line and/or the equilibriumline will be curved. Taking the inert parts of the gas and liquid

streams L´ = L· (1-x) and V´= V · (1-y) and use these in equations for HTU and NTU etc. In distillation applications, the equilibrium

constant K changes strongly from place to place → HTU and HETP etc. depend on position and change dramatically at the feedpoint. In calculations, instead of K use the local

slope ς = dy/dx of the equilbrium curve

→ 1/NTUoG = 1/NTUG + ς/NTUL etc.More detail: see SH06 §6.9 & §7.5

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mars 2015 55Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo

Tray vs. packedcolumnsfor distillation

Source: King 1980, p. 605

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Sources #13 BMH99 Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. ”Transport phenomena”

Wiley, 2nd edition (1999) BSL60 Bird., R.B., Stewart, W.E., Lightfoot, E.N, ”Transport phenomena” Wiley

(1960) CRBH83 J.M. Coulson, J.F. Richardson , J.R. Blackhurst, J.H. Harker ”Chemical

engineering”, vol. 2, 3rd ed. Pergamon Press (1983) Ch. 11,12 dHB07 A.B. De Haan, H. Bosch ”Fundamentals of industrial separations” 2nd

Ed., TU Eindhoven / U Twente, the Netherlands (2007) Ch. 4 K71, K80 King, C.J. ”Separation processes” McGraw-Hill (1971, 2nd ed. 1980) MSH93 W.L. McCabe, J.C. Smith. P. Harriott ”Unit operations of chemical

engineering” 5th ed. McGraw-Hill (1993) Ch. 22 SH06 J.D. Seader, E.J Henley ”Separation process principles” John Wiley, 2nd

edition (2006) §6.1, §6.7-6.9, §7.5 T68 R.E. Treybal ”Mass transfer operations” McGraw-Hill 2nd edition (1968) WH92 J.A. Wesselingh, H.H. Kleizen ”Separation processes” (in Dutch:

Scheidingsprocessen) Delft University Press (1992) Z97 F. Zuiderweg ”Physical separation methods” (in Dutch: Fysische

Scheidingsmethoden) TU Delft 1987 (vol. 1, vol. 2) Ö96 G. Öhman ”Massöverföring” Åbo Akademi Värmeteknik (1996) Ch. 6

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