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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
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Packed tower columns /1
Random packing materials
Pictures:T68, SH06
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Packed tower columns /2
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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Packed tower columns /3
Structured packing materials Support plate
”Mellapak”http://www.sulzerchemtech.com
Pictures: T68
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Packed tower columns /4
Structured packings
Pictures: SH06
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Packed tower columns /5
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.
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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
RoNzmars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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)
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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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Packed tower columns /6
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
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RoNzmars 2015 Åbo 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
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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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RoNz
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
mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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
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Distribution coefficient β
RoNzmars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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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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
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RoNzmars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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
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Transfer units HTU, NTU /2
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!
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Transfer units HTU, NTU /3
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
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Transfer units HTU, NTU /4
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
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Transfer units HTU, NTU /5
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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Transfer units HTU, NTU /6
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)
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Transfer units HTU, NTU /7
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
sectionalarea A (m2)
packingwith contactsurface(m2/m3)
uG(m/s)
uL(m/s)
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a
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Transfer units HTU, NTU /8
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
sectionalarea A (m2)
packingwith contactsurface(m2/m3)
uG(m/s)
uL(m/s)
Z(m)
ΦG(m3/s)
ΦL(m3/s) cross
sectionalarea A (m2)
packingwith contactsurface(m2/m3)
uG(m/s)
uL(m/s)
31
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Transfer units HTU, NTU /9
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
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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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Transfer units HTU, NTU /10
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
35Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
RoNzmars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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
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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
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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.....
RoNzmars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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
mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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tekn
ikExample: exam question 341 answer
mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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912.3m 1 HETP HETP- 1-or HETP,hfor 1- ln(1/e)
HETP.hfor /ecc(h) : limitation . thermodynno: .
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13.5 Design calculations
42Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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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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Dimensioning packed columns /2
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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mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
Dimensioning packed columns /3
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)
Pic
ture
: http
://w
ww
.koc
hkni
ght.c
om/C
hem
ical
.htm
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Dimensioning packed columns /4
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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Dimensioning packed columns /5
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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Dimensioning packed columns /6
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
Dimensioning packed columns /7
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Dimensioning packed columns /8
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
Pic
ture
: http
://w
ww
.ras
chig
.de/
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mars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
Dimensioning packed columns /9
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
Pic
ture
: http
://w
ww
.trip
atra
.com
/Equ
ipm
entP
acka
ge/E
quip
men
tPac
kage
.htm
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Dimensioning packed columns /10
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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13.6 Concentrated solutions; Distillation
53Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
RoNzmars 2015 Åbo Akademi - kemiteknik - Värme- och strömningsteknik Biskopsgatan 8, 20500 Åbo
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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