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14 Examples for the design of packed columns
The relationships discussed in the previous chapters are important for the design andoperation of countercurrent packed columns. Examples taken from the fields of desorption,absorption, rectification, and liquid-liquid extraction shall now be given to demonstrate howthey can be applied in plant engineering. It is presumed that the reader is acquainted withthe principles underlying thermal separation techniques.
Fig. 14.1 shows a flow chart for a rectification column with a concentration x of low boil-ers in the feed F, overhead product D, and bottoms B. These fractions are presented quali-tatively in the ylx diagram. The example concerned is the separation of a binary mixture;the equilibrium curve ye = f(x) and the operating lines y = f (JC), viz. BI in the strippingzone and ID in the enrichment zone, are given.
Rectification
Fig. 14.1. Qualitative determination oftransfer units in rectification
Reboiler
B = F-D
hF = feed enthalpyhp = feed enthalpy at boiling temperature
Ahv=vaporization enthalpy at feed section
Packed Towers in Processing and Environmental Technology. Reinhard BilletCopyright © 1995 VCH Verlagsgesellschaft mbH, WeinheimISBN: 3-527-28616-0
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338 14 Examples for the design of packed columns
Fig. 14.2 illustrates the principles of absorption and desorption. The factors given are thegas volumetric flow rate V, the concentration of transfer components y in the gas, the liquidvolumetric flow rate L, and the concentration of transfer components x in the liquid. Thesubscript u indicates the bottom of the column; and the subscript o, the head of the column.
The gas load Y of transfer component is related to the mole fraction y by
Likewise, the liquid load X of transfer component is related to the mole fraction x by
X = T^ <14-2>
The number of transfer units NTUOv required for rectification between zones with molarfractions ya and yb is given by
NTUov = /a~y" In 4 ^ (14"3)Aya - Ayb Ayb
The terms Aya and Ayb in this equation represent the difference between the equilibriummole fraction ye>a or ye>b and the vapour mole fraction ya or yb, i.e.
&ya = ye,a - ya (14-4)
An = ye,b-yb (14-5)
The size of the zones should be selected so that the slope myx of the equilibrium curve ispractically the same in each.
In contrast, the slopes of the equilibrium curves in absorption and desorption processescan be regarded as constant over the entire range of concentrations between the head andthe foot of the column, particularly at low phase loads. In this case, if the gas load isadopted as a measure for the concentration, the number of transfer units in absorption pro-cesses will be given by
where Yu and Yo are the gas loads at the foot and head of the column. The terms AYU andAYO are then defined by
AYu = Yu-mYXXu (14-7)
AYo = Yo-mYXXo (14-8)
The corresponding equations for the liquid load are
NTUOL= f-f I n - ^ (14-9)AXU - AXn AXn
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14 Examples for the design of packed columns 339
A X M _ 1 Y x
mYX
AX0 = —Yo-XomYX
(14-10)
(14-11)
Likewise, the number of transfer units required in desorption processes will be given byEqns (14-12) to (14-14) if the gas load is adopted as the measure for the concentration:
Y -Y° u
AY° (14-12)
Absorption Desorption
ye-
x,y Mole fraction in liquid, gas or vapourmyx Slope of the equilibrium lineL/V Molar liquid-to-gas ratio or slope of
the operating lineX,Y Load fraction of transfer component in liquid /gasL,V Carrier stream of liquid /gas phase
r777'L _ Y0-Yu
V A n — A iiY Q = ' U + *W~
Fig. 14.2. Qualitative determination of transfer units in absorption and desorption
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340 14 Examples for the design of packed columns
Yo = mYXXo-Y0 (14-13)
Yu = mYXXu-Yu (14-14)
The corresponding equations for the liquid phase are
= xu
1
mYX
1
mYX
Yo
Yu
(14-16)
(14-17)
In a rectification process, as illustrated in Fig. 14.1, the reflux ratio is given by
r > r-*n = TTTT (14-18)
The molar flow rate at the head of the column would then be
V= F XF~XB (r + 1) (14-19)XD ~ %B
In an absorption process, as illustrated in Fig. 14.2 (left), the total liquid-vapour ratiomust be higher than the ratio of the carrier liquid, i.e. solvent, to the vapour, i.e.
Y > T" (14"2°)
The flow rate of solvent in this case is given by
Lm = V lu~*° (14-21)
In desorption, as illustrated in Fig. 14.2 (right), the total vapour/liquid ratio must behigher than the ratio of the stripping gas to the liquid, i.e.
V V
The flow rate of stripping gas in this case is given by
Vm = L ^ ^ - (14-23)x o,m x u
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14 Examples for the design of packed columns 341
In absorption and desorption, the total molar flow rate of vapour, consisting of the carriergas and the transfer components, is given by
V =V(\ + Y) where V = Vu (1 - yu) (14-24)
and the total molar flow rate of liquid, consisting of the carrier liquid and the transfer com-ponents, by
L = L(l+X) where L = Lo(1 -xo) (14-25)
The column diameter ds can be expressed in terms of the free cross-sectional area As, i. e.
ds = \ —As (14-26)y JI
If uv is the superficial gas velocity, As can be defined by
As = ^ ^ (14-27)Uy Qy
Thus the column diameter can be calculated from the operating parameters V [kmol/h],uv [m/s], T [K], and p [bar], i. e.
ds = 5.4 • 1(T3 \l V — — (14-28)17 p Uy
Fig. 14.3 illustrates liquid-liquid extraction in which the specific gravity of the continuousphase (Subscript C) is greater than that of the dispersed phase (Subscript D). In this case,the total number of transfer units NTUOc derived from the HTU-NTU model is given by
NTUoc = Cc'°~Cc'u In ̂ (14-29)Aco - Acu Acu
where Aco is the driving concentration difference at the head of the column, as defined byEqn (14-30), and Acu is the driving concentration difference at the foot of the column, asdefined by Eqn (14-31), i. e.
Aco = cc,o - ~ - cDtO (14-30)
AcM = cc,u - — cD>u (14-31)
Another case to consider is that in which mass transfer is in the C —> D direction and thedesired separation efficiency is given by
^E = 1-££JL (14-32)
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342 14 Examples for the design of packed columns
h CQ.O
CCo
Fig. 14.3. Qualitative determination of transfer units in liquid-liquid extraction
Rearranging this equation gives rise to the following:
CC,u = (1 - (14-33)
from which it can be derived that the concentration of transfer components in the continuousphase decreases from cc>o at the head of the column to cc>u at the foot.
If the phase ratio in the extractor is uc/uD, the concentration of transfer components inthe dispersed phase increases from a value of cD>u at the foot of the column to the followingvalue of cDo at the head of the column:
cD,o — (cC,o ~UD
cD,u (14-34)
Eqn (13-22) allows the effective height H to be determined from NTUOc and HTUOc\and, if the volume flow rate Vc of the continuous phase is given, the column diameter ds canbe obtained from the load uc < 0.7 uc, FI from the following equation:
(14-35)
14.1 Determination of the diameter of an off-gas absorption column withvarious types of packing
Polluted air flowing at a rate of 105 m3/h STP is to be scrubbed with water in a packedcolumn. The liquid-to-gas ratio must be varied between LIV = 1 and L/V = 10, dependingon the degree of contamination. The maximum permissible load is that at the loading point.
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14.1 Determination of the diameter of an off-gas absorption column with various types of packing 343
Under these conditions, what diameter would the column have to be if 50-mm packing ofthe following types had to be installed?Plastics packing: Intalox saddles, Bialecki rings, Pall rings, and Norpac ringsMetal packing: Pall rings, Hiflow rings, Bialecki rings, and VSP rings
Solution
Properties of the system
Density and dynamic viscosity of the gas
QV= 1.19 kg/m3
TJV = 18 • 10~6 kg/msDensity and dynamic viscosity of liquid
QL = 998 kg/m3
TU = 1 • 10"3 kg/ms
The values of Cs required to calculate the resistance coefficients §5 at the loading pointfrom Eqn (4-29) are listed in Table 4.1.
Gas, air 100000 m3/h, Liquid: water, STP, loading capacity
TD
5.4
5.2
5.0
4.8
4.6
4.4
4.2
4.0
3.8
//
/fI1
ijI!1
1
/- Bic
fA
/
j
/
- I N
ilec
/
/
TALOX 5
<i ring
;adc
/
le
y
A-Pall ring
/
/
Packing size 50 mm
-NOR-PAC ring
- Plast IC
•y
1H1
Hifl
/
- v
DW
/
AfSP
rOu 11
my
K
y
<y
>
Bialecki ring
ing
Mr tnl
0 2 6 8 104 6 8 10 0 2 4
Liquid/gas ratio L/V
Fig. 14.4. Column diameter as a function of the liquid-vapour ratio for the data given in Section 14.1
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344 14 Examples for the design of packed columns
The values of §s calculated from Eqn (4-29) for all the packings, together with the asso-ciated values for a and 8, which are also listed in Table 4.1, allow the corrresponding vapourflow rates at the loading point uv = uvs to be determined by iteration as functions of theliquid/vapour ratio LIV from Eqn (11-82).
The evaluated results are plotted in Fig. 14.4. It can be seen that the column diameter ds
required for a ratio LIV = 10 is about 25% greater than that for LIV = 1. The diagramalso shows that the differences in diameter associated with plastics packing are greater thanthose for metal packing.
14.2 Design of a packed column for the rectification of an isobutane/n-butanemixture
An isobutane/n-butane mixture flowing at a rate of 500 kmol/h has to be separated in arectification column. It consists of 60 % of the low-boiling component and 40 % of the high-boiling and enters the column at the boiling point. The overhead product has to consist of99.9% mol of isobutane, and the bottoms may contain only 0 .1% mol of this component.The packing is of the Montz Bl-200 sheet-metal type, and the column operates at an over-head pressure of 3.6 bar and a reflux ratio of 15 under the conditions at the loading point.
Solution
Average properties of the system
\iv = 58.124 kg/kmol \iL = 58.124 kg/kmol
QV = 8.398 kg/m3 QL = 561.3 kg/m3
Tjv = 7.76 • 10~6 kg/ms t\L = 0.13078 • 10"3 kg/msDv = 1.17 • 1(T6 m2/s DL = 8.44 • 10~9 m2/sam = 1.35 oL = 8.88-10-3kg/s2
Characteristic data for the packing derived from Tables 4.2, 4.4, and 5.2
a = 200 m2/m3 Cs = 3.116 Cv = 0.3908 = 0.979 m3/m3 CL = 0.971 CP = 0.355
Product streams designated as in Fig. 14.1
kmol 0.6-0.001D 5°° 3OD 5 ° ° i r 0.999-0.001 3O
L = 300.1 kmol/h • 15 = 4501.5 kmol/h= 4501.5 kmol/h • 58.124 kg/kmol = 261645 kg/h
V = 300.1 kmol/h (15 + 1) - 4801.6 kmol/h= 4801.6 kmol/h • 58.124 kg/kmol = 279088 kg/h
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14.2 Design of a packed column for the rectification of an isobutaneln-butane mixture
Resistance coefficient at the loading point according to Eqn (4-29)
9.806
345
261645 / 8.398 \1/2 /0.131-10'U [ 279088 \ 561.3 / \ 7.76 • 10"6
Relationship at the loading point according to Eqn (4-28)
= 0.514
uv,s =/ 9.806^0.514
0.9792001 2001 12 0.131-10-3wL
12 0.131-10-3wL
9.806
561.3 \1/2
8.398
561.3
9.806 561.3
Relationship at the loading point according to Eqn (11-82) for uL and uv = uVtS
uL =261645 8.398279088 ' 561.3uv,s
Vapour load at the loading point determined by solving Eqns (4-28) and (11-82) byiteration
uv>s = 0.489 m/s
Cross-sectional area of column according to Eqn (14-27)
279088/3600As 0.489 • 8.398
Column diameter according to Eqn (14-26)
= 18.88 m2
ds = \—' 18.88 = 4.9 m
Column diameter selected for planning: ds = 5 m.
Cross-sectional area of the planned column
A s = — • 52 = 19.6354
Vapour and liqud loads in the planned column
279088/3600uv =
uL =
8.398 • 19.635
261645/3600561.3 • 19.635
= 0.47 m/s
= 6.59 • 10-3 m3/m2s
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346 14 Examples for the design of packed columns
Theoretical liquid holdup according to Eqn (4-27)
0.131 • 1(T3 • 6.59 • 1(T3 • 200 2 U / 3
HL " I 1 2 ' 9.806 • 561.3
= 0.0422 m3/m2
Particle diameter according to Eqn (4-66)
Factor to allow for end effects {Eqn (4-57)}
4fs = 1 +
- i
= 0.996200 -5
Reynolds number for the stream of vapour {Eqn (4-65)}
0.47 • 0.63 • 10"3 • 8.398(1 - 0.979) 7.76 • 1C
Reynolds number for the stream of liquid {Eqn (4-69)}
6.59 • 10-3 • 561.3
0-996 = 15 200
L 200 • 0.131 • 10-3
Wetting factor for the packing {Eqn (4-68)}
- 141.42
W= exp141.42
= 2.028200
Resistance coefficient for the stream of vapour {Eqn (4-67)}
Pressure drop in the vapour stream {Eqn (4-58)}
200 8.398 - 0.472 1H ' (0.979-0.0422)3 ' 2 ' 0.996
= 0-564 • , n ^ x 3 • ~ T— TTT^T = 128.3 Pa/m
Hydraulic diameter {Eqn (5-15)}
A 0.979= 0.01958 m
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14.2 Design of a packed column for the rectification of an isobutaneln-butane mixture 347
Area of phase contact {Eqn (5-16)}
^ 3 0 979- ( 6-59-10^-561.3a \ 200- 0.13078 • 10~3
' (6.59- 1Q-3)2 561.3 \075 / (6.59 • 10'3)2 200 ̂ 0 4 5
0.00888 -200 / \ 9.806 ~ "
Height of a transfer unit in the liquid phase {Eqn (5-5)}
HTUL =1
0.9710.13078 • 10-3 \1/6 / 0.01958
561.3 • 9.806 8.44 • 10
6.59 • 10"
200 1.045
- 0.083 m
Height of a transfer unit in the vapour phase {Eqn (5-14)}
200 • 7.76 • 10"6 \3M /1.17 • 10~6 • 8.398 \1/3 1
0.47 • 8.398 / \ 7.76 • 10~b / 1.045
= 0.143 m
Determination of the concentration zones (Table 14.1)Calculation for the height of the uppermost zone
Xa = XD = 0.999 kmol/kmol to xb = 0.8 kmol/kmol
Characteristic concentration in the uppermost zonePhase equilibrium concentration corresponding to ya = xa = 0.999 kmol/kmol {Eqn (1-38)}
1 35 • 0 999 = °"93* • = 1 + (1.35-1)0.999
Vapour concentration corresponding to xb = 0.8 kmol/kmol on the operating line for theenrichment zone {Eqn (1-13)}
15 0 999yb = — • 0.8 + -z = 0.8124 kmol/kmol
Phase equilibrium concentration
1 / ( O S - I 0.8= ° - 8 4 3 8
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348 14 Examples for the design of packed columns
Differences in concentration {Eqns (14-4) and (14-5)}
Aya = 0.9993-0.999 = 0.0003 kmol/kmolAj b = 0.8438-0.8124 - 0.0314 kmol/kmol
Number of vapour-side transfer units {Eqn (14-3)}
0.999-0.8124 0.0003
NTUov = o.ooo3-o.o3i4 ' l n "b^TT = 2 8 '
Mean slope of equilibrium curve {Eqn (1-66)}
1.3yx [1 + (1.3 - 1) 0.867]2
Stripping factor {Eqn 1-30)}
X = 0.8184
= 0.8184
4501.5 kmol/h
Number of theoretical stages {Eqn (1-27)}
Height of an overall transfer unit {Eqn (1-62)}
HTUov = 0.143 + 0.8335 • 0.083 = 0.212 m
Height of uppermost concentration zone {Eqn (1-64)} if no allowance is made for endeffects
H = 0.212 • 28.8 = 6.106 m
The results thus obtained for all the concentration zones are listed in Table 14.2.
Total theoretical height if no allowance is made for end and distribution effects
2 H = 15 m.
14.3 Scaling up measurements performed in pilot plants
In mass transfer studies in a column with a new type of packing, a total of NP>a = 5.6transfer units was measured for a height of HPa = 1.5 m. Hence the number of transfer unitsper unit height of bed was (NIH)P>a = 3.733 m"1. In a control experiment with a bed heightof HPb = 2 m, the number of transfer units measured was NPft> = 6.8. What would the error
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14.3 Scaling up measurements performed in pilot plants 349
be if the efficiency determined for a height HPa = 1.5 m were to be taken as a basis in cal-culating the height HT of an industrial-scale column required to realize NT = 20 transferunits on the assumption that the inlet distribution remains unchanged? For the purpose ofthe calculation, assume that the height of the inlet zone Ht is less than HPa = 1.5 m.
Solution
If no allowance is made for end effects, the height of the column would be
H w _ _ g t _ = - | L = 5.36 m
If it is assumed that the separation efficiency remains constant in the H > Ht zone, thefollowing applies for Hi < HP
dN \ NPb-NPa 6 .8-5 .6 „ A -,:.4 mdHJH>Hi HP>b-HP>a 2 - 1 . 5
According to the model presented in Fig. 8.4, the end effects (Nd + ANt) can be deter-mined by rearranging Eqn (6-8) for the case of H = HPa, i. e.
)n lH>Hi
= 5 .6-1 .5 • 2.4 = 2
The relative difference in height required to compensate end effects can then be obtainedfrom Eqn (8-15), i.e.
7 ^1-2Q 1+JJ'2A
According to Eqn (6-35), this corresponds to an efficiency ratio of
rjr = 1 - 4 r " = X " °-285 = °-715HT
The actual column height required can then be calculated from Eqns (6-35) and (6-39),i. e.
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350 14 Examples for the design of packed columns
The same result can be obtained from Eqn (6-28). Thus
P_a Nd + AN(\H jP,a
1= 7.49 m
3.7331.5 2
Hence, the absolute value for the additional height required is
AH = 4 ^ " HT = 0.285 • 7.49 = 2.13 m.HT
This figure agrees with that for the difference between the values calculated for HT and
HT(P), i- e.
AH = HT-HT(P) = 7.49-5.36 = 2.13 m.
Hence, if the column design were to be based on the measured value for (N/H)P>a =3.733 m"1, the height of the bed would be too short by an amount
AH 100 = 28.5%.HT
14.4 Design of an absorber for removing acetone from process air
The spent air from a production plant contains 1 % mol of acetone and flows at a rate of2000 m3/h at 27 °C. The acetone has to be removed to a final concentration of 100 mg/m3 byabsorption in water in a column packed with 50-mm metal Pall rings. The main dimensionsof the column, which is operated at the loading point, have to be determined for the case inwhich the water consumption is 1.4 times higher than the minimum.
Solution
Average propert ies of the gas phase
\iv = 28.96 kg/kmol
QV = 1 . 1 6 2 kg /m 3
T]v = 18.13 • 10~6 kg /ms
Dv = 10.8 • K T 6 m 2 / s
Molar mass of acetone
1^ = 58.08 kg/kmol
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14.4 Design of an absorber for removing acetone from process air 351
Average properties of the liquid phase
\iL = 18.02 kg/kmolQL = 997 kg/m3
T)L = 0.857 • 10"3 kg/msDL = 1.18 • 10~9m2/soL =12- 10-3 kg/s2
Slope of equilibrium curve for the acetone-air/water absorption system
YYlYx = 2.314
Packing characteristics (Tables 4.1, 4.3, 4.6 and 5.1)
a = 110 m2/m3 C5 = 2.725 CP = 0.763
E = 0.952 m3/m3 CL = 1.192 Ch = 0.784Cy = 0.410
Product streams (Fig. 14.2, left)
Vu = 2 0 0 0 ^ • , 1 ^ 2v ^ m 3
| = 80.25 kmol/hft 28.96 kg/kmol
yM = 0.01 kmol/kmol
yM = ° ^ ^ = 0.01 kmol/kmol
V = 80.25 (1-0.01) = 79.45 kmol/h
= i n o28.96 kg/kmol
m3 58.08 kg/kmol -1.162 kg/m3
= 4.291 • 10"5 kmol/kmol
4 291 • 10~5
4 2 9 11 -4,291k m o l / k m o 1
= 4 3 2 2 " 1 0 3 k m o l / k m o 1
L = 1.4 • 183.04 = 256.26 kmol/h = Lo
xu = 0 + T 9 ; 4 ^ (0.01 - 4.291 • 10'5) = 3.087 • 10~3 kmol/kmol256.26
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352 14 Examples for the design of packed columns
Lo = 256.26 kmol/h • 18.02 kg/kmol = 4617.7 kg/h
Vo = 79.45 (1 + 4.291 • 1(T5) = 79.45 kmol/h
= 79.45 kmol/h • 28.96 kg/kmol = 2301 kg/h
Resistance coefficient {Eqn (4-29)}
9.806
2.7252 L 1 6 2
2301 \ 997 / \ 18.13 • 10~6
Vapour velocity at the loading point {Eqn (4-28)}
= 0.629
uv,s =9.8060.629
0.9521101 -no1 12 0.857 uL
12 0.857 • 10~3 uL
9.806 997
9.806 997
1 /6 / 9 9 7 \ i / 2
1.162
Liquid velocity at the loading point {Eqn (11-82)} for uL and uv = uv,s
4617.7 1.162
Gas velocity at the loading point determined by solving Eqns (4-28) and (11-82) byiteration
uv>s = 1.988 m/s
Cross-sectional area of column {Eqn (14-26)}
2300.9/3600 2A ° 2 7 71.162 • 1.988
Column diameter {Eqn (14-27)}
ds=\—' 0.277 = 0.594 m
Column diameter selected for planning: ^ = 0.6 m
Cross-sectional area of planned column
As = — • 0.62 = 0.2827 m2
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14.4 Design of an absorber for removing acetone from process air 353
Vapour and liquid loads in the planned column
2300.9/3600Uv= 1.162-0.2827 = 1 - 9 4 5 m / s
4617.7/3600 3 3 2
^ = = 4 5 5 ' 1 0 m / m S^ 997 - 0.2827 4 ' 5 5 1 0 m
Theoretical liquid holdup {Eqn (4-27)}
/ 0.857 • 10-3 • 4.55 • HT3 • HO2 \1/3
)
Particle diameter {Eqn (4-66)}
dp = 6 • 1 - ^ p 9 5 2 = 2.618 • 10-3 m
Wall factor {Eqn (4-57)}
/ A \"1
= 0.9429
Reynolds number for gas flow {Eqn (4-65)}
1.945 • 2.618 • 10~3 • 1.162v (1-0.952) 18.13 • 10-6
Reynolds number for liquid flow {Eqn (4-69)}
° 9 4 2 9 -
4.55 • I P 3 • 997
-3 " 4 8 1 2u 110.0 • 876 • 10
Wetting factor for the packing {Eqn (4-68)}
Resistance factor in gas flow {Eqn (4-67)}
= 277.2 Pa/m
Pressure drop per unit height in gas flow {Eqn (4-58)}
Ap 110 1.162 • 1.9452 1()R2
H (0.952 - 0.0387)3 2 0.9429
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354 14 Examples for the design of packed columns
Hydraulic diameter {Eqn (5-15)}
rf, = 4 . ^ f = 0.0346m
Phase contact area {Eqn (5-16)}
Eh. — a . n QS?0-5 -a " 3 ° - 9 5 2 1 110 • 0.857 • 10-3
(4^55 • 1 0 - y 997 \015 ( (4.55 • 10'3)2 110 ,— U.u/^0.072 -110 / \ 9.806
Height of transfer unit in the liquid phase {Eqn (5-5)}
L0.857 • 10-3
1.192 \ 997 • 9.806
= 0.539 m
0.0346 \1/2 /4.55 • 101.18 - KT9 / \ 110 / \ 0.672 /
Height of a transfer unit in the gas phase {Eqn (5-14)}
HTTJ l rnos? n ^ ^ i / 2 0-03461/2 1.945HTUv = ^^ (0'952 " ° ° 3 8 7 )^ T ^ 10.8 • 10-6
/IIP - 18.13 • IP"6 \3/4/l0.8 • IP'6 • 1.162 \1/3 / 1 \\ 1.945 • 1.162 ) \ 18.13 • 10~6 / \ 0.672/
= 0.456 m
Differences in gas loads {Eqns (14-7) and (14-8)}
AYU = 0.01-2.314 • 3.807 • 10~3 = 2.864 • 10~3 kmol/kmol
AYO = 4.291 • 10~5 -2.314 • 0 = 4.291 • 10"5 kmol/kmol
Overall number of gas-side transfer units {Eqn (14-6)}
MTTI 0.01-4.291 -HP5 2.864 - IP'3
NTUov = 2 .864-10- 3 -4 .291-10- 5 ' l n 4.291 • 10"5 = U ^
Stripping factor {Eqn (1-68)}
- 2.314 • I9*™1* ^ 0.7174256.26 kmol/h
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14.5 Design of a desorber for removing carbon dioxide from process effluents 355
Overall height of a transfer unit {Eqn (1-62)}
HTUov = 0-456 + 0.7174 • 0.539 = 0.843 m
Height of bed of packing without allowance for end and distribution effects
H = 0.843 • 14.826 = 12.49 m
14.5 Design of a desorber for removing carbon dioxide from process effluents
The carbon dioxide content of a process effluent flowing at a rate of 4 m3/h is to bereduced from 1.5 kg/m3 to 0.02 mol/m3 by desorption with air in a packed stripper columnoperating at 20°C and 1 bar. The carbon dioxide content of the inlet air is 0.03% vol., andthat of the air leaving the desorber outlet must not exceed 1 % vol. The desorption columnis to be packed with 35-mm plastics Pall rings. Determine its main dimensions.
Solution
Average properties of the gas phase
\iv = 28.96 kg/kmolQV = 1.188 kg/m3
r\v = 17.98 • 10~6 kg/msDv = 15.4 • 10~6 m2/s
Data for carbon dioxide
\iA = 44.01 kg/kmolQA = 1.818 kg/m3
Average properties of the liquid phase
\iL = 18.02 kg/kmolQL = 998 kg/m3
r|L = 0.998 • 10"3 kg/msDL = 1.82 • 10~9m2/soL = 72 • 10-3 kg/s2
Slope of equilibrium curve for the carbon dioxide-water/air system
rriyx = 1440
Characteristic data for the packing derived from Tables 4.1, 4.3, 4.6 and 5.1
a = 145 m2/m3 Cs = 2.654 CP = 0.927
8 = 0.906 nr7m3 CL = 0.856 Ch = 0.718Cv - 0.380
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356 14 Examples for the design of packed columns
Product streams corresponding to Fig. 14.2, right
to = 4 • 998 = 3992 kg/h
= 4 • ~ ^ - = 221.53 kmollh
X° = 1"5 ' 4408l'02998 = 6 1 5 4 " 10"4 k m o l / k m o 1
X° = 1 + 0541(Vo-4 = 6-1 5° • 10"4 kmol /kmo1
L = 221.53 (1 - 6.150 • 10"4) = 221.39 kmol/h
° 0 3 10~2 I li 3 ° 2y" = ° 0 3 10 4I01 • liiss = 3 -° 2
3 02 • 10~4
Yu = t "3 Q2 1Q_4 = 3.02 • 10"4 kmol/kmol
= 0.01 • ^ ; \-™ = 0.01007 kmol/kmol
k m o l / k m o 1
18 09= 0.02 • 10"3 • —^— = 3.614 • 10~7 kmol/kmol
Total mass balance
K = 13,80 (1 + 0.01017) = 13.94 kmol/h- 13.94 • 28.96 = 403.71 kg/h
K - 13.80 • 28.96 (1 + 3.02 • HT4) = 399.77 kg/h
3 9 9 - 7 7 = 336.50 m3/h1.188
Lu = 221.39 • 18.02 (1 + 3.614 • 10~7) = 3989 kg/h
Resistance coefficient {Eqn (4-29)}
9.806
2.6543989.45399.77
1.188 1;2 0.998 • 10"998 / 17.98 • 10"
0.4 1-0.652
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14.5 Design of a desorber for removing carbon dioxide from process effluents 357
Vapour velocity at the loading point {Eqn 4-28)}
9.806 \1/2 [ 0.906 1/2 / 12 0.998 • 10~3 uL
1.980 / [ 1451/6 \ 9.806 998
12 0.998 • 10~3 uL \1161 998
9.806 998 / \ 1.188
Liquid velocity at the loading point {Eqn (11-82)} for uL and uvs
3989.45 1.188399.77 998 Uy's
Gas velocity at the loading point determined by solving Eqns (4-28) and (11-82) byiteration
uVjS = 1.195 m/s
Cross-sectional area of column {Eqn (14-26)}
399.77/3600As= 1.188-1.195 =
Column diameter {Eqn (14-27)}
ds = \ — • 0.0739 = 0.316 m
Column diameter selected for planning: ds = 0.32 m
Cross-sectional area of planned column
As = — - 0.322 = 0.0804 m2
Vapour and liquid loads in the planned column
399.77/3600 ^ r n ,Uv= 1.188-0.0804 = 1 - 1 6 3 m / s
3989.45/3600 3
UL = 998 • 0.0804 = 1 3 ' 8 ' 1 0
Theoretical liquid holdup {Eqn (4-27)}
0.998 • 10"3 • 13.8 • 10~3 • 1452 \1/3 = 0.0708
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358 14 Examples for the design of packed columns
Particle diameter {Eqn (4-66)}
= 6 = 3.890 • 1(T3 rn
Wall factor {Eqn 4-57}
Reynolds number for gas flow {Eqn (6-65)}
1.163 • 3.890 • 10~3 • 1.188(1 - 0.906) 17.98 • 10~6
Reynolds number for liquid flow {Eqn (4-69)}
13.8 • 10-3 • 998
0.9206 = 2927
*«• =
145 • 0.998
Wetting factor for the packing {Eqn (4-68)}
Resistance factor in gas flow {Eqn (4-67)}
o-™-j^ - " • " ' ^ 2 9 2 7 ' 2927008M 0.906
Pressure drop per unit height in gas flow {Eqn (4-58)}
= 1.284
Ap 145_
H ~ ' ' (0.906 - 0.0708)3
Hydraulic diameter {Eqn (5-15)}
1.188 • 1.1632 1
0.9206= 240Pa/m
dh = 4 • ° ^ — = 0.02499 m
Phase contact area {Eqn (5-16)}
145
- ( 13-8 • 10-3 • 998 p= 3 • 0.906 (a \ 145 • 0.998 • 10"3
(13.8 • IP'3)2 - 998 \075 / (13.8 • 10"3)2 1450.072 • 145
"3)2 145 \'Q-45
9.806= 0.799
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14.5 Design of a desorber for removing carbon dioxide from process effluents 359
Height of transfer unit in the liquid phase {Eqn (5-5)}
_ 1 / 0.998 • 1(T3 \ 1 / 6 / 0.02499 \ 1 / 2 /13.8 • 10~3 \ 2 / 3 / 1 \L ~ 0.856 \ 998 • 9.806 j \1 .82 • 10~9 f \ 145 / \ 0.799 /
= 0.772 m
Height of a transfer unit in the gas phase {Eqn (5-14)}
145 • 17.98 • 10"6 \3M/15.4 • 10"6 • 1.188 \1/3 / 1 \ _1.163-1.188 / 1 17.98-10-6 / \ o . 799 /~ ° ' 1 8 7 m
Differences in gas loads {Eqns (14-16) and (14-17)}
AXO = 6.154 • 10"4 - ° ' 0 i 0 ^ 7 = 6.083 • 10~4 kmol/kmol1440
3 02 • 10~4
AXU = 3.614 • 10"7 - — = 1.516 • 10~7 kmol/kmol1440
Number of liquid-side transfer units {Eqn (14-15)}
NTUoL ~6.154 • 10~4 - 3.614 • 10~7 / 6.083 • 10"4
oL ~ 6 0 8 3 • 10-4 1516 • lO"76.083 • 10-4 - 1.516 • 10"' \ 1.516 • 10"
Stripping factor {Eqn (1-68)}
= 8.39
221.39 kmol/h
Height of an overall transfer unit {Eqn (1-63)}
HTUOL = 0.772 + X • 0.187 = 0.774 m
Height of bed if end and distibution effects are neglected {Eqn 1-65)}
H = 0.774 • 8.39 = 6.494 m
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360 14 Examples for the design of packed columns
14.6 Determination of the difference between the theoretical and the real liquidholdup
The purpose here is to determine the differences between the theoretical and the realliquid holdups applicable to the loading conditions under which the columns discussed inSections 14.4 and 14.5 were operated.
Solution
Known values in Section 14.4
Ch = 0.784 QL = 997 kg/m3 uL = 4.55 • 10~3 m3/m2s
a = 110 m2/m3 i\L = 0.857 • 10~3 kg/ms ReL = 48.12
Froude number for the liquid phase {Eqn (4-27)}
= HO (4-55 • I C Y = . 1Q_4L 9.806
Hydraulic area {Eqn (4-75)}
— = 0.85 • 0.784 • 48.1025 (2.322 • 10~4)01 - 0.760a
Real liquid holdup {Eqn (4-73)}
= 0.0322 m3/m3
Calculated difference between the theoretical and real holdups
AhL 0.0387-0.0322
"TiT= ^ ^ = °'m
Known values in Section 14.5
Ch = 0.718 QL = 998 kg/m3 uL = 13.8 • 10"3 m3/m2sa = 145 m2/m3 r\L = 0.998 • 10~3 kg/ms ReL = 95.17
Froude number for the liquid.phase {Eqn (4-27)}
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14.7 Additional column height required to compensate maldistribution 361
Hydraulic area {Eqn (4-75)}
— = 0.85 • 0.718 • 95.17025 (2.816 • 10~3)01 = 1.060a
Real liquid holdup {Eqn (4-73)}
= 0.0736 m3/m3
Calculated difference between the theoretical and real holdups
AhL 0.0708 - 0.0736hL 0.0708
= - 0.039
14.7 Additional column height required to compensate maldistribution
In a column of ds = 2250 mm diameter packed with 50-mm Pall rings, the liquid distribu-tor gives rise to maldistribution of M = 10%. The phase ratio is LIV = 1. If the maldistri-bution were to remain effective over the entire height of the column, what would be theadditional column height required to compensate the attendant loss in efficiency?
Solution
The loss in efficiency can be read off from the diagram in Fig. 8.12 against the value
4 _ 2250 _d ~ 50 ~ 4 5
The numerical value thus obtained is
AE = 15%.
In other words, the efficiency would be reduced by a factor {Eqn (8-24)}
Hence, the relative increase in height required to compensate the maldistribution is givenby
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362 14 Examples for the design of packed columns
14.8 Retrofitting a plate column with packing for steam distillation of crude oil
In order to avoid thermal decomposition, the partial pressure of the hydrocarbon vapourmust not be allowed to exceed 34 mbar at the lower end of the intermediate zone in avacuum column (pT =100 mbar) designed for fractionating crude oil by steam rectification.The plates previously used (subscript 2) gave rise to a pressure drop of 110 mbar betweenthe head of the column and the intermediate zone. After the column had been retrofittedwith low-pressure-drop packing (subscript 1), the pressure drop was only 70 mbar. Howmuch steam has thus been saved?
Solution
According to Eqn (12-50), the pressure drop after retrofitting is given by
p -PT= {i - l) | -^£-] = 70 mbar
The pressure drop before retrofitting is given by
p -PT= (i - 1) ( - ^ - 1 = 110 mbar
According to Eqn (12-63), the factor k for the difference in pressure drop is given by
A n ol57Ap_\ ~ 70
nt /i
The relative savings in steam achieved by the installation of packing follows from Eqn(12-67), i.e.
AS 1.57-1^ T =
1 5 ? 100-34 = °-226
L 5 7 + ~1Q~
The volume flow rate of steam per unit flow rate of the hydrocarbon vapours after retro-fitting is obtained from Eqn (12-64), i. e.
Sj_= 100 + 7 0 - 3 4
V ~ 34
The corresponding flow rate before retrofitting is obtained from Eqn (12-65), i.e.
$2_ _ 100 + 110-34
V " 34 " 7
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14.9 Reduction in height effected by redistributors 363
Therefore, the savings in steam per unit flow rate of product vapour is as follows:
The same figure can be obtained from Eqn (12-66), i. e.
- ^ - = ^ • • 7 0 (1.57-1) = 1.17
The savings in steam achieved by substituting packing for fractionating plates can bedetermined from the values obtained for AS IV and S2/V, i. e.
100 = v ^ ~ 10° = 22-6 %5.17
14.9 Reduction in height effected by redistributors
It is intended to separate a mixture in a packed column with 30 transfer units, 35-mmmetallic Pall rings, and optimum inlet distribution. How much height would be saved byinstalling three redistributors?
Solution
Assume that the inlet effect and the data describing the efficiency are identical to thoselisted in Table 9.1. In this case, the effective height of the column without redistributors isgiven by Eqn (9-20), i.e.
(HT)nr = o = [30 - (0.5 + 1.9)] • y j ^ = 13 m
The relative reduction in the height of the bed can be read off against nr = 3 in the upperdiagram shown in Fig. 9.7. Thus,
AH/HT = 0.26
The absolute reduction in height is therefore
AH = 0.26 • 13 = 3.4 m
Hence, the total effective height of the bed with three redistributors is
(HT)nr = 3 = 13-3 .4 - 9.6 m
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364 14 Examples for the design of packed columns
and the effective height between the redistributors is
r 3 + 1
This value can also be read off on the axis of abscissae against nr = 3 in Fig. 9.7.
Assume further that the height occupied by a redistributor in the column is hr = 0.4 m.According to Eqn (9-50), the relative reduction in the total height would then be
F ^ - H = 1 - — [2.4 (3 + 1) + 3 • 0.4] = 0.169
The absolute saving in the height of the column shell is therefore
AHS = 0.169 • 13 = 2.2 m
14.10 Advantages of geometrically arranged packing in fractionating fatty acids
A modern plant for processing palm-kernel oil consists of nine distillation stages intandem. The plans for process optimization included the improvement of product quality byan appropriate choice of column internals. A Hermann Stage method was adopted forthe design.
Solution
Since the feedstock is sensitive to heat, the distillation stages were designed with struc-tured packing and falling-film evaporators. As a consequence, the residence times were veryshort, and the bottom temperatures were comparatively low (Fig. 14.5). The effects on theproduct quality were as follows.
The overhead product in each stage ranging from Column 2 (octanoic acid) to Column 5(myristic acid) is the pure distillate of the unhydrogenated crude acid concerned. Its purity ishigher than 99 % during continuous operation and even higher than 99.5 % for lauric (Ci2)and myristic (Ci4) acids. Owing to the unsaturated Cw fraction, the purity of the palmiticacid in the sidestream distillate of Column 8 is only just higher than 97 %. Since the linoleicacid fraction (Ci8») in the feed is higher than that of the oleic acid (Ci8), the sidestream dis-tillate in the next column downstream (Column 9) contains only 85% of oleic acid (Ci8H);the remainder consists of 8-10% of linoleic acid (Ci8») and 4 - 5 % of stearic acid. If thetrans-fatty acid content is less than 0.5%, the oleic acid content in this fraction is higherthan that attained in the Henkel or Alpha-Laval detergent processes.
The flow chart shown in Fig. 14.5 for a fatty acid fractionating plant includes the contin-uous cracker, the glycerol-water evaporation unit, and the pure glycerol distillation column.The two latter also incorporate a Hermann Stage process and the associated engineering.The entire plant has been on stream in Malaysia since 1992.
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14.11 Recovery of acetone from production effluents by liquid-liquid extraction 365
O
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14.11 Recovery of acetone from production effluents by liquid-liquid extraction
A production effluent containing 0.241 kmol/m3 of acetone is to be purified by liquid-liquid extraction with toluene. The packed column is to be designed and operated for 80 %acetone recovery. The intended packing is 25-mm Bialecki rings. The column diameter andthe effective height of bed should permit the effluent to flow at a rate of 1557 m3/h, thecolumn load may not exceed 35 % of that at the flood point, and the extractant is tolueneflowing at a rate of 2829 m3/h.
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366 14 Examples for the design of packed columns
Solution
Characteristic data for systematically stacked Bialecki rings (cf. Table 13.2)
E - 0.928 Cm = 0.67
/ = 0.050 m Ch = 2.151
Cd = 1
Physical properties of the system
QD = 857 kg/m3 QC =998 kg/m3
T|D = 0.55 • 10~3 kg/ms r|c - 0.893 • 10"3 kg/msDD = 2.478 • 10~9 m2/s Dc = 1.3 • 10"9 m2/sa = 0.026 kg/s2 m = 0.64
Phase flow ratio
uD 2.829= 1.82
uc 1.557
Dispersed phase holdup {Eqn (13-20)}
u (1-822 + 8 • 1.82)1/2-3 • 1.82h = 3 3
4 ( 1 - 1 . 8 2 ) ° ' 3 7 5 m / m
Droplet velocity {Eqn (13-3)}
9.81 (998 - 857) 0.026= 0.1096 m/s
9982
Velocity of continuous phase at the flood point {Eqn (13-16)}
uc,m = 0.67 • 0.928 • 0.1096 (1-0.375)2 (1 - 2 • 0.375) = 6.6 • 10~3 m3/m2s
Velocity of continuous phase under operating conditions
uc = 0.35 • 6.6 • 10~3 = 2.31 • 10'3 m3/m2s
Diameter of extraction column {Eqn (14-35)}
4_ 1.557/3600 _
JT ' 2.31 • 10~3 ~ >488 m
Velocity of dispersed phase under operating conditions
uD = 1.82 • 2.31 • 10~3 = 4.21 • 10"3 m3/m2s
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14.11 Recovery of acetone from production effluents by liquid-liquid extraction 367
Dispersed phase holdup under operating conditions {Eqn (13-11)}
" *
Mass transfer constant {Eqn (13-39)}
Cm i v °-o5°Height of overall transfer unit {Eqn (13-47)}
1 0.0265'8 1TJrTJJ
oc ~ 37.32 ' [9.81 (998-857)]3 /8 • 9981/4' 1.82
1 1 1 998 . „ „ A
+ = 1.14(1.3 • 10"9)1/2 0.64 (2.478 • 10~9)1/2 857
Concentration of continuous phase after extraction {Eqn (14-33)}
cCu = (1 - 0.8) 0.241 - 0.0482 kmol/m3
Concentration of solvent after extraction {Eqn (14-34)}
cD,o = ~rzr (0.241 - 0.0482) + 0 = 0.1059 kmol/m3
1.82
Driving concentration difference at the head of the column {Eqn (14-30)}
Aco = 0.241 -—^— • 0.1059 - 0.0755 kmol/m3
0.64
Driving concentration difference at the foot of the column {Eqn (14-31)}
Acu = 0.0482 -—|— • 0 - 0.0482 kmol/m3
0.64
Number of overall transfer units in the continuous phase {Eqn (14-29)}
0.241 - 0.0482 0.0755 ^ 1 ?oc ~ 0.0755 - 0.0482 ' n
Effective height of bed {Eqn (13-22)}
H = 1.14 • 3.17 = 3.61 m
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