Exercise 6.1
Subject:
Given:
Simultaneous absorption and stripping.
Fig. 6.1 in which simultaneous absorption and stripping occur.
Find: Whether absorption and stripping is more important.
Analysis:
Stripping of water = 22 kmol/h
Absorption of acetone = 10.35 kmol/h
Therefore, more stripping than absorption occurs.
However, the operation is primarily absorption because a very large percentage, 99.5%, of the
acetone is absorbed. Only a very small fraction of the water is stripped.
Exercise 6.2
Subject:
Given:
Column packings since 1950.
History of the development of column packings.
Find: Advantages of new packings. Advances in packing design and fabrication. Need for
structured packings.
Analysis: The newer packings provide more surface area for mass transfer, a higher flow
capacity, and a lower pressure drop. They provide more effective through flow. Many are
plastics made from molds or thin metal strips that can be inexpensively fabricated into intricate
shapes. Structured packings largely eliminate the problems of channeling and at the same time
give improvement in efficiency, capacity, and pressure drop.
Exercise 6.3
Subject:
Advantages of bubble-cap trays
Given: Characteristics and performance of bubble-cap trays
Find: Characteristics that give bubble-cap trays a very high turndown ratio.
Analysis: Unlike sieve and valve trays, bubble-cap trays do not allow liquid to weep.
Therefore, bubble-cap trays can be operated at very low liquid flow rates. Also, bubble caps
force the vapor to flow out sideways, rather than vertically up, thus allowing a relatively high
vapor rate.
Exercise 6.4
Subject:
Given:
Selection of alternative absorbent for Example 6.3
Flow rate, density, and MW of three potential absorbents listed below.
Find: Select the best absorbent with reasons why. Are any of the absorbents unacceptable?
Analysis: In Example 6.3, the rich gas contains C1 to nC6 hydrocarbons, with mostly C3. Object
of absorber is to absorb most of the nC4. The absorbent is an oil of 250 MW and 21oAPI, at a
flow rate, L, of 368 lbmol/h. An alternative absorbent must also be of higher molecular weight
than nC4 and must have a flow rate of at least 368 lbmol/h. If its molecular weight is too low, a
significant amount of it will be stripped. This can be judged by its K-value and stripping factor,
S=KV/L. The entering gas rate, V, is 946 lbmol/h. Column operating conditions are about 94
psia and a maximum temperature of 126oF.
Pertinent properties and factors for the three potential absorbents are:
Absorbent
C5 s
light oil
medium oil
gpm
115
36
215
ρ, lb/gal
5.24
6.0
6.2
lb/h
36,200
13,000
80,000
MW
72
130
180
lbmol/h
500
100
440
K-value
0.9
0.005
0.0005
S=KV/L
2.3
0.013
0.0013
The light oil can not be used because its flow rate of 100 lbmol/h is much lower than the
necessary 368 lbmol/h. The C5s can not be used because their stripping factor is very high. The
only possible alternative is the medium oil.
Exercise 6.5
Subject:
Stripping of VOCs from water effluents by air and water
Given: Packed tower for stripping
Find:
Advantages and disadvantages of air over steam
Analysis:
Advantages:
Air is available anywhere.
Air is inexpensive.
Disadvantages: Air can form a flammable or explosive mixture with the VOC.
With steam, the exit gas can be condensed and the VOC
recovered as a liquid.
Exercise 6.6
Subject:
Preferred operating conditions for absorbers and strippers.
Find: Best conditions of temperature and pressure, and the trade-off between number of stages
and flow rate of separating agent, using equations.
Analysis: Absorbers:
For high performance, want a large absorption factor, A = L/KV
Therefore, want a small K-value.
Assume that:
γ iL Pi s
Ki =
P
s
Pi = vapor pressure, which increases with increasing temperature
Therefore, operate at a high pressure and a low temperature.
Strippers:
For high performance, want a large stripping factor, S = KV/L
Therefore, want a large K-value.
Therefore, operate at low pressure and a high temperature.
For the tradeoff between number of equilibrium stages, N, and flow rate of mass separating
agent, L or V. Consider absorption. The fraction of a component absorbed is given from a
modification of Eq. (5-48),
A −1
A N +1 − A
1 − φ A = 1 − N +1
= N +1
A −1 A −1
Thus, a large fraction absorbed can be achieved with either N or A = L/KV. The tradeoff is most
clearly shown in Fig. 5.9. For low fractions absorbed (i.e. high φA), N has little effect and (1 φA) is approximately equal to A. But for high fractions absorbed, the larger the value of N, the
smaller the required value of A, and, thus, the smaller the required value of the flow rate of the
liquid separating agent, L.
The tradeoff for stripping is similar. The fraction of a component stripped is given by a
modification of Eq. (5-51),
S −1
S N +1 − S
1 − φ S = 1 − N +1
= N +1
S −1 S
−1
Thus, a large fraction stripped can be achieved with either N or S = KV/L. The tradeoff is most
clearly shown in Fig. 5.9. For low fractions stripped (i.e. high φS), N has little effect and (1 - φS)
is approximately equal to S. But for high fractions absorbed, the larger the value of N, the
smaller the required value of S, and, thus, the smaller the required value of the flow rate of the
vapor separating agent, V.
xercise 6.7
Subject:
Absorption of CO2 from air at 25oC by 5-N aqueous triethanolamine
Given: Feed gas containing 10 mol% CO2 and 90 mol% air. Absorbent of 5N aqueous
triethanolamine containing 0.04 moles of CO2 per mole of amine solution. Column with 6
equilibrium stages. Exit liquid to contain 78.4% of the CO2 in the feed gas. Therefore, exit gas
contains 21.6% of the CO2 in the entering gas. Equilibrium data for CO2 at 25oC in terms of
mole ratios.
Assumptions: Negligible absorption of air and stripping of amine and water.
Find: (a) Moles of amine solution required per mole of feed gas.
(b) Exit gas composition.
Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Therefore, for CO2,
X0 = 0.04 mol CO2/mol amine solution
YN+1 = Y7 = 10/90 = 0.1111 mol CO2/mol air
Y1 = 0.216(10)/90 = 0.024 mol CO2/mol air
(b) Therefore, the exit gas composition is 0.024 mol CO2/mol air or 0.024/(1 + 0.024) x 100% =
2.34 mol% CO2 and 97.66 mol% air.
(a) A plot of the equilibrium data as Y vs. X is given below. The operating point (X0, Y1) at the
top of the column is included. A straight operating line through this point is found by trial and
error to give 6 equilibrium stages, when using Y7 = 0.1111. The resulting XN = X6 = 0.085.
From Eq. (6-3), the slope of the operating line = L'/V' = (0.1111 - 0.024)/(0.085 - 0.04) = 1.936
mol triethanolamine solution/mol air. The feed gas contains 9 mol air/10 mol feed gas.
Therefore, mols of amine solution/mol feed gas = 1.936(0.9) = 1.74.
See plot on next page.
Exercise 6.7 (continued)
Exercise 6.8
Subject:
Absorption of acetone from air by water at 20oC and 101 kPa (760 torr) in a
valve-tray column.
Given: 100 kmol/h of feed gas containing 85 mol% air and 15 mol% acetone. Pure water is
the absorbent. Overall tray efficiency is 50%. Absorb 95% of the acetone. Equilibrium p-x data
for acetone are given as listed below.
Assumptions: Negligible absorption of air and stripping of water.
Find: (a) Minimum ratio, L'/V' of moles of water/mole of air.
(b) Number of equilibrium stages for L'/V' = 1.25 times minimum.
(c) Concentration of acetone in the exit water.
Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Then, the operating line
will be straight. For acetone,
X0 = 0.0 mol acetone/mol entering water
YN+1 = 0.15/0.85 = 0.1765 mol acetone/mol air in entering gas
Flow rate of acetone in exit gas = (1 - 0.95)(15) = 0.75 kmol/h. With 85 kmol/h of air,
Y1 = 0.75/85 = 0.00882 mol acetone/mol air
Convert the p-x equilibrium data to mole ratio, Y-X data, using y = p/P, Y = y/(1- y), X = x/(1-x)
p, torr
x
y
X
Y
30.0 0.033 0.0395 0.0341 0.0411
62.8 0.072 0.0826 0.0776 0.0901
85.4 0.117 0.1124 0.1325 0.1266
103.0 0.171 0.1355 0.2063 0.1568
(a) With the type of curvature in the Y-X equilibrium curve, shown below, the minimum
absorbent rate is determined by a straight operating line that passes through the point (Y1 , X0 )
and is drawn tangent to the equilibrium curve, as shown. From Eq. (6-3), the slope of the
operating line = L'/V' = 1.06 mol water/mol of air on an acetone-free basis = minimum ratio.
(b) For 1.25 times minimum, L'/V' = 1.25(1.06) = 1.325. Now a straight operating line that
passes through the point (Y1 = 0.00882 and X0 = 0.0) is drawn as shown below, where
equilibrium stages are stepped off, as in Figs. 6.11(a) and 6.12, until the steps reach or exceed the
required value of YN+1 = 0.1765. As seen, in the figure below, the determined number of
equilibrium stages is approximately 9.
(c) The concentration of acetone in the exit liquid, XN is the point on the operating line for Part
(b), where YN+1 = 0.1765. From Eq. (6-3), for the operating line for an absorber,
Y −Y
0.1765 − 0.00882
X N = N +1 1 + X 0 =
+ 0.0 = 0.127 mol acetone/mol water
L′
1.325
V′
or an exit mole fraction of acetone in the water of 0.127/(1.127) = 0.113
Analysis: (continued)
Exercise 6.8 (continued)
Exercise 6.9
Subject:
Absorption-stripping system for benzene (B) removal from gas by oil and
subsequent stripping of benzene from oil by steam.
Given: Plate column absorber and stripper. Feed gas containing 0.06 mol B/mol B-free gas.
Absorbent oil containing 0.01 mol B/mol B-free oil. Liquid leaving absorber contains 0.19 mol
B/mol B-free oil. 90% of B in the gas stream is absorbed. Liquid leaving stripper contains 0.01
mol B/mol oil. Flow ratio of B-free oil-to-B-free steam = 2.0. Molecular weights are 200, 78,
and 32 for oil, benzene, and gas, respectively. Equilibrium X-Y data given for absorber (25oC)
and stripper (110oC).
Assumptions: Gas is not soluble in oil and oil is not volatile.
Find: (a) Molar flow rate ratio of B-free oil to B-free gas in the absorber.
(b) Number of equilibrium stages in the absorber.
(c) Minimum steam flow rate in the stripper per mol of B-free oil.
Analysis: A flow diagram of the absorber-stripper system is shown in Fig. 5.10(a), except that
there is no makeup absorbent oil because the oil is assumed to be non-volatile.
(a) Consider just the absorber. Take a basis of V ' = 1 mol/h of B-free gas entering the absorber.
Because YN+1 = 0.06 mol B/mol B-free gas, we have 0.06 mol/h of B in entering gas. With 90%
absorption, 0.06(0.9) = 0.054 mol/h of B absorbed, leaving 0.006 mol/h of B in the exiting gas.
In the entering absorbent oil, X0 = 0.01 mol B/mol B-free oil. In the liquid leaving the absorber,
XN = 0.19 mol B/mol B-free oil. A material balance around the absorber on the benzene gives:
′ + X N L′
YN +1V ′ + X 0 L′ = YV
Therefore,
1
0.06(1.0)+0.01L′ = 0.006 + 0.19 L′
(1)
Solving Eq. (1), L' = 0.30 mol/h of B-free oil, giving L'/V’' = 0.30 mol B-free oil to B-free gas.
(b) Use a graphical method to determine the number of equilibrium stages in the absorber, using
Y-X coordinates to give a straight operating line. Data for the equilibrium curve of the absorber
are given and the operating line is obtained from terminal values of Y and X. The results are
given in the plot below. As seen, between 9 and 10 equilibrium stages are needed.
(c) The equilibrium curve for the stripper is shown in the plot below. With the type of curvature
in the Y-X equilibrium curve the minimum stripping gas rate is determined by a straight
operating line that passes through the operating point (Y0 = 0 , X1 = 0.01) at the bottom of the
column, as shown in Fig. 6.11(b) and is drawn tangent to the equilibrium curve, as shown. From
Eq. (6-5), the slope of the operating line = L'/V' = 3.0 mol B/mol of steam on a B-free basis.
Therefore,
(V’')min/L' = 0.33. This is compared to the given operating value of L'/V' = 2.0 or V'/L' = 0.5.
Analysis: (continued)
(a) Plot of results:
(b) Plot of results:
Exercise 6.9 (continued)
Exercise 6.10
Subject:
Stripping of benzene (B) from straw oil by steam at 1 atm (101.3 kPa).
Given: Oil enters stripper with 8 mol% benzene. 75% of B is stripped. Pure steam enters.
Steam exits with 3 mol% B. Sieve-plate column. Henry's law holds with a B partial pressure of
5.07 kPa when benzene mole fraction in the oil is 10 mol%.
Assumptions: Straw oil is not volatile and steam does not condense. Dalton's law.
Find: (a) Number of equilibrium stages required.
(b) Moles of steam required per 100 moles of benzene-oil feed to stripper.
(c) Number of equilibrium stages needed if 85% of B is stripped with same amount of
steam as in Part (b).
Analysis: With reference to the stripper in Fig. 6.11(b),
XN+1 = 8/92 = 0.087 mol B/mol B-free oil
Y0 = 0.0 mol B/mol B-free steam
YN = 3/97 = 0.0309 mol B/mol B-free steam
(b) For 100 moles of benzene-oil feed mixture, have 8 moles of benzene. Amount of benzene
stripped = 0.75(8) = 6 moles. Because the exit gas contains 0.0309 mol B/mol B-free steam, the
steam rate = 6/0.0309 = 194 moles.
Therefore, moles of steam/100 moles of benzene-oil feed = 194
Also, X1 = (8 - 6)/92 = 0.0217 mol B/mol B-free oil
(a) Assume Henry's law is given by Eq. (4-32), pB = HxB.. Because pB = 5.07 kPa when xB =
0.10, H = 5.07/0.1 = 50.7 kPa. Convert this to an a Y-X equilibrium equation. For a pressure of
101.3 kPa, assuming Dalton's law, yB = pB/P = 50.7xB/101.3 = 0.5xB . From Eq. (6-1), dropping
the benzene subscript, B, and noting that KB = 0.5,
KB = 0.5 =
Y / (1 + Y )
X / (1 + X )
Solving, Y =
0.5 X
1 + 0.5 X
(1)
The Y-X plot includes the equilibrium curve given by Eq. (1), and a straight operating line that
connects the column terminal points (Y0 = 0.0, X1 = 0.0217) and (YN = 0.0309, XN+1 = 0.087).
The corresponding slope of the operating line is (0.0309 - 0.0)/(0.087 - 0.0217) = L'/G' = 0.474.
From the plot below, stepping off stages as in Fig. 6.11(b), required N = approximately 3 stages.
(c) For 85% stripping of B, amount of B stripped = 0.85(8) = 6.8 moles B. Not stripped is1.2
moles B. Therefore, for the same B-oil and steam feeds, X1 = 1.2/92 = 0.01304 mol B/mol oil
and YN =6.8/194 = 0.03505 mol B/mol steam. The Y-X plot for this case is also given below,
where the equilibrium curve is the same as in Part (a), and a straight operating line connects the
column terminal points (Y0 = 0.0, X1 = 0.01304) and (YN = 0.03505, XN+1 = 0.087). The
corresponding slope of the operating line is (0.03505 - 0.0)/(0.087 - 0.01304) = L'/G' = 0.474
(same as in Part (a)). From the plot below, stepping off stages as in Fig. 6.11(b), required N =
between 5 and 6 stages.
Exercise 6.10 (continued)
Analysis: (continued)
(a) Plot of results:
(b) Plot of results:
Exercise 6.11
Subject:
Stripping of VOCs from water by air in a trayed tower to produce drinking water.
Given: 1,500 gpm of groundwater containing ppm amounts of DCA, TCE, and TCA given
below. Stripping at 1 atm and 25oC to reduce the ppm amounts to the very low levels below.
K-values of VOCs given below.
Assumptions: No stripping of water and no absorption of air. System is dilute with respect to
the VOCs.
Find: Minimum air flow rate in scfm (60oF and 1 atm).
Number of equilibrium stages for 2 times the minimum air flow.
Composition in ppm for each VOC in the resulting drinking water.
Analysis: Flow rate of water = L = (1,500 gpm)(8.33 lb/gal)/18.02 lb/lbmol = 693 lbmol/min
% stripping of a VOC = (inlet ppm - outlet ppm)/inlet ppm)
VOC
K-value Inlet ppm Outlet ppm % stripping
DCA
60
85
0.005
99.994
TCE
650
120
0.005
99.996
TCA
275
145
0.200
99.862
Because of its high % stripping and a K-value that is much lower than for the other two VOCs,
the stripper is likely to be controlled by DCA. So base the calculations on DCA and then check
to see that the % stripping for TCE and TCA exceed the above requirements. Because of the
dilute conditions, use Kremser's method.
693
L
From Eq. (6-12),
Vmin = (fraction stripped) =
(0.99994) = 11.55 lbmol./min
K
60
or minimum air flow rate = 11.55 (379 scf/lbmol) = 4,377 scfm at 60oF and 1 atm
For 2 times the minimum value, V = 2(11.55) = 23.1 lbmol/min.
K V 60(23.1)
The stripping factor, given by Eq. (6-16), is for DCA, S DCA = DCA =
= 2.0
L
693
S N +1 − S 2 N +1 − 2
From Eq. (6-14), Fraction stripped = 0.99994 = N +1
=
(1)
S
− 1 2 N +1 − 1
Solving Eq. (1), N = 13 stages.
Now, for V = 11.55 lbmol/min and N = 13 stages, compute the fraction stripped for TCE and
TCA from Eq. (6-14) with the stripping factor from Eq. (6-16). The results are:
VOC
S Fraction
stripped
DCA
2.0 0.99994
TCE
21.6 1.00000
TCA
9.17 1.00000
The drinking water contains 0.005 ppm of DCA and essentially zero ppm of TCE and TCA.
Exercise 6.12
Subject:
Stripping of a solution of SO2, butadienes, and butadiene sulfone with nitrogen.
Given: 120 lbmol/h of liquid containing in lbmol/h,10.0 SO2, 8.0 1,3-butadiene (B3), 2.0 1,2butadiene (B2), and 100.0 butadiene sulfone (BS). Liquid effluent to contain < 0.05 mol% SO2
and < 0.5 mol% (B3 + B2). Stripping agent is pure nitrogen. Stripping at 70oC and 30 psia. Kvalues are 6.95 for SO2, 3.01 for B2, 4.53 for B3, and 0.016 BS.
Assumptions: Negligible stripping of BS because of its low K-value. No absorption of N2.
Find: Flow rate of nitrogen. Number of equilibrium stages.
Analysis: Assuming no stripping of BS, the exiting liquid will contain 0.05 mol% SO2, 0.50
mol% (B3 + B2), and 99.45 mol% BS, or in lbmol/h, 100.0 BS, 0.0503 SO2, and 0.503 (B3 +
B2). Therefore, % stripping of SO2 = (10.0 - 0.0503)/10.0 = 0.99497 and % stripping of (B3 +
B2) = (8.0 + 2.0 - 0.503)/(8.0 + 2.0) = 0.9497.
The minimum stripping agent flow rate is given by Eq. (6-12). Assume that because the Kvalue of B2 is the lowest of SO2, B2, and B3, that B2 controls. Further assume that essentially
all of the B3 is stripped so that the 0.5 mol% of (B2 + B3) is entirely B2. The solute-free liquid
rate = flow rate of BS = L' = 100 lbmol/h and the fraction of B2 stripped = (2.0 - 0.503)/2.0 =
0.748
L'
100
'
(fraction stripped) =
(0.748) = 24.9 lbmol/h
Vmin =
KSO2
3.01
Now use this value to compute the values of fraction stripped of SO2 and B3.
KV ' 6.95(24.9)
=
> 1.0
100
L'
KV ' 4.53(24.9)
Fraction stripped of B3 = ' =
> 1.0
L
100
Fraction stripped of SO 2 =
Therefore, the assumption that B2 controls is correct at the minimum stripping gas rate.
Assume an operational nitrogen flow rate of 1.5 times minimum. Therefore V’ =1.5(24.9) = 37.4
lbmol./h. The stripping factor, given by Eq. (6-16), is for B2,
S B2 =
K B2V ' 3.01(37.4)
=
= 1.13
L'
100
Using Eq. (6-14), Fraction B2 stripped = 0.748 =
Solving, N = 2.4 equilibrium stages.
S N +1 − S 113
. N +1 − 113
.
=
N +1
N +1
S
−1
113
.
−1
Analysis: (continued)
Exercise 6.12 (continued)
Fraction B3 stripped =
S N +1 − S 1.7 2.4 +1 − 17
.
=
= 0.863
N +1
2 . 4 +1
−1
S −1
17
.
Fraction SO 2 stripped =
S N +1 − S 2.612.4 +1 − 2.61
=
= 0.936
S N +1 − 1
2.612.4 +1 − 1
These fractions do not meet specifications. Therefore, the number of stages and/or the nitrogen
flow rate must be increased.
First try increasing the number of stages.
S = KV/L
Fraction stripped:
N=5
N=7
N = 10
SO2
2.61
B2
1.13
B3
1.7
0.995
0.9993
0.99996
0.880
0.922
0.954
0.970
0.990
0.998
For N = 5 stages, have in the exit liquid in lbmol/h, 0.05 SO2, 0.24 B2, and 0.24 B3. These
values correspond to a total of 100.53 lbmol/h of exit liquid. This results in 0.05 mol% SO2 and
0.48% (B2 + B3). Therefore, N = 5 stages is sufficient with a feed gas rate, V' , of 37.4 lbmol/h.
Other combinations of N and V' could be used..
Exercise 6.13
Subject:
Absorption of a light hydrocarbon gas mixture by n-decane as a function of
pressure and number of equilibrium stages at 90oF.
Given: Light hydrocarbon gas containing in lbmol/h, 1,660 C1, 168 C2, 96 C3, 52 nC4, and 24
nC5 for a total of 2,000 lbmol/h = V. Absorbent of L = 500 lbmol/h of nC10. K-value of nC10 =
0.0011.
Find: Component flow rates in the lean gas and rich oil for:
(a) N = 6 and P = 75 psia.
(b) N = 3 and P = 150 psia.
(c) N = 6 and P = 150 psia.
Analysis: Use the Kremser method with Eqs. (5-48) for fraction not absorbed, φA, (5-50 for
fraction not stripped, φA ,(5-38) for absorption factor, A, (5-51) for stripping factor, S , (5-45) and
(5-53) for component flow rates in the exit gas, υ1 , and (5-44) and (5-52) for component flow
rates in the exit liquid, lN, since no light hydrocarbons enter with the absorbent. Use Fig. 2.8 or
other source for K-values of C1 through nC5. For nC10 , assume the K-value is inversely
proportional to pressure.
(a)
N = 6 and P = 75 psia:
Component
C1
C2
C3
nC4
nC5
nC10
Total
(b)
K-value
2.9
6.5
1.95
0.61
0.19
0.0011
A
0.0086
0.0385
0.1282
0.4098
1.3160
S
0.0044
φA
0.9914
0.9615
0.8718
0.5916
0.0541
φS
0.9956
υ1
l6
lbmol/h
14.0
6.5
12.3
21.2
22.7
497.8
574.5
υ1
l3
lbmol/h
30.0
12.0
22.6
33.2
22.9
498.9
619.6
lbmol/h
1,646.0
161.5
83.7
30.8
1.3
2.2
1,925.5
N = 3 and P = 150 psia:
Component
K-value
A
C1
C2
C3
nC4
nC5
nC10
Total
14.0
3.5
1.05
0.33
0.105
0.00055
0.0179
0.0714
0.2381
0.7576
2.3810
S
0.0022
φA
0.9821
0.9284
0.7644
0.3615
0.0443
φS
0.9978
lbmol/h
1,630.0
156.0
73.4
18.8
1.1
1.1
1,880.4
Exercise 6.13 (continued)
Analysis: (continued)
(c)
N = 6 and P = 150 psia:
Component
C1
C2
C3
nC4
nC5
nC10
Total
K-value
14.0
3.5
1.05
0.33
0.105
0.00055
A
0.0179
0.0714
0.2381
0.7576
2.3810
S
0.0022
φA
0.9821
0.9284
0.7619
0.2829
0.0032
φS
0.9978
υ1
lbmol/h
1,630.0
156.0
73.1
14.7
0.1
1.1
1,875.0
l6
lbmol/h
30.0
12.0
22.9
37.3
23.9
498.9
625.0
Exercise 6.14
Subject:
Absorption of a light hydrocarbon gas by nC10.
Given: 1,000 kmol/h of rich gas at 70oF with, in kmol/h, 250 C1, 150 C2, 200 C3, 200 nC4, and
150 nC5. 500 kmol/h of nC10 at 90oF. Absorber operates at 4 atm. K-value of nC10 at 80oF and
4 atm = 0.0014.
Assumptions: Using the Kremser method, use V = 1,000 kmol/h and L = 500 kmol/h, with an
average temperature of 80oF for K-values from Fig. 2.8 or other source.
Find: Percent absorption of each component for 4, 10, and 30 equilibrium stages.
Analysis: The K-values and absorption factors, from Eq. (6-15), A = L/KV, are:
Component
C1
C2
C3
nC4
nC5
K-value
38
7.6
2.25
0.64
0.195
A
0.0132
0.0658
0.2222
0.7813
2.564
For each component in the feed gas, the percent absorption is given by Eq. (6-13) x 100%,
A N +1 − A
Percent absorbed = N +1
× 100%
A −1
(1)
Using Eq. (1) with the values of A from the above table, the results are:
Component
C1
C2
C3
nC4
nC5
Percent absorbed:
for N = 4
for
1.32
6.58
22.03
65.14
96.30
N = 10
1.32
6.58
22.22
76.58
99.995
for N = 30
1.32
6.58
22.22
78.12
100.00
The number of stages affects only the heavier hydrocarbons that are absorbed to the greatest
extent.
Exercise 6.15
Subject: Back calculation of tray efficiency from performance data on absorption of propane.
Given: Large commercial absorber with Na = 16 actual plates. From Example 6.3, average
total liquid rate, L, is 446.7 lbmol/h, average total gas rate, V, is 866.5 lbmol/h. Propane in gas
feed = 213.8 lbmol/h. Propane in exit liquid = 43 lbmol/h. P = 93.7 psia. Average T = 111oF.
Find: Tray efficiency from performance data. Estimated tray efficiency from DrickamerBradford and O'Connell correlations.
Analysis: Fraction of propane absorbed = 43/213.8 = 0.20
From Fig. 2.8, for propane, K-value = 2.0
From Eq. (6-15), A = L/KV = 446.7/(2.0)(866.5) = 0.258
From Eq. (6-13),
A N +1 − A 0.258 N +1 − 0.258
Fraction absorbed = 0.20 = N +1
=
A −1
0.258 N +1 − 1
(1)
Solving Eq. (1), N = Nt = 0.93
From Eq. (6-1), Eo = Nt /Na = 0.93/16 = 0.058 or 5.8%
From Drickamer-Bradford correlation, Eq. (6-22), with µ L = 1.4 cP from Example 6.3,
Eo , % = 19.2 - 57.8 log µL = 19.2 - 57.8 log(1.4) = 10.8 %
From O'Connell correlation, Eq. (6-23), with ML = 250, ρL = 57.9 lb/ft3 from Example 6.3,
log E o = 1587
.
− 0.199 log C − 0.0896 log C
where, C =
2
KM L µ L
(2.0)(250)(1.4)
=
= 12.1
ρL
57.9
Therefore, from Eq. (2),
2
log E o = 1587
.
− 0.199 log 12.1 − 0.0896 log 12.1 = 1277
.
which gives Eo = 18.9%
The propane tray efficiency is lower than predicted.
(2)
Exercise 6.16
Subject:
Oil absorption of a gas to produce 95% hydrogen
Given: 800,000 scfm (0oC, 1 atm) of fuel gas containing 72.5% H2, 25% CH4, and 2.5% C2H6.
Absorbent of nC8. Absorber operates at 400 psia and 100oF. Exiting gas is to contain at least
80% of the entering H2 at a H2 purity of 95 mol%.
Assumptions: Ideal gas law.
Find: (a)
(b)
(c)
(d)
(e)
(f)
(g)
Minimum absorbent rate in gpm.
Actual absorbent rate at 1.5 times minimum.
Number of theoretical stages.
Stage efficiency of each component from O'Connell correlation.
Number of actual trays needed.
Composition of exit gas, accounting for stripping of octane.
Annual cost of lost octane.
Analysis: At 0oC and 1 atm, have 359 scf/lbmol. Therefore, entering V = 800,000/359 = 2,230
lbmol/min, with 0.725(2,230) = 1,616 lbmol/min H2, 558 lbmol/min CH4, 56 lbmol/min C2H6 .
From Fig. 2.8, at 400 psia, 100oF, K-values are: 30 for H2 , 7 for CH4 , and 1.75 for C2H6 .
(a) Assume the key component is CH4. If we neglect stripping of octane and assume
complete absorption of C2H6, with 20% absorption of H2, then exit gas will be 0.8(1,616) =
1,293 lbmol/min H2 and the exit liquid will contain 1,616 - 1,293 = 323 lbmol/min H2. For a
purity of 95 mol% H2, then will have 5 mol% CH4 or 1,293(5/95) = 68 lbmol/min of CH4 in the
exit gas and 558 - 68 = 490 lbmol/min of CH4 in exit liquid. Can not use Eq. (6-11) to compute
the minimum absorbent rate because we do not have a dilute system. Instead, for infinite stages,
assume the exiting rich oil is in equilibrium with the inlet gas. Then, for CH4, K = 7 = y/x. But y
for the inlet gas = 0.25. Therefore, x in the exiting oil = 0.25/7= 0.0357. Therefore, with 490
lbmol/min of CH4 in this oil, the total exiting oil = 490/0.0357 = 13,700 lbmol/min. Therefore,
the entering oil is 13,700 - 323 - 490 - 56 = 12,831 lbmol/min of nC8 = Lmin. For nC8 , MW =
114 and from Perry's Handbook, SG = 0.703. Therefore, liquid density = 0.703(8.33) = 5.86
lb/gal. Therefore, Lmin = 12,831(114)/5.86 = 250,000 gpm.
(b) For 1.5 times minimum, inlet absorbent rate = 19,200 lbmol/min = 375,000 gpm.
(c) Calculate the number of equilibrium stages from the Kremser equation based on CH4
absorption, inlet vapor and liquid flow rates. From Eq. (6-15),
ACH 4 = L / KV = 19,200 / (7)(2,230) = 123
.
From Eq. (6-13), for CH4, with a fraction absorbed = 490/558 = 0.878,
fraction of CH4 absorbed = 0.878 =
N +1
ACH
− ACH 4
4
N +1
ACH
−1
4
123
. N +1 − 123
.
=
N +1
1.23 − 1
(1)
Analysis: (c) (continued)
Exercise 6.16 (continued)
Solving Eq. (1), N =4.1 stages.
Check absorption of H2. Absorption factor = A = 19,200/(30)(2,230) =0.287
AHN2+1 − AH 2 0.287 4.1+1 − 0.287
From Eq. (6-13, fraction of H2 absorbed =
=
= 0.286
0.287 4.1+1 − 1
AHN2+1 − 1
This exceeds the minimum 20% specified. Actually, this problem is overspecified. The
absorber can be designed for 95 mol% purity of H2 in the exit gas or 20% absorption of H2. In
this problem, we can not achieve both specifications, unless the plate efficiency for H2 is much
less than that for CH4. Check this next.
(d) From the O'Connell correlation, Eq. (6-23), with ML = 114, ρL = 44 lb/ft3, and µL= 0.47 cP
from Perry's Handbook,
log E o = 1587
.
− 0.199 log C − 0.0896 log C
where, for H2, C =
2
(2)
KM L µ L
(30)(114)(0.47)
=
= 36.5
ρL
44
Therefore, from Eq. (2),
2
log E o = 1587
.
− 0.199 log 36.5 − 0.0896 log 36.5 = 1057
.
which gives Eo = 11.4 % for H2. Similarly, Eo = 21.1 % for CH4, and Eo = 32.5 % for C2H6.
(e) From Eq. (6-21), using CH4, Na = Nt /Eo = 4.1/0.211 = 19.4 or say 20 trays.
(f) For the stripping of nC8, from Fig. 2.8 by extrapolation, K = 0.003. From Eq. (2), Eo = 0.35.
Therefore, if 20 trays, have 20(0.35) = 7 equilibrium stages. From Eq. (6-16),
S = KV/L =0.003(2,230)/19,200 = 0.00035. From Eq. (6-14) for 7 stages,
S N +1 − S 0.000357 +1 − 0.00035
fraction stripped = N +1
=
= 0.00035
0.000357 +1 − 1
S −1
Therefore, 0.00035(19,200) = 7 lbmol/h of nC8 leaves in the exit gas.
For H2 , with 20 trays and Eo = 0.114, we have (0.114)(20) = 2.3 equilibrium stages. From Eq.
AHN2+1 − AH 2 0.287 2.3+1 − 0.287
(6-13), with A = 0.287, fraction H2 absorbed =
=
= 0.275
AHN2+1 − 1
0.287 2.3+1 − 1
Therefore, we can not meet the specification of less than 20% absorbed.
For C2H6, with 20 trays and Eo = 0.325, we have (0.325)(20) = 6.5 equilibrium stages.
From Eq. (6-13), with A = (19,200)/(1.75)(2,230) = 4.92,
A N +1 − A 4.92 6.5+1 − 4.92
fraction C2H6 absorbed =
=
= 0.999975
A N +1 − 1
4.92 6.5+1 − 1
Composition of exit gas in lbmol/min is 1,172 H2 , 68 CH4 , and 7 nC8 . H2 purity = 94%
(g) Cost of lost oil = 7(114)(60)(7,900)($1.00)/(5.86 lb/gal) = 64.6 million $/year.
Exercise 6.17
Subject:
Scale-up of absorber using Oldershaw column efficiency.
Given: Absorption operation of Examples 6.1 and 6.4 with a column diameter of 3 ft. New
column with 11.5 ft diameter. New tray with Oldershaw-column efficiency of 55% (EOV = 0.55).
Liquid flow path length from Fig. 6.16. Value of u/DE = 6 ft-1. From Example 6.4, λ = 0.68.
Assumptions: Straight operating line and equilibrium line as stated in Example 6.4
Find: Efficiencies EMV and Eo .
Analysis: Consider three cases: (a) complete mixing, (b) plug flow, and (c) partial mixing.
(a) From Eq. (6-31), EMV = EOV = 0.55. From Eq. (6-37),
log [1 + EMV (λ − 1) ] log [1 + 0.55(0.68 − 1) ]
=
Eo =
= 0.50
log λ
log 0.68
(b) From Eq. (6-32),
1
1
EMV = eλEOV − 1 =
( e(0.68)(0.55) − 1) = 0.67
λ
0.68
(
)
From Eq. (6-37),
log [1 + EMV (λ − 1) ] log [1 + 0.67(0.68 − 1) ]
=
Eo =
= 0.63
log λ
log 0.68
(c) To use Fig. 6.16, we need an estimate of the liquid flow rate. For the 3 ft diameter
column of Examples 6.1 and 6.4, liquid rate = 151.5 kmol/h of water. Assume the liquid rate is
proportional to the column cross-sectional area. Thus, for the 11 ft diameter, the liquid rate =
151.5(11/3)2 = 2,037 kmol/h = 162 gpm. From Fig. 6.16, a single-pass tower is adequate.
Assume liquid flow path = ZL = 7 ft. Eq. (6-34) is used to compute EMV, which requires the
Peclet number, given by Eq. (6-36),
Z u
N Pe = L = 7(6) = 42
DE
N
From Eq. (6-35), η = Pe
2
4 λEOV
1+
N Pe
1/ 2
−1 =
42
2
1+
4(0.68)(0.55)
42
1/ 2
− 1 = 0.37
From Eq. (6-34), EMV/EOV = 1.20. Therefore, EMV = 1.20(0.55) = 0.66
From Eq. (6-37),
log [1 + EMV (λ − 1) ] log [1 + 0.66(0.68 − 1)]
Eo =
= 0.62
=
log λ
log 0.68
From these results, we see that the column is operating closely to plug flow for the liquid.
Exercise 6.18
Subject: Estimation of column diameter based on conditions at bottom tray of a reboiled
stripper.
Given: Vapor and liquid conditions at bottom tray given in Fig. 6.48. Valve trays with 24-inch
spacing.
Find: Column diameter for 80% of flooding.
Analysis: Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
1/ 2
LM L ρG
FLV =
VM G ρ L
From Fig. 6.48, V = 546.2 lbmol/h and L = 621.3 lbmol/h
(1)
Average molecular weights of the gas and liquid are computed from M =
C
i =1
zi Mi ,
Component
M
y
My
x
Mx
Ethane
30.07 0.000006
0.0 0.000010
0.0
Propane
44.10 0.004817
0.2 0.001448
0.1
n-Butane
58.12 0.602573 35.0 0.391389 22.7
n-Pentane
72.15 0.325874 23.5 0.430599 31.1
n-Hexane
86.18 0.066730
5.8 0.176563 15.2
Total
64.5
69.1
Therefore, MV = 64.5 and ML = 69.1
Gas volumetric flow rate = QV = 6.192 ft3/s.
Therefore, ρV = VMV /QV = (546.2)(64.5)/(6.192)(3,600) = 1.58 lb/ft3
Liquid volumetric flow rate = 171.1 gpm = (171.1)(60)/7.48 = 1,372 ft3/h
Therefore, ρL = LML /QL = (621.3)(69.1)/1,372 = 31.3 lb/ft3
1/ 2
621.3(69.1) 158
.
From Eq. (1), FLV =
= 0.274
546.2(64.5) 31.3
Assume 24-inch tray spacing. From Fig. 6.24, CF = 0.24 ft/s
From relations below Eq. (6-44), since FLV is between 0.1 and 1.0,
F − 0.1
0.274 − 01
.
Ad
= 0.1 + LV
= 01
. +
= 0.119
A
9
9
From Perry's Handbook, surface tension = σ = 7 dyne/cm
From relation below Eq. (6-42), FST = (σ/20)0.2 = (7/20)0.2 = 0.8. FF = 1.0 , FHA = 1.0
From Eq. (6-42), C = FST FF FHA CF = (0.8)(1.0)(1.0)(0.24) = 0.192 ft/s
From Eq. (6-40), U f = C ( ρ L − ρV ) / ρV
1/ 2
= 0.192 ( 31.3 − 1.58 ) /1.58
1/ 2
= 0.833 ft/s
Exercise 6.18 (continued)
From Eq. (6-44),
4VM V
DT =
fU f π (1 − Ad / A ) ρV
1/ 2
4(546.2 / 3, 600)(64.5)
=
0.80(0.833)(3.14)(1 − 0.119)(1.58)
1/ 2
= 3. 56 ft
Exercise 6.19
Subject:
Estimation of column diameter based on conditions at the top tray of an absorber.
Given: Vapor and liquid conditions at the top tray. Valve trays with 24-inch spacing.
Find:
Flooding velocity and column diameter for 85% of flooding (f = 0.85)
Analysis: Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
FLV
LM L ρV
=
VM V ρ L
1/ 2
889(109) 1.924
=
530(26.6) 41.1
1/ 2
= 1.487
(1)
For 24-inch tray spacing, from Fig. 6.24, CF = 0.082 ft/s
From relations below Eq. (6-44), since FLV is < 0.1, Ad/A = 0.1
From relation below Eq. (6-42), FST = (σ/20)0.2 = (18.4/20)0.2 =0.98. FF = 0.75 , FHA = 1.0
From Eq. (6-42), C = FST FF FHA CF = (0.98)(0.75)(1.0)(0.082) = 0.0603 ft/s
From Eq. (6-40), U f = C ( ρ L − ρV ) / ρV
1/ 2
4VM V
From Eq. (6-44), DT =
fU f π (1 − Ad / A ) ρV
= 0.0603 ( 41.1 − 1.924 ) /1.924
1/ 2
1/ 2
= 0.272 ft/s
4(530 / 3, 600)(26.6)
=
0.85(0.272)(3.14)(1 − 0.1)(1.924)
1/ 2
= 3. 53 ft
Exercise 6.20
Subject:
Hydraulic calculations for bottom tray of a sieve-tray column
Given: Absorber of Exercise 6.16 with an absorbent flow rate of 40,000 gpm of nC8. Sieve
trays on 24-inch spacing , with a weir height of 2.5 inches and 1/4-inch holes. Foaming factor =
0.80 and fraction flooding = 0.70.
Find: (a)
(b)
(c)
(d)
(e)
(f)
Column diameter at bottom.
Vapor pressure drop/tray.
Whether weeping will occur.
Entrainment rate.
Fractional decrease in Murphree efficiency due to entrainment.
Froth height in downcomer.
Analysis: From Exercise 6.16, V = 800,000/359 = 2,230 lbmol/min of entering gas.
Average MW of gas = 0.725(2.016) + 0.25(16.04) + 0.025(30.05) = 6.22
Assuming ideal gas law for gas at 400 psia and 100oF,
ρV = PM/RT = (400)(6.22)/(10.73)(560) = 0.414 lb/ft3
For the liquid, neglect absorbed components. ρL = 43.9 lb/ft3 = 5.86 lb/gal
MW of nC8 = 114. Therefore, L = 40,000(5.86)/114 = 2,056 lbmol/min
(a) Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
FLV
LM L ρV
=
VM V ρ L
1/ 2
=
2, 056(114) 0.414
2, 230(6.22) 43.9
1/ 2
= 1.64
(1)
From Fig. 6.24, for 24-inch tray spacing, CF = 0.08. Because FLV > 1, Ad /A = 0.2.
FHA = 1.0, FF = 0.8, and since σ = 20 dynes/cm, FST = 1.0.
From Eq. (6-24), C = FSTFFFHACF = (1)(0.8)(1)(0.08) = 0.064 ft/s
From Eq. (6-40), U f = C ρ L − ρV / ρV
1/ 2
= 0.064 43.9 − 0.414 / 0.414
4VM V
From Eq. (6-44), DT =
fU f π (1 − Ad / A ) ρV
1/ 2
1/ 2
= 0.66 ft / s
4(2, 230 / 60)(6.22)
=
0.70(0.66)(3.14)(1 − 0.2)(0.414)
1/ 2
= 44 ft
This is a very large diameter. Would probably use two parallel columns of 31 ft each, but
continue with 44 ft diameter.
(b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ
(2)
From the continuity equation, m = uAρ , hole velocity for 10% hole area =
uo = (2,230/60)(6.22)/(3.14/4)(44)2(0.414)(0.1) = 3.68 ft/s and superficial velocity = 0.368 ft/s
uo2 ρV
3.682 0.414
From Eq. (6-50), hd = 0186
.
=
0186
.
= 0.045 in. of nC8
Co2 ρ L
0.732
43.9
Analysis: (continued)
Exercise 6.20 (continued)
Active bubbling area for Ad /A = 0.2 is Aa = A - 2Ad = 0.6 A. So, Ua = 0.368/0.6 = 0.61 ft/s
From Eq. (6-53), Ks = U a
ρV
ρ L − ρV
1/ 2
1/ 2
0.414
= 0.61
43.9 − 0.414
= 0.06 ft/s
From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.06)0.91] = 0.72
From Eq. (6-54), C = 0.362 + 0.317 exp( −35
. hw ) = 0.362 + 0.317 exp[ −3.5(2.5)] = 0.362
Take weir length, Lw = 0.73DT = 0.73(44) = 32.1 ft = 385 in.
2/3
qL
From Eq. (6-51), hl = φ e hw + C
Lw φ e
40,000
= 0.72 2.5 + 0.362
385 0.72
2/3
= 9.0 in. of nC8
From Eq. (6-55), with maximum bubble size of 1/4 inch = 0.00635 m,
hσ = 6σ / gρ L DBmax = 6(20 / 1000) / (9.8)(703)(0.00635) = 0.00274 m = 0.11 in. nC8
From Eq. (2), ht = hd + hl + hσ = 0.045 + 9.0 + 0.11 = 9.2 in. nC8 = 0.23 psi/tray. Excessive!
(c) Apply criterion of Eq. (6-64). hd + hσ = 0.045 + 0.11 = 0.155 in. < hl = 9.0 in.
Therefore, weeping will occur.
(d) From Fig. 6.28, since FLV = 1.64, entrainment will be very low.
(e) Because entrainment is very low, EMV will not decrease.
(f) From Eqs. (6-70) and (6-72), hdf =(ht + hl + hda) / 2
(3)
Area of downcomer opening = Ada = Lwha Take ha = 2 in., Ada = (32.1)(2/12) = 5.4 ft2
qL
From Eq. (6-71), hda = 0.03
100 Ada
2
40, 000
= 0.03
100(5.4)
2
=164 in.
Very excessive
Exercise 6.21
Subject: Column performance for 40% of flooding.
Given: Data in Examples 6.5, 6.6, and 6.7.
Find: (a) Column diameter in Example 6.5 for f = 0.4.
(b) Vapor pressure drop in Example 6.6 for f = 0.4.
(c) Murphree vapor-point efficiency in Example 6.7 for f = 0.4
Analysis: (a) Example 6.5:
In this example, a value of f = 0.80 was used, giving DT = 2.65 ft
From Eq. (6-44), by ratioing values of f,
0.80
DT = 2.65
0.40
1/ 2
= 3.75 ft = 1.15 m
(b) Example 6.6:
In this example, a tower diameter of 1 m gives a vapor pressure drop = 0.093 psi/tray,
with a vapor hole velocity of 47.9 ft/s, a weir length of 0.73 m, an active area vapor velocity of
5.99 ft/s, and Ks = 0.265 ft/s.
Vapor hole velocity varies inversely with the square of the column diameter. Therefore,
uo = 14.6(1/1.15)2 = 11.0 m/s = 36 ft/s
From Eq. (6-50), hd is directly proportional to hole velocity squared. Therefore,
hd = 1.56(36/47.9)2 = 0.88 in. of liquid
Weir length is proportional to column diameter. So, Lw = 0.73(1.15/1) = 0.84 m = 0.33 in.
Vapor velocity based on active area varies inversely with the square of column diameter.
Therefore, Ua = 5.99(1/1.15)2 = 4.53 ft/s.
From Eq. (6-53), Ks is proportional to Ua. Thus, Ks = 0.265(4.53/5.99) = 0.20 ft/s
From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.20)0.91] = 0.37
The value of C in Eq. (6-51) remains at 0.362.
From Eq. (6-51),
qL
hl = φ e hw + C
Lw φ e
2/3
12.9
= 0.37 2 + 0.362
33 0.37
2/3
= 0.88 in.
hσ = 0.36 in. (no change from Example 6.6)
From Eq. (6-49), ht = hd + hl + hσ = 0,88 + 0.88 + 0.36 = 2.12 in.
Tray pressure drop = htρL = 2.12(0.0356) = 0.076 psi/tray
(c) Example 6.7:
In this example, a tower diameter of 1 m gives a EOV = 0.77. Must redo all calculations.
DT = 1.15 m, A = 1.038 m2 = 10,380 cm2, Aa = 0.8(1.038) = 0.83 m2 = 8,300 cm2
Lw = 33 in. = 0.84 m, φe = 0.37, hl = 0.88 in. = 2.24 cm
Ua = 4.53 ft/s = 137 cm/s, Uf = 10.2 ft/s, f = Ua/Uf = 4.53/10.2 = 0.44
Analysis: (c) (continued)
Exercise 6.21 (continued)
F = UaρV0..5 = 1.37(1.92)0.5 = 1.90 (kg/m)0.5/s ,
From Eq. (6-64), t L =
From Eq. (6-65), t G =
qL = 812 cm3/s
hl Aa 2.24(10,380)
=
= 28.6 s
qL
812
1 − φ e hl (1 − 0.37)2.24
=
= 0.028 s
φ eU a
0.37(137)
From Eq. (6-67), k L a = 78.8 DL0.5 ( F + 0.425) = 78.8(181
. × 10−5 )(190
. + 0.425) = 0.78 s-1
From Eq. (6-66),
kG a =
1,030 DV0.5 f − 0.842 f 2
0.5
l
h
=
1,030(7.86 × 10 −2 ) 0.5 0.44 − 0.842 0.44
2.24
0.5
2
= 53.4 s-1
From Eq. (6-63), N L = k L at L = 0.78(28.6) = 22.3
From Eq. (6-62), N G = k G at G = 53.4(0.28) = 15
.
From Example 6.7, KV/L = 0.662
1
1
1
From Eq. (6-61), N OG =
=
=
= 144
.
1
KV / L
1 0.662 0.667 + 0.030
+
+
NG
NL
15
.
22.3
Mass transfer is controlled by the vapor phase.
From a rearrangement of Eq. (6-52), EOV = 1 − exp(− N OG ) = 1 − exp(−1.44) = 0.76 or 76%.
Exercise 6.22
Subject:
Sizing, hydraulics, and mass transfer for an acetone absorber
Given: Acetone absorber of Fig. 6.1, using sieve trays with 10% hole area, 3/16-inch holes, and
an 18-inch tray spacing. Foaming factor = 0.85.
Find: (a) Column diameter for 75% of flooding.
(b) Vapor pressure drop per tray.
(c) Number of transfer units.
(d) Number of overall transfer units.
(e) Controlling resistance to mass transfer.
(f) Murphree point vapor efficiency
If 30 trays are adequate.
Analysis: (a) Base column diameter on the top tray, where pressure is the lowest and gas flow
rate is the highest. Use data in Fig. 6.1. L = 1,943 kmol/h, ML = 18, ρL = 1,000 kg/m3
V = 6.9 + 144.3 + 536.0 + 22.0 + 0.1 = 709.3 kmol/h, MV = 28.8, P = 90 kPa, T = 25oC
From the ideal gas law, ρV = PM/RT = (90)(28.8)/(8.314)(298) = 1.05 kg/m3
Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
1/ 2
1/ 2
LM L ρV
1,943(18) 105
.
FLV =
=
= 0.056
VMV ρ L
709.2(28.8) 1,000
From Fig. 6.24, for 18-inch tray spacing, CF = 0.28 ft/s. Because FLV < 1, Ad /A = 0.1.
FHA = 1.0, FF = 0.85, and since σ = 70 dynes/cm, FST = (70/20)0.2 = 1.285
From Eq. (6-24), C = FSTFFFHACF = (1.285)(0.85)(1)(0.28) = 0.306 ft/s
From Eq. (6-40), U f = C ρ L − ρV / ρV
From Eq. (6-44),
4VM V
DT =
fU f π (1 − Ad / A ) ρV
1/ 2
1/ 2
= 0.306 1,000 − 105
. / 105
.
1/ 2
4(709.2 / 3, 600)(28.8)
=
0.75(9.44 / 3.28)(3.14)(1 − 0.1)(1.05)
= 9.44 ft / s
1/ 2
= 1.89 m = 6.2 ft
(b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ
(1)
From the continuity equation, m = uAρ , hole velocity for 10% hole area =
uo = (709.2/3,600)(28.8)/(3.14/4)(1.9)2(1.05)(0.1) = 19.0 m/s =62.3 ft/s
Superficial velocity = (0.1)(19.0) = 1.90 m/s = 6.23 ft/s
uo2 ρV
62.32 105
.
From Eq. (6-50), hd = 0186
.
=
0186
.
= 142
. in. of liquid
2
2
Co ρ L
0.73 1,000
Active bubbling area for Ad /A = 0.1 is Aa = A - 2Ad = 0.8 A. So, Ua = 6.23/0.8 = 7.8 ft/s
From Eq. (6-53), Ks = U a
ρV
ρ L − ρV
1/ 2
1.05
= 7.8
1,000 − 105
.
1/ 2
= 0.253 ft/s
From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.253)0.91] = 0.30
Assume a 2-inch weir height = hw
Exercise 6.22 (continued)
Analysis: (b) (continued)
. hw ) = 0.362 + 0.317 exp[ −35
. (2.0)] = 0.362
From Eq. (6-54), C = 0.362 + 0.317 exp( −35
Take weir length, Lw = 0.73DT = 0.73(6.2) = 4.52 ft = 54.3 in.
Volumetric liquid rate = qL = m/ρL = (1,943/60)(18)/(1,000)(0.003785 m3/gal) = 154 gpm
2/3
qL
From Eq. (6-51), hl = φ e hw + C
Lw φ e
154
= 0.30 2 + 0.362
54.3 0.30
2/3
= 1.09 in. of liquid
From Eq. (6-55), with maximum bubble size of 3/16 inch = 0.00476 m,
hσ = 6σ / gρ L DBmax = 6(70 / 1000) / (9.8)(1,000)(0.00476) = 0.009 m = 0.35 in. of liquid
From Eq. (2), ht = hd + hl + hσ =1.42 + 1.09 + 0.35 = 2.86 in. liquid = 0.103 psi/tray =
0.7 kPa/tray
(c) DT = 1.89 m, A = 2.804 m2 = 28,040 cm2, Aa = 0.8(2.804) = 2.24 m2 = 22,400 cm2
φe = 0.30, hl = 1.09 in. = 2.77 cm
Ua = 7.8 ft/s = 238 cm/s, Uf =9.44 ft/s, f = Ua/Uf = 7.8/9.44 = 0.826
F = UaρV0..5 = 2.38(1.05)0.5 = 2.44 (kg/m)0.5/s , qL = 154 gpm = 9,716 cm3/s
hA
2.77(22,400)
From Eq. (6-64), t L = l a =
= 6.39 s
qL
9,716
1 − φ e hl (1 − 0.30)2.77
=
= 0.027 s
φ eU a
0.30(238)
From the Wilke-Chang Eq. (3-39), DL = 1.12 x 10-5 cm2/s
. × 10−5 )(2.44 + 0.425) = 0.756 s-1
From Eq. (6-67), k L a = 78.8 DL0.5 ( F + 0.425) = 78.8(112
From Perry's Handbook, with a temperature correction based on Eq. (3-36), DV = 0.127 cm2/s
From Eq. (6-66),
From Eq. (6-65), t G =
kG a =
1,030 DV0.5 f − 0.842 f 2
hl0.5
=
1,030(0127
. ) 0.5 0.826 − 0.842 0.826
2.77 0.5
2
= 55.4 s-1
From Eq. (6-63), N L = k L at L = 0.756(6.39) = 4.83
From Eq. (6-62), N G = kG atG = 55.4(0.027) = 1.5
(d) From p. 271, for acetone, A = L/KV = 1.38. Therefore, KV/L = 1/1.38 = 0.725
1
1
1
From Eq. (6-61), N OG =
=
=
= 122
.
1
KV / L
1 0.725 0.667 + 0150
.
+
+
NG
NL
15
.
4.83
(e) Mass transfer is controlled by the vapor phase.
(f) From a rearrangement of Eq. (6-56), E OV = 1 − exp( − N OG ) = 1 − exp( −122
. ) = 0.705 or 70 %
Now, the separation requires 10 equilibrium stages. Assume the liquid on a tray is well mixed.
Then, from Eq. (6-31), EMV = EOV = 0.70. From below Eq. (6-33), take λ=KV/L=0.725.
log 1 + E MV ( λ − 1) log 1 + 0.70(0.725 − 1)
From Eq. (6-37), Eo =
=
= 0.66 (worst case)
log λ
log 0.725
Therefore, at most, we need from Eq. (6-21), Na =Nt /Eo = 10/0.66 = 15. Therefore, 30 okay.
Exercise 6.23
Subject: Design of a column to strip VOCs from water by air at 15 psia and 70oF.
Given: Conditions in Example 6.2 except that entering flow rates are twice as much.
Assumptions: Dilute system such that changes in vapor and liquid rates in the column are
negligible. Sieve trays on 24-inch spacing, with 10 % hole area of 3/16-inch holes, FF = 0.9,
FHA = 1.0 and f = 0.80.
Find: (a)
(b)
(c)
(d)
(e)
Number of equilibrium stages required.
Column diameter for sieve trays.
Vapor pressure drop per tray.
Murphree vapor-point efficiency from Chan-Fair method.
Number of actual trays.
Analysis: (a) Need 3 equilibrium stages, as determined in Example 6.2 and which is unchanged
when flow rates are doubled.
(b) liquid rate = 2(500) = 1,000 gpm or 1,000(60)(8.33)/18.02 = 27,800 lbmol/h = L
vapor rate = 2(3,400) = 6,800 scfm or 6,800(60)/379 = 1,077 lbmol/h = V
ρL = 62.4 lb/ft3. From ideal gas law, ρV = PM/RT = (15)(29)/(10.73)(530) = 0.0765 lb/ft3
Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
FLV
LM L ρV
=
VMV ρ L
1/ 2
27,800(18) 0.0765
=
1,077(29) 62.4
1/ 2
= 0.56
From Fig. 6.24, for 24-inch tray spacing, CF = 0.18 ft/s.
Because 0.1 < FLV > 1, from below Eq. (6-44), Ad /A = 0.1 + (FLV - 0.1)/9 = 0.15
FHA = 1.0, FF = 0.9, and since σ = 80 dynes/cm, FST = (80/20)0.2 = 1.32
From Eq. (6-24), C = FSTFFFHACF = (1.32)(0.9)(1)(0.18) = 0.21 ft/s
From Eq. (6-40), U f = C ρ L − ρV / ρV
From Eq. (6-44), for f = 0.80,
4VM V
DT =
fU f π (1 − Ad / A ) ρV
1/ 2
1/ 2
= 0.21 62.4 − 0.0765 / 0.0765
1/ 2
= 6.0 ft / s
4(1, 077 / 3, 600)(29)
=
0.80(6.0)(3.14)(1 − 0.15)(0.0765)
(c) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + hσ
From the continuity equation, m = uAρ . Hole velocity for 10% hole area =
uo = (1,077/3,600)(29)/(3.14/4)(6)2(0.0765)(0.1) = 40 ft/s
Superficial velocity = (0.1)(40) = 4 ft/s
1/ 2
= 6.0 ft
(1)
Exercise 6.23 (continued)
Analysis: (c) (continued)
uo2 ρV
402
0.0765
=
0186
.
= 0.68 in. of liquid
2
2
Co ρ L
0.73
62.4
Active bubbling area for Ad /A = 0.15 is Aa = A - 2Ad = 0.7 A. Therefore, Ua = 4/0.7 = 5.7 ft/s
From Eq. (6-50), hd = 0186
.
ρV
ρ L − ρV
From Eq. (6-53), Ks = U a
1/ 2
0.0765
= 5.7
62.4 − 0.0765
1/ 2
= 0.20 ft/s
From Eq. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.20)0.91] = 0.37
Assume a 2-inch weir height = hw
. hw ) = 0.362 + 0.317 exp[ −35
. (2.0)] = 0.362
From Eq. (6-54), C = 0.362 + 0.317 exp( −35
Take weir length, Lw = 0.73DT = 0.73(6)(12) = 53 in.
qL
From Eq. (6-51), hl = φe hw + C
Lw φe
2/3
1, 000
= 0.37 2 + 0.362
( 53) 0.37
2/3
= 2. 58 in. of liquid
From Eq. (6-55), with maximum bubble size of 3/16 inch = 0.00476 m,
hσ = 6σ / gρ L DBmax = 6(80 / 1,000) / (9.8)(1,000)(0.00476) = 0.0103 m = 0.40 in. of liquid
From Eq. (2), ht = hd + hl + hσ =0.68 + 2.58 + 0.40 = 3.66 in. liquid = 0.13 psi/tray
(d) DT = 6 ft, A = 28.3 ft2, Aa = 0.7(28.3) = 19.8 ft2 = 18,390 cm2
φe = 0.37, hl = 2.58 in. = 6.55 cm, ρG = 0.0765 lb/ft3 = 1.23 kg/m3
Ua = 5.7 ft/s = 1.74 m/s , Uf =6.0 ft/s, f = Ua/Uf = 5.7/6.0 = 0.95
F = UaρV0..5 =1.74(1.23)0.5 = 1.93 (kg/m)0.5/s , qL = 1,000 gpm = 63,100 cm3/s
hA
6.55(18,390)
From Eq. (6-64), t L = l a =
= 19
. s
qL
63,100
1 − φ e hl (1 − 0.37)6.55
=
= 0.064 s
φ eU a
0.37(174)
From the Wilke-Chang Eq. (3-39), DL = 0.96 x 10-5 cm2/s
From Eq. (6-67), k L a = 78.8 DL0.5 ( F + 0.425) = 78.8(0.96 × 10−5 )(193
. + 0.425) = 0.58 s-1
From Perry's Handbook, with T and P corrections based on Eq. (3-36), DV = 0.086 cm2/s
From Eq. (6-65), t G =
From Eq. (6-66),
kG a =
1, 030 DV0.5 ( f − 0.842 f 2 )
0.5
l
h
=
1, 030(0.086)0.5 0.95 − 0.842 ( 0.95 )
From Eq. (6-63), N L = k L at L = 0.58(1.9) = 1.1
6.55
0.5
2
= 2. 56 s-1
Analysis: (continued)
Exercise 6.23 (continued)
From Eq. (6-62), N G = kG atG = 2.56(0.064) = 0.16
(d) From Example 6.2, for benzene, S = KV/L = 9.89
1
1
1
From Eq. (6-61), N OG =
=
=
= 0.066
1
KV / L
1
9.89 6.25 + 8.99
+
+
NG
NL
016
.
11
.
From a rearrangement of Eq. (6-52), EOV = 1 − exp(− N OG ) = 1 − exp(−0.066) = 0.064 or 6.4%
(e) From Example 6.2, the separation requires 3 equilibrium stages.
If we assume the liquid on a tray is well mixed. Then, from Eq. (6-31), EMV = EOV = 0.064.
From Eq. (6-21), Na =Nt /Eo = 3/0.064 = 47 trays.
If we assume plug flow of liquid on a tray, then, if we take, below Eq. (6.33), λ=KV/L=9.89,
from Eq. (6-37),
log 1 + E MV ( λ − 1) log 1 + 0.064(9.89 − 1)
=
= 0197
.
log λ
log 9.89
Then we need from Eq. (6-21), Na =Nt /Eo = 3/0.197 = 15 trays. Sufficient information is not
given to establish the partial mixing prediction. Therefore, the number of trays required ranges
widely from 15 to 47.
Eo =
Exercise 6.24
Subject: Absorption of SO2 from air into water in an existing packed column.
Given: Feed gas flow rate of 0.062 kmol/s containing 1.6 mol% SO2. Absorbent is 2.2 kmol/s
of pure water. Packed column is 1.5 m2 in cross sectional area and packed with No. 2 plastic
super Intalox saddles to a 3.5-m height. Exit gas contains an SO2 mole fraction of 0.004.
Operating pressure is 1 atm. At operating temperature, equilibrium curve for SO2 is y = Kx =
40x
Assumptions: No stripping of water. No absorption of air.
Find: (a)
(b)
(c)
(d)
L/Lmin
NOG and Nt
HOG and HETP
KGa
Analysis: Compute material balance. SO2-free inlet air rate = 0.062(1-0.016) = 0.061 kmol/s
SO2 inlet rate in feed gas = 0.062(0.016) = 0.001 kmol/s = V'
SO2 outlet rate in gas = 0.061(0.004/0.996) = 0.00025 kmol/s
SO2 rate in outlet water = 0.001 - 0.00025 = 0.00075 kmol/s
Fraction absorbed = 0.00075/0.001 = 0.75 or 75%
(a) From Eq. (6-11), L'min = V'K(fraction absorbed) = 0.061(40)(0.75) = 1.83 kmol/s
Therefore, L/L'min = 2.2/1.83 = 1.20
0.887 N +1 − 0.887
(b) Take A = L/KV = 2.2/[(40)(0.062)] = 0.887, From Eq. (6-13), 0.75 =
0.887 N +1 − 1
Solving, Nt = 4. For NOG, use Eq. (6-89) with yin = 0.016, yout = 0.004, xin =0.0,
ln
N OG =
0.887 − 1 0.016
1
+
0.887
0.004 0.887
= 3.78
(0.887 − 1) / 0.887
(c) Given height of packing = 3.5 m = lT
From Eq. (6-73), HETP = lT /Nt = 3.5/4 = 0.875 m
From Eq. (6-89), HOG = lT /NOG = 3.5/3.78 = 0.926 m
V
K G aPS
3
= 0.045 kmol/s-m -atm
(d) From Table 6.7, HOG =
Therefore, K G a =
V
H OG PS
=
0.062
(0.926)(1)(1.5)
Exercise 6.25
Subject: Operating data for absorption of SO2 from air into water in a packed column.
Given: Column operates at 1 atm (760 torr) and 20oC. Solute-free water enters at 1,000 lb/h.
Mole ratio of water to air is 25. Liquid leaves with 0.6 lb SO2/100 lb of solute-free water.
Partial pressure of SO2 in exit gas is 23 torr (0.0303 atm). Equilibrium data are given as partial
pressures of SO2 in air as a function of lb SO2 dissolved/ 100 lb H2O.
Assumptions: No stripping of water. No absorption of air. Density of liquid taken as water.
Find: (a) % of SO2 absorbed.
(b) Concentration of SO2 in the liquid at the gas-liquid interface in lbmol/ft3 at a point
where the bulk liquid concentration is 0.001 lbmol SO2/lbmol of water and:
kL = 1.3 ft/h
kp = 0.195 lbmol/h-ft2-atm
Analysis: (a) Inlet water rate = 1,000/18.02 = 55.5 lbmol/h
Inlet air rate = water rate/25 = 55.5/25 = 2.22 lbmol/h
Partial pressure of air in exit gas = 760 - 23 = 737 torr
By partial pressure ratio, SO2 flow rate in exit gas = 2.22(23/737) = 0.0693 lbmol/h
Molecular weight of SO2 = 64.06
SO2 flow rate in exit liquid = 0.6(1,000/(100)(64.06) = 0.0937 lbmol/h
By material balance, SO2 flow rate in entering air = 0.0693 + 0.0937 = 0.1630 lbmol/h
Partial pressure of SO2 in entering gas = 0.1630/(0.1630 + 2.22) = 0.0690 atm
% of SO2 absorbed = 0.0937/0.1630 x 100% = 57.5%
(b)
At the point, the rate of mass transfer for SO2 as a flux across the gas-liquid interface, can
be written by the two-film theory as,
kp(pb - pi) = kL(ci - cb)
(1)
As shown in Fig. 6.31, the equilibrium interface composition can be determined in terms of the
ratio of mass-transfer coefficients. However, here, instead of compositions in mole fractions, the
gas composition is in partial pressure and the liquid is in concentration. From Eq. (1),
kL
1.3
p − pi
=
= 6.67 = b
k p 0.195
ci − cb
(2)
At the point (column height location), bulk liquid concentration = 0.001 lbmol SO2/lbmol H2O
and the flow rate is 0.001(55.5) = 0.0555 lbmol/h for SO2 in the liquid phase. The flow of SO2 in
the gas phase at that location is obtained by a material balance around the top of the column:
SO2 in entering liquid + SO2 in gas at the point = SO2 in liquid at the point + SO2 in exit gas
Therefore, 0 + SO2 flow rate in gas at the point = 0.0693 + 0.0555
Analysis: (b) (continued)
Exercise 6.25 (continued)
SO2 flow rate in gas at the point = 0.1248 lbmol/h. Therefore, the partial pressure of SO2 in the
bulk gas at the point = 0.1248/(0.1248 + 2.22) x 1 atm = 0.0532 atm = pb .
The density of water is 62.4 lb/ft3 or 62.4/18.02 = 3.46 lbmol/ft3. Therefore,
the concentration of SO2 in the bulk liquid at the point = 0.001(3.46) = 0.00346 lbmol/ft3 = cb .
We now need an algebraic equilibrium relationship between pi and ci .
Convert the given equilibrium data to partial pressures and concentrations in the vicinity of the
values at the point:
The SO2 concentration in the liquid is obtained from,
c,
lbmol SO 2
lb SO 2
=
3
100 lb H 2 O
ft
lb SO2 / 100 lb H2O
0.30
0.50
0.70
1.00
ci ,
lbmol SO2 / ft3
0.00292
0.00487
0.00682
0.00974
62.4
64.06
1
100
Partial pressure SO2 ,
torr
14.1
26.0
39.0
59.0
pi of SO2 ,
atm
0.01855
0.03421
0.05132
0.07763
These equilibrium data fit the curve, pi = 5.04536ci + 511.783ci2 - 21714.4ci3
(3)
From Eq. (2),
(4)
Solving Eqs. (3) and (4),
0.0532 - pi = 6.67(ci - 0.00346)
ci = 0.00550 lbmol/ft3 and pi = 0.0396 atm
The same result can be obtained by constructing a plot of SO2 partial pressure versus SO2
concentration in the liquid, similar to Fig. 6.31. First, the operating line is drawn as a straight
line connecting the column end points (0.0303, 0.0) and (0.0690, 0.00584), as (p , c). Then the
equilibrium curve is drawn, using Eq. (3). Then the point (pb , cb) is marked on the operating
line. A straight line is extended from this point, with a slope of (-kL/kp) = -6.67, to the point of
intersection on the equilibrium line, giving the same result as above
Exercise 6.26
Subject: Stripping of benzene from wastewater with air in a packed column.
Given: Column operation at 2 atm (1,520 torr) and 25oC. Wastewater enters at 600 gpm
containing 10 ppm by weight of benzene. Suggested air rate is 1,000 scfm (60oF and 1 atm).
Exit water to contain just 0.005 ppm benzene. Vapor pressure of benzene = 95.2 torr at 25oC.
Solubility of benzene in water = 0.180 g/100 g water at 25oC. Packing is 2-inch polypropylene
Flexirings. Mass transfer coefficients are: kLa = 0.067 s-1 and kGa = 0.80 s-1 (both driving
forces are in concentration units).
Assumptions: No stripping of water. No absorption of air.
Find: (a)
rate.
(b)
(c)
(d)
(e)
Minimum air stripping rate. Is it less than suggested? If not, use 1.4 times minimum
Stripping factor
NOG
KGa in units of s-1 and mol/m3-s-kPa, and which phase controls mass transfer
Volume of packing in m3
Analysis: Water flow rate = 600(8.33)(60)/18.02 = 16,640 lbmol/h
For benzene, xin = (10/1,000,000)(18.02/78) = 2.31 x 10-6
xout = (0.005/1,000,000)(18.02/78) = 1.16 x 10-9
Fraction of benzene stripped = 1 - (2.31 x 10-6/1.16 x 10-9) = 0.9995
(a) System is very dilute with respect to benzene, therefore use Kremser equation to determine
the minimum gas rate. From Eq. (6-12),
Vmin = L (fraction stripped)/ K
(1)
For the K-value, use Eq. (4) in Table 2.3, a modified Raoult's law, K = γLPs/P
(2)
To obtain the liquid-phase activity coefficient, use the given benzene solubility data. For liquidliquid equilibrium with respect to benzene, which is distributed between a water-rich phase (2)
and a benzene-rich phase (1), using Eq. (2-30), γ ( 2 ) = x (1) / x ( 2 ) γ (1) . But if the benzene-rich
phase is nearly pure benzene, γ(1) = 1 and x(1) = 1. Therefore, γ(2) = 1/x(2)
018
. (18.02)
From the given benzene solubility in water, x =
= 4.2 × 10 −4
100(78)
1
Therefore from Eq. (3), γ L =
= 2,380
4.2 × 10 − 4
(3)
From Eq. (2), K = (2,380)(95.2)/1,520 = 149
From Eq. (1), Vmin = 16,640(0.9995)/150 = 111.6 lbmol/h
At 60oF and 1 atm, there are 379 scf/lbmol. Therefore, Vmin = 111(379)/60 = 705 scfm.
This is less than the 1,000 scfm suggested by the expert. Therefore, use the value suggested by
the expert, rather than revise it.
Exercise 6.26 (continued)
Analysis: (continued)
(b) Operating V = 111.6(1,000/705) = 158.3 lbmol/h
From Eq. (5-51), S = KV/L = 149(158.3)/16,640 = 1.417 (a good value)
(c) For NOG, use Eq. (6-93), with A = 1/S = 1/1.417 = 0.7057
Benzene stripped = Linxin(fraction stripped) = 16,640(2.31 x 10-6)(0.9995) = 0.0384 lbmol/h
Therefore, yout = 0.0384/158.3 = 0.0002427. Also, xin = 2.31 x 10-6 and yin = 0.0. Therefore,
ln
N OG =
0.7057 − 1
0.7057
0.0 − 149(2.31× 10−6 )
1
+
−6
0.0002427 − 149(2.31× 10 )
0.7057
= 14.2
(0.7057 − 1) / 0.7057
(d) Note that both kL and kG are given for concentration driving forces. Therefore, we write the
rate of mass transfer of benzene as:
r = kG acG ( yb − yi ) = k L acL ( xi − xb ) = K G acG ( yb − y* )
(4)
where cG and cL are total gas and liquid concentrations, respectively, and y* is the vapor mole
fraction in equilibrium with the bulk liquid mole fraction, as given by y* = Kxb .
Solving Eq. (4) for the driving vapor and liquid phase driving forces,
Adding Eqs. (5) and (6) to eliminate yi and solving for r and equating to the last term in Eq. (4),
r=
yb − y*
=
1
K
+
kG acG k L acL
Solving Eq. (7) for KGa,
KG a =
yb − y *
1
K G acG
1
KcG
1
+
kG a k L acL
The total gas concentration is obtained from the ideal gas law,
(7)
(8)
Analysis: (d) continued)
xercise 6.26 (continued)
cG =
P
2(101.3)
=
= 81.8 mol/m3
−3
RT 8.314 × 10 (298)
ρ L 106 g / m3
=
= 55,500 mol/m3
ML
18.02
1
1
KG a =
=
= 0.22 s-1
1
150(81.8)
1.25
+
3.29
+
0.80 0.067(55,500)
This is with a concentration driving force.
cL =
The liquid phase controls with a relative resistance of 3.29 compared to 1.25 for the gas phase.
To obtain KGa with a partial pressure driving force, we write the rate of mass transfer as,
r = K G acG ( yb − y* ) = ( K G ) p aP ( yb − y* )
(9)
where KGa is with the concentration driving force = 0.22 s-1 and (KG)pa is with the partial
pressure driving force. Solving Eq. (9), and applying the ideal gas law in the form, cG=P/RT,
(K G ) p a = K G a
cG K G a
0.22
=
=
= 0.089 mol/kPa-m3-s
P
RT (8.314 × 10−3 )(298)
(e) From Table 6.7, with a partial pressure driving force,
H OG =
G
(K G ) p aPS
However, the cross sectional area of the tower, S, is not known. Therefore, can not compute HOG
From Eq. (6-89), lT = HOGNOG . Therefore, can not compute the height because HOG is
unknown. However, we can compute the packed volume.
Packed volume = lTS = NOGHOGS = N OG
G
(149)(454)
= (14.2)
= 14.8 m3
(K G ) p aP
0.089(3, 600)(2)(101.3)
Exercise 6.27
Subject: Absorption of GeCl4 from air into dilute caustic solution in an existing packed column.
Given: Column operates at 25oC (77oF) and 1 atm. Gas enters at 23,850 kg/day containing 288
kg/day of GeCl4 and 540 kg/day of Cl2. Dissolved GeCl4 and Cl2 react with the caustic so that
neither has a vapor pressure. Packed tower is 2-ft diameter with 10 ft of 1/2-inch ceramic
Raschig ring packing of given characteristics. Liquid rate is to give 75% of flooding. Equation
is given for estimating Kya.
Assumptions: No stripping of water. No absorption of air.
Find: (a) Entering dilute caustic flow rate.
(b) Required packed height. Which controls, GeCl4 or Cl2? Is 10-ft height adequate?
(c) % absorption of GeCl4 and of Cl2 for 10 ft of packing. If necessary, select an
alternative packing.
Analysis: (a) Determine entering dilute caustic flow rate from the 75% of flooding
specification, using the flooding curve of Fig. 6.36(a), with the correction factors of Figs.
6.36(b,c). The air rate in the entering gas = 23,850 - 288 - 540 = 23,022 kg/day.
MW of Cl2 = 71. MW of GeCl4 = 214.6. MW of air = 29.
23,022 288 540
Molar gas rate = V =
+
+
= 803 kmol/day or 73.7 lbmol/h or 2189 lb/h
29
214.6 71
Average molecular weight of gas = 23,850/803 = 29.7
Tower cross sectional area for 2-ft diameter = 3.14(2)2/4 = 3.14 ft2
The continuity equation for flow through the tower based on the gas superficial velocity is,
m = uoSρ. Therefore, the superficial velocity is, uo = m/Sρ
(1)
3
From the ideal gas law, ρ = PM/RT = (1)(29.7)/(0.7302)(460+77) = 0.076 lb/ft
Therefore, from Eq. (1), uo = (2189)/(3.14)(0.076) = 9,170 ft/h or 2.55 ft/s.
From Eq. (6-102), noting that for the dilute caustic solution at 25oF, f{ρL} = 1 and f{µL} = 0.98
for a liquid viscosity of 0.95 cP at 25oC,
Y=
uo2 FP ρV
g ρH 2O
f {ρ L }f {µ L }=
(2.55) 2 (580) 0.076
(1.0)(0.98)=0.14
32.2
62.4
(uo)flood = uo/0.75 =2.55/0.75 = 3.40 ft/s.
Therefore, Yflood = 0.14/(0.75)2 = 0.25
LM L
From Fig. 6.36,
VM V
ρ
LM L = X (GM G ) L
ρG
1/ 2
62.4
= 0.009(2,191)
0.076
ρV
ρL
1/ 2
= 0.009
1/ 2
= 565 lb/h or 0.071 kg/s of entering liquid
Exercise 6.27 (continued)
Analysis:
(b) Assume that GeCl4 is the controlling species, with mass transfer controlled in the gas phase.
From Table 6.7, column height = lT = HGNG where, HG = V/kyaS
(1)
(1 − y ) LM dy
(1 − y )( y − yI )
and, N G =
(2)
At the interface, yI = 0 because it is given that dissolved GeCl4 has no vapor pressure
In entering gas, y = (288/214.6)/803 = 0.00162. Therefore (1 - y)bottom = (1 - 0.00162) = 0.9984
In exiting gas, y = 0.01 of y in entering gas = 0.0000162. Therefore (1 - y)top = 1.0000
Therefore, (1 - y)LM = 0.9992. Also, on the average, (1 - y) is approximately 0.9992.
Therefore, Eq. (2) for NG simplifies to,
NG =
0.00162
dy
0.00162
= ln
= 4.61
y
0.0000162
0.0000162
V
DPV ′
Equation given for ky is, k y =
1.195
S
µ (1 − ε o )
−0.36
( NSc )
−2 / 3
(3)
V = 803/24(3,600) = 0.0093 kmol/s
S = 3.14 ft2 or 0.292 m2
V' = (V/S)MG = (0.0093/0.292)(29.7) = 0.944 kg/s-m2
DP = 0.01774 m
At 25oC, from Perry's Handbook, for air, µ = 0.018 cP or 1.8 x 10-5 kg/m-s
εο = ε - hL
ε = 0.63
hL = 0.03591(L')0.331
(4)
L' =(LML) /S= 0.071/0.292 = 0.243 kg/s-m2 From Eq. (4), hL = 0.0359(0.243)0.331 = 0.0225
Therefore, εo = 0.63 - 0.0225 = 0.608 NSc = µ/ρDGeCl4
(5)
3
3
From above, ρ of the gas = 0.076 lb/ft = 1.21 kg/m
DGeCl4 = 6 x 10-6 m2/s
From Eq. (5), NSc = (1.8 x 10-5)/(1.21)(6 x 10-6) = 2.48
0.0093
(0.01774)(0.944)
From Eq. (3), k y =
1195
.
0.292
(18
. × 10 −5 ) 1 − 0.608
Equation given for interfacial area is,
Equation given for exponent,
Therefore, from Eq. (6), a =
−0.36
2.48
−2 / 3
14.69(808 V ′ / ρ1/ 2 ) n
a=
0.111
( L′ )
= 126
. × 10 −3 kmol/s-m2
(6)
n = 0.01114 L' + 0.148 = 0.01114(0.243) + 0.148 = 0.151
(808)(0.944)
14.69
(121
. )1/ 2
0.243
0.111
0.151
= 46.2 m2/m3
Exercise 6.27 (continued)
Analysis: (b) (continued)
Kya = kya = (1.26 x 10-3)(46.2) = 0.058 kmol/s-m3
From Eq. (1), HG = HOG = (V/S)/kya = (0.0093/0.292)/0.058 = 0.55 m or 1.8 ft
Packed height = HGNG = 0.55(4.61) = 2.53 m or 8.3 ft.
Therefore, a 10 foot height is sufficient.
Now check the assumption that GeCl4 controls. Because gas is dilute in Cl2 and 99% of it is
absorbed, NG is the same as for GeCl4. Note that in Eq. (3),
k y is proportional to ( N Sc ) −2 / 3 or Di2 / 3 . Since the given gas diffusivity of Cl2 is about twice that
of GeCl4, the mass transfer coefficient for Cl2 is higher and the corresponding HG is lower.
Therefore, the assumption that GeCl4 controls is correct.
(c) If a packed height of 10 ft rather than the computed 8.3 ft is used, then for GeCl4 ,
NG = lT /HG = 10/1.8 = 5.55 rather than 4.61.
y
From above, N G = ln in = 5.55
yout
y
Solving, in = 258
Therefore, yout = 0.00162/258 = 0.0000063
yout
% absorption of GeCl4 = (1 - 0.0000063/0.00162) = 0.996 or 99.6% compared to 99% specified
For Cl2, from above, HG = HG of GeCl4 (DGeCl4/DCl2)2/3 = 1.8(0.000006/0.000013)2/3 = 1.08
Therefore, NG = 10/1.08 = 9.26 = = ln
yin
yout
Solving,
yin
= 10,500
yout
% absorption of Cl2 = (1 - 1/10,500) = 0.9999 or 99.99% compared to 99% specified
Exercise 6.28
Subject: Packed tower diameter and pressure drop for conditions of Exercise 6.26.
Given: Suggested tower diameter of 0.80 m and pressure drop of 500 N/m2-m of packed height
(0.612 in. H2O/ft).
Find: (a)
(b)
(c)
(d)
Fraction of flooding if FP = 24 ft2/ft3.
Pressure drop at flooding.
Pressure drop at operating conditions using GPDC chart.
Pressure drop at operating conditions using Billet-Schultes correlation.
Analysis: (a) Using data from Exercise 6.26 and Fig. 6.36,
LML = 600(8.33) = 5,000 lb/min
VMV = (1,000/379)(29) = 76.5 lb/min
From ideal gas law, ρV = PM/RT = (2)(29)/(0.7302)(460+77) = 0.148 lb/ft3
ρL =62.3 lb/ft3
1/ 2
1/ 2
LM L ρV
5, 000 0.148
From Fig. 6.36, X =
=
= 3.19
VM V ρ L
76.5
62.3
From Fig. 6.36, at flooding, Y = 0.0053
For the actual operation, by the continuity equation, m = uoSρ
Area for flow = S = πD2/4 = (3.14)(0.8/0.3048)2/4 = 5.41 ft2
Superficial velocity = uo = m/Sρ = (76.5/60)/(5.41)(0.148) = 1.59 ft/s
From Eq. (6-102), noting that for the dilute aqueous solution at 25oF, f{ρL} = 1 and f{µL} = 0.98
for a liquid viscosity of 0.95 cP at 25oC,
uo2 FP ρV
Y=
g ρH 2O
(1.59) 2 (24) 0.148
(1.0)(0.98)=0.0044
f {ρ L }f {µ L }=
32.2
62.3
1/ 2
u
Y
0.0044
Fraction of flooding = o =
=
uflood
Yflood
0.0053
This is too high. Should increase tower diameter.
1/ 2
= 0.91 or 91%
(b) From Eq. (6-104), ∆Pflood = 0.115 FP0.7 = 0.115(24)0.7 = 1.06 in. H2O/ft of packed height
(c) From Fig. 6.36, ∆P at design = 1.50 in. H2O/ ft of packed height (seems high)
(d) From Table 6.8, for 2-inch (50-mm) plastic Pall rings, the packing characteristics are:
a = 111.1 m2/m3 or 33.9 ft2/ft3 , ε = 0.919 m3/m3 , Ch = 0.593 , Cp = 0.698, Cs = 2.816
The pressure drop per unit height of packed bed is given by Eq. (6-115),
∆P
ε
=
∆Po
ε − hL
3/ 2
exp
1/ 2
13300
N
3/ 2 ( FrL )
a
(1)
Exercise 6.28 (continued)
Analysis: (d) (continued)
where, hL is given by (6-97) and is in m2/m3 and N FrL is given by (6-99).
From Eq. (6-110),
∆Po
a u2ρ 1
= Ψo 3 V V
lT
ε 2 KW
(2)
1− ε
1 − 0.919
=6
= 0.0143 ft
a
33.9
1
2 1 DP
2
1
0.0143
From Eq. (6-111),
= 1+
= 1+
= 1045
.
KW
3 1 − ε DT
3 1 − 0.919 2.62
Therefore, KW = 0.957 .
From Eq. (6-112), DP = 6
From Perry's Handbook, µ for air at 25oC = 0.018 cP, and uo = uV = 1.59 ft/s
From Eq. (6-114), N ReV =
uV DP ρV
(1.59)(0.0143)(0.148)(0.957)
KW =
= 3, 287
(1 − ε ) µV
(1 − 0.919 ) [(0.018)(0.000672)]
From Eq. (6-113), Ψ o = C p
64
1.8
64
1.8
+ 0.08 = 0.698
+
= 0.671
3, 287 3, 287 0.08
N ReV N ReV
From Eq. (2), with the introduction of gc because of the use of American
Engineering units,
∆Po
a u2ρ 1
33.9 1.592 (0.148)
1
= Ψo 3 V V
= 0.671
= 0.164 lbf/ft3 or 0.00114 psi/ft
3
lT
ε 2 g c KW
0.919
2(32.2)
0.957
From Eq. (6-97), hL = 12
N FrL
N Re L
1/ 3
ah
a
2/3
Superficial liquid velocity = uL = m/SρL = (5,000/60)/(5.41)(62.4) = 0.247 ft/s
From Perry's Handbook, liquid viscosity = 0.95 cP
u L2 a (0.247) 2 (33.9)
From Eq. (6-99), N FrL =
=
= 0.064
g
32.2
uρ
(0.247)(62.4)
From Eq. (6-98), N Re L = L L =
= 712
aµ L (33.9)(0.95)(0.000672)
a
0.25
From Eq. (6-101), h = 0.85Ch N Re
N Fr0.1L = 0.85(0.593)(712) 0.25 (0.247) 0.1 = 2.27
L
a
(3)
Exercise 6.28 (continued)
Analysis: (d) (continued)
0.064
From Eq. (3), hL = 12
712
1/ 3
2.27 2 / 3 = 0.177 m3/m3 or ft3/ft3
From Eq. (1),
∆P
ε
=
∆Po
ε − hL
3/ 2
exp
1/ 2
13300
N
3/ 2 ( FrL )
a
∆P
0.919
= ( 0.00114 )
lT
0.919 − 0.103
3/ 2
exp
13300
1/ 2
0.064 )
= 0.0241 psi/ft or 0.669 in. H2O/ft
3/ 2 (
111.1
This is only 10% higher than the suggested value of 0.612 in. H2O/ft
Exercise 6.29
Subject: Mass-transfer coefficients for conditions of Exercise 6.26.
Given: Suggested mass-transfer coefficients in Exercise 6.26.
Find: Mass-transfer coefficients kLa and kGa from Billet-Schultes correlation.
Analysis: Coefficient kLa is obtained from a rearrangement of the left-most part of Eq. (6-131),
uL
kL a =
(1)
H L ( aPh / a )
with HL from Eq. (6-132) and (aPh /a) from (6-136).
From Table 6.8, CL = 1.239 , ε = 0.919, and a = 111.1 m2/m3. From the results of Exercise 6.28,
come the following parameters and properties:
hL = 0.177 ft2/ft3, uL = 0.247 ft/s, ρL = 62.4 lb/ft3, µL = 0.95 cP
Take the surface tension, σ = 0.005 lbf/ft or 0.005(32.2) = 0.161 lb/s2
To compute (aPh /a), we need for the liquid phase, the Reynolds, Weber, and Froude numbers
from equations (6-138), (6-139), and (6-140), respectively, based on the packing hydraulic
diameter from (6-137).
d h = packing hydraulic diameter = 4
From Exercise 6.28,
Reynolds number = N ReL ,h =
ε
0.919
=4
= 0.0331 m = 0.108 ft
a
111.1
uL d h ρ L (0.247)(0.108)(62.4)
=
= 2, 607
[(0.95)(0.000672)]
µL
Weber number = N WeL ,h
( 0.247 ) (62.4)(0.108) = 2.55
u 2ρ d
= L L h =
σ
0.161
Froude number = N FrL , h
( 0.247 ) = 0.0175
u2
= L =
gd h 32.2(0.108)
2
2
From (6-136),
(
aPh
−1/ 2
= 1.5 ( ad h )
N Re L ,h
a
) (N
−0.2
= 1.5 [ (111.1)(0.0331) ]
−1/ 2
) (N )
0.75
We L , h
−0.45
FrL , h
( 2607 ) ( 2.55 ) ( 0.0175 )
−0.2
0.75
−0.45
= 2.02
Exercise 6.29 (continued)
Analysis: (continued)
From Exercise 6.23, correcting for temperature, DL = 9.6 x 10-6(298/294) = 9.73 x 10-6 cm2/s or
1.05 x 10-8 ft2/s
From (6-132), using American Engineering Units,
1 1
HL =
CL 12
1/ 6
1
1
=
1.239 12
From (1), k L a =
4hL ε
DL au L
1/ 6
1/ 2
uL a
a a ph
4(0.177)(0.919)
(1.05 × 10−8 )(33.9)(0.247)
1/ 2
0.247 1
= 5.23 ft
33.9 2.02
uL
0.247
=
= 0.0234 s-1
H L ( aPh / a ) 5.23 ( 2.02 )
This is 35% of the suggested value of 0.067 s-1
Coefficient kGa is obtained from the gas-phase analog of (1), kG a =
uV
H G ( aPh / a )
(2)
with HG obtained from (6-133), using dimensionless groups from (6-134) and (6-135).
From Exercise 6.28, uV = 1.59 ft/s, ρV = 0.148 lb/ft3, µV = 0.018 cP = 1.21 x 10-5 lb/ft-s
From Eq. (6-134), N ReV =
uV ρV
(1.59)(0.148)
=
= 574
(33.9)(1.21× 10−5 )
aµV
From Exercise 6.23, the diffusivity of benzene in air = DV = 0.086 cm2/s at 70oF and 15 psia.
Using Eq. (3-36) to correct for temperature and pressure,
DV = 0.086
15
(2)(14.7)
From Eq. (6-135), NScV =
77 + 460
70 + 460
1.75
= 0.045 cm2/s = 4.84 x 10-5 ft2/s
µV
1.21× 10−5
=
= 1.69
ρV DV (0.148)(4.84 × 10 −5 )
Exercise 6.29 (continued)
Analysis: (continued)
From Eq. (6-133)
1
1/ 2 4ε
HG =
( ε − hL )
CV
a4
1
4(0.919)
( 0.919 − 0.177 )1/ 2
0.368
33.94
From (2),
kG a =
1/ 2
1/ 2
(N ) (N )
−3/ 4
ReV
ScV
( 574 )−3 / 4 (1.69 )−1/ 3
−1/ 3
uV
DV
a
aPh
1.59
4.84 ×10−5
1.59
= 1.73 s-1
0.455 ( 2.02 )
This is approximately twice the suggested value of 0.80 s-1.
=
1
= 0.455 ft
2.02
Exercise 6.30
Subject: Absorption of NH3 from air into water in a packed tower.
Given: Column operation at 68oF and 1 atm. Inlet gas mass velocity = 240 lb/h-ft2. Inlet water
mass velocity = 2,400 lb/h-ft2. Tower packed with 1.5-inch ceramic Berl saddles. Henry's law
applies for NH3 with p (atm) = 2.7x.
Assumptions: No stripping of water. No absorption of air. Dilute system.
Find: (a) Packed height for 90% absorption of NH3.
(b) Minimum water mass velocity for 98% absorption of NH3.
(c) KGa, pressure drop, maximum liquid rate, KLa, packed height, column diameter, HOG,
and NOG for 1.5-inch ceramic Hiflow rings.
Analysis: (a) Assuming Dalton's law, the equilibrium equation for 1 atm is,
p = yP = 2.7x atm or with P = 1 atm, y = Kx = 2.7x
(1)
Because the system is dilute, Eq. (6-89) can be used to compute NOG .
Then the packed height is given by Eq. (6-85),
lT = NOGHOG (2)
The absorption factor from Eq. (6-15) is, A = L/KV = (2,400/18)/2.7(240/29) = 6.0
Entering and exiting ammonia mole fractions are: yin = 0.02, yout = 0.002, xin = 0.0
The ammonia mole fraction in the exiting liquid is obtained by an overall ammonia balance:
240
2,400
240
2,400
yin +
xin =
yout +
xout
29
18
29
18
(3)
Solving Eq. (3), xout = 0.00112.
From Eq. (6-89),
N OG =
ln ( A − 1) / A ( yin − Kxin ) / ( yout − Kxin ) + (1 / A)
( A − 1) / A
=
ln (6 − 1) / 6 0.02 / 0.002 + (1 / 6)
(6 − 1) / 6
= 2.57
To obtain an estimate of HOG, the correlations of Billet and Schultes in Chapter 6 or literature
data can be used. The latter is found on page 16-38 of Perry's Chemical Engineers' Handbook,
6th edition or on page 663 of "Equilibrium-Stage Separation Operations in Chemical
Engineering" by Henley and Seader, as shown below.
Analysis: (a) (continued)
Exercise 6.30 (continued)
Air-Ammonia-Water system with 1.5-inch Berl Saddles, AE units
From the above figure, HOG = 0.85 ft. From Eq. (2), lT = 2.57(0.85) = 2.2 ft
(b) From Eq. (6-11), (L/S)min = (V/S) K (fraction absorbed) = 240(2.7)(0.98) = 635 lb/h-ft2
(c) Consider the differences between 1.5-inch ceramic Berl saddles (Saddles) and 1.5-inch
ceramic Hiflow rings (Rings). In Table 6.8, data are given for the rings, but not for the saddles
of the 1.5-inch size. To make the comparison, take the near-1 inch size. Then, the following
parameters apply:
Packings
Saddles
Rings
FP
110
a
260
261.2
ε
0.68
0.779
Ch
0.62
1.167
Cp
0.628
CL
1.246
1.744
CV
0.387
0.465
Analysis: (c) (continued)
Exercise 6.30 (continued)
Note that for the NH3-air-water system, the above plot shows that HOG is approximately equal to
HG. Therefore, the liquid-phase resistance is small and KL a need not be considered.
Consider KGa and HOG:
In the above table, values of a are essentially the same. From Eqs. (6-136) and (6-140),
1/ 2
a / a for Rings
0.779
=
= 1.07
aPh/a is proportional to ε /a. Therefore, Ph
aPh / a for Saddles
0.68
From Eq. (6-133), if we ignore holdup, hL, in the term (ε - hL), and differences in a, then,
1 4 a
H G is proportional to
ε
CV
aPh
1/2
1
(0.779) 4
H G for Rings
Therefore,
= 0.465
(1.07 ) = 1.53
1
H G for Saddles
4
(0.680)
0.387
This ratio should be about the same for HOG .
K G a for Rings
1
From Table 6.7, KG is inversely proportional to HG. Therefore,
=
= 0.653
K G a for Saddles 1.53
Consider pressure drop:
The pressure drop is given by Eq. (6-106). If we ignore holdup, hL, in the term (ε - hL), and
differences in a, and combine Eqs. (6-99) and (6-110) to (6-115), then,
Cp
∆P is approximately proportional to 3
ε
However, Table 6.8 does not contain Cp for Berl saddles, so no comparison can be made.
Consider Column NOG and height:
The value of NOG is independent on packing material. Therefore, the height is shorter for
the saddle packing because the HOG is smaller.
Consider column diameter:
The column diameter is related to the flooding velocity, as given in Fig. 6.36(a). For a
given value X, the value of Y is fixed and the product (uo)2FP is a constant. The packing factor
for the ceramic Hiflow rings is not given in Table 6.8. However, if a value is extrapolated from
data for larger rings, the packing factor for the Rings is much less than for the Saddles.
Therefore, the flooding velocity for the Rings will be higher and the column diameter smaller.
Consider maximum liquid rate:
For a fixed gas velocity, the value of Y in Fig. 6.36(a) is smaller for the Rings because the
packing factor is less. Therefore, as shown in the same figure, the value of X is greater for
flooding, which gives a larger maximum liquid rate.
Exercise 6.31
Subject: Absorption of CO2 from air into a dilute caustic solution with a packed column.
Given: 5,000 ft3/min at 60oF and 1 atm, containing 3 mol% CO2. Recovery of 97% of CO2.
Equilibrium curve is Y = KX = 1.75X (i.e. mole ratios) at column operating conditions.
Assumptions: No stripping of water. No absorption of air. Caustic solution has properties of
water. Initial estimate of column diameter is 30 inches. Try 2-inch Intalox saddles packing.
Find: (a)
(b)
(c)
(d)
(e)
(f)
(g)
Minimum caustic solution-to-air molar flow-rate ratio.
Maximum possible concentration of CO2 in caustic solution.
Nt at L/V = 1.4 times minimum.
Caustic solution rate.
Pressure drop per ft of column packed height.
NOG
Packed height for KGa = 2.5 lbmol/h-ft3-atm.
Analysis: Gas flow rate = 5,000(60)/379 = 792 lbmol/h.
Therefore, air in entering gas = V' = 0.97(792) = 768 lbmol/h
The CO2 in the entering gas = 0.03(792) = 24 lbmol/h
The CO2 in the exiting gas = 0.03(24) = 0.72 lbmol/h
The CO2 absorbed in the leaving liquid = 24 - 0.72 = 23.28 lbmol/h
The mole ratios are Yin = 24/768 = 0.03125, Yout = 0.72/768 = 0.0009375,
Xin = 0.0
(a) From Eq. (6-11), L'min = V'K(fraction absorbed) = 768(1.75)(0.97) = 1,304 lbmol/h
(b) The maximum possible CO2 concentration in the caustic solution occurs at infinite stages
with the minimum liquid rate. The leaving liquid is in equilibrium with the entering gas.
Therefore,
Xout = Yin/K = 0.03125/1.75 = 0.0179 mol CO2 / mol caustic solution.
(c) Let L/V = L'/V' . Then, L = 1.4 (1,304) = 1,826 lbmol/h and L/G = 1,826/768 = 2.38.
For this liquid flow rate, a material balance for CO2 is,
768(0.03125 - 0.0009375) = 23.28 = 1,826(Xout - 0.0). Solving, Xout = 0.01275
In the Y-X plot on the next page, the equilibrium line, Y = 1.75X, and the straight operating line,
passing through the column Y-X end points {Yin = 0.03125, Xout = 0.01275} and {Yout =
0.0009375, Xin = 0.0} are shown, with the equilibrium stages stepped off, giving Nt = 7.3.
(d) From Part (c), caustic rate = 1,826 lbmol/h.
(e) Assume the use of 2-inch of ceramic Intalox saddles. From Table 6.8, FP = 40 ft2/ft3.
Use Fig. 6.36(a). Assume for the dilute caustic solution, f{ρL} = 1.0 and f{µL} = 1.0.
Take MV = 29, ML = 18, ρL = 62.4 lb/ft3, ρV = 29/379 = 0.0765 lb/ft3
Using conditions at the bottom of the column,
Exercise 6.31 (conditions)
Analysis: (d) and (e) (continued)
LM L ρV
X=
VM V ρ L
1/ 2
(1,849)(18) 0.0765
=
(792)(29)
62.4
1/ 2
= 0.051
From Fig. 6.36(a), at flooding, Y = 0.18
Now compute Y for the suggested column diameter of 30 inches = 2.5 ft.
From the continuity equation, uo = m/SρV = [(792)(29)/3600]/[3.14(2.5)2/4](0.0765) = 17 ft/s
From Table 6.8, the packing factor, FP = 40 ft2/ft3.
uo2 FP ρV
17 2 (40) 0.0765
Y=
=
= 0.440 , which is much greater than 0.18
g ρH 2O
32.2
62.4
Therefore, a diameter of 30 inches places the operation badly into the flooding region and a
pressure drop calculation is meaningless. A larger diameter is necessary.
Exercise 6.31 (continued)
(f) From Eq. (6-95), with A = L/KV = (1,826)/1.75(768) = 1.36,
NOG = Nt = A ln(1/A)/(1- A) = 7.3(1.36) ln (1/1.36)/(1 - 1.36) = 8.5
(g) From Table 6.7, HOG = V/KGaPS, with V = 792 lbmol/h,
KGa = 2.5 lbmol/h-ft3-atm, P = 1 atm
For column cross sectional area, S, assume 50% of flooding. At flooding, with Y = 0.18,
by ratio with the Y of 0.440 for a 2.5 ft diameter, need uo at flooding = 17(0.18/0.44)0.5 = 10.9 ft/s
At 50% of flooding, u = 5.45 ft/s. By ratio, column diameter = 2.5(17/5.45)0.5 = 4.4 ft
Column cross sectional area = S = 3.14(4.4)2/4 = 15.2 ft2.
Therefore, HOG = (792)/(2.5)(1)(15.2) = 20.8 ft, which is very high.
From Eq. (6-89), lT = HOG NOG = 20.8(8.5) = 177 ft. Given value of KGa seems very low.
Exercise 6.32
Subject: Absorption of NH3 from air into water with a packed column.
Given: Conditions at a point in the column where P = 101.3 kPa, T = 20oC, 10 mol% NH3 in
bulk gas, 1 wt NH3 in bulk liquid, NH3 partial pressure of 2.26 kPa at the interface, and ammonia
absorption rate of 0.05 kmol/h-m2.
Find: (a) X, Y, Yi , Xi , X*, Y*, KY, KX, kY, and kX in Fig. 6.49.
(b) % of mass transfer resistance in each phase.
1
1 H′
=
+
(c) Verify that
KY kY k X
Analysis: (a) At the point, as shown in Fig. 6.49, the mole ratios in the bulk are
X = (1/17)/(99/18) = 0.0107 and Y = 0.1/0.9 = 0.111.
At the interface, Yi = 2.26/(101.3 - 2.26) = 0.0229.
Then, using the equilibrium curve in Fig. 6.49, Xi = 0.028
The values of the mole fractions in equilibrium with the bulk values are obtained from the
equilibrium curve in Fig. 6.49. Thus, in equilibrium with Y = 0.111 is X* = 0.114.
In equilibrium with X = 0.0107 is Y* = 0.007
The given NH3 mass-transfer flux can be written with the following combinations of masstransfer coefficients and driving forces, using Table 6.7 as a guide:
N = 0.05 = KY Y − Y * = K X X * − X = kY Y − Yi = k X X i − X
(1)
From Eq. (1), KY = 0.05/(Y - Y*) = 0.05/(0.111 - 0.007) = 0.480 kmol/h-ft2-mole ratio
From Eq. (1), KX = 0.05/(X* - X) = 0.05/(0.114 - 0.0107) = 0.484 kmol/h-ft2-mole ratio
From Eq. (1), kY = 0.05/(Y - Yi) = 0.05/(0.111 - 0.0229) = 0.568 kmol/h-ft2-mole ratio
From Eq. (1), kX = 0.05/(Xi - X) = 0.05/(0.028 - 0.0107) = 2.89 kmol/h-ft2-mole ratio
(b) To determine the relative mass-transfer resistances in each phase, Eq. (1) can be
rearranged as in Eq. (6-78):
kX
(Y − Yi )
2.89
=−
=−
= −5.08
kY
( X − Xi )
0.568
Thus, the lowest coefficient and the highest driving force is that associated with the gas phase.
From the ratio of 5.08, the largest mass-transfer resistance is in the gas phase. That resistance is
about five times the resistance in the liquid phase, with 84% in the gas phase, 16% in the liquid.
(c) From a mole ratio form of Eq. (6-80),
1
1
1 Yi − Y *
=
+
(2)
KY kY k X X i − X
Therefore,
Therefore,
H′ =
Yi − Y *
0.0229 − 0.007
=
= 0.919
0.028 − 0.0107
Xi − X
1 H′
1
0.919
1
1
+
=
+
= 2.08 compared to
=
= 2.08
kY k X 0.568 2.89
KY 0.480
Exercise 6.33
Subject: Stripping of ammonia from an aqueous solution with air in a packed column.
Given: Aqueous solution of 20 wt% NH3. Exit liquid to contain no more than 1 wt% NH3.
Exit gas to be 1,000 ft3/h of a 10 mol% NH3 in air. Equilibrium data in Fig. 6.49.
Temperature = 20oC = 68oF.
Assumptions: No stripping of water. No absorption of air. Operation at 1 atm.
Find: Volume of packing for KGa = 4 lbmol/h-ft3-atm.
Analysis: First, determine the molar flow rates of components in the exit gas.
From the ideal gas law applied to the exiting gas at 1 atm and 68oF,
ρ = P / RT = 1 / (0.7302)(68 + 460) = 0.00259 lbmol/ft3
Flow rate of exit gas = 1,000(0.00259) = 2.59 lbmol/h
Gas contains 0.10(2.59) = 0.259 lbmol/h of NH3 and 0.90(2.59) = 2.331 lbmol/h of air.
Next, determine the mole fractions in the inlet and exit liquid streams.
Take as a basis, 100 lb of inlet liquid.
Component
H2O
NH3
Total
Inlet liquid:
lb
lbmol
80
4.44
20
1.18
100
5.62
mole fraction
0.79
0.21
1.00
Outlet liquid:
lb
lbmol
80.000 4.4400
0.808 0.0475
80.808 4.4875
mole fraction
0.9894
0.0106
1.0000
Determine the water flow rate, L', in and out by an ammonia balance:
0.21
0.0106
(1)
L′ + 0 =
L ′ + 0.259
0.79
0.9894
Solving Eq. (1), L' = 1.015 lbmol/h of H2O.
NH3 in entering liquid = (0.21/0.79)1.015 = 0.270 lbmol/h
NH3 in leaving liquid = 0.270 - 0.259 = 0.011 lbmol/h
Because of the concentrated solution, take into account the bulk-flow effect.
For a partial pressure driving force, can write the rate of mass transfer, V for volume, as,
nNH 3 = 0.259 =
Rearranging Eq. (1),
V=
KG aV
*
pNH
− pNH 3
3
( yair ) avg
0.259( yair ) avg
p
*
NH 3
− pNH 3
LM
KG a
LM
(2)
(3)
Exercise 6.33 (continued)
Analysis: (continued)
First consider the bulk-flow effect.
At the top of the column, in the bulk exiting air, yair = 0.90
To obtain the corresponding value of y in the vapor in equilibrium with the entering liquid at the
top of the column, use Fig. 6.49. In entering bulk liquid, moles NH3/moleH2O =1.18/4.44=0.266.
From Fig. 6.49, by a rough extrapolation, moles NH3/mole H2O in equilibrium gas = 0.266.
Therefore, y*air = 1 - 0.266/1.266 = 1 - 0.210 = 0.79. The average yair = (0.90 + 0.79)/2 = 0.845.
At the bottom of the column, in the bulk entering air, yair = 1.00
To obtain the corresponding value of y* in the vapor in equilibrium with the leaving liquid at the
column bottom, use Fig. 6.49.
In leaving bulk liquid, moles NH3/mol H2O = 0.0475/4.4 = 0.0108.
From Fig. 6.49, moles NH3/mole H2O in equilibrium gas = 0.007 = mole fraction of NH3
Therefore, y*air = 1 - 0.007 = 0.993. The average yair = (1.000 + 0.993)/2 = 0.997
The (yair)avg = (0.845 + 0.997)/2 = 0.921
Now, determine the log mean partial pressure driving force for ammonia, using Dalton's law
with the above mole fractions.
At the top of the column, for ammonia, (p*- p)top = P (y*- y)top = (1.0)(0.210 - 0.10) = 0.11 atm
At the column bottom, for ammonia, (p*- p)bottom = P (y*- y)bottom = (1.0)(0.007 - 0.0) = 0.007 atm
Therefore, for ammonia, (p*- p)LM = 0.037 atm
From Eq. (3), packed volume = V =
0.259(0.921)
= 1.61 ft3
(0.037)(4)
Exercise 6.34
Subject: Absorption of NH3 from air into water in a packed column.
Given: Entering gas is 2,000 lb/h of air (dry basis) containing NH3 with a partial pressure of 12
torr and saturated with water vapor at 68oF and 1 atm. Equilibrium data in Fig. 6.49. Absorb
99.6% of the NH3.
Assumptions: No absorption of air. No stripping of water.
Find: (a) Minimum water rate.
(b) Column diameter and height for water rate of 2 times minimum and 50% of flooding
gas velocity if 38-mm ceramic Berl saddles are used.
(c) Same as Part (b) for 50-mm Pall rings.
(d) Packing recommendation.
Analysis: First compute the ammonia material balance.
Flow rate of NH3-free air = 2,000/29 = 69 lbmol/h
Flow rate of NH3 in the entering gas = 12(69)/(760-12) = 1.107 lbmol/h
Flow rate of NH3 in the exiting liquid = 0.996(1.107) = 1.1026 lbmol/h
Flow rate of NH3 in the leaving gas = 1.107 - 1.1026 = 0.0044 lbmol/h
(a) Operation in dilute region, where we can apply Henry's law. From Fig. 6.49, using results
from Exercise 6.33, when x=0.0108, y=0.007. Therefore, the equilibrium equation is y = 0.65x.
Therefore, K = 0.65. Entering gas rate = V = 69 + 1.107 = 70.1 lbmol/h
From Eq. (6-11), Lmin = VK(fraction absorbed) = 45.4 lbmol/h
(b) L = 2Lmin = 2(45.4) = 90.8 lbmol/h
Compute column diameter for 50% of flooding, using Fig. 6.36.
Take properties of gas and liquid as those of air and water, respectively.
From the ideal gas law, ρG = PM/RT = (1)(29)/(0.7302)(528) = 0.0752 lb/ft3
LM L ρV
In Fig. 6.36(a), X =
VM V ρ L
1/ 2
(45.4)(18) 0.0752
=
(70.1)(29) 62.4
1/ 2
= 0.014
Take f{ρL} and f{µL} =1.0. From Fig. 6.36(a), at flooding, Y =
uo2 FP ρV
= 0.22
g ρH 2O
(1)
For 38-mm Berl saddles, the packing factor, FP, is not included in Table 6.8. However, it can be
found in Perry's Chemical Engineers' Handbook, which gives FP = 65 ft2/ft3.
Yg ρH2 O
(0.22)(32.2) 62.4
Solving Eq. (1), uo2 =
=
= 90.4
FP ρV
65
0.0752
Therefore, uo = 9.5 ft/s. The fraction of flooding = f = 0.5.
From Eq. (6-103), the column diameter is,
4VM V
DT =
fuo πρV
1/ 2
4(70.1/ 3, 600)(29)
=
(0.5)(9.5)(3.14)(0.0752)
1/ 2
= 1.42 ft and S=πDT2/4 =3.14(1.42)2/4=1.58 ft2
Exercise 6.34 (continued)
Analysis: (b) (continued)
For column height, use Eq. (6-89), because mass-transfer is controlled by the gas phase.
For NOG , use Eq. (6-88) with xin = 0. Absorption factor = A = L/KV = (90.8)/(0.65)(70.1) = 2
From Eq. (6-93), using yin = 1.107/70.1 = 0.0158, yout = 0.0044/69 = 0.0000638
N OG =
ln ( A − 1) / A ( yin ) / ( yout ) + (1 / A)
( A − 1) / A
=
ln (2 − 1) / 2 0.0158 / 0.0000638 + (1 / 2)
(2 − 1) / 2
= 9.65
For HOG, literature data can be used, as found on page 16-38 of Perry's Chemical Engineers'
Handbook, 6th edition or on page 663 of "Equilibrium-Stage Separation Operations in Chemical
Engineering" by Henley and Seader, as shown below.
To use this plot, L" = 90.9(18)/1.58 = 1,040 lb/h-ft2 and V”=G" = 70.1(29)/1.58 = 1,290 lb/h-ft2
Then, HOG = 1.9 ft and column height = lT = NOGHOG = (9.65)(1.9) = 18.3 ft.
Air-Ammonia-Water system with 1.5-inch Berl Saddles, AE units
Exercise 6.34 (continued)
Analysis: (c) (continued)
(c) Assume the 50-mm Pall rings are metal. From Table 6.8, FP = 27 ft2/ft3.
Solving Eq. (1), uo2 =
Yg ρH 2O
(0.22)(32.2) 62.4
=
= 235
FP ρV
27
0.0752
Therefore, uo = 15.3 ft/s. The fraction of flooding = f = 0.5.
From Eq. (6-103), the column diameter is,
4VM V
DT =
fuo πρV
1/ 2
4(70.1/ 3, 600)(29)
=
(0.5)(15.3)(3.14)(0.0752)
1/ 2
= 1.12 ft and S = 3.14(1.12)2/4 = 0.985 ft2
The value of NOG = 9.65, the same as in Part (b)
For HOG , can use Figs. 6.42 and 6.43 for 50 mm Pall rings.
Gas capacity factor = F = uG(ρV)0.5 = f uo(ρV)0.5 = (0.5)(15.3)(0.0752)0.5 = 2.1 lb1/2-s-1-ft-1/2
or 5.12 kg1/2-s-1-m-1/2
By the continuity equation, liquid load = QL/S = uL = m/SρL =
(90.8/3600)(18)/(0.985)(62.4) =0.0074 ft/s or 0.00225 m/s or 2.25 x 10-3 m3/m2-s
In Fig. 6.42, from the slope of the correlating line, kGa is proportional to F0.7
In Fig. 6.43, from the slope of the correlating line, kGa is proportional to uL0.45
Using the value of kGa = 3.1 s-1 at uL = 2.25 x 10-3 m3/m2-s and F = 1.16, then the two plots are
represented by the equation, kGa = 43.4 F0.7 uL0.45
For our case, kGa = 43.4 (5.120.7)(0.002250.45) = 8.8 s-1
Note that this kGa is in concentration units for the driving force.
Therefore, from Table 6.7 by analogy to the liquid phase case,
HG = VMV/kGaSρV = (70.1/3,600)(29)/(8.8)(0.985)(0.0752) = 0.87 ft
The column height is lT = 9.65(0.87) = 8.4 ft.
Exercise 6.34 (continued)
Analysis:
(d) Now compare the two packings on the basis of flooding and mass transfer:
Packing
38-mm ceramic Berl saddles
50-mm metal Pall rings
Column diameter, ft
1.42
1.12
Column height, ft
18.3
8.4
Clearly the 50-mm metal Pall rings are superior in performance. However, the selection would
have to be made on the basis of cost. Because of the much smaller column and volume of
packing, it is likely that the 50-mm Pall rings would be favored.
From page 14-59 of Perry's Chemical Engineers' Handbook, 7th edition (1997), the cost of 38mm ceramic Berl saddles is approximately $21/ft3. The cost of 50-mm Pall rings depends on
whether they can be made of carbon steel or must be made of stainless steel, with the former
$19.9/ft3 and the latter $99.0/ft3. So the material of construction is a very important factor.
Exercise 6.35
Subject: Absorption of Cl2 from air with water in a packed column.
Given: 100 kmol/h of feed gas containing 20 mol% Cl2. Column operation at 20oC and 1 atm.
Table of y-x data for Cl2. Exit gas to contain 1 mol% Cl2.
Assumptions: No stripping of water. No absorption of air.
Find: (a) Minimum water rate in kg/h.
(b) NOG for a water rate of 2 times minimum.
Analysis: Solve this problem using mole ratios. Feed gas is 80 kmol/h of air = G' , and 20
kmol/h of Cl2 with Yin = 20/80 = 0.25. Exit gas has Yout = 1/99 = 0.0101. Xin = 0.0.
Highest value of Y in the table of equilibrium data is 0.06/0.94 = 0.0638. Therefore, data for
higher values of Y are needed. Can obtain this from Perry's Handbook, page 3-102 in the 6th
edition and page 2-126 in the 7th edition. The expanded table of equilibrium data with
conversion to mole ratios includes the following values added from Perry's Handbook, using the
conversions:
y = pCl2, torr/760 torr
x = [(grams Cl2/L)/71]/[1,000/18 + (grams Cl2/L/71)]
pCl2, torr
100
150
200
grams Cl2/L
1.773
2.27
2.74
y
0.132
0.197
0.263
x
0.000450
0.000576
0.000695
These equilibrium data and those given in the exercise are now converted to mole ratios using,
Y = y/(1 - y) and X = x/(1 - x)
y
0.006
0.012
0.024
0.040
0.060
0.132
0.197
0.263
x
0.000100
0.000150
0.000200
0.000250
0.000300
0.000450
0.000576
0.000695
Y
0.00604
0.01215
0.02459
0.04167
0.06383
0.15207
0.24533
0.35685
X
0.000100
0.000150
0.000200
0.000250
0.000300
0.000450
0.000576
0.000695
(a) To determine the minimum water rate, it is best to do this graphically because of the high
degree of curvature of the equilibrium curve. The pinch region of infinite stages at minimum
absorbent rate occurs at the bottom of the column, such that the leaving liquid is in equilibrium
with the entering gas, with a Y = 0.25. With mole-ratio coordinates, the operating line is straight
and passes through the following point at the top of the column {X = 0.0, Y = 0.0101}.
Exercise 6.35 (continued)
Analysis: (a) (continued)
The point at the bottom of the column must pass through the point {X* in equilibrium with
Y = 0.25}. This is shown in the following plot, which is similar to Fig. 6.9.
From the solute material balance, Eq. (6-6), the slope of the operating line is given by L'/V'.
From the above plot, Xbottom = 0.000581, and, thus, the slope of that line is,
L'/V' = (0.25 - 0.0101)/(0.000581 - 0.0) = 413. Since V' = 80 kmol/h, L' = 80(413) = 33,000
kmol/h. Alternatively, since the Cl2 absorption rate = V'(Ybottom - Ytop) = 80(0.25 - 0.0101) =
19.19 kmol/h, then L' = 19.19/ Xbottom = 19.19/0.000581 = 33,000 kmol/h. Take the result as
33,000 kmol/h or 33,000 (18) = 594,000 kg/h.
(b) Twice the minimum water rate = 2(33,000) = 66,000 kmol/h. This gives a mole ratio of
chlorine in the leaving water of Xbottom =19.19/66,000 = 0.000291 moles Cl2 / mole water.
This operating line is shown in the plot below. To obtain NOG , using mole ratios, the integral in
Eq. (6-138) or Table 6.7 can be used,
Analysis: (b) (continued)
Exercise 6.35 (continued)
N OG =
Y = 0.25
dY
(Y − Y *)
Y = 0.0101
(1)
1
(Y − Y *)
as a function of Y, where for a given value of X, the value of Y is obtained from the operating line
and Y* is obtained from the equilibrium curve, with values as follows. The values of X are taken
from the above table, where the values of Y are the corresponding values of Y*. For given
values of X, values of Y for the operating line are obtained from the following equation of the
straight operating line passing through the points { X = 0.0, Y = 0.0101} and { X = 0.000291,
Y = 0.25},
L′
66, 000
Y = 0.0101 + X = 0.0101 +
X = 0.0101 + 825 X
V′
80
Eq. (1) can be solved graphically or numerically. Both require a table of values of
Note that several values of X are added by interpolation to increase the accuracy.
Analysis: (b) (continued)
X
0.000000
0.0000125
0.000025
0.000050
0.000075
0.000100
0.000150
0.000200
0.000250
0.000291
Y*
0.00000
0.00075
0.00150
0.00301
0.00452
0.00604
0.01215
0.02459
0.04167
0.06030
Exercise 6.35 (continued)
Y
0.0101
0.0204
0.0307
0.0514
0.0720
0.0926
0.1339
0.1751
0.2164
0.2500
1/(Y-Y*)
99.01
50.86
34.22
20.69
14.82
11.55
8.22
6.64
5.72
5.27
∆Y
Avg 1/(Y-Y*)
Area
0.0103
0.0103
0.0207
0.0206
0.0206
0.0413
0.0412
0.0413
0.0336
74.9
42.5
27.5
17.8
13.2
9.88
7.43
6.18
5.50
0.772
0.438
0.569
0.367
0.272
0.408
0.306
0.256
0.185
1
versus Y. The area under the curve is the value of the
(Y − Y *)
integral for NOG . In the above table, the trapezoidal method is used to break the curve into 9
segments of ∆Y each, varying in width from 0.0103 to 0.0413. For each segment (width), the
average value of 1/(Y-Y*) is listed and the product = Area = ∆Y times 1/(Y-Y*) is given in the
last column. The sum of these areas is NOG = 3.573.
The following is a plot of
Exercise 6.36
Subject: Absorption of ammonia from air into water with a packed column.
Given: Operating conditions in Example 6.15. Entering gas is 100 lbmol/h containing 40
mol% NH3. Absorbent is 488 lbmol/h of water. Absorb 95% of the NH3. Equilibrium data in
Fig. 6.44. Column operates at 1 atm and 25oC.
Assumptions: No stripping of water. No absorption of air. Liquid has properties of water.
Find: Column diameter and height if packing is 1.5-inch metal Pall rings, using conditions at
bottom of tower.
Analysis: From Example 6.15, the conditions at the bottom of the tower are:
Entering gas: 60 lbmol/h of air and 40 lbmol/h of NH3
Exiting liquid: 488 lbmol/h of water and 38 lbmol/h of NH3
Operating temperature is 298 K (536oF) and pressure is 1 atm
To compute column diameter use Fig. 6.36 to obtain flooding gas velocity, assume operation at
50% of flooding to obtain operating gas velocity, and use continuity equation to obtain column
cross sectional area and column diameter.
Average gas molecular weight = [(60)(29) + (40)(17)]/100 = 24.2
Gas density = ρV = PM/RT = (1)(24.2)/(0.7302)(536) = 0.0618 lb/ft3
From the abscissa of Fig. 6.36(a),
LM L ρV
X=
VM V ρ L
From Fig. 6.36(a), Y = 0.105 =
1/ 2
=
(526)(17.9) 0.0618
(100)(24.2) 62.4
1/ 2
= 0.123
uo2 FP ρV
f {ρ L } f {µ L }
g ρH2 O
(1)
For 1.5-inch (35 mm) metal Pall rings, from Table 6.8, FP = 40 ft2/ft3
At 25oC, from Figs. 6.36(b) and (c), f{ρL} = 1.0 and f{µL} = 0.98
From a rearrangement of Eq. (1),
uo2 =
Yg ρH2 O
FP ρV
1
(0.105)(32.2) 62.4
1
=
= 87 ft2/s2
f {ρ L } f {µ L }
40
0.0618 (1)(0.98)
Then, uo = 9.3 ft/s
Take the operating superficial velocity = uV = fuo = 0.5 uo = 0.5(9.3) = 4.65 ft/s
From Eq. (6-99),
4GM G
DT =
fuo πρG
1/ 2
4(100 / 3, 600)(24.2)
=
(0.5)(9.3)(3.14)(0.0618)
1/ 2
= 1.73 ft
Analysis: (continued)
Exercise 6.36 (continued)
(2)
From Eq. (6-127), column height = lT = HOGNOG
From Example 6.15, NOG = 3.46 and from above, column cross section = S = πD2/4 = 2.35 ft2.
Mass-transfer data for 2-inch metal Pall rings for the absorption of NH3 from air into water are
presented in Figs. 6.42 and 6.43. In the solutions to Exercise 6.34, these data are fitted to the
(3)
equation,
kGa = 43.4 F0.7 uL0.45 , s-1
0.5
0.5
1/2 -1
where the gas capacity factor = F = uV(ρV) = f uo (ρV)
in kg -s -m-1/2 and from the
continuity equation, uL = m/SρL in m/s or m3/m2-s
For our case, F = (0.5)(9.3)(0.0618)0.5 = 1.16 lb1/2-s-1-ft-1/2 or 2.83 kg1/2-s-1-m-1/2
and uL = [(526)(17.9)/3,600)]/(2.35)(62.4) = 0.018 ft/s = 0.00547 m3/m2-s
Substitution of these values into Eq. (3) gives, kGa = 43.4(2.83)0.7(0.00547)0.45 = 8.6 s-1
Note that this kGa is in concentration units for the driving force.
Therefore, from Table 6.7 by analogy to the liquid phase case,
HOG = HG = VMV/kGaSρV = (100/3,600)(24.2)/(8.6)(2.35)(0.0618) = 0.54 ft
Now, we must correct this value to 1.5-inch metal Pall rings.
From Table 6.8, using interpolation when necessary, we have the following characteristics:
Packing
1.5-inch Pall rings
2.0-inch Pall rings
a, m2/m3
139.4
112.6
Ch
0.644
0.784
ε
0.965
0.951
CV
0.373
0.410
From Eqs. (6-136) to (6-140), aPh/a is proportional to ε1/2/a . Therefore,
aPh / a for 1.5-inch Rings
0.965
=
aPh / a for 2.0-inch Rings
0.951
1/ 2
112.6
= 0.814
139.4
From Eq. (6-133), if we ignore holdup in the term (ε - hL), then,
1 ε
a
H G is proportional to
1.25
CV a
aPh
1
0.965
H G for 1.5-inch Rings 0.341 139.41.25
1
Therefore,
=
= 1.15
1
0.951
H G for 2.0-inch Rings
0.814
0.410 112.61.25
This ratio should be about the same for HOG .
Therefore, for the 1.5-inch rings, HOG = 0.54(1.15) = 0.62 ft.
From Eq. (2), column height = 3.46(0.62) = 2.1 ft.
Exercise 6.37
Subject: Absorption of acetone from air with water in a packed column.
Given: Column operates at 60oF and 1 atm. Entering air is 50 ft3/min at 60oF and 1 atm,
containing 3 mol% acetone. 97% of the acetone is to be absorbed. Maximum allowable gas
superficial velocity is 2.4 ft/s. Equilibrium equation is Y = 1.75 X
(1)
Assumptions: No stripping of water. No absorption of air.
Find: (a)
(b)
(c)
(d)
(e)
(f)
Minimum water-to-air ratio.
Maximum acetone concentration possible in the water.
Nt for absorbent flow rate equal to 1.4 times the minimum.
Number of overall gas transfer units.
Height of packing for Kya = 12.0 lbmol/h-ft3-mole ratio difference.
Height of packing as a function of molar flow ratio for constant V and HTU.
Analysis: First compute material balance. Total gas in = 50(60)/379 = 7.916 lbmol/h
Acetone in entering gas = 0.03(7.916) = 0.237 lbmol/h
Air in entering gas = 7.916 - 0.237 = 7.679 lbmol/h
Acetone in exiting gas = 0.03(0.237) = 0.007 lbmol/h
Acetone in entering liquid = 0.0 lbmol/h
Acetone in exiting liquid = 0.237 - 0.007 = 0.230 lbmol/h
Xin = 0.0
Yin = 0.03/0.97 = 0.03093
Yout = 0.007/7.679 = 0.000912
(a) For minimum water rate, the exiting liquid will be in equilibrium with the entering air.
Therefore, using the equilibrium equation, Eq. (1), Xout = Yin /1.75 = 0.03093/1.75 = 0.01767
Therefore, water rate out and in = 0.230/0.01767 = 13.02 lbmol/h
Minimum water-to-air molar ratio = 13.02/7.679 = 1.70
(b) Maximum possible acetone concentration in the liquid is at min. water rate, X = 0.01767
This corresponds to an acetone mole fraction of 0.01767/(1 + 0.01767) = 0.01736.
(c) Use Lin = L' = 1.4 Lmin = 1.4(13.02) = 18.23 lbmol/h
Using mole ratios with no acetone in entering liquid, and K' = constant = 1.75, can apply
Eq. (5-29), which was derived for extraction but applies to absorption using:
Yout
=
Yin
1
N
En
where E =
L′
18.23
=
= 1.356
K ′G ′ (175
. )(7.679)
n=0
Therefore,
Yout 0.000912
=
= 0.0295 =
Yin
0.03093
1
N
n =0
1356
.
Solving, N = 7.5 equilibrium stages
n
Exercise 6.37 (continued)
Analysis: (c) (continued)
Could also solve this using Kremser's equation or by a graphical construction like Fig. 6.12.
(d) Because operating line and equilibrium line are straight, use Eq. (6-95) with A = E = 1.356
N OG = N t
ln(1/ A)
ln(1/1.356)
= 7.5
= 8 .7
(1 − A) / A
(1 − 1.356) /1.356
(e) From Eq. (6-141), lT = HOGNOG
From Table 6.7, H OG =
V′
7.679
0.64
=
=
, where S = cross sectional area of tower in ft2
KY aS (12.0) S
S
The allowable gas superficial velocity = uV = 2.4 ft/s. Take this at the bottom of the tower where
gas rate is the highest. From the continuity equation, Q = uV S = gas volumetric rate
Therefore, S = Q/uV = (50/60)/2.4 = 0.347 ft2. Therefore, HOG = 0.64/0.347 = 1.84 ft
Packing height = (1.84)(8.7) = 16 ft.
(f) To obtain packed height as a function of L'/V' with V' and HOG constant, the only change will
be the value of NOG . Use Eq. (6-93), which for xin = 0 and yin/yout = 0.03/0.000911 = 32.9, is:
ln 32.9
N OG =
A −1
1
+
A
A
A −1
A
(2)
where for a dilute system, we can take A = L'/K'V' = (L'/V')/K' = (L'/V')/1.75
The results obtained from Eq. (2) are as follows:
L'/V'
1.75
2.0
2.5
3.0
5.0
10.0
A
1.00
1.143
1.429
1.714
2.857
5.714
1/A
1.00
0.875
0.700
0.583
0.350
0.175
NO'G
Height, ft
12.9
7.86
6.38
4.74
4.01
23.6
14.5
11.7
8.72
7.38
∞
∞
Exercise 6.38
Subject: Absorption of NH3 from air into water with a packed column.
Given: Column operating at 68oF and 1 atm. Gas enters at 2,000 ft3/min at 68oF and 1 atm,
containing 6 mol% NH3 in air. Entering water rate is twice the minimum. Absorb 99% of the
NH3. Gas velocity at 50% of flooding velocity.
Assumptions: No stripping of water. No absorption of air.
Find: Tower diameter and packed height.
Analysis: Entering gas rate is
2,000 60 + 460
60 = 312 lbmol /h
379 68 + 460
Compute material balance. NH3 in entering gas = 0.06(312) = 18.72 lbmol/h
NH3 in exiting gas = 0.01(18.72) = 0.19 lbmol/h
Air in entering and exiting gas = V' = 312 - 18.72 = 293.28 lbmol/h
NH3 in exiting water = 18.72 - 0.19 = 18.53 lbmol/h
Therefore, Xin = 0.0
Yin = 18.72/293.28 = 0.06383
Yout = 0.19/293.28 = 0.000648
At the minimum water rate, the exiting liquid is in equilibrium with the entering gas.
From Fig. 6.49, X*out = 0.072 for Yin = 0.06383. Therefore, L'min = 18.53/0.072 = 257 lbmol/h
Operating liquid rate = 2 times minimum = 2(257) = 514 lbmol/h
Now, compute column diameter using Fig. 6.36. Corrections for liquid density and viscosity are
neglected. From the ideal gas law on the entering gas, with an average molecular weight = 28.3,
ρV =PM/RT = (1)(28.3)/(0.7302)(68+460) = 0.0734 lb/ft3
From Table 6.8, packing factor for 1-inch metal Pall rings = 56 ft2/ft3.
LM L ρV
The abscissa in Fig. 6.36 is, X =
VM V ρ L
From Fig. 6.36, Y = 0.18. Therefore, uo2 =
1/ 2
(514 + 18.5)(18) 0.0734
=
(312)(28.3)
62.4
1/ 2
Yg ρH 2O
(0.18)(32.2) 62.4
=
= 88
FP ρV
56
0.0734
Therefore, flooding velocity = uo = 9.4 ft/s
For 50% of flooding, f = 0.5
4GM G
From Eq. (6-103), DT =
fuo πρG
1/ 2
= 0.0373
4(312 / 3, 600)(28.3)
=
(0.5)(9.4)(3.14)(0.0734)
1/ 2
= 3.0 ft.
Exercise 6.38 (continued)
Analysis: (continued)
Assume gas phase controls mass transfer rate. Then, from Eq. (6-89), lT = HOGNOG
From Eq. (6-93), for xin = 0, yin = 0.06, and yout = 0.00065,
ln 92.3
N OG =
A −1
1
+
A
A
A −1
A
(1)
For the absorption factor, A = L/KV. Take L = 514 lbmol/h. Take V = 312 lbmol/h.
From Fig. 6.49, K = 0.82. Therefore, A = (514)/(0.82)((312) = 2.0
From Eq. (1),
2 −1 1
ln 92.3
+
2
2
= 7.7
N OG =
2 −1
2
Mass-transfer data for 2-inch metal Pall rings for the absorption of NH3 from air into water are
presented in Figs. 6.42 and 6.43. In the solutions to Exercise 6.34, these data are fitted to the
equation,
(3)
kGa = 43.4 F0.7 uL0.45 , s-1
where the gas capacity factor = F = uV(ρV)0.5 = f uo (ρV)0.5 in kg1/2-s-1-m-1/2 and from the
continuity equation, uL = m/SρL in m/s or m3/m2-s. Here, S = 3.14(3)2/4 = 7.07 ft2
For our case, F = (0.5)(9.4)(0.0734)0.5 = 1.27 lb1/2-s-1-ft-1/2 or 3.10 kg1/2-s-1-m-1/2
and uL = [(532.5)(18)/3,600)]/(7.07)(62.4) =0.006 ft/s = 0.00182 m3/m2-s
Substitution of these values into Eq. (3) gives, kGa = 43.4(3.10)0.7(0.00182)0.45 = 5.6 s-1
Note that this kGa is in concentration units for the driving force.
Therefore, from Table 6.7 by analogy to the liquid phase case,
HOG = HG = VMV/kGaSρV = (312/3,600)(28.3)/(5.6)(7.07)(0.0734) = 0.84 ft
Now, we must correct this value to 1-inch metal Pall rings.
From Table 6.8, using interpolation when necessary, we have the following characteristics:
Packing
1.0-inch Pall rings
2.0-inch Pall rings
a, m2/m3
223.5
112.6
ε
0.954
0.951
Ch
0.719
0.784
CV
0.336
0.410
Analysis: (continued)
Exercise 6.38 (continued)
From Eqs. (6-136) to (6-140), aPh/a is proportional to ε1/2/a . Therefore,
aPh / a for 1.0-inch Rings
0.954
=
aPh / a for 2.0-inch Rings
0.951
1/ 2
112.6
= 0.505
223.5
From Eq. (6-133), if we ignore holdup in the term (ε - hL), then,
H G is proportional to
Therefore,
1 ε
a
1.25
CV a
aPh
1
0.954
H G for 1.0-inch Rings 0.336 223.51.25
=
1
0.951
H G for 2.0-inch Rings
0.410 112.61.25
This ratio should be about the same for HOG .
Therefore, for the 1.5-inch rings, HOG = 0.84(1.03) = 0.87 ft.
From Eq. (2), column height = 7.7 (0.87) = 6.7 ft.
1
= 1.03
0.505
Exercise 6.39
Subject: Absorption of SO2 from air by water in a packed column.
Given: Mass velocity of entering gas = 6.90 lbmol/h-ft2 on SO2-free basis, containing 80 mol%
air and 20 mol% SO2. Water enters at a mass velocity of 364 lbmol/h-ft2. Gas exits with 0.5
mol% SO2. Tower operates at 30oC and 2 atm. Equilibrium y-x equation given for SO2. Tower
is packed with Montz B1-200 metal structured packing.
Assumptions: No stripping of water. No absorption of air.
Find: (a) Molar material balance operating line.
(b) NOG
Analysis: (a) A material balance for the solute, SO2, from the top of the column to an
intermediate point gives:
Yn +1V ′ + X in L′ = YoutV ′ + X n L′
(1)
Solving Eq. (1), the operating line is given by,
Y=
L′
L′
X + Yout − X in
V′
V′
(2)
From the given data, L'/V' = 364/6.90 = 52.75. Xin = 0.0 and Yout = 0.005/0.995 = 0.005025
Rearrange Eq. (2), letting Y = y/(1-y) and X = x/(1-x), which is approximately x because of the
dilute condition,
x
V′ y
V′
y
X=
≈x=
− Yout = 0.01896
− 0.000095 (3)
L′ 1 − y
L′
1− x
1− y
which is very close to the given operating line.
(b) From Eq. (6-138), N OG =
Y = 0.25
dY
Y −Y *
Y = 0.00503
(4)
1
(Y − Y *)
as a function of Y, where for a given value of Y, the value of X is obtained from Eq. (3). Then, x
is computed from X. Then, y = y* is computed from the given equilibrium equation, and Y* is
1
then obtained from the equilibrium curve. Below is a plot of
versus Y. The area under
(Y − Y *)
the curve is the value of the integral for NOG . In the table below, the trapezoidal method is used
Eq. (4) can be solved graphically or numerically. Both require a table of values of
Exercise 6.39 (continued)
Analysis: (b) (continued)
to break the curve into segments of ∆Y. each, varying in width from 0.0103 to 0.0413. For each
segment (width), the average value of 1/(Y-Y*) is used to compute the product = Area = ∆Y times
1/(Y-Y*), which is given in the last column. The sum of these areas is NOG = 5.16.