PROBLEM WORKBOOK
Control of Gaseous Emissions
Conducted by the
AIR FOLIATION TRAINING INSTITUTE
The Control of Gaseous Emissions Workbook has been prepared
specifically for the trainees attending the course and should
not be included in reading lists or periodicals as generally
available.
United States Environnental Protection Agency
Office of Air and Water Programs
Control Programs Development Division
Air Pollution Training Institute
Research Triangle Park, North Carolina 27711
January, 1974
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US EPA
This is not an official policy and standards
document. The opinions, findings, and conclusions
are those of the authors and not necessarily those
of the United States Environmental Protection Agency.
Every attempt has been made to represent the
present state of the art as well as subject areas
still under evaluation. Any mention of products,
or organizations, does not constitute endorsement
by the United States Environmental Protection Agency.
ill
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TABLE OF CONTENTS
Page Number
Section I 1
Examination 2
Section II COMBUSTION 15
Tables 16
Figures 19
C-I Comprehensive Combustion Problem 20
C-II Theoretical Flame Temperature 30
C-III Plan Review of a Catalytic Afterburner 36
Section III ADSORPTION 42
AD-I Heat of Adsorption 43
Ad-II Surface Area Determination 50
AD-III Adsorption Dynamics 59
AD-IV Adsorption System 61
AD-V Design Problem 67
AD-VI Design Problem 77
AD-VII Plan Review of an Adsorption System 85
Section IV ABSORPTION 100
AB-I Basic Concepts 101
AB-II Material Balance 107
Information Necessary for Reviewing Absorber Plans 115
Procedure for Reviewing Absorber Plans 116
AB-III Packed Tower 117
AB-IV Packed Tower 122
AB-V Spray Tower I29
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Section I
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Examination
DIRECTIONS: There are three parts to this examination. Circle the "best
answer" to each question or problem on the Answer Sheet. There is one
and only one "best answer" to each question or problem. l\"hen you have
completed the examination, turn in only the Answer Sheet.
Part I Multiple Choice
(Each question is worth 2 points)
1. The principle commercial adsorbent used for adsorption of organic
vapors or gases is:
a. Silica Gel
b. Alumina
c. Zeolite
d. Carbon
2. The principle advantage of catalytic combustion over furnace combustion
is that oxidation is accomplished at a much lower:
a. Flow Rate
b. H/C Ratio
c. Gross Heating Value
d. Temperature
e. Pressure
3. A rule of thumb for non-catalytic afterburner operation ranges between
and seconds at between and °F.
a.
b.
c.
d.
0.1
1.0
0.1
5.0
- 0.3 ;
- 5.0 ;
- 1.5 ;
- 10.0:
1000 -
1000 -
1000 -
1000 -
1200
1200
1700
1700
-------
4. A rule of thumb for catalytic afterburner operating temperatures
and °F.
d. 1000 - 1200
5. The representation of adsorption data at a constant temperature which
indicates the amount adsorbed versus the partial pressure of the
adsorbate is called an .
a. Isobar
b. Isostere
c. Isotherm
d. Isokinetic
6. The process by which the large surface area of an adsorbent is
created is known as .
a. Regeneration
b. Atomization
c. Desorption
d. Activation
7. The two basic types of fans or blowers are and .
a. Induced draft, forced draft
b. Propellor, axial
c. Centrifugal, axial
d. Radial, reciprocating
8. During gas adsorption, as the temperature of the system increases, the
amount of material adsorbed .
a. Increases
b. Decreases
c. Remains constant
-------
9. The mole fraction of one component of a multi-component solution
is defined as:
a. The number of moles in the solution.
b. The number of moles of that component in the
solution divided by the number of moles of
the other components in the solution.
c. The number of moles of that component in the
solution divided by the total number of moles
of all the components in the solution.
d. The weight of that component in the solution
divided by the total weight of the solution.
10. In the following diagram of fan curves, Curve "A" represents
represents , and Curve "E" represents .
a. Horsepower; total pressure; static pressure
b. Mechanical efficiency; static pressure; horsepower
c. Total pressure; horsepower; static efficiency
d. Static pressure; horsepower; mechanical efficiency
, Curve "C"
100
100
% wide open
11. Dalton's Law of partial pressures is commonly represented by the
following equation:
a.
b.
c.
d.
PV = nRT
pi = PT Xi
Pi = Hi xi
-------
12. For an ideal gas mixture, the mole fraction and the volume fraction
are .
a. Functions of the void fraction
b. Identical
c. Unrelated
d. Both equal to the weight fraction
13. The variables which control absorption equilibrium are:
a. Temperature
b. Pressure
c. Concentrations of the contaminant in the gas
and liquid phases
d. All of the above
e. None of the above
14. A rule of thumb for flare operation indicates that a sooty emission
will occur with:
a. A premixed air-fuel system
b. Blue flame combustion
c. A H/C ratio > 0.28
d. A H/C ratio < 0.28
15. The equation describing the column packed height for a gas scrubber
is commonly represented by the following equation:
a. Pi = ^ Xi
b. Z = HQG^x NQG ^ ^
c• HOG =
KGa C1 - olm ave.
d- NQG =
(1
^i2
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16. The Mass Transfer Zone (MTZ) for an adsorber may be defined as
follows:
a. That zone in an adsorption bed where the
adsorbate exists in the highest loading on
the adsorbent.
b. That portion of the length of an adsorption
bed which is only partially saturated and
progresses in a downstream direction through
the
c . That portion of the adsorbent bed where the
Coefficient of Diffusion is the greatest.
d. None of the above.
17. The basic difference between a regenerative and a non-regenerative
adsorption system is that the regenerative type is usually
while the non-regenerative type is usually .
a. A thick bed; a thin bed.
b. For low concentrations; for high concentrations.
c. Activated carbon; activated alumina.
d. Heated; cooled.
Part II True-False
(Each question is worth 2 points)
1. The fan laws apply to fans from the same homologous series
at the same point of rating.
2. The regeneration time for an adsorber in an adsorption system may
be considerably less than the adsorption phase of the operation cycle.
3. In combustion processes the available heat for useful heating purposes
is usually identical to the gross heating value of the fuel.
-------
4. The volumes delivered by similar fans at the same rpm vary directly
as the square of the fan blade diameter.
5. The Mass Transfer Zone (MTZ) for an adsorption system may theoretically
be longer than the adsorption bed itself.
6. The heat of physical adsorption is usually about equal to the latent
heat of vaporization of the adsorbate.
7. The relative direction of gas to liquid flow in a packed tower is
usually cocurrent rather than countercurrent.
Part III Problems
4 pts. 1. If Henry's Law constant for C02 in the C02 -Air-Water -System is
1420 (atm/mole fraction) at 78°F, what is the partial pressure
of C02 when the mole fraction of C02 in the liquid is 0.0005?
a. 7.1
b. 0.028
c. 0.71
d. 0.28
4 pts. 2. Given the following information for a packed countercurrent gas scrubber,
determine the outlet liquid concentration, x^.
Gas Flow Rate: 10 Ib. moles/hr .ft2
Liquid Flow Rate: 40 Ib. moles/hr. ft2
mole fraction of contaminant in inlet gas: yj = 0.02
mole fraction of contaminant in outlet gas: y2 = 0.004
mole fraction of contaminant in inlet liquid: x2 = 0
a. 0.400
b. 0.004
c. 0.064
d. 0.640
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4 pts. 3. 49,000 BTU/hr. are transferred in an air preheater, calculate the outlet
temperature of the air stream using the following information:
For air, Cp = 0.245 BTU/lb. °F
Air Flow Rate = 1000 Ib./hr.
Air Inlet Temperature = 200°F
a. 220°F
b. 440°F
c. 400°F
d. 600°F
10 pts. 4. Type G II carbon was used in a two-foot adsorption bed for solvent
recovery. In this application the carbon capacity was 36%, the
MTZ was two inches, and the heel was 2.2%. Determine the working
charge.
a. 34.5%
b. 32.3%
c. 36.7%
d. 33.8%
10 pts. 5. Determine the scfm of dry natural gas (Gross Heating Value,
HVG = 1200 BTU/scf) required to heat 3790 scfm of a contaminated
gas stream (assume air) from 200°F to 1500°F. Assume no excess
air and no heat losses.
a. 156
b. 137
c. 92
d. 81
10 pts. 6. Fan A at 1622 rpm has a blade diameter of 46 inches, delivers 15,120 cfm,
and requires 45.9 bhp. If Fan B operates at 1600 rpm, has a blade diameter
of 50 inches, delivers a gas having the same density as for Fan A, and is
from the same homologous series as Fan A, what is the power requirement for
Fan B? a. 66.8 bhp
b. 58.1 bhp
c. 49.2 bhp
d. 69.6 bhp
8
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Under the following conditions, determine the packed height of a
countercurrent packed scrubber using a contaminant-free liquid
rate (assume water) of (55.5 Ib.moles/hr.ft2):
Gas Flow Rate: 500 (lb/hr.ft2) or, 17.2 (Ib.moles/hr.ft2)
Inlet contaminant gas mole fraction, y± = 0.020
Outlet contaminant liquid mole fraction, x{ = 0.0059
Slope of equilibrium line, m = 0.972
Packing "B"
HINT: First, use a material balance to find y2-
a. 4.7 ft.
b. 5.5 ft.
c. 7.1 ft.
d. 8.0 ft.
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ENTHALPIES OF COMBUSTION GASES
Btu/lb.moie
AIR
°F N M.W. 28.97
32 0 000
60 194.9 194.6 243.1 224.2
77 312.2 312.7 392.2 360.5
100 473.3 472.7 597.9 545.3
200 1,170 1,170 1,527 1,353
300 1,868 1,870 2,509 2,171
400 2,570 2,576 3,537 3,001
500 3,277 3,289 4,607 3,842
600 3,991 4,010 5,714 4,700
700 4,713 4,740 6,855 5,572
800 5,443 5,479 8,026 6,460
900 6,182 6,227 9,224 7,364
1,000 6,929 6,984 10,447 8,284
1,200 8,452 8,524 12,960 10,176
1,500 10,799 10,895 16,860 13,140
2,000 14,840 14,970 23,630 18,380
2,500 19,020 19,170 30,620 23,950
3,000 23,280 23,460 37,750 29,780
" Kobe, Kenneth A., and Long, Ernest G., "Thermochemistry for the
Petroleum Industry," Petroleum Refiner, Vol. 28, No. II, November,
1949, page 129, table 9.
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r wo.ooo
6
-5 120.000
s 100.000
I I I I I I I I I I I I
AVAILABLE HEATS "FOR-
SOME TYPICAL FUELS -
en
-i
d
300 600 900 1200 1500 1800 2100 2400 2700 3000
FLUE GAS EXIT TEMPERATURE "F
11
-------
j3 A Packing B :
G = Gas flow LB/HR FT2
12
-------
en
-P
•H
C
D
0)
U-l
W
(a
1
A.P. Colburn
Tram. AlChE J5.2l6(l939)
10 20
50 100 200 500 IJDOO
" m X
- m
13
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Section II
Combustion
15
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TABLE 1
Combustion Constants
No. Subsuuwe Formula
1 Carbon* C
2 Hydrogen Hi
3 Oaygcn Ot
4 Mnrogcn (Mm) Ni
S Carbon monoxide CO
6 Carbon dioxide COj
Paraffin series
7 Methane CHi
8 Ethane CiHi
9 Propane CiH«
10 n-Bulane C«Hn
1 1 hobutane CiHi.
12 n-Penlane CiHn
13 liopcmane CiHu
14 Ncopentane C,Hn
19 n-Heune OHn
Okfin tenet
16 Elhylcnc CiH,
IT Propylene CiHi
18 n-Butene C.H.
19 liobutene C.H.
20 n-Pcnlenc CiHio
Aromilic wriei
21 Benunc C.H.
22 Toluene C-H,
2) Xylene C.H,,
Mitcellaneoui g»e>
24 Acetylene C,Hi
23 Naphthalene Ci.H.
26 Methyl alcohol CHiOH
27 hthyl alcohol CiHtOH
28 Ammonia NHi
29 Sulfur* S
JO Hydrogen tulfide H:S
31 Sulfur dioude SO:
32 Water vapor H-O
33 Air . ..
Molecu-
lar Lb per
Weight Cu Ft
12.01
2016 00053
32000 00846
28016 00744
2801 00740
4401 0.1170
16041 00424
30067 00803
44092 01196
58.118 01582
58 118 0.1582
72 144 0 1904
72 144 0 1904
72 144 0 1904
86169 02274
28051 00746
42077 OHIO
56 102 0 1480
56 102 0 1480
70 128 0 1852
78 107 0 2060
92132 02431
106158 02803
26036 00697
128162 03384
32041 00846
46067 01216
17031 00456
3206
34076 00911
6406 01733
18016 00476
:«» 00766
CuFl
perLb
187 723
11.819
13.443
13506
8548
23565
12.455
8365
6321
6.321
5252
5252
5252
4.398
13412
9007
6756
6756
5400
4852
4 113
3567
14344
2955
11 820
8221
21914
10979
5770
21017
13.063
SpGr
Air =
10000
0.0696
1.1053
09718
09672
1.5282
0.5543
1.0488
1 5617
20665
7.0665
2.4872
24872
24872
29704
09740
14504
19336
19336
24190
26920
31760
36618
09107
44208
1 1052
1 5890
05961
1 1898
22640
06215
1.0000
For 100% Total Air
Molei per mole of Combustible
Heal of Combustion or For I007r Total Air
Blu per Cu Ft Blu per Lb Cu Ft per Cu Ft of Combustible Lb per Lb of Combustible
Cross Nei Cross Net Required for Combustion Flue Products Required for Combustion Flue Products
(High) (Low) (High) (Low) O, N, Air CO, H,O N, O, N, Air CO, H,O N:
14.093 14.093 1.0 3.76 476 10 . . 376 266 886 1153 366 H 86
325 275 61,100 51.623 05 1.88 238 . 10 188 794 2641 3434 .. 894 2* 41
322 322 4,347 4.347 0.5 1 88 2 38 10 ... 1 88 0 57 1 90 2 47 1 57 . 1 '*>
1013 913 23.879 21.520 2.0 7.53 953 10 20 753 399 1328 1727 274 225 H 2*
1792 1641 22.320 20.432 35 13 IS 1668 20 30 1318 373 1239 1612 293 80 12 W
2590 2385 21.661 19.944 50 1882 2382 30 40 1882 363 1207 1570 299 63 1207
3370 3113 21,308 19,680 65 24.47 3097 40 50 2447 358 1191 '1549 303 55 ll«l
3363 3105 21.257 19.629 65 24.47 3097 40 50 24 47 3 58 1191 1549 303 55 1191
4016 3709 21.091 19.517 80 30.11 38.11 50 60 3011 355 1181 1535 305 50 1181
4008 3716 21.052 19.478 80 30.11 38.11 50 60 3011 355 1181 1535 305 50 1181
3993 3693 20.970 19.396 8.0 30.11 38.11 50 60 3011 355 1181 1535 305 50 1181
4762 4412 20,940 19,403 9.5 3576 4526 60 70 3576 3.53 1174 1527 306 46 1174
1614 1513 21.644 20.295 30 11.29 14.29 20 20 1129 342 1139 !' 314 29 1139
2336 2186 21.041 19.691 4.5 16.94 2144 30 30 16.94 342 1139 1481 314 29 1139
3084 2885 20.840 19.496 60 2259 2859 40 40 22.59 3.42 11.39 1481 314 .29 1139
3068 2869 2Q.730 19,382 60 22.59 28.59 40 40 2259 342 1139 1481 314 29 1139
3836 3586 20.712 19,363 7.5 2823 3573 SO 50 28.23 3.42 1139 1481 314 29 1139
3751 3601 18,210 17,480 75 28.23 3573 60 30 2823 307 1022 1330 338 069 1022
4484 4284 18.440 17.620 90 3388 4288 7.0 40 33.88 3.13 1040 1353 334 078 1040
5230 4980 18.650 17,760 10.5 3952 5002 80 5.0 39.52 3.17 1053 1370 332 085 1053
1499 1448 21.500 20.776 2.5 9.41 11.91 2.0 10 9.41 3.07 1022 13.30 3.38 069 1022
5854 5654 17.298 16.708 120 45.17 5717 100 4.0 45.17 3.00 997 1296 343 056 «97
868 768 10.259 9.078 1.5 565 715 10 20 561 1.50 498 648 137 113 498
1600 1451 13,161 11.929 30 11.29 1429 2.0 30 11.29 2.08 693 902 1.92 117 69J
441 365 9.668 8.001 0.75 2.82 3.57 .. 15 3.32 141 469 610 ... 159 J 51
SO: SO:
. . ... 3 983 3 983 10 37£ 4 76 10 YTCinn 190 A y» t nn I-M
. . ... -r.^w* ^,7«J IV J fv ~r IV IV ,. J fQ | ml J ft ^ ft i UU . . 3 ZV
647 596 7.100 6,545 1.5 565 715 10 1.0 565 1.41 469 610 188 053 469
......
•
'Carbon and sulfur arc considered as gases for rnolal calculations only.
Note: This table is reprinted from Fuel Flue Gases, 1941 Edition,
courtesy of American Gas Association.
All gas volumes corrected to 60 F and 30 in. Hg dry.
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TABLE 2 — ENTHALPIES OF COMBUSTION GASES ::
Btu/lb.mole
AIR
oF N2 M.W. 28.97 C02 H20
32 0 000
60 194.9 19^.6 243.1 224.2
77 312.2 312.7 392.2 360.5
100 473.3 472.7 597.9 545.3
200 1,170 1,170 1,527 1,353
300 1,868 1,870 2,509 2,171
400 2,570 2,576 3,537 3,001
500 3,277 3,289 4,607 3,842
600 3,991 4,010 5,714 4,700
700 4,713 4,740 6,855 5,572
800 5,443 5,479 8,026 6,460
900 6,182 6,227 9,224 7,364
1,000 6,929 6,984 10,447 8,284
i 200 8,452 8,524 12,960 10,176
i 500 10,799 10,895 16,860 13,140
2 000 14,840 14,970 23,630 18,380
2,500 19,020 19,170 30,620 23,950
3,000 23,280 23,460 37,750 29,780
:: Kobe, Kenneth A., and Long, Ernest G., "Thermochemistry for the
Petroleum Industry," Petroleum Refiner, Vol. 28, No. II, November,
1949, page 129, table 9.
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TABLE T — MEAN MOLAL HEAT CAPACITIES OF GASES ABOVE 0°F
t = °F C= Btu/(lbmole)(F°)
t
0
200
400
600
800
1000
1200
1400
1600
1800
9nnn
99no
2400
9ROO
28QO
•znnn
7900
Ti. r\n
3"UU
•zCnn
JOUU
^800
unnn
u.9nn
ttUU
uunn
uftnn
URnn
N2
6.94
6.96
6.98
7.02
7.08
7.15
7.23
7.31
7 39
7 46
7 R"?
7 fin
7 66
7 79
7 78
7 JH
/ . o j
7 a?
7 09
/ • j£
7 OR
/ . JO
Son
& n/j.
807
8 in
81 Z
Q Ifi
o2
6.92
7.03
7.14
7.26
7.39
7.51
7.62
7.71
7 80
7 88
7 OR
8 09
8 08
8 14
8 19
8 24
8 29
8 T,£I
8 38
8 U9
8 tig
850
8 S4
8 58
8 62
H20
7.93
8.04
8.13
8.25
8.39
8.54
8.69
8.85
q oi
9 17
9 33
9 48
9 64
9 79
9 93
10 07
10 20
10 ?2
10 44
10 56
10 67
10 78
10 88
10 97
U08
C02
8.50
9.00
9.52
9.97
10.37
10.72
11.02
11.29
11 53
11.75
11 94
12 12
12.28
12 42
12 55
12 67
12 79
12 89
12 98
13 08
13 16
13 23
13 31
13 38
13 44
H2
6.86
6.89
6.93
6.95
6.97
6.98
7.01
7.03
7 07
7.10
7 15
7 20
7.24
7 28
7 33
7.38
7 43
7 48
7 53
7 57
7 62
7 66
7 70
7 75
7 79
CO
6.92
6.96
7.00
7.05
7.13
7.21
7.30
7.38
7 47
7.55
7 62
7.68
7.75
7 80
7.86
7.91
7 95
8 00
8.04
8.08
8 11
8 14
8 18
8 20
8 23
CH4
8.25
8.42
9.33
10.00
10.72
11.45
12.13
12.78
13.38
S02
9.9'
10.0
10.3
10.6
10.9
11.2
11.4
11.7
11.8
12.0
12.1
12.2
12.3
12.4
12.5
12.5
NH3
8.80
8.85
9.05
9.40
9.75
10.06
10.43
10.77
HC1
6.92
6.96
7.01 '
7.05
7.10
7.15
7.19
7.24
7.29
7.33
7.38
7.43
7.47
7.52
7.57
7.61
NO
7.1
7.2
7.2
7.3
7.3
7.4
7.5
7.6
7.7
7.7
7.8
7.8
7.9
8.0
8.0
8.1
AIR
6.94
6.97
7.01
7.07
7.15
7.23
7.31
7.39
7.48
7.55
7.62
7.69
7.75
7.81
7.86
7.92
7.96
8.01
8.05
8.09
8.13
8.16
8.19
8.22
8.26
CO
Williams, E. T., and Johnson, R. C., Stoichiometry for Chemical Engineers,
(New York: Me Graw-HiI I Book Company, 1958), page 321.
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. 140,000
01
to
ro
C3
3
-t-J
CO
ro
01
200
* 100
eC
f*fcA
"=C
300 600 900 1200 1500 1800 2100 2400 2700 3000
Flue gas exit temperature °F
Figure 1 . AVAILABLE HEATS FOR SOME
TYPICAL FUELSU) (Refer to 60°F)
This chart is only applicable
to cases in which there is no
unburned fuel in the products
of combustion.
The average temperature of the
hot mixture just beyond the end
of the flame may be read at the
point where the appropriate %
excess air curve intersects the
zero available heat line.
800
1200
1600 2000 2400 2800 3200
Flue gas temperature F
Figure 2. GENERALIZED AVAILABLE HEAT CHART FOR ALL
FUELS AT VARIOUS FLUE GAS TEMPERATURES AND VARIOUS
EXCESS COMBUSTION AIR^ (Refer to 60°F)
19
-------
C-l
COMPREHENSIVE COMBUSTION PROBLEM
PART I
s
For a natural gas of the following composition, find the gross heating
value by using the data given in Table I. The dry natural gas is delivered
at 60°F.
Natural Gas Analysis
(Volume or Mole Fraction)
N2 0.0515
CH4 0.81 I I
C2H6 0.0967
C3H8 0.0351 *^~V
0.0056
I .0000
PART 2
For the natural gas specified above, determine the available heat at
I200°F by two methods. Method a. should utilize Table 2, and Method b.
shou Id use Figure I . . ,
y\?fh
-------
PART I — SOLUTION*
I. The natural gas analysis was given in volume fractions.
Therefore, for each standard cubic foot of natural gas there
is 0.0515 scf of nitrogen, 0.81 I I scf of methane, and etc.
2. From Table I, fill in the following gross heats of
.combustion:
Gross Heat of Combustion
(Btu/scf )
N2
CH4
C2H6
C3H8
C4H|Q
3. The gross heating value for natural gas is:
HV~ = (0.81 I IMIOI3) + (0.09675(1792) + (0.0351X2590) + (0.0056) (3370)
b
HVo = Btu/scf natural gas
*Answers are on page 29.
21
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PART 2 — SOLUTION
Method a.
I. Balanced chemical combustion equations are written for
each component of the natural gas. Based on (either) one
mole or one standard cubic foot of natural gas, the required
amount of each constituent is written under that constituent
in each equation.
CH4
0.81 I I
2 02
2 (0.811 I)
C02
0.81 I I
2 H20
2 (0.81 I I )
C2H6
0.0967
7/2 02
7/2 (0.0967)
2 C02
2 (0.0967)
3 H20
3 (0.0967)
C3H8
0.0351
5 02
5 (0.0351 )
3 C02
3 (0.0351)
4 H20
4 (0.0351)
0.0056
13/2 02
13/2 (0.0056)
4 C02
4 (0.0056)
5 H20
5 (0.0056)
2. Calculate the number of standard cubic feet for each of the
following constituents of combustion:
C02
H20
scf/scf natural gas
scf/scf natural gas
scf/scf natural gas
scf/scf natural gas
22
-------
* Assuming the natural gas is burned with air (21$ 02 and
79/6 N2), the total amount of nitrogen In the products of
combustion is the sum of that l\|2 in the natural gas
(0.0515 scf) and that amount traveling with the required
02 (? x scf 02).
3. How many cubic feet of products are there per scf of natural gas
burned (assuming that a stoichiometric quantity of air is used)?
scf C02 + scf H20 + scf N2 _ scf Products
scf Gas scf Gas scf Gas scf Gas
4. The available heat (HAj) at any temperature is the gross
heating value (HVg) minus the a.mount of heat required to take
products of combustion to that temperature (E AH):
HAT = HVg - I AH
For the natural gas, the HV~ at 60T ir, 1105 Btu/scf. It is therefore
necessary to find Z AH for the products of combustion. Enthalpies of
combustion gases are given in Table 2, and the latent heat of vapori-
zation of water at 60°F Is I06Q Btu/lb,
1.1322 (scf C07/scf Gas) (12960 - 243) (Btu/lb. mole)
AH = . £ ,
2 379 (Scf C02/lb. mole)
AHCQ = (Btu/scf Gas)
8.2246 (scf N7/scf Gas) (8452 - 195) (Btu/lb. mole)
AHN = ±
2 379 (scf N2/lb. mole)
AHN = (Btu/scf Gas)
First the liquid water must be vaporized, and then the vapor is
heated to I200°F. ^ «u-'
idS) (Btu/lb. H20) 18 (Ib. H20/lb. mole) 2.0807 (scf H20/scf Gas)
379 (scf H20/lb. mole)
= (Btu/scf Gas)
23
-------
AH = 2.0807 (scf H20/scf Gas)(10176-224)(Btu/Ib. mole)
2 379 (scf H20/lb. mole)
AHH n = (Btu/scf Gas)
H20
E AH = (Btu/scf Gas)
HAI200°F = HVG ~ E AH
HA,
200
op = I 105 - Z AH
(Btu/scf natural gas)
24
-------
Part 2 — Solution
Method b.
I. Use Figure I and the following equation:
LHVSJ
HA-,
fuel in HV,
Fig. I L
2. The unknown is HA|2QO°F f°r a natural gas with an HVp of
1105 But/scf. From Figure I for a natural gas with an HVQ of
1059 Btu/scf the HA|2oo°F is 695 Btu/scf natural gas.
3. HA|200op - HVG
HAI200°F 1
HVr J Fuel in
Fig. I
HA
I200°F
or = 1105 x
695
1059
HA
I200°F
(Btu/scf natural gas)
-------
PART 3 — SOLUTION
Method a.
I. From Table 2 the enthalpy of air at 200°F and I200°F is
listed as 1170 and 8524 Btu/lb. mole respectively.
2. The heat rate required to heat 5000 scfm from 200°F to
I200°F is given as follows:
5000 (scfm) x (8524 - I 170)(Btu/lb. mole)
Q (Btu/min) =
379 (scf/lb. mole)
Q = Btu/min
3. If the amount of available heat per scf of natural gas is
725 Btu/scf (refer to PART 2), what is the natural gas rate that
is required to heat 5000 scfm of air from 200°F to I200°F?
97,000 Btu/min
Natural Gas Rate =
725 Btu/scf
Natural Gas Rate = scfm
26
-------
Part 3 — Solution
Method b.
I. From Table 3 the C_ for air is 6.97 (Btu/lb.mole F°)
_ P200
and the Cn for air is 7.31 (Btu/lb.mole F°).
P| 200
2. Using the following equation, the heat required to heat
5000 scfm from 200°F to I200°F is calculated.
0 = m [Cpf (t2 - tb) - Cpt| (t, - tb>]
Q = m [(7.3l)(Btu/lb.mole°F) x (1200 - 0)(F°) -
6.97 (Btu/lb.mole°F) x (200 - 0) F°]
Q = m [ (Btu/lb.mole)]
5000 (scfm)
m =
379 (scf/lb.mole)
m =
Ib. moIe/mi n.
Q = Btu/min.
3. If the amount of available heat per scf of natural gas is
725 Btu/scf (refer to PART 2), what is the natural gas rate that
is required to heat 5000 scfm of air from 200°F to I200°F?
Btu/min
Natural Gas Rate = —-
Btu/scf
Natural Gas Rate = scfm
27
-------
PART 4 — SOLUTION
I. Assuming the natural gas flow rate is 135 scfm, what is the
volumetric rate of products of combustion? (Remember from PART 2
that there are 11.44 scf of products for scf of natural gas burned.)
I I .44 (scf/scf gas) x (scfm gas) = scfm
2. The total flue gas flow rate is the total of the contaminated
gas stream flow rate and the flow rate of the products of combustion,
scfm + scfm = scfm
3. What is the flue gas flow rate at I200°F in cubic feet per
second ?
CJ- , I (1200 + 460) (°R)
6544 cfm x x = cfs
60 sec/min (60 + 460) (°R)
4. Assuming a linear gas velocity of 20 fps, what is the diameter of the
afterburner?
TT D2 2
Area = = (cfs)/ fps = (ft)
D =
v[ 17.4 (ft2) x 4
\I TT
5. Assuming a length-to-diameter design ratio of 2/0, how long
should the afterburner be?
2.0 x ft. = ft.
6. What is the residence time for the gases in the afterburner?
Length
Residence Time = — —; rr—
Linear Gas Velocity
Residence Time = ft./ fps
Residence Time = sec.
7. Is the residence time calculated in (6.) acceptable with
respect to the minimum residence time for air pollution control?
(Assume that the minimum design criteria for this afterburner
application requires the gases to be at I200°F for 0.3 second.)
28
-------
ANSWERS
PART
No. 2 1013, 1792, 2590, 3370
No. 3 I 105
PART 2
Method a.
No. 2 2.1726, 1.1322, 2.0807, 8.2246
No. 3 1.1322, 2.0807, 8.2246, 11.4375
No. 4 38, 179, 105, 55, 377, 728
Method b.
No. 3 725
PART 3
Method a.
No. 2 97,000
No. 3 133.8
Method b.
No. 2 7378; 13.2; 97,500
No. 3 97,500; 725; 134.5
PART 4
No. I 135, 1544
No. 2 1544, 5000, 6544
No. 3 348
No. 4 348, 20, 17.4, 4.71
No. 5 4.71, 9.42
No. 6 9.42, 20, 0.47
No. 7 Yes
29
-------
C-ll
Theoretical Flame Temperature
Calculate the theoretical flame temperature of a gas
stream with the following analysis (volumetric):
C02 IQ%
CO \0%
02 5%
N0 15%
The gas stream enters the afterburner at 200°F at
a flow rate of 6350 cfm.
The heat of combustion of CO at 60°F is 121,745
(Btu/lb. mole).
Mean heat capacity data are available in Table 3.
Assume that for the given concentrations the system
is within the flammability range.
30
-------
SOLUTION*
Calculate the flow rate in standard cfm (at 60°F).
_ (460 + 60) °R
^0 cfm x = Scfm
(460 + 200) °R
2. Calculate the number of Ib. moles of each constituent in the gas
stream per minute.
5000 (scf) x 0.10 (scf C02/scf)
nCQ = - = 1.32 I b. mole C02
2 379 (scf/lb. mole)
Ib. moles CO
Ib. mo Ies 0-
n = Ib. moles N
No ^^~~~~~~^~^~ L.
3. Write the balanced chemical equation for the oxidation of carbon
monoxide and write the number of Ib. moles of reactants available
under each reactant.
CO + 02
4. Is there enough oxygen present for the reaction to theoretically go
to completion or must additional air be added?
5. If you determined that no additional oxygen is theoretically required,
you are correct.
*Answers are on page 35.
31
-------
6. In order to calculate the theoretical flame temperature, the. combustion
process must be carried out (on paper) in such a manner as to use the
available data:
a. Cool all the gases from 200°F to 60°F.
b. React the materials at 60°F.
c. By using the heat available from (a) and (b),
heat the products to the final flame
temperature.
7. The above procedure can be written in the following form as a heat
balance: results (6a) + results(6b) = results(6c) or,
[E n. Cpj] (200 - 60) * n AHcomb> 6Qo = [E n, Cpi] (t - 60)
8. Evaluate [E n; C .] (200 - 60) for all the gases.
n C (From Table 3)
co2
CO
°2
N_
2
(Ib. mole)
1 .32
1 .32
0.66
9.90
E n, Cni
(Btu/lb.
9.00
6.96
7.03
6.96
(Btu/
[E n, C .](I40) = 13,246 Btu
9. Evaluate n AHcomb_ 6Qop.
(Ib. mole) x 121,745 (Btu/lb. mole)] = 160,703 Btu
32
-------
10. The equation from Part 7 may now be rewritten:
173,949 = [E n. C"pj] (t - 60)
This equation must be solved by trial and error for t.
II. Evaluate [E n. C .] (t - 60) when t -- I800°F.
C . (From Table 3)
(Ib. mole) (Btu/lb. mole)
C02 2.64
CO 0
07 0
N2 9.90
E n. C . = (Btu/F°)
i pi
[E MJ C j] (1800 - 60) = 182,481 Btu
This is too high, assume I700°F.
12. Evaluate [E n, C j] (t - 60) when t = I700°F.
n. C . (From Table 3)
(Ib. mole) (Btu/lb. mole)
2.64
9.90
E n. C . = (Btu/F°)
[E n. C".] (1700 - 60) = 171,030 Btu
This is close enough, therefore the flame temperature is I700°F,
33
-------
13. If a closer approximation for t is desired, the equation in (10) can
be solved for t:
[E n. C ;] (t - 60) = 173,949
173,949
t = =— + 60
1 ni Cpi
t = 1728 °F
The flame temperature can now be solved for explicitly because the
mean heat capacities can be considered to be approximately the same
at I728°F as at I700°F.
34
-------
ANSWERS
I. 5000
2. I.32, 0.66, 9.90
3. h
I .32, 0.66
4. There is enough oxygen.
8. 94.6
9. I.32
II. II .75, 7.55, 7.88, 7.46
104.874
12. I I .64, 7.43
104.287
35
-------
C-lll
PLAN REVIEW OF A CATALYTIC AFTERBURNER
Plans have been submitted for a catalytic afterburner. The
afterburner is to be installed to incinerate a 3000 cfm contaminated
gas stream discharged from a direct-fired paint baking oven at 350°F.
The following table summarizes the data taken from the plans.
,-„,-- ?
Data Sheet
Exhaust Flow Rate from Oven at 350°F
Exhaust Gas Temperature from Oven
Solvent Emission to Afterburner
Final Temperature in Afterburner
Gross Heating Value of Natural Gas
Total Heat Requirement
Natural Gas Requirement
Furnace Volume
Exhaust Flow Rate from Afterburner at IOOO°F
Gas Velocity Through Catalyst Bed
No. of Type A 19 x 24 x 33/4 in. Catalyst
Elements
3000 cfm
350 °F
0.3 Ib./m i n,
1000 °F
I 100 Btu/scf
26884 Btu/min
35.0 scfm
46.0 ft3
6530 cfm
8.6 fps
4
The following information and rules of thumb may be required in order
to review the plans for the catalytic afterburner.
a. Heat will be recovered from the afterburner effluent,
but that process will not be considered in this
p rob I em.
b. Catalytic afterburner operating temperatures of
approximately 950°F have been found sufficient
to control emissions from most process ovens.
c. Preheat burners are usually designed to increase
the temperature of the contaminated gases to the
required catalyst discharge gas temperature without
regard to the heating value of the contaminants
(especially if considerable concentration variation
occurs).
36
-------
d. A 10 per cent heat loss is usually a reasonable estimate
for an afterburner.
e. The properties of the contaminated effluent may usually
b° considered identical to those of air.
f. A design value for heat release rates for catalytic
afterburners is 50,000 (Btu/hr.ft3).
g. The catalyst manufacturer's literature suggests a super-
ficial gas velocity through the catalyst element of 10 fps.
37
-------
SOLUTION*
In order to review these plans, the following questions
must be answered.
I. Is the operating temperature adequate for sufficient control?
The plans indicate a combustion temperature of °F;
this is acceptable when compared to the °F rule of
thumb.
2. Is the fuel requirement adequate to maintain the operating temperature?
a. Calculate the number of lb,mo!es/min of gas to be heated
from 350*F to IOOO°F.
(460 + 60)°R
3000 (cfm) x = scfm
(460 + 350)°R
, , . Ib. mo Ie
(scfm) x = |b. moies/min.
379 scf
b. Use data for air from a table of enthalpy data (Table 2) to
find the heat requirement to heat the gas stream (air) from
350°F to IOOO°F.
(Ib. moles/mini x ( - KBtu/lb. mole) = 24196 Btu/min.
c. Calculate the total heat requirement, remembering the
10$ heat loss rule of thumb.
Total Heat Requirement = 24196 T 0.90
Total Heat Requirement = Btu/min
"Answers are on page 41.
38
-------
d. Calculate the heat available at IOOO°F by using Figure I
and assuming theoretical air.
HA
IOOO°F
HV
G
HAIOOO°F
HV
G
Fuel in Fig. I
HA|000°F =
J
1059
= 774 Btu/scf natural gas
e. By using (c) and (d), determine the natural gas requirement
and compare it with that given on the data sh°ct.
Total Heat Requirement
Natural Gas Requirement - -
Available Heat in Fuel
Btu/mi n
Natural Gas Requirement = _^^^i
Btu/scf
Natural Gas Requirement = 34.7 scfm
3. Is the catalyst section sized properly?
a. Determine the total volumetric gas rate at the afterburner
conditions (IOOO°F).
(I) One scf of natural gas yields 11.45 scf of
combustion products.
Volume of Combustion Products |OQO°F =
(460 + )°R
= 11.45 (scf/scf gas) x (scfm gas) x
(460 + )°R
Volume of Combustion Products |QOO°F =
39
-------
(2) Volume of Contaminated Gases |QOO°F =
(460 + IOOO)°R
(scfm) x
(460 + 60)°R
Volume Contaminated Gases |0oo°F =
(3) Total Volumetric Gas Rate IQQQOF = Volume of Combustion
Products |QOO°F + v°lume °f Contaminated Gases |000°F'
Total Volumetric Gas Rate IQQQOF = c^m +
Total Volumetric Gas Rate |QQO°F = 6530 cfm
b. How many 19 x 24 x 3 '4 in. catalyst elements are required
if a linear gas velocity of 10 fps (or 600 fpm) is used?
(cfm) x 144 (in.2/ft.2)
No. of Elements =
(fpm) x 19 (in.) x 24 (in.)
No. of Elements = 3.44
It is seen that four elements have been specified; this is a
conservative design allowing a slightly slower gas flow through
the elements.
40
-------
ANSWERS
I. 1000, 950
2. a. 1925
1925, 5.08
b. 5.08, 6984, 2222
c. 26884
d. 745
e. 26884, 774
3. a. (I) 35, 1000, 60
I 125
(2) 1925, 5405
(3) 1125, 5405
b. 6530, 600
41
-------
Section III
Adsorption
-------
AD-1
Heat of Adsorption
This problem is an exercise in plotting data and calculating the heat of
adsorption using the Clausius - Clapeyron equation. This equation is usually
stated as follows:
d(ln p) _
- q/R
Where
P
T
q
R
equilibrium adsorption pressure in mm Hg.
absolute temperature in °K
heat of adsorption in cal
g mole
ideal gas law constant = 2 cal
'K g mole
We will plot an NH^ adsorption isostere from the NH^ isotherms given in Figure 1.
An adsorption isostere is a representation of equilibrium data relating pressure
and temperature for a constant volume of material adsorbed per gram of carbon.
Calculate the heat of adsorption for adsorbing 35 cc of NH3 per gram of carbon.
The system temperature is 0°C and the pressure is 760 mm Hg.
Solution
1. Use Figure 2 to convert T to 1/T.
2. The Clausius - Clapeyron equation tells us that if we plot In p vs 1/T we
will get a straight line with a slope - q/R.
3. Complete the following table.
Data for 35 cc NH~ adsorbed per gram carbon
p (mm Hg)
T (°K) .
1/T (l/'K)
/
2¥VT
*,-v\.
*>*/*" 1
?D
re
t 15"
1<)3
r ^^-=
b /b
3^3
(answer is on next page)
43
-------
Answer to (3)
Data for Adsorption Isostere
(35 cc NH~ adsorbed per gram
carbon)
p (mm Hg)
T (°K)
1/T (l/'K)
15
249.5
.004
40
273
.00366
122
303
.0033
725
353
.00284
4. The next steps are to plot the values of p ?nd 1/T on the semi -log graph
paper provided (Figure 3) and to draw the best straight nine. Recall that
we can plot In p by plotting the value p on the log scale.
5. Our objective here is to obtain a value of the slope d * npvd (1/T). We
need to recall what d Inp is in terms of p. Since we aic plotting values of
p on a log scale, d Inp is really a very smal^ iicrsment of In p or (In
In
For example, In 10 - In 5 = 2.303 - 1.610 = 0.692
or In 10 - In 5 = In 10/5 = In 2 = 0.692
emulate the slope
which Is
'l •
Do this in the space below and use the values p2 = 600 mm Hg and PI = 200 mm Hg,
(answer is on next paqe)
44
-------
Answer to (5)
d In p _ 1n P2/P1
1/T2 -
In 600/200 = 1.100
.00278 - .003H» - .00036
d jn P, = .
d (1/T)
6. Calculate the value of q in units of cal/g mole.
In p _ _ /R
~ q/R
- q/R = -3050 and R = 2 cal/g mole °K
then q = 2 x 3050= - cai/g mole
(answer is on next page)
7. Use the ideal gas law (PV = n RT) to calculate the number of
gram moles of NH., in 35 cc.
P, atm; V, liters; n, g moles; T, °K; R = .08205
_.. atm pressure x liters NH_
R
.08205 x °K
< )
)
.08205
= (
(answer is on next page)
8. Now calculate the heat released by the adsorption of 35cc of NH^ per
gram of carbon.
q = 6100 x 0.00156 = ( ) cal
(answer is on next page)
45
-------
Answer to (6)
q = 6100 cal/g mole
Answer to (7)
(.035)
(.08205) (273)
n = 0.00156 g moles NH.
Answer to (8)
q = 9-52 cal
46
-------
-------
400
350
300
250
.0025
0045
-------
.0030
,0035
.0040
-------
AD-11
.Surface Area Determination
Using the NH3 adsorption isotherm at -23.5°C and the following BET
equation, calculate the surface area of the carbon.
V = V x where x = p/ps; Vm = Volume of
(1-x) [1+ (C-l) x] NH^ adsorbed per gram of carbon
when surface is covered by one
layer of NH, molecules.
By rearranging the previous equation, the following equation is obtained:
V (1-x) VmC A VmC
Figure 1 is given, and p , the saturation equilibrium pressure, is given as
1216 mm Hg. s
Tabulate the following values: p, x, 1-x, V, and
Using Figure 2 plot x vs. x/V (1-x). Find the intercept (1/V C) and the
slope (C-l/VmC), and solve for Vm.
The following information may be important:
2
The area occupied by one NH3 molecule is 13 A° .
p is expressed in mm Hg.
V is expressed in cubic centimeters, cc.
Avogadro's number = 6.028 x 1023
1 Angstrom = 1A° = 1 1 __ _ ln-8
10,000 u = cm - 10 cm
(A0)2 = 10~16 cm2
R - R? 057 > CC
R b^-Ub/ g mole °K
50
-------
Solution
1. By using the given value of ps and Figure 1 calculate values of x
= p/ps for specific values of volume adsorbed V. Enter these values
in the table below.
0.0074
•cf'*
V n ' ^
• , v
l-x
0.9926
€ f
25
50
75
100
125
V (1-x)
.000298
(answer is on next page)
2. Plot this data in the form of x vs
on Figure 2.
Obtain the intercept which is equal to ^-^ and the slope which is equal
C-l
to
V C.
m
J_
' I
m
rr > ~ \
M C
m
C-1
V C
m
(answer is on next page)
3. Solve these equations for V by substitution
m
(work space is on next page)
51
-------
Answer to (l)
p x 1-x V
9
22
*5
78
128
0. 007^0
0.01809
0.03700
Q.06k]k
0.10526
0.9926
0.9819
0.9630
0.9359
0.89^7
25
50
75
100
125
X
V (1-x)
.000298
.000368
.000512
.000685
.0009^1
Answer to (2)
= 0.00027
0.0064
(answer to (3) is on next page)
52
-------
Answer to (3)
=0.00027 and ^1=0.0064
LI / _L - r i - 0.0064
VmC ' VmC L~' " 0.00027
C-l = 23.7
C = 24.7
Substitution into gives :
Vm= ] - 1
C x 0.00027 24.7 x 0.00027
Vm = 150 cc NH3
g carbon
53
-------
Now we are ready to convert this volume of ammonia t<-> an equivalent
area which we must assume is the surface area of '-lie carbon. The
ammonia molecule has a molecular area of 13 A°2 (Angstrom Units).
We must visualize these molecules as covering the surface in a close
packed monolayer. Calculating the surface area is as simple as
counting the molecules of ammonia and multiplying by the area of
each molecule.
'First we will count the ammonia molecules. Use the ideal gas law to
calculate the gram moles of NH3 in the 150 cc (Vm). Then use Avogadro s
number to calculate the total number of molecules. Avogadro s number,
remember, is the number of molecules in a gram mole of a substance.
In the ideal gas law.PV = nRT, what values «f Ps T, and V must be used?
(answer is on next page)
5. n = PV.
RT
g moles NH. /
g carbon
) atm (
) cc/g carbon
82,06
atm cc
g moles K
(answer is on next page)
6. Now we can "count" the molecules by using Avogadro's number.
number molecules NH, g moles NH-
UM - n( „ ^*)x Avogadro's number(
gram carbon
g carbon
number molecules
t
g mole
number molecules NH.
g carbon
) x (
(answer is on next page)
54
-------
Answer to (4)
We must use those values that apply to the calculation of Vm<
P = 1 atm pressure
T = -23.5°C (249.5°K)
V = V ,cc/g carbon
Answer to (5)
n =
(1) x (150)
(82.06) x (249. bj
= 7 3 x 10'3 9 mole NH3
g carbon
Answer to (6)
number molecules NH,
g carbon
number molecules NH.
g carbon
= 7.3 x 10"3 x 6.028 x 1023
= 4.40 x 10
21
7. The surface area of the carbon is easily calculated by adding the cross-
sectional area of all the NH3 molecules. The only complication of this
simple operation is that the cross-sectional area of the molecules is in
square Angstrom units but the area of the carbon is normally quoted in
square centimeters.,,.
Note: 1 (A0)2 = 10~lb
cm
Surface Area of carbon = number NH3 molecules x cross section area per
hi I t HMM 1 M M • I 1 A •
cm
Area carbon = ( ) molecules x ( ) molecule
NH3 molecule
(answer is on next page)
55
-------
Answer to (7)
?i 16 cm
Area carbon = 4.40 x 10 molecules ( 13 x 10" ) molecule NH"3
6 cm
Area carbon = 5.7 x 10 g carbon
56
-------
-------
-------
Adsorption Dynamics
The breakthrough capacities were obtained on a test board using 4x6 mesh pelleted
carbon in bed depths of 4" and 9" removing benzene vapor at an inlet concentration
.of 2000 ppm by volume and 100 fpm superficial gas velocity.
D. = 4" bed C, = 15.00 g benzene/100 g carbon
D2 = 9" bed C2 = 17.77 g benzene/100 g carbon
1. Use the following equation to calculate the saturation capacity, Cs:
r
"
D2 - DI
2. Substitute C into the following equation to find the MTZ length:
MTZ = Y~£- DI 1 - ^1 1
L Cs J
The benzene adsorption breakthrough curve is symmetrical and "S" shaped, yielding at
x = 0.5. This implies 50% saturation in the MTZ.
3. Calculate the temperature rise of the above described carbon bed using the following
equation:
At = 6.1
(Sg/C) x 105 + 0.51 (SA/W)
C = 2000 ppm
Sg = 0.021 BTU/Cft3°FD
Sft = 0.25 BTU/Clb °F)
Since W is not known correctly until the temperature of the bed is known, the result
has to be obtained by successive approximations.
At is first calculated for t, where W is equal to Cs from the previous calculations.
A new W for temperature t + At, is obtained from the adsorption isobar.
For the type of carbon used and benzene vapor, the adsorption isobar is such that
a 10° C temperature rise results in a 10 percent loss of capacity. [Example: If W
is 40% at 20°C it will be 36% at 30°CJ
59
-------
Answers to AD-III
1. C = 17.77 (9) - 15.00 (4) on _ ,
s *-* *• J =20.0 g benzene
9-4 100 g carbon
2. MTZ = I 1 la /1-15\] = 2.0 inches
i ]p (1-15
I - 0.5JI >. 20
3. At = 6.1_
(0.021/2000) x 10s + 0.51 (0.25/20)
At = 5.8°F = 3.2°C
The resulting At should cause a neglibible loss in capacity. No
additional trials are necessary.
60
-------
AD-IV
Adsorption System
A solvent recovery system was designed to recover benzene from an air stream
having a concentration of 1316 ppm. This represents a vapor concentration of
1.0% Relative Saturation at 26°C which is the temperature of the carbon. The
plant capacity has increased; the question now regards the need to enlarge
the capacity of the adsorption unit by adding another adsorber, changing the
type of carbon, or changing the adsorption cycle.
The new load on the system is estimated to be a 75% increase in benzene. The
properties of carbons available are as follows:
Carbon properties I II
Apparent Density 23 26
Lbs/ft3
Particle Size
U.S. Standard Sieve 4x6 4x6
Information on the existing system is given below:
Carbon used: Type II
Gas Velocity: 100 fpm (STP)
Air Volume: 12,000 CFM
Bed area: 120 ft2
Bed depth: 24 inches
Carbon charge per adsorber: 5832 Ibs.
Working charge, wt. per cent: 31.0 Ib. benzene/100 Ib. carbon
Working charge, pounds of solvent: 1808 Ibs.
Steam solvent ratio of 3:1
Bed pressure drop: 7.2 inches w.g.
Adsorbent temperature: 26°C
Residual charge of benzene: 2.2%
Standard Temperature and Pressure (STP): 0°C and 760 mm Hg
Determine:
1. Benzene vapor content at 1% Relative Saturation @ 26°C.
2. Benzene vapor content at the new concentration (75% increase).
3. Carbon capacities for carbons I and II at each benzene concentration.
4. The break through carbon loading at new condition.
5. The working charge at new condition.
6. Whether to alter the system.
7. What changes take place in the unit due to the new conditions.
61
-------
SOLUTION
1. Calculate the benzene vapor content of the gas in Ibs per cu. ft. at
1% Relative Saturation using the formula W = MPV/RT (P:mm Hg, V: ft3, T: 539
°R, R: 555 mm Hg ft3
Ib mole "RJ-
i
i i '
At first we will calculate the benzene vapor content at 100% RS and then at
1% RS. Note that the vapor pressure of benzene at 26°C is 100 mm Hg, and that
the molecular weight of benzene is 78.11.
M flbs Benzene .. P (mm Hg)
W ......3 Llb mole J
V lb/ft =
, mm Hg ft .
R e Ib mole UR JX
W = _( _ ) lbs/lb mole x ( ) mm Hg
f i """ HE f t Y r 1 °R
( ) Ib mo1 3 u" X l J R
( ) lbs/ft3 @ 100% RS
( ) lbs/ft3 @ 1% RS (answer is on next page)
2. Next we want to adjust the data to correspond to 75% increase in pollutant
concentration. This 75% increase will be a multiplying factor of _ .
(answer is on next page)
3. What will be the new benzene concentration after we multiply by this factor?
Benzene concentration at 1% RS x factor = Benzene cone, at new condition.
^—^^^—^—^— X ^—^—— — ^-^— = — ^— —
(answer is on next page)
4. Next you must obtain the carbon capacities from Figure 1 for each benzene
concentration or % R S.
62
-------
Answer to (1)
W/V = (78.11) (100)/(555) (539)
W/v = 0.0261 Ibs/ft3 @ 1QO% RS
W/V = .000261 lb/ft3 @ 1% RS
Answer to (2)
Factor will be 1.75 or 100% + 75% = 175%
Answer to
(3)
,*, lb
3
ft5
lb
ft3
(# 4 cont.)
Concentration of Benzene
1.0% RS
1.75% RS
Carbon
Type I
39.0
Capacities wt
Type II
34.5
(answer is on next page)
5. Now that we have the saturation capacity we can calculate the adsorber
charge at breakthrough. To do this we will assume that the MTZ length is,
as calculated previously, 2 inches,and that the average concentration in
the MTZ is 1/2 Saturation charge.
This diagram should help.
22"
Loading:
Saturation loading
Loading MTZ:
Saturation loading r_2
2 24
63
-------
Answer to (4)
Carbon Capacity, wt
i nt RS Type I Type "
1>0% RS 39.0 34.5
1.75% RS 41.5 36.0
(# 5 cont.)
The breakthrough charge is the sum of these two:
+ _ % or Ibs benzene
100 :.'„ carbon
(answer is on next page)
6. Now calculate the working charge. The working charge is the difference
between the breakthrough charge and the "heel" or residual charge.
Working charge = - = % or Ibs Benzene
100 Ib carbon
(answer is on next page)
7. The qiestion now is - do we need to alter the system to handle the
additional load on the system ?
yes
no (answer is on next page)
8. The next question concerns what changes do occur in the system. Check
those factors which you think will change in the adsorber.
Pressure drop
Regeneration time
Adsorption time
(answer is on next page)
64
-------
Answer to (5)
36 ( ) + 18(-i) = 34.5% or Ibs Benzene
24 100 Ib carbon
Answer to (6)
Working charge =34.5 - 2.2 =32.3% or Ibs Benzene
100 Ib carbon
Answer to (7)
No, since the new working charge is still greater than the 31.0
percent as designed.
Answer to (8)
Adsorption time will be shortenedand as a result the time between
regeneration will shorten. The actual regeneration time may not be
shortened however.
65
-------
-------
AD-V
Design Problem
An electrolytic process generates 5000 CFM of high pressure (1500 psi) hydrogen
with contaminants consisting of oil vapor and Mercury vapor.
Temp: 38 °C
Hg inlet concentration: 7.5 ppm (Volume)
RH: 50%
Oil vapor concentration: 1 ppm (Volume)
Given Figures 1-4, select a carbon, size an adsorber for 6 months life using a
20% safety factor, and calculate the pressure drop across the bed.
Carbon Type Density
G215 (10 x 20) 30 lbs/ft3
G352 (10 x 20) 27 lbs/ft3
Since regeneration of this carbon will be far more complicated than a steam
regeneration we will not consider a multiple bed unit. Carbon replacement will
be made at the breakthrough time.
67
-------
Solution
1 First inspect the adsorption curves and select a carbon you feel is most
suitable for the adsorption of mercury vapors.
a Consider the effects of aging on the capacity of the carbon.
Calculate the percent reduction at breakthrough for each
conditioning experiment.
b For carbon aged 20 days @ 65°C, 50% RH (Figure 2):
% Reduction =
c For carbon aged 30 days @ 30°C, 100% RH (Figure 2):
% Reduction =
T is on next page)
d The reduction expected would then be between 15% -r.d 30%. A 20%
reduction would be a good estimate. This will be in addition to
the suggested safety factor of 20%, suggested in cne statement of
the problem.
e Notice that Figure 1 indicate" the effect of added Iodine on
mercury adsorption. Select the mosc suitable carbon from this
figure, and from Figure 3.
Carbon _ having _ % 1 2
(answer is on next page)
2 The next step is to size the adsorber using the carbon you selected. The
problem requires a 20% safety factor to be used. This factor can be applied
only to the carbon required, not to the gas flow rate since this would affect
pressure drop calculations.
a If the adsorption bed is to last six months we must calculate the
total pounds of mercury to be removed in that period.
(l.)The total gas volume to be treated is:
(answer is on next page)
( 2.)The Hg concentration is 7.5 ppm(volume). Now calculate the volume
of Hg vapor in the gas.
Hg (vol) = ^- x ( ) ft5 = ( ) ft3 Hg
10^ 6 mo. 6 mo.
(answer is on next page)
68
-------
Answer to (1- b§c)
1-b % Reduction =(49^9g 420jlOO = 14.3%
1-c % Reduction =('490;l" 54° ) 100 = 30.6%
\ 490 /
Answer to (1-e)
You should have selected carbon G352 having 15% I_
Answer to (2-a(l))
5000 x 60 x 24 x 183 = 1.318 x 109 ft3 /6 mo.
Answer to (2-a(2))
x (1.318 x 109) = 9882 ft3 Hg
106
3 From Figure 3, note that the breakthrough time for G352 with 15% I. is 800
minutes.
a We can introduce the 20% factor for carbon conditioning at this
point by reducing the breakthrough time by 20%. The breakthrough
time becomes: ( )min.x( ) = ( ) min.
(answer is on next page)
b Next calculate the treated volume of gas during this breakthrough
period for the laboratory data on Figure 3 for carbon G352.
Remember we have adjusted the breakthrough time for conditioning
the carbon.
Flow rate (ml/min) x breakthrough time (min) = gas volume treated (ml)
( ) ml/min x ( ) min = ( ) ml
(answer is on next page)
c Now you can calculate the volume of mercury vapor captured on each
gram of carbon.
(ppm Hg in H2) x (Vol. HZ) ( ) ml Hg Vap°r
= g carbon
(106) x (grams carbon tested)
(answer is on next page)
69
-------
Answer to (3-a)
800 min. breakthrough time.
800 x 0.80 = 640 minutes for conditioned carbon.
This is 10 hours and 40 minutes.
Answer to (3-b)
64000 ml/min x 640 min. =4.1 x 107 ml
Answer to ( 3-c)
7.5ppm x 4.1 x 107 ml H2
10b x 42 gram carbon
ml Hg vapor I
~ g carbon !
d Express the volume of Hg vapor to be adsorbed (refer 2-a-(2)) as ml
) cu ft Hg x 28,316
= ( ) ml Hg vapor
(.answer is on next page)
4 Now calculate the grams of carbon rtquiicu. To do this assume that the length
of the MTZ (mass transfer zone) is insignificant. A rough estimate would place it
at about 0.375 inches.
) ml Hg vapor _ ,
v J ml Hg
g carbon
b Calculate the pounds of carbon
Ib carbon =
) g carbon
(answer is on next page)
454 g/lb
) g carbon _ ,
) Ib carbon
(answer is on next page)
5 With a flow rate of only 5000 CFM we can use a thick bed of carbon and still
use little power for moving the gas. The bed depth selected is 10 feet.
70
-------
Answer to (3-d)
9882 x 28,316 = 280 x 106 ml Hg vapor
Answer to (4-a)
(280 x 1Q6 ml Hg) = (38 2 x 10^) g carbon
. ml Hg
*• ' •* g carbon
Answer to (4-b)
38!?,x lp =8.42 x 104 Ibs carbon
454
a Calculate the volume of the carbon in the system.
{., . ) -3 C _ ) Ib carbon _
(Vol.) ft3 = ^ - ) * --
(answer is on next page)
b Now calculate the area of the carbon bed for a 10 ft bed depth. Introduce
the 20% safety factor here on the volume of carbon in the adsorber.
(Area) ft* - -^ - TfHh ' < > ***
(answer is on next page)
6 The superficial gas velocity is needed to determine the pressure drop across
the adsorption bed.
a
f ~\ PFM
(gas velocity) ft/min = ± - y-ftz = C D ft/min.
(answer is on next page)
b Now determine the pressure drop for 10 x 20 mesh carbon from Figure 4.
AP = ( ) in H20/inch carbon
For the carbon bed, the pressure drop is:
AP = f ) . n 2 x ( ) in carbon
a*Bed m carbon
= ( ) in H20 (answer is on next page)
71
-------
Answer to (5-a)
8.42 x 10 Ib carbon = 3 ^ x ^ ft=3
27 Ib/ft3
Answer to (5-b)
(1.2) (3120) ft3 = 375 ft2
10 ft
Answer to (6-a)
5000 CFM
375 ft2
13.3 ft/min.
Answer to (6-b)
= (0.275) in HZ°— x 120 in carbon = 33.0 in H.O
APBed
in carbon
72
-------
z
o
u
ofi
o
I
0.30
0.25
0.20
0.15
0.10
0.05
0.0
100
300 500
BREAKTHROUGH TIME, MINUTES
700
800
-------
200
300
400 500
BREAKTHROUGH T" '., MINUTES
700
800
-------
CIA:
us
-IN
jpn
til
50
Bi)
RA
OOC)
/mi
Tit
SEE
2D
irr
10
3J2
•Or
7
100
200
300
400
500 600 700
BREAKTHROUGH TIME MIN.
800
900
1000
-------
ACTIVATED CARBON
PRESSURE DROP CURVES
FIGURE 1
30
50 60 70 80 90
1.0
0.9
OS
^0,7
MEASUREMENTS MADE AT
MAXIMUM PACKING DENSITY
20 30 40 60 60 70 00 90
VELOCITY (STANDARD AIRJ-FPM.
76
-------
AD-VI
Design Problem
The exhaust from a coating operation amounts to 12,000 CFM of gas
containing:
7 Ibs/hour of 2-nitro propane
22 Ibs/hour of MEK, methyl ethyl ketone
90 Ibs/hour of MIBK, methyl isobutyl ketone
The temperature will vary between 70°F and 95°F, averaging 82.5°F (28°C).
The pressure is atmospheric. The blower is capable of developing 10 inch
w.g. There is no visible evidence of solvent droplets or particulate carry
over.
V.P.,mm Hg Mol. WT. B.P.,°C
2-NP 20.0 89.09 120.3
MEK 115.0 72.10 76.6
MIBK 9.0 100.16 118.0
Solution
1. Using the ideal gas law, determine the saturation concentration for each
component at 28°C.
Saturation concentration:
W Ibs solvent _ MP_
V cu ft RT
W = weight solvent in gas, Ib.
V = volume, ft
M = molecular weight of solvent, Ib./lb mole
T = absolute temperature, °R
P = vapor pressure of solvent component, mm Hg
R = 555 (mm Hg ft3)/(lb moles °R)
'«"-«• y ° (sss x^' L-( > H*
r ucv W ( ) X ( ) , , IbS
for MEK V ' 555 x 543 = ( 5 ^3
W ( ) x ( ) Ibs
for MIBK TT- = —c c c c A T = C J —T"
V 555 x 543 ~ 3
(answer is on next page)
77
-------
Answer to (1) 3 lb "?
,, MD Wi 89.09 x 20.0 = 5.91 x 10~ /ft:
2-NP V' = 55S x 543
W 72.1 x 115 _ j ., ,n-2 lb/f 3
V = 555 x 543 2.75x10 /ft
MTDI, W. 100.2 x 9.0 = 2.99 x 10"3 lb/ft3
MIBK V = 555 x 543
2. The next step is to determine the actual concentration of each solvent
from the stated solvent flow rates.
For example, 2-NP is emitted at 7 Ibs/hour:
7 Ibs 2-NP/hr. = (9 72 x lQ-6llb/ft3
12,000 CFM x 60 Min/hour * ' J '
Now calculate the values for MEK and MIBK.
( _ ] Ibs MEK/hr. = . b/f 3
TmTn/hour < ' 1D/It
( _ ) Ibs MIBK/hr. _, T
( ) CFM ( ) min/hour v '
(answer is on next page)
3. Now you can calculate the percent relative saturation, %RS, of each solvent
component .
actual concentration
saturation concentration
= -£ - 2. x 100 =
%RS.. (MEK) = — £ - j- x 100 =
%RS,!(MIBK) = — £ - 1 x 100 =
(answer is on next page)
4. At these concentrations 3 carbons show the following equilibrium (isothermal)
capacities at 28° C. These capacities are given in wt. % units.
Carbon I II III
2-NP 45 42 40
MEK 27.0 27.0 32
MIBK 35.0 37.0 33
Surface Area 1600 m2/g 1400 m2/g 1200 m
-------
Answer to (2)
MEK 22
12,000 x 60
= (3.05 x 10"5) lb/ft3
MIBK 90
12,000 x 60
= (1.25 x 10~4) lb/ft3
Answer to (3)
%RS (2-NP) _ 9.72 x 10
5.91 x 10"3
-6
x 100 = 0.164%
3.05 x 10
%RS (MEK)
2.75 x 10"
%RS (MIBK) _ 1.25 x 10
2.99 x 10"
-5
-4
x 100 = 0.111%
x 100 = 4.17%
\ Look at the capacities each carbon has for the solvents and select a carbon.
Checkyour selection:
Carbon I
Carbon II
Carbon III
Give your reason here:
(answer is on next page)
5. Laboratory tests run on the solvent mixture supplied the following information
for Type II carbon:
At a gas velocity of 100 FPM,a working capacity of 11.0% is
obtained using a 3:1 steam-solvent ratio.
At a gas velocity of 80 FPM, a working capacity of 14.0% is
obtained using a 4:1 steam-solvent ratio.
Does increasing the steam: solvent ratio increase or decrease the cycle
time available for adsorption? Why?
Increase
Decrease
(answer is on next page)
79
-------
Answer to (4)
Carbon II should be selected since it has the highest capacity for the
component present in the highest concentration which is MIBK. Breakthrough
would occur first with this component.
Answer to (5)
Increases. The higher steam to solvent ratio reduces the "heel" or
residual solvent on the carbon. This increases the working charge which
increases the time of the adsorption step for the cycle. The additional
steam required also increases the time of the regeneration step.
6. The remainder of the problem is to size the adsorbers required to remove:
O.--a.—640 Ibs solvent mixture
\?*. b. 1020 Ibs solvent mixture
per regeneration using the 100 FPM value for (a.) and the 80 FPM value for (
7. Determine the adsorbent volume for a gas velocity of 100 FPM.
a The working capacity is % (see part 5).
(answer is on next page)
b Calculate the pounds of carbon required.
0.11 Ibs solvent
Ib carbon
Ibs solvent
x Ibs carbon
x = ( ) Ibs carbon , . .
' (answer is on next page)
c What is the pressure drop per inch of 4 x 6 mesh carbon (see Figure 1)?
Ap = ( ) in H20
in carbon (answer is on next page)
d Calculate the bed depth for the previously specified lOin H20 pressure drop.
_ , , .. APR , Jin HJ3 r •)
Bed depth. Bed! 2 ( J_ _ ^ )in carbon
AP in H20/in carbon ( )
(answer is on next page)
e Calculate the cross-sectional area of the adsorber and the diameter for
100FPM and 64Q# solvent adsorbed.
80
-------
Answer to (7-a)
Working capacity is 11% for 100 FPM
Answer to(7-b)
x = n4?!— = 581° lbs carbon
VJ t -L. J-
Answer to (7-c)
Ap = 0.62 in
in carbon
@ 100 FPM 4x6 mesh
Answer to (7-d)
Bed depth = 10 = 16 inches carbon
u. oZ
Area = volume = lbs carbon/bulk density carbon
dePth depth of carbon
Area _ _( / )_ _ «• -. £t2
( )
(answer is on next page)
f Now calculate the diameter of a cylindrical adsorber having this area.
_ /Area x 4 /
V n V/~
Diameter _ /Area x 4 / 167 x 4
n
Diameter = /
(answer is on next page)
8. Determine the same values for 80 FPM and 1020 Ib solvent adsorbed. This
time go through all the calculations, then check the answers on page 83.
a The working capacity is % .
b lbs of carbon required:
( ) lbs solvent } lbs splvent
Ib carbon = lbs carbon
x = ( ) lbs carbon
c Pressure drop per inch of 4 x 6 mesh carbon:
Ap = ( ) in H^O
in carbon
d Bed depth _ ( ) in H20 across bed _ ^ ) in carbon
( ) in H2°/in carbon
81
-------
Answer to (7-e)
Area = 5810 Ibs carbon/26 lbs/ft
_ (16/12)ft carbon
ft.
Answer to (7-f)
Diameter =/212.5 = 14.58 ft
e Cross-sectional area of adsorber:
Area = ( ) Ibs carbon in bed/( ) carbon bulk density
( ) carbon bed depth
2
(answer is on next page)
f Diameter of cylindrical adsorber:
Area = (
)
Diameter =>
Diameter = (
/Area x 4 _ I ( )_
V n N/ "
(answer is on next page)
82
-------
Answer to part (8)
a Working capacity is 14% for 80 FPM.
b Ib carbon required:
0.14 = ; x = 7290 Ibs carbon
A
c Ap per inch of 4 x 6 mesh carbon @ 80 FPM:
Ap = 0.44 in t^O/inch carbon
d Bed depth- £04" S°H20/ln carbon = ".7 in carbon
e Cross- sectional area:
Area = C7290/26) = 143 ft2
(22.7/12)
f Diameter of cylindrical adsorber:
Diameter = X 4 =i89 = 13.7
=/148
V "
83
-------
ACTIVATED CARBON
PRESSURE DROP CURVES
70 80 90
1,0
0,9
OS
gg^gf^»it^f^|pgp;
MEASUREMENTS MADE AT
[MAXIMUM PACKING
20 30 40 60
VELOCITY (STANDARD AIR)-FPM.
60 70 00 90
84
-------
AD-VII
Plan Review of an Adsorption System
This solvent recovery plant must recover acetone from an air stream having a
volume flow of 30,000 CFM, containing 0.15% (volume) acetone. This solvent re-
covery plant operates 24 hours per day, 365 days per year. The permit to construct
this unit has been applied for and you are asked to review the plans for the
adsorption system.
The diagram for this unit is presented here. Note that the solvent separation
is done by a distillation process which will be reviewed separately.
Acetone
1
sti: i
T
+ Air
The available information is summarized:
Storage Water
Valve
B Blower
1. Air flow:30,000 CFM @ 1 ATM
2. Air temperature:20°C (293°K or 527°R)
3. Concentration of acetone in air'. 0.15%(volume)
4. Operation: 24 hours per day,365 days per year .
5. Carbon charge per adsorber.'15,000 Ibs. Carbon "B" ,(4X6 mesbj
6. Adsorber diameter:18 feet (254.1 ft2)
7. Adsorption bed depth:1.96 feet
8. Vapor pressure acetone @ 20°C;170 mm Hg
9. Blower: centrifugal straight blade
60 inch wheel diameter
180 HP motor, 1600 RPM
10. Plant steam available, minimum, 5,000 Ibs (5 PSIG) per hour
11. Two of the three adsorbers are in the adsorption phase at all times.
In reviewing these plans you will check:
1. The time to run each adsorber to breakthrough.
85
-------
2. The Ibs. of steam required to regenerate the carbon.
3. The size of the blower.
1. The first step is to check the adsorption time per adsorber.
a First we must calculate the saturation loading of the carbon bed and
the fraction of the carbon saturated.
b The relative saturation of acetone in the air stream is expressed as
follows:
gg _ partial pressure acetone in air
saturation partial pressure of acetone @ 20°C
This is usually expressed as the percent relative saturation by mutliplying RS by
a factor of 100.
RS = (-) X 100 = 0.67
c Determine the saturation capacity of carbon B for acetone at the calculated
% RS from Figure 3.
Saturation capacity = ( )% wt (answer is on next page)
This is an ideal capacity, but in reality we can adsorb only 75-85% of the
saturation capacity at break through because the mass transfer zone (MTZ) is
less than saturated.
d Now we must know the length of the MTZ. Why?
In order to calculate adsorber length?
(answer is on next page)
In order to calculate adsorber run time?
In order to calculate the break through solvent loading?
e The length of the MTZ is supplied by the carbon manufacturer based upon
laboratory data (see Figure 5). The MTZ length is a function of gas velocity, time,
adsorbate concentration, temperature of gas and carbon, system pressure, and type
and size of the carbon. In this case the length of the MTZ is 2 inches.
f Now we can calculate the breakthrough capacity as follows:
Breakthrough capacity =
[(Ave. loading in MTZ)x(MTZ length)] + [(Saturation loading)x(bed length-MTZ length)]
bed length
86
-------
Breakthrough capacity =
Breakthrough capacity
)J+ [(0.
23) (1.96
)J
C )
115) ( ) + ( ) ( )
[(0.
]
) + (
_ . .
Breakthrough capacity
v
)
Ibs Acetone
lb carbon
(answer is on next page)
Answer to (c)
Saturation capacity = 23% wt
Answer to (d)
In order to calculate the carbon breakthrough loading. The adsorber length is
fixed, and the run time depends upon the solvent loading on the carbon.
g Next we calculate the working capacity of the carbon. The working capacity
of the carbon is lower than the saturation capacity or the breakthrough capacity
and results from unrecoverable solvent in the carbon and from a lower packing density
than laboratory specifications. Based upon previous experience* the unrecoverable
solvent or (heel) is 2% wt, and the lower capacity due to lower packing density is
estimated to be 3% wt.
The working capacity is then calculated as follows :
% W. C. = (Break capacity) % - (heel) % - 3%
% W. C. = 22% -2.0% -3.0%= ( )% (answer is on next page)
h Calculate the mass flow rate of the acetone.
15 cu ft acetone
™ nnn r,™. • „
30,000 CFM air X 10>000 cu ft air
,
= (
cu ft acetone
- 55 -
The ideal gas law is used to calculate the mass flow rate.
(P)(V) (MW) _ ( ) ATM x ( ) cu ft/min x ( ) Ib/mole
"
00(T)
W = ( ) Ib/min
ATM cu ft
lb mole "R X
(answer is on next page)
87
-------
Answer to (f)
Breakthrough capacity
Breakthrough capacity
fi^23)(0.166l+((0.23)(1.96 -0.166)]
L 2
(1.96)
[(0.115) (0.166)1+ £b.23) (1.794)1
(1.96)
Breakthrough capacity = (0.0190)+(0.413) = 0.2204
(1.96)
Or 22%
Answer to (g)
% W. C. = 17%
Answer to (h)
30000 CFM air x
10000 cu ft air
= (45) cu ft acetone
min
w = (1) (45) (58) = (6.77) Ib acetone/min
(0.73) (527)
i Now we calculate the adsorption time per adsorption bed.
NOTE: 2 adsorbers on stream, each cleaning 1/2 the total gas volume
15 T OOP Ib carbon
adsorber
(
Ib acetone
Ib carbon
( }
v '
min
(
} minutes
Ib acetone
(answer is on next page)
j For regeneration of each adsorption bed we will have how much time to
complete the regeneration cycle?
( ) hours (answer is on next page)
The time for regeneration should be less than 1/2 the time for breakthrough.
88
-------
Answer to (i)
15,000 x 0.17 x 2/6.77 = 753 minutes
This is 12.55 hours or 12 hrs 33 min.
Note: Each adsorber carries
1/2 the flow.
Answer to (j)
If your answer is 6.275 hours or 6 hours 16 minutes you are correct.
k The steam should be supplied during the first 3 hours of the regeneration cycle.
Why not supply the steam for the full 6 hours and 16 minutes? What else must be done
before returning the adsorption bed to the adsorb cycle?
(answer is on next page)
2. Now calculate the total pounds of steam required to regenerate the carbon bed
using Figure 4.
a Since the percent working charge is about 17% we can refer to Figure
4 and obtain the steam requirement for carbon B.
Ib steam required
Ib solvent removed
(answer is on next page)
b Now we can calculate the Ibs of steam required.
,, 2.4 Ib steam .17 Ib acetone
lb steam Ib solvent X Ib carbon
= ( ) Ibs steam
' j lb carbon
(answer is on next page)
c If we wish to reduce the required Ibs of steam we would change
the amount of "heel" on the carbon. By deciding to operate with
a lower working charge we increase the percent "heel" and require
less steam.
89
-------
Answer to (k)
Before returning the adsorber to the adsorb cycle we must:
r. remove moisture from the carbon
2. reduce the temperature of the carbon
Both are accomplished by blowing air through the hot, moisture-laden
carbon bed.
Answer to (a)
If you answered 2.4 you are correct. If not, look again at Figure A.
Answer to (b)
lb steam = M "» steam . 17 lb Acetone x Q ^ carbon = lfc
Ib solvent lb carbon
d How would this increase of the percent increase
"heel" affect the adsorption time?
decrease
e How would this affect the regeneration time (assuming a constant
steam supply rate)?
increase
decrease (answer is on next page)
f The regeneration time depends upon the steam supply rate to heat the
carbon plus the time to cool the carbon to the operating temperature. This is
not usually a critical factor. Four hours is usually ample time to regenerate
the largest adsorber bed, but we may allow a longer time depending upon avail-
able steam and breakthrough time.
3. The last step is to check the pressure drop across the adsorber bed to
determine blower capacity.
a Calculate the superficial velocity through the adsorber bed.
2
Ns (fpm) = volume flow (CFM) * adsorber area (ft )
= ( ) CFM i ( )ft2
= ( ). fpm (answer is on next page)
b Use Figure 1 to determine the pressure 'drop per inch of* bed depth.
Then calculate the pressure drop across the bed.
90
-------
Answer to (d) & (e)
If you answered "decrease" in both cases you are correct. What we have done
is to decrease the effective size of the system.
Answer to (a)
Ns (fpm) = 15,000 cfm 4 254.1 ft2 = 59 fpm
) (in H20/in of bed) x 12 (in/ft) x (
( ) (in H20/bed)
) (ft/bed) =
(answer is on page 94)
c The next logical step is to check the fan or blower size based upon the
calculated pressure drop. This is easily checked using the fan multi rating table
(Figure 2) and the "fan laws" .
(1.) We will use information for the 15,120 cfm fan.
Summarizing the fan conditions from the multirating
table:
Vtl =
15,120 cfm
46 in.
1622 rpm
45.9 bhp
7.0 in. H20
(2.) The fan laws are as follows:
(a.) Vt = k D3 N or
(b.) hr = k D2 N2 p or
hr2 " hrl
D2
N
1 J
(c.) P - k D5 N3 p or
|-D21
J
91
-------
(3.). Summarizing the specifications for the proposed fan.
D2 = 60 in.
N2 = 1600 rpm
Vt2 must be
h must be
30,000 cfm
6.5 in. H20
must be 4 180 bhp
(4.) Calculate the volumetric flow for the proposed fan using
the proper equation from (3c(2)(a)) above.
't2
V
tl
D2
DIJ
NI
) cfm
Does this meet the requirement?
Yes No
(answer is on page 94)
(5.) Calculate the static pressure, hr , for the proposed fan
using the proper equation from (3c(2)(b)).
hr2
ir2
D2
D
No 1
£. £.
- X —
•l Nl J
( K )
Ir2
) in. H20
Does this meet the requirement?
Yes No
92
(answer is on page 94)
-------
(6.) If we need only 6.5 in. 1^0 pressure drop, how will
we control the pressure?
a. Install a valve in the duct to the adsorber?
b. Recycle the gas through the blower at a
controlled rate?
(answer is on next page)
(7.) Calculate the horsepower required by the proposed fan
using the proper equation from (3c(2)(c)).
*1
D2 - '
if
P2 = ( ) bhp
Does this meet the requirement?
Ves No (answer is on next page)
4 You have now completed all the calculations necessary to review the plans for
this adsorber. The actual approval is subject to engineering judgement as to
whether the data supplied or assumptions made are valid, or whether the design is
conservative. This problem is difficult to simulate in an exercise, but it is the
main problem in a real situation when plans have been submitted for review.
Based upon the calculations made in this problem would you conclude that this design
is:
critical f
r\
conservative r
ridiculous?
Do you approve this plan for a construction permit? yes, no.
(answer is on next page)
93
-------
Answer to (3-b)
(0.277) x (12) x (1.96) = 6.5 in H20/bed
Answers to (3c(4)-(7))
(4) (15,120) |-60f riMOJ = 33>100 cfm
(5
yes
2
(7.0) f60 16001 _
L46 x I622"J - 11<6 in' H2°
yes
(6 ) Recycling gas through the blower is the usual control
method since the blower can operate most efficiently
at constant speed and constant power.
(7 ) (45.9)
yes
Answers to (4 )
conservative
yes
94
-------
ACTIVATED CARDON
PRESSURE DROP CURVES
10
SO CO 70 80 90
1.0
FIGURE 1
T j'Vi f -T-,--m i ~^.1 Ji__. 77; . ; _ , : : • , ~ ^ , ,
C^'.'^'l'.'.I'.t' "f. '.".'.'.': 1
—— ^/ ': ""r -^~\-: ' ] 1 :w^~^
f_..f..^\ i\ ,...:;:,.!/..•.:..!.._
-/• j :. j/ I ——-
/ / -
MEASUREMENTS MADE AT
iMAXIMUM PACKING DENSITY 1;
2O 30 40 60
VELOCITY (STANDARD AIR)-FPM.
60 70 00 90
95
-------
FIGURE
TYPICAL FAN MULTIRATING TABLE
Volume,
dm
2.52C
3, 120
3,530
4,030
4,530
5,040
5,540
6,040
6,550
7,060
7,560
8,060
8, 560
9,070
9,570
10,080
10,580
11, 100
11,600
12,100
12,600
15, 120
Outlet
velocity,
fpm
1,000
1,200
1,400
1.600
1,800
2,000
2,200
2,400
2,600
2,800
3,000
3,200
3,400
3,600
3,800
4,000
4,200
4,400
4,600
4,800
5,000
6,000
Velocity
pressure,
in. WC
0.063
0.090
0.122
0.160
0.202
0.250
0.302
0.360
0.422
0.489
0.560
0.638
0.721
0.808
0.900
0.998
1.100
1.210
1.310
1.450
1.570
2.230
1 in
rpm
437
459
483
513
532
572
603
637
670
708
746
. SP
bhp
0.63
0.85
1.05
1.33
1.61
2.00
2.36
2.79
3.27
3.81
4.42
2 in.SP
rpm
595
610
626
642
666
688
712
746
762
795
833
866
900
bhp
1.27
1.55
1.87
2. 18
2.56
2.97
3.43
3.99
4.62
5.32
6.05
6.96
7.93
3 in. SP
rpm
728
735
746
759
774
797
816
840
866
892
920
943
964
1,010
1.038
1,162
bhp
2.00
2.3C
2.72
3.17
3.63
4. 12
4.66
5.33
6.05
6.72
7.70
8.71
9.80
11.00
12.25
13.60
4 in. SP
rpm
837
842
847
858
876
890
910
926
954
963
993
1,020
1,053
,078
,108
,138
1, 168
1,198
1,232
1,270
1,301
bhp
2.66
3. 10
3.57
4.12
4.63
5.30
5.93
6.73
7.83
8.78
9.32
10.40
11.48
12.70
14. 15
15.40
16.90
18.58
20.30
21.00
24. 20
5 in. SP
rpm
943
950
964
976
999
1,017
1,032
1,050
1,068
1,097
1,120
1,148
1,170
1,200
1,230
1,258
1,290
1,321
1,355
bhp
4.60
5.21
5.82
6.50
7.38
8.17
9.08
9.97
11.00
12.10
13.30
14.65
14.90
17.35
19.05
20.55
22.50
24.40
26,40
6 in. SP
rpm
1,030
,040
,052
,068
',088
,095
, 125
,142
,168
1, 188
1,213
1,240
1,270
1,283
1.32Z
1,355
1,383
1,410
bhp
6.29
6.92
7.75
8.60
9.50
10.50
11.60
12.75
14.02
15.35
16.70
18.80
19.70
21.50
22.50
23.80
25.65
28.80
7 in. SP
rpm
1, 125
1, 134
1,145
1, 160
1, 171
1, 188
1,210
1,228
1,248
1,270
1,292
1,320
1,348
1,373
1,405
1,432
1,462
1,622
bhp
8. 18
8.96
9.93
10.88
11.98
13.06
14.28
15.50
16.93
18.42
19.46
21.70
23.50
25.40
27.40
29.60
31.80
45.90
6 In. SP
rpm
1,208
1,210
1,230
1,245
1,257
1,277
1,292
1,310
1,335
1,355
1,380
1,405
1,430
1,450
1,482
1,513
1,670
bhp
10. 15
11.18
12.25
13.50
14.70
15.98
17.36
19.00
20.75
22.35
23. 15
26. 10
27.95
30, 15
32.40
34.60
49.00
9 in. SP
rpm
1.270
1,279
1,288
1,298
1,310
1,328
1,340
1,360
1,380
1,405
1,430
1,450
1,478
1,500
1,528
,555
1,702
bhp
11.67
12.82
13.92
15. 10
16,48
17.80
19-15
20.90
22.60
24.40
26.40
28.45
30.60
32.90
35.20
37.80
51.50
For straight blade centrifugal blower having 46 inch diameter
96
-------
7891
0.1
% RELATIVE SATURATION
-------
QBfODIQ.
RkEI
CHRRt
OS
E
ZHIS1S.
SODiVINT
3231AM
WCOTEff
S
HAEH
HI IE
20
irlon
10
7
7-
-------
SOOTHSAYER CONSULTING LABORATORY
"We guarantee the answer you want"
13 Stargazer Plaza
Anywhere, U.S.A.
Test Performed: Measurement of length of MTZ
Material Tested: Carbon "B"
Test Number: B-17
Conditions:
Carbon bed depth
Saturation capacity
Temperature of test
Breakthrough capacity
Calculations:
62.0 cm, 2.0 feet
26% @ 20°C
20°C
24.9% calculated from test
Breakthrough
Capacity
0.249 =
j, 1 Saturation!
h 2 L Capacity J
^(.26") (MTZ) +
Q.26
| MTZ j + F Saturation!
|_ Length] L Capacity]
Bed Length
(62.0 - MTZ)
[ Bed
L Length
MTZ 1
Length]
62.0 cm
MTZ
= 5.3 cm
MTZ = 2.09 in
99
-------
Section IV
Absorption
100
-------
AB-I
Basic Concepts
The following data are from Perry's Chemical Engineering Handbook, Fourth
Edition, Page 14-6.
WEIGHTS OF
HCI PER 100
WEIGHTS OF H20
25.0
19.05
13.64
8.70
7.32
4.17
2.04
HYDROGEN CHLORIDE (HCl)
PARTIAL PRESSURE OF HCl, mm Hg @ 68°F
0.205
0.0428
0.0088
0.00178
0.000928
0.00024
0.000044
TABLE 1.
The information in this table is to be transformed to other units, as
indicated in Table 2. Atmospheric pressure is assumed to be 760 mm. Hg,and
the temperature is assumed to be 68°F.
SOLUTION *
1. Convert the partial pressure data to volume fractions
by dividing by the total pressure (760 mm.). Remember
Dalton's law of partial pressures.
*Answers are on page 106.
101
-------
a. Volume fraction is converted to volume % by multiplying
by 100. This is done in Column (3).
b. Is volume per cent of a gas the same as mole per cent
of a gas? If the answer is yes, write "Mole % of HC1
in the Gas" in above Column (3) in the space provided.
2. Next convert the liquid concentration of HC1 in Column (1) to
mole units.
a. Use the molecular weight of HC1 to convert
the weight of HC1 to moles.
b. Molecular weight HC1 is 36.46. The units are expressed
as either grams/gram mole or Ibs./lb. mole.
3. Now determine the moles of water in the (100 parts) referred
to in Column (1).
a. Molecular weight water is 18.0.
b. Total moles of water is 100 T- 18.0.
c. The total moles of water and HC1 is then (100 ± 18.0) +
moles HC1.
d. This is calculated in Column (5) by adding the moles
water to the moles MCI from Column (4).
4. The mole per cent of HC1 in the liquid can now be determined by
dividing the moles HC1 by the total moles liquid or Moles HC1/
Moles Liquid.
a. This is Column (4) divided by Column (5).
b. Calculate the Mole % of HC1 in the Liquid and enter in Column (6)
5. Plot the values for Mole % gas (y) and Mole % liquid (x) on the
graph provided (Figure 1). From this plot of the data, does it
appear that Henry's Law applies? Yes No
6. Plot the data on Figure 2. Does Henry's Law apply here?
Yes No
102
-------
HYDROGEN CHLORIDE (HC1) EQUILIBRIUM DATA
CD
V.EIGHTS OF
HC1 PER 100
WEIGHTS OF H20
25. r)
19.05
13.64
8.70
7.32
4.17
2.04
C2)
PARTIAL PRESSURE
HC1 mm. Hg
@ 68°F
0.205
. O^m
A /*\ ifc?
•• (/ C? 0 W
. odn'^
, ^^P-fT
, 0(j<> :> '/
, 0&&O*i<°\
(3)
C v )
/
COL (2) X 100
760 mm. Hg
0.0269
, 0£?G6J2
~ 0 O / / ^
,"> )^> '-u:- j>
w ,-•" ,**'«f- , •;• (_.
,^.^'^v/
,y:v,.;3 •••
. ^^ . .:; - • •
C4)
MOLES HC1 IN
MIXTURE
COL (1) f 36.461
i
0.685 •
. <; '? ^
. ?v^/
*") f*f\
• <&
C5)
MOLES H20 IN
MIXTURE + MOLES
HC1 IN MIXTURE
(100 T 18.0)+ COL (4)
5.55 + 0.685 = 6.235
^0~)-2
£1^- -
$^°]
^^
Gifel
e5ct\
C6)
MOLE % OF HC1
IN LIQUID
COL C4) 10Q
COL C5) x
60
v- y
11.0
^,r9£
- >, -^:i'
H, /IJ
Ix^'^/
^. t ' '
. ?v *;
o
CO
TABLE 2.
-------
-------
-------
Ib. Yes
ANSWERS
4.
Col. (2)
Col. (3)
Col. (4) Col. (5) Col. (6)
5.
6.
0.2Q5
Q.Q428
0.0088
0.00178
0.000928
0.00024Q
0.000044
NO
NO
0.0269
0.00564
0.00115
0.000234
0.000122
0.0000316
0.00000578
0.685
0.521
0.374
0.238
0.201
O.I 14
0.056
6.235
6.071
5.924
5.788
5.751
5.664
5.606
1 1 .0
8.6
6.3
4.1
3.5
2.0
1 .0
106
-------
AB-II
Material Balance
You are given the basic data for an absorption system to be used for scrubbing
ammonia (NH ) from air with water. The water rate will be 300 lb/(hr ft ) and
the gas rate will be 250 lb/(hr ft2) at 72°F. The equilibrium data for the
ammonia-air system is shown in the following table.
EQUILIBRIUM DATA NH3 - WATER 72°F
pe mm Hg
Equilibrium Pressure
Concentration of NH3
# NH3/100 # H20
3.4
O.S
7.4
1.0
9.1
1.2
12.0
1.6
1 ATM PRESSURE
15.3
2.0
19.4
2.5
23.5
3.0
PERRY'S HANDBOOK
FOURTH EDITION 1963
The air to be scrubbed has \.5% wt. NH3 at 72°F and I atmosphere pressure and
is to be vented with 95% of the ammonia removed. The inlet scrubber water will
be ammonia free.
The exercise in this problem is to:
a. Plot the equilibrium data in mole fraction units.
b. Calculate the material balance and plot the operating line on the
equiIibrium plot.
c. Set up the mass flux equations for both the gas and liquid driving forces,
107
-------
SOLUTION
A. Convert the equilibrium data to mole fraction units.
I. Remember that in an ideal gas mixture the partial pressure of
a gas is equal to its volume fraction in the gas mixture. This
is Dalton's Law. The volume fraction is also the mole fraction
for an ideal gas mixture. Since the atmospheric or system pres-
sure (P) is 760 mm Hg, the volume fraction or mole fraction of
NH, for a partial pressure (pe) of 3.4 mm Hg is:
Pe
P
3.4
760
0.00447
Calculate the remaining gas mole fraction values and
enter i n Table I.
GAS
pe mm Hg
3:4
7.4
9.1
12.0
15.3
19.4
23.5
GAS
Mole Fraction
Y
0.00447
, 00^7l
'fl/O^
,o^^3>
.D^^J
.0^2^
LIQUID
Ibs. NH3
100 Ibs. H20
0.5
1.0
1.2
1.6
2.0
2.5
3.0
LIQUID
Mole Fraction
X
0.0052
. #10^"
,0I*T
. 01 bt
.0^1
*<&
• &1X
Table I
^Answers are on page 114.
108
-------
2. The factor for converting the liquid concentration data
to mole fraction units depends upon the molecular weights
of the substances.
i- -j j. A. j. * e LBS. NH,
For a liquid concentration of 0.5 ± the
100 IBS. H20
calculation is as follows:
Ib. moles of Nhk
mole fraction NH3 i n H20 = °r»
Total Ib. moles of solution
0.5
M.W. -NH3
mole fraction NH3 in H20 = oT3 TOO"
M.W.- NH3 M.W. H20
NOTE: M.W. NH3 = 17.03 (use 17.0)
M.W. H20 « 18.01 (use 18.0)
M.W. Air = 28.95 (use 29.0)
Calculate the liquid mole fraction values and
enter in Table I.
B. Plot the mole fraction values on the following graph. Liquid
concentrations should be plotted on the x-axis and gas concen-
trations on the y-axis.
1. Does the plot indicate that Henry's Law applies?
YES or NO? \f& 5
2. Why?
109
-------
: Q.Q1
3. Calculate the slope of the equilibrium line.
' 0 .0 i
Slope = ±L =
Ax
110
-------
C. Label the following schematic of the system.
"M,
• 1
2
1
. It G . v
•? M- -
y
*\
D. Calculate the outlet liquid concentration using the material
balance. It is assumed that for air pollution applications
LM and GM are essentially constant throughout the column.
LM (x, - x2> = GM (y, - y2)
All values should be expressed in LB. Mole units or mole fraction units:
t
x. = outlet NH, concentration in liquid = ?
y. = inlet NH-, concentration in gas = *
y2 = outlet NH, concentration in gas = *
x2 = inlet NH, concentration in liquid = *
2
LM = Liquid flow, LB Moles/CHour Ft ) ?
2
GM = Gas flow, LB Moles/(Hour Ft ) ?
* NOTE: The concentration units which were given in
wt. % must be converted to mole fraction units.
Solve for x
. L
M
300 LB.
18
LB. H20
LB. Mole H20
LB. Moles
Hour Ft2
111
-------
250 LB. Alr/(Hr. ft2) ~ , LB. Moles Air
2- G = _ =
2g 0 LB. Air Hour ft2
LB. Mole Air
GM
X| = —
LM
4. X = g.fe (.oTxr-brfn + (7
5. x, = /Z LB- Moles NH3
LB. Mole Liquid
E. • Plot the inlet and outlet concentrations on the graph, Fig. I.,
i.e. plot points x y and x_, y?.
I. Connect them with a I i ne. This Is caI led the
operating line. The operating line defines a
boundary for operating conditions within the
absorption unit. The operating line states the
concentrations for adjacent streams throughout
the column.
2. Any I ine on a coordinate system such as this
graph may be expressed as an equation.
a. Write the equation for this straight
line explicit for Ay or (y| - y2).
NOTE: Remember the material balance
equation!
b. What is the value for the slope of
the operating line?
F. Indicate the gas phase concentration driving force between the
operating line and the equilibrium line for the gas inlet conditions,
I. The driving force In the gas phase is
expressed as a function of the inlet ammonia
concentration (y|) and the concentration of
ammonia in equilibrium with the outlet Iiquid
(y*).
driving force = (y| - y*)
112
-------
2. Identify the gas phase driving force on the graph.
_ Ib. moles
G. The mass flux, N/\ 7—5- can be expressed as a function
of the driving force and the mass transfer coefficient (Kg).
I. This is analogous to Ohm's Law.
FLOW (I) = Driving Force (E)/Resistance (R)
2. The reciprocal of the mass transfer coefficient is analogous to
electrical resistance. The mass transfer coefficient is therefore
analogous to (electrical) conductance.
.'.Mass Flow (NA> = Kg x (driving forcej
NA = KG
-------
ANSWERS
Gas
Mole Fraction, y
0.00447
0.00973
0.0120
0.0158
0.0201
0.0255
0.0309
Liquid
2. Mole fraction, x
0.0052
0.0105
0.0125
0.0167
0.0206
0.0258
0.0307
B. I. Yes
2. The equilibrium data describe a straight line.
3. I .0
y2
D. 0.0253, 0.0013, 0
I. 16.67
2. 8.62
4. 0.517, Q.0253, 0.0013, 0
5. 0.0124
E. 2. a. Ay = mAx = I.93 Ax
b. 1.93
3.
I b. mo Ies
hr. ft2
or
Ib. moles
hr. ft2 mole fraction
or
Ib. mo Ies
hr. ft2 atm.
114
-------
INFORMATION NECESSARY FOR REVIEWING ABSORBER PLANS
1. Equipment Description
a. Drawings
b. Specifications
c. Make, Model, Size
2. Process Description
a. Flow diagram
b. Chemical compositions of all streams
c. Temperatures
d. Volumes
e. Gas velocities through equipment
f. Flow rates for materials entering and leaving
the equipment
g. Pressure drop data for the equipment
3. Design Information
a. Data used in the design
(1) Equilibrium data
(2) Mass transfer or height of a transfer
unit data
(3) Number of transfer unit data
b. Calculations made in the design and selection
of the equipment
4. Description of Gas Mover
a. Make, Model, Size, Speed, Capacity
b. Cubic feet per minute to be handled
c. Static pressure
d. Motor description
115
-------
Procedure for Reviewing Absorber Plans
1. Study the chemical feasibility of the scrubbing system.
2. Perform a material balance to determine the emission
quantities.
3. Compare emissions with standards or laws.
a. If comparision indicates a violation: deny
b. If comparision is favorable: proceed to 4.
4. Calculate tower diameter.
5. Evaluate column's performance by calculating:
a. Height of tower, or
b. Number of transfer units, or
c. Outlet emission concentration
6. Compare 4. and 5. with submitted information.
7. Recommendation for approval or denial.
8. Plan Review Air Mover
116
-------
AB-IM
Packed Tower
XYZ Company has submitted plans for a H2S scrubber. Hydrogen
sulfide is to be removed from a waste air discharge by scrubbing the
gas stream with a triethanolamine —water solution in a packed
absorption tower at atmospheric pressure.
Specification sheets indicate that the gas and liquid flow
rates are 290 lb/(Hr.ft2) and 40 Ib. moles/(Hr.ft2), respectively.
The inlet liquid is indicated to be contaminant free. XYZ Company
pilot plant data indicate that Henry's Law applies (m = 2.0) and
-the HQG is I .94 ft.
From your preliminary calculations you have determined that
the scrubber must reduce the h^S from 17 lb/(Hr.ft2) to 0.64
lb/(Hr.ft2).
Is the 12 foot tower (which the XYZ Co. proposes) adequate?
117
-------
SOLUTI ON *
I. Draw and label a schematic of the tower:
Lm' X2 Gm- V2
Lm> x| Gm' V,
2. Convert all units to a Ib. mole basis:
M.W. H2S = 34 Ib/lb. mole
M.W. Air = 29 Ib/lb. mole
I ">. moles
= 290
m
Hr.ft2 M.W. Air Hr.ft2
. _ 40 Ib. moles
m —
Hr.ft2
17 Ib. H2S I |
Vi = — X x
Hr.ft2 M.W. H2S Gm
0.64 Ib. H7S | |
" x x
M.W. H2S Gm
*Answers are on page 121.
118
-------
3. By using a material balance, find x
Lm (x, - x2) = Gy (y, - y2:
x, = r
m
4. By using the Col burn Chart, find the N :
OG
X2
m
y2
Lm
G
m
Gm
Lm
~m
y2
• • NOG
= NOG X HOG
Z = ft
119
-------
tfl
-p
-H
c
D
M
0)
IM
CO
c
ra
M
u-i
O
i-l
CD
2
I
O
O
A.P. ColBurn
Troni. AtChE 35.216(1939)
50
100 200 500 IJ300
- m
y2.-mx2
Colburn's Chart Permits an Easy Solution for
NOG.
120
-------
ANSWERS
2. 10
0.05
0.00188
0
3. 0.012
4. 0, 2.0, 0.05, 0.00188, 40, 10, 0.5, 26.6, 5.1
5. 9.9 Yes, the 12 foot design Is adequate.
121
-------
AB-IV
Packed Tower
Pollution Unlimited, Inc. has submitted plans for a packed ammonia
scrubber on 1575 cfm air stream containing 2% NH-^. The emission
regulation is Q.\% NH-j.
The following information is provided:
I. Tower Diameter - 3.57 feet
2. Packed Height of Column - 8 feet
3. Gas and liquid temperature - 75°F
4. Operating Pressure - 1.0 atm.
5. Ammonia-free liquid flow rate - 1000 lb/(hr.ft2)
6. Gas flow rate - 1575 cfm
7. Inlet NH3 gas concentration - 2.0 volume %
8. Outlet NH-j gas concentration - O.lvolume %
9. Figure I
10. Packing 'A' is used
II. Air density - 0.0743 Ib/ft3
12. Molecular weight air - 29 Ib/lb mole
13. Henry's Law constant - m = 0.972
14. Molecular weight water - 18 Ib/lb mole
You remember approving plans for a nearly identical scrubber for Pollution
Unlimited, Inc. in 1968. After consulting your old files you find all the
conditions were identical —except the gas flow rate was 1121 cfm.
What is your recommendation?
122
-------
SOLUTION*
I. Draw and label a schematic of the tower:
Lm' X2 -
Lm' x| ' Gm«
2. Calculate the cross-sectional area of the tower:
•n D2 ,-rx
A = = h , ft2
3. Calculate the gas molar flow rate:
G _ /^"')Cft3/min x 60 min/hr x «£7:;3l b/ft3
m
f/A ft2 x _J23llb/lb. mole
G " lb- mole
~ hr. ft2
m
4. Calculate the liquid molar flow rate:
Lm = 1000 lb' x '
hr. ft2 /^ Ib./lb. mole
c: " - *•** I b. mo I e
Lm ~ !-i
hr. ft2
^Answers are on page 128.
123
-------
5. Determine NQQ graphically from Figure 2:
*2
m
Y2 =
Lm =
Gn
• " \ "^^V"1 %
m
m
y2-m x2
.. NOG - -f J?
6. a. For the previously specified gas flow rate, liquid flow rate,
and packing type, what is the best HQG (see Figure I)?
b. What is the packed height needed to do the job?
Z = NQG x HQG
z =
7. Part (b) tells us that the packed column height necessary to give
4.3 transfer units 9.47 feet. The plans which Pollution Unlimited
submitted specified a packed column height of 8 feet. What is your
recommendation?
_ Approve _ // Deny
124
-------
8. Company- off hcbals ask if there hs any modrf i-catron which
would allow the scrubber to operate. They plead that
because of a pri'or shut down of an adjacent gas cleaning
operation, this scrubber already exists in the specified
location and would only require minor piping changes to
be put on stream.
From your experience thus far, and by inspection of
Figure I, list process modifications and rank them
in probable order of increasing feasibility.
a.
b.
d. U <;!- / ' - ' ''
9. hnstead of buying a taller tower or new, more efficient
packing, we will Investigate manipulating the liquid
flow rate rn order to attain the desired separation.
Calculate the tower height needed for each L rate:
L m Gm N u Z ft
L OG "OG '
m
A * 2.2 9.47
1000 °-423 4-3
1200 ''•'- '" r'^
1400 - \ •*' ^ 't^'
, t
1600 .'2V\ '•- ' I -)
From these calculations we see that the liquid rate must
be increased to JD c V> lb/(hr. ff2) for the 8' tower-
to perform according to specifications. If a safety factor
is to be incorporated into the recommendation, the liquid
rate must be increased beyond this value.
125
-------
0 Pack1ng A
Packing *B
G = Gas flow LB/HI^ FT2
Ftg. 1
Air - Ammonta - Water System
Ref: Perry's Chemical Engineer's Handbook
pg. 18^43 Ftg. 18-82.
126
-------
en
-P
•H
d
P
M
01
C
ti
M
OJ
O
O
A.P. Colburn
Trant. AlChE 35,216(1939)
10 20
• m
- m x
Fig. 2
Colburn's Chart Permits an Easy Solution for
NOG.
127
-------
ANSWERS
2. 10.0
3. 1575, 0.0743, 10.0, 29
24.2
4. 18
55.6
5. 0, 0.972, 0.02, 0.001, 55.6, 24.2, 0.423, 20, 4.3
6. a. 2.2.
b. 4.3, 2.2, 9.47
7. Deny
8. a. Engineer a new scrubbing system (buy a new tower).
b. Lengthen existing tower.
c. Increase liquid flow rate.
d. Purchase new packing.
9. m Gm
— NOG HOG z
m
0.423 4.3 2.2 9.47
0.353 4.0 2.0 8.00
0.302 3.9 1.85 7.21
0.265 3.75 1.70 6.38
1200
128
-------
AB-V
Spray Tower
A steel pickling operation emits HCI fumes (hydrochloric acid) of
300 ppm average with peak values of 500 ppm for 1556 of the time. The
air flow is a constant 25,000 cfm at 75°F and I atm pressure. Regulations
limit emissions to no more than 25 ppm HCI at any time. Only sketchy
information was submitted with the scrubber permit application. The
plans show a 14-foot diameter counter current water spray tower.
By using the following rules of thumb, determine if the spray
unit is satisfactory.
I. Gas velocity through the tower is not to exceed 3 ft/sec.
2. For HCI, Henry's Law does not apply.
3. For a very soluble gas (Henry's Law constant close to
zero), the number of transfer units (NQG) can be determined
by the following equation:
NQG
where y. = concentration of inlet gas
^ = concentration of outlet gas
4. In a spray tower, the number of transfer units (NQQ) for
the first or top spray will be about 0.7. Each lower spray
will have only about 60% of the (NQQ) of the spray above it.
This is due to the mixing of liquids with that from upper
sprays and back mixing of liquids and gases. The final spray,
if placed in the inlet duct, adds no height and has 0.5 NQG.
5. The spray sections of a tower are normally spaced at three
foot i ntervaIs.
129
-------
SOLUTION*
Calculate the diameter of the tower. We do this using the gas
velocity (ft/sec) and the gas flow rate (cu. ft/sec).
First, set up the equation for the gas velocity.
. .. gas flow rate
velocity = 3
area of cross section
Remember that the area of the cross section is a function of the
tower diameter.
_
Area =
4
25,000/60 ftVsec
3 ft/sec = —
. 21
4
What are the units of D?
Solve for D.
1 25,000/60
0 ' 1 3 ,/4
2. Calculate the Number of Transfer Units required.
NQG = In y!/y2 = ln_/__ = In
Remember, we need the natural log of the ratio of the concentrations.
This can be obtained directly from the lne scales of your slide rule,
or calculated by lne = 2.303 log,0, or from natural log tables.
Take your pick!
^Answers are on page 133.
130
-------
NU~ = transfer units
Ub -
3. Now, determine the number of transfer units provided by a tower
with several spray sections (take 5 or more). Remember, each
lower spray has only 60$ the efficiency of the section above it.
Spray Section N
Top 0.700 (this is a starter)
2nd
3rd
4th
5th
Inlet
TOTAL
Remember, we need only enough spray sections to get the required
•v
4 Calculate the efficiency of a spray tower with as many spray sections
as you chose to give it (based on your answer to part 3). That is,
calculate the outlet concentration with the relationship:
N_r = I n — =
°G V
Y2
You should have no trouble with this calculation now.
— = ant I lne NQG
Y2
Yl
= X =
y-,
131
-------
5. Does this meet the specification?
132
-------
ANSWERS
I . ft., 13.3
2. 500, 25, 20
3.0
3. 0.420, 0.252, 0.151, 0.091, 0.500
2.1 14
4. 2.114
8.3
60 ppm
5. NO
133
-------
APPENDIX C: HYPOTHETICAL AVAILABLE HEATS FROM NATURAL GAS
Burners for combustion devices such as after-
burners frequently use the oxygen present in the
contaminated effluent stream. An example would
be a natural gas-fired afterburner that takes in
60 percent of its combustion air from the atmo-
sphere, and 40 percent from an air containing
contaminated effluent stream.
One step in checking afterburner design is the cal-
culation of the natural gas flow rate required to
raise an effluent stream to a given temperature.
A calculation such as this normally makes use of
the available heat from natural gas. Available
heat ia the amount of heat remaining after the
products of combustion from a cubic foot of natural
gas are raised to the afterburner temperature.
Available heat from natural gas is shown in Table
D7.
If the afterburner gas burner takes a portion of
the combustion air from the effluent stream,
then the calculation of the gas flow rate becomes
a trial-and-error procedure. By the method of
hypothetical available heats given here, the trial-
and-error solution is eliminated.
The natural gas used in illustrating this calcula-
tion procedure requires 10. 36 cubic feet of air
for theoretical combustion of 1 cubic foot of gas
(Los Angeles area natural gas). Products of
complete combustion evolved from this process
are carbon dioxide, water, and nitrogen. If
the combustion of 1 cubic foot of natural gas is
thought of as taking place at 60°F, then a portion
of the heat released by combustion must be used
to raise the products of combustion from 60"F to
the temperature of the device. The remaining
heat is called available heat. This quantity repre-
sents the heat from natural gas that can be used
to do useful work in the combustion device, such
as heating an effluent stream in an afterburner.
Consider a gas-fired aftej-burner adjusted to pro-
vide a fraction, X, of theoretical air through the
burner. If the contaminated effluent contains air,
then the remaining air for combustion, 1-X, is
taken from the effluent stream. This means that
a smaller quantity of effluent has to be heated by
the natural gas, since a portion of the effluent is
involved in the combustion reaction. Thus, A
burner taking combustion air from an effluent
stream can be fired to raise the temperature of
the effluent at a natural gas input lower than that
of a burner firing with all combustion air taken
from the atmosphere.
Let the heat content of an effluent stream, at the
desired final temperature, be H Btu/lb. Since
10.36 cubic feet of air is required for combus-
tion of 1 cubic foot of natural gas, the weight of
air taken from the effluent would be
W = (10.36)(l-X)(p)
(Cl)
The heat contents of this secondary combustion
air would be
Q = WH = (10. 36Hl-X)(p)(H)
where
-------
HYPOTHETICAL AVAILABLE HEATS FROM NATURAL GAS
«
.200
.300
.400
,500
,600
.700
.800
t721 + 227 (1-X)
-r—t' *
693 + 249 (1-X)
669 + 270 (1-X)
643 + 292 (1-X)
615 + 314 (1-X)
590 + 336 (1-X)
562 + 358 (1-X)
.
3. The gases in the afterburner will consist of:
a. Products of combustion from 1,230 cfh
natural gas with theoretical air - 1,230
x 11. 45 scfh,
b. the portion of the effluent not used for
combustion air = effluent volume rate -
(1. 230)(10. 36)(1-X).
X
as the burner's primary air.
Hypothetical available heats are given in Table
Cl for varying temperatures and percentages
of primary air.
The use of this concept is illustrated in the follow-
ing examples.
Example Cl:
An afterburner is used to heat an effluent stream
to 1.200°F by using 1 x 10& Btu/hr. The burner
is installed and adjusted so that 60% of the theoret-
ical combustion air is furnished through the buriit-r,
and the remainder is taken from the effluent. De-
termine the required natural gas rate.
1. The percent primary air is 60%, the required
temperature is 1,200'F, the hypothetical
available heat from Table Cl is 812 Btu/ft3
of gas.
2. Burner flow rate = 106/812 = 1,210 cfh gas.
Example C2:
An afterburner is used to heat an effluent stream
to 1,200°F by using 1 x 10& Btu/hr. The burner
is installed and adjusted so that all the combustion
air is taken from the effluent stream. Determine
the natural gas rate.
This is equivalent to the burner's operating at 0%
primary air.
1. At 1,200°F the hypothetical available heat is
948 Btu/ft3 for 0% primary air.
2. Burner flow rate = 106/948 = 1,058 cfh.
3. Gasi-s in aflcrlmrner will consist of-
a. Combustion products from 1,058 cfh nat-
ural gas with theoretical air - 1,058 x
II. 45,
b. the portion of the cfllucnt not used for
secondary combustion air = effluent
volume - (1.058)(10. 36)(1-X>.
Table Cl. HYPOTHETICAL AVAILABLE HEATS
Temp,
"F
600
700
800
900
,000
, 100
,200
,300
,400
, 500
,600
,700
1,800
Hypothetical a\ailable heats. Btu/ft3 gas
% primary air through the burner
0
975
970
965
965
958
953
.948
942
939
935
929
926
920
10
965
958
950
948
939
933
926
917
912
906
897
892
885
20
954
945
936
931
921
912
903
892
885
976
866
859
949
30
944
933
922
915
902
891
880
867
858
847
834
825
813
40
933
921
907
898
884
371
858
842
831
818
803
791
777
50
923
908
893
881
865
850
835'
818
804
789
772
758
741
60
913
396
878
865
847
830
812
793
777
760
740
724
706
70
902
S83
864
848
828
809
789
768
750
730
709
691
670
HO
892
971
850
831
810
78S
767
743
723
701
677
657
634
90
831
859
835
814
791
768
744
718
696
672
646
623
598
-------
MASS TRANSFER
GAS ABSORPTION THEORY
There are many theories describing the mechanism of mass transfer across
an interface. The Lewis and Whitman Two-Film theory assumes a turbulent
gas phase and a turbulent liquid phase, between which the turbulence dies
out forming a laminar gas film and a laminar liquid film at the gas-liquid
interface. The entire resistance to mass transfer is considered to exist
in these two films through wnich the mass transfer is effected by molecular
diffusion. «,
GAS LIQUID
I
Film 1 r
^A*--,
I
v. . _<
'Ai 1
I
1
- — —Liquid Film
I
! XA
! v
1 Ai
1
z+
Zl
Figure 1.
Schematic of the Gas-Liquid Interfacial Region
For the steady state diffusion of gaseous material A in a binary mixture
of A and B:
Total flux of
material A with
respect to fixed
axes
[Flux from j | Flux from |
diffusion bulk flow
(11
~ P
dyA
i A
A -
A dZ
(la)
-------
Let
VNA
Then
1
- ' C IT * yA ll +
dz
P zr-
(2)
Integrating across the distance from the gas film to the gas-liquid
interface (z^ to z'.):
\
2i
rAi
dz = Az
Uy
P =~
N.
A
yA (1 *
(2a)
^ r i
/.= Ll *
Y) - 1
(3)
or in the alternate Conn of P. and
N,
tt'
RT
(1 + Y) - D?
(3a)
For"the specific case of gas absorption, unimolal unidijrectional diffusion
occurs in the gas (diffusion through a stagnant gas), Nfi = 0, and therefore,
Y = 0. The general expression for the flux of A may be written as follows:
(4)
Let (1 - yA)
yAi) - (i - yA+)
rAi
LM
(5)
-2-
-------
Then,
or in terms of P. and IP,
"
and
RT
It is assumed that local equilibrium prevails at the interface. If the
gas and liquid interfacial concentrations can be related by a form of
Henry's Law, then:
yAi = »XA. (7)
If the diffusivity is constant, equations (6) and (6a) may approximate
the flux at any point as a constant times a concentration driving force:
\ - kg &A - y^ • Y
where
- • """A (8b)
8 RT iz (IP - P)
or
kg - 3E kg' (8c)
-3-
-------
The individual gas phase mass transfer coefficient for the transfer
of material A, k (sometimes written as k. ), is analogous to the
conductance for material A across the laminar gas film.
The expression for mass flux could have just as easily been developed
considering the mass flux across the liquid film (since steady state
conditions are assumed). Equation (8) would then become:
*A ' ke
where
\ • V'e (9a)
Interfacial concentrations are difficult if not impossible to measure;
the flux equations are therefore written for any point in terms of
overall mass transfer coefficients (which account for the resistance
in both the gas and liquid films) and overall concentration driving
forces .
"A = Kg Cio)
or analogously,
NA = K£ (XA* - XA) (10a)
The expressions relating the individual to the overall mass transfer
coefficients are as follows:
1 1 , m
_ - — (11)
kg ke
111
— = — + — (lla)
Ke ke m kg
-4-
-------
Whenever an overall process consists of several sequential steps,
the overall rate of the process may be governed by the rate of the
slowest step. If the resistance to mass transfer is high in the gas
film relative to the liquid film (1/k » m/k ), then mass transfer is
said to be gas phase controlled and equation (11) becomes:
K
g
g
(lib)
Similarly when mass transfer is liquid phase controlling (1/k « m/k )
equation (11) becomes: 8 £
K
g
m
k
(lie)
The same arguments could be applied to equation (lla) with similar
equations resulting.
DESIGN CONSIDERATIONS FOR A OOUNTERCURRHNT CONTACTOR
a represent interfacial contact area per unit tower volurre.
Let V represent the empty tower volume.
Consider a tower of differential height having a unit cross- sectional
area (see Figure 2), then,
dV
CD dZ
dZ
Liquid Phase
Gas Phase
Figure 2.
Differential Element of Tower
-5-
-------
and for the differential element,
The rate of
transfer of
material A per
unit tower cross
section
Flux of A
(see Eon. (10))
X
"interfacial
area per unit
tower volume
X
Tower volume
per unit tower
cross section
(12)
(Gm yA)
(a) (dZ)
(12a)
If the left side of equation (12a) is expanded,
But from a material balance on the transferring component in the gas phase:
dG
m
d G
m
and
d G.
m
then, d (Gm /A) =
- y,
(13)
Substituting equations (10) and (13) into (12a) the following is
obtained:
Jm 'A
g
(yA - yA*) (a) (dZ)
(14)
Integrating equation (14) across the tower gives the most general form
of the design equation.
yAl
V »-V<*A-v*
-6-
(15)
-------
Empirical results indicate that k and ke are proportional to Gra°'8
and Lm°'8. Equations (6) and (8a) suggest that [Gm/Kga (1 - yA)LM]
should be constant. The average of this term evaluated at the top
and bottom of the tower is then taken from the integral:
m
V
ave.
(16)
Although equation (16) is based on overall gas phase concentrations, an
analogous equation can be written based on overall liquid phase concen-
trations .
m
Kea d -
LM
i
ave.
Equations (16) and (16a) are of the following form :
Z
Z
H
OG
(17)
(17a)
Mumber of Transfer Units
Qualitatively the number of transfer units is a measure of the degree
of separation required. Evaluation of the number of transfer units is
generally accomplished by a graphical integration of the second term
in equation (16), or in equation form,
N
OG
i
yA2
(18)
Although the development of equations is straight forward based on
either gas phase or liquid phase concentrations, only the gas phase
concentrations will be considered throughout the remainder of this
discussion. -7-
-------
For the following conditions, simplifications to equation (18) can be
made.
(1) Contaminant concentration, y , is low throughout the contactor.
• • t\
(From the material balance on A for the non-reacting system, one
implication is a straight operating line with slope L/G .)
(2) Straight equilibrium line with slope m. Condition (1) yields
equation (19) which can be evaluated by graphical integration.
N
OG
(I-»A> ''W
(19)
Under conditions (1) and (2), equation (19) can be further simplified
to the form,
N
OG
1-
m
r, -s."
L ^.
Lm
(20)
This equation is plotted in Figure 3 in the form N__ vs Sin
OG
for different values of mG
m
L .
m
Ihis plot, the Colburn Diagram, allows easy solution for N
OG.
- mXA2)
-8-
-------
in
-u
•H
C
D
M
0)
14-t
I/)
c
a)
u-i
0
(U
D
2
o
*.P. Coiburn
Tronj. »IC»iE 35.Z'6(i939)
50 100 200 500 IpOO
- m
- m
FIGURE 3 CDLBUFN DIAGRAM
-9-
-------
If condition (1) exists and, instead of condition (2), yft* approaches
zero, as is the case for a very rapid chemical reaction occurring in
the liquid phase or for an extremely soluble contaminant (m •+• o), then
equation (19) can be simplified as follows:
YAI <»*_ » p
y_. M \y» y» j * y»*^ JT * _y * o
AJ ^*»JA. * AS ** •* A/
fib ^X *» •MB *» «»fc
dyA = \ dyA = 4n yAl (21)
Application of equation (21) has been referred to as the "Transfer Unit
Approach." The underlying assumptions should always be considered before
this equation is applied. When equation (21) is applicable, the following
results are easily shown:
% Efficiency NOG
98 4
95 3
86 2
63 1
Height of a Transfer Unit
Qualitatively the height of a transfer unit is a measure of the height
of a contactor required to effect a standard separation, and it is a
function of the gas flow rate, the liquid flow rate, the type of packing,
and the chemistry of the system. In equation form,
HOG ~ | "m (22)
ave.
-10-
-------
This type of information appears in the literature as H^., K a, or k a
OG g g
k^a and can be used in previously presented equations for tower design.
If the data are not available, pilot plant studies are run or approxima-
tions are made.
Individual mass transfer coefficient data can be roughly estimated fron
available data using th« following formula:
(23)
-,- * '-"-1
tt Note that although the Two Film Theory would have predicted an
exponent of unity, more sophisticated surface renewal theories and
empirical results support the exponent of 0.5 as in equations (23)
and (23a). Note also that in order to apply this equation the pa-
rameters should all be referring to films of the same phase, with
the same thicknesses. If kff is the mass transfer coefficient for
« through a gas film of « and an inert, 6, then k_ should be the
similar quantity for 8 through a gas film of B and 6. V^ and VR
should be the respective gas phase diffusivities for = and B through
their mixtures with inert, 6.
For the limiting cases of either gas or liquid phase controlling the
rate of mass transfer, height of transfer unit data may be similarly
approximated.
- f'-l
- hrj
(23a)
Consider the following chemical reaction,
A (in the gas) + dD (in the liquid) —- Products (24)
-11-
-------
For the case where a chemical reaction occurs in addition to physical
absorption, the relative rates of chemical reaction versus mass transfer
must be considered. When the following condition is met, the special
case exists in which mass transfer controls the overall rate of the
process, the chemical reaction is extremely rapid, and the reaction
zone is located at the gas-liquid interface:
AgyA (25)
When the above condition is met at both ends of the contactor, there is
no longer a liquid film resistance, and individual and overall gas phase
mass transfer coefficients become approximately equal (refer to equations
(11) and (lib)).
Kg = kg (26)
The appropriate change in design equations (15), (16), or (22) should be
made before their application to design. The cases for which equation
(25) dees not apply are much more complex and will not be considered in
this discussion.
Comments
The remaining task of choosing the tower diameter will be left to the
design engineer. The methods involved are outlined in the literature
and are straightforward based on pressure drop, flooding velocity data,
and correlations for the type of packing to be used.
Although the design treatment of this discussion applies only to counter-
current contactors, simple modifications are required to extend its
applicability to (less efficient) concurrent towers.
Remembering that generally the best efficiency is obtained with counter-
current contactors, the air pollution plan review engineer would do well
to consider their performance as a bound with which to determine the
reasonableness of the stated performance of submitted non-countercurrent
designs.
J. E. Sickles, II
-12- May 1972
-------
NOMENCLATURE
A material A, the gaseous contaminant
a interfacial area per unit tower volume
= material «
fi material B
B material 6
c molar concentration, (moles/volume)
D reactant material D
d stoichiometric coefficient
V diffusivity
PA diffusivity of material A in B
6 inert material 5
G molar gas flow rate, [moles/(time x tower cross-sectional area)]
Y ratio of N. / Nfi
H height of a transfer unit
K overall mass transfer coefficient
k individual mass transfer coefficient
-L molar liquid flow rate, [moles/(time x tower cross-sectional area)]
m slope of the equilibrium line when plotted on an x - y diagram
N number of transfer units
N molar flux, [moles/(time x interfacial area)]
IP total pressure
P partial pressure
R ideal gas law constant
p total molar density, (total moles/volume)
T absolute temperature
-13-
-------
V empty tower volume
x mole fraction concentration in liquid phase
x * mole fraction concentration in the liquid which would be
in equilibrium with the existing gas at that point in the
contactor
y mole fraction concentration in gas phase
y * mole fraction concentration in the gas which would be in
equilibrium with the existing liquid at that point in the
contactor
z distance through which diffusion occurs
Az film thickness, z. - z+
Z tower height
Subscripts
A material A, the gaseous contaminant
« material «
B material B
6 material 8
D reactant material D
6 inert material 6
g gas
i located at the interface
LM log-mean
£ liquid
OG based on the overall gas phase concentrations
OL based on the overall liquid phase concentrations
-14-
-------
1 located at the gas inlet end of a countercurrent contactor
2 located at the liquid inlet end of a countercurrent contactor
+ located at the commencement of the laminar gas film
Superscript
based on concentrations expressed in partial pressures for
gases, or molar density for liquids
-15-
-------
I. Combustion
1. J. A. Danielscn, "Air Pollution Engineering Manual," U.S.P.H.S.
Publication No. 999-AP-40, 1967.
2. J. H. Perry, "Chemical Engineers' Handbook," 4th ed., McGraw--
Hill, New York, 1963.
3. "North American Combustion Handbook," North American Manu-
facturing Co., Cleveland, 1965.
4. R. W. Rolke, R. J. Hawthorne, C. R. Garbett, E. R. Slater,
T. T. Phillips, and G. D. Towell, "Afterburner Systems Study,"
Shell Development, EPA Contract EHSD-71-3, 1972.
II. Absorption
1. A. H. P. Skelland, "Mass Transfer," in Vol. 13, Kirk-Othirer
Ency. of Chemical Technology, 2nd ed., Interscience, New
York, 1967, pp. 99-159.
2. Danielson, "Engineering Manual."
3. O. Levenspiel, "Chemical Reaction Engineering," John Wiley
& Sons, New York, 1962.
4. Perry, "Handbook."
5. R. B. Bird, W. E. Stewart, and E. N. Lightfoot, "Transport
Phenomena," John Wiley & Sons, New York, 1966.
6. S. Calvert, J. Goldshmid, D. Leith, and D. Mehta, "Wet
Scrubber Systems Study," Ambient Purification Technology,
EPA Contract CPA-70-95, 1972.
16
-------
415 - CONTROL OF GASEOUS EMISSIONS
Lecture Handout - Condensation and Combustion
by
L. Theodore
Frequently, It becomes necessary to handle a condensable vapor and
a noncondensable gas. The most common mixture is, of course, water vapor
and air. A mixture of organic vapors and air is another such example;
and this mixture can cause pollution problems. Condensers can be used to
collect organic emissions discharged to the atmosphere. This is accomplished
by lowering the temperature of the gaseous stream although an increase in
pressure will produce the same result. There are two basic types of con-
densers used for control - contact and surface. In contact condensers,
the gaseous stream is brought into direct contact with a cooling medium so
that the vapors condense and mix with the coolant. The most commonly used
unit is the surface condenser (heat exchanger). Here, the vapor and the
cooling medium are separated by the exchanger wall. Since high removal
efficiencies cannot be obtained with low condensable vapor concentrations,
condensers are used as a preliminary unit that can be followed by a more
efficient control device such as a reactor, absorber or adsorber.
The phase equilibrium constant is employed in both chemical (combustion)
reaction operations and organic removal (condensation) calculations. The
phase equilibrium constant for component i, Kj, is defined by
K, = y1/x1
where y. = mole fraction of i in the vapor phase
x^ = mole fraction of i in the liquid phase.
If the vapor phase behaves as an ideal gas and the liquid phase is assumed
to be an ideal solution
k- = P'/P
-------
where p1 = vapor pressure of i
p • total pressure of the system.
This equilibrium constant is therefore seen to be a strong function of
temperature and pressure. Numerical values for K can be obtained
directly from the attached figure.
We now apply phase equilibrium principles to a typical condenser.
.
A material balance for component i gives
where 3r. = mole fraction of feed
F = moles of feed
L = moles of liquid
V = moles of vapor
some manipulation of the above two equations yields
Fj*i __ ,
^ = UAO+V
Me sum this equation to obtain
/w
3'v (0
where tf\ = number of components in the system.
The usual energy (enthalpy) balance calculations can be applied to determine
the rate of heat transfer in the condenser. Standard equations are employed
to design the heat exchanger.
-------
§ 8 § § 8 g § S
.........i. ......I....I....1 i.i.i.i.i.i.i
Prnum. pill
8883888 S
••'•'••••••I
V Mcthm
& S S 8 8 S
S S S 3
8 § § I 8 2
S 88S*Sg 8 8
Mtthrat'
l-t
-------
Example. A 9,700 ft /min (80°F, 1 atm) process gas stream containing
85% air, 10% heptane, 3.1% hexane, and 1.9% pentane Is to be condensed
in a unit so that only 0.5% combined hydrocarbons remain 1n the vapor phase.
Determine the temperature and pressure of the unit necessary to carry out
this operation. Assume the minimum temperature that can be achieved in the
condenser is 20°F.
Solution: Assume all the air remains in the vapor phase since
An overall material balance calculation giyes
F = 1.0
L = 0.145
V = 0.855
The values of each of the mole fractions in the vapor phase are calculated
from Eq. (1). The following table provides the results of a trial and error
calculation for the pressure at the minimum temperature:
K<320°F -&
"Component 300 psia y_
air oo 0.995
pentane 0.02 0.00234
hexane 0.0058 0.00120
heptane 0.0021 0.00144
2. = 0.99998
The operating conditions are 20°F and 300 psia. The result- clearly shows
that condensation is not economically feasible for separating light hydro-
carbons from air.
-------
Example. A 23,150 ft3/irrin stream at 530°F and 14.9 psia and containing
6056 air, 4% re-octane, 24% n-nonane, and 12% n-decane is to be condensed
in a heat exchanger operating with a discharge temperature of 150°F.
Calculate the mole fraction of each hydrocarbon in the vapor phase of
the exit stream.
Solution. The values of each of the mole fractions in the vapor
phase are calculated from Eq. (1) for assumed values of L and V. The
following table provides the results of the calculation with V = 0.63
and L = 0.37.
Component
air
octane
nonane
decline
K @ 150°F
14.9 psia
00
0.15
0.05
0.015
y
0.953
0.0129
0.0295
0.005
2. = 1.0004
The fraction of the inlet hydrocarbons remaining in the vapor phase is
therefore
= 0.075 or 7.5%
-------
Example. In order to meet recently updated pollution specifications
for discharging hydrocarbons to the atmosphere, the pollutants in the gas
stream in the previous example must be reduced to 99.5% of Its present
concentration. Due to economic considerations, it is proposed to meet
the above requirement by combusting the hydrocarbons in a thermal reactor
operating at 1500°F, rather than further increasing the pressure or de-
creasing the temperature of the condenser. The gas and methane (fuel) are
to be fed to the reactor at 80°F and 1 atm. Calculate fuel requirements
and design the proposed reactor.
Solution. The fuel requirements are .obtained using the calculational
procedures outlined in class. Using the data in the manual and solving
the enthalpy balance equation for a reaction (discharge) temperature of
1500°F gives a methane to gas ratio of
0.0326; ft3 CH4/ft3 gas or mole CH4/mole gas
We now calculate the volumetric flow rate in the reactor at 1500°F
neglecting the volume change associated with both the combustion of the
fuel and the hydrocarbons.
9,700' (^H§) = 35,200 cfm
A velocity of 20 ft/sec is suggested in the reactor section. (A slightly
higher velocity is usually employed in the burner.) The cross-sectional
area required is then
35.200 = 2g ,5 f 2
(60)120) "••" Tt
The reactor diameter is therefore 6 ft.
-------
Kinetic considerations dictate the length (and volume) of the reactor.
It Is conmon practice to treat a mixture of reactive hydrocarbons in terms
of a single hydrocarbon component undergoing a first order irreversible
reaction. A reaction velocity constant for the mixture of 7.8 (sec)
Is recommended at this temperature. We employ the plug-flow model, i.e.,
For 99.5% removal, we have
AC,
dL-^ - _
1 ^ i -A
k~) cf - *-*' CA
Integrating gives
where
£*_ .. o. ooS. ... Of
Therefore,
CAO A0
L= -(^?K'&i| ' °r
--13. GfT
Note that this corresponds to an average residence time of 0.68 sec.
-------
REVIEW OF FA1B & BLOUERS
1. TO CALCULATE A SYSTEM CURVE FROM AUY ONE KNOWN OPERATING POINT
FOR All EXHAUST SYSTEM
SP -
2. TO CALCULATE FAN CHARACTERISTICS FOR ANY SPEED FROM A FAN RATING
TABLE FCR 70°F
- spi
3» TO CORIIECT FAN CHARACTERISTICS FOR OPERATION AT TKIPERArURES OTrtiSR
THAN 70°F
CFMg «
SP2 " SP,
HP2 - HP.
•'(I)
-------
10-8
INDUSTRIAL VENTILATION
a
Volume
Volume
Fan with
variation
POOR SELECTION GOOD SELECTION
flat pressure curve gives wide volume . Fan with steep pressure curve gives small volume
with pressure change. variation with pressure change.
EFFECT OF FAN CURVE SLOPE
' I
Volume
POOR SELECTION
Small fan used with system curve crossing fan
too for to right of peak
Excessive horsepower
Low efficiency
EFFECT OF FAN SIZE
Volume
GOOD SELECTION
Large fan used with curve crossing fan curve to
right of peak.
Low horsepower
High efficiency
I
Volume of air delivered less than design flow
To obtain design flow:
increase fan speed -*-
••'- increase fan SP ~
Volume increase fan BHP (gj-J
EFFECT OF VARIATION BETWEEN DESIGN AND ACTUAL RESISTANCE
' AMERICAN CONFERENCE OF
GOVERNMENTAL INDUSTRIAL KYGIENISTS
FAN SELECT/ON
DATE lf-68
Fifj. IO-7
-------
INDUSTRIAL VENTILATION
Volume in cfm
Two idenficof fans
Recommended for best efficiency
t: ,/:
Volume ia cfm
Two different fans
Satisfactory
Notes:
/
FGf.T-individuak.fan-air
&. To establish system curve, include
losses in individual fan connections.
3. System curve must intersect combined
fan curve or higher pressure fan
may handle more air alone.
Voluma ia cfm - .
Two different fans J- -
Unsatisfactory
«
When system curve does not cross combined fan
curve, or crosses projected combined curve
before fan B, fan B will handle more air than
fans A and B in parallel.
. AMERICAN CONFERENCE OF
GOVERNMENTAL INDUSTRIAL HYGIEN1STS
FANS
PARALLEL OPERATION
DATE
\
Fig. /O-8
------- |