V. Sulfur and Sulfur Compounds sulfur – one of the important basic raw materials in the chemical industry sulfur compounds – used in the manufacture of wood pulp, insecticides, bleaching agents, vulcanized rubber, detergents, pharmaceutical products, dyes A.
Sources of Sulfur
1. Mined, as either a. raw sulfur – pure sulfur plus inert materials b. sulfide ores, e.g. pyrite (FeS2), sphalerite (ZnS), chalcopyrite (CuFeS2) Frasch method – process for extracting sulfur developed by Herman Frasch and first used in 1894 – three pipes, one inside another, are sunk to the bottom of the sulfur bed. Water heated under pressure to a temperature well above the melting point of the sulfur (>160°C) is conducted down the outer pipe, and air under pressure down through the innermost pipe. The heated water melts the sulfur and the compressed air forces it through the middle pipe to the surface.
2. Obtained as H2S from the desulfurization of crude oil and natural gas. The H2S is converted to pure S using the Claus process. At present, more than half of the world’s supply of sulfur is recovered from this process. B.
Processes
Sulfur is usually burned to SO2 and SO3 first before it is converted into other compounds.
air sulfur (raw sulfur or pyrite)
SO2 SO3 O2 N2
burner
catalytic converter
SO2 more SO3 O2 N2
absorption tower
contact tower
bisulfites (Ca(HSO3)2, Mg(HSO3)2) sulfates (CaSO4, MgSO4)
sulfuric acid solution (H2SO4, H2O) oleum (H2SO4, SO3)
Summary of processes: 1. 2. 3. 4. 5.
Burning of raw sulfur roasting of pyrites Bisulfite and sulfate production Conversion of SO2 to SO3 Sulfuric acid production
Table 6-1, page 167: heats of formation of sulfur compounds Figure 6-1: page 168: average molal heat capacities C.
Burning of Raw Sulfur
Under normal furnace conditions, much of the sulfur is oxidized to SO2. A small percentage (2 to 10%) may be further oxidized to SO3.
S + O2 → SO2 3 S + O2 → SO3 2 During analysis, the SO3 in the product gases condenses in the presence of H2O.
SO3 + H2O → H2SO4 The resulting analysis is usually reported on an SO3-free basis.
theoretical/excess oxygen – based on the conversion of S to either SO2 or SO3, which must be clearly stated in the problem e.g. Raw sulfur analyzing 95% S and 5% inerts is burned with 65% excess air (S→SO2). Analysis of the cinders shows 10% S and 90% inerts. 88% of the S gasified burns to SO2, the rest to SO3. Determine the analysis of the burner gases on an SO3-free basis. Also, determine its complete analysis. Given: air 65% x’s (S→SO2) raw sulfur 95% S 5% inerts
burner
burner gases 88% S→SO2
cinders 10% S 90% inerts Required: a. analysis burner gases, SO3-free b. complete analysis burner gases Solution: Basis: 100 lb raw sulfur S inerts total
lb 95 5 100
mol theo O 2 = 2.97 lbmol S ⋅
MW 32 -
1 lbmol O2 lbmol S
= 2.97 lbmol O2 mol O2 supplied = 1.65 ⋅ 2.97 lbmol O2 = 4.90 lbmol O2 Inerts balance:
5 lb = 0.90 ⋅ mass cinders mass cinders = 5.56 lb
mol 2.97 -
mass S in cinders = 0.10 ⋅ 5.56 lb = 0.56 lb S mass S gasified = 95 lb − 0.56 lb = 94.44 lb S 1 lbmol 32 lb = 2.95 lbmol S
mol S gasified = 94.44 lb S ⋅ Burner gases:
mol SO2 = 0.88 ⋅ 2.95 lbmol S ⋅
1 lbmol SO2 lbmol S
= 2.60 lbmol SO2 mol SO3 = 0.12 ⋅ 2.95 lbmol S ⋅
1 lbmol SO3 lbmol S
= 0.35 lbmol SO3 O2 balance:
4.90 lbmol O2 = 2.60 lbmol SO2 ⋅
1 lbmol O2 3 lbmol O2 + 0.35 lbmol SO3 ⋅ + lbmol SO2 2 lbmol SO3
mol free O2 mol free O2 = 1.78 lbmol O2 N2 balance:
4.90 lbmol O2 ⋅
0.79 lbmol N2 = mol N2 in burner gas 0.21 lbmol O2
18.43 lbmol N2 = mol N2 in burner gas Orsat analysis burner gas, SO3-free: SO2 O2 N2 total
mol 2.60 1.78 18.43 22.81
mol % 11.40 7.80 80.80 100
a
mol % 11.23 1.51 7.69 79.58 100.01
b
Complete Orsat analysis burner gas: SO2 SO3 O2 N2 total
mol 2.60 0.35 1.78 18.43 23.16
e.g. Burner gases resulting from the burning of sulfur have the following dry analysis: 9.14% SO2, 10.66% O2, and 80.20% N2. Calculate the percent conversion of S to SO2 and the percent excess air used (S→SO2). Given: air
S
burner gases dry, SO2-free basis: SO2 9.14% O2 10.66% N2 80.20%
burner
cinders Required: a. % conversion S→SO2 b. % x’s air (S→SO2) Solution: Basis: 100 lbmol dry SO3-free burner gas SO2 O2 N2 total
mol 9.14 10.66 80.20 100
mol O2 9.14 10.66 19.80
No N2 from feed. N2 balance:
mol N2 from air = 80.20 lbmol N2 mol O2 supplied = 80.20 lbmol N2 ⋅
0.21 lbmol O2 0.79 lbmol N2
= 21.32 lbmol O2 mol O2 unaccounted for = 21.32 lbmol O2 − 19.80 lbmol O2 = 1.52 lbmol O2 = mol O2 in SO3 mol SO3 = 1.52 lbmol O2 ⋅ = 1.01 lbmol SO3 Assume all S in feed gasified.
2 lbmol SO3 3 lbmol O2
S balance:
mol S in feed = 9.14 lbmol SO2 ⋅
1 lbmol S 1 lbmol S + 1.01 lbmol SO3 ⋅ lbmol SO2 lbmol SO3
= 10.15 lbmol S 1 lbmol S lbmol SO2 % S → SO2 = ⋅ 100 10.15 lbmol S = 90.05% a 9.14 lbmol SO2 ⋅
mol theo O 2 ( S → SO2 ) = 10.15 lbmol S ⋅
1 lbmol O2 lbmol S
= 10.15 lbmol O2 % x's O2 ( S → SO2 ) =
21.32 lbmol O2 − 10.15 lbmol O2 ⋅ 100 10.15 lbmol O2
= 11 0.05%
b
e.g. A plant burns sulfur which is 99.4% pure at the rate of 680 lb per hour. The average air temperature is 60°F. The gases emerge from the burner at 760°C and are found to contain 17.4% SO2 and 2.7% O2 when analyzed in the usual way. The gases then pass to a cooler, which reduces their temperature to 70°F by means of water which rises from 58°F to 90°F. Calculate: a. the percent of the sulfur burned to SO3 b. the pounds per hour of SO2 leaving the burner c. the air consumption in cubic feet per minute d. the cubic feet per minute of gas leaving the burner and the cubic feet per minute leaving the cooler e. the heat dissipated from the burner in BTU per hour f. the water used in the cooler in gallons per hour Given:
raw sulfur 680 lb/hr 99.4% S
air, 60°F
water, 58°F
burner
cooler
cinders
burner gases 760°C SO2 17.4% O2 2.7% water, 90°F
burner gases 70°F
Required: a. % S burned to SO3 b. lb/hr SO2 c. ft3 /min air d. ft3/min burner gases leaving the burner, ft3/min burner gases leaving the cooler e. heat loss from burner, BTU/hr f. gal/hr water Solution: Basis: 100 lbmol dry, SO3-free burner gases SO2 O2 N2 total
mol 17.4 2.7 79.9 100
mol O2 17.40 2.70 20.10
No N2 from raw sulfur. N2 balance:
mol N2 from air = 79.90 lbmol N2 mol O2 supplied = 79.90 lbmol N2 ⋅
0.21 lbmol O2 0.79 lbmol N2
= 21.24 lbmol O2 mol O2 unaccounted for = 21.24 lbmol O2 − 20.10 lbmol O2 = 1.14 lbmol O2 = mol O2 in SO3 mol SO3 = 1.14 lbmol O2 ⋅
2 lbmol SO3 3 lbmol O2
= 0.76 lbmol SO3 Assume all S in raw sulfur gasified. S balance:
mol S in raw sulfur = 17.4 lbmol SO2 ⋅
1 lbmol S 1 lbmol S + 0.76 lbmol SO3 ⋅ lbmol SO2 lbmol SO3
= 18.16 lbmol S 1 lbmol S 0.76 lbmol SO3 ⋅ lbmol SO3 % S → SO3 = ⋅ 100 18.16 lbmol S = 4.19% a
Basis: 1 hr operation
total mass raw sulfur = 680 lb Raw sulfur: lb 675.92 4.08 680
S inerts total
MW 32 -
mol 21.12 -
% S → SO2 = 100% − 4.19% = 95.81% mol SO2 in burner gases = 0.9581 ⋅ 21.12 lbmol S ⋅
1 lbmol SO2 lbmol S
= 20.24 lbmol SO2 mass SO2 = hr
20.24 lbmol SO2 ⋅
64 lb lbmol
1 hr 1,295.36 lb = b hr
mol burner gases (SO3 -free) = 20.24 lbmol SO2 ⋅ = 116.32 lbmol Burner gases: SO2 O2 N2 total
mol 20.24 3.14 92.94 116.32
mol N2 from air = 92.94 lbmol N2 mol dry air = 92.94 lbmol N2 ⋅
1 lbmol dry air 0.79 lbmol N2
= 117.65 lbmol Assume: Air supplied is dry, PT = 1 atm
100 lbmol burner gases 17.4 lbmol SO2
vol dry air = min =
117.65 lbmol ⋅ 744 ft 3 min
359 ft 3 ( 60 + 460°R ) ⋅ 1 hr lbmol 492°R ⋅ 1 hr 60 min
c
mol SO3 in burner gases = 0.0419 ⋅ 21.12 lbmol S ⋅
1 lbmol SO3 lbmol S
= 0.88 lbmol SO3 total mol burner gases = 116.32 lbmol + 0.88 lbmol = 117.20 lbmol Tburner gases = 760°C ⋅
9 + 32 5
= 1400°F vol burner gases = min
117.20 lbmol ⋅
359 ft 3 (1400 + 460°R ) ⋅ 1 hr lbmol 492°R ⋅ 1 hr 60 min
2651 ft 3 d min vol cooled burner gases 2651 ft 3 70 + 460°R = ⋅ min min 1400 + 460°R 3 755 ft = d min =
Energy balance around burner: air, 60°F
raw sulfur
burner
cinders
ΔHformation = ΔHburner gases + Qloss Qloss = ΔHformation − ΔHburner gases
heat loss
burner gases 1400°F
Heats of formation from (Table 6-1):
S + O2 → SO2 3 S + O2 → SO3 2
70,920 cal gmol 93,900 cal gmol
Basis: 1 hr
mol SO2 = 20.24 lbmol SO2 mol SO3 = 0.88 lbmol SO3 454.5 gmol 70,920 cal 1 BTU ⋅ ⋅ + lbmol gmol 252 cal 454.5 gmol 93,900 cal 1 BTU 0.88 lbmol SO3 ⋅ ⋅ ⋅ lbmol gmol 252 cal = 2,737,917 BTU
ΔHformation = 20.24 lbmol SO2 ⋅
Enthalpy burner gases at 1400°F: Figure 6-1: cpave of SO2, SO3 Figure 1-3: cpave of O2, N2 SO2 SO3 O2 N2 total
n 20.24 0.88 3.14 92.94 117.2
cpave, 1400°F 11.7 17.0 7.7 7.3 -
ΔHburner gases, 1400°F = 1,278,907 BTU Qloss = 2,737,917 BTU − 1,278,907 BTU =
1,459,010 BTU hr
Energy balance around cooler: Assume: No heat loss
e
n⋅cpave(1400-60) 317,323 20,046 32,399 909,139 1,278,907
water 58°F
burner gases 1400°F
cooler
burner gases 70°F
water 90°F
ΔHburner gases, 1400°F + ΔHwater , 58°F = ΔHburner gases, 70°F + ΔHwater , 90°F ΔHwater , 90°F − ΔHwater , 58°F = ΔHburner gases, 1400°F − ΔHburner gases, 70°F Assume: cp of water is constant
ΔHwater , 58°F→90°F = ΔHburner gases, 1400°F − ΔHburner gases , 70°F Enthalpy burner gases at 70°F: n 20.24 0.88 3.14 92.94 117.2
SO2 SO3 O2 N2 total
ΔHburner
gases , 70° F
cpave, 70°F 9.4 11.8 7.0 7.0 -
= 8,733 BTU
ΔHwater , 58 °F→90°F = 1,278,907 BTU − 8,733 BTU = 1,270,174 BTU = mwater ⋅ cp ⋅ ΔT 1,270,174 BTU BTU 1 ⋅ ( 90°F − 58°F ) lb ⋅ °F = 39, 693 lb
mwater =
vol water = hr =
39,693 lb ⋅ 4,766 gal hr
ft 3 7.48 gal ⋅ 62.3 lb ft 3 1 hr f
n⋅cpave(70-60) 1,903 104 220 6,506 8,733
