Presentation on theme: "1 Control of Sulfur Oxides 朱信 Hsin Chu Professor Dept. of Environmental Engineering National Cheng Kung University."— Presentation transcript:
1 Control of Sulfur Oxides 朱信 Hsin Chu Professor Dept. of Environmental Engineering National Cheng Kung University
2 This chapter and the next concern pollutants-sulfur oxides and nitrogen oxides-cannot be economically collected by physical means nor rendered harmless by combustion as the control of particulates and VOCs. Their control is largely chemical rather than physical.
3 SO 2, SO 3, and NO 2 are strong respiratory irritants that can cause health damage at high concentrations. These gases also form secondary particles in the atmosphere. They are the principal causes of acid rain.
4 1.The Elementary Oxidation-Reduction Chemistry of Sulfur and Nitrogen The sources and control methods are significantly different for sulfur oxides and nitrogen oxides, but their chemistry is quite similar. Table 11.1 (next slide) shows the oxidation and reduction products of nitrogen and sulfur.
6 Both hydrogen sulfide and ammonia are very strong-smelling substances, gaseous at room temperature (-60 o C and -33 o C boiling points, respectively), and toxic in high concentrations. NO, NO 2, SO 2, and SO 3 are all gases at room temperature or slightly above room temperature (boiling points 21 o C, 34 o C, -10 o C, and 45 o C, respectively). The estimated concentrations of these materials in unpolluted parts of the world’s atmosphere are SO 2, 0.2 ppb; NH 3, 10 ppb; NO 2, 1 ppb.
7 2. An Overview of the Sulfur Problem Fig (next slide) shows in part how sulfur moves in the environment as a result of human activities.
9 Sulfur is the sixteenth-most abundant element in the earth’s crust, with an abundance of about 260 ppm. The vast majority of this sulfur exists in the form of sulfates, mostly as gypsum, CaSO 4 2H 2 O, or anhydrite, CaSO 4.
10 All organic fuels used by humans (oil, coal, natural gas, peat, wood, others) contain some sulfur. Fuels like wood have very little (0.1% or less), whereas most coals have 0.5% to 3%. Oils generally have more sulfur than wood but less than coal.
11 If we burn the fuels, the contained sulfur will mostly form sulfur dioxide: If we wish to prevent this SO 2 from getting into the atmosphere, we can use any of the methods described in this chapter, all of which have the effect of capturing the sulfur dioxide in the form of CaSO 2 2H 2 O that will then be returned to the earth, normally in a landfill.
12 Most often the overall reaction will be In natural gas most of the sulfur is in the form of H 2 S, which is easily separated from the other constituents of the gas.
13 In oil and also in oil shales and tar sands, the sulfur is chemically combined with the hydrocarbon compounds; normally it cannot be removed without breaking chemical bonds. In oils the sulfur is concentrated in the higher- boiling fraction of the oil, so the same crude oil can yield a low-sulfur gasoline (average 0.03% S) and high-sulfur heavy fuel oil (e.g., 0.5% to 1% S).
14 In coal much of the sulfur is also in the form of chemically bound sulfur, but some coals have a large fraction of their sulfur in the form of small (typically 100µ) crystals of iron pyrite (fools gold, FeS 2 ). When the fuel is burned, almost all of the sulfur in the fuel is converted to SO 2. Some small fraction is captured in the ash, and some is converted to SO 3.
15 Mixtures of SO 2 and SO 3 are sometimes called SO X. The other important source of SO 2 attributable to humans is the processing of sulfur-bearing ores. The principal copper ore of the world is chalcopyrite, CuFeS 2.
16 The basic scheme for obtaining copper from it is the overall high-temperature smelting reaction: CuFeS O 2 → Cu + FeO + 2 SO 2 (2) The principal ores of lead, zinc, and nickel are also sulfides, whose processing is similar to above reaction. Table 11.2 (next slide) shows the emission sources for SO 2 in the U.S. in 1997.
18 The sulfur-containing gas streams most often dealt with in industry belong to three categories-reduced sulfur, concentrated SO 2 streams, and dilute SO 2 streams-each with its own control method.
19 3.The Removal of Reduced Sulfur Compounds from Petroleum, Natural Gas Streams, and Fuel Gases of Coal Gasification Example 1 A flow of 10 8 scf per day (32.8 sm 3 /s) of natural gas (≈ 0.2% of average U.S. consumption), which contains 1% (10,000 ppm) of H 2 S, is treated in the apparatus sketched in Fig (absorption –stripping) to reduce the H 2 S concentration to 4 ppm (the maximum allowed in commercial natural gas in the U.S.).
20 The gas is scrubbed at a pressure of 100 atm and 20 o C. Assuming that we will use water as the scrubbing agent, estimate the required water flow rate. Solution: From a handbook we find the Henry’s law constant for H 2 S at 20 o C = 483 atm, so
21 As in Example 14 of last chapter, we arbitrarily specify that the outlet liquid shall have y i * = 0.8 y i. Thus we can calculate that x i bottom = 0.8 0.01 = 1.66 (mol fraction )
22 The molar flow rate of gas is So the required liquid flow rate is This is a large liquid flow rate, one must find a solvent that can absorb much more H 2 S than can the water in this example.
23 H 2 S, SO 2, SO 3, NO 2, HCl, and CO 2 are acid gases, which form acids by dissolving in water. For H 2 S the process is If we can add something to the scrubbing solution that will consume either the H + or the HS -, then more H 2 S can dissolve in the water, and much less water is needed.
24 For acid gases, the obvious choice is some alkali, a source of OH - that can remove the H + by: Removing the H + on the right side of Eq. (3) drives the equilibrium to the right, greatly increasing the amount of H 2 S absorbed.
25 If the solution is to be regenerated in the rightmost column in Fig , then the alkali should be a weak alkali that can easily give back the acid gas on heating or pressure reduction. If only the leftmost column of Fig is used and the resulting solution is discarded, then a strong alkali, which could not be easily regenerated, can be used.
26 The most common choices of alkali for H 2 S removal are ethanolamines (monoethanolamine, diethanolamine, and triethanoamine) and also the sodium or potassium salts of weak acids like carbonic or phosphoric.
27 Example 2 Repeat Example 1, using as the absorbent a 2N (12.2 wt%, 3.94 mol%) solution of monoethanolamine (MEA), HO(CH 2 ) 2 NH 2, as the scrubbing solution, at 77 o F = 25 o C. Solution: For this solution strength and temperature, Kohl and Nielsen give a plot of p i = Py i * as a function of mols H 2 S per mol of MEA that can be approximated by:
28 Following Example 1, we choose the outlet liquid concentration to have y i * = 0.8 y i and solve Eq. (4) for the corresponding x, finding If we also require that at the top of the column y i * = 0.8 y i, we can compute the maximum permitted concentration of H 2 S in the regenerated solution as:
29 and The required liquid flow rate is that in Example 1 multiplied by (0.0250/6.04) = 0.041, or 13.5 lb/s = 6.1 kg/s. #
30 3.1The Uses and Limitations of Absorbers and Strippers for Air Pollution Control The system in Fig also works extremely well for removing ammonia or SO 2 from a gas stream. NO and NO 2 are not readily removed from gas streams by the process shown in Fig
31 Although NO 2 is an acid gas that produces nitric acid by reaction with water: the reaction rate is slow. NO is not an acid gas, so that although we can remove NO 2 from a gas stream with an alkaline solvent, we cannot remove NO with the same solvent.
Sulfur Removal from Hydrocarbons Once H 2 S has been separated from the other components of the gas, it is normally reacted with oxygen from the air in controlled amounts to oxidize it only as far as elemental sulfur (the Claus process): H 2 S = ½ O 2 → S + H 2 O Sulfur in hydrocarbon fuels is normally converted to SO 2 during combustion and then emitted to the atmosphere.
33 Large oil-burning facilities can have equipment to capture that SO 2, but autos, trucks, and airplanes do not. The only way to limit the SO 2 emissions from these sources is to limit the amount of sulfur in the fuel. Most of that sulfur is removed by catalytic hydrodesulfurization:
34 The mixture leaving the reactor is cooled, condensing most of the hydrocarbons. The remaining gas stream, a mixture of H 2 and H 2 S, is one of the streams treated in a refinery for H 2 S removal by the process shown in Fig
35 4. Removal of SO 2 from Rich Waste Gases The SO 2 concentrations in off-gases from the smelting of metal sulfide ores depend on which process is used and vary with time within the batch smelting cycle. However, they generally range from 2% to 40% SO 2. Such gases can be economically treated in plants that produce sulfuric acid by the following reactions:
36 Example 3 One of the largest copper smelters in the U.S. produces 320,000 tons of copper per year. The coppor ore smelted is principally chalcopyrite. If all the sulfur were emitted to the atmosphere as SO 2, how much would be emitted? If all the sulfur in the ore were converted to sulfuric acid, how much sulfuric acid per year would that smelter produce?
37 Solution: From Eq. (2) we know that we would expect to produce 2 mols of SO 2 per mol of copper. The molecular weights are 64 for SO 2 and 63 for copper, so we would expect to produce 320,000 ton/yr (2 64/63) = 650,000 ton/yr of SO 2. If this were all converted to H 2 SO 4 (MW = 98), it would be 650,000 ton/yr (98/64) = 996,000 ton/yr. #
38 Eq. (5) is an equilibrium reaction, which does not go to completion. Furthermore, this reaction is exothermic, so that the percent conversion at equilibrium is higher at low temperatures than at high temperatures. For this reason the reaction is customarily carried out in three or four separate catalyst beds with intercoolers between them. The course of the reaction is shown in Fig (next slide).
40 From Fig. 11.2, about 98% of the incoming SO 2 can be converted to H 2 SO 4. In a simple acid plant, the remaining 2% of the SO 2 is vented to atmosphere. In response to legal pressure to reduce this SO 2 emission, plants have been developed that are, in effect, two plants in series, as shown in Fig (next slide).
42 Comparing this flow diagram to Fig , we see an SO 3 absorber but no stripper. There is no need to regenerate the absorbing liquid, because the solution of SO 3 in water, H 2 SO 4, is the saleable product. The Henry’s law constant for SO 3 in water at 20 o C is roughly atmosphere, so this absorption is very rapid and easy.
43 A limitation is that production of H 2 SO 4 is uneconomic when the concentration of SO 2 in the waste gas is too low. Most analysts believe that with SO 2 concentrations more than 4%, the acid plant can show a profit if there is a nearby market for the acid.
44 5. Removal of SO 2 from Lean Waste Gases The major source of SO 2 is the stacks of large coal- or oil-burning facilities. For them, the typical SO 2 content of the exhaust gas is about 0.1% SO 2, or 1000 ppm, which is much too low for profitable recovery as H 2 SO 4.
45 The most widely used procedure for controlling SO 2 emissions from these sources is scrubbing with water containing finely ground limestone; the overall effect is shown by Eq. (1). The whole process is called flue gas desulphurization, and the acronym FGD is widely used.
46 Example 4 A power plant produces 10 6 scfm (472 sm 3 /s) of exhaust gas with 0.1% SO 2. We are required to remove 90% of it before the gas is discharged into the atmosphere. We propose to do so by dissolving the gas in Fig How large a flow of water will be needed if we make the same requirement as in Example 1, that at the bottom of the absorber y i * = 0.8y i ?
47 Solution: This situation is clearly similar to Example 1. However, here the Henry’s law constant for SO 2 is much smaller, 9 atm, leading to a greater solubility. The pressure is 1 atm instead of 100 atm. The outlet liquid will have the following characteristics:
49 As we saw in Example 2, the amount of scrubbing water required can be substantially reduced if we add a regent to the water that increases the solubility of the gas being removed. Example 5 The power plant in Example 4 wishes to remove 90% of the SO 2 by scrubbing the exhaust gas with a dilute solution of sodium hydroxide, NaOH. How much NaOH will they need? What problems will they encounter?
50 Solution: The overall reaction (including the oxidation of sulfite to sulfate) will be: 2NaOH + SO 2 + ½ O 2 → Na 2 SO 4 + H 2 O (7) From Example 4, we know that we must remove (0.9)(0.001)(43.3 lb mol/s) = (lbmol/s) of SO 2. Therefore, we will need, as a minimum, 2(0.039) = lbmol/s of NaOH.
51 One lbmol of NaOH weighs 40 lb, so the annual NaOH requirement will be 40 3.15 10 7 s/yr = 98.4 10 6 lb/yr = 49,200 t/yr = 44,700 tonne/yr. The prices of industrial chemicals fluctuate, but the price of NaOH is about $500/ton (dry basis), so that the NaOH for this plant would cost about $22 million per year. #
52 For the high-pressure treatment of H 2 S, either plate or packed towers are used, with little problem. For the SO 2 problem, three plausible arrangements are sketched in Fig (next slide).
54 The first of these is a simple bubbler. If the liquid is deep enough and the bubbles are small enough, this kind of device will bring the gas close to chemical equilibrium with the liquid. However, it has a high pressure drop.
55 Plate-type distillation and absorption columns are, in effect, a series of such bubblers, stacked one above the other. At high pressures, where pressure drops are unimportant, they are the most widely used device.
56 The second arrangement is a spray chamber. In this arrangement the gas pressure drop is small, but it is difficult to approach equilibrium because the gas also not contact the liquid as well as it does in the bubbler. Nonetheless, it is widely used because of its simplicity, low pressure drop, and resistance to scale deposition and plugging.
57 The third arrangement is a packed column, which is similar to the spray chamber except that the open space is filled with some kind of solid material that allows the liquid to coat its surface and run down over it in a thin film. This kind of contactor can be designed to have a better mass transfer per unit of gas pressure drop than either of the other two kinds.
58 The gas velocities in such devices range from about 1 ft/s in a packed tower to 10 ft/s for a spray chamber. If we assume we are going to treat the gas in Example 5 in a spray chamber at a gas velocity of 10 ft/s, the cross-sectional area perpendicular to the gas flow will be:
59 Such devices are almost always cylindrical. For this example the diameter would be 14 m. a typical length in the flow direction would be 15 m. That is a very large diameter for any piece of chemical plant equipment, but not for a power plant.
60 What problems might power plant operators encounter? First, there is the question of what to do with the sodium sulfate produced. The sodium sulfate produced in this process would be contaminated with fly ash from the coal.
61 So that although a few power plants might sell their sodium sulfate, most could not. Because of its wates solubility, it is not generally acceptable in landfills they are well protected from water infiltration.
62 But the real difficulty is with carbon dioxide. Normally the concentration of CO 2 in the exhaust gas is about 12%, or 120 times that of the SO 2. We are not generally concerned with the fate of CO 2, but if it gets into solution it will use up sodium hydroxide by the reaction: 2NaOH + CO 2 → Na 2 CO 3 + H 2 O
63 Any sodium hydroxide used up this way is not available to participate in Reaction (7). The real problem is how to absorb one acid gas while not absorbing another acid gas that is present in much higher concentration!
64 Fortunately, this is possible because SO 2 forms a much stronger acid than does CO 2. The reactions that occur in the liquid phase are these: If we find the right concentration of H + in solution, it may be possible to drive the equilibrium in Eq. (8) to the left while driving the equilibrium in Eq. (9) to the right.
65 That is indeed possible if the concentration of hydrogen ion is between and mols per liter (pH = 4 to 6). To remove SO 2 without absorbing CO 2, we must use a scrubbing solution that is a weak acid. Furthermore, we must be careful to control the pH of our solution so that it is acid enough to exclude CO 2 but not acid enough to exclude SO 2.
66 5.1Forced-Oxidation Limestone Wet Scrubbers The most widely used process to deal with these problems is forced-oxidation limestone wet scrubbing because limestone is much cheaper than NaOH. There are a variety of flowsheets and of mechanical arrangements for this process; Figs (next slide) and 11.6 (second slide) and Table 11.3 (third slide) show one of the most commonly used varieties.
70 Entrainment separators, often called demisters, cause the fine droplets carried with the gas to collect on their surfaces, coalesce, and fall back into the scrubber as drops large enough to fall counter to the upward-flowing gas. The slurry of water and solid particles (CaCO 3, CaSO 4 2H 2 O, and CaSO H 2 O) is pumped from the sump at the bottom of the module to the sprays, where it forms drops that fall through the rising flue gas and do the actual SO 2 removal.
71 Finely ground limestone is added to the sump. A small stream of slurry is sent by the solid removal pump to a hydroclone from which the underflow passes to a belt filter, from which a semidry gypsum product leaves the system.
72 The scrubber operates at or near the adiabatic saturation temperature of the entering flue gas, which is about 125 o F (52 o C). In some installations the cleaned flue gas is reheated to about 175 o F (80 0 C) to restore plume buoyancy and prevent acid corrosion of the ducts and stack downstream of the reheater.
73 Example 6 The spray nozzles in Fig are normally designed to produce droplets with an average diameter of about 3 mm. (a)How fast do those fall, relative to the gas and relative to the container? (b)For the slurry flow rate shown in Table 11.3, what is the equivalent rainfall rate in in./s?
74 (c)What fraction of the droplets evaporates to bring the gas from its inlet temperature to its adiabatic saturation temperature? (d)What fraction of the CaCO 3 will react as the slurry makes one pass through the scrubber?
75 Solution: (a)From Fig. 8.7 we read that 3 mm sphere with a specific gravity of 1.0 would be expected to fall about 20 ft/s. This velocity is relative to the gas. Relative to the walls of the scrubber, the velocity will be ( ) = 10 ft/s.
76 (b)The equivalent rainfall rate is This is far beyond any outdoor rainfall rate. Cloudbursts sometimes delive a few inches (≈ 100 mm) of rain an hour.
77 (c)From the combustion textbook, we know that the evaporated water ratio is given by where λ = the latent heat of vaporization c p = the heat capacity of the gas at constant pressure The liquid-gas ratio passing up and down the scrubber is So that about 0.4% of the water is evaporated inside the scrubber.
78 (d)We see that 1000 acf at scrubber inlet temperature corresponds to 614 scf, which corresponds to 1.71 lbmol, containing lbmol of SO 2. The 90% of that which is captured corresponds to lbmol, which equals the number of mols of CaCO 3 reacted. The corresponding slurry is 100 gal8.3 lb/gal ≈ 830 lb, of which 15% or lb is solids, of which 11.6% (typical FGD) or 14.4 lb is CaCO 3. Thus the fraction of the contained CaCO 3 that could be converted in one pass through the scrubber is ( lbmol 100 lb/lbmol)/14.4 lb ≈ 1%.
79 This example shows that liquid circulation rates in these scrubbers are very large. As a consequence, even thought they remove most of the SO 2 from the gas, the scrubbing slurry passes through them practically unchanged.
80 Most of the chemical reactions take place in the effluent hold tank. The slurry spends about 3 seconds per pass in the scrubber and about 8 minutes between passes through the scrubber in the hold tank.
The Development Problems with Limestone Scrubbers (1)Corrosion: the exhaust gases from coal combustion contain small amounts of many chemicals, e.g., chlorides. In an acid environment these proved much more corrosive to metals, including stainless steels, than the designers of the first systems had anticipated.
82 (2)Solids deposition, scaling, and plugging: Calcium sulfate and its near chemical relatives are slightly soluble in water and can precipitate on solid surfaces to form hard, durable scales that are very difficult to remove.
83 (3)Entrainment separator plugging: The spray nozzles shown in Fig do not produce totally uniform drops. Some of the drops are small enough to be carried along with the gas and must be removed from the gas in the entrainment separator. The early entrainment separators were plugged by the solids contained in those small drops.
84 (4)Poor reagent utilization: The product sulfates and sulfites can precipitate on the surface of the limestone particles, thus blocking their access to the scrubbing solution. (5)Poor solid-liquid separation: CaSO 3 0.5H 2 O tends to form crystals that are small, flat plates. These are very good at trapping and retaining water.
85 If the solid product has too many CaSO 3 0.5H 2 O it will have the consistency of toothpaste and not be acceptable for landfills. CaSO 4 2H 2 O forms larger, rounder crystals that are much easier to settle and filter.
86 The solution to these problems has been found by careful attention to engineering and chemical detail. The rate of liquid rejection to waste water (Fig. 11.5) is chosen to control the chloride content of the circulating liquid. The most widely used metal for lining the surfaces of the modules is alloy C-276, 55% Ni, 17% Mo, 16% Cr, 6% Fe, and 4% W. it costs roughly 15 times as much as ordinary steels.
Other Approaches As the technical difficulties with the limestone scrubber were wroked out, it became the clear economical choice for scrubbing stack gas from the combustion of medium-or high-sulfur coal. The other processes are not being used now for new installations, and some of those installed 20 years ago are being converted to forced oxidation limestone scrubbers. Table 11.4 (next slide) compares these processes.
Other Wet Systems Ca(OH) 2 (hydrated lime, quicklime) is an alternative to limestone in wet throwaway processes. Normally, CaO (lime, burned lime) is added to the oxidation tank and hydrates there to Ca(OH) 2.
90 CaO is more chemically reactive than limestone, mostly because it has a much higher surface area. CaO is prepared by heating limestone and driving off the CO 2. The result is a porous structure with typical surface areas of 15 m 2 /g.
91 In the early days of scrubber development this extra reactivity seemed necessary, but as the problems with wet limestone scrubbers have mostly been solved, the additional reactivity of lime has seemed less likely to repay its extra cost. The other regents shown have similar situations.
Dry Systems The solids handling and wet sludge handling and disposal difficulties that are integral to wet throwaway processes induced engineers to develop dry throwaway processes that would have fewer corrosion and scaling difficulties and would produce a waste product much easier to handle and dispose of.
93 All of these systems inject dry alkaline particles into the gas stream, where they react with the gas to remove SO 2. The SO 2 –containing particles are then captured in the particle collection device that the plant must have to collect fly ash. The flow diagrams for such systems are sketched in Fig (next slide).
95 The desired reaction is CaO + SO 2 → CaSO 3 CaSO 3 would then oxidize to CaSO 4. In principle this should work, but most tests have shown that to get high SO2 collection efficiencies one must put a large excess of lime or limestone into the system, thus increasing reagent costs, the load on the particle collector, and the volume of solid wastes to be disposed of.
Wet-Dry Systems The most widely used wet-dry systems are spray dryers. The flow diagram of this arrangement is shown in Fig (next slide).
98 Spray dryers are widely used in the process industries, e.g., powdered milk, instant coffee, laundry detergents, etc. In all such spray dryers a liquid (almost always water) containing dissolved or suspended solids is dispersed as droplets into a hot gas stream.
99 The dispersion can be done by a high- pressure gas-atomizing nozzle or a rapidly rotating (about 10,000 rpm) atomizing wheel. The hot gas is well above the boiling temperature of water, so that the water in the droplets evaporates rapidly.
100 The particles formed from the evaporating drops are dry before they reach the wall or bottom bin of the dryer, so they form a free-flowing powder that is easily removed. The reagent slurry is dispersed as 10 to 50 µ drops, containing about 30 wt% solids.
101 The resulting dry particles are small enough that most are carried along with the gas stream. This feature is different from most spray dryers for consumer products, in which the particles are large enough that most of the particles settle to the bottom of the dryer. Some test results from a large pilot plant spray dryer designed to control SO 2 are shown in Fig (next slide).
103 Example 7 On Fig. 11.9, at a reagent ratio of 1.1 (a)What is the percent efficiency of the spray dryer alone for SO 2 capture? (b)What is the percent efficiency of the filter alone for SO 2 capture, based on the inlet concentration to the filter?
104 (c)How much water is admitted per pound of inlet gas? (d)How much solids are fed, and what fraction of those are recycled? (e)In the mix of gas and particles in the spray dryer, what is the average spacing between drops as they turn into particles?
105 Solution: (a)Reading Fig. 11.9, we find an efficiency of 68% in the spray dryer and 17% in the filter. (b)For the filter, the incoming gas has 100% - 68% = 32% of the original SO 2, so that its SO 2 collection efficiency is η=17% / 32% = 53% Viewing these as penetrations, p overall = p dryer p filter = = 0.15
106 (c)The inlet temperature is 325 o F and the approach to adiabatic saturation is 20 o F. The inlet gas moisture content is not stated, but it is assumed to be 8 mol% ≈ lb/lb. According to a psychrometric chart, one finds that the corresponding adiabatic saturation temperature is about 128 o F. Thus the outlet temperature is 148 o F.
107 Checked again with the psychrometric chart, one finds that the outlet moisture content is about lb/lb (corresponding to 148 o F), so that the moisture added is ( ) = lb/lb of air.
108 (d)The slurry is 30 wt% solids, so the solid feed rate is
109 The lime feed rate is Thus (0.0056/0.020) = 28% of the solids is fresh Ca(OH) 2 and the remaining 72% is recycled solids from the baghouse or from the bottom of the spray dryer.
110 (e)For 1.0 lb of gas there are lb of water and lb of solids, or lb of slurry. Estimating the specific gravity of the slurry as 1.2, we estimate the volume of slurry per pound of gas as
111 For an assumed average droplet size of 20µ, we calculate the number of drops per lb of gas as The volume of 1.0 lb of gas at the average temperature of the dryer as
112 Dividing this value by the corresponding number of particles, we find that each particle would occupy a volume of 8.2 m 3, corresponding to a cube with an edge length of 0.4 mm. #
113 Comparing this device to the wet limestone scrubber, we see that the drops are much smaller (20 µ / 3 mm = 1/150). The time the gas spends in the scrubbing environment is roughly twice as large.
114 However, once the particles become dry, their reactive capacity is greatly reduced compared to the drops in wet scrubbers. In addition, the co-flow parttern is much less efficient.
o F approach temperature indicates that the amount of water fed in the slurry was limited to keep the gas temperature 20 o F above its adiabatic saturation temperature. Test data show that the collection efficiency improves as one approaches saturation, presumably because much of the reaction takes place before the droplet is completely converted to a solid and a close temperature approach keeps the drops wet longer.
116 However, if more water is introduced than is needed to cool the gas to the adiabatic saturation temperature, then not all the drops will evaporate, the resulting particles will be wet and sticky, and the dryer and downstream equipment will suffer severe plugging and corrosion.
117 6.Alternatives to “Burn and Then Scrub” 6.1Change to a Lower Sulfur Content Fuel If the management of a power plant can replace a high-sulfur coal with a low-sulfur coal, it reduces the SO 2 emissions quickly, simply, and without having to install expensive SO 2 control devices or to deal with their solid effluent.
Remove Sulfur from the Fuel Coal Cleaning Pyritic sulfur can be removed by grinding the coal to a small enough size that the pyrites are mostly present as free pyrite particles. Gravity methods are then used to separate the low-density coal (s.g. = 1.1 to 1.3) from high-density pyrites (s.g. = 5.0).
Solvent-Refined Coal It is also possible to dissolve coal in strong enough solvents and then to treat the solution by the same kind of catalytic hydrogenation processes that are used to remove sulfur from petroleum products. The mineral (ash-forming) materials do not dissolve, so they are rejected by filtration or settling.
120 When the solvent is then removed for reuse, the remaining product is a very clean-burning combustible solid. Considerable development work on this process showed that it can be done, but so far not at a price comparable to “burn and then scrub”.
Modify the Combustion Process The standard way of burning large amounts of coal (pulverized-coal furnace) is to grind the coal to about 50- to 150-µ size and blow it with hot air into a large combustion chamber. Fluidized bed combustion (FBC) is an alternative way to burn coal that is currently in the demonstration plant stage.
122 In FBC, coal is burned in gravel-sized pieces by injecting them into a hot fluidized bed of limestone particles instead of as a finely dispersed powder in air. In such a fluidized bed combustor SO 2 is formed in the presence of a large number of limestone particles and has a high probability of reacting with one of them in the combustion bed.
123 The fluidized bed has tubes full of water and steam projecting into the bed. The heat transfer between the hot bed in which the coal is burned and the tubes is much better than that between the flames and the walls of an ordinary coal-fired boiler.
124 For this reason fluidized bed combustors are smaller and operate at lower temperatures than ordinary coal-fired boilers. This saves on some costs and greatly reduces the formation of nitrogen oxides.
125 A second combustion modification alternative, also in the demonstration plant stage, is to convert the coal to a synthetic fuel gas and then burn that in combination gas-turbine steam-turbine power plants. Fig (next slide) shows a schematic flow diagram of an integrated gasification combined cycle power plant.
Don’t Burn at All Nuclear, wind, solar, tidal, geothermal, and hydroelectric power plants will reduce our emissions of SO 2. More efficient electric devices (lights, refrigerators, motors) are, in effect, SO 2 control devices.