163 C HAPTER 8 Gaseous Emission Control Be it known to all within the sound of my voice, Whoever shall be found guilty of burning coal Shall suffer the loss of his head. King Edward II 8.1 INTRODUCTION Limiting gaseous emissions into the air is technically difficult as well as expensive. Although it is true that rain is “Nature’s vacuum cleaner” — the only natural air-cleansing mechanism available — it is not very efficient. Good air quality depends on pollution prevention (limiting what is emitted) and sound engineering policies, procedures, and practices. The control of gaseous air emissions may be realized in a number of ways. In this chapter, we discuss many of these technologies, as well as the sources of gaseous pollutants emitted from various sources and their control points (see Figure 8.1). The applicability of a given technique depends on the properties of the pollutant and the discharge system. In making the difficult and often complex decision of which gaseous air pollution control to employ, follow the guidelines based on experience and set forth by Buonicore and Davis (1992) in their prestigious engineering text, Air Pollution Engineering Manual . Table 8.1 summa- rizes the main techniques and technologies used to control gaseous emissions. In the following, we discuss the air control technologies given in Table 8.1. Much of the information contained in this chapter is heavily adapted from Spellman’s The Science of Air (1999) and USEPA-81/12 (1981), Control of Gaseous Emissions, Course 415. The excerpted materials are rearranged and edited to make the materials more accessible for the reader. 8.2 ABSORPTION Absorption (or scrubbing) is a major chemical engineering unit operation that involves bringing contaminated effluent gas into contact with a liquid absorbent so that one or more constituents of the effluent gas are selectively dissolved into a relatively nonvolatile liquid. Key terms used when discussing the absorption process include: • Absorbent : the liquid, usually water mixed with neutralizing agents, into which the contaminant is absorbed • Solute : the gaseous contaminant being absorbed, such as SO 2 , H 2 S, and so forth L1681_book.fm Page 163 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 164 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK • Carrier gas : the inert portion of the gas stream, usually flue gas, from which the contaminant is to be removed • Interface : the area where the gas phase and the absorbent contact each other • Solubility : the capability of a gas to be dissolved in a liquid Absorption units are designed to transfer the pollutant from a gas phase to a liquid phase, accomplished by providing intimate contact between the gas and the liquid, which allows optimum diffusion of the gas into the solution. The actual mechanism of removal of a pollutant from the gas stream takes place in three steps: (1) diffusion of the pollutant gas to the surface of the liquid; (2) transfer across the gas–liquid interface; and (3) diffusion of the dissolved gas away from the interface into the liquid (Davis and Cornwell, 1991). Figure 8.1 Air pollution control points. (From Spellman, F.R., 1999, The Science of Air: Concepts and Appli- cations . Boca Raton, FL: CRC Press.) Table 8.1 Comparison of Air Control Technologies Treatment technology Concentration and efficiency Comments Absorption (<200 ppmv) 90–95% efficiency (>200 ppmv) 95%+ efficiency Can blowdown stream be accomplished at site? Carbon adsorption (>200 ppmv) 90%+ efficiency (>1000 ppmv) 95%+ efficiency Recovered organics may need additional treatment; can increase cost. Incineration (<100 ppmv) 90–95% efficient (>100 ppmv) 95–99% efficient Incomplete combustion may require additional controls. Condensation (>2000 ppmv) 80%+ efficiency Must have low temp. or high pressure for efficiency. Source : Spellman, F.R., 1999, The Science of Air . Boca Raton, FL: CRC Press, p. 221. Cooling Treatment Collection Source Dispersion L1681_book.fm Page 164 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC GASEOUS EMISSION CONTROL 165 Several types of scrubbing towers are available, including spray chambers (towers or columns); plate or tray towers; packed towers; and Venturi scrubbers. Pollutant gases commonly controlled by absorption include sulfur dioxide; hydrogen sulfide; hydrogen chloride; ammonia; and oxides of nitrogen. The two most common absorbent units in use today are the plate and packed tower systems. Plate towers contain perforated horizontal plates or trays designed to provide large liquid–gas interfacial areas. The polluted airstream is usually introduced at one side of the bottom of the tower or column and rises up through the perforations in each plate; the rising gas prevents the liquid from draining through the openings rather than through a downpipe. During continuous operation, contact is maintained between air and liquid, allowing gaseous contaminants to be removed, with clean air emerging from the top of the tower. The packed tower scrubbing system (see Figure 8.2) is predominantly used to control gaseous pollutants in industrial applications, where it typically demonstrates a removal efficiency of 90 to 95%. Usually vertically configured (Figure 8.2), the packed tower is literally packed with devices (see Figure 8.3) of large surface-to-volume ratio and a large void ratio that offer minimum resistance to gas flow. In addition, packing should provide even distribution of both fluid phases; be sturdy enough to support itself in the tower; and be low cost, available, and easily handled (Hesketh, 1991). The flow through a packed tower is typically countercurrent, with gas entering at the bottom of the tower and liquid entering at the top. Liquid flows over the surface of the packing in a thin film, affording continuous contact with the gases. Though highly efficient for removal of gaseous contaminants, packed towers may create liquid disposal problems, may become easily clogged when gases with high particulate loads are introduced, and have relatively high maintenance costs. Figure 8.2 Typical countercurrent-flow packed tower. (From USEPA, Control Techniques for Gases and Par- ticulates, 1971.) Clean gas out Liquid in Packing material Dirty gas in Liquid and pollutant out L1681_book.fm Page 165 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 166 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 8.2.1 Solubility Solubility is a function of system temperature and, to a lesser extent, system pressure. As temper- ature increases, the amount of gas that can be absorbed by a liquid decreases (gases are more soluble in cold liquids than in hot liquids). Gas phase pressure can also influence solubility; by increasing the pressure of a system, the amount of gas absorbed generally increases. However, this is not a major variable in absorbers used for air pollution control because they operate at close to atmospheric pressure (USEPA-81/12, 1981). 8.2.2 Equilibrium Solubility and Henry’s Law Under certain conditions, Henry’s law can express the relationship between the gas phase concen- tration and the liquid phase concentration of the contaminant at equilibrium. This law states that for dilute solutions in which the components do not interact, the resulting partial pressure ( p ) of a component A in equilibrium with other components in a solution can be expressed as (8.1 ) where p = partial pressure of contaminant in gas phase at equilibrium H = Henry’s law constant x A = mole fraction of contaminant or concentration of A in liquid phase Figure 8.3 Various packing used in packed tower scrubbers. (Adapted from Air Pollution Control Equipment, Part II, American Industrial Hygiene Association, 1968.) Raschig Ring — most popular type Berl saddle — efficient but costly Pall rings — good liquid distribution Tellerette — very low unit weight Intalox saddle — efficient but expensive pHx A = L1681_book.fm Page 166 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC GASEOUS EMISSION CONTROL 167 Equation 8.1 is the equation of a straight line where the slope ( m ) is equal to H . Henry’s law can be used to predict solubility only when the equilibrium line is straight — when the solute concentrations are very dilute. In air pollution control applications this is usually the case. For example, an exhaust stream that contains a 1000-ppm SO 2 concentration corresponds to a mole fraction of SO 2 in the gas phase of only 0.001. Another restriction on using Henry’s law is that it does not hold true for gases that react or dissociate upon dissolution. If this happens, the gas no longer exists as a simple molecule. For example, scrubbing HF or HCl gases with water causes both compounds to dissociate in solution. In these cases, the equilibrium lines are curved rather than straight. Data on systems that exhibit curved equilibrium lines must be obtained from experiments. The units of Henry’s law constants are atmosphere per mole fractions. The smaller the Henry’s law constant is, the more soluble the gaseous compound is in the liquid. The following example from USEPA-81/12 (1981) illustrates how to develop an equilibrium diagram from solubility data. Example 8.1 Problem : Given the following data for the solubility of SO 2 in pure water at 303 K (30°C) and 101.3 kPa (760 mm Hg), plot the equilibrium diagram and determine if Henry’s law applies. Solution : Step 1. The data must first be converted to mole fraction units. The mole fraction in the gas phase y is obtained by dividing the partial pressure of SO 2 by the total pressure of the system. For the first entry of the data table: The mole fraction in the liquid phase x is obtained by dividing the moles of SO 2 by the total moles of liquid: For the first entry ( x ) of the data table: Solubility of SO 2 in Pure Water Equilibrium data Concentration SO 2 (g of SO 2 per 100 g/H 2 O) p (Partial pressure of SO 2 ) 0.5 6 kPa (42 mm Hg) 1.0 11.6 kPa (85 mm Hg) 1.5 18.3 kPa (129 mm Hg) 2.0 24.3 kPa (176 mm Hg) 2.5 30.0 kPa (224 mm Hg) 3.0 36.4 kPa (273 mm Hg) Yp/P 6kPa/101.3 kPa 0.9592 (0.06)== = x = moles SO in solution moles SO in solutio 2 2 nnmolesHO 2 + moles of H O (100 g H O 100 g H 0/18 g 222 == HHOper mole) 5.55 2 = x 0.0078/(0.0078 5.55)=+ = 0.0014 L1681_book.fm Page 167 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 168 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Step 2. The following table is completed. The data from this table are plotted in Figure 8.4. Henry’s law applies in the given concentration range, with Henry’s law constant equal to 42.7 mole fraction SO 2 in air per mole fraction SO 2 in water. 8.2.3 Material (Mass) Balance The simplest way to express the fundamental engineering concept or principle of material or mass balance is to say, “Everything has to go somewhere.” More precisely, the law of conservation of mass says that when chemical reactions take place, matter is neither created nor destroyed. This important concept allows us to track materials — in this case, pollutants, microorganisms, chem- icals, and other materials — from one place to another. The concept of materials balance plays an important role in environmental treatment technol- ogies, where we assume a balance exists between the material entering and leaving the treatment process: “What comes in must equal what goes out.” The concept is very helpful in evaluating process operations. In air pollution control of gas emissions using a typical countercurrent flow absorber, the solute (contaminant compound) is the material balance. Figure 8.5 illustrates a typical countercurrent flow absorber in which a material balance is drawn (USEPA-81/12, p. 4-14). The following equation can be derived for material balance: (8.2) Figure 8.4 SO 2 absorption data for Example 8.1. (Adapted from USEPA-81/12, p. 4-8, 1981.) Solubility data for SO 2 C = (g of SO 2 )/(100 g H 2 O) p (kPa) y = p /101.3 x = (C/64)/(C/64 + 5.55) 0.5 6 0.06 0.0014 1.0 11.6 0.115 0.0028 1.5 18.3 0.18 0.0042 2.0 24.3 0.239 0.0056 2.5 30 0.298 0.007 3.0 36.4 0.359 0.0084 50 y, mole fraction of SO 2 in water 30 10 0.004 0.008 0.012 x, mole fraction of SO 2 absorption data Y–Y (L /G)(X – X ) 12 mm1 2 = L1681_book.fm Page 168 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC GASEOUS EMISSION CONTROL 169 where Y 1 = inlet solute concentration Y 2 = outlet solute concentration X 1 = inlet composition of scrubbing liquid X 2 = outlet composition of scrubbing liquid L m = liquid flow rate, gram-moles per hour G m = gas flow rate, gram-moles per hour Equation 8.2 is the equation of a straight line. When this line is plotted on an equilibrium diagram, it is referred to as an operating line (see Figure 8.5 ). This line defines operating conditions within the absorber, that is, what is going in and what is coming out. The slope of the operating line is the liquid mass flow rate divided by the gas mass flow rate, which is the liquid-to-gas ratio or ( L m / G m ). When absorption systems are described or compared, the liquid-to-gas ratio is used extensively. The following example (using Henry’s law) illustrates how to compute the minimum liquid rate required to achieve desired removal efficiency. Example 8.2 Problem : Using the data and results from Example 8.1, compute the minimum liquid rate of pure water required to remove 90% of the SO 2 from a gas stream of 84.9 m 3 /min (3000 actual cubic feet per minute [acfm] ) containing 3% SO 2 by volume. The temperature is 293 K and the pressure is 101.3 kPa (USEPA-81/12, p. 4-20). Figure 8.5 Operating line for a countercurrent flow absorber. (From USEPA-81/12, p. 4-17.) Y X Slope of operating line = L m G m X 2 Y 2 X 1 Y 1 G m2 L m1 L m2 G m1 Equilibrium curve Operating line Driving forces L1681_book.fm Page 169 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 170 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Given: Inlet gas solute concentration ( Y 1 ) = 0.03 Minimum acceptable standards (outlet solute concentration) ( Y 2 ) = 0.003 Composition of the liquid into the absorber ( X 2 ) = 0 Gas flow rate ( Q ) = 84.9 m 3 /min Outlet liquid concentration ( X 1 ) = ? Liquid flow rate ( L ) = ? H = Henry’s Constant Solution : Step 1. Sketch and label a drawing of the system (see Figure 8.6). Step 2. At the minimum liquid rate, Y 1 and X 1 will be in equilibrium. The liquid will be saturated with SO 2 . From Figure 8.4, Figure 8.6 Material balance for absorber. (From USEPA-81/12, p. 4-20.) Q = 84.9 m 3 /min Y 1 = 0.03 Y 2 = 0.003 X 1 = ? L = ? X 2 = 0 Y3%byvolume 0.03 1 == Y 90% reduction from Y or only 10% of Y 21 = 112 ;therefore Y (0.10)(0.03) 0.003== YHX 11 = L1681_book.fm Page 170 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC GASEOUS EMISSION CONTROL 171 Step 3. The minimum liquid-to-gas ratio is Step 4. Compute the minimum required liquid flow rate. First, convert cubic meters of air to gram-moles: In mass units: Step 5. Figure 8.7 illustrates the graphical solution to this problem. The slope of the minimum operating line × 1.5 = the slope of the actual operating line (line AC): H = 42.7 (mole fraction SO in air)/(mole f 2 rraction SO in water) 2 0.03 42.7= X 1 X 1 = 0.000703 mole fraction YY(L/G)(X X ) mm12 1 2 − = − (L / G ) (Y Y ) / (X X ) mm = −− 12 1 2 = −−(. . )/(. )0030003 0 000703 0 = 38 4.(gmolwater)/(g mol air) At 0C(273 K) and 101.3 kPa, there are 0.0° 2224 m /g mole of an ideal gas 3 At 20 C (293 K): 0.0224(293 K/273 K) 0.02°=44m/gmol 3 G 84.9 (m /min)(g-mol air/0.024 m ) m 33 = = 3538 g-mol air/min L/G 38.4 (g-mol water/g-mol of air) at mm = mminimum conditions L 38.4(3,538) m = = 136.0 kg-mol water min L 136,000 g-mol/min (18 kg/kg-mol)= = 2448 kg/min 38.4 1.5 57.6×= L1681_book.fm Page 171 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 172 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 8.2.4 Sizing Packed Column Diameter and Height of an Absorber 8.2.4.1 Packed Tower Absorber Diameter The main parameter that affects the size of a packed tower is the gas velocity at which liquid droplets become entrained in the existing gas stream. Consider a packed tower operating at set gas and liquid flow rates. By decreasing the diameter of the column, the gas flow rate (meters per second or feet per second) through the column will increase. If the gas flow rate through the tower is gradually increased by using increasingly smaller diameter towers, a point is reached at which the liquid flowing down over the packing begins to be held in the void spaces between the packing. This gas-to-liquid flow ratio is termed the loading point . The pressure drop over the column begins to build, and the degree of mixing between the phases decreases. A further increase in gas velocity causes the liquid to fill the void spaces in the packing completely. The liquid forms a layer over the top of the packing, and no more liquid can flow down through the tower. The pressure drop increases substantially, and mixing between the phases is minimal. This condition is referred to as flooding, and the gas velocity at which it occurs is the flooding velocity. Using an extremely large diameter tower eliminates this problem. However, as the diameter increases, the cost of the tower increases (USEPA-81/12, p. 4-22). Normal practice is to size a packed column diameter to operate at a certain percent of the flooding velocity. A typical operating range for the gas velocity through the towers is 50 to 75% of the flooding velocity, assuming that by operating in this range, the gas velocity will also be below the loading point. A common and relatively simple procedure to estimate the flooding velocity (and thus minimum column diameter) is to use a generalized flooding and pressure drop correlation. One version of the flooding and pressure drop relationship in a packed tower is shown in Figure 8.8. This correlation is based on the physical properties of the gas and liquid streams and tower packing characteristics. We summarize the procedure to determine the tower diameter in the following set of calculations: Step 1. Calculate the value of the horizontal axis (abscissa) of Figure 8.8 using Equation 8.3: (8.3) Figure 8.7 Solution to Example 8.2. (From USEPA-81/12, p. 4-21.) 0.03 0.02 0.01 0 0.0002 0.0004 0.0006 0.0008 C B 38.4 Operating Line 57.6 A Minimum operating line Abscissa (L/G)(P /P ) g1 0.5 = L1681_book.fm Page 172 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC [...]... equilibrium relationship 8. 3.4 Factors Affecting Adsorption According to USEPA -8 1 /12, p 5-1 8, a number of factors or system variables influence the performance of an adsorption system These include: • Temperature • Pressure • Gas velocity © 2005 by CRC Press LLC L1 681 _book.fm Page 188 Tuesday, October 5, 2004 10:51 AM 188 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 120 100 80 60 40 20 0 50 100 150... B (Note: Lines AB–BC are one theoretical plate Need a total of 2.3 plates) X Figure 8. 11 © 2005 by CRC Press LLC Graphical determination of the number of theoretical plates (From USEPA -8 1 /12, p 4-3 5.) L1 681 _book.fm Page 182 Tuesday, October 5, 2004 10:51 AM 182 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Equation 8. 14 is a simplified method of estimating the number of plates It can only be used if... 8. 7 Example 8. 7 Problem: A dry cleaning process exhausts a 15,000-cfm air stream containing 680 -ppm carbon tetrachloride Given Figure 8. 13 and assuming that the exhaust stream is at approximately 140°F and 14.7 pounds per square inch absolute (psia), determine the saturation capacity of the carbon (USEPA81/12, p 5 -8 ) © 2005 by CRC Press LLC L1 681 _book.fm Page 186 Tuesday, October 5, 2004 10:51 AM ENVIRONMENTAL. .. viscosity of liquid (for water = 0 .8 cP = 0.00 08 Pa-sec (use pascal-seconds in this equation) Step 4 Calculate the actual gas flow rate per unit area as a fraction of the gas flow rate at flooding (Equation 8. 5) © 2005 by CRC Press LLC L1 681 _book.fm Page 174 Tuesday, October 5, 2004 10:51 AM 174 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK G*operating = ƒG*flooding (8. 5) Step 5 Cross-sectional area of tower A... 28 2 1.4 1.2 1.1 1.0 0.95 0.9 0 .85 24 0 .8 20 0.75 0.7 16 NOG 0.65 0.6 0.55 12 0.5 0.4 0.3 0.2 0.1 8 4 0 1 2 5 10 100 1000 (Y1 − mX2)/(Y2 − mX2) Figure 8. 9 m X2 Y2 Gm Lm = = = = = Colburn diagram (From EPA -8 4 /03, p.104; USEPA -8 1 /12, p 4-3 0.) slope of the equilibrium line mole fraction of solute entering the absorber in the liquid mole fraction of solute in exiting gas molar flow rate of gas, kilogram-moles... cases, Equation 8. 11 reduces to Equation 8. 12: NOG = In(Y1 /Y2 ) (8. 12) The number of transfer units depends only on the inlet and outlet concentration of the solute (contaminant or pollutant) For example, if the conditions in Equation 8. 12 are met, 2.3 transfer © 2005 by CRC Press LLC L1 681 _book.fm Page 1 78 Tuesday, October 5, 2004 10:51 AM 1 78 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK units are... ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Capacity weight, % (1 lb CCl4/100 lb C) 186 100 32°F 77°F 140°F 10 212°F 300°F 3.0 2.0 1.5 1.0 Figure 8. 13 0.001 0.01 Partial pressure, psia 0.1 1.0 Adsorption isotherm for carbon tetrachloride on activated carbon (Adapted from USEPA -8 1 /12, p 5 -8 .) Solution: Step 1 In the gas phase, the mole fraction Y is equal to the percent by volume: Y = % volume = 680 ppm = 680 /(10)6... October 5, 2004 10:51 AM 180 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Tray spacing, inches 15 18 21 24 27 30 1.5 Correction factor for dt 1.4 1.3 1.2 1.1 1.05 1.0 0.9 0.4 0.53 0.6 0 .8 Tray spacing, meters Figure 8. 10 Tray spacing correction factor (Adapted from USEPA -8 1 /12, p 4-3 3.) Example 8. 5 Plate Tower Diameter Problem: For the conditions described in Example 8. 2, determine the minimum acceptable... carbon per 4-h cycle per adsorber © 2005 by CRC Press LLC L1 681 _book.fm Page 190 Tuesday, October 5, 2004 10:51 AM 190 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Carbon capacity (weight %) 100 77°F 212°F 10 1 0.0001 0.001 0.01 0.1 1 Partial pressure, psia Figure 8. 16 Adsorption isotherm for toluene (Adapted from USEPA -8 1 /05, p 1 8- 9 .) Example 8. 9 Note: The following example is based on a number of adsorber... handle 1 .89 m3/sec (4000 acfm) The area required for a limiting velocity of 0.5 08 m/sec is: Area to handle half flow A = (1 .89 m 3 /sec)/(0.5 08 m/sec) © 2005 by CRC Press LLC L1 681 _book.fm Page 192 Tuesday, October 5, 2004 10:51 AM 192 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK = 3.72 m 2 This cross-sectional area corresponds to a vessel diameter of: d = (4A/π)0.5 = [4(3.72)/π)0.5 = 2. 18 m (7 ft) . ) m 33 = = 35 38 g-mol air/min L/G 38. 4 (g-mol water/g-mol of air) at mm = mminimum conditions L 38. 4(3,5 38) m = = 136.0 kg-mol water min L 136,000 g-mol/min ( 18 kg/kg-mol)= = 24 48 kg/min 38. 4 1.5 57.6×= . (abscissa) of Figure 8. 8 using Equation 8. 3: (8. 3) Figure 8. 7 Solution to Example 8. 2. (From USEPA -8 1 /12, p. 4-2 1.) 0.03 0.02 0.01 0 0.0002 0.0004 0.0006 0.00 08 C B 38. 4 Operating Line 57.6 A Minimum. in Equation 8. 12 are met, 2.3 transfer Figure 8. 9 Colburn diagram. (From EPA -8 4 /03, p.104; USEPA -8 1 /12, p. 4-3 0.) N OG 28 1000 24 20 16 12 8 4 0 0.95 0.92 1.4 1.2 1.1 1.0 0 .85 0 .8 0.75 0.7 0.65 0.6 0.55 0.5 0.4 0.3 0.2 0.1 12510