Standard Methods for Examination of Water & Wastewater_8 docx

TX249_frame_C09.fm Page 419 Friday, June 14, 2002 4:37 PM Aeration, Absorption, and Stripping Aeration, absorption, and stripping are unit operations that rely on flow of masses between phases When a difference in concentration exists between two points in a body of mass, a flow of mass occurs between the points When the flow occurs between two phases of masses, a transfer of mass between the phases is said to occur This transfer of mass between phases is called mass transfer Examples of unit operations that embody the concept of mass transfer are distillation, absorption, dehumidification, liquid extraction, leaching, and crystallization Distillation is a unit operation that separates by vaporization liquid mixtures of miscible and volatile substances into individual components or groups of components The separation of water and alcohol into the respective components of liquid air into nitrogen, oxygen, and argon; and the separation of crude petroleum into gasoline, oil, and kerosene are examples of the distillation unit operation Absorption is a unit operation that removes a solute mass or masses from a gas phase into a liquid phase Aeration of water dissolves air into it; thus, aeration is absorption Another example of absorption is the “washing” of ammonia from an ammonia-polluted air In this operation, ammonia is removed from the air by its dissolution into the water The reverse flow of masses from the liquid phase into the gas phase is called stripping In stripping, the solute molecule is removed from its solution with the liquid into the gas phase Dehumidification is the removal of a solute liquid vapor from a gas phase by the solute condensing into its liquid phase The removal of water vapor in air by condensation on a cold surface is dehumidification The reverse of dehumidification is humidification In this unit operation, the flow of the solute is from the liquid phase evaporating into the gas phase The end result of this movement is saturation of the gas For example, during heavy rains, the atmosphere may become saturated with water vapor, the degree of this saturation being measured by the relative humidity Liquid extraction is the removal of a solute component from a liquid mixture called the raffinate using a liquid solvent In this operation, the solvent preferentially dissolves the solute molecule to be extracted Leaching is a unit operation where a solute molecule is removed from a solid using a fluid extractor This is similar to liquid extraction, except that the solute to be removed comes from a solid rather than from a liquid as in the case of liquid extraction Also, the fluid extractor may be a fluid or a gas For example, pollutants can be leached out from solid wastes in a landfill as rain percolates down the heap Crystallization is a unit operation where solute mass flows toward a point of concentration forming crystals The driving force for the transfer of mass from liquid © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 420 Friday, June 14, 2002 4:37 PM 420 into the solid phase is the affinity of the solute to form into a solid An example of this operation is the making of ice from liquid water The formation of snow from water vapor in the atmosphere is also a process of crystallization Of all the unit operations of mass transfer, this chapter will only discuss aeration, absorption, and stripping As mentioned, aeration is a form of absorption Because this operation plays a very important and significant role in water and wastewater treatment, however, we will give it a separate heading and call it specifically aeration 9.1 MASS TRANSFER UNITS The major purpose of dissolving air is to provide oxygen to be used by microorganism in the process of wastewater treatment This is exemplified by the aeration employed in the activated sludge process Aeration may also be employed for the removal of iron and manganese from groundwaters In the removal of hardness, the presence of high concentrations of carbon dioxide may result in high cost for lime, as CO2 reacts with lime Thus, excess concentrations of this gas may be removed from the water by stripping or spraying the water into the air H2S is another compound that may be removed by stripping as benzene, carbon tetrachloride, p-dicholorobenzene, vinyl chloride, and trichloroethylene may also be removed by stripping The discussions that follow address the units or method used in aeration, absorption, and stripping Figure 9.1 illustrates how a pollutant may be stripped by spraying the water into the air As the water is sprayed, droplets are formed This creates the condition for the pollutant to transfer from the droplet phase to the air phase, in addition to the direct liberation of the pollutant as the bulk mass of water breaks up into the smaller size droplets Figures 9.2a through 9.2d show the various types of nozzles that may be used in sprays Figure 9.2e is an inclined apron which may be studded with riffle plates At the air–water interface at the surface of the flowing water, the air transfers between the water and air phases The studding creates turbulence which, in aeration, transports the water exposed at the surface to the main body or bulk of the flowing water The whole mass of water is aerated this way, because the mass of water transported to the main body carries with it any air that was dissolved when it was exposed at the surface In stripping, the turbulent flowing water exposes the solute at the surface triggering the process of stripping The rate of aeration or stripping depends upon how fast the surface is renewed (i.e., how fast the water mass from the main body is transported to the surface for exposure to the air) Figure 9.2f is a stack of perforated plates The water is introduced at the top and allowed to trickle down the plates; the trickling water is met by a countercurrent flow of air The process creates a droplet phase and the air-gas phase inducing a mass transfer between the droplets and the air Figure 9.2g is a spray tower The water is sprayed using spray nozzles at the top of the tower forming droplets These droplets are then met by a countercurrent flow of air creating the two phase for mass transfer as in the case of the perforated plates Figure 9.2h is a cascade aerator or deaerator as the case may be The cascade operates on the same principle as the inclined apron, only that this is more effective because of the steps © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 421 Friday, June 14, 2002 4:37 PM 421 Nozzle FIGURE 9.1 Water spray Figures 9.3 and 9.4 show various types of aeration devices used in wastewater treatment plants Figure 9.3a is a turbine aerator with an air sparger at the bottom As the air emerges from the sparger, the larger bubbles that are formed are sheared into small pieces by the turbine blade above Figure 9.3b is a porous ceramic diffuser Because of the small openings through which the air passes, this type of diffuser creates tiny bubbles Tiny bubbles are more effective for mass transfer, since the many bubbles produced create a large sum total areas for transfer Figure 9.3c is a surface aerator Water is drawn from the bottom of the aerator and sprayed into the air creating droplets, thus, aerating the water Figure 9.4 shows a dome-type bubble diffuser The dome is porous, can have a diameter of 18 cm, and may be constructed of © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 422 Friday, June 14, 2002 4:37 PM 422 Perforated pipe discharging water downward Inlet conduit Central cone Outlet conduit Palm Beach nozzle (a) Perforated pipe discharging air or gas upward (e) Collecting pan and outlet Sacramento nozzle (c) (g) Perforated inlet pipe Removable mouth piece Inlet conduit Outlet conduit Vanes Outlet Berlin nozzle New York nozzle Collecting pan (b) (d) (f) (h) FIGURE 9.2 Nozzles for sprays and units for aeration or stripping: (a–d) nozzle types; (e) inclined apron that may be studded with riffle plates; (f) perforated plates; (g) spray tower; and (h) cascade Porous ceramic diffuser units (tubes) Liquid level Rotor Air bubbles Compressed air Air inlet Bottom (a) (b) Manifold (c) FIGURE 9.3 Aeration units: (a) turbine aerator with an air sparger; (b) porous ceramic diffuser; and (c) surface aerator © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 423 Friday, June 14, 2002 4:37 PM 423 FIGURE 9.4 Dome-type diffusers (Courtesy of Aerocor Co.) FIGURE 9.5 An activated sludge aeration tank at Back River wastewater treatment plant, Baltimore, MD aluminum oxide This diffuser produces very tiny bubbles As indicated, the diffusers are mounted on rows of pipes Figure 9.5 shows an actual aeration in action This happens to be one of the activated sludge process tanks at the Back River Wastewater Treatment Plant, Baltimore, MD © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 424 Friday, June 14, 2002 4:37 PM 424 A common device used in gas absorption and stripping is the packed tower, the elevational section of which is shown in Figure 9.6i The device consists of a column or tower equipped with a gas inlet and distributor at the bottom and a liquid inlet and distributor at the top It also consists of a liquid outlet at the bottom and a gas outlet at the top and a supported mass of solid shapes called tower packing or filling The liquid trickles down through the packing while the gas goes up the packing The packing causes a thin film of liquid to be created on the surfaces which are contacted by the gases flowing by Two phases and an interface between liquid and gas are therefore created inducing mass transfer Figures 9.6a to Figure 9.6h show the various shapes of packings used in practice Packings are either dumped randomly into the tower or are stacked manually Dumped packings consists of units that are 0.6 cm to cm in major dimensions; they are mostly utilized in small towers Stacked packings are cm to 20 cm in major dimensions and are used in large towers The spiral partition rings single, double, and triple are stacked The Berl and Intalox saddles, the Raschig, Lessing, and crosspartition rings are normally dumped packings Large Raschig rings to cm in diameter are often stacked 9.2 INTERFACE FOR MASS TRANSFER, AND GAS AND LIQUID BOUNDARY LAYERS Figure 9.7a shows the formation of boundary layers in absorption operations, and Figure 9.7b shows the formation of boundary layers in stripping operations The righthand side in each of these figures represents the liquid phase as in the liquid phase of a droplet and the left-hand side represents the gas phase as in the gas phase of the air Consider the absorption operation Imagine the two phases being far apart initially As the phases approach each other, a point of “touching” will eventually be reached This point then determines a surface; being a surface, its thickness is equal to zero This surface is identified as the interface in the figure This figure shows the section cut across of the interface surface The line representing the interface must have a zero thickness From fluid mechanics, when a fluid flows parallel to a plate, a boundary layer is formed closed to the plate surface At the surface itself, the velocity is zero relative to the plate, because of the no-slip condition Considering the two phases mentioned previously, either the liquid or the gas may be considered as analogous to the plate Taking the liquid phase as analogous to the plate, the gas phase would be the fluid The interface would then represent the surface of the plate Because of the no-slip condition, the relative velocity of the phases parallel to the interface at the interface is equal to zero In other words, the two phases are “glued” together at the interface and they not move relative to each other As in the case of the fluid flow over a plate, a boundary layer is also formed With the liquid phase considered as analogous to the plate, a gas phase boundary layer is formed; with the gas phase considered as analogous to the plate, the liquid phase boundary layer is formed (see figures) These boundary layers are also called films © 2003 by A P Sincero and G A Sincero Liquid inlet Liquid distributor Packed section (a) (b) (c) (d) Packed section (e) (f) (g) (h) Gas inlet Liquid outlet (i) © 2003 by A P Sincero and G A Sincero 425 FIGURE 9.6 Various types of packings (left) and an elevational cross section of a packing tower (right): (a) cross-partition ring; (b) single-spiral ring; (c) double-spiral ring; (d) triple-spiral ring; (e) Berl saddles; (f) Intalox saddles; (g) Raschig ring; (h) Lessing ring TX249_frame_C09.fm Page 425 Friday, June 14, 2002 4:37 PM Aeration, Absorption, and Stripping Gas outlet TX249_frame_C09.fm Page 426 Friday, June 14, 2002 4:37 PM 426 i i Gas-phase boundary layer Liquidphase boundary layer (a) [x] [x ] i [y ] i Gas-phase boundary layer Buld liquid-phase concentration [x ] [y ] Bulk gas-phase concentration [y] Interface Buld liquid-phase concentration Bulk gas-phase concentration Interface Liquidphase boundary layer (b) FIGURE 9.7 Formation of interfacial boundary layers: (a) absorption; (b) stripping Thus, the transfer of mass between the two phases must pass through the gas and liquid films In environmental engineering literature, the term film is normally used The same discussions would apply to the stripping operation represented by Figure 9.7b, with the difference that the direction of flow of mass transfer is from the liquid phase to the gas phase In absorption operations, the concentration in the gas phase is larger in comparison to the concentration in the liquid phase Thus, the flow of mass transfer is from gas to liquid The reverse is true in the case of stripping, and the direction of mass transfer is from liquid to gas In other words, the liquid phase is said to be “stripped” of its solute component, decreasing the concentration of the solute in the liquid phase and increasing the concentration of the solute in the gas phase In absorption, the solute is absorbed from the gas into the liquid, increasing the concentration of the solute in the liquid phase and, of course, decreasing the concentration of the solute in the gas phase 9.3 MATHEMATICS OF MASS TRANSFER Between liquid and gas phases, the transfer of mass from one phase to the other must pass through the interfacial boundary surface Call the concentration of the solute at this surface as [ yi] referred to the gas phase The corresponding concentration referred to the liquid phase is [xi] [xi] and [ yi] are the same concentration of © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 427 Friday, June 14, 2002 4:37 PM 427 the solute only that they are referred to different basis; in effect, they are equal Because they are equal and because the thickness of the interface is zero, xi and yi must be in equilibrium with respect to each other Consider the process of absorption If [ y] is the concentration in the bulk gas phase, the driving force toward the interfacial boundary is [ y] − [ yi] and the rate of mass transfer is ky([ y] − [ yi]), where ky is the gas film coefficient of mass transfer For this rate of mass transfer to exist, it must be balanced by an equal rate of mass transfer at the liquid film The liquid phase mass transfer rate is kx([xi] − [x]), where kx is the liquid film coefficient of mass transfer and [x] is the bulk concentration of the solute in the liquid phase Thus, ky([ y] − [ yi]) = kx([xi] − [x]) (9.1) What is known is that [xi] and [ yi] are in equilibrium But, determining these values experimentally would be very difficult Thus, instead of using them, use [ x * ] and [ y * ], respectively [ x * ] is the concentration that [x] would attain if it were to reach equilibrium value By parallel deduction, [ y * ] is also the concentration that [ y] would attain if it were to reach equilibrium value The corresponding driving forces are now [ y] − [ y * ] and [ x * ] − [x], respectively To comprehend the physical meaning of the driving forces, refer to Figure 9.8 As shown, [xi], [ yi] is on the equilibrium curve The equilibrium curve is the relationship between the concentration [x] in the liquid phase and the concentration [ y] in the gas phase when there is no net mass transfer between the phases For any given pair of values [x] and [ y] in the liquid and gas phase, respectively, the point ([xi], [ yi]) represents the “distance” that ([x], [ y]) will have to “move” to attain their equilibrium values concurrently at the equilibrium curve This distance, represented by the line segment ([x], [ y] → [xi], [ yi]), is the actual driving force for mass transfer; Operating line [xi ], [y] [y] [x* ], [y] Equilibrium curve [xi ], [yi ] [x], [y* ] [x] FIGURE 9.8 Relationship among the various mole fractions © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 428 Friday, June 14, 2002 4:37 PM 428 however, as mentioned before, ([xi], [ yi]) is impossible to determine experimentally Thus, locate the point ([x], [ y * ]) to view the surrogate driving force in the gas film and locate point ([ x * ], [ y]) to view the surrogate driving force in the liquid film As seen, [ y] − [ y * ] is greater than [y] − [yi]; however, it is not the true driving force for transfer Also [ x * ] − [x] is greater than [xi] − [x], but, again, it is not the true driving force for transfer When the transfer equation is written, however, it is prefixed with a proportionality constant This situation is therefore taken advantage of by using a different proportionality constant for the case of the surrogate driving forces Thus, using Ky as the proportionality constant for the gas-side mass transfer equation in the surrogate situation, k y ( y – yi ) = K y ( y – y * ) (9.2) On the liquid side, using Kx as the proportionality constant, k x ( [ xi ] – [ x ] ) = K x ( [ x* ] – [ x ] ) (9.3) Ky and Kx are called overall mass transfer coefficients for the gas and liquid sides, respectively To differentiate, ky and kx are called individual mass transfer coefficients for the respective sides It is instructive to determine the equation relating the overall and the individual mass transfer coefficients Equation (9.2) may be rearranged to obtain ( [ y ] – [ y i ] ) + ( [ y i ] – [ y * ]) [ yi ] – [ y * ] [ y ] – [ y* ] - = - = = - + Ky k y ( [ y ] – [ yi ] ) k y ( [ y ] – [ yi ] ) k y k y ( [ y ] – [ yi ] ) (9.4) Replacing ky([y] − [yi]) using Equation (9.1), [ yi ] – [ y * ] 1 - = - + Ky k y k x ( [ xi ] – [ x ] ) (9.5) Letting ([yi] − [ y * ])/([xi] − [x]) equal m, 1 m - = - + Ky ky kx (9.6) 1 1 - = - + = -k x mk y mK y Kx (9.7) A parallel derivation for Kx yields The coordinates [xi] and [yi] are the coordinates of the point ([xi], [yi]) on the equilibrium curve On the other hand, the coordinates [x] and [y] are coordinates of © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 450 Friday, June 14, 2002 4:37 PM 1 1 G  –  = ( 1.5 ) ( 28 ) ( 0.39 )  – -   – [ y f ] – 0.03  – [ x f ] – 0.001 0.39y f x f = 16.38 – 15.99y f From yi = 30xi, y * = 30xf f yf xf y f* 1/(yf − y f* ) 1/(1 − yf) ∆yf ∆yf /(1 − yf)(yf − yf) 0.0031 0.0050 0.0080 0.0110 0.0130 0.0160 0.0190 0.0210 0.0240 0.0270 0.03 0.0000 0.00012 0.00019 0.00026 0.00031 0.00038 0.00044 0.00049 0.00056 0.00063 0.00060 0.0000 0.0036 0.0057 0.0078 0.0092 0.011 0.013 0.0147 0.0168 0.0188 0.021 — 714.3 429.19 309.9 261 210.7 172 159 138 122 109 — 1.0050 1.0081 1.0111 1.0132 1.0163 1.0194 1.0215 1.0245 1.0277 1.0309 — 0.002 0.003 0.003 0.002 0.003 0.003 0.002 0.003 0.003 0.003 — 1.44 1.38 0.94 0.53 0.64 0.53 0.32 0.28 0.38 0.34 ∑ = 6.78 ZT = 1.5(6.78) = 10.17 m Ans 9.6.4 AMMONIA STRIPPING (OR ABSORPTION) The reaction of NH3 with water can be represented by NH + H O + NH + OH − (9.55) − From this reaction, raising the pH (as represented by the OH ) will drive the reaction to the left, increasing the concentration of NH3 This makes ammonia more easily removed by stripping In practice, the pH is raised to about 10 to 11 using lime Stripping is done by introducing the wastewater at the top of the column and allowing it to flow down countercurrent to the flow of air introduced at the bottom The stripping medium inside the column may be composed of packings or fillings, such as Raschig rings and Berl saddles, or sieve trays and bubble caps The liquid flows in thin sheets around the medium, thereby, allowing more intimate contact between liquid and the stripping air Let G be the moles per unit time of ammonia-free air, L be the moles per unit time of ammonia-free water, [Y1] be the moles of ammonia per mole of ammoniafree air at the bottom, [Y2] be the moles of ammonia per mole of ammonia-free air at the top, [X1] be the moles of ammonia per mole ammonia-free water at the bottom, and [X2] be the moles of ammonia per mole of ammonia-free water at the top At any section between the bottom and the top of the column, the [X]s and [y]s are simply [X] and [Y ] By mass balance, the ammonia stripped from the wastewater is © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 451 Friday, June 14, 2002 4:37 PM equal to the ammonia absorbed by the air Thus, G([Y 2] – [Y 1]) = L([ X 2] – [ X 1]) (9.56) * Let the equilibrium concentrations corresponding to [X] and [Y ] be [X ] and [ Y ], respectively From the equilibrium between ammonia in water and ammonia * * in air, the following equations for the relationships between [X ] and [Y ] at various temperatures and one atmosphere of total pressure may be derived (Metcalf and Eddy, 1991): * at at at at at * 10°C:[Y ] * 20°C:[Y ] * 30°C:[Y ] * 40°C:[Y ] * 50°C:[Y ] = = = = = * 0.469[X ] * 0.781[X ] * 1.25[X ] * 2.059[X ] * 2.692[X ] * The coefficients of [X ] in these equations plot a straight line with the temperatures Letting these coefficients be K produces the regression equation, K = 0.0585T – 0.338 with T in °C K = 0.0585T k – 16.31 with T k in degrees Kelvin * * Therefore, in general, the relationship between [X ] and [ Y ] using the temperature in degrees Kelvin, is * [ Y ] = ( 0.0585T k – 16.31 ) [ X * ] (9.57) This is the equation of the equilibrium line for ammonia stripping or absorption It may be expressed in terms of the respective mole fractions [ y * ] and [ x * ] Performing f f the operation and solving for [ y * ] produces f [x*] f ( 0.0585T k – 16.31 ) -1 – [x*] f * ] = [y f *] [x f + ( 0.0585T k – 16.31 ) -1 – [x*] (9.58) f L, [X1], [X2], and [Y1] are known values Assuming the ammonia in the air leaving the top of the tower is in equilibrium with the ammonia content of the entering wastewater, [Y2] may be written as * [ Y ] = [ Y ] = ( 0.0585T k – 16.31 ) [ X ] (9.59) Thus, the ratio G/L, the moles of air needed for stripping per mole of water all in solute-free basis is ([ X2] – [ X1]) G - = -( 0.0585T k – 16.31 ) [ X ] – [ Y ] L © 2003 by A P Sincero and G A Sincero (9.60) TX249_frame_C09.fm Page 452 Friday, June 14, 2002 4:37 PM In terms of mole fractions, (1 – [ y f 1])[(1 – [ x f 1])[ x f 2] – (1 – [ x f 2])[ x f 1]] G - = - (9.61) ( – [ x f ] ) [ ( 0.0585T k – 16.31 ) ( – [ y f ] ) [ x f ] – ( – [ x f ] ) [ y f ] ] L G is the theoretical amount of air needed to strip ammonia in the wastewater from a concentration of [X2] down to a concentration of [X1] In practice, the minimum theoretical G is often multiplied by a factor of 1.5 to 2.0 Once G has been established, the cross-sectional area of the tower may be computed using the equation of continuity The superficial velocity through the tower should be less than the velocity that will cause flooding or boiling up of the incoming wastewater The method for estimating tower height was previously discussed Operating line for ammonia The equation G([Y ] − [Y1]) = L([X] − [X1]) is the equation for the operating line for ammonia Expressing in terms of the equivalent mole fractions [yf], [yf1], [xf], and [xf1], respectively, and solving the resulting expression for [yf], [y f ] G[ y f 1](1 – [ x f 1])(1 – [ x f ]) + L[[ x f ](1 – [ y f 1])(1 – [ x f 1]) – [ x f 1](1 – [ y f 1])(1 – [ x f ])] -(1 – [ y f 1])(1 – [ x f 1])(1 – [ x f ]) = G[ y f 1](1 – [ x f 1])(1 – [ x f ]) + L[[ x f ](1 – [ y f 1])(1 – [ x f 1]) – [ x f 1](1 – [ y f 1])(1 – [ x f ])] G + -(1 – [ y f 1])(1 – [ x f 1])(1 – [ x f ]) (9.62) Example 9.13 A wastewater containing 25 mg/L of NH3−N is to be stripped The temperature of operation is 15°C and the flow is 0.013 m /s Determine the amount of air needed assuming a multiplying factor of 1.5 for the minimum theoretical G Calculate the cross-sectional area of the tower Assume a superficial velocity of 0.30 m/s Solution: To get the minimum G, assume the operation at the top of the tower is at equilibrium Thus, ([ X2] – [ X1]) G - = -L ( 0.0585T k – 16.31 ) [ X ] – [ Y ] 1L of water = 1000 g = 1,000,000 mg Therefore, 25/14 25/N [ X ] Ӎ = ( 1,000,000 – 25 )/18 ( 1,000,000 – 25 )/H O mol ( NH −N ) –5 = 3.2 ( 10 ) -mol H O (ammonia free) [ Y ] = T k = 15 + 273 = 288°K –5 G 3.2 ( 10 ) – - = = 1.86 –5 L [ 0585 ( 288 ) – 16.31 ] ( 3.2 ) ( 10 ) – 1000 L Ӎ 0.013 m /s = 0.013  - = 0.722 kgmol/s  18  © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 453 Friday, June 14, 2002 4:37 PM Therefore, G = 1.5 ( 1.86 ) ( 0.722 ) = 2.01 kgmol/s = 1.74 ( 10 ) gmol/d 288 = 1.74 ( 10 ) ( 22.4 )    - = 4.11 ( 10 ) m /d  273  1000 Ans 4.11 ( 10 ) Overall cross-sectional area of towers = = 158.56 m 0.3 ( 60 ) ( 60 ) ( 24 ) 158.56 Using three towers, cross-sectional area = - = 52.85 m Ans GLOSSARY Absorption—A unit operation that removes a solute mass or masses from a gas phase into a liquid phase Actual oxygenation rate—The rate of transfer of oxygen to water at actual operating conditions Aeration—The unit operation of absorption where the gas involved is air Carrier gas—The gas phase that contains the solute Carrier liquid—The liquid phase that contains the solute Contact time—The time allowed for the process to take place Control volume—The volume in space that contains the materials addressed in any particular problem setting Crystallization—A unit operation where solute mass flows toward a point of concentration forming crystals Dehumidification—The removal of a solute liquid vapor from a gas phase by the solute condensing into its liquid phase Distillation—A unit operation that separates by vaporization liquid mixtures of miscible and volatile substances into individual components or groups of components Destination medium—The overall entity inside the control volume from, through, and into which the gas and liquid phases flow Equilibrium curve—The relationship between the concentration x in the liquid phase and the concentration y in the gas phase when there is no net mass transfer between phases Eulerian derivative—The total rate or rate of change of a quantity that reflects the effect of convection Humidification—The reverse of dehumidification where the solute flow from the liquid phase into the gas phase as in stripping, although in humidification nothing is stripped, because the liquid phase is composed of the same molecule as the solute molecule Hydraulic detention time—The average time that the particles of water in an inflow to a basin are retained in the basin before outflow Kjeldahl nitrogen—The sum of organic and ammonia nitrogens © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 454 Friday, June 14, 2002 4:37 PM Leaching—A unit operation where the solute molecule is removed from a solid using a fluid extractor Lagrangian derivative—The total rate of change of a quantity as if convection is absent in the process Liquid extraction—The removal of a solute component from a liquid mixture called the raffinate using a liquid solvent Mass velocity—Mass flow rate divided by cross-sectional area Mass transfer—The transfer of mass between phases Mean cell retention time—The average time that cells of organisms are retained in the basin Minimum flow rate—The liquid flow rate corresponding to the condition when the operating line and the equilibrium line intersect each other Molar mass velocity—Mass velocity expressed in terms of mole unit Operating line—The plot between the concentration in the liquid phase and the concentration in gas phase Packing or filling—A supported mass of solid shapes in towers and similar structures Standard oxygen rate, SOR—The rate of transfer of oxygen from air into tap water or distilled at a pressure of one atmosphere and a dissolved oxygen concentration of mg/L Stripping—The flow of masses from the liquid phase into the gas phase SYMBOLS a AOR B BOD BOD5 [C ] Interfacial contact area per unit bulk volume of tower Actual oxygenation rate Temperature lapse rate Biochemical oxygen demand Five-day biochemical oxygen demand Concentration of dissolved oxygen in water at any given moment at a corresponding temperature and pressure CBOD Carbonaceous oxygen demand [C e ] Concentration of dissolved oxygen in the effluent of the secondary sedimentation basin [C o ] Concentration of dissolved oxygen in Qo [Cos] Equilibrium concentration of oxygen in water at a particular temperature and pressure This equilibrium concentration is also called the saturation concentration of the dissolved oxygen (DO) [Cos,sp] [Cos] at standard pressure [Cos,w] [Cos] of the wastewater at field conditions [Cos,20,sp] Saturation DO at 20°C and standard pressure [C w ] Concentration of dissolved oxygen in the wasted sludge DO Saturation dissolved oxygen f Factor to convert BOD5 concentrations to equivalent CBOD concentrations in terms of S © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 455 Friday, June 14, 2002 4:37 PM fn fN fs g G G′ G1 H Hx Hy KLa (KLa)20 (KLa)w (K′La)w (KLa)w,20 kx Kx Kx a ky Ky Ky a L L′ m Nt Nty Ntx [N] [No] NBOD P Pb Ps Pbo Qo Qw r (r )n (r )s Factor for converting the nitrogen in the wasted sludge to oxygen equivalent Factor for converting nitrogen concentrations to oxygen equivalent Factor that converts the mass of microorganisms in the waste sludge to the equivalent oxygen concentration Acceleration due to gravity Mole flow rate of solute-free gas phase (carrier gas); the moles of ammonia per unit time of ammonia-free air Gas flow rate as a mixture of carrier gas and solute Peebles number Height of transfer unit Height of transfer unit based on the liquid side Height of transfer unit based on the gas side Dissolved oxygen overall mass transfer coefficient of tap or distilled water Dissolved oxygen overall mass transfer coefficient of tap or distilled water at standard conditions Dissolved oxygen overall mass transfer coefficient of wastewater at field conditions Apparent overall mass transfer coefficient of wastewater Dissolved oxygen overall mass transfer coefficient of wastewater at standard conditions Individual liquid film coefficient of mass transfer Overall mass transfer coefficient for the liquid side Overall mass transfer coefficient for the liquid side Individual gas film coefficient of mass transfer Overall mass transfer coefficient for the gas side Overall mass transfer coefficient for the gas side Mole flow rate of solute-free liquid phase (carrier liquid); the moles ammonia per unit time of ammonia-free water Liquid flow rate as mixture of carrier liquid and solute Slope of operating line if it were a straight line Number of transfer units Number of transfer units based on the gas side Number of transfer units based on the liquid side Nitrogen concentration in the effluent The nitrogen concentration in the influent Nitrogenous oxygen demand Pressurization pressure Barometric at a given elevation Standard pressure Barometric pressure at elevation z = Inflow to reactor Outflow of wasted sludge Respiration rate; average radius of bubbles Respiration due to NBOD Respiration due to CBOD © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 456 Friday, June 14, 2002 4:37 PM R Re (syn)s [S] [So] SOR t T To vb V VMy VMx [xi] [ x* ] [X] [X1] [X2] [ Xu ] [y] [yi] [ y *] [Y ] [Y1] [y2] Z ZT α β θ µ ρg ρl σ Gas constant for air ( = 286.9 N⋅m/kg⋅°Κ) Reynolds number Portion of S consumed for synthesis for growth or to replace dead cells Concentration of CBOD; superficial area Influent CBOD concentration Standard oxygen rate Time of contact for aeration or mass transfer Temperature in degree Celsius or Kelvin Temperature in °K at z = = ambient temperature corresponding Pb Bubble rise velocity Volume of the control volume or volume of reactor Gas side molar mass velocity Liquid side molar mass velocity Concentration of solute at the interface surface referred to the liquid phase The concentration that x would attain if it were to reach equilibrium value Mole of solute per unit mole of the solute-free liquid Moles of ammonia per mole ammonia-free water at the bottom Moles of ammonia per mole of ammonia-free water at the top Concentration of organisms in the waste sludge Concentration of solute in the bulk gas phase Concentration of solute at the interface surface referred to the gas phase The concentration that y would attain if it were to reach equilibrium value Mole units of solute per unit mole of solute-free gas Moles of ammonia per mole of ammonia-free air at the bottom Moles of ammonia per mole of ammonia-free air at the top Tower height Tower height (KLa)w,20 /(KLa)20 [Cos,w]/[Cos] Temperature correction factor Absolute viscosity of fluid Mass density of the gas phase (air) Mass density of fluid Surface tension of fluid PROBLEMS 9.1 9.2 The value of (KLa)20 for a certain industrial waste is 2.46 per hour and the value of (KLa)w at 25°C = 2.77 per hour What is the value of the temperature correction factor? The value of (KLa)w at 25°C = 2.77 per hour If the value of the temperature correction factor is 1.024, what is the value of (KLa)20? © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 457 Friday, June 14, 2002 4:37 PM 9.3 9.4 9.5 9.6 9.7 9.8 9.9 The value of (KLa)w = 2.77 per hour If the value of the temperature correction factor is 1.024 and the value of (KLa)20 = 2.46 per hour, at what temperature is the value of (KLa)w for? A chemical engineer proposes to purchase an aerator for an activated sludge reactor In order to so, he performs a series of experiments obtaining the following results: AOR = 1.30 kg/m ⋅ day, and β = 0.94 The aeration is to be maintained to effect a dissolved oxygen concentration of 1.0 mg/L at 25°C in the reactor The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day Determine the value of α A chemical engineer proposes to purchase an aerator for an activated sludge reactor In order to so, he performs a series of experiments obtaining the following results: α = 0.91, and β = 0.94 The aeration is to be maintained to effect a dissolved oxygen concentration of 1.0 mg/L at 25°C in the reactor If the SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day, what was the AOR? A chemical engineer proposes to purchase an aerator for an activated sludge reactor In order to so, he performs a series of experiments obtaining the following results: α = 0.91, and AOR = 1.30 kg/m ⋅ day The aeration is to be maintained to effect a dissolved oxygen concentration of 1.0 mg/L at 25°C in the reactor The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day Determine the value of β A chemical engineer proposes to purchase an aerator for an activated sludge reactor In order to so, he performs a series of experiments obtaining the following results: α = 0.91, β = 0.94, and AOR = 1.30 kg/m ⋅ day The aeration is to be maintained to effect a dissolved oxygen concentration of 1.0 mg/L at 25°C in the reactor The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day and the concentration of dissolved oxygen at 20°C and one atmosphere pressure of tap water is 9.2 mg/L The Arrhenius temperature correction factor is 1.024 From these data, calculate the saturation dissolved oxygen concentration of tap water corresponding to the temperature of the proposed reactor A chemical engineer plans to purchase an aerator for a proposed activated sludge reactor In order to so, he performs a series of experiments obtain3 ing the following results: α = 0.91, β = 0.94, and AOR = 1.30 kg/m ⋅day The average temperature of operation of the reactor is expected to be 25°C The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅day and the concentration of dissolved oxygen at 20°C and one atmosphere pressure of tap water is 9.2 mg/L The Arrhenius temperature correction factor is 1.024 The saturation dissolved oxygen concentration of tap water corresponding to the temperature of the proposed reactor at 25°C is 8.4 mg/L Using these data, calculate the dissolved oxygen concentration at which the reactor can be maintained A chemical engineer plans to purchase an aerator for a proposed activated sludge reactor In order to so, he performs a series of experiments obtain3 ing the following results: α = 0.91, β = 0.94, and AOR = 1.30 kg/m ⋅ day The average temperature of operation of the reactor is expected to be 25°C © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 458 Friday, June 14, 2002 4:37 PM 9.10 9.11 9.12 9.13 9.14 9.15 9.16 The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day The Arrhenius temperature correction factor is 1.024 The saturation dissolved oxygen concentration of tap water corresponding to the temperature of the proposed reactor at 25°C is 8.4 mg/L The proposed reactor is to be maintained at a dissolved oxygen concentration of 1.0 mg/L Using these data, calculate the dissolved oxygen at 20°C and one atmosphere pressure of tap water A chemical engineer plans to purchase an aerator for a proposed activated sludge reactor In order to so, he performs a series of experiments obtain3 ing the following results: α = 0.91, β = 0.94, and AOR = 1.30 kg/m ⋅ day The average temperature of operation of the reactor is expected to be 25°C The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day The saturation dissolved oxygen concentration of tap water corresponding to the temperature of the proposed reactor at 25°C is 8.4 mg/L The proposed reactor is to be maintained at a dissolved oxygen concentration of 1.0 mg/L The dissolved oxygen at 20°C and one atmosphere pressure of tap water is 9.2 mg/L Using these data, calculate the Arrhenius temperature correction factor θ A chemical engineer plans to purchase an aerator for a proposed activated sludge reactor In order to so, he performs a series of experiments obtain3 ing the following results: α = 0.91, β = 0.94, and AOR = 1.30 kg/m ⋅ day The SOR specified to the manufacturer of the aerator is 2.05 kg/m ⋅ day The saturation dissolved oxygen concentration of tap water corresponding to the temperature of the proposed reactor is 8.4 mg/L The proposed reactor is to be maintained at a dissolved oxygen concentration of 1.0 mg/L The dissolved oxygen at 20°C and one atmosphere pressure of tap water is 9.2 mg/L The Arrhenius temperature correction factor θ is 1.024 Using these data, calculate the average temperature that the proposed reactor will be operated A civil engineer performs an experiment for the purpose of determining the values of some aeration parameters of a particular wastewater (KLa)20 and α were found, respectively, to be 2.46 per hr and 0.91 Calculate (KLa)w,20 A civil engineer performs an experiment for the purpose of determining the values of some aeration parameters of a particular wastewater (KLa)w,20 and α were found, respectively, to be 2.25 per hr and 0.91 Calculate (KLa) 20 An environmental engineer performs an experiment for the purpose of determining the value of β of a particular wastewater If β is found to be equal to 0.89 and the temperature of the wastewater is 25°C, calculate [Cos,w] An environmental engineer performs an experiment for the purpose of determining the value of β of a particular wastewater If β is found to be equal to 0.89 and the [Cos,w] of the wastewater after shaking the jar thoroughly is 7.5 mg/L , calculate [Cos ] An environmental engineer performs an experiment for the purpose of determining the value of β of a particular wastewater in the mountains of Allegheny County, MD The prevailing barometric pressure is 92,974 N/m © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 459 Friday, June 14, 2002 4:37 PM 9.17 9.18 9.19 9.20 9.21 9.22 9.23 at an average atmospheric temperature of 30°C At what elevation in the county is the experiment conducted? An environmental engineer performs an experiment for the purpose of determining the value of β of a particular wastewater in the mountains of Allegheny County, MD The prevailing barometric pressure is 92,974 N/m and the elevation where the experiment is conducted is 756 m above mean sea level What is the average temperature of the atmosphere? An activated sludge reactor is aerated using a turbine aerator located 5.5 m below the surface of the mixed liquor The pressurizing pressure is cal2 culated to be 101,436 N/m If the water temperature is 20°C, calculate the prevailing barometric pressure An activated sludge reactor is aerated using a turbine aerator The pressur2 izing pressure is calculated to be 101,436 N/m at a prevailing barometric pressure of 760 mm Hg If the water temperature is 20°C, determine the depth of submergence of the aerator An activated sludge reactor is aerated using a turbine aerator The pres2 surizing pressure is calculated to be 101,436 N/m at a prevailing barometric pressure of 760 mm Hg The depth of submergence of the aerator is 5.5 m What is the specific weight of the water in the reactor? An activated sludge reactor is aerated using a turbine aerator The [Cos] corresponding to the dissolution pressure is 9.2 mg/L and the prevailing barometric pressure is 761 mm Hg The water temperature is 20°C Calculate the depth of submergence An activated sludge reactor is aerated using a turbine aerator The [Cos] corresponding to the dissolution pressure is 9.2 mg/L and the prevailing barometric pressure is 761 mm Hg The water temperature is 20°C and the depth of submergence is 5.5 m Using the calculated value of the pressuring pressure, calculate [Cos,sp] An experiment to determine the overall mass transfer coefficient of a tap water is performed using a settling column m in height The result of the unsteady state aeration test is shown below The experiment is performed in Allegheny County, MD For practical purposes, assume mass density of water = 1000 kg/m Assume an ambient temperature of 28°C Calculate KLa Tap Water at 5.5°C Time (min) 12 15 18 21 © 2003 by A P Sincero and G A Sincero [ C ](mg/L) 0.5 1.7 3.1 4.4 5.5 6.1 7.1 TX249_frame_C09.fm Page 460 Friday, June 14, 2002 4:37 PM 9.24 9.25 An activated sludge reactor is aerated using a turbine aerator The [Cos] corresponding to the dissolution pressure is 9.2 mg/L and the prevailing barometric pressure is 761 mm Hg The water temperature is 20°C and the depth of submergence is 5.5 m Using the calculated value of the pressuring pressure and using a value of 8.6 for [Cos,sp], calculate the standard pressure Ps An experiment to determine the overall mass transfer coefficient of a tap water is performed using a settling column m in height The result of the unsteady state aeration test is shown below The experiment is performed in Allegheny County, MD For practical purposes, assume mass density of water = 1000 kg/m Assume an ambient temperature of 28°C Calculate (KLa)20 Tap Water at 5.5°C Time (min) 12 15 18 21 9.26 [ C ] ,(mg/L) 0.5 1.7 3.1 4.4 5.5 6.1 7.1 A settling column m in height is used to determine the overall mass transfer coefficient of a wastewater The wastewater is to be aerated using a fine-bubble diffuser in the prototype aeration tank The laboratory diffuser releases air at the bottom of the tank The result of the unsteady state aeration test is shown below Assume β = 0.94, r = 1.0 mg/L ⋅ hr and the plant is in the mountains of Allegheny County, MD For practical purposes, assume mass density of water = 1000 kg/m Assume an ambient temperature of 28°C Calculate the apparent overall mass transfer coefficient, overall mass transfer coefficient, and the overall mass transfer coefficient corrected to 20°C Wastewater at 25°C Time (min) 12 15 18 21 © 2003 by A P Sincero and G A Sincero [ C ](mg/L) 0.9 1.8 2.4 3.3 4.0 4.7 5.3 TX249_frame_C09.fm Page 461 Friday, June 14, 2002 4:37 PM 9.27 9.28 9.29 9.30 9.31 9.32 9.33 The influent of 10,000 m /day to a secondary reactor has a BOD5 of 150 mg/L It is desired to have an effluent BOD5 of mg/L, an MLVSS (mixed liquor volatile suspended solids) of 3000 mg/L, and an underflow concentration of 10,000 mg/L The effluent suspended solids concentration is mgL at 71% volatile suspended solids content The sludge is wasted at the rate of 43.3 m /day Assume the aerator to be of the fine-bubble diffuser type with an α = 0.55; depth of submergence equals 2.44 m Assume β of liquor is 0.90 The influent TKN is 25 mg/L and the desired effluent NH3 − N concentration is 5.0 mg/L ⋅ f = 1.47 The average temperature of the reactor is 25°C and is operated at an average of 1.0 mg/L of dissolved oxygen The AOR is calculated to be 3.36 kg/d ⋅ m Calculate the volume of the reactor The volume of a secondary reactor is 1611 m It receives a primarytreated wastewater containing a BOD5 of 150 mg/L It is desired to have an effluent BOD5 of mg/L, an MLVSS (mixed liquor volatile suspended solids) of 3000 mg/L, and an underflow concentration of 10,000 mg/L The effluent suspended solids concentration is mgL at 71% volatile suspended solids content The sludge is wasted at the rate of 43.3.m /day Assume the aerator to be of the fine-bubble diffuser type with an α = 0.55; depth of submergence equals 2.44 m Assume β of liquor is 0.90 The influent TKN is 25 mg/L and the desired effluent NH3 − N concentration is 5.0 mg/L ⋅ f = 1.47 The average temperature of the reactor is 25°C and is operated at an average of 1.0 mg/L of dissolved oxygen The AOR is calculated to be 3.36 kg/d ⋅ m Determine the volume of inflow to the reactor The volume of a secondary reactor is 1611 m It receives an inflow of 10,000 m /day from primary-treated wastewater It is desired to have an effluent BOD5 of mg/L, an MLVSS (mixed liquor volatile suspended solids) of 3000 mg/L, and an underflow concentration of 10,000 mg/L The effluent suspended solids concentration is mg/L at 71% volatile suspended solids content The sludge is wasted at the rate of 43.3 m /day Assume the aerator to be of the fine-bubble diffuser type with an α = 0.55; depth of submergence equals 2.44 m Assume β of liquor is 0.90 The influent TKN is 25 mg/L and the desired effluent NH3 − N concentration is 5.0 mg/L f = 1.47 The average temperature of the reactor is 25°C and is operated at an average of 1.0 mg/L of dissolved oxygen The AOR is calculated to be 3.36 kg/d ⋅ m What is the influent BOD5? Using the data in Problem 9.29, calculate the effluent BOD5 from the secondary reactor if the influent BOD5 is 150 mg/L Using the data in Problem 9.29, calculate the rate of sludge wasting from the process if the influent BOD5 is 150 mg/L Using the data in Problem 9.29, calculate the underflow concentration from the secondary clarifier if the influent BOD5 is 150 mg/L An experiment was performed on a trickling filter for the purpose of determining the overall mass transfer coefficient Raw sewage with zero dissolved oxygen was introduced at the top of the filter and allowed to © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 462 Friday, June 14, 2002 4:37 PM 9.34 9.35 9.36 9.37 9.38 9.39 9.40 9.41 9.42 9.43 9.44 flow down the bed A tracer was also introduced at the top to determine how long it takes for the sewage to trickle down the bed At the bottom, the resulting DO was then measured From this experiment, the overall mass transfer coefficient (KLa)w was found to be 2.53 per hour This information is then used to design another trickling filter The β of the waste = 0.9 and the respiration rate r = 1.0 mg/L ⋅ hr Assume the average temperature in the filter is 25°C and the effluent should have a DO = 1.0 mg/L What void volume should be provided in the interstices of the bed to effect this detention time? The inflow rate is 10,000 m /d The detention time of the sewage in the proposed filter to effect this DO at the effluent of 1.0 mg/L is calculated to be 0.06 hour Calculate the overall mass transfer coefficient of the wastewater Using the data in Problem 9.33, calculate β if the overall mass transfer of the wastewater is 2.53 per hour Using the data in Problem 9.33, calculate the saturation dissolved oxygen concentration at the dissolution pressure if the overall mass transfer of the wastewater is 2.53 per hour Using the data in Problem 9.33, calculate the effluent DO if the overall mass transfer of the wastewater is 2.53 per hour Using the data in Problem 9.33, calculate the respiration rate if the overall mass transfer of the wastewater is 2.53 per hour An aerator in an activated sludge reactor is located below the surface The temperature inside the reactor is 29°C The rise velocity of the bubbles is determined to be 0.26 m/s at a time of rise of 13.46 sec The approximate average diameter of the bubbles at mid-depth is 0.25 cm Calculate the depth of submergence An aerator in an activated sludge reactor is located 3.5 m below the surface The temperature inside the reactor is 29°C The time of rise of the bubbles to the surface is 13.46 sec The approximate average diameter of the bubbles at mid-depth is 0.25 cm Calculate the rise velocity A packed absorption tower is designed to removed SO2 from a coke oven stack The water flow rate L = 16.64 kgmols/s and the gas flow rate G = 0.39 kgmols/s The mole fraction of SO2 in the air at the bottom of the tower is 0.03 and its mole fraction in the absorption water at the bottom of the tower is 0.0007 If the mole fraction of SO2 in the air at a point in the tower meters from the bottom is 0.016, what is the corresponding mole fraction of the SO2 in the downward flowing scrubbing water Using the data in Problem 9.40, calculate the corresponding mole fraction of SO2 in air if its mole fraction in water at the 7-m point is 0.0004 Using the data in Problem 9.40, calculate the gas flow rate G if the mole fraction of SO2 in water at the 7-m point is 0.0004 Using the data in Problem 9.40, calculate the water flow rate L if the mole fraction of SO2 in water at the 7-m point is 0.0004 Using the data in Problem 9.40, calculate the mole fraction of SO2 in the air at the bottom of the tower if the mole fraction of SO2 in water at the 7-m point is 0.0004 © 2003 by A P Sincero and G A Sincero TX249_frame_C09.fm Page 463 Friday, June 14, 2002 4:37 PM 9.45 Using the data in Problem 9.40, calculate the mole fraction of SO2 in the water at the bottom of the tower if the mole fraction of SO2 in water at the 7-m point is 0.0004 BIBLIOGRAPHY Benyahia, F., et al (1996) Mass transfer studies in pneumatic reactors Chem Eng Technol., 19, 5, 425–43 Boswell, S T and D A Vaccari (1994) Plate and frame membrane air stripping Proc 21st Annu Conf on Water Policy and Manage.: Solving the Problems, May 23–26, 1994, Denver, CO, 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Wilson, D J and R D Norris (1997) Groundwater cleanup by in-situ sparging, engineered bioremediation with aeration curtains Separation Science Technol 32, 16, 2569–2589 © 2003 by A P Sincero and G A Sincero ... affinity of the solute to form into a solid An example of this operation is the making of ice from liquid water The formation of snow from water vapor in the atmosphere is also a process of crystallization... performs an experiment for the purpose of determining the value of β of a particular wastewater The [Cos,w] of the wastewater after shaking the jar thoroughly is 7.5 mg/L The temperature of the... Therefore, 7.5 β = = 0.89 Ans 8.4 Example 9.5 An environmental engineer performs an experiment for the purpose of determining the value of β of a particular wastewater in the mountains of Allegheny