4 Dispersion DSP
DSP.l ADVANTAGES AND DISADVANTAGES
Discuss the advantages and disadvantages of employing atmospheric dispersion as an ultimate disposal method of waste gases.
Solution
The advantages are listed below:
1. Dispersion of the waste gases leads to the dilution of the pollutants in the atmosphere. Selfpurification mechanisms of atmospheric air also assists the process.
2. Tall stacks emit gas into the upper layer of the atmosphere and lower the ground concentration of the pollutants.
3. The method is commonly used, cheap and easily applicable.
4. By selecting the proper location of stacks through the use of different models for dispersion, it is possible to significantly reduce the concentration of waste gases in the atmosphere.
The disadvantages are:
1. Any particulate matter contained in the dispersed gases have a tendency to settle down to the ground level.
2. The location of the industrial source may prohibit dispersion as an option.
3. Plume rise can significantly vary with ambient temperature, stability conditions, molecular weight, and exit velocity of the stack gases.
4. The models of atmospheric dispersion are rarely accurate. They should only be used for estimation and comparative analysis.
DSP.2 GROUND SURFACE DEPOSITION
An incinerator stack is emitting fly ash at the rate of 2 tons per hour. Natural processes are capable of removing these particles from the affected ground surface at a steady rate, provided no more than 0.02% of the ground is covered by them per hour. Assume the particles are, on the average, spheres of radius 10~4 ft and have an average density of 120 lb/ft3. The wind speed is 5mph.
If L is the distance through which an average particle is carried by the wind, the particles will settle out uniformly over a wedgeshaped area the central angle of which is 20°, at distances ranging from 0.5L to 20L. Determine the minimum stack height H required to prevent ground level accumulation. Assume monolayer deposition, with the ground area covered by a particle equal to its crosssectional area [i.e. 7r(i/p)2/4],
Solution
Select 1.0 h as a basis for the calculations: Volume of one particle =  nr3
Mass (weight) of one particle = ppV = (120)(4.19 x 10"12)
= 5.027 x 10'° lb Number of particles emitted per hour = (2)(2000)/(5.027 x 10"10)
Area covered by particles per hour = (number)(area) = (number)(7tr2) = (7.958 x 1012)(7r)(104)2 = 2.5 x 105ft2
Set a (maximum deposition distance) = 20Z. and b (minimum deposition distance) = 0.5L.
Area of deposit = n(a2  ¿>2)(20/360) = jr[(20L)2  (0.5I)2](20/360) = 69.8I2
0.02% of deposited area = (2 x 10"4)(69.8)Z,2 = 1.396 x 10"2!2 Substitution yields
1.39 x 102Z,2 =2.5 x 105 L2 = 1.79 x 107 L = 4230 ft
The Stokes' law region applies for the settling particles. Therefore,
H 0.698
H = (0.0952)(4230) = 403 ft The minimum height of the stack is approximately 400 ft.
DSP.3 PLUME RISE
If a waste source emits a gas with a buoyancy flux of 50m4/s3, and the wind averages 4m/s, find the plume rise at a distance of 750 m downward from a stack that is 50 m high under unstable atmospheric conditions. Use the equation proposed by Briggs.
Solution
Several plume rise equations are available. Briggs used the following equations to calculate the plume rise:
= l.6Fl/3u1x2fp ifx^x{ x* = 14F5/s when F < 55 m4/ s3 = 34F2/5 when F ^ 55 m4/ s3 Xf = 3 .Sx where Ah = plume rise, m
F = buoyancy flux, m4/s3 = 3.7 x 10~5QH u = wind speed, m/s x* = downward distance, m
Xf = distance of transition from first stage of rise to the second stage of rise, m
Qh = heat emission rate, kcal/s If the term QH is not available, the term F may be estimated by
F = (g/n)q(Ts  T)/Ts where g = gravity term 9.8 m/s2
q = stack gas volumetric flowrate, m3/s (actual conditions) Ts, T = stack gas and ambient air temperature, K, respectively
Calculate xf to determine which plume equation applies.
x* = 14Fs/s since F is less than 55 m4/s3 = (14)(50)5/8 = 161.43m x{ = 3.5x*
The plume rise is therefore h = 1.6Fi/3m'x2/3 since x ^ x{ = (1.6)(50),/3(4)1(565)2/3 = 101 m
Many more plume rise equations may be found in the literature. The Environmental Protection Agency (EPA) is mandated to use Brigg's equations to calculate plume rise. In past years, industry has often chosen to use the Holland or DavidsonBryant equation. The Holland equation is
where ds = inside stack diameter, m vs = stack exit velocity, m/s u = wind speed, m/s P = atmospheric pressure, mbar TS,T = stack gas and ambient temperature, respectively, K AT=TS  T Ah = plume rise, m
The DavidsonBryant equation is
The reader should also note that the "plume rise" may be negative in some instances due to surrounding structures, topography, etc.
DSP.4 POWER PLANT EMISSION
A power plant burns 12 tons of 2.5% sulfur content coal per hour. The effective stack height is 120 m and the wind speed is 2 m/s. At one hour before sunrise, the sky is clear. A dispersion study requires information on the approximate distance of the maximum concentration under these conditions. {Hint: Calculate concentrations for downward distances of 0.1, 1.0, 5, 10, 20, 25, 30, 50 and 70 km.)
Solution
The coordinate system used in making atmospheric dispersion estimates of gaseous pollutants, as suggested by Pasquill and modified by Gifford, is described below. The origin is at ground level or beneath the point of emission, with the x axis extending horizontally in the direction of the mean wind. The y axis is in the horizontal plane perpendicular to the x axis, and the z axis extends vertically. The plume travels along or parallel to the x axis (in the mean wind direction). The concentration, C, of gas or aerosol at (x,y, z) from a continuous source with an effective height, He, is given by:
C(x,y,z,He) = (m/2^(T^zM)[e<1/2)<>^,)2][e</2)((ZHc)/)2 +e(i/2)((z+He)/.)2]
where Hc = effective height of emission (sum of the physical stack height, Hs and the plume rise, Ah), m u = mean wind speed affecting the plume, m/s m — emission rate of pollutants, g/s oy, az — dispersion coefficients or stability parameters, m
C = concentration of gas, g/m3 x, y, z = coordinates, m
The assumptions made in the development of the above equation are: (1) the plume spread has a Gaussian (normal) distribution in both the horizontal and vertical planes, with standard deviations of plume concentration distribution in the horizontal and vertical directions of <t,„ and az, respectively; (2) uniform emission rate of pollutants, m; (3) total reflection of the plume at ground z = 0 conditions; and (4) the plume moves downstream (horizontally in the x direction) with mean wind spead, u. Although any consistent set of units may be used, the cgs system is preferred. For concentrations calculated at ground level (z = 0), the equation simplifies to
C(x,y, 0, He) = (ml2navazii)\e(Xll){yl'!^\e(XI'Z)(H'ta^\
If the concentration is to be calculated along the centerline of the plume (y — 0), further simplification gives
In the case of a groundlevel source with no effective plume rise (He — 0), the equation reduces to
It is important to note that the two dispersion coefficients are the product of a long history of field experiments, empirical judgments, and extrapolations of the data from those experiments. There are few knowledgeable practitioners in the dispersion modeling field who would dispute that the coefficients could easily have an inherent uncertainty of ±25%.
The six applicable stability categories for these coefficients are shown in the following table:
Night
Surface Wind
Incoming Solar Radiation
Thinly Overease or >4/8
Surface Wind
Incoming Solar Radiation
Thinly Overease or >4/8
Speed at 10 m (m/s) 
Strong 
Moderate 
Slight 
Low Cloud 
<3/8 
2 
A 
AB 
B 
— 
— 
23 
AB 
B 
C 
E 
F 
35 
B 
BC 
C 
D 
E 
56 
C 
CD 
D 
D 
D 
6 
C 
D 
D 
D 
D 
Note that A, B,C refer to daytime with unstable conditions; D refers to overcast or neutral conditions at night or during the day; E and F refer to night time stable conditions and are based on the amount of cloud cover. "Strong" incoming solar radiation corresponds to a solar altitude greater than 60° with clear skies (e.g., sunny midday in midsummer); "slight" insolation (rate of radiation from the sun received per unit of Earth's surface) corresponds to a solar altitude from 15° to 35° with clear skies (e.g., sunny midday in midwinter). For the AB, BC, and CD stability categories, one should use the average of the A and B values, B and C values, and C and D values, respectively. Figures 125 and 126 provide the variation of a and az with stability categories and distances.
First calculate the sulfur dioxide emission rate, m, in grams/second. The molecular weights of S and S02 are 32 and 64, respectively.
10000 5000 3000
1000
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50 30
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100 400 1000 4000 10000 40000 100000 Distance, x, meters Figure 125. Dispersion coefficients, y direction.
1000 4000 10000 40000 100000 Distance, x, meters Figure 126. Dispersion coefficients, z direction.
Since it is night time, the sky is clear, and the wind speed is 2 m/s. Therefore, the stability category is F (see table above).
Determine the values of ay and crz for each downward distance for stability category F from Figures 125 and 126.
x (km) 
ay (m) 
<T2 (m) 
0.1 
4 
2.3 
1.0 
34 
14.0 
5 
148 
34.0 
10 
275 
46.5 
20 
510 
60.0 
25 
610 
64.0 
30 
720 
68.0 
50 
1120 
78.0 
70 
1500 
84.0 
Since the maximum groundlevel (z = 0) concentration along the x axis is desired and the maximum concentration occurs along the centerline (y = 0) of the plume,
The concentrations of sulfur dioxide at the downward distances are obtained by completing the table below.
x (km) 
C (g/m3) 
0.1 
0 (infinitely small) 
1.0 
5.61 x 10"18 
5 
9.41 x 10"6 
10 
6.72 x 10"5 
20 
1.06 x 10"4 
25 
1.06 x 10"4 
30 
1.03 x 10"4 
50 
8.41 x 10"5 
70 
7.03 x 10~5 
C(1.0, 0, 0,He) = [(151.3)/(7r)(34)( 14)(2)][e"1 /2< 120/14>2] = 5.61 x 10"18 g/m3
From the table above, the maximum concentration is approximately 1.06 x 10"4 g/m3 and the corresponding location is between 20 and 25 km.
DSP.6 OBSERVATION TOWER CONCENTRATION 845
DSP.5 MAXIMUM GROUNDLEVEL CONCENTRATION
Determine the required recovery of an air pollution control device knowing that the maximum concentration along the groundlevel centerline of the plume must not exceed 1.05 x 10~6g/m3 at a downward distance of 50 km. The wind speed is 6 m/s and the effective height of emission is 150 m. The concentration was measured during the day, when incoming solar radiation was very intense. The incoming rate of pollutant to the control device is known to be 4189g/s.
Solution
Based on the problem conditions, the stability category is C. From Figures 125 and 126, at x = 50km, oy = 3300m, and cxz = 1500m:
Using the Pasquill Gifford equation,
Solving for m yields m = 94.44 g/s
Thus, the collection efficiency required is
¿■(required) = (4189  98.44)/4189 = 0.9765 = 97.65%
DSP.6 OBSERVATION TOWER CONCENTRATION
What is the concentration of pollutants at the top of a recently constructed ranger's observation tower that is 2 km north and 100 m west of a stack emitting pollutants at a rate of 8.0 g/s. The stack is known to have a centerline ground zero concentration of 3 mg/m3 at 1.0 km. The volumetric flowrate of the stack gas is 40,000 scfm, the diameter of the stack is 6.0 ft, the temperature of the emission is 300°F, the wind is out of the south at a rate of 10.0 miles/h, and the height of the ranger's tower is 12 m. Assume Holland's equation to apply, both the standard and ambient temperature to be 68°F (298 K), and the stability category to be D.
846 DISPERSION (DSP) Solution
A topview schematic of the system is provided below in Figure 127. Pertinent data follow:
qs = 40,000 scfm u = lOmph = 14.67 ft/s = 4.47 m/s Ts = 300°F = 422 K T = 68°F = 298 K P= latin = 1013.25 mbar ds = (stack diameter) = 6ft = 1.8288 m
Key calculations for the actual volumetric flowrate (q.d), stack area (A s), stack velocity (vs), and plume rise (Ah) are listed below.
qa = 40,000(300 + 460)/528 = 57,575.8 acfrn As = 7r(6)2/4 = 28.234 ft2
vs = qJAs = 57,575.8/28.234 = 33.94 ft/s = 10.35 m/s Ah = [us(í/s/m)][1.5 + 0.00268P(rs  T)(dJT)]
= [(10.35)(1.83)/(4.47)][1.5 + (0.00268)(1013.28)[(432  298)/298](l.83)]
15.1m
O tower 12 ml
12 m high
2 km
— 100 m O stack wind, 10 mi/h Figure 127. Schematic for Problem DSP.6.
wind, 10 mi/h Figure 127. Schematic for Problem DSP.6.
The PasquillGifford equation is used to estimate the effective stack height, He, noting that m is 8 g/s.
From Figures 125 and 126, at x = 1 km, a = 70 m, and a2 — 31 m. Thus,
Therefore, the stack height is
(l/2)(lO0/13O)2^e(l/2)(l092.7/50)2 j. ^(1/2X10+92.7/50)
_(27r)( 130)(50)(4.47)_ = 4.38 x 10"5(0.7439)(0.2719 +0.11165)
DSP.7 LINE SOURCE APPLICATION
A sixstory hospital building is located 300 m east and downwind from an expressway. The expressway runs northsouth and the wind is from the west at 4m/s. It is 5:30 in the afternoon on an overcast day. The measured traffic flow is 8000 vehicles per hour during this rush hour and the average vehicle, traveling at an average speed of 40mph, is expected to emit 0.02 g/s of total hydrocarbons. Concentrations at the hospital are required as part of a risk assessment study. How much lower, in percent, will the hydrocarbon concentration be on top of the building (where the elderly patients are housed) as compared with the concentration estimated at ground level? Assume a standard floor to be 3.5 m in height.
Solution
Line source applications are generally confined to roadways and streets along which there are welldefined movements of motor vehicles. For these types of line sources, data are required on the width of the roadway and its center strip, the types and amounts per unit time per unit length [g/(s • m)] of pollutant emissions, the number of lanes, the emissions from each lane, and the height of emissions. In some situations, e.g., a traffic jam at a tollbooth, or a series of industries located along a river, or heavy traffic along a stright stretch of highway, the pollution problem may be modeled as a continuous emitting infinite fine source. Concentrations downwind of a continuous emitting infinite line source, when the wind direction is normal to the line, can be calculated from
where q = source strength per unit distance, g/(s • m) Hc = effective stack or discharge height, m u = wind speed, m/s az = vertical dispersion coefficients, m
Note that the horizontal dispersion coefficient, ay, does not appear in this equation since it is assumed that lateraldispersion from one segment of the line is compensated by dispersion in the opposite direction from adjacent segments. Also, y does not appear since the concentration at a given x is the same for any value of y.
Since it is early evening and the wind speed is 4 m/s, the stability category is D. The number of vehicles per meter, n, is
= 0.1243 vehicles/m
Calculate the source strength per unit length, q, from n and the emission rate per vehicle:
q = (n)(0.02) = (0.1243)(0.02) = 2.49 x 10~3 gHC/(s • m)
The vertical dispersion coefficient, <rz, for stability category D from Figure 126 is az = 12matx = 300 m
The height of the building, Hb, is
The concentration 300 m downwind at ground level conditions is then
C(x,y, 0, He) = (2q/(2n)l/2<Jzu)(e]/2{H^) C(300, 0, 0, 0) = [(2)(2.49 x 10"3)/(27t)1/2(12)(4)](l) = 4.14 x 10~5 g/m3
Note that the exponential term becomes 1.0 if He = 0.
The concentration at 300 m downwind at the top of the building may also be calculated.
C(300, 0, 0, Hb) = (2q/{2nyi2ozu)(eyi2(H*>,a)1) C(300, 0, 0, 21) = [(2)(2.49 x 10"3)/(2tt) 1 /2( 12)(4)](e"1 /2(21 /12>2) = 8.95 x 10~6 g/m3
Determine how much lower the concentration, in percent, will be at the top of the building:
% = 100 (ground conc. — top conc.)/(ground conc.) = 100(4.14 x 10"5  8.95 x 10"6)/(4.14 x 10"5) = 78.4%
Concentrations from infinite line sources, when the wind is not perpendicular to the line, can also be approximated. If the angle between the wind direction and the line source is 9
This equation should not be used when 9 is less than 45°.
When the continuously emitting line source is reasonably short in length or "finite," one can account for the edge effects caused by the two ends of the source. If the line source is perpendicular to the wind direction, then it is convenient to define the x axis in the direction of the wind and also passing through the sampling point downwind. The ends of the line source are then at two positions in the crosswind direction, j, and y2, where v, is less than y2. The concentration along the x axis at ground level is
C(x,y, 0, He) = [2q/(2n)i/2azu](e(l/2W^)2) V[(\l(2n){l2)e(Xl2)p2]dp
Once the limits of integration are established, the value of the integral may be determined from standard tables of integrals.
DSP.8 AREA SOURCE APPLICATION
An inventory of emissions has been made in an urban location by square areas, 1524 m on a side, as part of an emergency planning and response study. The emissions from two such adjacent areas (source areas B and D) are estimated to be 6.0 g/s for each area. The effective stack height of the sources within each area is approximately 20 m. The wind is 2.5 m/s on a thinly overcast night and perpendicular to the two adjacent square areas (i.e., source areas B and D are aligned perpendicular to the wind). What is the percentage contribution of emissions from the two sources to the center point of the square area (point A) immediately downwind of the source area D?
Solution
Area sources include the multitude of minor sources with individually small emissions that are impractical to consider as separate point or line sources. Area sources are typically treated as a grid network of square areas, with pollutant emissions distributed uniformly within each grid square. Area source information required are types and amounts of pollutant emissions, the physical size of the area over which emissions are prorated and representative stack height(s) for the area.
In dealing with dispersion of pollutants in areas having large numbers of sources, e.g., as in fugitive dust from (coal) piles, a large number of automobiles in a parking lot, or a multistack situation, there may be too many sources to consider each source inividually. Often an approximation can be made by combining all of the emissions in a given area and treating this area as a source having an initial horizontal standard deviation, aM. A virtual distance, xy, that will give this standard deviation can be found. This virtual distance is the distance that will yield the appropriate value for a from Figure 125 provided in Problem DSP.4. Values of x,, will vary with stability. The equation for point sources may then be used, determining <ry as a function of x + xy. This procedure effectively treats the area source as a crosswind line source with a normal distribution across the area source. Finally, the initial standard deviation for a square area source can be approximated by cv<) = s/4.3, where 5 is the length of a side of the area.
Since it is a thinly overcast night and the wind speed is 2.5 m/s, stability category E applies. Calculate the initial standard deviation, <7,,0.
Obtain the virtual point source distance, xy using <t):() of 354 m from Figure 125.
x — 8.4 km for oy, = 354 m and stability category E
The total distance, x„ is therefore x, = 8.4+1.524 = 9.9 km
Also obtain a which corresponds to an x, of 9.9 km, from Figure 125.
Since the area source can be assumed to be well mixed, az may be obtained at x = 1524 m from Figure 126.
Calculate the concentration at point A because of point D (directly downward):
C(1524,0, 0, H) = (m/nOyCzu)(e{Xl2)iyla>)2)(e^l2)(HJaJ) — {(6 0)/ [(7i)(410)(26)(2.5)]} (e~f 1 /2'<20/26)2) = 5.33 x 10"5g/m3
Calculate the concentration at point A (directly downward from point D).
C(1524, 1524, 0,H) = (m/nayazu)(e{>/2)(H'/c,'}')
= {(6.0)/[(7T)(410)(26)(2.5)]} 1 1524/41)(e( 1 /2)(20/26)2 } = 5.33 x 108 g/m3
Finally, calculate the ratio of the contribution to point A from points B and D.
DSP.9 STACK DESIGN RECOMMENDATIONS
List some of the key rules of thumb of stack design.
Solution
As experience in designing stacks has accumulated over the years, several guidelines have evolved.
1. Stack heights should be at least 2.5 times the height of any surrounding buildings or obstacles so that significant turbulence is not introduced by these factors.
2. The stack gas exit velocity should be greater than 60 ft/s so that stack gases will escape the turbulent wake of the stack. In many cases, it is good practice to have the gas exit velocity on the order of 90 or 100 ft/s.
3. A stack located on a building should be set in a position that will assure that the exhaust escapes the wakes of nearby structures.
4. Gases from the stacks with diameters less than 5 ft and heights less than 200 ft will hit the ground part of the time, and the ground concentration will be excessive. In this case, the plume becomes unpredictable.
5. The maximum ground concentration of stack gases subjected to atmospheric dispersion occurs about 510 effective stack heights downwind from the point of emission.
6. When stack gases are subjected to atmospheric diffusion and building turbulence is not a factor, groundlevel concentrations on the order of 0.0011% of the stack concentration are possible for a properly designed stack.
7. Ground concentrations can be reduced by the use of higher stacks; the ground concentration varies inversely as the square of the effective stack height.
8. Average concentrations of a contaminant downwind from a stack are directly proportional to the discharge rate; an increase in the discharge rate by a given factor increases groundlevel concentrations at all points by the same factor.
9. In general, increasing the dilution of stack gases by the addition of excess air in the stack does not affect groundlevel concentrations appreciably. Practical stack dilutions are usually insignificant in comparison to the later atmospheric dilution by plume diffusion. Addition of diluting will increase the effective stack height, however, by increasing the stack exit velocity. This effect may be important at low wind speeds. On the other hand if the stack temperature is decreased appreciably by the dilution, the effective stack height may be reduced.
10. Stack dilution has an appreciable effect on the concentration in the plume close to the stack.
These 10 guidelines represent the basic design elements of a stack "pollution control" system. An engineering approach suggests that each element be evaluated independently and as part of the whole control system. However, the engineering design and evaluation must be an integrated part of the complete pollution control program.
Renewable Energy 101
Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of nonrenewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a nonrenewable supply, the nonrenewable energy sources release emissions into the air, which has an adverse effect on the environment.
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