dispersion handout

8/22/2019 Dispersion Handout

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CE 524

n r 2010

1



• Air pollution law in most industrial countries

based on concentration of contaminants – NAAQS in US

• ee me o o pre c concen ra ons a anygiven location

– An iven set of ollutant

– Meteorological conditions

– At any location

– For any time period

• But even best currently available concentration

2



Concentration

• Commonly express concentration as ppm or /m3

• Parts per million (ppm) = 1 volume of

– __ __

106 volumes (pollutant + air)

• μg m = m crograms cu c me er

3



Factors that determine Dispersion

• Physical nature of effluents•

• Meteorology• ocat on o t e stac

• Nature of terrain downwind from the stack

4



• Gas and particulate matter

• Particles < 20 μm behave same as gas – Low settling velocity

• ar c e > μm ave s gn can se ng ve oc y

• Only gases and Particles < 20 μm are treated in

• Others are treated as particulate matter

•momentum and buoyancy

– Hot gases continue to rise

5



• Effluents leave the stack with sufficient

momentum and buoyancy – Hot ases continue to rise

• Plume is deflected along its axis in

6



Gaussian or Normal Distribution

•

• Dispersion in y and z directions uses a

u u u -- u

• Dispersion in (x, y, z) is three-dimensional• Used to model instantaneous puff of

emissions

7




• Pollution dispersion follows a distributionfunction

• Theoretical form: gaussian distribution

8




• = m n f h i ri i n

• = standard deviation

context formula used to predict steady state concentration

at a point down stream

9



Gaussian or Normal DistributionWhat are some properties of the normal

10f(x) becomes concentration, maximum at center of plume



Gaussian or Normal distribution

• 68% of the area fall within 1 standard deviation of the mean (µ ± 1 σ).

• 95% of area fall within 1.96 standard deviation of the mean (µ ± 1.96 σ).

. of the mean (µ ± 3 σ)

11



Gaussian dispersion model

• Dispersion in y and z directions aremodeled as Gaussian

• Becomes double Gaussian model

’

distribution in the x direction?

– rec on o w n

12



Gaussian Dispersion Model

• For localized point sources – stacks

• General appearance

• Plume exits at height, hs• Rises an additional distance, Δh

–

– called plume rise

– reaches distance where buoyancy and upward momentum cease• x ve oc y, s

• Plume appears as a point source emitted at height H =hs + Δh

• Emission rate Q (g/s)• Assume wind blows in x direction at speed u

13

– , ,




14




• Stack gas transported downstream• Dispersion in vertical direction governed by

• Dispersion in horizontal plane governed by

molecular and eddy diffusion• x-axis oriented to wind direction

• z-axis oriented vertically upwards

• y- rec on or en e ransverse o e w n• Concentrations are symmetric about y-axis and z-

axis

15




Z-axis through stack

-

to wind

X-axis in

direction of wind

16



increase so doesdispersion

17Image source: Cooper and Alley, 2002



18Image source: Cooper and Alley, 2002



Point Source at Elevation H

• Assumes no interference or limitation to

x0 an z0 are ocat on o center ne o p ume

y0 taken as base of the stack

z0 is H

Q = emission strength of source (mass/time) – g/su = average wind speed thru the plume – m/s

C = concentration – g/m3 (Notice this is not ppm)

19

y and z are horizontal and vertical standard deviations in meters



• Wind speed varies by height• International standard hei ht for wind-s eed

measurements is 10 m

• Dis ersion of ollutant is a function of windspeed at the height where pollution isemitted

• But difficult to develop relationship between height and wind speed

20



Point Source at Elevation H without Reflection

• 3 terms

– gives concentration on the centerline of the plume

– gives concentration as you move in the sideways direction ( ydirection), direction doesn’t matter because ( y)2 gives a positivevalue

– gives concentration as you move in the vertical direction ( z’ 2, –

positive value

• Concentrations are symmetric about y-axis and z-axis

• = =

21

- -

• Close to ground symmetry is disturbed



reflection

• Equation 4-6 reduces to

Note in the book there are 2 equation 4-8s

22

This is the first one



• A factory emits 20 g/s of SO2 at height H

(includes plume rise)

• Wind speed = 3 m/s (u)

• At a distance of 1 km downstream, y and z are

30 m and 20 m (given, otherwise we would haveto oo up

• What are the SO2 concentrations at the centerline

o e p ume an a a po n me ers o e s eand 20 meters below the centerline

23



Gaussian Plume Example

0

Z – H = 0 2

• u = 3 m/s (u)

• y and z are 30 m and 20 m

o secon a

of equation

goes to 0• y = 0 and z = H

• So reduces to:

C x,0,0 = 20 /s = 0.00177 /m3 = 1770 /m3 __ __

2(Π*3*30*20)

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At centerline y and Z are 0



What are the SO2 concentrations at a point 60 meters to the side and 20

c = ____Q____ exp-1/2[(-y2) + ( (z-H)2)]

2u y z [y2 z

2]

= ___20 g/s___ exp-1/2 [(-60m)2 + (-20m)2] =

2 3*(30)(20) [(30m)2 (202m)]

(0.00177 g/m3) * (exp –2.5) = 0.000145 g/m3 or 145.23 µ g/m3

25

At 20 and 60 meters



Evaluation of Standard Deviation

• Horizontal and vertical dispersioncoefficients -- are a function

– downwind position x

–

• many experimental measurements –charts

– Correlated y and z to atmospheric stability

26



Pasquill-Gifford Curves

• Concentrations correspond to sampling times of approx. 10 minutes

• Regulatory models assume that the concentrations

predicted represent 1-hour averages• Solid curves represent rural values

• Dashed lines represent urban values

• Estimated concentrations represent only the lowestseveral hundred meters of the atmosphere

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-• z less certain than y

– Especially for x > 1 km

• For neutral to moderately unstable

atmospheric conditions and distances out to

a few kilometers, concentrations should bewithin a factor of 2 or 3 of actual values

• Tables 3-1: Ke to stabilit classes

28



Example

For stability class A, what are the values

of y and z at 1 km downstream

(assume urban)

- -

29



200

σy ~ 220 m

30



σz ~ 310 m

31



Example

For sta ty c ass A, w at are t e va ues o

y and z at 1 km downstream

rom a es - an -

= 220 m

z = 310 m

32



Empirical Equations

• Often difficult to read charts• Curves fit to em irical e uations

y = cxd

= axb

Where

x = ownw n s ance ome ersa, b, c, d = coefficients from Tables 4-1 and 4-2

33



xamp e: w at are va ues o y an z at mdownstream for stability class A using equations rather

than charts?

y

= cxd

z = ax b

Usin table 4-1 for stability class A

c = 24.1670

d = 2.5334

34



y z downstream for stability class A using equations rather

than charts?

y = cx

z = ax b

Using table 4-2 where x = 1 km

a = 453.850

= .

35



y z for stability class A using equations rather than charts?

y = cxd

z = ax = .

b = 2.11660d = 2.5334

= .

Solution

y = cxd = 24.1670(1 km)2.5334 = 24.17 m

= b = 2.11660 =z . .

36



Point Source at Elevation H with Reflection

• Previous equation for concentration of lumes a considerable distance above

ground

•

• Pollutants “reflect” back up from ground

37




pollutants back into the atmosphere

• e ect on at some stance x s

mathematically equivalent to having am rror mage o t e source at –

• Concentration is equal to contribution of

both plumes at ground level

38



39




Notice this is also equation 4-8 in text,

it is the second e uation 4-8 on the

bottom of page 149

40



Nitrogen dioxide is emitted at 110 g/s from stack with H = 80 m

Wind speed = 5 m/s

Plume rise is 20 m

Calculate ground level concentration 100 meter from centerlineof plume (y)

41

Assume stability class D so σy = 126 m and σz = 51 m



Q = 110 g/s H = 80 m u = 5 m/s Δh = 20 m y = 100 m

= =y z

Effective stack height =80 m + 20 m = 100 m

σy = 126 m and σz = 51 m

=

42

_____ ____ .

2Π*5*126*51



Q = 110 g/s H = 80 m u = 5 m/s Δh = 20 m y = 100 m

= =y z

Solving in pieces exp -[__1002 ] = 0.726149

[2*1252]

- - 2 =

43

.

[2*51

2

]



Q = 110 g/s H = 80 m u = 5 m/s Δh = 20 m y = 100 m

= =y z

Solving in pieces both sides of z portion are same so add

c = 0.000496 * 0.726149 * (2 * 0.14625) = 0.000116 g/m3 or 116.4 µg/m3

44



reflection

• Often want ground level – Peo le ro ert ex osed to ollutants

• Previous eq. gives misleadingly low results

• Pollutants “reflect” back up from ground

45



• Equation for ground level concentration

• Z = 0

Reduces to at ground1 + 1 cancels 2

46

G d L l E l



Ground Level Example

C- stability class

H = 5 0 m

Q = 95 g/sWind s eed is 3 m/s

What is ground level concentration at 0.5 km downwind,

along the centerline?From Figure 4-6, y = 90 m,

From Figure 4-7, z = 32 m

C = 95 x 106 µg/s * exp[-(502)] exp [0] = 1023.3 µg/m3

47



Concentration

• Effect of ground reflection increases groundconcentration

• Does not continue indefinitely

-

(crosswind) and z-direction decreases

48



Concentration

Values for a, b, c, d are in Table 4-5

49



Alternative to Eq. 4-15

• For moderately unstable to neutralconditions

z = 0.707H

Cmax, reflection = 0.1171Q

u y z

50



.What is maximum ground level concentration and where is it

located downstream for the followin ?

•Wind speed = 2 m/s

•H = 71 m

•Q = 2,500,000 µg/s

So ut on:

z = 0.707H = 0.707(71m) = 50.2 m

From Fi ure 4-7, this occurs at x = 500 m

51



z = 50.2 m

- ,

occurs at x = 500 m

52



At 500 m, σy =

120 m

53



.What is maximum ground level concentration and where is it

located downstream for the followin ?

•Wind speed = 2 m/s

•H = 71 m

•Q = 2,500,000 µg/s

So ut on:

z = 0.707H = 0.707(71m) = 50.2 m

From Fi ure 4-7, this occurs at x = 500 m

From Figure 4-6, y = 120 mCmax, reflection = 0.1171Q = 0.1171(2500000) = 24.3 µg/m3

54

u y z (2)(120)(50.2)



Calculation of Effective Stack Hei ht

• H = hs + Δh•

– Stack characteristics

–

– Physical and chemical nature of effluent

• ar ous equa ons ase on erencharacteristics, pages 162 to 166

55



Carson and Moses

• Equation 4-18

Where:

=Vs = stack gas exit velocity (m/s)

d = stack exit diameter meters

us = wind speed at stack exit (m/s)Qh = heat emission rate in kilojoules per second

56



Other basic equations

• Holland•

57



From text

ea em ss on ra e = s

Wind speed = 5 mph

Stack gas velocity = 15 m/s

Stack diameter at top is 2 mEstimate plume rise

58

Concentration Estimates for Different




amp ng mes• Concentrations calculated in previous examples based

-

• Current regulatory applications use this as 1-hour

• For other time periods adjust by:

– 3-hr multiply 1-hr value by 0.9 – 8-hr multiply 1-hr value by 0.7

– 24-hr multiply 1-hr value by 0.4

– annua mu p y - r va ue y . – .

59





amp ng mes— xamp e• For other time periods adjust by:

– - r mu p y - r va ue y .

– 8-hr multiply 1-hr value by 0.7 – 24-hr multi l 1-hr value b 0.4

– annual multiply 1-hr value by 0.03 – 0.08

Conversion of 1-hr concentration of previous example to an 8-hour average =

c = 36.4 /m3 x 0.7 = 25.5 /m3-

60



Line Sources

• Imagine that a line source, such as ahi hwa consists of an infinite number of

point sources

•elements, each representing a point source,

summed to predict net concentration

61



• When wind direction is normal to line of emission

• roun eve concentrat on ownw n

x, = ___ q ___ exp - .

(2Π)0.5 z u z2

q = source strength per unit distance (g/s * m)

Concentration should be uniform in the y-direction ata iven x

62



• For ground level (H = 0), could also use breathing

2

1

, ___ ___ - .

(2Π)0.5 z u z2

63



64From Guensler, 2000

u (wind direc

Instantaneous Release of a Puff



Instantaneous Release of a Puff

• Pollutant released quickly

• xp os on

• Accidental spill

• Release time << transport time

• Al n i n i ri i n function

65



Instantaneous Release of a Puff • E uation 4-41 to redict maximum round level

concentration

Cmax = _____2Qp____

(2Π)3/2 x

y

z

Receptor downwind would see a gradual increase inconcen ra on un cen er o pu passe an en

concentration would decrease

=

66

x y



Figure 4-9 and Table 4-7

67




68



Puff Example, .

What exposure will vehicles directly behind the tanker (downwind)receive if x =100 m? Assume very stable conditions.

From Table 4-7,

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70



Puff Example, .


From Table 4-7, y = 0.02(100m)0.89 = 1.21

rom a e - , z = . m . = .

x = y = 1.21

71



Puff Example, .


From Table 4-7, y = 0.02(100m)0.89 = 1.21

rom a e - , z = . m . = .

Cmax

= _____2Q p

____ = ____2(400000 g)_____ = 42,181 g/m3

(2Π)3/2 x y z (2Π)3/2(1.21)(1.21)(0.83)

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dispersion handout

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