Dispersion2pasquill Gifford

8/13/2019 Dispersion2pasquill Gifford

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Pasquill-Gifford

Model


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3

nx utC

222

2

1

Pasquill-Gifford Model

Cases 1 through 10 described previously depend on the

specification of a value for the eddy diffusivity, Kj. In general, Kjchanges with position, time, wind velocity, and

prevailing weather conditions and it is difficult to determine.

Sutton solved this difficulty by proposing the following

definition for a disp ersion coeff ic ient

with similar relations given for yand z. The dispersion coefficients, x, y, and zrepresent the

standard deviations of the concentration in the downwind,

crosswind and vertical (x,y,z) directions, respectively. Values

for the dispersion coefficients are much easier to obtain

experimentally than eddy diffusivities


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Table 2 A tmospheric Stabi l i ty Classes for Use

w ith the Pasqui l l -Gi fford Dispersion Model

Day radiation intensity Night cloud cover Wind

speed (m/s) Strong Medium Slight CloudyCalm &

clear

< 2 A A B B

2 3 A B B C E E

3 5 B B C C D E

5 6 C C D D D D

> 6 C D C D D

Stability class for puff model :

A,B : unstable

C,D : neutral

E,F : stable


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5

Figure 10 Horizontal dispersion coefficient for Pasquill-Gifford

plume model. The dispersion coefficient is a function of distance

downwind and the atmospheric stability class.


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6

Figure 11 Vertical dispersion coefficient for Pasquill-Gifford plume

model. The dispersion coefficient is a function of distance downwind

and the atmospheric stability class.


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Figure 13 Vertical dispersion coefficient for puff model. This data is

based only on the data points shown and should not be considered

reliable at other distances.


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Table 3 Equations and data for Pasqui l l -

Gifford Dispersion Coeff ic ients

Equations for continuous plumes

Stability class y(m)

A y= 0.493x

0.88

B y= 0.337x0.88

C y= 0.195x0.90

D y= 0.128x0.90

E y= 0.091x0.91

F y= 0.067x0.90


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10

Stability

classx (m) z(m)

A100

300

3003000Z= 0.087x0.88log10z= -1.67 + 0.902 log10x+ 0.181(log10x)

B100500

5002 104

Z= 0.135x0.95log10z= -1.25 + 1.09 log10x+ 0.0018(log10x)

C 100105 Z= 0.112x0.91

D100500

500105

Z= 0.093x0.85log10z= -1.22 + 1.08 log10x- 0.061(log10x)

E 100

50050010

5

Z= 0.082x0.82

log10z= -1.19 + 1.04 log10x- 0.070(log10x)

F100500

500105

Z= 0.057x0.80log10z= -1.91 + 1.37 log10x- 0.119(log10x)


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Data for puff releases

x= 100 m x= 4000 mStability

conditiony(m) z(m) y(m) z(m)

Unstable 10 15 300 220

Neutral 4 3.8 120 50

Very stable 1.3 0.75 35 7


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12

This case is identical to Case 7. The solution has a form similar to

Equation 33.

(38)

The ground level concentration is given atz= 0.

(39)

2

2

2

22

23

*

2

1exp

2,,,

zyxzyx

m zyutxQtzyxC

2

22

23

*

2

1exp

2,0,,

yxzyx

m yutxQtyxC

Case 11 Puff. Instantaneous point sorce at ground

level. Coordinates fixed at release point. Conatant wind

in x direction only with constant velocity u


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The ground level concentration along thex-axis is given aty=z= 0.

(40)

The centre of the cloud is found at coordinates (ut,0,0). The

concentration at the centre of this moving cloud is given by

(41)

The total integrated dose,Dtidreceived by an individual standing

at fixed coordinates (x,y,z) is the time integral of the concentration.

(42) dttzyxCzyxD ,,,,,

0tid

2

23

*

21exp

2,0,0,

xzyx

m utxQtxC

zyx

mQtutC

23

*

2,0,0,


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The total integrated dose at ground level is found by integrating

Equation 39 according to Equation 42. The result is -

(43)

The total integrated dose along thex-axis on the ground is

(44)

Frequently the cloud boundary defined by a fixed concentrationis required. The line connecting points of equal concentration

around the cloud boundary is called an isopleth.

2

2*

tid2

1exp0,,

yzy

m y

u

QyxD

u

QxD

zy

m

*

tid 0,0,


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For a specified concentration, *, the isopleths at ground level are

determined by dividing the equation for the centreline concentration,

Equation 40, by the equation for the general ground level concentration,

Equation 39. This equation is solved directly fory.

(45)

The procedure is

1. Specify *,u, andt.

2. Determine the concentrations, (x,0,0,t), along thex-axis using

Equation40. Define the boundary of the cloud along thex-axis.

3. Set (x,y,0,t) = * in Equation 45 and determine the values

ofyat each centreline point determined in step 2.

The procedure is repeated for each value oftrequired.

tyxC

txCy y

,0,,

,0,0,ln2


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This case is identical to Case 9. The solution has a form similar

to Equation 35.

(46)

The ground level concentration is given atz= 0.

(47)

2

2

2

2

21exp,,

zyzy

zyu

QzyxC

2

2

1exp0,,

yzy

y

u

QyxC

Case 12- Plume. Continuous, steady state, source

at ground level, wind moving in x direction at

constant velocity u


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The concentration along the centreline of the plume directly

downwind is given aty=z= 0.

(48)

The isopleths are found using a procedure identical to theisopleth procedure used for Case 1.

For continuous ground level releases the maximum

concentration occurs at the release point.

u

QxC

zy

0,0,


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This case is identical to Case 10. The solution has a form

similar to Equation 36.

(49)

2

2

2

1

exp2

1

exp

2

1exp

2,,

z

r

z

r

yzy

m

HzHz

y

u

QzyxC

Case 13 Plume. Continuous, Steady State

Source at Heignt H, above ground level, wind

moving in x direction at constant velocity u


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The maximum ground level concentration along the x-axis,

max, is found using.

(52)

The distance downwind at which the maximum ground level

concentration occurs is found from

(53)

The procedure for finding the maximum concentration and thedownwind distance is to use Equation 53 to determine the distance

followed by Equation 52 to determine the maximum concentration.

y

z

r

m

uHeQC

2max 2

2

r

z

H


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For this case the centre of the puff is found at x= ut. The

average concentration is given by

(54)

22

2

23

2

1exp2

1exp

2

1exp

2,,,

z

r

z

r

yzyx

m

HzHz

yQtzyxC

Case 14 Puff. Instantaneous point source at

height H, above ground level. Coordinate

system on ground moves with puff


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The time dependence is achieved through the dispersion

coefficients, since their values change as the puff moves

downwind from the release point. If wind is absent (u= 0),

Equation 54 will not predict the correct result.

At ground level,z= 0, and the concentration is computed using

(55)

22

23

*

2

1

2

1exp

2,0,,

z

r

yzyx

m HyQtyxC


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The concentration along the ground at the centreline is given at

any y=z= 0,

(56)

The total integrated dose at ground level is found by application

of Equation 42 to Equation 55. The result is

(57)

2

23

*

2

1exp

2,0,0,

z

r

zyx

m HQtxC

22*

tid2

1

2

1exp0,,

z

r

yzy

m Hy

u

QyxD


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For this case, the result is obtained using a transformation of

coordinates similar to the transformation used for Case 7. The

result is

(58)

wheretis the time since the release of the puff.

22

2

1

2

1exp

)P(,,,

56through54Equationssystem,

coordinatemovingwithequationsuff

z

r

y

Hy

tzyxC

Case 15 Puff. Instantaneous point source at

height H, above ground level. Coordinate

system fixed on ground at release point


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(59)pt

t

n

Comparison of Plume and Puff Model

Plume is based on steady state, Puff is based on

transient state The puff model can also be used for continuous

releases by representing the release as a succession

of puffs.

For leaks from pipes and vessels, iftpis the time toform one puff, then the number of puffs formed, n, is

given by


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where t is the duration of the spill. The time to form one puff,tp,

is determined by defining an effective leak height,Heff. Then,

(60)

whereuis the wind speed. Empirical results show that the best

Heffto use is

(61)

For a continuous leak,

(62)

u

Htp

eff

5.1leakofheighteff H

pmm tQQ *


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and for instantaneous release divided into a number of smaller

puffs,

(63)

where (Qm*)totalis the release amount.

This approach works for liquid spills, but not for vapor releases.

For vapor releases a single puff is suggested.

The puff model is also used to represent changes in wind speed

and direction.

nQQ mm

total**


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On an overcast day, a stack with an effective height of 60 meters is

releasing sulphur dioxide at the rate of 80 grams per second. The

wind speed is 6 meters per second.

Determine

a. The mean concentration of SO2 on the ground 500 meters

downwind.

b. The mean concentration on the ground 500 meters downwind

and 50 meters crosswind.

c. The location and value of the maximum mean concentration on

ground level directly downwind.


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a. This is a continuous release. The ground concentration directly

downwind is given by Equation 51.

(51)

From Table 2, the stability class is D. the dispersion coefficients are

obtained from Figures 10 and 11. The resulting values are y= 36

meters and z= 18.5 meters. Substituting into Equation 51

2

2

1exp0,0,

z

r

zy

m H

u

QxC

35

2

mgm1031.3

m18.5

m60

2

1exp

sm6m18.5m3614.3

sgm800,0,m500

C


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b. The mean concentration 50 meters crosswind is found using

Equation 50 and setting y = 50. The results from part aare applied

directly,

35

2

35

2

mgm1026.1

m36

m50

2

1expmgm1031.3

2

1exp0,0,m005m,0m,50500

y

yCC


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c. The location of the maximum concentration is found from

Equation 53,

From Figure 11, the dispersion coefficient has this value atx= 1500

m. At x= 1500 m, from Figure 10, y= 100 m. The maximum

concentration is determined using Equation 52,

m42.42

m60

2 rz

H

34

2

2max

mgm103.68

m100

m42.4

m60sm63.142.72

sgm802

2

y

z

r

m

uHe

QC


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Chlorine is used in a particular chemical process. A source model study

indicates that for a particular accident scenario 1.0 kg of chlorine will be

released instantaneously. The release will occur at ground level. A

residential area is 500 m away from the chlorine source. Determine

a. The time required for the centre of the cloud to reach the residential

area. Assume a wind speed of 2 m/s.

b. The maximum concentration of chlorine in the residential area.

Compare this with a TLV for chlorine of 0.5 ppm. What stability

conditions and wind speed procedures the maximum concentration?

c. Determine the distance the cloud must travel to disperse the cloud to a

maximum concentration below the TLV. Use the conditions of Part b.

d. Determine the size of the cloud, based on the TLV, at a point 5 km

directly downwind on the ground. Assume the conditions of Part b.


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a. For a distance of 500 m and a wind speed of 2 m/s, the timerequired for the centre of the cloud to reach the residential area is

This leaves very little time for emergency warning.

min2.4s250sm2

m500

u

xt


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b. The maximum concentration will occur at the centre of the cloud

directly downwind from the release. The concentration is given by

Equation 41.

(41)

The stability conditions are selected to maximize in Equation

41. This requires dispersion coefficients of minimum value. From

Figures 12 and 13, this occurs under stable condition. From Table

2, this will occur at night with a 2 - 3 m/s wind.

zyx

mQtutC

23

*

2,0,0,


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Assume a slow moving cloud of 2 m/s. from Figures 12 and 13, at

500 m, y= 5.2 m and z= 2.2 m. also assume x= y. From

equation 41,

Assuming a pressure of 1 atm and a temperature of 298K, the

concentration in ppm is 737 ppm. This is much higher than the TLVof 0.5 ppm. Any individuals within the immediate residential area,

and any personnel within the plant will be excessively exposed if

they are outside and downwind from the source.

3mmg

3mkg

mm

kg21403102.14

2.22

5.223

3.142

1.0C


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c. From Table 2 - 8, the TLV of 0.5 ppm is 1.45 mg/m or 1.4510-6

kg/m. The concentration at the centre of the cloud is given by

Equation 41. Substituting the known values,

This equation is satisfied at the correct distance from the release

point. A trial and error procedure is required. The procedure is

1. Select a distance,x.

2. Determine x, y, and zusing Figures 12 and 13.

3. Check if dispersion coefficients satisfy above equation.

342

223

36

m10768

1432

kg01mkg1045.1

.

.

.

zy

zy


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Figure 6 indicates that the release characteristics of a puff or

plume are dependent on the initial release momentum and

buoyancy. The initial momentum and buoyancy will change the

effective height of release. A release that occurs at ground level but

in an upward spouting jet of vaporizing liquid will have a greater

effectiveheight than a release without a jet. Similarly, a release ofvapor at a temperature higher than the ambient air temperature will

rise due to buoyancy effects, increasing the effectiveheight of the

release.

Both of these effects are demonstrated by the traditionalsmokestack release shown in Figure 14. The material released from

the smokestack contains momentum, based on its upward velocity

within the stack pipe, and it is also buoyant, since its temperature is

higher than the ambient temperature.


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Figure 14 Smokestack plume demonstrating initial buoyant rise of hot

gases.


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Thus, the material continues to rise after its release from the

stack. The upward rise is slowed and eventually stopped as the

released material cools and the momentum is dissipated.

For smokestack releases, Turner suggests using the empirical

Holland formula to compute the additional height due to the

buoyancy and momentum of the release,

(64)

s

ass

rT

TTPd

u

du 31068.25.1


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Building and structures provide barriers to vapor clouds and

ground releases. The behaviour of vapor clouds moving around

buildings and structures is not well understood.

Dispersion2pasquill Gifford

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