1. Section 5 Consequence Analysis 5.1 Dispersion Analysis
Process Control & Safety Group
Institute of Hydrogen Economy
Universiti Teknologi Malaysia
Dr. Arshad Ahmad
innovative ● entrepreneurial ● global

2. Accident Happens Spills of materials can lead to disaster
toxic exposure
Fire
explosion
Materials are released from holes, cracks in various plant components
Tanks, pipes, pumps
Flanges, valves,
Most accidents in chemical plants result in spills of toxic, flammable, and explosive materials. For example, material is released from holes and cracks in tanks and pipes, from leaks in flanges, pumps, and valves, and a large variety of other sources.
Source models represent the material release process. They provide useful information for determining the consequences of an accident, including the rate of material release, the total quantity released, and the physical state of the material. This information is valuable for evaluating new process designs, process improvements and the safety of existing processes. Alternatives must be considered if the source models predict unacceptable release characteristics.
Source models are constructed from fundamental or empirical equations representing the physico-chemical processes occurring during the release of materials. For a reasonably complex plant, many source models are needed to describe the required to fit the specific situation.

3. Source Model Arshad Ahmad Professor of Process Control & Safety
Director, Institute of Hydrogen Economy, UTM

4. Various types of limited aperture releases.

5. Release Mechanism Wide Aperture Limited Aperture
Release of substantial amount in short time
example: Overpressure of tank and explosion
Limited Aperture
Release from cracks, leaks etc
Relief system is designed to prevent over- pressure

6. Source Models Source models represent the material release process
Provide useful information for determining the consequences of an accident
rate of material release, the total quantity released, and the physical state of the material.
valuable for evaluating new process designs, process improvements and the safety of existing processes.

7. Release of Gasses Gas / Vapour Leak
Gasses/vapours Disperse to atmosphere
Gas / Vapour Leak
Gas / Vapour

8. Release of Liquids Vapour or Two Phase Liquid
Gasses/vapours Disperse to atmosphere
Vapour or Two Phase Liquid
Vapour
Liquid
Liquid flashes into vapour
Liquid collected as in a pool

9. Basic Models Flow of liquids through a hole
Flow of liquids through a hole in a tank
Flow of liquids through pipes
Flow of vapor through holes
Flow of vapor through pipes
Flashing liquids
Liquid pool evaporation or boiling

10. Mechanical Energy Balance
General mechanical energy balance Equation
(1)
P is the pressure (force/area),
r is the fluid density (mass/volume)
ū is the average instantaneous velocity of the fluid (length/time)
gc is the gravitational constant (length mass/force time²),
a is the unitless velocity profile correction factor - (0.5 for laminar flow, 1.0 for plug flow, >1.0 for turbulent flow)
g is the acceleration due to gravity (length/time²)
z is the height above datum (length)
F is the net frictional loss term (length force/mass)
Ws is the shaft work (force length)
m is the mass flow rate (mass/time)

11. Mechanical Energy Balance
Typical Simplifications
Incompressible Fluid
Density is constant
No elevation difference (Dz = 0)
No shaft work, Ws = 0
Negligible velocity change (small aperture), Du = 0

12. 1. Flow of Liquid Through Holes
Liquid pressurized within process units
P=Pg
Uave= 0
Dz=0
Ws=0
r = liquid density
External Surrounding
P = 1 atm
Uave, 2 = Uav
A= leak area
(2)
Liquid escaping through a hole in a process unit. The energy of the liquid due to its pressure in the vessel is converted to kinetic energy with some frictional flow losses in the hole.

13. Discharge Coefficients
The following guidelines are suggested to determine discharge coefficients
For sharp-edged orifices and for Reynolds number greater than 30,000, Co approaches the value For these conditions, the exit velocity of the fluid is independent of the size of the hole.
For a well-rounded nozzle the discharge coefficient approaches unity.
For short sections of pipe attached to a vessel (with a length-diameter ratio not less than 3), the discharge coefficient is approximately 0.81.
For cases where the discharge coefficient is unknown or uncertain, use a value of 1.0 to maximize the computed flows.

14. 2. Flow of Liquid Through a Hole in a Tank
A hole develops at a height hL below the fluid level. The flow of liquid through this hole is represented by the mechanical energy balance
assumptions: fluid is incompressible, he shaft work, Ws is zero and the velocity of the fluid in the tank is zero.
The mass discharge rate at any time t.
The time for the vessel to empty to the level of the leak, te, is

15. 3. Flow of Liquid Through Pipes
A pressure gradient across the pipe is the driving force
Frictional forces between the liquid and the wall of the pipe converts kinetic energy into thermal energy. This results in a decrease in the liquid velocity and a decrease in the liquid pressure.

16. Summation of Friction Elements
The friction term, F, is the sum of all of the frictional elements in the piping system. For a straight pipe, without valves or fitting, F is given by
(12)
where
ƒ is the Fanning friction factor (no units)
L is the length of the pipe
d is the diameter of the pipe (length)

17. Fanning Friction Factor
The Fanning friction factor, ƒ, is a function of the Reynolds number, Re, and the roughness of the pipe, e.
For laminar flow, the Fanning friction factor is given by
For turbulent flow, the data shown in Figure 6 are represented by the Colebrook equation
Refer to Figure 6 and roughness in Table 1

18. Roughness
Table 1 Roughness factor, e, for clean pipes.

19. Fanning Friction Factor
Figure 6 Plot of Fanning friction factor, f, versus Reynolds number

20. Figure 7 Plot of 1/ ƒ, versus Re  ƒ
Figure 7 Plot of 1/ ƒ, versus Re  ƒ. This form is convenient for certain types of problems. (see Example 2.)

21. Fittings, elbows etc
For piping systems composed of fittings, elbows, valves, and other assorted hardware, the pipe length is adjusted to compensate for the additional friction losses due to these fixtures. The equivalent pipe length is defined as
The summation of all of the valves, unions, elbows, and so on, are included in the computation of overall piping equivalent length (see Table 2))

22. Table 2 Equivalent pipe lengths for various pipe fittings (Turbulent flow only).

23. Table 2 Equivalent pipe lengths for various pipe fittings (Turbulent flow only)- cont’d

24. 4. Flow of Vapour Through Holes
For flowing liquids the kinetic energy changes are frequently negligible and the physical properties (particularly the density are constant.
For flowing gases and vapor these assumptions are only valid for small pressure changes (P1/P2 < 2)and low velocities ( 0.3 × speed of sound in gas).
Gas and vapor discharges are classified into throttling and free expansion releases.
For throttling releases, the gas issues through a small crack with large frictional losses; very little of energy inherent with the gas pressure is converted to kinetic energy.
For free expansion releases, most of the pressure energy is converted to kinetic energy; the assumption of isentropic behavior is usually valid.
Source models for throttling releases require detailed information on the physical structure of the leak; they will not be considered here. Free expansion release source models require only the diameter of the leak.

25. A free expansion leak is shown in Figure 9
A free expansion leak is shown in Figure 9. The mechanical energy balance, Equation 1, describes the flow of compressible gases and vapors. Assuming negligible potential energy changes and no shaft work results in a reduced form of the mechanical energy balance describing compressible flow through holes.
A discharge coefficient, C1, is defined in a similar fashion to the coefficient defined in the first section.
Equation 32 is combined with Equation 31 and integrated between any two convenient points. An initial point (denoted by subscript o) is selected where the velocity is zero and the pressure is Po. The integration is carried to any arbitrary final point (denoted without a subscript). The result is
(33)

26. Figure 9 A free expansion gas leak
Figure 9 A free expansion gas leak. The gas expands isentropically through the hole. The gas properties (P, T) and velocity change during the expansion

27. Flow of Vapour Through Holes
The resulting equation is:
The maximum flowrate is at the choke,
Here:

28. 5. Flow of Vapour Through Pipes
Vapor flow through pipes is modeled using two special cases : adiabatic or isothermal behavior.
The adiabatic case corresponds to rapid vapor flow through an insulated pipe.
The isothermal case corresponds to flow through an uninsulated pipe maintained at a constant temperature; an underwater pipeline is an excellent example.
Real vapor flows behave somewhere between the adiabatic and isothermal cases.

29. For both the isothermal and adiabatic cases it is convenient to define a Mach number as the ratio of the gas velocity to the velocity of sound in the gas at the prevailing conditions.
(41)
where a is the velocity of sound. The velocity of sound is determined using the thermodynamic relationship.
(42)
which, for an ideal gas, is equivalent to
(43)
which demonstrates that, for ideal gases the sonic velocity is a function of temperature only. For air at 20°C the velocity of sound is 344 m/s (1129 ft/s)

30. Adiabatic Flows
An adiabatic pipe containing a flowing vapor is shown in Figure For this particular case the outlet velocity is less than the sonic velocity. The flow is driven by a pressure gradient across the pipe. This expansion leads to an increase in velocity and an increase in the kinetic energy of the gas. The kinetic energy is extracted from the thermal energy of the gas; a decrease in temperature occurs. However, frictional forces are present between the gas and the pipe wall. These frictional forces increase the temperature of the gas. Depending on the magnitude of the kinetic and frictional energy terms either an increase or decrease in the gas temperature is possible.

31. Figure 11 Adiabatic, non-choked flow of gas through a pipe
Figure 11 Adiabatic, non-choked flow of gas through a pipe. The gas temperature might increase or decrease, depending on the magnitude of the frictional losses

32. the following assumptions are valid for this case :
The mechanical energy balance, Equation 1, also applies to adiabatic flows. For this case it is more conveniently written in the form
(44)
the following assumptions are valid for this case :
is valid for gases, and
From Equation 23, assuming constant f, and

33. since the pipe is adiabatic, since no mechanical linkages are present.
Since no mechanical linkages are present. An important part of the frictional loss term is the assumption of a constant Fanning friction factor, f, across the length of the pipe. This assumption is only valid at high Reynolds number.
A total energy balance is useful for describing the temperature changes within the flowing gas. For this open, steady flow process it is given by
(45)
Where h is the enthalpy of the gas and q is the heat. The following assumptions are invoked.
for an ideal gas,
is valid for gases,
since the pipe is adiabatic,
since no mechanical linkages are present.

34. The above assumptions are applied to Equations 45 and 44
The above assumptions are applied to Equations 45 and 44. The equations are combined, integrated (between the initial point denoted subscript o and any arbitrary final point), and manipulated to yield, after considerable effort,
(46)
(47)
(48)
(49)

35. Where G is the mass flux with units of mass/(area time), and
(50)
Equation 50 relates the Mach numbers to the frictional losses in the pipe. The various energy contributions are identified. The compressibility term accounts for the change in velocity due to the expansion of the gas.
Equations 49 and 50 are converted to a more convenient and useful form by replacing the Mach numbers with temperatures and pressures using Equations 46 through 48.
(51)
(52)
kinetic
energy
compressibility
pipe
friction

36. For most problems the pipe length (L), inside diameter (d), upstream temperature (T1) and pressure (P1), and downstream pressure (P2) are known. To compute the mass flux, G, the procedure is as follows.
1. Determine pipe roughness, e from Table 1. Compute e/d.
2. Determine the Fanning friction factor, f, from Equation 27. This assumes fully developed turbulent flow at high Reynolds numbers. This assumption can be checked later, but is normally valid.
3. Determine T2 from Equation 51.
4. Compute the total mass flux, G, from Equation 52.
For long pipes, or for large pressure differences across the pipe, he velocity of the gas can approach the sonic velocity. This case is shown in Figure 12. At the sonic velocity the flow will be choked. The gas velocity will remain at the sonic velocity, temperature, and pressure for the remainder of the pipe. For choked flow, Equations 46 through 50 are simplified by setting Ma2 = 1.0. The results are :

37. (53)
(54)
(55)
(56)
(57)
Choked flow occurs if the down stream pressure is less than Pchoked. This is checked using Equation 54.

38. Figure 12 Adiabatic, choked flow of gas through a pipe
Figure 12 Adiabatic, choked flow of gas through a pipe. The maximum velocity reached is the sonic velocity of the gas

39. For most problems involving choked, adiabatic flows, the pipe length (L), inside diameter (d), and upstream pressure (P1) and temperature (T1) are known. To compute the mass flux, G, the procedure is as follows.
1. Determine the Fanning friction factor, f, using Equation 27. This assumes fully developed turbulent flow at high Reynolds number. This assumption can be checked later, but is normally valid.
2. Determine Ma1, from Equation 57.
3. Determine the mass flux, Gchoked, from Equation 56.
4. Determine Pchoked from Equation 54 to confirm operation at choked conditions.

40. Isothermal Flow
Isothermal flow of gas in a pipe with friction is shown in Figure 13. For this case the gas velocity is assumed to be well below the sonic velocity of the gas. A pressure gradient across the pipe provides the driving force for the gas transport. As the gas expands through the pressure gradient the velocity must increase to maintain the same mass flowrate. The pressure at the end of the pipe is equal to the pressure of the surroundings. The temperature is constant across the entire pipe length.
Isothermal flow is represented by the mechanical energy balance in the form shown in Equation 44. The following assumptions are valid for this case.

41. is valid for gases, and
form Equation 23, assuming constant f, and
since no mechanical linkages are present. A total energy balance is not required since the temperature is constant.

42. Figure 13 Isothermal, non-choked flow of gas through a pipe.

43. Applying the above assumptions to Equation 44, and, after considerable manipulation
(58)
(59)
(60)
(61)

44. where G is the mass flux with units of mass/(area time), and,
(62)
The various energy terms in Equation 62 have been identified.
A more convenient form of Equation 62 is in terms of pressure instead of Mach numbers. This form is achieved by using Equations 58 through 60. The result is
(63)
kinetic
energy
compressibility
pipe friction

45. A typical problem is to determine the mass flux, G, given the pipe length (L), inside diameter (d), and upstream and downstream pressures (P1 and P2). The procedure is as follows.
1. Determine the Fanning friction factor, f, using Equation This assumes fully developed turbulent flow at high Reynolds number. This assumption can be checked later, but is usually valid.
2. Compute the mass flux, G, from Equation 63.
Levenspiel has shown that the maximum velocity possible during the isothermal flow of gas in a pipe is not the sonic velocity as in the adiabatic case. In terms of the Mach number, the maximum velocity is
(64)

46. This result is shown by starting with the mechanical energy balance and rearranging it into the following form.
(65)
the quantity -(dP/dL)--> when Ma --> 1/g. Thus, for choked flow in an isothermal pipe, as shown in Figure 14, the following equations apply.

47. (66)
(67)
(68)
(69)
(70)
where Gchoked is the mass flux with unit of mass/(area/time), and
(71)

48. Figure 14 Isothermal, choked flow of gas through a pipe
Figure 14 Isothermal, choked flow of gas through a pipe. The maximum velocity reached is a/.

49. For most typical problems the pipe length (L), inside diameter (d), upstream pressure (P1), and temperature (T) are known. The mass flux, G, is determined using the following procedure.
1. Determine the Fanning friction factor, f, using Equation 27. This assumes fully developed turbulent flow at high Reynolds number. This assumption can be checked later, but is usually valid.
2. Determine Ma1 from Equation 71.
3. Determine the mass flux, G, from Equation 70.

50. 6. Flashing Liquids
Liquids stored under pressure above their normal boiling point will partially flash into vapour following a leak, sometimes explosively.
Flashing occurs so rapidly that the process is assumed to be adiabatic.
The excess energy contained in the superheated liquid vaporizes the liquid and lower the temperature to the new boiling point.
m = the mass of original liquid,
Cp = heat capacity of the liquid (energy/mass deg),
To = temperature of the liquid prior to depressurization
Tb = boiling point of the liquid
Q = excess energy contained in the superheated liquid

51. Fraction of Liquid Vaporised

52. 7. Liquid Pool Evaporation or Boiling
The case for evaporation of volatile from a pool of liquid has already been considered in Chapter 3. The total mass flowrate from the evaporating pool is given by
Qm is the mass vaporization rate (mass/time),
M is the molecular weight of the pure material,
K is the mass transfer coefficient (length/time),
A is the area of exposure,
Psat is the saturation vapor pressure of the liquid,
Rg is the ideal gas constant, and
TL is the temperature of the liquid.

53. Mass Dispersion Models

54. Dispersion models
Dispersion models describe the airborne transport of toxic materials away from the accident site and into the plant and community.
After a release, the airborne toxic is carried away by the wind in a characteristic plume or a puff
The maximum concentration of toxic material occurs at the release point (which may not be at ground level).
Concentrations downwind are less, due to turbulent mixing and dispersion of the toxic substance with air.

55. Plume

56. Instantaneous release
Puff
Wind Direction
Concentrations are the Same on All Three Surfaces
Puff at time
t1> 0
Puff at time
t2> t1
Initial puff formed by
Instantaneous release
of Materials
Puff moves down wind and dissipates
By Mixing with fresh air

57. Factors Influencing Dispersion
Wind speed
Atmospheric stability
Ground conditions, buildings, water, trees
Height of the release above ground level
Momentum and buoyancy of the initial material released

58. Wind speed
As the wind speed increases, the plume becomes longer and narrower; the substance is carried downwind faster but is diluted faster by a larger quantity of air.

59. Atmospheric stability
Atmospheric stability relates to vertical mixing of the air.
During the day the air temperature decreases rapidly with height, encouraging vertical motions.
At night the temperature decrease is less, resulting in less vertical motion.
Temperature profiles for day and night situations are shown in Figure 3.
Sometimes an inversion will occur. During and inversion, the temperature increases with height, resulting in minimal vertical motion. This most often occurs at night as the ground cools rapidly due to thermal radiation.

60. Figure 3: Day & Night Condition
Air temperature as a function of altitude for day and night conditions. The temperature gradient affects the vertical air motion.

61. Ground conditions
Ground conditions affect the mechanical mixing at the surface and the wind profile with height. Trees and buildings increase mixing while lakes and open areas decrease it. Figure 4 shows the change in wind speed versus height for a variety of surface conditions.

62. Figure 4: Effect of Ground Condition
Effect of ground conditions on vertical wind gradient.

63. Height of the release above ground level
The release height significantly affects ground level concentrations.
As the release height increases, ground level concentrations are reduced since the plume must disperse a greater distance vertically. This is shown in Figure 5.

64. Figure 5 - Effect of Release Height
Continuous Release Source
Wind Direction
Plume
As Release Height Increases, This Distance
Increases. This leads to Greater Dispersion and
Less Concentration at Ground level

65. Momentum and buoyancy of the initial material released
The buoyancy and momentum of the material released changes the “effective” height of the release.
Figure 6 demonstrates these effects. After the initial momentum and buoyancy has dissipated, ambient turbulent mixing becomes the dominant effect.

66. Figure 6- Effect of Momentum and Buoyancy
Dominance of Internal Buoyancy
Dominance of Ambient Turbulence
Transition from Dominance of internal buoyancy to dominance on Ambient Turbulent
Initial Acceleration and dilution
Release Source
The initial acceleration and buoyancy of the released material affects the plume character. The dispersion models discussed in this chapter represent only ambient turbulence.

67. Neutrally Buoyant Dispersion Model
Estimates concentration downwind of a release
Two types
Plume Model
Puff Model
The puff model can be used to describe a plume; a plume is simply the release of continuous puffs.

68. Neutrally Buoyant Dispersion Model
Consider the instantaneous release of a fixed mass of material, Qm*, into an infinite expanse of air (a ground surface will be added later). The coordinate system is fixed at the source. Assuming no reaction or molecular diffusion, the concentration, C, of material due to this release is given by the advection equation.
(Eq 1)
where uj is the velocity of the air and the subscript j represents the summation over all coordinate directions, x, y, and z.

69. Case 1: Steady state continuous point release with no wind
The applicable conditions are –
Constant mass release rate, Qm = constant,
No wind, <uj> = 0,
Steady state, <C>/t = 0, and
Constant eddy diffusivity, Kj = K* in all directions.
Analytical Solution

70. Case 2: Puff with No Wind The applicable conditions are -
- Puff release, instantaneous release of a fixed mass of material, Qm* (with units of mass),
- No wind, <uj> = 0, and
- Constant eddy diffusivity, Kj = K*, in all directions.

71. Case 2: Puff With No Wind
The initial condition required to solve Equation 17 is
(18)
The solution to Equation 17 in spherical coordinates is
(19)
and in rectangular coordinates is
(20)

72. Case 3: Non Steady State, Continuous Point Release with No Wind
The applicable conditions are
- Constant mass release rate, Qm = constant,
- No wind, <uj> = 0, and
- Constant eddy diffusivity, Kj = K* in all directions
Analytical Solution in rectangular coordinates is
(22)
As t , Equations 21 and 22 reduce to the corresponding steady state solutions, Equations 15 and 16.

73. Case 4: Steady State, Continuous Point Release with No Wind
The applicable conditions are
Continuous release, Qm = constant,
Wind blowing in x direction only, <uj> = <ux> = u = constant, and
Constant eddy diffusivity, Kj = K* in all directions.
For this case, Equation 9 reduces to
(23)

74. Case 5: Puff with no wind. Eddy diffusivity a function of direction
This is the same as Case 2, but with eddy diffusivity a function of direction. The applicable conditions are -
- Puff release, Qm* = constant,
- No wind, <uj> = 0, and
- Each coordinate direction has a different, but constant eddy diffusivity, Kx, Ky and Kz.

75. Case 6 – Steady state continuous point source release with wind
Case 6 – Steady state continuous point source release with wind. Eddy diffusivity a function of direction
This is the same as Case 4, but with eddy diffusivity a function of direction. The applicable conditions are –
Puff release, Qm* = constant,
Steady state, <C>/t = o,
Wind blowing in x direction only, <uj> = <ux> = u = constant,
Each coordinate direction has a different, but constant eddy diffusivity, Kx, Ky and Kz, and.

76. Case 7- Puff with no wind
This is the same as Case 5, but with wind. The applicable conditions are –
Puff release, Qm* = constant,
Wind blowing in x direction only, <uj> = <ux> = u = constant, and
Each coordinate direction has a different, but constant eddy diffusivity, Kx, Ky and Kz,.

77. Case 8 – Puff with no wind with source on ground
This is the same as Case 5, but with the source on the ground. The ground represents an impervious boundary. As a result, the concentration is twice the concentration as for Case 5. The solution is 2 times Equation 29.
(34)

78. Case 9 – Steady state Plume with source on ground
This is the same as Case 6, but with the release source on the ground, as shown in Figure 9. The ground represents an impervious boundary. As a result, the concentration is twice the concentration as for Case 6. The solution is 2 times Equation 31.
(35)

79. Case 10 – continuous steady state source
Case 10 – continuous steady state source. Source as height Ht, above the ground
For this case the ground acts as an impervious boundary at a distance H from the source. The solution is
(36)

80. Pasquill-Gifford Model
Dr. AA
Department of Chemical Engineering
University Teknology Malaysia

81. Pasquill-Gifford Model
Cases 1 through 10 described previously depend on the specification of a value for the eddy diffusivity, Kj.
In general, Kj changes 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 dispersion coefficient
with similar relations given for sy and sz.
The dispersion coefficients, sx, sy, and sz represent 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

82. Table 2 Atmospheric Stability Classes for Use with the Pasquill-Gifford Dispersion Model
Stability class for puff model :
A,B : unstable
C,D : neutral
E,F : stable

83. Figure 10 Horizontal dispersion coefficient for Pasquill-Gifford plume model. The dispersion coefficient is a function of distance downwind and the atmospheric stability class.

84. Figure 11 Vertical dispersion coefficient for Pasquill-Gifford plume model. The dispersion coefficient is a function of distance downwind and the atmospheric stability class.

85. Figure 12 Horizontal dispersion coefficient for puff model
Figure 12 Horizontal dispersion coefficient for puff model. This data is based only on the data points shown and should not be considered reliable at other distances.

86. Figure 13 Vertical dispersion coefficient for puff model
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.

87. Table 3 Equations and data for Pasquill-Gifford Dispersion Coefficients

90. Case 11 – Puff. Instantaneous point source at ground level
Case 11 – Puff. Instantaneous point source at ground level. Coordinates fixed at release point. Constant wind in x direction only with constant velocity u
This case is identical to Case 7. The solution has a form similar to Equation 33.
(38)
The ground level concentration is given at z = 0.
(39)

91. The ground level concentration along the x-axis is given at y = 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, Dtid received by an individual standing at fixed coordinates (x,y,z) is the time integral of the concentration.
(42)

92. 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 the x-axis on the ground is
(44)
Frequently the cloud boundary defined by a fixed concentration is required. The line connecting points of equal concentration around the cloud boundary is called an isopleth.

93. For a specified concentration, <C>
For a specified concentration, <C>*, 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 for y.
(45)
The procedure is
1. Specify <C>*, u, and t.
2. Determine the concentrations, <C> (x,0,0,t), along the x-axis using Equation40. Define the boundary of the cloud along the x-axis.
3. Set <C> (x,y,0,t) = <C>* in Equation 45 and determine the values of y at each centreline point determined in step 2.
The procedure is repeated for each value of t required.

94. Case 12- Plume. Continuous, steady state, source at ground level, wind moving in x direction at constant velocity u
This case is identical to Case 9. The solution has a form similar to Equation 35.
(46)
The ground level concentration is given at z = 0.
(47)

95. The concentration along the centreline of the plume directly downwind is given at y = z= 0.
(48)
The isopleths are found using a procedure identical to the isopleth procedure used for Case 1.
For continuous ground level releases the maximum concentration occurs at the release point.

96. Case 13 – Plume. Continuous, Steady State Source at Heignt H, above ground level, wind moving in x direction at constant velocity u
This case is identical to Case 10. The solution has a form similar to Equation 36.
(49)

97. The ground level concentration is found by setting z = 0.
(50)
The ground centreline concentrations are found by setting y = z= 0.
(51)

98. The maximum ground level concentration along the x-axis, <C>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 the downwind distance is to use Equation 53 to determine the distance followed by Equation 52 to determine the maximum concentration.

99. Case 14 – Puff. Instantaneous point source at height H, above ground level. Coordinate system on ground moves with puff
For this case the centre of the puff is found at x = ut. The average concentration is given by
(54)

100. 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)

101. 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)

102. Case 15 – Puff. Instantaneous point source at height H, above ground level. Coordinate system fixed on ground at release point
For this case, the result is obtained using a transformation of coordinates similar to the transformation used for Case 7. The result is
(58)
where t is the time since the release of the puff.

103. End of Section 5.1