Section 5
Consequence Analysis
5.1 Dispersion Analysis
Process Control & Safety Group
Institute of Hydrogen Economy
Universiti Teknologi Malaysia
Dr. Arshad Ahmad
Email: arshad@cheme.utm.my
www.utm.my
innovative ● entrepreneurial ● global
1
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,
Source Model
Arshad Ahmad
Professor of Process Control & Safety
Director, Institute of Hydrogen Economy, UTM
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Various types of limited aperture releases.
Release Mechanism
• Wide 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 overpressure
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.
Release of Gasses
Gasses/vapours Disperse
to atmosphere
Gas / Vapour Leak
Gas / Vapour
Release of Liquids
Gasses/vapours Disperse
to atmosphere
Vapour or Two Phase Liquid
Vapour
Liquid
• Liquid flashes into vapour
• Liquid collected as in a pool
Basic Models
1.
2.
3.
4.
5.
6.
7.
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
Mechanical Energy Balance
General mechanical energy balance Equation

•
•
•
•
•
•
•
•
•
•
 u²
 
 2ag
r
c

dP

Ws
g


z

F


 g
m
c

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)
(1)
Mechanical Energy Balance
Typical Simplifications
• Incompressible Fluid
– Density is constant
dP
 r
• No elevation difference (z = 0)
• No shaft work, Ws = 0
• Negligible velocity change (small
aperture), u = 0

P
r
1. Flow of Liquid Through Holes
Liquid pressurized
within process
units
P=Pg
Uave= 0
z=0
Ws=0
r = liquid density
Qm  ru ave A  AC O
External Surrounding
P = 1 atm
Uave, 2 = Uav
A= leak area
2 rg c Pg
(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.
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 0.61. 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.
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.
Qm
2


 g c Pg

r
gC
o
o A²
 rC o A 2
t
 r  ghL 
 


A
t

 

The time for the vessel to empty to the level of the leak, te, is
 g c Pg
1  At  
o 


te 
2

gh



L 


Co g  A  
 r


2 g c Pg 

r 

3. Flow of Liquid Through Pipes
P
2
ws
u
g


z  F  
r
2ag c
gc
m
•
(11)
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.
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
2 fLu
F 
gcd
2
where
ƒ is the Fanning friction factor (no units)
L is the length of the pipe
d is the diameter of the pipe (length)
(12)
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 f  16
Re
• For turbulent flow, the data shown in Figure 6
are represented by the Colebrook equation
 1 
1
1.25 5
 4 lo g

 3.7 d
f
Re
f





• Refer to Figure 6 and roughness in Table 1
Roughness
Table 1 Roughness factor, , for clean pipes.
Pipe material
, mm
Riveted steel
1 – 10
Concrete
0.3 – 3
Cast iron
0.26
Galvanized iron
0.15
Commercial steel
Wrought iron
Drawn tubing
0.046
0.046
0.0015
Glass
0
Plastic
0
Fanning Friction Factor
Figure 6 Plot of Fanning friction factor, f, versus Reynolds number
Figure 7 Plot of 1/ ƒ, versus Re  ƒ. This form is convenient for certain
types of problems. (see Example 2.)
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
Lequiv  Lstraight   Lequiv
total
pipe
• 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))
Table 2 Equivalent pipe lengths for various pipe
fittings (Turbulent flow only).
Pipe fitting
Lequiv/d
Globe valve, wide open
~300
Gate valve, wide open
¾ open
½ open
¼ open
90º elbow, standard
45º elbow, standard
Tees
Used as elbow, entering the stem
Used as elbow, entering one of two
sides
Straight through
Pipe connections to vessels
Ordinary, pipe flush with wall
Borda, pipe protruding into vessel
Rounded entrance, union, coupling
~7
~40
~200
~900
30
15
90
60
20
16
30
~0
Table 2 Equivalent pipe lengths for various
pipe fittings (Turbulent flow only)- cont’d
Sudden enlargement from d to D
Laminar flow in d:
Re
32

 d ² 
1




D
²



2
Turbulent flow in d:
f ind  
 d ² 
1




4
D
²



2
Sudden contraction from D to d (except
choked gas flow)
Re 
 d ² 
Laminar flow in d
1
.
25



160 
D
²



Turbulent flow in d:
f ind  
 d ² 
1
.
25



10 
D
²



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.
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.
 u² 
 r   2ag c   F  0
A discharge coefficient, C1, is defined in a similar fashion to the
dP
coefficient defined in the first section.

dP 

 F  C12 



r
r 


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

dP
2
1
C
P
dP
Po
r


u²
0
2ag c
(33)
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
Flow of Vapour Through Holes
• The resulting equation is:
Qm  C o APo
2
 1 


 P 
2 g c M   P 








R g To   1  Po 
Po 



Here:
  C P CV
• The maximum flowrate is at the choke,
Qm choked
 C o APo
g c M 
Rg To
2 

  1



 1  1




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.
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.
u
Ma 
a
(41)
where a is the velocity of sound. The velocity of sound is
determined using the thermodynamic relationship.
a
 P 
gc 
 r 


S
(42)
which, for an ideal gas, is equivalent to
a 
g c R g T M
(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)
Adiabatic Flows
An adiabatic pipe containing a flowing vapor is shown in Figure
11. 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.
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
The mechanical energy balance, Equation 1, also applies to
adiabatic flows. For this case it is more conveniently written in the
form
dP
r

Ws
ud u
g

dz  dF  
ag c
gc
m
the following assumptions are valid for this case :
g
dz  0
gc
is valid for gases, and
dF 
2 f u ² dL
gc d
From Equation 23, assuming constant f, and
Ws  0
(44)
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
Ws
ud u
g
dh 

dz  q 
ag c
gc
m
Where h is the enthalpy of the gas and q is the heat. The
following assumptions are invoked.
dh  C p dT for an ideal gas,
g
dz  0
gc
q  0
WS  0
is valid for gases,
since the pipe is adiabatic,
since no mechanical linkages are present.
(45)
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,
T2
Y1
γ 1

whereY1  1 
Mai2
T1
Y2
2
(46)
P2
Ma1

P1
Ma 2
Y1
Y2
(47)
r2
Ma1

r1
Ma 2
Y2
Y1
(48)
G  r u  Ma1 P1
g c M
R g T1
 Ma 2 P2
g c M
R g T2
(49)
Where G is the mass flux with units of mass/(area time), and
 1
2
 Ma22Y1
ln
 Ma 2Y
1
2

kinetic
energy
  1
1 
 4 fL 








0
2 
  Ma 2
Ma2 
 d 
1
 
compressibility
(50)
pipe
friction
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.
  1 P1T2   1  P12T22  P22T12

ln


P2T1
2 
T2  T1

G
 1
1



2
 P 2T
P
2 T1
 1 2
2g c M 
T2  T1
R g   1 T1 P1 2  T2 P2 2
 4 fL(51)

 d 0

(52)
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,  from Table 1. Compute
/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 :
(53)
Tchoked
2Y1

T1
 1
Pchoked
 Ma1
P1
2Y1
 1
(54)
r choked
 1
 Ma1
r1
2Y1
(55)
Gchoked  r u  Ma1 P1
 1
2
g c M
R g T1
 Pchoked
g c M
(56)
R g Tchoked

  1

2Y1
 4 fL 


ln 


1



0
2 
2


 d 
   1Ma1   Ma1

(57)
Choked flow occurs if the down stream pressure is less than
Pchoked. This is checked using Equation 54.
Figure 12 Adiabatic, choked flow of gas through a pipe. The maximum velocity
reached is the sonic velocity of the gas
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.
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.
g
dz  0
gc
is valid for gases, and
dF 
2 f u ² dL
gc d
form Equation 23, assuming constant f, and
WS  0
since no mechanical linkages are present. A total energy
balance is not required since the temperature is constant.
Figure 13 Isothermal, non-choked flow of gas through a pipe.
Applying the above assumptions to Equation 44, and, after
considerable manipulation
T2  T1
(58)
P2
Ma1

P1
Ma 2
(59)
r2
Ma1

r1
Ma 2
G  r u  Ma1 P1
(60)
g c M
Rg T
(61)
where G is the mass flux with units of mass/(area time), and,
2 ln
Ma2
1 1
1  4 fL

 

0
2
2 

Ma1
  Ma1
d
Ma2 
kinetic
energy
(62)
pipe
friction
compressibility
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)
2 ln


g M
P1
4 fL
 2c
P12  P22 
0
P2
d
G Rg T
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
27. 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
Machoked 
1

(64)
This result is shown by starting with the mechanical energy
balance and rearranging it into the following form.
dP
2 fG ²


dL
g c rd


1
2 fG ²


g c rd
1  u ² r / g c P 




1
1  Ma² 


(65)
the quantity -(dP/dL)--> when Ma --> 1/. Thus, for choked
flow in an isothermal pipe, as shown in Figure 14, the following
equations apply.
Tchoked  T1
Pchoked
 Ma1
P1
(66)

(67)
r choked
Ma1 
r1
(68)
(69)
u choked
1

u1
Ma1 
Gchoked  r u  r 1 u 1  Ma1 P1
g c M
Rg T
 Pchoked
gc M
Rg T
(70)
where Gchoked is the mass flux with unit of mass/(area/time), and
 1
ln
 Ma 2
1

  1
 4 fL


   Ma 2  1
 d 0
1
 

(71)
Figure 14 Isothermal, choked flow of gas through a pipe. The maximum
velocity reached is a/.
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.
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.
Q  mCp To  Tb 
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
Fraction of Liquid Vaporised

f v  1  exp  C p To  Tb  v

C p is themean heat capacity
H V is themean latentheat of vaporisation
mV
fV 
is thefractionof liquid vaporised
m
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
sat
MKAP
Qm 
RgTL
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.
Mass Dispersion
Models
53
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.
54
Plume
55
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
56
Factors Influencing Dispersion
• Wind speed
• Atmospheric stability
• Ground conditions, buildings, water, trees
• Height of the release above ground level
• Momentum and buoyancy of
released
the initial material
57
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.
58
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.
59
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.
60
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.
61
Figure 4: Effect of Ground Condition
Effect of
gradient.
ground conditions on vertical
wind
62
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.
63
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
64
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.
65
Figure 6- Effect of Momentum and Buoyancy
Initial Acceleration
and dilution
Release
Source
Dominance of
Internal Buoyancy
Dominance of
Ambient
Turbulence
Transition from Dominance
of internal buoyancy to
dominance on Ambient
Turbulent
The initial acceleration and buoyancy of the released material
affects the plume character. The dispersion models discussed in
this chapter represent only ambient turbulence.
66
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.
67
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.
C

u j C   0

t
x j
(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.
68
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.
2 C
x
2

2 C
y
2

2 C
z
2
0
• Analytical Solution
C x, y, z  
Qm
4K *
x2  y2  z 2
69
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.
 C
 C
 C
1  C



*
2
2
t
K
x
y
z 2
2
2
2
70
Case 2: Puff With No Wind
The initial condition required to solve Equation 17 is
(18)
C x, y, z   0 at t  0
The solution to Equation 17 in spherical coordinates is
C r , t  

Qm*
8 K t
*

3
2

r2 

exp 
* 
 4K t 
(19)
and in rectangular coordinates is
C  x, y , z , t  

Qm*
8 K t
*

3

2

 x2  y2  z2 
exp

*
4
K
t


(20)
71
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
C
r , t 

Qm

erfc
*

4K r
2



K *t 
r
Analytical Solution in rectangular coordinates is
C x, y, z, t  

erfc

x2  y2  z 2

Qm
4K *
x 2  y 2  z 2  (22)

2 K *t

As t , Equations 21 and 22 reduce to the corresponding
steady state solutions, Equations 15 and 16.
72
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
 C
 C
 C
u  C



*
2
2
x
K
x
y
z 2
2
C  x, y , z  
2
2
(23)
Qm
4K *
x2  y2  z2
u

exp 
*
 2K



x2  y2  z2  x 

73
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.
 C
t
 Kx
2 C
x
C x, y, z, t  
2
 Ky
8t 
3 2
2 C
y
2
 Kz
2 C
z 2
2
2 
 1  x2
y
z


exp


Ky
K z 
KxKyKz
 4t  K x
Qm
74
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.
u
 C
x
 Kx
2 C
x
2
 Ky
2 C
y
2
 Kz
2 C
z 2
2 
 u  y2
z


C x, y, z  
exp

K z 
4x K y K z
 4 x  K y
Qm
75
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,.
C  x, y , z , t  
8t 
3 2
Qm*
KxKyKz
2

y2
z 2 
 1   x  ut 

exp




4t  K x
Ky
K z  



76
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.
C  x, y , z , t  
4t 
3 2
Qm*
KxKyKz
2
2 
 u  x2
y
z


exp


Ky
K z 
 4t  K x
(34)
77
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.
C x, y, z  
Qm
2x K x K y
2 
 u  y2
z


exp

K z 
 4 x  K y
(35)
78
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

uy 2 

C  x, y , z  
exp 
4x K y K z
 4K z x 
Qm



u
u

2
2 
z  H r    exp
z  H r   
 exp

 4K z x

 4K z x


(36)
79
Pasquill-Gifford
Model
Dr. AA
Department of Chemical Engineering
University Teknology Malaysia
80
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

2
x
1

C
2
2
ut 2 n
• with similar relations given for y and z.
• The dispersion coefficients, x, y, and z 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
81
Table 2 Atmospheric Stability Classes for Use with
the Pasquill-Gifford Dispersion Model
Day radiation intensity
Night cloud cover
Wind
speed (m/s)
Strong
Medium
Slight
<2
A
A–B
B
2–3
A–B
B
3–5
B
5–6
>6
Cloudy
Calm &
clear
C
E
E
B–C
C
D
E
C
C–D
D
D
D
C
D
C
D
D
Stability class for puff model :
A,B : unstable
C,D : neutral
E,F : stable
82
Figure 10 Horizontal dispersion coefficient for Pasquill-Gifford
plume model. The dispersion coefficient is a function of distance
downwind and the atmospheric stability class.
83
Figure 11 Vertical dispersion coefficient for Pasquill-Gifford plume
model. The dispersion coefficient is a function of distance downwind
and the atmospheric stability class.
84
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.
85
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.
86
Table 3 Equations and data for Pasquill-Gifford
Dispersion Coefficients
Equations for continuous plumes
Stability class
y (m)
A
0.88
=
0.493x
y
B
y = 0.337x
0.88
C
y = 0.195x
0.90
D
0.90
=
0.128x
y
E
y = 0.091x
0.91
F
y = 0.067x
0.90
87
Stability
class
x (m)
A
100 – 300
300 – 3000
Z = 0.087x0.88
log10z = -1.67 + 0.902 log10x + 0.181(log10x)²
B
100 – 500
500 – 2 × 104
Z = 0.135x0.95
log10z = -1.25 + 1.09 log10x + 0.0018(log10x)²
C
100 – 105
Z = 0.112x0.91
D
100 – 500
500 – 105
Z = 0.093x0.85
log10z = -1.22 + 1.08 log10x - 0.061(log10x)²
E
100 – 500
500 – 105
Z = 0.082x0.82
log10z = -1.19 + 1.04 log10x - 0.070(log10x)²
F
100 – 500
500 – 105
Z = 0.057x0.80
log10z = -1.91 + 1.37 log10x - 0.119(log10x)²
z (m)
88
Data for puff releases
Stability
condition
x = 100 m
x = 4000 m
y (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
89
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.
C  x, y , z , t  
2




y2
z 2 
 1 x  ut





exp



(38)

2
2 


3 2
2   x 
 y  z  
2  x y z




Qm*
The ground level concentration is given at z = 0.
C  x, y,0, t  
2


Q
y 2 
 1  x  ut 





exp



2 
 

2

2 3 2 x y  z

x

y 




*
m
(39)
90
The ground level concentration along the x-axis is given at y =
z= 0.
C x,0,0, t  
 1  x  ut  2 
 
exp 
3 2
2  x y z
 2   x  
Qm*
(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
C ut ,0,0, t  
Qm*
2
 x y  z
(41)
3 2
The total integrated dose, Dtid received by an individual standing
at fixed coordinates (x,y,z) is the time integral of the concentration.
Dtid x, y, z  


0
C x, y, z, t dt
(42)
91
The total integrated dose at ground level is found by integrating
Equation 39 according to Equation 42. The result is  1 y2 

Dtid x, y,0 
exp 
2
 2 
 y z u
y 

Qm*
(43)
The total integrated dose along the x-axis on the ground is
Dtid x,0,0 
Qm*
 y z u
(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.
92
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.
y y
 C x,0,0, t  

2 ln



C
x
,
y
,
0
,
t


(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.
93
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.
2 
 1  y2
z
C x, y, z  
exp  2  2 
 y z u
 2   y  z 
Q
(46)
The ground level concentration is given at z = 0.
2

 y  
Q
1
 
C  x, y ,0  
exp 
 y  z u
 2   y  


(47)
94
The concentration along the centreline of the plume directly
downwind is given at y = z= 0.
C x,0,0 
Q
(48)
 y z u
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.
95
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.
2

 y  
Qm
1
 
C  x, y , z  
exp 

2 y z u
 2   y  



 1  z  Hr

 exp 

 2 z

 1 z H

r
  exp 
 2   z

(49)



2



 

96
The ground level concentration is found by setting z = 0.
 1 y
Qm
C  x, y , z  
exp 
2 y z u
 2   y

2
2

 Hr  
1
  
 


2z  


(50)
The ground centreline concentrations are found by setting y =
z= 0.
 1H
Qm
C x,0,0 
exp  r
 y z u
 2   z



2



(51)
97
The maximum ground level concentration along the x-axis,
<C>max, is found using.
C
max
2Qm

euH r2
z


 y




(52)
The distance downwind at which the maximum ground level
concentration occurs is found from
z 
Hr
(53)
2
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.
98
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
2




Qm
1
y
 
C  x, y , z , t  
exp 
32
 2   y  
2   x y z


2
2








1 z  Hr
1 z  H r 


   exp 
  
 exp 
 2   z  
 2   z   



(54)
99
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
 1 y
Q
C x, y,0, t  
exp 
32
 2   y
2  x y z

*
m
2
 1  Hr  
  
 

 2 
 z  

2
(55)
100
The concentration along the ground at the centreline is given at
any
y = z = 0,
 1H
Q
r


C x,0,0, t  
exp


2
2 3 2 x y z
  z
*
m



2



(56)
The total integrated dose at ground level is found by application
of Equation 42 to Equation 55. The result is
 1 y
Q
Dtid x, y,0 
exp 
 y z u
 2   y

*
m
2
 1  Hr  
  
 

 2 
 z  

2
(57)
101
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 C  x, y, z , t   (Puff equationswith movingcoordinate
system,Equations54 through56)
 1 y
 exp 
 2   y

2


  1  Hr


2
 z




2




(58)
where t is the time since the release of the puff.
102
End of Section 5.1
103