Consequence Analysis
2009_2nd semester
En Sup Yoon
Introduction
Objective



To review the range of models available for consequence
analysis
Provide an overview of consequence and effect models
commonly used in CPQRA
Consists of source models, dispersion models and effect models
Overall logic diagram for
the consequence model for
release of volatile
hazardous substance
Discharge model
Define the release scenario by estimating discharge rate,
total quantity released, total release duration, extent of
flash and evaporation from a liquid pool and aerosol
formation
Objective

Prediction of the final state of the release as the material emerges
into the atmosphere
Input

Temperature, pressure, phase, liquid fraction etc.
Output

Mass flow rate, duration, pseudo-velocity, discharge velocity,
temperature, liquid fraction, droplet trajectories and size
Logic diagram for
discharge and
dispersion model
Discharge Rate Model-1
First step in this procedure

Determine an appropriate scenario
 Liquid discharge


Hole in atmospheric storage tank or other atmospheric pressure vessel or
pipe under liquid head
Hole in vessel or pipe containing pressurized liquid below its normal
boiling point
 Gas discharge





Hole in equipment containing gas under pressure
Relief valve discharge
Boiling-off evaporation from liquid pool
Relief valve discharge from top of pressurized storage tank
Generation of toxic combustion products as a result of fire
 Two-phase discharge


Hole in pressurized storage tank or pipe containing a liquid above its
normal boiling point
Relief valve discharge
Discharge Rate Model-2
Release Phase

Depend on the release process and be determined by using
thermodynamic diagram, data or vapor-liquid equilibrium
model
Thermodynamic path and endpoint

Play a important role to the development of the source model
Hole size

A primary input to discharge calculate
Leak duration

10-min leak duration (LNG Federal Safety Standards) or 3-min
are generally used
Hole Size Selection
Methodologies for hole size selection



Generally hole having 20% or 100% of pipe (World Bank(1995))
Generally 2 and 4-inch hole are proposed, regardless of pipe size
More detail procedure
 For small bore piping up to 11/2”, use 5-mm
 For 2-6” piping, use 5-mm, 25-mm
 For 8-12” piping, use 5-mm, 25-mm, 100mm
Governing Equation
Energy balance equation
P2
Ws
1
dP g
2
2
(z 2 − z1 ) +
(ν 2 − ν1 ) + ∑ e f +
+
=0
∫

m
2g c
p1 ρ g c
P is the pressure (force/area)
Ρ is the density (mass/volume)
g is the acceleration due to gravity (length/time2)
gc is the gravitational constant (force/mass-acceleration)
z is the vertical height from some datum (length)
v is the fluid velocity (length/time)
f is the frictional loss term (length2/time2)
Ws is the shaft work (mechanical energy/time)
m is the mass flow rate (mass/time)
(1)
Frictional loss term
 ν2 

e f = K f 
 2g c 


Kf is the head loss due to th pipe fitting
v is the fluid velocity
4fL
Kf =
D



f is the fanning friction factor(unitless)
L is the flow path length
D is the flow path diameter
2-K Methods-1
Define excess head loss using Reynolds number and the
pipe internal diameter



inches 

1

K
=
+
+
K N K 1 ID

1
∞
f
Re
 K1 and K ∞ are constants(dimensionless)
 NRe is the Reynolds number(dimensionless)
 IDinches is the internal diameter of the flow path(inches)

Metric form


25.4


K
K = N + K 1 + ID 


1
∞
f
Re
mm
 IDmm is the internal diameter in mm

Table 2.4 contains a list of K value
2-K Methods-2
Fanning friction factor
1
f
ε D
5.0452 log
= − 4 log
−
3.7065
N



10


A


=



Re
(ε D)
A 
10



0.8981
 7.149 


2.8257  N Re 
1.1098
+





At high Reynolds number(fully developed turbulent
flow), the friction factor is independent of the Reynolds
number
1 = − log 
D 
 3.7

4

Ε is the roughness factor
ε 
f

10
Table 2.5
Liquid Discharge-1
For liquid discharge, the driving force for the discharge
is normally pressure

Pressure energy being converted to kinetic energy
P − P + g ( − )+ 1  −  + ∑ e + W = 0
ρ
g z z 2g  v2 v1 
m
2
2
1
2
1
•
f
C

2
s
C
If the flow is considered frictionless and no shaft work, the result
equation is called the Bernoulli equation

P − P + g ( − )+ 1 
−v =0
z
z
v
2
1

ρ
g
2g 
2
2
1
2
C
1
C
2
Liquid Discharge-2
Discharge pure liquid(non-flashing) through sharp-edges orifice

Eq.(1) by assuming that the frictional loss term is represented by a
discharge coefficient, CD
m = AC 2ρρ (P − P )
•
D







C
1
2
m is the liquid discharge rate(mass/time)
A is the area of the hole(length2)
CD is the discharge coefficient(dimensionless)
ρ is the density of fluid(mass/volume)
gc is the gravitational constant(force/mass-acceleration)
P1 is the pressure upstream of the hole(force/area)
P2 is the pressure downstream of the hole(force/area)
Discharge Coefficient, CD
Guidance for selection of discharge coefficient(Lee, 1986)




For sharp-edged orifice and for Reynolds numbers greater than
30,000 the discharge coefficient approaches the value 0.61. For
these condition the exit velocity is independent of the hole size
For a well-rounded nozzle the discharge coefficient approaches
unity
For short sections of pipe attached to a vessel with a lengthdiameter 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 flow to
achieve a conservative result
Liquid Discharge-3
Discharge coefficient calculation using 2-K method
C
D


∑K
=
1
1 + ∑K
f
is the sum of all excess head loss term including entrances,
exits, pipe lengths and fitting
For a simple hole in a tank, with no pipe connections or fitting,
the friction is cause only by the entrance and exit effect the hole
f
 Example

For Reynolds numbers greater than 10,000, Kf=0.5 for the entrance and
Kf=1.0 for the exit, thus the sum of Kf is 1.5. So CD is 0.63
Solution Procedure
Solution procedure to determine the mass flow rate of
discharged material from a piping system









Given ; length, diameter, type of pipe, pressure and elevation
changes across the piping system
Specify the initial point and final point
Determine the pressure and elevation at point 1 and 2
Determine the initial fluid velocity at point 1
Guess a value for the velocity at point 2
Determine the friction factor for the pipe
Determine the excess head loss terms for the pipe
Compute values for all of the terms in governing eqn.
Determine the mass flow rate using eqn. m = ρvA
Liquid Discharge-4
(Hole in Tanks)
Tank drainage time
v dV(h)
1
t=
∫
AC 2g v h
1
D




2
Assume a constant leak area and a constant discharge coefficient
CD
t is the time to drain the tank from volume V2 volume V1 (time)
V is the liquid volume in the tank above the leak (length3)
h is the height of the liquid above the leak (length)
Liquid Discharge-5
(Hole in Tanks)
Mass discharge rate of liquid from a hole in a tank(Crowl and
Louvar, 1990)





gP+
=
=
m ρvA ρAC 2 ρ gh
•
D









C
q



L


m is the mass discharge rate (mass/time)
v is the fluid velocity(length/time)
A is the area of the hole(length2)
CD is the mass discharge coefficient(dimensionless)
gc is the gravitational constant(force/mass acceleration)
Pg is the gauge pressure at the top of the tank(force/area)
ρ is the liquid density(mass/volume)
g is the acceleration due to gravity(length/time2)
hL is the height of liquid above the hole(length)
Gas Discharge-1
Source



From a hole at or near a vessel
From a long pipeline
From relief valve or process vent
The majority of gas discharge from process plant leaks will initially
be sonic or choked
Rate equation for sonic and subsonic discharge through an orifice






AIChE/CCPS (1987, 1995)
API (1996)
Crane Co. (1986)
Crowl and Louvar (1990)
Fthenakis (1993)
Gerry and Green (1984)
Gas Discharge-2
Gas discharge through holes
2g M k
m C AP R T k −1
•
=
C
D
1
q








1






(k −1 ) k 
2 k
 P2 
 
 P1 
−
 P2 
 
 P1 





m is the mass flow rate of gas through the hole(mass/time)
CD is the discharge coefficient(dimensionless)
A is the area of the hole(length2)
P1 is the pressure upstream of the hole(force/area)
P2 is the pressure downstream of the hole(force/area)
gc is the gravitational constant(force/mass acceleration)
M is the molecular weight of the gas(mass/mole)
k is the heat capacity ratio, CP/CV(unitless)

Table 2.6 Heat capacity ratio k for selected gas
 Rg is the ideal gas constant(pressure-volume/mole-deg)
 T1 is the initial upstream temperature of the gas(deg)
Gas Discharge-3
Sonic(choked flow)
2g M  2
=
mchoked C AP R T 
 k +1
•
C
1
D
q
1
(k +1) (k −1)




The pressure ratio required to achieve choking
P
P
choked
1
 2
=
 k +1








k
(k −1) 
 An upstream pressure of greater than 13.1 psig for an ideal gas is
adequate to produce choked flow for a gas to atmosphere
 For real gas, a pressure of 20 psig is typically used
Two-Phase Discharge-1
Two-phase flow are classified as following

Reactive flow
 Typical case of emergency reliefs of exothermic chemical reaction

Nonreactive flow
 The flashing of liquid as they are discharged from containment
⋅
m= A
2
GSUB
2
G ERM
+
N
 m is the two-phase mass discharge rate(mass/time)
 A is the area of the discharge(length2)
 GSUB is the subcooled mass flux(mass/area time)
 GERM is the equilibrium mass flux(mass/ area time)
 N is a non equilibrium parameter
Two-Phase Discharge-2
The subcooled mass flux
G





SUB
=
C 2ρ g  P − P
sat 
D
f
c


CD is the discharge coefficient(unitless)
ρf is the density of the liquid(mass/volume)
gc is the gravitational constant(force/mass acceleration)
P is the storage pressure(force/area)
Psat is the saturation vapor pressure of the liquid at ambient
temperature)force/area)
Two-Phase Discharge-3
The equilibrium mass flux
G
ERM
=
h
v
fg
fg




g
TC
c
P
hfg is the enthalpy change on vaporization(energy/mass)
vfg is the change in specific volume between liquid and
vapor(volume/mass)
T is the storage temperature(absolute degrees)
CP is the liquid heat capacity(energy/mass deg)
Available Computer Code
Pipe flow



AFT Fathom (Applied Flow Technology)
Crane Companion (Crane ABZ)
INPLANT (Simulation Science Inc.)
Two-phase flow


DEERS Klein two-phase flashing discharges (JAYCOR Inc.)
SAFIRE (AIChE)
Several integrated analysis package containing discharge module







ARCHIE (Environmental Protection Agency)
EFFECT-2 (TNO)
FOCUS+ (Quest Consultants)
PHAST (DNV)
QRAWorks (Primatech)
AUPERCHEMS (Arthur D. Little)
TRACE (Safer Systems)
Flash and Evaporation-1
Purpose

To estimate the total vapor or vapor rate that forms a cloud, for
use as input to dispersion models
Equation for flash fraction
(T − Tb )
FV = C P
h fg





CP is the heat capacity of the liquid(energy/mass deg)
T is the initial temperature of the liquid(deg)
Tb is the atmospheric boiling point of the liquid(deg)
hfg is the latent heat of vaporization of the liquid at
Tb(energy/mass)
FV is the mass fraction of released liquid vaporized(unitless)
Flash and Evaporation-2
Effect on cloud dispersion of aerosol entrainment




The cloud will have a larger total mass
There will be an aerosol component
Evaporating aerosol can reduce the temperature below the
ambient atmospheric temperature
The colder cloud temperature may cause additional
condensation of atmospheric moisture
Evaporation

Evaporation from liquid spill onto land and water has received
substantial attention
Flash and Evaporation-3
Evaporation from a pool
⋅
dT
mC P
= H − Lm
dt







m is the mass of the pool(mass)
CP is the heat capacity of the liquid(energy/mass deg)
T is the temperature of the liquid in the pool(deg)
t is the time(time)
H is the total heat flux into the pool(energy/time)
L is the heat of vaporization of the liquid(energy/mass)
m is the evaporation rate(mass/time)
Flash and Evaporation-4
Modeling divided into two classes


Low volatile liquid
High volatile liquid
For the high volatile case (assuming steady state)
⋅
H
m=
L



(1)
m is the vaporization rate(mass/time)
H is the total heat flux to the pool(energy/time)
L is the heat of vaporization of the pool(energy/mass)
Flash and Evaporation-5
For liquids having normal boiling points near or above
ambient temperature
m mass =
Mk g AP sat
(3)
R g TL
 mmass is the mass transfer evaporation rate(mass/time)
 M is the molecular weight of the evaporating material(mass/mole)
 kg is the mass transfer coefficient(length/time)
 A is the area of the pool(area)
 Psat is the saturation vapor pressure of the liquid(force/area)
 Rg is the ideal gas constant(pressure volume/mole deg)
 TL is the temperature of the liquid(abs. Deg)
Flash and Evaporation-6
Direct evaporation model includes an evaporation rate
due to solar radiation
msolar





Qsol MA
=
HV
(2)
msol is the evaporation rate(mass/time)
Qsol is the solar radiation(energy/area-time)
M is the molecular weight(mass/mole)
A is the pool area(area)
HV is the heat vaporization of the liquid(energy/mole)
Flash and Evaporation-7
Combined equation (1), (2) represent evaporation due to
mass transfer
m tot



 β 
 1 

 + m mass 
= m sol 
1+ β 
1+ β 
mtot is the net evaporation rate(mass/time)
β is a parameter which is a function of vapor
pressure(dimensionless)
mmass is the mass transfer evaporation rate given by eqn. (3)
(mass/time)
Flash and Evaporation-7
Parameter β is given by
2

kJ − Pa  0.67 U grd R g T  R g T
β =  3650
 N Sc +
 sat 2
mol
K
k
−


 P Hv







Nsc is the dimensionless Schmidt number given by Eq.(2.1.43)
Ugrd is the overall heat transfer coefficient of the
ground(energy/area-time-deg)
Rg is the ideal gas constant(pressure volume/mole deg)
T is the absolute temperature(deg)
k is the mass transfer coefficient(length/time)
Psat is the saturation vapor pressure(pressure)
Hv is the heat of vaporization of the liquid(energy/mole)
Available Computer Codes
Several integrated analysis packages contain evaporation
and pool model







ARCHIE (Environmental Protection Agency)
EFFECT-2 (TNO)
FOCUS+ (Quest Consultants)
PHAST (DNV)
QRAWorks (Primatech)
SUPERCHEMS (Arthur D. Little)
TRACE (Safer Systems)
Dispersion Models-1
Provide an estimate of the area affected and the average
vapor concentrations expected
Three kinds of vapor cloud behavior



Neutrally buoyant gas
Positively buoyant gas
Dense(or negatively) buoyant gas
Different release-time modes



Instantaneous(Puff)
Continuous release(Plumes)
Time varying continuous
Dispersion Models-2
Parameters affect the dispersion of gases







Atmospheric stability
Wind speed
Local terrain effects
Height of the release above the ground
Release geometry, that is, from a point, line, or area source
Momentum of the material released
Buoyancy of the material released
Atmospheric Stability
Weather conditions




Major influence on extent of dispersion
Stable atmospheric conditions conditions lead to the least
amount of mixing and unstable conditions to the most
Atmospheric conditions are classified according to six Pasquill
stability classes(denoted by the letters A through F)
Commonly defined in terms of the atmospheric vertical
temperature gradient
Night-Stable
Cool Air
Warm air
Height above ground
Height above ground
Day-Unstable
Warm air
Cool Air
Effect of stability on dispersion
Wind Speed
As the wind speed is increased, the material is carried
down-wind faster but the material is also diluted faster
by a larger quantity of air
Wind speed can calculate using power law relation
 z
u z = u10  
 10 


P
z is the height
P is a power coefficient (table 2.11)
Local Terrain Effects
Terrain characteristics affect the mechanical mixing of
the air as it flows over the ground(Table 2.10)
Depend on the size and number of the surface feature on
the terrain


When surface feature are smaller, so is the ground roughness
The smaller the roughness, the faster the cloud is dispersed
Height of the Release Above the
Ground

As the release height increase, the ground concentration
decreases since the resulting plume has more distance to mix
with fresh air prior to contacting the ground(Figure 2.26)
Momentum of the Material
Released and Buoyancy
Neutral and Positively buoyant
Plume and Puff Model-1
Puff Model

Describes near instantaneous release of material

G
1  y

C (x, y, x, t ) =
exp −
3/2
(2π ) σ x σ y σ z  2  σ y
*




2
2
 
 1  z + H  2  


−
1
z
H

 × exp − 
  + exp − 
  

   2  σz  
 2  σ z  






 <C> is the time average concentration(mass/volume)
 G* is the total mass of material released(mass)
 σx σy σz are the dispersion coefficient in x,y,z direction(length)
 y is the cross-wind direction(length)
 z is the distance above the ground(length)
 H is the release height above the ground(length)
Neutral and Positively buoyant
Plume and Puff Model-2
Plume Model


Describe a continuous release of material
The solution depend on the rate of release

G
1  y

C (x, y, x ) =
exp −
(2π )σ y σ z u  2  σ y





2
2
 
 1  z + H  2  


−
1
z
H

 × exp − 
  + exp − 
  

   2  σz  
 2  σ z  






 <C> is the average concentration(mass/volume)
 G* is the continuous release rate(mass/time)
 σx σy σz are the dispersion coefficient in x,y,z direction(length)
 u is the wind speed(length/time)
 y is the cross-wind direction(length)
 z is the distance above the ground(length)
 H is the release height above the ground(length)
Dense Gas Dispersion-1
Commonly mathematical approach

Mathematical model
 Box model which estimates overall features of the cloud such as mean radius,
mean height and mean cloud temperature without calculating details
features of the cloud
 More rigorous computational fluid dynamics(CFD)

Dimensional analysis
 Britter and Mcquaid model

Physical model
 Dispersion simulation employing wind tunnels or water channel
Refined model




UDM (1987)
SLAB (1990)
DEGADIS 2.1 (1990)
HEGADAS 5.0 (1990)
Dense Gas Dispersion-2
Britter and McQuaid(BM) model




Set of correlation based on an assessment of many laboratory
and field studies
Applicable only to ground-level instantaneous and continuous
release having negligible momentum
Applicable to stability classes C or D
Applicable only to flat terrain with no obstacle (i.e., building or
structure)
Dense Gas Dispersion-3
Hoot, Meroney and Peterka(HMP) model




Set of equation for elevated, vertically directed continuous
release of dense gas based on wind tunnel experiments
Plume is assumed to be a single-phase gas
Applicable to stability classes E or F
Applicable to low wind speeds and low surface roughness
values
Dense Gas Dispersion-4
UDM(Unified Dispersion) Model



Generalized Oom’s Model(the change of power)
Integration between momentum or density dominated
entrainment regime(dense gas) and ambient entrainment
regime( passive gas)
Treat all regimes dispersion with smooth transitions
Dense Gas Dispersion-5
DEGADIS 2.1 model



Estimates cloud dispersion based on assumed similarity profiles for
the concentration distribution of a released heavy gas
simulates ground-level, steady or time-dependent evaporating pool
with low momentum
The released materials may be diluted
Dense Gas Dispersion-6
HEGADAS 5.0 model



Estimates cloud dispersion based on assumed similarity profiles
for the concentration distribution of a released heavy gas
Has a special thermodynamics model for hydrogen fluoride(HF)
Contained in the HGSYSTEM modeling package
Example Problem-1
Example 2.1 : Liquid discharge through a hole
Calculate the discharge rate of a liquid through a 10-min
hole, if the tank head space is pressurized to 0.1 barg.
Assume a 2-m liquid head above the hole

Data : Liquid density = 490 kg/m3
Solution


No pumps or compressor, So Ws = 0
v1 = 0
P − P + g ( − )+ 1  −  + ∑ e + W = 0
ρ
g z z 2g  v2 v1 
m
2
2
1
2
1
f
C

2
•
s
C
So above equation reduces to following
g c (P − P )
1 2
2
1
+ g z −z +
v2 + g c ∑ e = 0
2
1
f
ρ
2
(
)
 Assume Nre > 10,000. Then the excess head loss for the fluid
entering the hole is Kf=0.5. For the exit Kf=1.0
 So sum of the Kf = 1.5
∑e
f
v22
= 1.5
2gc
From eqn. 2.1.2

P1=0.10, P2=0 barg
The hole area

3.14(10 × 10 −3 m) 2
=
= 7.85 × 10 −5 m 2
A=
2
4
The first term

πD 2
 kg − m / s 2

1N
g c ( P2 − P1 ) 
=
ρ

 1N / m 2
(0bar − 0.10bar )(100,000 Pa / bar )

 Pa


490kg / m 3
g ( z 2 − z1 ) = (9.8m / s 2 )(0m − 2m) = −19.6m 2 / s 2

So Substituting the terms into eqn. (2.1.10)
− 20.4 − 19.6 +
1 2 1.5 2
v2 +
v2 = 0
2
2



 = −20.4m 2 / s 2


So obtains v2 = 5.7m/s
m = ρv2A = (490 kg/m3)(5.7 m/s)(7.85*10-5 m2) = 0.22 kg/s
Example Problem-2
Boiling pool vaporization

Calculation the vaporization rate due to heating from the
ground at 10 s after an instantaneous spill of 1000 m3 of LNG on
a concrete dike of 40 m radius
 Data





Thermal diffusivity of soil, αs : 4.16*10-7 m2/s
Thermal conductivity of soil, ks : 0.92 W/m K
Temperature of liquid pool, T : 109 K
Temperature of soil, Ts : 293 K
Heat of vaporization of pool, L : 498 kJ/kg at 109 K
Solution

The total pool area
πr 2 = (3.14)(40m) 2 = 5024m 2
The liquid depth in the pool is thus (1000m3)/(5024m2)=0.2 m
 The heat flux from the ground is given by Eq. (2.1.39)
k s (Tg − T)
(0.92W/mK)(293K − 109K)
4
2
4.68
10
J/m
s
=
=
×
qg =
1/2
1/2
−
7
2
(ππ s t)
(3.14)(4.16 × 10 m /s)(10s)

[


]
The evaporative flux
⋅
H 4.68 × 10 4 J/m 2s
2
m= =
=
0.094kg/m
s
5
L
4.98 × 10 J/kg
Total evaporation rate for the entire pool area
(0.094 kg/m2s)(5024m2) = 472
kg/s
Example Problem-3
Plume release


Determine the concentration in ppm 500 m down-wind from a
0.1 kg/s ground release of a gas. The gas has a molecular weight
of 30. Assume a temperature of 198K, a pressure of 1 atm, F
stability, with a 2m/s wind speed. The release occur in a rural
area
Data
 Use equation (2.1.61) for a plume
 Assuming a ground level release(H=0)
 Assuming a location on the ground(z=0) and a position directly
downwind from the release(y=0)
Solution

For a location 500 m downwind from either table 2.12 or figure
2.28, for F stability conditions, σy = 19.52, σz =6.96
C (500m,0,0 ) =

G
πσ yσ z u
(0.1kg / s )
= 1.17 × 10 − 4 kg / m 3
(3.14)(19.52m )(6.96m )(2m / s )
To convert to ppm, the following equation is used
C ppm

=
 0.08206 L − atm  T 
3

= 
 × mg / m
 mg − moleK  PM 
So the result is 95.55 ppm