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 forthe consequence model forrelease of volatilehazardous 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 formationObjective Prediction of the final state of the release as the material emerges

into the atmosphereInput Temperature, pressure, phase, liquid fraction etc.

Output Mass flow rate, duration, pseudo-velocity, discharge velocity,

temperature, liquid fraction, droplet trajectories and size

Logic diagram fordischarge anddispersion 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

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)

0m

sWfe)2

1ν22(ν

c2g1)1z2(z

cgg2P

1p ρdP

=+∑+−+−+∫

Energy balance equation

(1)

Frictional loss term

=

c

2

ff 2gνKe

Kf is the head loss due to th pipe fitting v is the fluid velocity

D4fLKf =

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

K1 and K ∞ are constants(dimensionless) NRe is the Reynolds number(dimensionless) IDinches is the internal diameter of the flow path(inches)

Metric form

IDmm is the internal diameter in mm Table 2.4 contains a list of K value

++=

∞ ID11KN

KKinchesRe

1f

++=

∞ ID25.41KN

KKmmRe

1f

2-K Methods-2

Fanning friction factor

At high Reynolds number(fully developed turbulent flow), the friction factor is independent of the Reynolds number

−−=

045254 N

log.3.7065log

f1

Re

1010

ADε

( )

+=

N7.149

2.8257Dε

ARe

0.89811.1098

−=

εD3.7log4f

110 Ε is the roughness factor

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

If the flow is considered frictionless and no shaft work, the result equation is called the Bernoulli equation

( ) 0mWevv2g

1zzgg

ρPP s

f

22

C

12

C

1212 =++

−+−+

−∑ •

( ) 0vv2g1zzg

gρ

PP12

22

C

12

C

12 =

−+−+

−

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 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)

( )PP2ρρACm 21CD−=

•

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 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 flow to achieve a conservative result

Liquid Discharge-3

Discharge coefficient calculation using 2-K method

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 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

∑+=

K11C

fD

∑ fK

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

Tank drainage time

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-4(Hole in Tanks)

∫=v

v

1

2 hdV(h)

2gAC1tD

Liquid Discharge-5(Hole in Tanks)

Mass discharge rate of liquid from a hole in a tank(Crowl and Louvar, 1990)

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)

+==•

ghρPg

2ρACρvAm L

qCD

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 chokedRate 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

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)

( )

−=

−

−•

PP

PP

1kk

TM

R2g

APCm1

2k1k

1

2k2

1q

C1D

Gas Discharge-3

Sonic(choked flow)

The pressure ratio required to achieve choking

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

( ) ( )

+

−+

=•

1k2

TM

R2g

APCm1k1k

choked1q

C1D

( )

+

−

=

1k2

PP

1kk

1

choked

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 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

NGGAm ERM

SUB

22 +=

⋅

Two-Phase Discharge-2

The subcooled mass flux

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)

−= PPg2ρCG sat

cfDSUB

Two-Phase Discharge-3

The equilibrium mass flux

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)

TCg

vhG

P

c

fg

fgERM

=

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 modelsEquation for flash fraction

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)

fg

bPV h

)T(TCF

−=

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 moistureEvaporation Evaporation from liquid spill onto land and water has received

substantial attention

Flash and Evaporation-3

Evaporation from a pool

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)

⋅−= mLH

dtdTmCP

Flash and Evaporation-4

Modeling divided into two classes Low volatile liquid High volatile liquid

For the high volatile case (assuming steady state)

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)

LHm =

⋅(1)

Flash and Evaporation-5

For liquids having normal boiling points near or above ambient temperature

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)

Lg

satg

mass TRAPMk

m = (3)

Flash and Evaporation-6

Direct evaporation model includes an evaporation rate due to solar radiation

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)

V

solsolar H

MAQm = (2)

Flash and Evaporation-7

Combined equation (1), (2) represent evaporation due to mass transfer

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)

+

+

+

=β1

βmβ1

1mm masssoltot

Flash and Evaporation-7

Parameter β is given by

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)

2v

sat

2gggrd0.67

Sc HP

TRk

TRUN

KmolPakJ3650β

+

−−

=

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 expectedThree 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

Hei

ght a

bove

gro

und

Hei

ght a

bove

gro

und

Warm air

Cool Air Warm air

Cool Air

Day-Unstable Night-Stable

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 airWind speed can calculate using power law relation

z is the height P is a power coefficient (table 2.11)

P

10z 10zuu

=

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

<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)

( )( )

+−+

−−×

−=

2

z

2

z

2

yzyx3/2

*

σHz

21exp

σHz

21exp

σy

21exp

σσσ2πGtx,y,x,C

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

<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)

( ) ( )

+−+

−−×

−=

2

z

2

z

2

yzy σHz

21exp

σHz

21exp

σy

21exp

σσ2πGxy,x,C

u

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 channelRefined 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 holeCalculate 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

So above equation reduces to following

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

( ) 0mWevv2g

1zzgg

ρPP s

f

22

C

12

C

1212 =++

−+−+

−∑ •

( ) 0e2

1zz

ρ

)PP(f

2212

12 =∑++−+−

cgvgcg

cf g

ve2

5.122∑ = From eqn. 2.1.2

P1=0.10, P2=0 barg The hole area

The first term

So Substituting the terms into eqn. (2.1.10)

25232

1085.74

)1010(14.32

mmDA −−

×=×

==π

223

22

12 /4.20/490

/1)/000,100)(10.00(1

/)( sm

mkg

PamNbarPabarbar

Nsmkg

PPgc −=

−

−

=−

ρ

22212 /6.19)20)(/8.9()( smmmsmzzg −=−=−

025.1

216.194.20 2

222 =++−− vv

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

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)

The evaporative flux

Total evaporation rate for the entire pool area(0.094 kg/m2s)(5024m2) = 472 kg/s

222 5024)40)(14.3( mmr ==π

[ ] sJ/m104.68/s)(10s)m106(3.14)(4.1

109K)(293K(0.92W/mK)t)(ππ

T)(Tkq 24

1/2271/2s

gsg ×=

×

−=

−=

−

s0.094kg/mJ/kg104.98

sJ/m104.68LHm 2

5

24=

××

==⋅

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

To convert to ppm, the following equation is used

So the result is 95.55 ppm

( ) ( )( )( )( )( )

34 /1017.1/296.652.1914.3

/1.00,0,500 mkgsmmm

skgu

GmCzy

−×===σπσ

3/08206.0 mmgPMT

KmolemgatmLC ppm ×

−

−=