Case Studies

This chapter presents two example CPQRA studies to aid in understanding and apply-ing the techniques reviewed in these guidelines. This first example (Section 8.1) esti-mates risk from a chlorine rail tank car loading facility due to potential exposure totoxic vapors. The second, more complicated example (Section 8.2) estimates the riskfrom a distillation column in flammable liquid service. The emphasis in both examplesis to demonstrate the calculations involved and the judgments exercised during the exe-cution of a CPQRA, not to determine accurately the risk from an actual operation.Thus, the risk estimated should not be taken as representing the true risks of a realinstallation, which would be much more complex in design, operation, and the topog-raphy and demographics of the surroundings.

Disclaimer: Readers should note that the worked examples in this chapter areintended to illustrate the type of calculations required for a CPQBA study. The facili-ties described have been greatly simplified so that the various CPQBA calculations canbe demonstrated with a variety of model types. The described facilities do not necessar-ily represent good practice for a real plant design, and the risk estimates in the exampleproblems have no "real world55 significance for the systems studied. Likewise, the fail-ure rate data, consequence models, and modeling parameters have been contrived forpurposes of illustrating calculation techniques, and must not be used or referenced forother studies where such data are required. In many cases, simple models suitable forhand or simple computer calculation have been selected so that the calculations caneasily be understood and reproduced by the reader. For a real study, more realistic (andcomplex) models may be more appropriate. Other factors which introduce complexityto the calculations, such as sheltering, escape, and time varying population, have notbeen considered, so the example calculations do not become more complex. Readersmust select models, modeling parameters, and failure rate data based on a thoroughunderstanding of the system being studied, and the current state of the art in modeling.Chapters 2-6 of this book provide extensive guidance to aid in this selection.

Readers should also note that many of the calculation results in these examples arereported with more significant figures than are appropriate, given the error and uncertaintyin the models. This has been done intentionally so that readers attempting to reproduce thecalculations will know the exact result. Readers should recognize that, in general, CPQRAresults should not be reported with more than 1 or 2 significant figures.

8.1. Chlorine Rail Tank Car Loading Facility

8.1.1. Introduction

This case study demonstrates the application of CPQRA to a simplified chlorine railtank car loading facility. It uses the simplified methodologies for calculation of individ-ual risk contours and societal risk F-N curves (Section 4.4). All the steps involved in aCPQRA are demonstrated. An individual risk contour plot, a societal risk F-N curve,and several index measures of risk are calculated. Steps in the analysis are

1. System description (Section 8.1.2)2. Incident identification, enumeration, and selection (Section 8.1.3)3. Consequence estimation (Section 8.1.4), including

a. Discharge calculationsb. Toxic effect calculationsc. Dispersion calculations

4. Frequency estimation (Section 8.1.5)5. Risk estimation (Section 8.1.6)

a. Individual riskb. Societal riskc. Risk index measures

6. Conclusions (Section 8.1.7)

The objective of this study is to estimate the risk of fatality in the neighboringcommunity. The representative set of incidents selected for this study does not includelocalized incidents, which would cause only on-site casualties.

For orientation purposes, Figure 8.1 locates the depth of study selection forthis example on the Study Cube (Section 1.3.1 and Figure 1.5). Consider each of threeaxes of the cube in turn. In order to satisfy study objectives, risk estimation requires useof all the major CPQRA component techniques (Figure 1.1). To limit the example'scomplexity, relatively simple models are used for consequence estimation. As alreadystated, a representative set of incidents is developed.

So that readers may follow the details of all calculations, the number of inci-dents considered is kept small and the weather conditions are limited to a single typicalwind speed/stability class. For a real study, it would be preferable to provide a morerepresentative sample of the weather conditions. The atmospheric dispersion modelselected is simple neutral buoyancy Gaussian plume model rather than a complex densegas dispersion model. A realistic study would improve the analysis in these areas asdemonstrated in the second case study (Section 8.2). The level of treatment in thisexample might correspond to a project in the early stages of design.

8.1.2. Description

The design of the chlorine loading system is well developed, and has been the subject ofsubstantial review. It normally follows the recommendations of the Chlorine Institute.Designs usually do not vary significantly from standard practice, but for this example

FIGURE 8.1. Depth of study for CPQRA Example 8.1.

substantial modifications to a typical design are assumed in order to simplify calcula-tions.

A line diagram of the simplified liquid chlorine rail tank car loading system isshown in Figure 8.2. The supply tank is mounted on weigh scales and liquid chlorine istransferred to a rail car using pressurized nitrogen. Liquid fill and vapor return lines areconnected to the rail car with short sections of braided stainless-steel hose withthreaded connections. Two remotely actuated emergency shutoff valves are located asshown, and the storage tank has an emergency vent. The chlorine supply tank and plantnitrogen systems are protected from back flow by check valves.

The 10,000 gal (50 ton) ambient temperature rail tank car is fitted with apressure relief valve and four connection points (two each for vapor and liquid) and isvented to the atmosphere. The liquid lines extend to the bottom of the tank car and arefitted with excess flow valves. The chlorine rail tank car is insulated with 4 in. ofcorkboard or urethane foam. The rail car, after inspection and required maintenance, isspotted on a weigh scale and properly positioned with brakes set and wheels chocked.The car and loading system are then electrically grounded. The procedure for loadingthe rail tank car is given in Figure 8.3.

To simplify calculations, only one typical weather condition is considered: awind speed of 4 m/s (13 ft/s) and neutral (D) atmospheric stability. To further simplifycalculations, the wind is assumed uniformly distributed in all directions.

A map of the chlorine loading facility and its surroundings is shown inFigure 8.4. The chlorine loading facility is located 100 m west of a populated area

Depth of studyfor Example 8.1

RISK ESTIMATION TECHNIQUE.̂Consequence Frequency Risk

Complex

Intermediate

Simple

ORIGIN

BoundingGroup

RepresentativeSet

ExpansiveList

INCR

EASI

NG N

UMBE

ROF

SELE

CTED

INCI

DENT

S

FIGURE 8.2. Diagram of liquid chlorine rail tank car loading installation.

Load cell scaleLoad cell scale

Chlorinebatchsupplytank

Chlorine ventcollection andvapor destructionsystem

Chlorinefeed line

Nitrogenpadding

Brakes set

Wheels chockedand tankcar grounded

50 tonChlorine

rail tank car

Hose with screwconnectionExcess flowvalveAngle valveCheck valve

Relief valve

Manual vaiveValve in liquidserviceValve in vaporserviceEmergency shutoff valvePressure controlvalve

LEGEND:

4" insulation

FIGURE 8.3. Simplified chlorine rail car loading procedure.

400 meters (about 1A mile) square with a uniformly distributed population of 400people. A nearby flammable-materials pipe rack and flammable liquid rail cars stored onneighboring rail lines have the potential to cause a pool fire in the chlorine loading area.

8.1.3. Identification, Enumeration, and Selection of Incidents

The CPQRA is based on a list of specific chlorine release incidents. Consistent with thedepth of study defined in the problem specifications, the grouping of incidents must berepresentative of the full list of incidents, but need not be exhaustive. An exhaustive listwould be unmanageable. The study objectives (estimation of risk to the public) allowthe full list of incidents to be screened based on the following criteria:

RAIL TANK CARS

Empty rail tank cararrives at plant

Stored on asiding

Moved to suitablelocation for venting

and inspection. Smallmaintenance required

Moved toloading berth

Liquid chlorinetransferred torail tank car

Moved tosiding

Full rail tankcar dispatched

PRESSURIZED CHLORINESTORAGE

Liquid chlorinetransferred to batch

supply tank

Nitrogen pad used totransfer chlorine to

rail tank car

Batch supplytank vented

FIGURE 8.4. Layout of chlorine rail loading facility.

Chlorine Loading FacilityN.T.S.

Residential Area

LEGEND

Chlorine rail tank car

Flammable liquids rail tank car

Rail line

Plant fence line

Chlorine pipe rack

Flammables pipe rackResidential area: 400 people uniformlydistributed ( - 10 people / acre). Otherareas are vacant.

• Localized incidents whose consequences do not extend beyond the boundaryfence need not be evaluated for purposes of estimating public risk.

• Major and catastrophic incidents of similar scale can be grouped and representedby single incidents with frequencies determined from contributions of all indi-vidual incidents in each group.

The representative set of incidents should cover a range of incident outcomes capa-ble of causing consequences in the community (Section 7.1). The set should includenot only large, rare incidents, but also more frequent, medium-scale incidents, whichcan often represent the largest contribution to the total risk.

Hazards and specific incidents in the chlorine loading facility can be identified in anumber of ways, as reviewed in the HEP Guidelines (AIChE/CCPS, 1985, 1992).Engineering judgment, incorporating experience and the historical record, can identifymore frequent incidents and random selection of rare ones. While this information isessential, it is not sufficient for the enumeration of all incidents.

A structured method such as the Hazard and Operability Study (HAZOP) or Fail-ure Mode and Effect Analysis (FMEA) is often used to enumerate a more complete listof incidents. The HAZOP and FMEA procedures are reviewed in a number of sourcesincluding the HEP Guidelines (AIChE/CCPS 1985, 1992). The HAZOP should beconducted for the entire loading procedure and also for any related activity (e.g. sam-pling) or interconnected utility (e.g. nitrogen padding).

A partial output from a HAZOP review is shown in Table 8.1. This is only anextract; a full HAZOP review would produce many pages of similar notes and addressmany more issues, including existing safeguards and suggested actions for systemimprovements. From Table 8.1, the incidents listed in Table 8.2 are identified for fur-ther study. The incident description (size and duration) is based on historical data andengineering judgment and is intended to represent a spectrum of possible realincidents.

Pinhole leaks from any pipe or equipment item are not analyzed because they havebeen determined to be too small to cause public impact. Spontaneous catastrophic fail-ure of the chlorine supply tank or rail tank car, although theoretically possible, has beenjudged to be too rare to contribute any significant risk for this example. However, theconsequences of this incident would be very large, and such incidents may need to beconsidered in the analysis of an actual facility.

Again it should be noted that the list of incidents considered in this example hasbeen severely restricted in order to simplify calculations for demonstration purposes.Other incidents might be identified that could be important contributors to risk, suchas tank car pull away while connected, larger pipeline failures, and longer durationleaks.

For this example, the incidents and incident outcomes are identical. Each chlorinerelease incident has only one incident outcome, a toxic cloud blowing downwind. Thelocation of the toxic effect zone depends on the weather conditions and wind direction.Thus, each combination of incident, weather conditions, and wind direction results in aseparate incident outcome case. Because only one weather condition is considered inthis example, the incident outcome cases differ only in the direction of the wind. Theeffect zones have the same physical dimensions for all incident outcome cases.

Guideword

I. No

II. High

III. High

IV. High

V. High

VL Other

Deviation

Flow

Flow

Level

Temperature

Pressure

Corrosion

Possible causes

A. Manual valves shut

B. ESD valve shut

C. Hose blocked

D. Pipe blocked

A. Flange leak

B. Valve leak

C. Pipe leak due toimpact

D. RV malfunctionon rail car

E. Hose leak

A. Weigh scale error

A. External fire fromneighboring

(1) rail line handlingflammable material;

(2) elevated pipelinewith flammablematerial

A. Nitrogen supplyoverpressure

A. Internalcorrosion of tanksor pipe fittings

Consequences

A-D Operational delay

A. Toxic hazard

B. Toxic hazard

C. Toxic hazard

D. Toxic hazard

E. Toxic hazard

A. Chlorine passesthrough relief valve-toxic hazard

A.I Relief valve liftspasses large vapor flow

A.2 Shell failurecatastrophic rupture-toxic hazard

A. Relief valve liftsand emergency vent-toxic hazard

A. Liquid or vaporleak; toxic hazard

Action required

A-D Confirm operationinstructions suitable forsafe corrections of all NoFlow cases

A.I. Minimize flangeconnections

A.2. Check suitability ofgasket specification

A. 3. Ensure ESDactivation point is atconvenient location,consider 2 actuation points

A.4. Check location ofbreathing apparatus

B . Similar to flange leak

C. Minimize activitiesnear chlorine line

D. Inspect RV beforeloading.

E. Develop preventivemaintenance program forhoses

A. Unlikely, existingdesign and weigh scalesystem consideredadequate

A. Fire protection facilitiesadequate, approximately60 min to control poolfire. Catastrophic ruptureunlikely before firebrought under control

A. No action required;pressure control system onnitrogen supply and PCVto emergency ventadequate

A. Periodic internalinspections (1-5 yearintervals) should detectany incipient corrosion

TABLE 8.1. Output from a HAZOP Review of Chlorine Rail Car Loading FacilityFlowsheet (Extract Only)

FLOWSHEET: CHLORINE LOADING SYSTEM

TABLE 8.2. Representative Set of Incidents Extracted from Table 8.1

Incident Reference fromNo. Incident description Table 8.1

1 Small liquid leakage (equivalent to 1A", 12.7 mm hole) IIA, UB

Duration =10 min (estimated) IIE

Causes IIC

Valve leak (7 valves and associated flanges)

Hose leak

Impact failure of liquid connecting pipe

2 Small vapor leakage (equivalent to W, 12.7 mm hole) IIA, IIB

Duration =10 min (estimated) IIE

Causes UC

Valve leak (5 valves and associated flanges) UD

Hose leak

Impact failure of vapor connecting pipe

Relief valve leak

3 Large vapor leakage IVA

Duration = 60 min (estimated time for fire fighting measures to

cool chlorine car and stop release)

Cause

External fire lifts relief valve

8.1.4. Incident Consequence Estimation

A combination of models is used to determine the effect zones in which fatalities couldresult for each incident in Table 8.2. The discharge rate for each incident is calculated toprovide input to the Gaussian atmospheric dispersion model A toxic effect probitmodel is used to estimate the chlorine LC50 concentration for each incident. Atmo-spheric dispersion calculations provide the concentration-distance relationship thatpermits the effect zone to be estimated.

8.1.4.1. DISCHARGERATECALCULATIONSThe three representative incidents (1, 2, and 3, Table 8.2) require three different dis-charge rate calculations. Models for discharge rate calculations are reviewed in Section2.1.1. The calculations are straightforward for leaks of liquid or vapor from holes near atank or large diameter pipe. Leak rates from longer lengths of piping will be reduced bypipe friction. However, for this case study, leaks are assumed to be unaffected by pipelength and geometry.

Incident No. 1: Liquid Discharge, Vi-in (12.7-mm) Hole. The liquid chlorinesystem is specified to be under slight nitrogen pressure at 6.3 bar (6.3 X 105 N/m2

absolute). The equation for liquid release is

I (gcP.\ (2.1.15)mL =pvA=pACDJ2\—*-\ + ghL

wherew L = liquid discharge rate (kg/s)p = chlorine density (1420 kg/m3)A — hole cross-sectional area (for 12.7 mm, 1.27 X ICT4 m2)CD = discharge coefficient (for liquids use 0.61) (dimensionless)Pg = upstream pressure (5.3 X 10"5 N/m2 gauge)g = acceleration due to gravity (9.8 m/s2)hL = liquid head (assume O m)Sc ~ gravitational constantWith these data, a liquid discharge rate of mL = 3.0 kg/s is calculated. This rate

must be checked against the excess flow valve shut off rating. For purposes of thisexample we assume the excess flow valve does not close.

As the pressurized liquid chlorine discharges through the hole, a fraction flashes.This can be estimated using Eq. (2.1.36).

(T -Tb)F* -Cp-T^ (2.1.36)

"%

whereFv = fraction of liquid flashed to vaporC = average liquid heat capacity over the range T to Tb (0.950 kj/kg 0C)T = initial temperature (180C)Tb = final temperature = atmospheric boiling point (-340C)hf& = heat of vaporization (at -340C, 285 kj/kg)With these* data, the flash fraction is calculated to be 0.17.A rule-of-thumb (Section 2.1.2.2) suggests that the aerosol fraction (liquid drop-

lets suspended in the vapor) my be assumed as equal to the flash fraction. Thus, in thisexample, the cloud is 34% (17% vapor and 17% aerosol) of the release, and 66% rainsout as liquid onto the ground. For small releases, most of this liquid immediately evap-orates on contacting warm ground. The actual rate of evaporation can be estimatedbased on the chlorine temperature (-340C), the local ground temperature, and thequantity of liquid in relation to the ground heat capacity (Section 2.1.2.2). In this case,for a leak lasting 10 min, the liquid pool consists of about 1200 kg of chlorine (600 s X0.66 X 3.0 kg/s). It can be shown that all of this material evaporates very quickly, andthus the source terms for the dispersion calculation should be based on the full dis-charge rate of 3.0 kg/s. Larger or longer duration releases should be more rigorouslymodeled. For this example, the Incident 1 release rate for input the data dispersion cal-culations is 3.0 kg/s.

Incident No. 2: Vapor Discharge, Vi-in. (12.7-mm) Hole. The vapor dischargecalculation must also account for compressibility. If the pressure difference betweenthe chlorine system and the atmosphere exceeds the critical pressure ratio, the flowthrough the orifice will be limited by the sonic or critical velocity. The critical pressureratio is given by

P / 9 \k/(k~l}5T-(^) <2X18)

wherePchoked = maximum down stream pressure resulting in maximum flow

P1 = upstream pressure (6.3 bar abs)P2 = downstream pressure (1.01 bar abs, atmospheric)k = heat capacity ratio (1.32 for chlorine).

The upstream pressure is 6.3 bar resulting in a choked pressure of Pchoked = (6.3bar) (0.542) = 3.42 bar. The discharge downstream is to atmospheric pressure which isless than the calculated choked flow, thus sonic flow is expected through the hole.

The equation for sonic or choked flow through the hole is given by

J ~r -A*! ~ \( *+!)/(*-!)

Sf(*Ti)

where

^choked = Sas discharge rate, choked flow (kg/s)Cn = discharge coefficient (approximately 1.0 for gases)A = hole cross-section area (for 12.7 mm, 1.27 x lO^m2)P1 = upstream pressure (6.3 X 105 N/m2 absolute)M = Molecular weight (kg/kg-mol) (for chlorine, 71)R = Gas constant (8314 J/kg-mole/°K)T = upstream temperature (180C, 2910K)

The Incident 2 vapor release rate for entry to the dispersion model is 0.29 kg/s.

Incident No. 3: Vapor Discharge from Rail Car Relief Valve. The vapor gen-erated in a pressure vessel engulfed in an external fire can be estimated using the for-mula from NFPA 58. Fire heat input is

JO, = 34,500E4082 (2.1.30)

To calculate Q^ in SI units a conversion factor must be applied.

Qf = 34,500£4082[2.91 x 10"4 (kJ/s)/(Btu/hr)] (8.1.1)

The relief valve gas discharge rate is

m=Q{/h(g (2.1.31)

wherem = gas discharge rate (kg/s)

Qf = heat input through vessel wall (kj/s)A - total surface area (approximately 650 ft2)F = environment factor (from API RP-520, use F = 0.3 for insulated tank)Afg = latent heat of vaporization at relief pressure (257 kj/kg)

thus(Qf = (34,500)(0.3)(650)082(2.93 x IO"4) = 614 kj/s

andm = (614 kj/s)(257 kj/kg) = 2.4 kg/s

The duration of release is estimated to be 60 min, based on the estimated response timefor firefighters to cool the vessel by the application of water streams. The chlorine

TABLE 8.3. Summary of Representative Incident Release Rate Estimates

Estimated release rateIncident Description (kg/s)

1 Liquid leak 3.0

2 Vapor leak 0.29

3 Relief valve discharge 2.4

vapor discharging from the relief valve is not buoyant, as the boiling inside the vesselkeeps it cool and it has a higher molecular weight than air. The chlorine release rate forIncident 3 is 2.4 kg/s.

Table 8.3 summarizes the estimated release rate for the three representativeincidents.

8.1.4.2. CHLORINETOXICITYCALCULATIONBefore undertaking the dispersion analysis, it is necessary to determine the toxicity rela-tionship to be used for estimating fatalities from the exposure to chlorine vapor. TheProbit Method (Section 2.3) is often used to estimate fatal effects. Withers and Lees(1985) propose a Probit equation for chlorine fatal effects for an average population.From Table 2.32:

Pr = -8.29 + 0.92 In(C2/:)

where Pr is the Probit function value (see Table 2.28); C is the chlorine concentration(ppm); and t is the duration of exposure (min).

For the purposes of this example, assume that all persons within a cloud whoseboundaries are defined by LC50 (concentration fatal to 50% of those exposed) of chlo-rine are killed and all persons outside that boundary survive. In reality, not all peoplewithin a cloud defined in this manner are killed, and some people outside the cloud maybe killed. The assumption is that the survivors within the LC50 boundary are balancedby the fatalities outside this boundary. Actual CPQRAs can include a determination ofthe probability of fatality for each individual within the effect zone, but this calculationis beyond the scope of this example. This assumption allows simplified methods forestimation individual and societal risk (Section 4.4.1.3) to be used.

In order to determine the size of the chlorine cloud for each incident, it is necessaryto determine the LC50 of chorine for the exposure time resulting from the incident. Weassume that the time required for the cloud to pass any location is equal to the incidentrelease duration. In a more detailed study, the dispersion model could be used to esti-mate the exact toxic dose at any location.

In this example, Incidents 1 and 2 have durations of 10 min, and Incident 3 lasts 60min. The chlorine concentrations corresponding to LC50 for t = 10 and 60 min can bedetermined by setting Pr =5.0 (the LC50 condition) and solving the above equationfor C. This calculation yields the results in Table 8.4.

For Incidents 1 and 2 (10 min liquid or vapor release) the effect zone is defined bya chlorine concentration of 433 ppm. For Incident 3 (60 min relief valve discharge) theeffect zone is defined by a chlorine concentration of 177 ppm.

TABLE 8.4. Estimated LC50 for Chlorine Exposures

Exposure time Estimated LCso(min) Incident (PPm)

10 1,2 433

60 3 177

8.1.4.3. DISPERSIONCALCULATIONSThe dispersion of vapor clouds is reviewed in Section 2.1.3. The chlorine releases inthis case study are initially denser than air. However, it can be shown that transition toneutral buoyancy occurs long before reaching the populated areas 100 m to the east. Insuch circumstances, estimates of adequate accuracy may be obtained from the Pascal-Gifford Gaussian plume model (Section 2.1.3.1.2). The appropriate form of thePascal-Gifford equation for a continuous source of nonabsorbed material is

F ( \2~(C)(X,y, z) = —— exp -^ —

2jtOyOzU 2 \ 0 y \

L J (2.1.61)f f l(z-H\2] \ l(z + H\2]}

"H-il—J \ + ̂ [-2(— J Jj

where(C)(x^z) is the average concentrationG is the continuous release rateOx, Oy> and GZ are the dispersion coefficients in the x, y, and z directionsu is the wind speedy is the cross-wind directionz is the distance above the groundH is the height of the source above ground level plus plume riseFor this example, releases are assumed to occur at ground level, so H = O. We are

interested in ground level chlorine concentrations, so z = O. Furthermore, we calculatethe maximum concentration (i.e., the centerline concentration) of chlorine in the cloudfor each downwind distance, soy = O.

Thus, for a maximum centerline concentration for a ground level source and recep-tor, Eq. (2.1.61) simplifies to

(c}™=^-u (8-L2)Equation (8.1.2) gives (C) in kg/m3. To obtain (C) in ppm, a conversion factor

must be included:

(C)ppm =—— R^XlO6I (8.1.3)pp noyazu \_MP J v '

whereR = gas constant (0.082057 atm-m3/kg-mol 0K)T = temperature (0K)M = molecular weight (kg/kg-mol)P = pressure (atm)The following relationships (McMullen, 1975) are used for the spreading parame-

ters ay and az for D atmospheric stability. Note these are different from the equationsutilized in Chapter 2.

oy ^exp^S + O^ln^j-O.OOS/^)]2] (8.1.4)

az ^exp^.^^O^Syiln^j-O.OS^ln^)]2] (8.1.5)

These relationships are equations fitted to the Gaussian dispersion spread parame-ter relationships found in Turner (1970).

The following values are substituted into Eqs. (8.1.3), (8.1.4), and (8.1.5) to cal-culate centerline concentration in ppm as a function of distance form the source (x):

w L = 3.0 kg/s (Incident No. 1)mv = 0.29 kg/s (Incident No. 2)ww = 2.4 kg/s (Incident No. 3)T = 180C = 2910Ku = 4 m/sAf =71 kg/kg-molP= 1 atm

Table 8.5 summarizes the results of these calculations. The distance at which thechlorine concentration reaches the LC50 (433 ppm for Incidents 1 and 2,177 ppm for

TABLE 8.5. Downwind Center Line Ground Level Chlorine Concentrations for theThree Representative Incidents

Incident 1 Incident 2 Incident 3

Liquid leak Vapor leak Relief valve discharge

(3.0 kg/s 10 min) (0.29 kg/s 10 min) (2.4 kg/s 60 min)

x (m) C (ppm) x (m) C (ppm) x (m) C (ppm)

TABLE 8.6. Distance at Which Chlorine Concentration Reaches LC50

Downwind distance atDuration Chlorine LCs0 which concentration

Incident Description (min) (PPm) = LC50 (m)

1 Liquid leak 10 433 244

2 Vapor leak 10 433 68

3 Relief valve discharge 60 177 358

Incident 3) is highlighted for each incident. From the data in Table 8.5, the distancefrom the release at which the chlorine concentration reaches the LC50 can be deter-mined. These results are summarized in Table 8.6.

Downwind distance alone does not fully characterize the effect zone for a vaporcloud. A measure of cloud width, or spread, is also necessary. The Gaussian plumemodel [Eq. (2.1.61)] may be solved at a number of intermediate and crosswind dis-tances to plot a complete isopleth corresponding to the LC50. A simpler approach, ade-quate for this example, is to approximate the plume as a pie-shaped segment defined byits length and plume arc. The plume arc can be estimated by determining the sidewaysdistance to the concentration of interest at some intermediate distance downwind (e.g.half-way). This calculation gives plume arcs of roughly 15° for the three incidents(actual range 14-18°).

8.1.5. INCIDENT FREQUENCY ESTIMATION

The likelihood of each representative incident identified can be estimated. If the designis sufficiently similar to facilities represented in the historical record, an incident fre-quency can be derived from historical statistics (Section 3.1); this is determined to betrue for Incidents 1 and 2 in this example. Where the design is substantially different orrelevant historical data do not exist, the fault tree method (Section 3.2.1) can beemployed to estimate the failure frequency; this method is applicable to Incident 3.

Failure data for process equipment items (e.g., flanges, valves, hoses) can beobtained from various reliability data bases derived from industry historical records(Section 5.1). Such data must be reviewed to determine if local conditions (e.g., tem-perature, pressure, corrosivity, vibration) differ substantially from the source data. Forthis example, the historical record is judged adequate for estimating the frequency ofleaks, equivalent to a l/2-'m. hole from liquid and vapor pipes, valves, and hoses. The fre-quency estimates in Table 8.7 for leaks represented by a Vi-in. hole are derived from anumber of sources, including Rijnmond (1982).

Using these data, the frequencies of Incidents 1 and 2 can be estimated. The fre-quency of each incident is equal to the sum of the failure frequencies of all of the indi-vidual components whose failure is included in that representative incident (Table 8.2).

^=J/, (8-1-6)y=i

where P- is the overall frequency of the representative incident i and^ is the failure fre-quency of componenty, which is included in representative incident i.

TABLE 8.7. Estimated Failure Frequency for Chlorine SystemComponents

Failure frequency,average service

Failure description (events/year)

Valve leak 1 X IQ-5

Hose leak 5 X 1(H

Impact failure of pipe"1 1 X 10~5

Relief valve leak at normal operating pressure 1 X 1(H

3It should be noted that among the many factors that must be considered whenestimating pipe failure rate are pipe length and pipe size.

Hence for Incident 1, the frequency of the representative vapor leak (7 valves, 1hose, 1 impact pipe failure) (using data from Table 8.7) is

F1 =(1 XlO"5)+ (Ix 10~5)+ (1XlO'5)-T-(IxIO'5)

+(1 x 10"5) + (1 x 10~5)+ (1 x 10~5)+ (5 x 10"4) + (1 x 10"5)

= 5.8 XlO'4 per year

For Incident 2, the frequency of the representative vapor leak (5 valves, 1 hose, 1impact pipe failure, and 1 relief valve leak) (using data from Table 8.7) is

F2 =(1 XlO'5)+ (IxIO'5)+ (Ix HT5)+ (1XlO'5)

+(1 XlO"5)+ (5 XlO'4)+ (IxIO'5)+ (IxIO'4)

= 6.6 XlO'4 per year

For Incident 3 (a large vapor leak caused by an external fire), historical data are notsuitable for frequency estimation. External fire frequencies are strongly dependent onthe features of each site. A simple fault tree model of the external fire scenario (Figure8.5) is developed to calculate the frequency from basic causative factors. Using thegate-by-gate procedures described in Section 3.2.1, an incident frequency estimate of3 x 1(T6 per year is calculated.

Table 8.8 summarizes the results of the frequency estimation step. These failurefrequencies for the three representative incidents will be combined with consequenceeffect zones to estimate risk.

TABLE 8.8. Summary of Representative Incident Frequency Estimates

Incidents Description Estimated frequency (yr"1)

1 Liquid leak 5.8 x 1(H

2 Vapor leak 6.6 x 1(H

3 Relief valve discharge 3.0 X IQ-6

FIGURE 8.5. Fault tree for external fire around chlorine loading facility, leading to relief valvedischarge of chlorine (Incident 3).

Incident No. 3Vapor discharge from

CL2 Railcar relief valve3x1(T6/yr

External fire aroundCL2 Railcar3x10'6fyr

Flammable liquid spillnear CL2 railcar

6x10"6/yr

Flammable liquid poolspreads to CL2

railcar2xlO"5/yr

Operator does notcontrol flammable

liquid spillp s 0.3

8.1.6. Risk Estimation

8.1.6.1. INDIVIDUALRISKIndividual risk estimates can be calculated around the chlorine loading facility from thethree representative incidents, their likelihood of occurrence, their effect zones, and thewind direction distribution. No mitigation factors such as sheltering or evacuation areconsidered. For this example, the assumptions listed for the simplified approach toindividual risk calculation in Section 4.4 apply, and the simplified approach of Figure4.8 can be used.

All steps described in Figure 4.8 through the listing of incident outcomes (inci-dents for this example) with associated effect zones and frequencies have been com-pleted. Table 8.9 summarizes these data, as extracted from Tables 8.2 through 8.8. Thefollowing steps in the generation of the individual risk contour map should be reviewedwith reference to Figure 8.6.

The next step (Figure 4.8) is to select the incident outcome with the longest effectzone. For this example, that is Incident 3. The wind direction does affect the location ofthe effect zone (it extends downwind from the leak), so, following Figure 4.8, the inci-dent frequency must be reduced by a direction factor before assigning a value to theindividual risk contour. This direction factor is given by Equation (4.4.4):

/M =/^] (4-4.4)

where £ d is the frequency at which incident outcome case i affects a point in any partic-ular direction assuming a uniform wind direction distribution;/J is the estimated fre-quency of occurrence of incident outcomes case i\ and O1 is the angle enclosed by theeffect zone for incident outcome case i.

For Incident ZJ3 = 3 x 1(T6 yr1 and 93 = 15

/^ = ( 3 x 1(T6 yr^ 15/360)

= 1.2 x IQ-7 yr1

TABLE 8.9. Summary of Representative Incidents with Associated Effect Zones andFrequencies

Effect zone

Cl2 Leak Distance Plume Frequency ofDischarge duration LC50 to LC50 arc occurrence

Incident Description rate (kg/s) (min) (ppm) (m) (deg) (yr'1)

1 Liquid leak: V2- 3.0 10 433 244 15 5.8 x IQ-4

in. equivalenthole

2 Vapor leak: 1X2- 0.29 10 433 68 15 6.6 x 10"4

in. equivalenthole

3 Vapor discharge 2.4 60 177 358 15 3.0 x 1(Hfrom relief valvedue to fire

The next step is to draw a circle (risk contour) around the chlorine facility with aradius equal to the effect zone radius (358 m, in this case). An individual risk value isthen assigned to this contour using Eq. (4.4.5):

IRC1 =./i(or/w) +IRQ. t (4.4.5)

where IRC1 is the value of individual risk at the contour of the incident outcome caseunder consideration (yr'1); IRQ-1 is the value of individual risk at the next further riskcontour; and f$fiA are as defined for Eq. (4.4.4).

Because the Incident 3 risk contour is the first contour plotted, IR1-1 = O and theindividual risk for the contour plotted is

IRClncident 3 C0111011, =/w = 1.2 x 10-7 yr1 + O (8.1.7)

Following the procedure of Figure 4.8, we select the incident with the next longesteffect zone (Incident 1). The location of the effect zone is affected by the wind direc-tion, so the frequency must again be reduced by the direction factor as given by Equa-tion (4.4.4):

/M =/i(0i/360) = (5.8 x 1(T4 yr'XIS/SeO)

/M = 2.4Xl(T 5Yr- 1

A risk contour of radius equal to the Incident 1 effect zone distance of 244 m isthen drawn around the chlorine facility. Equation (4.4.5) can be written as follows togive the frequency to be assigned to the next risk contour:

!•^Incident 1 Contour ~ f\,d + ^Incident 3 Contour

= 2.4 x IQ-5 + 1.2 x IQ-7

= 2.4 x 1(T5Yr-1 (8.1.8)

To complete the analysis, the frequency of the final incident (Incident 2), isadjusted by the direction factor [Eq. (4.4.5)]:

f2>d = f2 (0^360) = (6.6 x 10^(15/36O)

fa = 2.8 x ID'5 yr1

A risk contour of radius equal to the Incident 2 effect zone distance of 68 m is thendrawn around the chlorine facility and a frequency assigned to it using Eq. (4.4.5) inthe following form:

-^Incident 2 Contour ~~ J 2,4 ~*~ -^Incident 1 Contour

= 2.8 x ID"5 + 2.4 x IQ-5

= 5.2 x 1(T5YT1 (8>L9)

Figure 8.6 is the final individual risk contour map for this example problem.Examination of the map shows that Incident 2 (vapor leak) does not reach the popu-lated area, but Incidents 1 and 3 do. Incident 3, the relief valve discharge, impacts a sig-nificant portion of the populated area, but at a lower frequency than the otherincidents.

FIGURE 8.6. Individual risk contours around chlorine loading facility.

Chlorine Loading FacilityN.T.S.

Residential Area

LEGEND

Chlorine rail tank car

Flammable liquids rail tank car

Rail line

Plant fence line

Chlorine pipe rack

Fiammables pipe rackResidential area: 400 people uniformlydistributed ( ~ 10 people / acre). Otherareas are vacant.

8.1.6.1. SOCIETALRISK

Societal risk calculation requires an estimate of the number of people killed by eachincident outcome case, rather than an estimate of the likelihood of fatality at a particu-lar location. The methodology described in Section 4.4 and Figure 4.14 is used todevelop the F-N curve. The population distribution surrounding the example problemsite is specified in Figure 8.4. Table 8.9 lists the representative set of incidents that canpotentially affect the population. For this example, the list of incidents is identical tothe list of incident outcomes because each incident (chlorine leak) has only one out-come (toxic cloud). Thus, referring the Figure 4.14, the list of incidents in Table 8.9represents the list of potential incident outcomes.

The next step in the analysis (Figure 4.14) is the development of a list of incidentoutcomes cases. For this example only one weather class (D atmospheric stability and 4m/s wind speed) is considered, but the wind direction will vary. For purposes of thefollowing calculation, the wind direction is divided into an 8-point wind rose (N, NE,E, SE, S, SW, W, NW). Thus each incident in Table 8.9 has 8 incident outcome casesassociated with it, one for each wind direction. The problem statement specifies thatthe wind is equally likely to blow in any direction. Thus, the probability that the windwill blow in any one of the 8 possible direction is 1/8, and the frequency of each inci-dent outcome case is equal to 0.125 times the corresponding incident frequency.

The selection of an 8-point wind rose is arbitrary and gives results with a certainlevel of resolution. Better resolution would be obtained by using a 12-point or 16-point wind rose, but at the expense of additional calculations. However, the basic shapeof the F-N curve would be the same. A wind rose with an uneven distribution couldalso be used. In that case, the frequency of each incident outcome case (wind direction)would be determined by multiplying the incident frequency by the probability thewind blowing in that direction.

Table 8.10 list all the incident outcome cases in this study, each with an estimatedfrequency of occurrence. Most incident outcome cases (indicated by comment "B") donot affect the populated area and are not considered further.

Referring to Figure 4.14, we now have the list of all incident outcome cases. Wealso have to determine the frequency of each incident outcome case (Table 8.10) and itseffect zone (Table 8.9). Now we can determine the number of fatalities for each inci-dent outcome case. The simplified method described in Section 4.4 using graphicaltechniques and Eq. (4.4.10) will be used.

The simplified procedure begins by superimposing each incident outcome caseeffect zone on the local population map. Figure 8.7 shows this for Incident OutcomeCases 3SW, 3W, and 3NW. The number of people in each effect zone is counted. Forexample, the effect zone from Incident Outcome Case 3W covers an area of about15,460 m2 of the populated area. Given the population density of 25 persons per10,000 m2 this effect zone affects 39 people (15,460 m2 X 25 people/10,000 m2). Thenumber of fatalities is then determined from Equation (4.4.10).

Nf = Pifja (4.4.10)

where Ni is the number of fatalities resulting from incident outcome case i\ P1

is the total number of people within the effect zone for incident outcome case i; and pffi

is the probability of fatality within the effect zone for incident outcome case i.

For example, it has been assumed in the individual risk calculation that all personswithin the effect zone defined by the LC50 concentration are killed, sopfi = 1. There-fore, for all incident outcome cases in this example

N1 = P, (8.1.10)

For incident outcome case 3W, N^ = Pi = 39 fatalities. A similar analysis of Inci-dent Outcome Cases 3SW and 3NW (Figure 8.7) shown that these incident outcomecases each corresponding to 20 fatalities. The analysis of incident outcome cases ISW,IW and INW is done in a similar manner. Analysis of the remaining incidents in Table8.10 shows that they do not impact on the populated area. The results of analysis of theincident outcome cases are summarized in Table 8.11. Note that incident outcomecases that do not affect any people have been omitted.

TABLE 8.10. List of Incident Outcome Cases Assuming an 8-Point Wind Rose

Incident outcome case

WindIncident frequency direction Frequency

Incident (yr'1) No. probability (y*"1) Comments*

4A, Effect zone affects populated area; B, effect zone does not affect populated area.

FIGURE 8.7. Effect zones for Incident 3.

Chlorine Loading FacilityN.T.S.

LEGEND

Chlorine rail tank car

Flammable liquids rail tank car

Rail line

Plant fence line

Chlorine pipe rack

Flammables pipe rack

Incident Outcome Case

Residential area: 400 people uniformlydistributed ( ~ 10 people / acre). Otherareas are vacant.

Residential Area

The next step in the procedure according to Figure 4.14 is to put the data in cumu-lative frequency from using Eq. (4.4.9),

FN = 2^ FI f°r all incident outcome case i for which Ni > N (4.4.9)i

where FN is the frequency of all incident outcomes cases affectingN or more people; Fi

is the frequency of incident outcome case i\ and .N1. is the number of people affected byincident outcome case i.

Table 8.12 shows the application of Eq. (4.4.9) to the data from Table 8.11,giving the cumulative frequency FN of incident outcome cases resulting in N or morefatalities. The data in the first 2 columns of Table 8.12,.FN and N, are plotted as a loga-rithmic F-N curve in Figure 8.8.

8.1.6.3. SINGLE NUMBER RISK MEASURES AND INDICESSeveral single number risk measures and risk indices can also be calculated for thisexample problem.

1. Maximum Individual Risk. The person incurring the maximum individual risk islocated at the center of the west edge of the populated area (Figure 8.6). Theindividual risk of fatality at this location is the maximum individual risk (2.4 X10~5 yr"1, for this example).

TABLE 8.11. Estimated Number of Fatalities for Incident Outcome Cases Affectingthe Populated Area

Estimated number ofIncident outcome case Frequency i7 (yr'1) fatalities

ISW 7.3 x IO-5 13

IW 7.3 x 10-5 16

INW 7.3 x 10~5 13

3SW 3.8 x 10-7 20

3W 3.8 x 10-7 39

3NW 3.8 x 10-7 20

AU others — O

TABLE 8.12. Societal Risk Calculation and F-N Curve Data

Estimated number Cumulative frequency of Nof fatalities'* or more fatalities, FN (yr'1) Incident outcome cases included

N > 39 O None

20 < N < 39 3.8 x 10~7 3W

1 6 < N < 2 0 1.1 x 10-6 3W, 3SW, 3NW

N = 16 7.3 x 10-5 3W, 3SW, 3NW, IW

N < 13 2.2 x 10^ 3W, 3SW, 3NW, IW, ISW, INW

aN must be an integer value.

Number of fatalities, N

FIGURE 8.8. Societal risk F-N curve for chlorine rail tank car loading example.

2. Average Rate of Death. The average rate of death (ROD) is calculated by Equa-tion (4.4.11):

ROD = JiV^ (4.4.11)i=\

Table 8.11 gives JF- and N1 for the six incident outcomes case which result infatalities.

ROD = (7.3 x 10-5)(6) + (7.3 x Hr5)(17)

+ (7.3 x 10-5)(6) + (3.8 x 10-7)(21)

+ (3.8 x 1(T7)(39) + (3.8 x 1(T7)(21)

= 2.1 x 10~3 fatalities / year

Freq

uenc

y of

N o

r M

ore

Fata

litie

s (pe

r yea

r)

3. Equivalent Social Cost Index. This index is calculated as»

Equivalent Social Cost = ]£ /. Nf (4.4.12)1=1

where p = risk aversion power factor (p > 1).Using Okrent's suggested value off =1.2:

ESC0k = (7.3 x l(r5)(6)L2 -h (7.3 x 10-5)(17)L2

+ (7.3 x Kr5)(6)L2 + (3.8 x 1(T7)(21)L2

+ (3.8 x 10-7(39)12 + (3.8 x KT7)(21)L2

= 3.5 x IQ-3

Using the Netherlands government's suggested value of p = 2:

ESCN = (7.3 x 1(T5)(6)2 + (7.3 x I(r5)(17)2

+ (7.3 x 1(T5)(6)2 + (3.8 x 1(T7)(21)2

+ (3.8 x 10-7)(39)2 + (3.8 x I(r7)(21)2

= 2.7 x ID'2

Units of equivalent social cost are not meaningful.4. Average Individual Risk. Equation (4.4.6) is used to calculate average individual

risk from individual risk estimates at each location around the facility. It can beshown that average individual risk can also be calculated from the average rateof death (ROD) by

IRAV = ROD/PR (8.1.11)

where PR = the total population exposed to some risk from the facility (i.e., thetotal population within the most distant risk contour).

For this example, the most distant risk contour enclosed slightly more than Viof the populated area (Figure 8.7). Give the uniform population density, thisarea is estimated to contain 240 people (PR = 240).

IRAV = ROD/PR = (2.1 x 10-3)/240

= 8.8 x KT6Yr-1

The risk can also be averaged over the total population of 400.

IRAV (total population) = (2.1 x 10'3) /400.

IRAV (total population) = 5.2 X IO^6 / yr

This average is not a good measure of risk because 160 people who incur no riskfrom the chlorine facility are included in the population base over which risk isaveraged. This gives an artificially low estimate of risk.

5. Fatal Accident Rate. Fatal Accident Rate (FAR) for the exposed population iscalculated by

FAR = IRAV (1.14 x 104) fatalities/108 exposure hours (4.4.13)

TABLE 8.13. Summary of Single Number Risk Measure and Risk Indices

Risk measure Value

Maximum individual risk 2.4 x 10~5 / yr

Average individual risk

Exposed population 8.8 X 1(T6 / yr

Total population 7.5 x 1O-6 / yr

Fatal accident rate 0.10 fatalities / 108 man-hr exposure

Average rate of death 2.1 X 10~3 fatalities / yr

Equivalent social cost

Okrent 3.5 x 1(T3

Netherlands 2.7 x 10~2

FAR is normally used to measure risk to on-site personnel. If we assume thatthe populated area represents an occupied part of the plant, and that the peopleare present at all times, then the FAR can be calculated for this example.

FAR = (8.8 x lO^yr1) (1.14 x 104)= 0.10 fatalities / 108 exposure hours

Table 8.13 summarizes the various single number measures of risk and riskindices calculated for this example.

8.1.7. Conclusions

This case study illustrates a simple CPQRA using a representative set of incidents tocalculate risk for a simplified chlorine rail tank car loading facility. Both individual andsocietal risk measure are estimated and presented. These can be compared with com-pany or other risk targets. Alternatively, risk reduction measures that would reduce theconsequences of incidents or the frequency of occurrence, as well as more fundamentaldesign parameters such as facility location, can all be evaluated quantitatively. The cost-benefit for each option can be developed and rational basis ensured for consideration ofrisk reduction measures.

8.2. Distillation Column

8.2.1. Introduction

This second case study addresses the risks associated with a system containing flamma-ble materials. The depth of study for this case (Figure 8.9) represents an intermediatelevel of complexity using a representative set of incidents in the risk plane.

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