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Pollution Prevention & Environmental Essentials Conference Paul Haas CSP, CIH University of South Florida SafetyFlorida Consultation Program

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Emergency Planning and Disaster Control Safety and Health Planning for the Trades

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Exposure Modeling (Spill Models) Exposure Assessment Using Modeling To Determine Air Concentration After Chemical Releases

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Exposure Modeling (Spill Models) The use of models to describe employee exposures is not new, but the Occupational Safety and Health Administration (OSHA) has proposed a simple methodology to use for calculation of air concentrations from spills. OSHA has not determined if these will apply in all cases where employees may be exposed. Chemical reactivity is not considered in these models.

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Exposure Modeling (Spill Models) More than 20 chemical accidents are reported each day in the U.S, according to data collected by the U.S. Environmental Protection Agency. Responding to these accidents is a dangerous but essential job. In the U.S., this job is usually handled by firefighters from local fire departments. gov/photos/gallery.html#db

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Exposure Modeling (Spill Models) RIO NEUQUEN chemical incident, Houston, Texas, July One of many containers of the culprit substance is pictured. Models can be used for release of chemicals. Web?quickfind=chemical&catal og

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Constant Decay Model

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EXPOSURE MODELING Principles from physical chemistry are applied Applications will be presented in examples Graphical presentations from Dr. Mark Nicas – UC Berkeley are provided

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Elements of the Exposure Model A chemical substance release Determine the air concentration from a release into a room And Estimate if the release may pose a probable risk inhalation health hazard.

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Elements of the Exposure Model An airborne exposure model uses the following elements:* The contaminant mass emission rate The contaminant dispersion in a room The employee exposure pattern *Source – OSHA TECHNICAL MANUAL ON PHYSICAL – CHEMICAL MATHEMATICAL EXPOSURE MODELS

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EXPOSURE ASSESSMENT When must an employer conduct an exposure assessment? –When there is a substance specific standard (e.g. lead, methylene chloride) –When employees notice symptoms or complain of respiratory effects –When the workplace contains visible emissions (e.g. fumes, dust aerosols)

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EXPOSURE ASSESSMENT OSHA Regulations for methods to determine employee exposure can be found for the following: –HAZWOPER – 29 CFR –RESPIRATORS – 29 CFR –SUBSTANCE SPECIFIC STANDARDS – 29 CFR – 1052 (e.g. Formaldehyde)

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EXPOSURE ASSESSMENT OSHA Regulations do not specify how the employer is to make a reasonable estimation for the purposes of selecting respirators for example (osha.gov/SLTC/respiratory_advisor) OSHA Substance Specific Standards allow for the use of ‘objective evidence’ to estimate exposure.

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EXPOSURE ASSESSMENT When? What? How much employee exposure is there in the workplace? –Sampling – Personal exposure monitoring –Objective information – Data –Variation – Sampling + Data + Safety Factors

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Air Monitoring Equipment

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Personal Samplers

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Personal Samplers- Media

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Other Types of Site Monitoring HAZCAT Kit Geiger Counter - radiation - radiation Specialty Monitors - Passive monitoring badges - Passive monitoring badges - TIFF TIFF 5000

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EXPOSURE ASSESSMENT BASIC TERMS –Exposure Model –Air Contaminants Parts Per Million Milligrams Per Cubic Meter –Employee Exposure

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EXPOSURE ASSESSMENT BASIC TERMS –Exposure Model An exposure model is the description of a: –Air contaminant –Room or space volume –Employee exposure

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EXPOSURE ASSESSMENT BASIC TERMS –Air Contaminants Parts Per Million (PPM) –PPM is a ‘dimensionless number’ –A 1 PPM concentration is $1 in a $1,000,000 Milligrams Per Cubic Meter –Is in weight per unit volume –Expressed in milligrams per cubic meter for gases, mists, vapors

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EXPOSURE ASSESSMENT BASIC TERMS –Employee Exposure Who, What, When, Why, How Exposed? ‘Typical’ or ‘Emergency’ release What is an ‘Incidental’ release as defined in the HAZWOPER regulation?

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Chemical Identity and Form If the chemical is a gas – The molecular weight and the gas density If the chemical is a liquid – The molecular weight and the vapor pressure If the chemical is a solid – The molecular weight

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Material Release Parameters The room or space volume (V) in cubic meters (m 3 ) The room supply/exhaust air rate (Q) in cubic meters per minute (m 3 /min) The contaminant emission rate function (G) in milligrams per minute (mg/min)

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Room Ventilation and Volume V = Room volume determined by (Length X Width X Height) Q = Air supply. It is assumed to be the room’s entire supply/exhaust air exchange rate from a mechanically – driven system. note: If room air supply is not known use the following assumptions** **Air speed (s) = m/min in a room with no strong air motion Air speed (s) = 7.6 m/min in a room with strong air currents

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Mass Emission Q = The product of the air speed times the room area (Speed X Length X Width) Gt = The emission rate function (Gt) is expressed in a release rate of mass-per-time or milligrams per minute (mg/min). Air Concentration in a room after a release (C 0 ) is a ‘worst case’ scenario of the Emission Rate Function over the Room Exhaust Rate or (Gt/Q)

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‘Worst Case Scenerio’ The air concentration (C 0 ) of a release of a material in a room using the Exposure Model Gt/Q is a ‘worst case’ scenario model Use the ideal gas law equation – (PV/nRT) to determine the concentration C 0

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‘Worst Case Scenerio’ The chemical is continually exposed to room air There is no initial air dispersal Room temperature is constant There is sufficient time to reach equilibrium Enough chemical mass exists The ideal gas law holds

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Ideal Gas Law Background A variation of Boyle’s and Dalton’s laws P1 V1 = P2 V2 T1 T2 P TOTAL = P 1 + P 2 + … + P k, for k constituents Application includes converting a mass of liquid evaporating per minute to the vapor volume evaporating per minute

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Dalton’s Law The total pressure of a gaseous mixture is the sum of the partial pressures exerted by each constituent of the mixture –P TOTAL = P 1 + P 2 + … + P k, for k constituents According to the ideal gas law, the mole fraction (Y i ) of a gas constituent is –Y i = P i / P TOTAL, expressed in ppm (parts per million)

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Ideal Gas Law Constants P = Pressure in mm Hg V = Volume in M 3 T = Temperature in K R = Gas Constant mm Hg M 3 MOL -1 K -1 n = number of moles of gas

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Ideal Gas Law PV = nRT At NTP (298.3 K and 760 mm Hg), one mole of gas generates a gas volume V = M 3 (24.45 Liters) What about gas in containers? –One mole of gas introduced into a rigid container = ? V

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Ideal Gas Law One mole of gas introduced into a 1 M 3 container will occupy 1 M 3 not M 3 However, a constant temperature will yield a gas partial pressure of 18.6 mm –P = nRT/V = (1 mol)(0.623 mm Hg M 3 MOL - 1 K -1 )(298.3 k)/1 M 3 = 18.6 mm

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Vapor Volume Mass per time is converted to volume per time using the following equation: G(t), M 3 /min = (G(t), mg/min)(0.001 g/mg) X RT A /Mol. Wt. P A This equation will be explained in examples to illustrate the gas conversion relationship

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Vapor Pressure (Eq) When the rate of evaporation = rate of condensation in the headspace of containers –If the system is at equilibrium and the headspace air is saturated with chemical vapor The partial pressure (P v ) of a chemical is related to the temperature, e.g. –P v of benzene at 20 C is 75 mm Hg and 96 mm Hg at 25 C

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Saturation Concentration (C sat ) C sat in ppm = P V in mm Hg X mm Hg This unifies Boyle’s and Dalton’s Law into the ideal gas law

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Saturation Concentration (C sat ) C sat in mg/M 3 = (C sat in ppm) X Mol. Wt This is expressed at NTP conditions (298.3 K, 760 mm Hg)

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Saturation Concentration (C sat ) C sat in mg/M 3 Can also be determined by the following: = P V X Mol. Wt. X 1000 RT This is the product of the vapor pressure of n moles of gas and the chemical’s molecular weight

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Spill Modeling Example 1 – Estimate the air concentration using a ‘worse case scenario using an identified chemical and mass emission rate Example 2 – Determine the air concentration of a spill after one minute knowing the room volume and dispersal rate

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Example #1 (C 2 Cl 4 Release) A container of perchloroethylene (C 2 Cl 4 ) is left open in a unventilated cabinet. An individual opens the door and is exposed What is the perchloroethylene concentration in ppm that the employee is exposed to?

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Example (C 2 Cl 4 Release) Use the following values in the C sat equation: T = 20 C; Perchloroethylene P V = 14 mm Hg at 20 C; Cabinet Pressure = 760 mm Hg C sat = 14 mm Hg X mm Hg C sat = 18,421 ppm

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Dalton’s Law (Remember!!!) To be rigorous, one must add the partial pressure of the perchloroethylene to the total pressure in the cabinet. –This yields a C sat of 18,090 ppm (A 2% difference) This estimate is based on a Q intial of 0, any air movement (Q) > 0 would revise the estimate of concentration downward.

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Exposure Limits for C 2 Cl 4 For perchloroethylene (C 2 Cl 4 ) the acceptable OSHA maximum peak is 300 ppm* * 29 CFR , Table Z-2 The initial exposure concentration of 18,090 ppm C 2 Cl 4 is 60 times greater than the acceptable ceiling limit.

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Example #2 (Gas Release) Carbon Dioxide (CO 2 ) is released into a 10 X 20 square meter (33 X 65 foot 2 ) room from a 10 - liter container. How many parts per million (ppm) of CO 2 are released? Assume that the air speed in the room is 3 meters/minute.

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Solution First, determine the molecular weight of Carbon Dioxide (CO 2 ) in a molar unit (mole) of gas. It is the product of the gas molecules of carbon (C) and oxygen (O): C + O + O 2 = or 44 milligrams CO 2

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Solution Then, assume that carbon dioxide gas occupies all of the container and all of the gas is released at once. The total volume of the container is 10 liters (L). The amount of gas released in cubic meters is as follows: 10 liters X cubic meter (M 3 )/ 1000 liters or 0.01 m 3

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Solution In a ‘normal’ temperature (T) and pressure (P) environment, a mole of gas occupies a volume (V) 0f M 3 (24.45 L) The number (n) of moles of CO 2 released is: 0.01 M 3 CO 2 / M 3 /mole = 0.4 moles CO 2

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Answer To determine a ‘worst-case’ scenario, we first determine the partial pressure of the gas release in the room air.

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Answer The partial pressure of the CO 2 released is determined using the ideal gas equation: P = nRT/V

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Temperature & Pressure Room temperatures must be converted to degrees Kelvin (K) using the ideal gas law, so room temperature in C is added to Pressure is expressed in milliliters of mercury (mm Hg)

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Answer P CO2 = nRT/V P = (0.4 mole CO 2 )( mm Hg. m3. mole -1. K -1 )(298.3 K)/ 0.01 M 3 P CO2 = / 0.01 mm Hg = 2.55 mm Hg

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Answer P 1 nRT/V/ P 2 nRT/V Partial Pressure = P 1/ P 2 Concentration is then expressed as the vapor volume of a chemical in its ratio to the vapor volume of air at atmospheric pressure (V/V).

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Answer The V/V application model is a ‘best-guess’ assumption of exposure. This is the estimated vapor volume of a contaminant in a million parts of air (ppm). V/V – vapor pressure of contaminant X 10 6 vapor pressure of air vapor pressure of air

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Concentration (C 0 ) The ‘worst-case concentration (C 0 ) of carbon dioxide released into the room is determined by the following equation: (C 0 ) = Partial Pressure/ Total Atmospheric Pressure in a million parts of air (ppm)

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Concentration (C 0 ) (C 0 ) = 2.5 mm Hg Carbon Dioxide / 760 mm Hg X 10 6 = X 10 6 (C 0 ) is 3279 ppm of carbon dioxide from the release of the 10 liter container.

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Concentration (C t ) (C t ) is the concentration at a time from the initial release C 0 to a time t This takes into account diffusion of the release into a space by airflow

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Concentration (C t ) What is the expected air concentration of CO 2 in the room after one minute or (Gt/Q)? Assume that in one minute the entire container is released or that 3279 ppm exists from the release. We need to determine the concentration of CO 2 throughout the room.

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Concentration (C t ) The room floor area is 200 meters square and the air speed is assumed to be 3 meters/minute so Q is expected to be 3 M/min X 200 M or 600 cubic meters/ minute ( 600 M 3 /min) Gt?

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Concentration (C t ) Convert 3279 ppm CO 2 to milligrams per cubic meter (mg/M 3 ) C 0 of carbon dioxide in mg/m 3 = 3279 ppm X 44 mg/mole / M 3 /mole = 5900 mg/M 3 of carbon dioxide

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Concentration (C t ) Answer: After one minute there are 5900 mg/M 3 / 600 M 3 or 9.83 mg/ M 3 of carbon dioxide remaining in the room. The concentration is then reduced by a factor of 5900/10 or 59 times less

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Constant Decay Theory by Dr. Mark Nicas Applications: –Objective Exposure Assessment –Spill models from container filling –Confined Spaces? Exponential decay model with an initial concentration (C o )

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Constant Decay Theory by Dr. Mark Nicas C IN -- mg/m3 V -- m3 G -- mg/min k L -- per min Q -- m3/min C 0 -- mg/m3

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Constant Decay

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Backpressure Effect on Contaminant Emission Applications: –Low vapor pressure chemicals < 1 mm Hg Pesticides Nerve Agents Net rate model with a steady state factor for chemicals with a low vapor pressure at equilibrium

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Backpressure Effect on Contaminant Emission C t -- mg/m3 V -- m3 G 0 -- mg/min t -- min Q -- m3/min C sat -- mg/m3

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Backpressure Effect on Contaminant Emission

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NEAR / FAR FIELD (NF/ FF) with constant contaminant emission rate Applications: –Operations with Dilution Ventilation –Assume a well mixed room dispersion pattern Model describes zonal concentration in the near hemispherical free surface area and an air flow rate to a far field

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Near Field / Far Field Model Dr.Nicas G – mg /min Q -- m3/min NF -- Volume m3 -- m3/min FF -- Volume m *( -b + SQRT((b 2 ) - 4*cc))

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NF/ FF with constant contaminant emission rate

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Spherical Turbulent Diffusion – Pulse Release Applications: –Operations with Local Exhaust Ventilation –Storage tank filling operations in buildings Models apply turbulent eddy diffusion for a continuous concentration near an emission source

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The Spherical Turbulent Diffusion Model without Advection - Pulse Release M 0 -- mgY -- m D T -- m2/minX -- m r -- SQRT(X 2 + Y 2 + Z 2 ) -- mZ -- m

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Spherical Turbulent Diffusion – Pulse Release

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