3. You determined that your workplace has fugitive o-cresol emissions with maximum annual o-cresol concentrations of 75 μg/m3 (in air), and you would like to assess worker exposure to o-cresol.

a) Calculate the worker’s daily o-cresol intake. b) Calculate the hazard quotient and say whether you think this exposure level merits further investigation or is not likely to be a

concern. c) If, in addition to o-cresol, your worker was also exposed to another hydrocarbon with an intake rate of 0.25 mg/kg-day and this

hydrocarbon had an RfD of 0.5 mg/kg-day, what would the hazard index be in this case, whether you think this exposure level merits further investigation or is not likely to be a concern.

Information you’ll need to solve 3: • Average adult body weight (BW) = 70 kg

• Air breathing rate (IR) = 20 m3/day

• Exposure frequency (EF)= 260 days/year

• Exposure duration (ED)= 30 years (working life)

• Averaging Time (Delta T) = 30 years

• Reference dose for the other hydrocarbon (part c ) is = 0.5 mg/kg-day

• Reference does of o-cresol = 0.06 mg/kg-day

• O-cresol and xylene both have the same target and cause weight loss.

Questions/RemindersHW2 Question #3

Reminders

• Mike Molenaar, Questar on Tuesday the 27th. You are welcome to help me walk him in or out.

• How can I study for the test?

• HW3 posted

VOC Control

Normal boiling point - temperature where compound’s vapor pressure = atmospheric pressure (liquid rapidly turns to vapor), Water this is 100 C (212 F). At room temperature (20 C), water’s vapor pressure is 0.23 atm

Fires Mobile Industrial Processes Solvent MiscellaneousFuel Combustion Biogenics

EPA, 2011

Generally, VOCs are organic liquids or solids whose room temperature vapor pressure is greater than about 0.01 psia (= 0.0007 atm) and whose atmospheric boiling points are up to about 500oF (=260oC). TypicallyC12 and under.

Methane is not a VOC. Sometimes NMOG

Raoult’s LawClosed container, vapor of a liquid will come into equilibrium with the vapor above it.

Vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present.

yi = xi* pi/P

yi = mol fraction (vapor fraction), assume perfect gas, of component i in the vapor

xi = mole fraction of component i in the liquid

pi = vapor pressure of pure component i at the temperature of interest (look up)

P = total pressure

Raoult’s Law

For a liquid mixture of 50% benzene and 50% toluene (mole %) in equilibrium with air in a closed tank at 20 C, estimate the concentration of benzene and toluene in the vapor in the tank. The vapor pressure of benzene & toluene at 20 C are 1.45 and 0.42 psi, respectively.

ybenzene = xbenzene * pbeneze/P = 0.5 * 1.45/14.7 = 0.049 vol fraction

ytoluene = xtoluene * ptoluene/P = 0.5 * 0.42/14.7 = 0.014 vol fraction

vol fraction of air = 1 - 0.049 - 0.014 = 0.937

Example from DeNevers 2000.

Estimating Vapor Pressure

log p = A - B/(T - C)

where,

A, B, and C are empirical constants (look up).

TanksFilling - Change in volume causes saturated vapor to be displaced.

Breathing - changes in temperature (and pressure) that cause changes in volume and displace saturated vapor. Tanks breathe in at night and out during the day.

Working - vapor escapes when adding or removing material from tanks, splashing can increase this.

Note - must vent when filling or emptying (otherwise overpressure, rupture, or under pressure, collapse).

VOCemissions = Volair-VOC mix * ConcVOCs in mix

Tanks

FillLiquid

Vapor Vapor out

Bottom fill or submerged pipeis typical

Estimating EmissionsFor all three types of losses

where, mi = mass emission of component i ci = concentration in the displaced gas

Replacing the vapor mol fraction by Raoult’s law, replacing the gas molar volume by the ideal gas law, and substituting

xi = mole fraction of component i in the liquid pi = vapor pressure of pure component i at the temperature of interest (look up) P = total pressure M = molecular weight T = Temperature (K or R) R = ideal gas constant

Filling Example A tank contains pure liquid benzene at 68oF ( 20 C) which is in equilibrium

with air-benzene vapor in its headspace. If we pump in liquid benzene (and don’t overfill the tank), how many pounds of benzene will be emitted in the vent gas per cubic foot of benzene liquid pumped in?

xi = 1 (mole fraction) pi = 1.45 psia (vapor pressure benzene) Mi = molecular weight (78 lb/lbmol) R = 10.73 psi - ft3/lbmol - R T = Temperature (R or K) (528 R)

Breathing Example The tank in the previous sample is now heated by the sun to

100oF; both vapor and liquid are heated to this temperature. How many pounds of benzene are expelled per cubic foot of tank? Assume that initially the tank was 50% by volume full of liquid, 50% by volume full of vapor.There are two contributions to the emissions: • Vapor expelled because of thermal expansion of the vapor and

liquid in the tank, • Vapor expelled because of the vaporization of benzene as the

liquid temperature is raised.

Simplify by assuming these processes take place in sequence-heating/thermal expansion then equilibration.

Example from Denevers 2000

To calculate volume change due to thermal expansion.

The fraction of volume change as a function of temperature is given by:

where,

alpha is the thermal expansion coefficient.

Breathing Example

Breathing Example

To solve:

Need alpha for the vapor, the liquid, and the tank.

For the vapor, if we assume a perfect gas (neglect intermolecular forces)

Breathing Losses

For organic liquids like benzene, alpha is approximately 6 x 10-4/oF.

For the tank, alpha (αtank) is the coefficient of the volume expansion, which is three times the coefficient of linear expansion for the material of which the tank is made; for a steel tank α is 3 x 6.5 x10-6/oF = 1.95 x 10-5/oF.

Breathing LossesSubstituting (remember 1/2 full of benzene)

Breathing Losses

Now consider the vaporization of benzene

• Next we look up the vapor pressure of benzene at 100oF and find it is 3.22 psia.

• For every cubic foot of benzene evaporated in this step, 1 cubic foot of benzene-air mixture is displaced from the container vent.

Breathing Losses

The volume of benzene vaporized = the volume of the vapor in the tank times the change in mol fraction (= volume fraction).

The total fraction of the tank volume expelled = 0.039 + 0.06 = 0.099 ft3/ft3 of tank from expansion and vaporization.

VOC Control

Substitution - VOC-based solvents with water-based paints and coatings. Fuel switching (to CNG).

Process modification - electric vehicles instead of gasoline (relocate emissions), vapor recovery, powder coating, floating-roof tanks

End-of-pipe control - flares, carbon canisters