Chapter 3. Industrial Hygiene

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Chapter 3. Industrial Hygiene

The learning objectives for this chapter are:

Learn the four steps in an industrial hygiene study.
Learn the Globally Harmonized System (GHS) for Safety Data Sheets (SDS) and labels.
Evaluate the magnitude of exposures and responses.
Develop and evaluate control techniques to prevent exposures.
Learn to use the National Fire Protection Association (NFPA) diamond.

Industrial hygiene is the science devoted to the anticipation, identification, evaluation, and control of occupational conditions that cause sickness and injury. Such exposures may involve chemicals, dusts, noise, and thermal radiation, to name a few possible conditions.

Typical projects involving industrial hygiene are monitoring toxic airborne vapor concentrations, reducing toxic airborne vapors through the use of ventilation, selecting proper personal protective equipment, developing procedures for the handling of hazardous materials, and monitoring and reducing noise, heat, thermal radiation, and other physical factors to ensure that workers are not exposed to harmful levels.

Chemical process technology is so complex that preventing hazardous workplace exposures requires the concerted efforts of industrial hygienists, engineers, process designers, operators, laboratory personnel, and management. The industrial hygienist works closely with these personnel as an integral part of a safety and loss prevention program. Engineers are very likely to work with industrial hygienists so they must have a basic understanding of industrial hygiene to be an effective member of the team.

After identifying and evaluating the hazards, the industrial hygienist makes recommendations relevant to control methods. The industrial hygienist and plant personnel work together to ensure that the control measures are applied and maintained.

The major responsibility of the industrial hygienist is to prevent hazardous workplace exposures. This is done using four traditional steps in an industrial hygiene study:

Anticipation: expectation of the presence of workplace hazards and worker exposures
Identification: determination of the presence of worker exposures
Evaluation: determination of the magnitude of the exposure
Control: application of appropriate technology to reduce workplace exposures to acceptable levels

3-1 Anticipating and Identifying Hazardous Workplace Exposures

Anticipating potential exposures takes into account potential hazards, entry modes of toxicants into the human body, and potential human injury. Table 3-1 lists some of these potential hazards; note that it is not inclusive.

Table 3-1 Identification of Some Potential Hazards

Potential hazards
Toxicity	Flammability
Chemical reactivity	Noise
Thermal radiation	Ionizing radiation
Mechanical	Temperature, either high or low
Entry mode of toxicants
Inhalation	Ingestion
Skin absorption	Injection
Potential human injury
Lungs	Skin
Ears	Eyes
Nervous system	Liver
Kidneys	Reproductive organs
Circulatory system	Other organs

Industrial hygiene studies are only as good as the information available. Table 3-2 provides a list of potential information that would be valuable for an industrial hygiene study. Of course, this information must be up to date and reliable. Other information will likely be required depending on the specific situation.

Table 3-2 Data Useful for an Industrial Hygiene Study

Chemicals

Physical state: solid, liquid, or gas

Threshold limit values (TLVs), permissible exposure levels (PELs)

Odor threshold for vapors

Vapor pressure of liquids

Chemical reactions

Reaction rate

Heat of reaction

Hazardous properties of by-products

Reactivity with other chemicals

Thermal stability

Sensitivity of chemical to temperature or impact

Flammability

Flammability limits

Flash point temperature

Auto-ignition temperature (AIT)

Minimum explosible concentration (MEC) for dusts

Minimum ignition energy (MIE)

Physical conditions

Noise

Heat

Humidity

Thermal radiation

Ionizing radiation

Equipment

Mechanical hazards

Operating limits

Design specifications

Plant layout

Materials of construction

Note: This list will vary based on the particular situation.

The anticipation step involves understanding the potential hazards that may exist in the workplace. For example, the industrial hygienist could develop a list of all chemicals used in a plant. However, only some of these chemicals will have hazardous properties, and even fewer will be handled or used in such a way to result in a hazardous exposure.

The identification step determines whether hazardous exposures actually exist. These hazards are numerous, and they can also act in combination. Table 3-1 provides a list of some potential hazards that are commonly found during the identification step of industrial hygiene projects.

One approach to identify the presence of chemical vapors in the workplace is to list the odor thresholds for the chemicals present (Table 3-3). Individuals vary greatly with respect to odor detection so variability in the actual levels is expected. Also, some chemicals, such as methyl ethyl ketone, anesthetize the olfactory organs with continued exposure, reducing the ability to detect the odor. In many cases, the odor threshold is below the threshold limit value (TLV). For instance, chlorine has an odor threshold of 0.05 ppm, while its TLV is 0.5 ppm (see Appendix E). In this case, the odor is noticed at a concentration well below the TLV. For other chemicals, the reverse case is true: Ethylene oxide has an odor threshold of 851 ppm, while its TLV is 1 ppm. In this case, by the time the odor is detected, the exposure limit has been greatly exceeded.

Table 3-3 Odor Thresholds for Selected Chemicals

Chemical species	Odor threshold (ppm)
Acetaldehyde	0.186
Acetic acid	0.016
Acrolein	0.174
Acrylic acid	0.4
Acrylonitrile	16.6
Ammonia	5.75
Aniline	0.676
Bromine	0.066
Butane	204
Butyraldehyde	0.009
Camphor	0.051
Chlorine	0.05
Chloroform	11.7
Cumene	0.024
Diethylamine	0.186
Ethyl alcohol	0.136
Ethylamine	0.324
Ethylene oxide	851
Ethyl ether	2.29
Ethyl mercaptan	0.001
Fluorine	0.126
Hydrogen chloride	0.77
Hydrogen sulfide	0.0005
Isopropyl ether	0.055
Methyl alcohol	141
Methylene chloride	0.912
Methyl ethyl ketone (MEK)	0.27
Methyl isocyanate	2.1
Methyl mercaptan	0.001
Ozone	0.061
Phenol	0.011
Phosgene	0.55
Styrene	3.44
Toluene	0.16
Trichloroethylene	1.36
Vinyl acetate	0.603
Vinyl chloride	0.253

Source: Data from 3M Respirator Selection Guide (St. Paul, MN: 3M Corporation, 2010).

The evaluation step obtains actual data to characterize the exposure. This could include chemical vapor concentrations, noise levels, temperatures, and thermal radiation levels, among other information. The industrial hygienist is invaluable in helping with these measurements. The evaluation step is discussed in detail in Section 3-3.

Control is the final step in industrial hygiene studies. This step is undertaken after potential health hazards are identified and evaluated, and then appropriate control techniques are developed and installed to reduce workplace exposures. The control step is discussed in detail in Section 3-4.

3-2 Globally Harmonized System

The Globally Harmonized System (GHS)¹ is an international system that the United Nations created for the unified development of Safety Data Sheets (GHS SDS) and for unified label development for substances and mixtures (GHS label). The GHS is intended to be a worldwide system that all countries can use to identify the hazardous properties of chemicals and to provide unified labeling to facilitate shipping chemicals between countries.

¹ Globally Harmonized System of Classification and Labelling of Chemicals (GHS), 7th rev. ed., Part 1, 2017. www.unece.org/trans/danger/publi/ghs/ghs_rev07/07files_e0.html.

In 2013, the U.S. Occupational Safety and Health Administration (OSHA) adopted the GHS for classification and labeling of chemicals.² The new GHS requirements are as follows:

² A Guide to The Globally Harmonized System of Classification and Labeling of Chemicals (GHS) (The Purple Book), pp. 40, 47. www.osha.gov/dsg/hazcom/ghsguideoct05.pdf.

Hazard classification: The standard requires chemical manufacturers and importers to determine the hazards associated with the chemicals they produce or import. Information on the specific health and physical hazards of these chemicals must be provided to the consumer.
GHS SDS: The standard requires SDSs to have 16 specific sections, ensuring consistency in the presentation of important safety information about a chemical.
GHS labels: Chemical labels must include the name of the chemical as well as a signal word, pictograms, hazard statements, and precautionary statements that describe the hazards associated with the chemical.

Globally Harmonized System for Safety Data Sheets

The SDS has 16 sections that give a clear description of the hazards—that is, classify the hazards of substances and mixtures according to their health, environmental, and physical hazards. The sections of the SDS are as follows:

Identification
Hazard identification
Composition/information on ingredients
First aid measure
Firefighting measures
Accidental release measures
Handling and storage
Exposure controls/personal protection
Physical and chemical properties and safety characteristics
Stability and reactivity
Toxicological information
Ecological information
Disposal consideration
Transport information
Regulatory information
Other informatio.

Sections 1 through 8 provide general information about the chemical. This includes the chemical’s name, hazards, composition, safe handling practices, and emergency response measures. These first eight sections are helpful when information is required quickly in an emergency situation.

Sections 9 through 11 and section 16 contain other technical and scientific information, such as the chemical’s physical and chemical properties, information about the chemical’s stability and reactivity, toxicological information, and the date the safety data sheet was prepared or last revised.

Sections 12 through 15 contain information related to the chemical’s effect on the environment, its proper disposal, transport of the chemical, and additional regulations governing its use. These sections are considered “nonmandatory” by OSHA because the content of these sections is enforced by other federal and state government agencies.

The minimum required information for the 16 sections of the GHS SDS is identified in the “Purple Book.”³ A detailed description of the GHS SDS and actual SDSs for chemicals are readily available on the Internet.

³Occupational Safety and Health Administration. “Purple Book” www.osha.gov/dsg/hazcom/ghsguideoct05.pdf.

The GHS includes nine pictograms and 29 hazard classes, as shown in Table 3-4. The hazard classes include 17 physical hazard classes, 10 health hazard classes and 2 environmental hazard classes.

Table 3-4 GHS Pictograms and Hazard Classes

Group pictograms

Health Hazard

Flammability

Compressed Gas

Corrosive

Explosive

Oxidizers

Environmental

Acute Toxicity

Other Hazards

Hazard classes – www.un.org for specific details on each class.

Physical Hazards (17 classes)

Explosives

Flammable gases

Flammable liquids – see Table 3-5

Flammable solids

Aerosols

Gases under pressure

Self-reactive substances

Pyrophoric liquids

Pyrophoric solids

Self-heating substances

Substances which, in contact with water emit flammable gases

Oxidizing gases

Oxidizing liquids

Oxidizing solids

Organic peroxides

Corrosive to metals

Desensitized explosives

Health Hazards (10 classes)

Acute toxicity (oral / dermal / inhalation) – see Table 3-6

Skin corrosion / irritation

Serious eye damage / eye irritation

Respiratory or skin sensitization

Germ cell mutagenicity

Carcinogenicity

Reproductive toxicology

Target organ systemic toxicity – single exposure

Target organ systemic toxicity – repeated exposure

Aspiration toxicity

Environmental Hazards (2 classes)

Hazardous to aquatic environment (acute / chronic)

Hazardous to ozone layer

Each of the 29 hazard classes has a detailed specification for that class. Table 3-5 provides the specification for flammable liquids and Table 3-6 provides the specification for acute toxicity – inhalation. Refer to the Purple Book for specifications on the other hazard classes.

Table 3-5 GHS SDS Information for One Hazard Class: Flammable Liquid

Signal word	GHS hazard statement/criteria	GHS hazard category	pictogram
Danger	Extremely flammable liquid and vapor Flash point < 23°C, boiling point < 35°C	1
Danger	Highly flammable liquid and vapor Flash point < 23°C, boiling point > 35°C	2
Warning	Flammable liquid and vapor 23°C < Flash point < 60°C	3
Warning	Combustible liquid 60°C <Flash point < 93°C	4	No symbol

Table 3-6 GHS SDS Information for Hazard Class: Acute Toxicity – Inhalation

Signal word	GHS hazard statement/criteria	GHS hazard category	pictogram
Danger	Fatal If Inhaled: Dusts and mists: LC₅₀ < 0.05 mg/L Gases: LC₅₀ < 100 ppm Vapors: LC₅₀ < 0.5 mg/L	1
Danger	Fatal If Inhaled: Dusts and mists: 0.05 mg/L < LC₅₀ < 0.5 mg/L Gases: 100 ppm < LC₅₀ < 500 ppm Vapors: 0.5 mg/L < LC₅₀ < 2 mg/L	2
Danger	Toxic If Inhaled: Dusts and mists: 0.5 mg/L < LC₅₀ < 1 mg/L Gases: 500 ppm < LC₅₀ < 2500 ppm Vapors: 2 mg/L < LC₅₀ < 10 mg/L	3
Warning	Harmful If Inhaled: Dusts and mists: 1 mg/L < LC₅₀ < 5 mg/L Gases: 2500 ppm < LC₅₀ < 5000 ppm Vapors: 2 mg/L < LC₅₀ < 10 mg/L	4
Warning	No Specified Label Elements: Dusts and mists: LC₅₀ > 5 mg/L Gases: LC₅₀ > 5000 ppm Vapors: LC₅₀ > 20 mg/L	5	No symbol

LC₅₀ 50% lethal concentration

Each hazard class specification has four elements: signal word, classification, hazard number, and pictogram, as shown in Figures 3-4 and 3-5. The hazard classes are numbered from 1 to 4 or 5, with the number 1 denoting the most severe hazard. As an example, gasoline has a signal word of “Danger” and a GHS hazard category of 2, while diesel fuel has a signal word of “Warning” and a GHS hazard category of 4 (see Table 3-5).

Example 3-1

A survey of a laboratory identifies the following chemical species: sodium chloride, sodium hydroxide (0.5M), and phenol. Using online GHS SDSs, identify for each chemical: (a) two hazard classes with corresponding categories, (b) signal words, (c) two hazard statements, (d) two precautionary statements, and (e) the manufacturer’s information. Note that a chemical may have more than two hazard classes.

Solution

SDS information for each chemical can be found on the Internet. The following table summarizes the results:

Chemical	Description and potential hazard
Sodium chloride	Not dangerous None May be harmful if swallowed and may be harmful if inhaled None BioLabs, 800-623-5227
Sodium hydroxide (0.5M)	Health hazard: serious eye damage / eye irritation, Category 1; and Physical hazard: corrosive to metals, Category 1 Danger May be corrosive to metals, and causes severe skin burns and eye damage When calling for medical assistance, have the label in hand, and keep out of reach of children Fisher Science Education, 800-535-5053
Phenol	Health Hazard: Acute toxicity: oral /dermal/ inhalation, Category 3; and Health Hazard: skin corrosion / irritation—1B, where 1B is a subcategory for exposure < 1 hour, and Health Hazard: serious eye damage / irritation, Category 1 Danger Toxic if swallowed, and toxic in contact with skin When calling for medical assistance, have the label in hand, and keep out of reach of children Fisher Science Education, 800-535-5053

Globally Harmonized System for Labeling

The GHS label has six elements, as shown in Figure 3-1. The label format is not specified in the GHS; however, it is recommended that the product name or identifier, signal word, and the GHS pictograms should be located together on the label.^4-7

Six elements of a GHS label are shown in a figure. — **Figure 3-1** Sample GHS label showing all the six required elements.

A sample GHS label is shown. The product name is Isobutyl alcohol. Signal words, "Danger," Hazard statement, Precautionary statement or first aid, Manufacturer information, and three pictograms are present. The pictogram symbols are, Other hazards, Flammability, and Corrosive.

⁴A Guide to The Globally Harmonized System of Classification and Labeling of Chemicals (GHS) (The Purple Book), pp. 33–38. www.osha.gov/dsg/hazcom/ghsguideoct05.pdf.

⁵Occupational Safety and Health Administration. “Labels.” www.osha.gov/Publications/OSHA3636.pdf.

⁶“Six Elements of GHS Label.” www.bradyid.com/en-us/applications/ghs-labeling-requirements.

⁷U.S. Department of Transportation. “Hazardous Materials Regulations.” https://hazmatonline.phmsa.dot.gov/services/publication_documents/howtouse0507.pdf.

Six Elements of the GHS Label

Although the SDS is the major source of information about the hazardous properties of the chemical, the GHS label is designed to give the worker the most important information related to safe handling of the chemical. Each GHS label includes six elements:

Product name or identifier. This provides information on the chemical substance, or mixture, that is in the container.
Signal word. There are only two words used as signal words: “Danger” and “Warning.” “Danger” is used for the most severe circumstances, and “warning” for less severe conditions
GHS pictograms. Refer to Table 3-4
Hazards statements. These phrases describe the nature of the hazardous material and the degree of the hazard. Hazardous statements include “highly flammable liquid and vapor,” “may be fatal if swallowed and enters airways,” “causes skin irritation,” “causes serious eye irritation,” and “may cause cancer,” to name a few.
Precautionary statements/first aid. These instructions (and/or pictograms) are intended to prevent or minimize the effects of exposure to the hazardous product. Examples include special instructions (from manufacturer) before use—for example, “do not handle until all safety precautions have been read and understood,” “keep away from heat/sparks/open flames/hot surfaces,” “no smoking in the area,” “wear protective gloves/eye protection/face protection,” and “if swallowed—immediately call a poison center or doctor.”
Manufacturer information. This element identifies the manufacturer’s company name, address, and telephone number.

Additional label regulations apply when shipping materials by ground, air, or water transportation.⁸

⁸U.S. Department of Transportation. “Hazardous Materials Regulations.” https://hazmatonline.phmsa.dot.gov/services/publication_documents/howtouse0507.pdf.

Primary and Secondary Containers

Primary containers are the bags, barrels, bottles, and cans that are received from the manufacturer. These containers should have GHS labels. The supplier labels cannot be removed, altered, or defaced. If the label needs to be replaced, then the new label must contain the same information as the original.

Secondary containers are usually smaller than primary containers; for example, they may include spray bottles, jugs, and jars. These containers hold the material taken from the primary container. Secondary containers must comply with the GHS label requirements except when the following criteria are met: (1) the material is used within the work shift of the person making the transfer, (2) the worker making the transfer remains in the work area the entire time during use, and (3) the container stays within the work area and in the possession of the worker who filled the container. These requirements are so rigorous that labeling of all chemicals at all times is the best approach.

3-3 Evaluate the Magnitude of Exposures and Responses

The industrial hygiene evaluation phase determines the extent and degree of employee exposure to toxicants and physical hazards in the workplace environment. The various types of existing control measures and their effectiveness are also studied during this phase. (The industrial hygiene control techniques are presented in more detail in Section 3-4.)

During the evaluation study, the likelihood of large and small leaks must be considered. Sudden exposures to high concentrations, through large leaks, may lead to immediate acute effects, such as unconsciousness, burning eyes, or fits of coughing. There is rarely lasting damage to individuals if they are removed promptly from the contaminated area. In this case, ready access to a clean environment is important.

Chronic effects, however, arise from repeated exposures to low concentrations, mostly by small leaks or volatilization of solid or liquid chemicals. Many toxic chemical vapors are colorless and odorless (or the toxic concentration might be below the odor threshold). Small leaks of these substances might not become obvious for months or even years, but such exposures may lead to permanent and serious impairments. Special attention must be directed toward preventing and controlling low concentrations of toxic gases. In these circumstances, some provision for continuous evaluation is necessary; that is, continuous or frequent periodic sampling and analysis is important.

To establish the effectiveness of existing controls, samples are taken to determine the workers’ exposure to conditions that may be harmful. If problems are evident, controls must be implemented immediately; temporary controls such as personal protective equipment can be used in these circumstances. Longer-term and permanent controls, however, must be subsequently developed.

After the exposure data are obtained, it is necessary to compare actual exposure levels to acceptable occupational health standards, such as TLVs, PELs, or IDLH concentrations. These standards, together with the actual concentrations, are used to identify the potential hazards requiring improved or additional control measures.

Evaluating Exposures to Volatile Toxicants by Monitoring

A direct method for determining worker exposures is continuous monitoring of the air concentrations of toxicants online in a work environment. For continuous concentration data C(t), the TWA (time-weighted average) concentration is computed using the equation

$\begin{matrix} TWA = \frac{1}{8} \int_{0}^{t_{w}} C (t) d t & (3-1) \end{matrix}$ $\begin{matrix} TWA = \frac{1}{8} \int_{0}^{t_{w}} C (t) d t & (3-1) \end{matrix}$

where

C(t) is the concentration (in ppm or mg/m³) of the chemical in the air and

t_w is the worker’s shift time in hours.

The integral is always divided by 8 hours, independent of the length of time actually worked in the shift. Thus, if a worker is exposed for 12 hours to a concentration of chemical equal to the TLV-TWA, then the TLV-TWA has been exceeded, because the computation is normalized to 8 hours.

Continuous monitoring is rarely performed because most facilities do not have the necessary equipment available. Instead, intermittent samples are typically obtained, representing worker exposures at fixed points in time. If we assume that the concentration C_i is fixed (or averaged) over the period of time t_i, the TWA concentration is computed by

$\begin{matrix} TWA = \frac{C_{1} t_{1} + C_{2} t_{2} + \dots + C_{n} t_{n}}{8 hours} & (3-2) \end{matrix}$ $\begin{matrix} TWA = \frac{C_{1} t_{1} + C_{2} t_{2} + \dots + C_{n} t_{n}}{8 hours} & (3-2) \end{matrix}$

All chemical monitoring methods have drawbacks because (1) the workers move in and out of the exposed workplace, (2) the concentration of toxicants may vary at different locations in the work area, and (3) the concentrations will change with time depending on work activities. Industrial hygienists play an important role in the selection and placement of workplace monitoring equipment and the interpretation of the data.

If more than one chemical is present in the workplace, one procedure is to assume that the effects of the toxicants are additive (unless other information to the contrary is available). The combined exposures from multiple toxicants with different TLV-TWAs is determined with the equation

$\begin{matrix} Σ_{i = 1}^{n} \frac{C_{i}}{({TLV-TWA)}_{i}} & (3-3) \end{matrix}$ $\begin{matrix} Σ_{i = 1}^{n} \frac{C_{i}}{({TLV-TWA)}_{i}} & (3-3) \end{matrix}$

where

n is the total number of toxicants,

C_i is the concentration of chemical i with respect to the other toxicants, and (TLV-TWA)_i is the TLV-TWA for chemical species i.

If the sum in Equation 3-3 exceeds 1, then the workers are overexposed.

The mixture TLV-TWA can be computed with the equation

$\begin{matrix} {(TLV-TWA)}_{mix} = \frac{Σ_{i = 1}^{n} C_{i}}{Σ_{i = 1}^{n} \frac{C_{i}}{{(TLV-TWA)}_{i}}} & (3-4) \end{matrix}$ $\begin{matrix} {(TLV-TWA)}_{mix} = \frac{Σ_{i = 1}^{n} C_{i}}{Σ_{i = 1}^{n} \frac{C_{i}}{{(TLV-TWA)}_{i}}} & (3-4) \end{matrix}$

If the sum of the concentrations of the toxicants in the mixture exceeds this amount, then the workers are overexposed.

For mixtures of toxicants with different effects (such as an acid vapor mixed with lead fumes), the TLVs cannot be assumed to be additive.

Example 3-2

Air contains 5 ppm of diethyl amine (TLV-TWA of 5 ppm), 20 ppm of cyclohexanol (TLV-TWA of 50 ppm), and 10 ppm of propylene oxide (TLV-TWA of 2 ppm). What is the mixture’s TLV-TWA, and has this level been exceeded?

Solution

From Equation 3-4,

$\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {(TLV-TWA))}_{mix} \end{matrix} \end{matrix} \end{matrix} \end{matrix} & = \frac{Σ_{i = 1}^{n} C_{i}}{Σ_{i = 1}^{n} \frac{C_{i}}{({TLV- TWA)}_{i}}} = \frac{5 + 20 + 10}{\frac{5}{5} + \frac{20}{50} + \frac{10}{2}} \\ = 5.5 ppm \end{matrix}$ $\begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} \begin{matrix} {(TLV-TWA))}_{mix} \end{matrix} \end{matrix} \end{matrix} \end{matrix} & = \frac{Σ_{i = 1}^{n} C_{i}}{Σ_{i = 1}^{n} \frac{C_{i}}{({TLV- TWA)}_{i}}} = \frac{5 + 20 + 10}{\frac{5}{5} + \frac{20}{50} + \frac{10}{2}} \\ = 5.5 ppm \end{matrix}$

The total mixture concentration is 5 + 20 + 10 = 35 ppm. The workers are overexposed under these circumstances.

An alternative approach is to use Equation 3-3:

$\sum_{i = 1}^{3} \frac{C_{i}}{{(TLV - TWA)}_{i}} = \frac{5}{5} + \frac{20}{50} + \frac{10}{2} = 6.4$ $\sum_{i = 1}^{3} \frac{C_{i}}{{(TLV - TWA)}_{i}} = \frac{5}{5} + \frac{20}{50} + \frac{10}{2} = 6.4$

Because this quantity is greater than 1, the TLV-TWA has been exceeded.

Example 3-3

Determine the 8-hour TWA worker exposure if the worker is exposed to toluene vapors as follows:

Duration of exposure (hr)	Measured concentration (ppm)
2	110
2	330
4	090

Solution

Using Equation 3-2,

$\begin{array}{l} TWA = \frac{C_{1} T_{1} + C_{2} T_{2} + C_{3} T_{3}}{8} \\ = \frac{(110) (2) + (330) (2) + (90) (4)}{8} = 155 ppm \end{array}$ $\begin{array}{l} TWA = \frac{C_{1} T_{1} + C_{2} T_{2} + C_{3} T_{3}}{8} \\ = \frac{(110) (2) + (330) (2) + (90) (4)}{8} = 155 ppm \end{array}$

Because the TLV for toluene is 20 ppm (Appendix E), the worker is overexposed. Additional control measures must be developed. On a temporary and immediate basis, all employees working in this environment need to wear the appropriate respirators.

Example 3-4

Determine the mixture TLV at 25°C and 1 atm pressure of a mixture derived from the following liquid:

Component	Mole percent	Species TLV (ppm)
Heptane	50	400
Toluene	50	20

Solution

The solution requires the concentrations of the heptane and toluene in the vapor phase. Assuming that the composition of the liquid does not change as it evaporates (the quantity is large), the vapor composition is computed using standard vapor–liquid equilibrium calculations. Assuming that Raoult’s and Dalton’s laws apply to this system under these conditions, the vapor composition is determined directly from the saturation vapor pressures of the pure components. Saturation vapor pressure equations are provided in Appendix C. From these equations the saturation vapor pressure is calculated at 25°C:

$\begin{matrix} P_{heptane}^{sat} = 46.4 mm Hg \\ \begin{matrix} \end{matrix} P_{toluene}^{sat} = 28.2 mm Hg \end{matrix}$ $\begin{matrix} P_{heptane}^{sat} = 46.4 mm Hg \\ \begin{matrix} \end{matrix} P_{toluene}^{sat} = 28.2 mm Hg \end{matrix}$

Using Raoult’s law, the partial pressures in the vapor are determined:

$\begin{matrix} p_{i} & = x_{i} P_{t}^{sat} \\ p_{heptane} & = (0.5)(46.4 mm Hg) = 23.2 mm Hg \\ p_{toluene} & = (0.5)(28.2 mm Hg) = 14.1 mm Hg \end{matrix}$ $\begin{matrix} p_{i} & = x_{i} P_{t}^{sat} \\ p_{heptane} & = (0.5)(46.4 mm Hg) = 23.2 mm Hg \\ p_{toluene} & = (0.5)(28.2 mm Hg) = 14.1 mm Hg \end{matrix}$

The total pressure of the toxicants is (23.2+14.1) = 37.3 mm Hg. From Dalton’s law, the mole fractions on a toxicant basis are

$\begin{matrix} y_{heptane} & = \frac{23.2 mm Hg}{37.3 mm Hg} = 0.622 \\ y_{toluene} & = 1 - 0.622 = 0.378 \end{matrix}$ $\begin{matrix} y_{heptane} & = \frac{23.2 mm Hg}{37.3 mm Hg} = 0.622 \\ y_{toluene} & = 1 - 0.622 = 0.378 \end{matrix}$

The mixture’s TLV is computed using Equation 3-4:

$\begin{matrix} \begin{matrix} \begin{matrix} {TLV}_{mix} \end{matrix} \end{matrix} & = \frac{Σ_{i = 1}^{n} C_{i}}{Σ_{i = 1}^{n} \frac{C_{i}}{{(TLV-TWA)}_{i}}} = \frac{1}{\frac{0.622}{400} + \frac{0.378}{20}} \end{matrix} = 48.9 ppm$ $\begin{matrix} \begin{matrix} \begin{matrix} {TLV}_{mix} \end{matrix} \end{matrix} & = \frac{Σ_{i = 1}^{n} C_{i}}{Σ_{i = 1}^{n} \frac{C_{i}}{{(TLV-TWA)}_{i}}} = \frac{1}{\frac{0.622}{400} + \frac{0.378}{20}} \end{matrix} = 48.9 ppm$

The TLVs for the individual species in the mixture are

$\begin{matrix} {TLV}_{heptane} & = (0.622) (48.9 ppm) = 30.4 ppm \\ {\begin{matrix} TLV \end{matrix}}_{toluene} & = (0.378) (48.9 ppm) = 18.5 ppm \end{matrix}$ $\begin{matrix} {TLV}_{heptane} & = (0.622) (48.9 ppm) = 30.4 ppm \\ {\begin{matrix} TLV \end{matrix}}_{toluene} & = (0.378) (48.9 ppm) = 18.5 ppm \end{matrix}$

If the actual species concentrations exceed these levels, more control measures will be needed.

Evaluating Worker Exposures to Dusts

Dusts are also a contaminant that may cause health exposures. According to toxicological theory, the dust particles that present the greatest hazard to the lungs are normally in the respirable particle size range between 2 and 5 μm (see Chapter 2). Particles larger than 5 μm are usually unable to reach the lungs, whereas those smaller than 0.2 μm settle out too slowly and are mostly exhaled with the air.

The main reason for sampling for atmospheric particulates is to estimate the concentrations that are inhaled and deposited in the lungs. Sampling methods and the interpretation of data relevant to health hazards are relatively complex; industrial hygienists, who are specialists in this technology, should be consulted when confronted with this type of problem.

Dust evaluation calculations are performed in a manner identical to that used for volatile vapors. Instead of using ppm as a concentration unit, mg/m³ or mppcf (millions of particles per cubic foot) is more convenient.

Example 3-5

Determine the TLV for a uniform mixture of dusts containing the following particles:

Type of dust	Concentration (wt.%)	TLV (mppcf)
Dust A	70	20
Dust B	30	2.7

Solution

From Equation 3-4:

$\begin{matrix} TLV of mixture = \frac{1}{\frac{C_{1}}{T L V_{1}} + \frac{C_{2}}{T L V_{2}}} \\ = \frac{1}{\frac{0.70}{20} + \frac{0.30}{2.7}} \\ = 6.8 mppcf \end{matrix}$ $\begin{matrix} TLV of mixture = \frac{1}{\frac{C_{1}}{T L V_{1}} + \frac{C_{2}}{T L V_{2}}} \\ = \frac{1}{\frac{0.70}{20} + \frac{0.30}{2.7}} \\ = 6.8 mppcf \end{matrix}$

Special control measures will be required when the actual particle count (of the size range specified in the standards or by an industrial hygienist) exceeds 6.8 mppcf.

Evaluating Worker Exposures to Noise

Noise problems are common in chemical plants; this type of problem is also evaluated by industrial hygienists. If a noise problem is suspected, the industrial hygienist should immediately make the appropriate noise measurements and develop recommendations.

Noise levels are measured in decibels. A decibel (dB) is a relative logarithmic scale used to compare the intensities of two sounds. If one sound is at intensity I and another sound is at intensity I_o, then the difference in intensity levels in decibels is given by

$\begin{matrix} Noise intensity(dB) = 10 \cdot \log_{10} (\frac{I}{I_{0}}) & (3-5) \end{matrix}$ $\begin{matrix} Noise intensity(dB) = 10 \cdot \log_{10} (\frac{I}{I_{0}}) & (3-5) \end{matrix}$

Thus, a sound 10 times as intense as another has an intensity level 10 dB.

An absolute sound scale (in dBA for absolute decibels) is defined by establishing an intensity reference. For convenience, the hearing threshold is set at 0 dBA. Table 3-7 contains dBA levels for a variety of common activities.

Table 3-7 Sound Intensity Levels for a Variety of Common Activities

Source of noise	Sound intensity level (dB)
Riveting (painful)	120
Punch press	110
Passing truck	100
Factory	90
Noisy office	80
Conventional speech	60
Private office	50
Average residence	40
Recording studio	30
Whisper	20
Threshold of good hearing	10
Threshold of excellent youthful hearing	0

Permissible noise exposure levels for single sources are provided in Table 3-8. Noise evaluation calculations are performed identically to calculations for vapors, except that dBA is used instead of ppm and hours of exposure is used instead of concentration.

Table 3-8 Permissible Noise Exposures

Sound level (dBA)	Maximum exposure (hr/day)
85	16
88	10.5
90	8
91	7
92	6
94	4.8
95	4
97	3
100	2
102	1.5
105	1
110	0.5
115	0.25

Source: “Permissible Noise Exposures,” www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=standards&p_id=9735.

Example 3-6

Determine whether the following noise level is permissible with no additional controls:

Noise level (dBA)	Duration (hr)	Maximum allowed (hr)
85	3.6	16
95	3.0	4
110	0.5	0.5

Solution

From Equation 3-3:

$\sum_{i = 1}^{3} \frac{C_{i}}{{(TLV - TWA)}_{i}} = \frac{3.6}{16} + \frac{3}{4} + \frac{0.5}{0.5} = 1.97$ $\sum_{i = 1}^{3} \frac{C_{i}}{{(TLV - TWA)}_{i}} = \frac{3.6}{16} + \frac{3}{4} + \frac{0.5}{0.5} = 1.97$

Because the sum exceeds 1.0, employees in this environment are immediately required to don hearing protection. On a longer-term basis, noise reduction control methods should be developed for the specific pieces of equipment producing excessive noise levels.

Evaluating Worker Exposures to Thermal Radiation

Thermal radiation also impacts workers. In this case, the effect on the worker is a function of both the thermal radiation intensity and the time of the exposure. Table 3-9 shows the effects on workers and Table 3-10 shows the effects on various materials. The effects may vary considerably depending on actual circumstances.

Table 3-9 Effects of Thermal Radiation on Workers

Thermal radiation intensity (kW/m²)	Time for severe pain (s)	Time for second-degree burn (s)
1	115	663
2	45	187
3	27	92
4	18	57
5	13	40
6	11	30
8	7	20
10	5	14
12	4	11

Source: Handbook of Chemical Hazard Analysis Procedures (Washington, DC: U.S. Federal Emergency Management Agency, U.S. Department of Transportation, and U.S. Environmental Protection Agency, 1988).

Table 3-10 Effects of Thermal Radiation on Materials

Thermal radiation intensity (kW/m²)	Effects
4	Glass ruined
12	Plastic melting
15	Wood firing, when there is an ignition source
18-20	Destruction of thermal insulation
25	Automatic firing of wood
37.5	Damage to process equipment
100	Damage and breaking of steel.

Sources: F. P. Lees, Loss Prevention in the Process Industries (Oxford, UK: Butterworth-Heinemann, 1980);

Methods for the Determination of Possible Damage to People and Objects Resulting from Releases of Hazardous Materials, 1st ed. (Radarweg, Netherlands: TNO, 1992).

First-degree burns are the least severe. With these burns, the skin is red and painful and swells slightly. Second-degree burns have blisters and are painful. Third-degree burns damage all the layers of the skin and the skin looks white or charred; these burns may cause little or no pain if the nerves are damaged.

Example 3-7

Flares are used to burn off vapors from a plant. The flare burns the vapors using a flame at the top of a stack. One design criterion for a flare is that the thermal radiation intensity in the vicinity of the flare on the ground should not exceed 1500 BTU/hr ft². What thermal effect will this have on the workers?

Solution

First convert to metric units:

$(1500 {BTU/hr ft}^{2}) (\frac{2.930 \times 10^{- 4} kW hr}{1 BTU}) {(\frac{1 ft}{0.3048 m})}^{2} = 4.73 k W/m$ $(1500 {BTU/hr ft}^{2}) (\frac{2.930 \times 10^{- 4} kW hr}{1 BTU}) {(\frac{1 ft}{0.3048 m})}^{2} = 4.73 k W/m$

From Table 3-9, this will cause second-degree burns in about 40 s, with severe pain occurring in 13 s. Access to the ground area around the flare must be controlled to prevent burns.

Estimating Worker Exposures to Toxic Vapors

The best procedure to determine exposures to toxic vapors is to measure the vapor concentrations directly. For design purposes, estimates of vapor concentrations are frequently required in enclosed spaces, above open containers, where drums are filled, and in the area of spills.

Consider the enclosed volume shown in Figure 3-2. This enclosure is ventilated by a constant-volume airflow. Volatile vapors are evolved within the enclosure. An estimate of the concentration of volatile in the air is required.

Mass balance in an enclosure is drawn in a diagram. — **Figure 3-2** Mass balance for volatile vapor in an enclosure.

A diagram illustrates the mass balance in an enclosure. The enclosure is represented as a cuboid, of volume V. The rate of the ventilation going in, Q subscript V, is expressed as Volume slash time. The evolution rate of Volatile, Q subscript m, is expressed as Mass slash time. The concentration of volatile, C, is expressed as Mass slash Volume. The volatile rate out, k times Q subscript V, times C, is expressed as Mass slash time. An equation below reads as follows. C subscript ppm, equals Q subscript m, times R subscript g, times T, over k times Q subscript V, times P, times M, times 10 to the power 6.

Where

C is the concentration of volatile vapor in the enclosure (mass/volume),

V is the volume of the enclosure (volume),

Q_v is the ventilation rate (volume/time),

k is the nonideal mixing factor (unitless), and

Q_m is the evolution rate of volatile material (mass/time).

The nonideal mixing factor k accounts for conditions in the enclosure that are less than well mixed. It follows that:

Total mass of volatile in volume =VC

Accumulation of mass of volatile $= \frac{d (V C)}{d t} = V \frac{d C}{d t}$ $= \frac{d (V C)}{d t} = V \frac{d C}{d t}$

Mass rate of volatile material resulting from evolution =Q_m

Mass rate of volatile material out = k Q_VC

Because accumulation equals mass-in minus mass-out, the dynamic mass balance on the volatile species is

$\begin{matrix} V \frac{d C}{d t} = Q_{m} - k Q_{v} C & (3-6) \end{matrix}$ $\begin{matrix} V \frac{d C}{d t} = Q_{m} - k Q_{v} C & (3-6) \end{matrix}$

At steady state, the accumulation term is 0, and Equation 3-6 is solved for C:

$\begin{matrix} C = \frac{Q_{m}}{k \cdot Q_{V}} & (3-7) \end{matrix}$ $\begin{matrix} C = \frac{Q_{m}}{k \cdot Q_{V}} & (3-7) \end{matrix}$

Equation 3-7 is converted to the more convenient concentration units of ppm by direct application of the ideal gas law. Let m represent mass, ρ represent mass density, and the subscripts v and b denote the volatile and bulk gas species, respectively. Then

$\begin{matrix} C_{ppm} = \frac{V_{v}}{V_{b}} \times 10^{6} = (\frac{m_{v} / ρ_{v}}{V_{b}}) \times 10^{6} = (\frac{m_{v}}{V_{b}}) (\frac{R_{g} T}{P M}) \times 10^{6} & (3-8) \end{matrix}$ $\begin{matrix} C_{ppm} = \frac{V_{v}}{V_{b}} \times 10^{6} = (\frac{m_{v} / ρ_{v}}{V_{b}}) \times 10^{6} = (\frac{m_{v}}{V_{b}}) (\frac{R_{g} T}{P M}) \times 10^{6} & (3-8) \end{matrix}$

where

R_g is the ideal gas constant,

T is the absolute ambient temperature,

P is the absolute pressure, and

M is the molecular weight of the volatile species.

The term m_v/V_b is identical to the concentration of volatile computed using Equation 3-7. Substituting Equation 3-7 into Equation 3-8 yields

$\begin{matrix} \begin{matrix} C_{ppm} = \frac{Q_{m} R_{g} T}{k Q {}_{v}P_{}^{} M} \times 10^{6} \end{matrix} & (3-9) \end{matrix}$ $\begin{matrix} \begin{matrix} C_{ppm} = \frac{Q_{m} R_{g} T}{k Q {}_{v}P_{}^{} M} \times 10^{6} \end{matrix} & (3-9) \end{matrix}$

Equation 3-9 is used to determine the average concentration (in ppm) of any volatile species in an enclosure given a source term Q_m and a ventilation rate Q_v. It can be applied to the following types of exposures: a worker standing near a pool of volatile liquid, a worker standing near an opening to a storage tank, or a worker standing near an open container of volatile liquid.

Equation 3-9 includes the following important assumptions:

The calculated concentration is an average concentration in the enclosure. Localized conditions could result in significantly higher concentrations; workers directly above an open container might be exposed to higher concentrations.
A steady-state condition is assumed; that is, the accumulation term in the mass balance is zero.

The nonideal mixing factor varies from 0.1 to 0.5 for most practical situations.⁹ For perfect mixing, k = 1.

⁹R. Craig Matthiessen. “Estimating Chemical Exposure Levels in the Workplace.” Chemical Engineering Progress (April 1986): 30.

Example 3-8

An open toluene container in an enclosure is weighed as a function of time, and it is determined that the average evaporation rate is 0.1 g/min. The ventilation rate is 100 ft³/min. The temperature is 80°F and the pressure is 1 atm. Estimate the concentration of toluene vapor in the enclosure and compare your answer to the TLV for toluene of 20 ppm.

Solution

Because the value of k is not known directly, it must be used as a parameter. From Equation 3-9,

$k C_{ppm} = \frac{Q_{m} R_{g} T}{Q_{v} P M} \times 10^{6}$ $k C_{ppm} = \frac{Q_{m} R_{g} T}{Q_{v} P M} \times 10^{6}$

From the data provided

$\begin{matrix} Q_{m} & = 0.1 g / \min = 2.20 \times 10^{- 4} {1b}_{m} / \min \\ R_{g} & = {0.7302 ft}^{3} atm / {lb-mole}^{o} R \\ T & = 80^{ο} F = 540^{ο} R \\ Q_{v} & = {100 ft}^{3} / \min \\ M & = {92 lb}_{m} / 1b-mole \\ P & = 1 atm \end{matrix}$ $\begin{matrix} Q_{m} & = 0.1 g / \min = 2.20 \times 10^{- 4} {1b}_{m} / \min \\ R_{g} & = {0.7302 ft}^{3} atm / {lb-mole}^{o} R \\ T & = 80^{ο} F = 540^{ο} R \\ Q_{v} & = {100 ft}^{3} / \min \\ M & = {92 lb}_{m} / 1b-mole \\ P & = 1 atm \end{matrix}$

Substituting into the equation for kC ppm:

$\begin{matrix} \begin{matrix} k C_{ppm} \end{matrix} & = \frac{Q_{m} R_{g} T}{Q_{v} P M} \times 10^{6} = \frac{(2.20 \times 10^{- 4} {1b}_{m} / \min)(0.7302 {ft}^{3} atm / {1b-mole}^{o} R {)(540}^{o} R)}{({100 ft}^{3} / \min)(1 atm)({92 1b}_{m} / 1b-mole)} \times 10^{6} \\ = 9.43 ppm \end{matrix}$ $\begin{matrix} \begin{matrix} k C_{ppm} \end{matrix} & = \frac{Q_{m} R_{g} T}{Q_{v} P M} \times 10^{6} = \frac{(2.20 \times 10^{- 4} {1b}_{m} / \min)(0.7302 {ft}^{3} atm / {1b-mole}^{o} R {)(540}^{o} R)}{({100 ft}^{3} / \min)(1 atm)({92 1b}_{m} / 1b-mole)} \times 10^{6} \\ = 9.43 ppm \end{matrix}$

Because k varies from 0.1 to 0.5, the concentration is expected to vary from 18.9 ppm to 94.3 ppm. Actual vapor sampling is recommended to ensure that the TLV of 20 ppm is not exceeded.

Estimating the Vaporization Rate of a Liquid

Liquids with high saturation vapor pressures evaporate more rapidly. As a result, the evaporation rate (mass/time) is a function of the saturation vapor pressure. In reality, for vaporization into stagnant air, the vaporization rate is proportional to the difference between the saturation vapor pressure and the partial pressure of the vapor in the stagnant air; that is,

$\begin{matrix} Q_{m} \propto (P^{sat} - p) & (3-10) \end{matrix}$ $\begin{matrix} Q_{m} \propto (P^{sat} - p) & (3-10) \end{matrix}$

where

P^sat is the saturation vapor pressure of the pure liquid at the temperature of the liquid and

p is the partial pressure of the vapor in the bulk stagnant gas above the liquid.

A more generalized expression for the vaporization rate is available:¹⁰

¹⁰Steven R. Hanna and Peter J. Drivas. Guidelines for the Use of Vapor Cloud Dispersion Models, 2nd ed. (New York, NY: American Institute of Chemical Engineers, 1996).

$\begin{matrix} Q_{m} = \frac{M K A (P^{sat} - p)}{R_{g} T_{L}} & (3-11) \end{matrix}$ $\begin{matrix} Q_{m} = \frac{M K A (P^{sat} - p)}{R_{g} T_{L}} & (3-11) \end{matrix}$

where

Q_m is the evaporation rate (mass/time),

M is the molecular weight of the volatile substance,

K is a mass transfer coefficient (length/time) for an area A,

R_g is the ideal gas constant, and

T_L is the absolute temperature of the liquid.

For many situations P^sat>>p, and Equation 3-11 is simplified to

$\begin{matrix} \begin{matrix} Q_{m} = \frac{M K A P^{sat}}{R_{g} T_{L}} \end{matrix} & (3-12) \end{matrix}$ $\begin{matrix} \begin{matrix} Q_{m} = \frac{M K A P^{sat}}{R_{g} T_{L}} \end{matrix} & (3-12) \end{matrix}$

Equation 3-12 is used to estimate the vaporization rate of a volatile from an open vessel or from a spill of liquid.

The vaporization rate or source term, Q_m, determined by Equation 3-12, is used in Equation 3-9 to estimate the concentration (in ppm) of a volatile in a ventilated enclosure resulting from evaporation of a liquid:

$\begin{matrix} C_{ppm} = \frac{K A T P^{sat}}{k Q_{v} P T_{L}} \times 10^{6} & (3-13) \end{matrix}$ $\begin{matrix} C_{ppm} = \frac{K A T P^{sat}}{k Q_{v} P T_{L}} \times 10^{6} & (3-13) \end{matrix}$

For most situations, T = T_L and Equation 3-13 is simplified to

$\begin{matrix} \begin{matrix} C_{ppm} = \frac{K A P^{sat}}{k Q_{v} P} \times 10^{6} \end{matrix} & (3-14) \end{matrix}$ $\begin{matrix} \begin{matrix} C_{ppm} = \frac{K A P^{sat}}{k Q_{v} P} \times 10^{6} \end{matrix} & (3-14) \end{matrix}$

As defined previously, P is the absolute pressure.

The gas mass transfer coefficient is estimated using the relationship¹¹

¹¹Louis J. Thibodeaux. Environmental Chemodynamics, 2nd ed. (New York, NY: Wiley, 1996), p. 85.

$\begin{matrix} K = a D^{2 / 3} & (3-15) \end{matrix}$ $\begin{matrix} K = a D^{2 / 3} & (3-15) \end{matrix}$

where

a is a constant and

D is the gas-phase diffusion coefficient.

Equation 3-15 is used to determine the ratio of the mass transfer coefficients between the species of interest K and a reference species K_o:

$\begin{matrix} \frac{K}{K_{0}} = {(\frac{D}{D_{0}})}^{2 / 3} & (3-16) \end{matrix}$ $\begin{matrix} \frac{K}{K_{0}} = {(\frac{D}{D_{0}})}^{2 / 3} & (3-16) \end{matrix}$

The gas-phase diffusion coefficients are estimated from the molecular weights M of the species:¹²

¹²Gordon M. Barrow. Physical Chemistry, 2nd ed. (New York, NY: McGraw-Hill, 1966), p. 19.

$\begin{matrix} \frac{D}{D_{0}} = \sqrt{\frac{M_{0}}{M}} & (3-17) \end{matrix}$ $\begin{matrix} \frac{D}{D_{0}} = \sqrt{\frac{M_{0}}{M}} & (3-17) \end{matrix}$

Equation 3-17 is combined with Equation 3-16, giving

$\begin{matrix} K = K_{0} {(\frac{M_{0}}{M})}^{1 / 3} & (3-18) \end{matrix}$ $\begin{matrix} K = K_{0} {(\frac{M_{0}}{M})}^{1 / 3} & (3-18) \end{matrix}$

Water is most frequently used as a reference substance; it has a mass transfer coefficient¹³ of 0.83 cm/s.

¹³R. Craig Matthiessen. “Estimating Chemical Exposure Levels in the Workplace.” Chemical Engineering Progress (April 1986): 33.

Example 3-9

A tank with a 1.5-m diameter opening contains toluene, C₇H₈. Estimate the evaporation rate from this tank assuming a temperature of 25°C and a pressure of 1 atm. If the ventilation rate is 85 m³/min, estimate the concentration of toluene in this workplace enclosure.

Solution

The molecular weight of toluene is 92. The mass transfer coefficient is estimated from Equation 3-18 using water as a reference:

$K = K_{0} {(\frac{M_{0}}{M})}^{1 / 3} = (0.83 cm/s) {(\frac{18}{92})}^{1 / 3} = 0.482 cm/s = 0.289 m/min$ $K = K_{0} {(\frac{M_{0}}{M})}^{1 / 3} = (0.83 cm/s) {(\frac{18}{92})}^{1 / 3} = 0.482 cm/s = 0.289 m/min$

The saturation vapor pressure is given in Example 3-4:

$P_{to1uene}^{sat} = 28.2 mm Hg = 0.0371 atm$ $P_{to1uene}^{sat} = 28.2 mm Hg = 0.0371 atm$

The pool area is

$A = \frac{π d^{2}}{4} = \frac{{(3.14)(1.5 m)}^{2}}{4} = {1.77 m}^{2}$ $A = \frac{π d^{2}}{4} = \frac{{(3.14)(1.5 m)}^{2}}{4} = {1.77 m}^{2}$

The evaporation rate is computed using Equation 3-12:

$\begin{array}{l} Q_{m} = \frac{M K A P^{sat}}{R_{g} T_{L}} = \frac{(92 kg/kg-mole)(0.289 m/ \min {)(1.77 m}^{2})(0.0371 atm)}{{(0.082057 m}^{3} {atm/kg-mole}^{o} {K)(298}^{o} K)} \\ = 0.0714 kg/min \end{array}$ $\begin{array}{l} Q_{m} = \frac{M K A P^{sat}}{R_{g} T_{L}} = \frac{(92 kg/kg-mole)(0.289 m/ \min {)(1.77 m}^{2})(0.0371 atm)}{{(0.082057 m}^{3} {atm/kg-mole}^{o} {K)(298}^{o} K)} \\ = 0.0714 kg/min \end{array}$

The concentration is estimated using Equation 3-14 with k as a parameter:

$\begin{array}{l} k C_{ppm} = \frac{K A P^{sat}}{Q {}_{v}P_{}^{}} \times 10^{6} \\ = \frac{(0.289 m/ \min {)(1.77 m}^{2})(0.0371 atm)}{{(85 m}^{3} / \min)(1 atm)} \times 10^{6} \\ = 223 ppm \end{array}$ $\begin{array}{l} k C_{ppm} = \frac{K A P^{sat}}{Q {}_{v}P_{}^{}} \times 10^{6} \\ = \frac{(0.289 m/ \min {)(1.77 m}^{2})(0.0371 atm)}{{(85 m}^{3} / \min)(1 atm)} \times 10^{6} \\ = 223 ppm \end{array}$

The concentration will range from 446 ppm to 2230 ppm, depending on the value of k (ranging from 0.1 to 0.5). Because the TLV for toluene is 20 ppm, additional ventilation is recommended, or the amount of exposed surface area should be reduced. The amount of ventilation required to reduce the worst-case concentration (2230 ppm) to 20 ppm is

$Q_{v} = ({85 m}^{3} / \min) (\frac{2230 ppm}{20 ppm}) {= 9480 m}^{3} / \min$ $Q_{v} = ({85 m}^{3} / \min) (\frac{2230 ppm}{20 ppm}) {= 9480 m}^{3} / \min$

This represents an impractical level of general ventilation. Potential solutions to this problem include containing the toluene in a closed vessel or using local ventilation at the vessel opening.

Estimating Worker Exposures during Vessel Filling Operations

For vessels being filled with liquid, volatile emissions are generated from two sources, as shown in Figure 3-3:

Evaporation of the liquid, represented by Equation 3-14
Displacement of the vapor in the vapor space by the liquid filling the vessel

A filling vessel's evaporation and displacement are represented in a diagram. — **Figure 3-3** Evaporation and displacement from a filling vessel.

A diagram illustrates the evaporation and displacement from a filling vessel. Liquid is filled into a drum or vessel via a pipe. The level of the liquid inside the vessel is higher than the mouth of the pipe. The liquid evaporates and fills the vapor space. The vapor is labeled inside the space between the mouth of the vessel and the liquid level. The total source coming out of the vessel equals evaporation plus displaced air.

The net generation of volatile is the sum of the two sources:

$\begin{matrix} Q_{m} = {(Q_{m})}_{1} + {(Q_{m})}_{2} & (3-19) \end{matrix}$ $\begin{matrix} Q_{m} = {(Q_{m})}_{1} + {(Q_{m})}_{2} & (3-19) \end{matrix}$

where

(Q_m)₁ represents the source resulting from evaporation and

(Q_m)₂ represents the source resulting from displacement.

The source term (Q_m)₁ is computed using Equation 3-12. (Q_m)₂ is determined by assuming that the vapor is completely saturated with the volatile. An adjustment is introduced later for less than saturated conditions.

Define

V_c as the volume of the container (volume),

r_f as the constant filling rate of the vessel (time⁻¹),

P^sat as the saturation vapor pressure of the volatile liquid, and

T_L as the absolute temperature of the container and liquid.

It follows that r_fV_c is the volumetric rate of bulk vapor being displaced from the drum (volume/time). Also, if ρ_v is the density of the volatile vapor, then r_fV_c ρ_v is the mass rate of volatile displaced from the container (mass/time). Using the ideal gas law,

$\begin{matrix} ρ_{V} = \frac{M P^{sat}}{R_{g} T_{L}} & (\begin{matrix} 3-20 \end{matrix}) \end{matrix}$ $\begin{matrix} ρ_{V} = \frac{M P^{sat}}{R_{g} T_{L}} & (\begin{matrix} 3-20 \end{matrix}) \end{matrix}$

and it follows that

$\begin{matrix} {(Q_{m})}_{2} = \frac{M P^{sat}}{R_{g} T_{L}} r_{f} V_{C} & (3-21) \end{matrix}$ $\begin{matrix} {(Q_{m})}_{2} = \frac{M P^{sat}}{R_{g} T_{L}} r_{f} V_{C} & (3-21) \end{matrix}$

Equation 3-21 can be modified for container vapors that are not saturated with the volatile. Let φ represent this adjustment factor; then,

$\begin{matrix} {(Q_{m})}_{2} = \frac{M P^{sat}}{R_{g} T_{L}} ϕ r_{f} V_{C} & (3-22) \end{matrix}$ $\begin{matrix} {(Q_{m})}_{2} = \frac{M P^{sat}}{R_{g} T_{L}} ϕ r_{f} V_{C} & (3-22) \end{matrix}$

For splash filling (filling from the top of a container with the liquid splashing to the bottom), φ = 1. For subsurface filling¹⁴ (by a dip leg to the bottom of the tank), φ = 0.5.

¹⁴R. Craig Matthiessen. “Estimating Chemical Exposure Levels in the Workplace.” Chemical Engineering Progress (April 1986): 33.

The net source term resulting from filling is derived by combining Equations 3-12 and 3-22 with Equation 3-19:

$\begin{matrix} Q_{m} = {(Q_{m})}_{1} + {(Q_{m})}_{2} = \frac{M P^{sat}}{R_{g} T_{L}} (ϕ r_{f} V_{c} + K A) & (3-23) \end{matrix}$ $\begin{matrix} Q_{m} = {(Q_{m})}_{1} + {(Q_{m})}_{2} = \frac{M P^{sat}}{R_{g} T_{L}} (ϕ r_{f} V_{c} + K A) & (3-23) \end{matrix}$

This source term is substituted into Equation 3-9 to compute the vapor concentration (in ppm) in an enclosure resulting from a filling operation. The assumption that T = T_L is also invoked. The result is

$\begin{matrix} \begin{matrix} C_{PPm} = \frac{P^{sat}}{k Q_{v} P} ({ϕ r}_{f} V_{c} + K A) \times 10^{6} \end{matrix} & (3-24) \end{matrix}$ $\begin{matrix} \begin{matrix} C_{PPm} = \frac{P^{sat}}{k Q_{v} P} ({ϕ r}_{f} V_{c} + K A) \times 10^{6} \end{matrix} & (3-24) \end{matrix}$

For many practical situations, the evaporation term KA is much smaller than the displacement term and can be neglected.

Example 3-10

Railroad cars are being splash-filled with toluene. The 10,000-gal cars are being filled at the rate of one every 8 hours. The filling hole in the tank car is 4 inches in diameter. Estimate the concentration of toluene vapor as a result of this filling operation. The ventilation rate is estimated at 3000 ft³/min. The temperature is 77°F and the pressure is 1 atm.

Solution

The concentration is estimated using Equation 3-24. From Example 3-9, K = 0.949 ft/min and P^sat = 0.0371 atm. The area of the filling hole is

$A = \frac{π d^{2}}{4} = \frac{{(3.14)(4 in)}^{2}}{{(4)(144 in}^{2} {/ft}^{2})} = {0.0872 ft}^{2}$ $A = \frac{π d^{2}}{4} = \frac{{(3.14)(4 in)}^{2}}{{(4)(144 in}^{2} {/ft}^{2})} = {0.0872 ft}^{2}$

Thus

$K A = (0.949 {ft/min)(0.0872 ft}^{2}) = 0.0827 {ft}^{3} / \min$ $K A = (0.949 {ft/min)(0.0872 ft}^{2}) = 0.0827 {ft}^{3} / \min$

The filling rate r_f

$r_{f} = (\frac{1}{8 hr}) (\frac{1 hr}{60 \min}) = 0.00208 \min^{- 1}$ $r_{f} = (\frac{1}{8 hr}) (\frac{1 hr}{60 \min}) = 0.00208 \min^{- 1}$

For splash filling, the nonideal filling factor ϕ is 1.0. The displacement term in Equation 3-24 is

$ϕ r_{f} V_{c} =(1.0)(0.00208 \min^{-1})(10,000 gal) (\frac{{ft}^{3}}{7.48ga1}) = {2.78 ft}^{3} / \min$ $ϕ r_{f} V_{c} =(1.0)(0.00208 \min^{-1})(10,000 gal) (\frac{{ft}^{3}}{7.48ga1}) = {2.78 ft}^{3} / \min$

As expected, the evaporation term is small compared to the displacement term. The concentration is computed from Equation 3-24, using k as a parameter:

$k C_{ppm} = \frac{P^{sat} ϕ r_{f} V_{c}}{Q_{v} P} = \frac{{(0.0371 atm)(2.78 ft}^{3} / \min)}{{(3000 ft}^{3} / \min)(1 atm)} \times 10^{6}$ $k C_{ppm} = \frac{P^{sat} ϕ r_{f} V_{c}}{Q_{v} P} = \frac{{(0.0371 atm)(2.78 ft}^{3} / \min)}{{(3000 ft}^{3} / \min)(1 atm)} \times 10^{6}$

$= 34.4 ppm$ $= 34.4 ppm$

The actual concentration could range from 69 ppm to 344 ppm, depending on the value of k. Sampling to ensure that the concentration is below 20 ppm is recommended. For subsurface filling, ϕ = 0.5, and the concentration range is reduced to 35–172 ppm.

With splash filling or filling with a dip pipe, if the addition rate is low, then the evaporation rate could be higher than the displacement rate.

3-4 Develop and Evaluate Control Techniques to Prevent Exposures

After potential industrial hygiene health hazards are identified and evaluated, the appropriate control techniques must be developed and installed. This requires the application of appropriate technology for reducing workplace exposures. The types of control techniques used in the chemical industry are described in Table 3-11.

Table 3-11 Industrial Hygiene Control Methods

Type and explanation	Typical control techniques
Inherently safer
Eliminate or reduce hazard	Eliminate chemical entirely. Reduce chemical inventories, including raw materials, intermediates, and products. Replace chemical with less hazardous chemical. Decrease temperature and pressure of chemical. Reduce pipeline size to reduce hold-up inventory.
Enclosures
Enclose room or equipment and place under negative pressure.	Enclose hazardous operations such as sample points. Seal rooms, sewers, ventilation, and the like. Use analyzers and instruments to observe inside equipment. Shield high-temperature surfaces.
Local ventilation
Contain and exhaust hazardous substances.	Use properly designed hoods. Use hoods for charging and discharging. Keep exhaust systems under negative pressure. Use ventilation at drumming station. Use local exhaust at sample points.
Dilution ventilation
Design ventilation systems to control low-level toxics.	Design locker rooms with good ventilation and special areas or enclosures for contaminated clothing. Design filter press rooms with directional ventilation. Design ventilation to isolate operations from rooms and offices.
Wet methods
Use wet methods to minimize contamination with dusts.	Clean vessels chemically versus sandblasting. Use water sprays for cleaning. Use water sprays to shield trenches or pump seals. Clean areas frequently.
Good housekeeping
Keep toxicants and dusts contained.	Use dikes around tanks and pumps. Provide water and steam connections for area washing. Provide lines for flushing and cleaning. Provide well-designed sewer system with emergency containment.
Personal protection
As last line of defense	Use aprons, arm shields, and space suits. Use safety glasses and face shields. Wear appropriate respirators; airline respirators are required when oxygen concentration is less than 19.5%.

Designing control methods is an important and creative task. During the design process, the engineer must pay particular attention to ensure that the newly designed control technique provides the desired control and that the new control technique itself does not create another hazard, sometimes even more hazardous than the original problem.

The two major control techniques are environmental controls and personal protection. Environmental control reduces exposure by reducing the concentration of toxicants in the workplace environment. This includes enclosures, local ventilation, dilution ventilation, wet methods, and good housekeeping, as discussed previously.

Personal protection prevents or reduces exposure by providing a barrier between the worker and the workplace environment. This barrier is usually worn by the worker—hence the designation “personal.” Typical types of personal protective equipment (PPE) are listed in Table 3-12. The major problem with personal protection is that it frequently compromises the workers’ abilities to move and is frequently uncomfortable. As a result, PPE should always be considered as a last line of defense.

Table 3-12 Personal Protective Equipment, Not Including Respirators

Type	Description
Hard hat	Protects head from falling equipment and bumps
Safety glasses	Impact-resistant lenses with side shields
Chemical splash goggles, gas-tight	Suitable for liquids and fumes
Steel-toed safety shoes	Protects feet against dropped equipment
Wraparound face shield	Provides additional face protection. Resistant to most chemicals
Vinyl apron	Resists most chemicals
Splash suit	Viton or butyl rubber for nonflammable exposures
Umbilical cord suit	Used with external air supply
Rubber gloves over sleeves	Protects forearms
PVC-coated gloves	Resists acids and bases
PVC and nitrile knee boots	Resists acids, oils, and greases
Ear plugs	Protects against high noise levels

Respirators

Respirators are routinely found in chemical laboratories and plants. Respirators should be used only under the following conditions:

On a temporary basis, until regular control methods can be implemented
As emergency response equipment, to ensure worker safety in the event of an incident
As a last resort, in the event that environmental control techniques are unable to provide satisfactory protection

Respirators always compromise worker ability, so that a worker with a respirator is unable to perform or respond as well as a worker without one. Various types of respirators are listed in Table 3-13.

Table 3-13 Respirators Useful to the Chemical Industry

Exposure	Type	Limitations
Dust	Mouth and nose dust mask	For dusts, fumes, and mists Not for chemicals Oxygen > 19.5% Replace when discolored or clogged
Chemical vapors	Mouth and nose with chemical cartridge	Concentrations less than IDLH Specified concentrations must not be exceeded Oxygen > 19.5% 10 times exposure limit Exposure limits must be known for chemicals
Chemical vapors	Full face mask with chemical cartridge	Concentrations less than IDLH Specified concentrations must not be exceeded Oxygen > 19.5 % 50 times exposure limit Exposure limits must be known for chemicals
Chemical vapors and dusts	Self-contained breathing apparatus (SCBA)	Used for: chemical, biological, radiological, and nuclear exposures Can be used for concentrations greater than IDLH and oxygen < 19.5 % Limited time use depending on air tank capacity

Source: Information from following and National Institute for Occupational Safety and Health. “Respirator Selection Logic.” 2004. www.cdc.gov/niosh/docs/2005-100/

Selection guide: https://multimedia.3m.com/mws/media/639110O/3m-respirator-selection-guide.pdf

General information about masks: www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=standards&p_id=12716

Gas mask description: www.osha.gov/dts/shib/respiratory_protection_bulletin_2011.html

Gas mask limitations: wwwn.cdc.gov/NIOSH-CEL/Limitations/G14#14G_1

Respirators can be used improperly and can be damaged to the extent that they do not provide the needed protection. OSHA and National Institute for Occupational Safety and Health (NIOSH) have developed standards for using respirators,¹⁵ including fit testing (to ensure that the device does not leak excessively), periodic inspections (to ensure that the equipment works properly), specified use applications (to ensure that the equipment is used for the correct job), training (to ensure that it is used properly), and record keeping (to ensure that the program is operating efficiently). All industrial users of respirators are legally bound to understand and fulfill these OSHA requirements. All respirators require training. Annual respirator medical exams are also required.

¹⁵National Institute for Occupational Safety and Health. “Respirator Selection Logic.” 2004. www.cdc.gov/niosh/docs/2005-100/.

Ventilation

For environmental control of airborne toxic material, the most common method is ventilation, for the following reasons:

Ventilation can quickly remove dangerous concentrations of flammable and toxic materials.
Ventilation can be highly localized, reducing the quantity of air moved and the equipment size.
Ventilation equipment is readily available and can be easily installed.
Ventilation equipment can be added to an existing facility.

The major disadvantage of ventilation is the operating cost. Substantial electrical energy may be needed to drive the potentially large fans, and the cost to heat or cool the large quantities of fresh air can be large. These operating costs need to be considered when evaluating alternatives.

Ventilation is based on two principles: (1) dilute the contaminant below the target concentration, and (2) remove the contaminant before workers are exposed. Thus, ventilation systems are composed of fans and ducts. The fans produce a small pressure drop (less than 0.1 psi) that moves the air. The best system is a negative-pressure system, in which the fans are located at the exhaust end of the system, pulling air out. This ensures that leaks in the dust system draw air in from the workplace rather than expel contaminated air from the ducts into the workplace. This arrangement is shown in Figure 3-4.

The positive and negative pressure ventilation systems are shown in a figure. — **Figure 3-4** The difference between a positive-pressure and a negative-pressure ventilation system. The negative-pressure system ensures that contaminants do not leak into workplace environments.

A figure compares the positive-pressure and negative-pressure ventilation systems. The positive-pressure ventilation contains two floors. The first floor has a blower and a hood intake. Leakage out of ducts happens in the second floor. Exhaust escapes out of the duct above the second floor. The negative-pressure ventilation contains two floors. The first floor has hood intake and leakage into ducts happens in the second floor. Blower is present above the second floor. Exhaust escapes out of the duct above the second floor.

Two types of ventilation techniques are used: local and dilution ventilation.

Local Ventilation

The most common example of local ventilation is the hood. A hood is a device that either completely encloses the source of contaminant or moves the air in such a fashion as to carry the contaminant to an exhaust device. Several types of hoods are available:

An enclosed hood completely contains the source of contaminant.
An exterior hood continuously draws contaminants into an exhaust from some distance away.
A receiving hood is an exterior hood that uses the discharge motion of the contaminant for collection.
A push–pull hood uses a stream of air from a supply to push contaminants toward an exhaust system.

The most widely encountered enclosed hood is the laboratory hood, like the standard laboratory utility hood shown in Figure 3-5. Fresh air is drawn through the window area of the hood and is removed out the top through a duct. The airflow profiles within the hood are highly dependent on the location of the window sash. It is important to keep the sash open a few inches, at a minimum, to ensure adequate fresh air. However, the sash should never be fully opened because contaminants might escape. The baffle, which may be present at the rear of the hood, ensures that contaminants are removed from the working surface and the rear lower corner.

Airflow into a laboratory hood in accordance with sash height is illustrated via two diagrams. — **Figure 3-5** Standard utility laboratory hood. Air flow patterns and control velocity are dependent on sash height. (Source: Adapted from N. Irving Sax. *Dangerous Properties of Industrial Materials*, 4th ed. (New York, NY: Van Nostrand Reinhold, 1975), p. 74.)

Two diagrams show the airflow into a standard utility laboratory hood. In the first diagram, the sash height is low and minimal air flows into the hood. The baffle redirects the air into the exhaust duct. In the second diagram, the sash height is higher and adequate fresh air flows through the hood. Baffle redirects the air into the exhaust duct.

Another type of laboratory hood is the bypass hood, shown in Figure 3-6. In this design, bypass air is supplied through a grille at the top of the hood. This ensures the availability of fresh air to sweep out contaminants in the hood. The bypass air supply is reduced as the hood sash is opened.

A pair of diagrams illustrate the flow of air into a Standard bypass laboratory hood. — **Figure 3-6** Standard bypass laboratory hood. The bypass air is controlled by the height of the sash. (Source: Adapted from N. Irving Sax. *Dangerous Properties of Industrial Materials*, 4th ed. (New York, NY: Van Nostrand Reinhold, 1975), p. 75.)

A pair of diagrams shows the flow of air into a Standard bypass laboratory hood, according to the sash height. The first diagram shows the flow of air into the hood via the bypass, the sash is fully closed. The second diagram shows the flow of air into the hood, when the sash blocks the bypass. In both the cases, the air enters into the hood and escape via the exhaust duct.

Enclosed hoods offer the following advantages:

Completely eliminate exposure to workers
Require minimal airflow
Provide a containment device in the event of fire or explosion
Provide a shield to the worker by means of a sliding door on the hood

However, hoods also have some disadvantages:

Limit the workspace available
Can be used only for small, bench-scale or pilot plant equipment

Most hood calculations assume plug flow. For a duct of cross-sectional area A and average air velocity ū (distance/time), the volume of air moved per unit time Q_v is computed as

$\begin{matrix} Q_{v} = A \bar{u} & (3-25) \end{matrix}$ $\begin{matrix} Q_{v} = A \bar{u} & (3-25) \end{matrix}$

For a rectangular duct of width W and length L, Q_v is determined using the equation

$\begin{matrix} Q_{v} = L W \bar{u} & (3-26) \end{matrix}$ $\begin{matrix} Q_{v} = L W \bar{u} & (3-26) \end{matrix}$

Consider the simple box-type enclosed hood shown in Figure 3-7. The design strategy is to provide a fixed velocity of air at the opening of the hood. This face or control velocity (referring to the face of the hood) ensures that contaminants do not exit from the hood.

A diagram shows a box-type hood. — **Figure 3-7** Determining the total volumetric airflow rate for a box-type hood. For general operation, a control velocity of between 80 and 120 feet per minute (fpm) is desired.

A diagram of a box-type hood is shown. The height of the box is W, and the length is L. The exhaust of the hood is labeled out. A text below the sketch reads as follows. Q subscript v, equals L times W times u bar. Q subscript v, equals Volumetric flow rate, Volume or time. L equals length. W equals Width. u bar equals Required control velocity.

The required control velocity depends on the toxicity of the material, the depth of the hood, and the evolution rate of the contaminant. Shallower hoods need higher control velocities to prevent contaminants from exiting the front. However, experience has shown that higher velocities can lead to the formation of a turbulent eddy from the bottom of the sash; backflow of contaminated air is possible. For general operation, a control velocity between 80 and 120 feet per minute (fpm)—that is, 24.4 to 36.6 m3/min— is suggested.

The airflow velocity is a function of the sash height and the blower speed. Arrows are frequently used to indicate the proper sash height to ensure a specified face velocity. Instruments are available for measuring the airflow velocity at specific points of the hood window opening. Testing of this velocity is an OSHA requirement. Design equations are available for a wide variety of hood and duct shapes.¹⁶

¹⁶Industrial Ventilation: A Manual of Recommended Practice, 27th ed. (Cincinnati, OH: American Conference of Governmental Industrial Hygienists, 2010).

Other types of local ventilation methods include “elephant trunks” and free-hanging canopies and plenums. The elephant trunk is simply a flexible vent duct with a diameter of 10 cm or more that is positioned near a source of contaminant. It is most frequently used for loading and unloading toxic materials from drums and vessels. Free-hanging canopies and plenums can be either fixed in position or attached to a flexible duct to enable movement. These methods will most likely expose workers to toxicants, but in diluted amounts.

Dilution Ventilation

If the contaminant cannot be placed in a hood and must be used in an open area or room, dilution ventilation is necessary. Unlike hood ventilation, where the airflow prevents worker exposure, dilution ventilation always exposes the worker to the chemical, but in amounts diluted by fresh air. Dilution ventilation always requires more airflow than local ventilation, so the operating expenses for such a system can be substantial.

Equations 3-9, 3-12, and 3-14 are used to compute the ventilation rates required. Table 3-14 1ists values for k, the nonideal mixing factor used with these equations.

Table 3-14 Nonideal Mixing Factor, k, for Various Dilution Ventilation Conditions

	Vapor concentration (ppm)
Ventilation condition	0–100	101–500	More than 500
Poor	1/11	1/8	1/7
Average	1/8	1/5	1/4
Good	1/7	1/4	1/3
Excellent	1/6	1/3	1/2

For exposures to multiple sources, the dilution air requirement is computed for each individual source. The total dilution requirement is the sum of the individual dilution requirements.

The following restrictions should be considered before implementing dilution ventilation:

The contaminant must not be highly toxic.
The contaminant must be evolved at a uniform rate.
Workers must remain a suitable distance from the source to ensure proper dilution of the contaminant.
Scrubbing systems must not be required to treat the air before exhaust into the environment.

Example 3-11

Xylene is used as a solvent in paint. A certain painting operation evaporates an estimated 10 liters of xylene in an 8-hour shift. The ventilation quality is rated as average. Determine the quantity of dilution ventilation air required to maintain the xylene concentration below 100 ppm, the TLV-TWA. Also, compute the air required if the operation is carried out in an enclosed hood with an opening of 5 m² and a face velocity of 30 m/min. The temperature is 25°C and the pressure is 1 atm. The specific gravity of xylene is 0.864, and its molecular weight is 106.

Solution

The evaporation rate of xylene is

$\begin{array}{l} Q_{m} = (\frac{{0.01 m}^{3}}{8 hr}) (\frac{1 hr}{60 \min}) (\frac{864 kg}{m^{3}}) \\ = 0.0180 kg/min \end{array}$ $\begin{array}{l} Q_{m} = (\frac{{0.01 m}^{3}}{8 hr}) (\frac{1 hr}{60 \min}) (\frac{864 kg}{m^{3}}) \\ = 0.0180 kg/min \end{array}$

From Table 3-14, for average ventilation and a vapor concentration of 100 ppm, k = 1/8 = 0.125. Using Equation 3-9, we solve for Q_v:

$\begin{array}{l} Q_{v} = \frac{Q_{m} R_{g} T}{k C_{ppm} P M} \times 10^{6} = \frac{(0.0180 kg/min) (0.082057 m^{3} atm/kg-mole K) (298 K)}{(0.125) (100) (1 atm) (106 kg/kg-mole)} \times 10^{6} \\ = 334 m^{3} /min \end{array}$ $\begin{array}{l} Q_{v} = \frac{Q_{m} R_{g} T}{k C_{ppm} P M} \times 10^{6} = \frac{(0.0180 kg/min) (0.082057 m^{3} atm/kg-mole K) (298 K)}{(0.125) (100) (1 atm) (106 kg/kg-mole)} \times 10^{6} \\ = 334 m^{3} /min \end{array}$

For a hood with an open area of 5 m², using Equation 3-25 and assuming a required control velocity of 30 m/min, we get

$Q_{v} = A \bar{u} = {(5 m}^{2}) ({30 m}^{3} /min) = {150 m}^{3} / \min$ $Q_{v} = A \bar{u} = {(5 m}^{2}) ({30 m}^{3} /min) = {150 m}^{3} / \min$

The hood requires significantly less airflow than dilution ventilation and prevents worker exposure completely.

3-5 National Fire Protection Association Diamond

Another method that is used to characterize the hazardous properties of chemicals is the National Fire Protection Association (NFPA) diamond. The NFPA is a professional society that was established in 1896 to reduce worldwide fatalities and injuries due to fires and other hazards. Its primary function is to promote consensus codes and standards, including the National Electrical Code (NEC).

The NFPA diamond frequently appears on chemical containers and storage vessels. The main purpose of the diamond is to provide a quick means for emergency response personnel to recognize the chemical hazards that they may face during a fire or other emergency. However, it is also useful for routine operations where hazards recognition is important. NFPA diamonds are frequently found on chemical containers and process storage vessels.

The NFPA diamond consists of four separate areas, as shown in Figure 3-8. These areas correspond to health, fire, stability, and special hazards. In the past, the word “reactivity” was used instead of “stability,” but chemical stability is a more accurate description of the hazard facing emergency response personnel during a fire incident.

NFPA 704 diamond is illustrated in a figure. — **Figure 3-8** The NFPA 704 diamond used to identify chemical hazards. (Source: Reprinted with permission from NFPA 704-2017, System for the Identification of the Hazards of Materials for Emergency Response, Copyright © 2016, National Fire Protection Association.)

A figure shows the NFPA 704 diamond. Flammability Rating is as follows. 4 - Rapidly or completely vaporize and burn readily. 3 - Ignite readily in ambient conditions. 2 Ignite when moderately heated. 1 Require preheating for ignition. 0 Will not burn under normal fire conditions. Instability rating is as follows. 4 May detonate or have explosive reaction. 3 Shock and Heat may detonate or cause explosive reaction. 2 Violent chemical change at elevated temperatures. 1 Unstable if heated. 0 Normally stable. The symbols for special hazards are as follows. OX Oxidizers. W Water Reactives. SA Simple Asphyxiants. No other hazards should be listed in this quadrant. In cases where a unique hazard symbol exists it must be placed outside of the Special Hazards quadrant. The Health rating is as follows. 4 Can be lethal. 3 Serious or permanent injury. 2 Temporary incapacitation or residual injury. 1 Significant irritation. 0 No hazard beyond ordinary combustibles.

The simplicity of the NFPA diamond is that the respective hazards are indicated by a number, with the number 0 representing minimal hazard and the number 4 representing the greatest hazard. Figure 3-8 shows how the numbers are assigned. Note that the NFPA hazard numbers are the reverse of the GHS system. Appendix E contains NFPA numbers for a variety of common chemicals.

Online Resources

National Institute for Occupational Safety and Health (NIOSH), www.cdc.gov/niosh.

U.S. Code of Federal Regulations, www.gpoaccess.gov.

U.S. Occupational Safety and Health Administration (OSHA), www.osha.gov.

Problems

3-1. Search the Internet and

Print a GHS label for phenol. If you are unable to find a label, then develop your own label.
Develop a few conclusions concerning the process of finding labels.

3-2. Copy the GHS label pictograms for (a) oxidizers, (b) flammables, (c) explosives, (d) acute toxicity, (e) corrosives, and (f) carcinogens.

3-3. An air mixture contains 4 ppm of carbon tetrachloride, 25 ppm of 1,1-dichloroethane, 5 ppm of diethyl amine, 20 ppm of cyclohexanol, and 10 ppm of propylene oxide. Determine the mixture TLV and determine if this TLV has been exceeded.

3-4. An open container was monitored and found to have a toluene evaporation rate of 0.1 g/min. The ventilation rate is 2.84 m³/min, the temperature is 27°C, and the pressure is 1 atm. Estimate the concentration of toluene vapor in the enclosure and compare it to the TLV of 50 ppm. Perform the calculation for a k of 1.0 and 0.1.

3-5. An ideal liquid mixture of benzene and toluene (50% by volume each) is used in a plant at 27°C and 1 atm. Determine (a) the mixture TLV, (b) the evaporation rate per unit area for this mixture, and (c) the ventilation rate required to keep the concentration below the TLV if the liquid is contained in a drum with a bung diameter of 5.1 cm. Assume the ventilation quality within the vicinity is average.

3-6. A drum contains 0.16 m³ of toluene. If the lid is left open (lid diameter is 0.92 m), determine the

Time required to evaporate all of the toluene.
Concentration of toluene (in ppm) near the drum if the local ventilation rate is 28.34 m³/min. The temperature is 30°C and the pressure is 1 atm.

3-7. A railcar is being splash filled with toluene. The car is 38 m³ in volume, and the filling hole is 0.10 m in diameter. It takes 30 min to fill the rail car. Estimate the concentration of toluene in this vicinity if the ventilation rate is 85 m³/min. The temperature is 25°C and the pressure is 1 atm.

3-8. Xylene is used in a hood that has a face area of 4.65 m², with a temperature of 25°C and pressure of 1 atm. The evaporation rate of xylene is 11.6 liters in 8 hours.

Determine the dilution ventilation required to keep the area concentration below the TLV-TWA.
Compare this rate to the recommended face velocity for hoods.

3-9. Determine the recommended air volumetric flow rate, in m³/min, that is required for a hood having face dimensions of 1.22 m (height) and 0.91 m (width).

3-10. If the hood in Problem 3-9 is used for handling trichloroethylene, determine the rate of air passing into the hood face to keep the concentration (a) below the TLV-TWA and (b) below the lower explosive limit. Then (c) compare these rates with the recommended rate shown in Problem 3-9. The temperature is 25°C and the pressure is 1 atm. In all cases, the evaporation rate is 0.04 kg/min.

3-11. Use Safety Data Sheets to determine the TLV and NFPA ratings for ethanol, chlorine, and phosgene.

3-12. A worker is exposed to multiple sound sources: $I = I_{1} + I_{2} + \dots + I_{n} = \sum I_{i}$ $I = I_{1} + I_{2} + \dots + I_{n} = \sum I_{i}$ . Derive the following equation for the net sound exposure from these multiple sources:

$d B A = 10 log (Σ_{i = 1}^{n} 10^{d B_{i} / 10})$ $d B A = 10 log (Σ_{i = 1}^{n} 10^{d B_{i} / 10})$

Two sound sources are at 90 dBA. What is the net exposure from these two sound sources?

3-13. Two liters of a liquid chemical is spilled on the laboratory floor. Assume 1 atm of pressure and a temperature of 25°C.

If the liquid pool is 2 mm deep, calculate the area of the pool.
What ventilation rate, in m³/s, is required to prevent the air concentration from exceeding the TLV for the chemical of 100 ppm?
Ventilation air is provided for the lab via a 0.5-m-square duct. The air velocity exiting the duct is measured with a velometer at 30 m/s. Calculate the volumetric flow of air from this duct in m³/s. Is this adequate ventilation for this spill?

Data for the liquid chemical:

Molecular weight: 100
Saturation vapor pressure of chemical at 25°C: 100 mm Hg

3-14. A large workshop uses a 2:1 mixture by volume of liquid xylene and hexane to strip varnish from antique furniture. A total of 15 liters is used per 8-hour shift. The workshop is maintained at 25°C and 1 atm and has a ventilation system that provides 1 m³/s of fresh air. Is the ventilation system adequate to maintain the concentrations below the TLV-TWAs?
Xylene: molecular weight: 106.16, specific gravity: 0.870, TLV-TWA: 100 ppm
Hexane: molecular weight: 86.18, specific gravity: 0.657, TLV-TWA: 50 ppm

Additional homework problems are available in the Pearson Instructor Resource Center.

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Table of Contents for
Chapter 3. Industrial Hygiene

Chapter 3. Industrial Hygiene

3-1 Anticipating and Identifying Hazardous Workplace Exposures

3-2 Globally Harmonized System

Globally Harmonized System for Safety Data Sheets

Globally Harmonized System for Labeling

Six Elements of the GHS Label

Primary and Secondary Containers

3-3 Evaluate the Magnitude of Exposures and Responses

Evaluating Exposures to Volatile Toxicants by Monitoring

Evaluating Worker Exposures to Dusts

Evaluating Worker Exposures to Noise

Evaluating Worker Exposures to Thermal Radiation

Estimating Worker Exposures to Toxic Vapors

Estimating the Vaporization Rate of a Liquid

Estimating Worker Exposures during Vessel Filling Operations

3-4 Develop and Evaluate Control Techniques to Prevent Exposures

Respirators

Ventilation

Local Ventilation

Dilution Ventilation

3-5 National Fire Protection Association Diamond

Online Resources

Suggested Reading

Industrial Hygiene

Ventilation

Problems

Table of Contents for Chapter 3. Industrial Hygiene

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Chapter 3. Industrial Hygiene

3-1 Anticipating and Identifying Hazardous Workplace Exposures

3-2 Globally Harmonized System

Globally Harmonized System for Safety Data Sheets

Globally Harmonized System for Labeling

Six Elements of the GHS Label

Primary and Secondary Containers

3-3 Evaluate the Magnitude of Exposures and Responses

Evaluating Exposures to Volatile Toxicants by Monitoring

Evaluating Worker Exposures to Dusts

Evaluating Worker Exposures to Noise

Evaluating Worker Exposures to Thermal Radiation

Estimating Worker Exposures to Toxic Vapors

Estimating the Vaporization Rate of a Liquid

Estimating Worker Exposures during Vessel Filling Operations

3-4 Develop and Evaluate Control Techniques to Prevent Exposures

Respirators

Ventilation

Local Ventilation

Dilution Ventilation

3-5 National Fire Protection Association Diamond

Online Resources

Suggested Reading

Industrial Hygiene

Ventilation

Problems

Table of Contents for
Chapter 3. Industrial Hygiene