# MOS 5201 CSU Introduction to Safety Engineering Discussion

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UNIT VI STUDY GUIDE
Control Strategies and
Quantitative Problem-Solving
Course Learning Outcomes for Unit VI
Upon completion of this unit, students should be able to:
5. Formulate control strategies based on quantitative problem-solving.
5.1 Determine the risks and control measures associated with pressure.
5.2 Determine the risks and control measures associated with airborne contaminants using
ventilation.
5.3 Determine the risks and control measures associated with ionizing radiation.
Course/Unit
Learning Outcomes
5.1
5.2
5.3
Learning Activity
Unit Lesson
Chapter 19, pp. 275–285
Unit VI Assessment
Unit Lesson
Chapter 25, pp. 357–371
Unit VI Assessment
Unit Lesson
Chapter 22, pp. 307–315
Unit VI Assessment
Chapter 19: Pressure, pp. 275–285
Chapter 22: Ionizing Radiation, pp. 307–315
Chapter 25: Ventilation, pp. 357–371
Unit Lesson
In order to provide control measures for employee risks, it is first necessary to understand the tools to
evaluate those risks. In Unit VI, we will discuss calculations to evaluate pressures inside containers, which
vary depending on volume and temperature. Also included in this unit will be an explanation on how to
calculate the release of chemical vapors, along with the calculations to control or reduce those vapors using
ventilation calculations. Finally, this unit will cover basic calculations to evaluate employee exposures to
radiation as well as the tools to evaluate the methods to reduce or eliminate an employee’s exposure to those
Pressure
In nearly every industrial, commercial, manufacturing, or mining operation today, the use of pressure to
perform work is critical. Examples of pressure uses to perform work include pressure vessels, such as boilers,
to create steam or other forms of energy or products; hydraulic systems; storage tanks; and associated piping
systems. The hazards associated with pressure can be serious and even fatal. Therefore, it is incumbent on
every safety professional and engineer to have a good basic understanding of gas laws to calculate pressure.
MOS 5201, Safety Engineering
1
The three basic gas laws that will be discussed and used in this chapter include
Boyle’s
law, Charles’s
UNIT
x STUDY
GUIDE law,
and Boyle’s-Charles’s law (otherwise known as the combined gas law). Each of
these will be discussed in
Title
greater detail, and examples will be provided as to how to use them in a practical application.
Boyle’s Law
Boyle’s law is used to determine the differences in pressure and volume when one or the other increases or
decreases. For example, when the pressure increases or decreases, the volume is inversely proportional. It is
represented mathematically, which is shown below.
𝑃1 𝑉1 = 𝑃2 𝑉2
Where:
P = pressure exerted
V = volume of gas
Example
A compressed gas cylinder has a volume of 18.2 liters, exerting a pressure of 650 mmHg. Determine the
volume of the cylinder if the pressure increases to 760 mmHg.
(18.2 𝑙𝑖𝑡𝑒𝑟𝑠)(650 𝑚𝑚𝐻𝑔) = (760 𝑚𝑚𝐻𝑔)(𝑉2 )
(18.2 𝑙𝑖𝑡𝑒𝑟𝑠)(650 𝑚𝑚𝐻𝑔)
(760 𝑚𝑚𝐻𝑔)(𝑉2 )
=
760 𝑚𝑚𝐻𝑔
760 𝑚𝑚𝐻𝑔
15.6 𝑙𝑖𝑡𝑒𝑟𝑠 = 𝑉2
According to Boyle’s law, when the pressure increases, the volume will decrease and vice versa.
Charles’s Law
Charles’s law is used to determine what occurs with a containerized gas when there is a change in volume or
temperature. Of special note in the following equation is that the temperature is expressed in terms of
degrees Kelvin. Expressed mathematically, the equation is shown below.
𝑉1
𝑉2
=
𝑇1
𝑇2
Where:
V = volume of gas
T = temperature of gas (Kelvin temperature scale)
Example
A compressed gas cylinder of oxygen has a volume of 18.2 liters at a temperature of 30 oC. The compressed
gas cylinder is removed from the transport truck and allowed to sit outside a maintenance shop for 3 hours in
the middle of a summer afternoon, where the temperature reaches 36oC. Determine the volume based on the
new temperature.
To perform the calculation using Charles’s law, it is first necessary to covert the oC to oK for both
temperatures. This is done using the calculation below.
𝑡𝑜𝐾 = 𝑡𝑜𝐶 + 273
MOS 5201, Safety Engineering
2
Where:
UNIT x STUDY GUIDE
Title
toK = temperature in Kelvin
toC = temperature in Celsius
𝑡𝑜𝐾 = 300 𝐶 + 273
𝑡𝑜𝐾 = 303
and
𝑡𝑜𝐾 = 36𝐶 + 273
𝑡𝑜𝐾 = 309
Now we can insert the temperatures for each into the equation of Charles’s law below.
18.2 𝑙𝑖𝑡𝑒𝑟𝑠
𝑉2
=
303𝑜 𝐾
309𝑜 𝐾
(309𝑜 𝐾)
18.2 𝑙𝑖𝑡𝑒𝑟𝑠
𝑉2
=
(309𝑜 𝐾)
𝑜
303 𝐾
309𝑜 𝐾
18.6 𝑙𝑖𝑡𝑒𝑟𝑠 = 𝑉2
Some important conversion factors to know when calculating and understanding pressures include the
different measurements compared to an atmosphere, which are listed below.
1 atm = 14.6959488 pounds per square inch (psi)
1 atm = 29.9246899 inches of mercury (in Hg)
1 atm = 760 mm mercury (mm Hg)
1 atm = 760 torr (torr)
1 atm = 101,325 pascals (Pa)
1 atm = 101.325 kilopascals (kPa)
1 atm = 1.01325 bar (bar)
Combined Gas Law
Occasionally, there is a need to use a combination of both Boyle’s law and Charles’s law, commonly referred
to as the combined gas law. The combined gas law is represented mathematically below.
𝑃1 𝑉1
𝑃2 𝑉2
=
𝑇1
𝑇2
Where:
P = absolute pressure (atm)
V = volume of gas (liters)
T = temperature of gas (kelvin temperature scale)
Example
A compressed gas cylinder sitting on a loading dock is under pressure at 1.8 atm with an initial temperature of
78oF. Determine what the pressure would be if the temperature rises to 97 oF. Note: Since there has been no
change in the volume, you can assume a volume of 1 on both sides of the equation.
MOS 5201, Safety Engineering
3
Using the combined gas law, we insert the information into the equation. Remember
convertGUIDE
the oF to oK.
UNIT xto
STUDY
1.8 𝑎𝑡𝑚(1 𝑙𝑖𝑡𝑒𝑟𝑠)
(𝑃) (1)
=
𝑜
298.6 𝐾
309.1𝑜 𝐾
Title
(𝑃)(1)309.1𝐾
309.1 𝐾 (1.8 𝑎𝑡𝑚)(1 𝑙𝑖𝑡𝑒𝑟𝑠)
=
298.6 𝐾
309.1 𝐾
556.4 𝑎𝑡𝑚
=𝑃
298.6 𝐾
1.87 𝑎𝑡𝑚 = 𝑃
Ventilation
There are three major types of ventilation, including general ventilation, dilution ventilation and local exhaust
ventilation (LEV). General ventilation is used primarily for the comfort of the occupants, such as temperature
and humidity control. Dilution ventilation is used to bring a contaminant of moderate toxicity to an acceptable
exposure level. LEV is used to capture and exhaust air contaminants of highly toxic substances that are
generated from a single source.
Calculating for Volumetric Air Flow
The most basic of all ventilation equations uses a cross-sectional area of a round or square duct, with the air
velocity traveling through the duct to calculate the overall volumetric flow rate, which is expressed in cubic
feet per minute.
𝑄=𝑉𝑥𝐴
Where:
Q = volumetric flow rate (expressed in CFM (cubic feet per minute))
V = Velocity of the air (expressed in FPM (feet per minute))
A = Cross-sectional area of duct (expressed in square feet of SF)
Example
A supply air duct measures 12 inches in diameter with an air velocity of 700 feet per minute. Determine the
overall volumetric flow rate.
The first step is to determine that the radius (1/2 the diameter) is equal to 6 inches of .5 feet. It is important to
make sure that the unit of measurements are consistent with the requirements of the equation. Using the
equation for the area of a circle (Area = πr2), the duct cross section is 0.79 ft2. The next step is to take the
area and the velocity and insert into the equation below.
𝑄 = 700 𝑓𝑝𝑚 𝑥 0.79 𝑠𝑓
𝑄 = 553 𝐶𝐹𝑀
How can this information be used? One use is to determine the amount of air that enters an area.
Contaminant Generation
When using dilution ventilation it is important to understand the flow rate of fresh air is determined by (a)
contaminant generation, (b) proper mixing and (c) target final concentration. The methods used apply to uniform
rates of generation and low to moderate toxicity. The accumulation of contaminants is equal to the generation
MOS 5201, Safety Engineering
4
minus the removal. The following equation is used to determine the concentration
buildup
that GUIDE
will occur over a
UNIT
x STUDY
given period of time: (Yates, 2015 Chapter 7)
Title
𝑙𝑛 (
𝐺 − 𝑄′ 𝐶2
𝑄1 (𝑡2 − 𝑡1 )
)
=
𝐺 − 𝑄′ 𝐶1
𝑉
Where:
G = rate of generation of contaminant (cfm)
Q’ = (Q/K)
K = design distribution constant (a constant factor (1-10))
C = concentration at a given time (ppm)
V = volume of room or enclosure (ft3)
Q = flow rate (cfm)
t2-t1 = time interval or ∆t
Example
Acetone is being generated under the following conditions: G = 1.4 cfm; Q’ = 2500 cfm; V = 75,000 ft3; C1 =
0; K = 2. How long will it take (in minutes) before the concentration reaches 175 ppm (parts per million)?
The question is straightforward and asks that we calculate for time, which is the unknown. We can use the
above equation, but must first manipulate it in order to obtain the desired outcome (time). We, therefore set
up the equation as follows: (Yates, 2015 Chapter 7)
∆𝑡 = −
𝑉
𝐺 − 𝑄′ 𝐶2
[ln
(
)]
𝑄′
𝐺
Now, insert the data from our example into the equation, and solve for time. See below.
∆𝑡 = −
75,000 𝑓𝑡 3
1.4 𝑐𝑓𝑚 − 2500 𝑐𝑓𝑚(0.000175)
[ln | (
) |]
2500 𝑐𝑓𝑚
1.4 𝑐𝑓𝑚
∆𝑡 = −30 [ln | (
0.9625
) |]
1.4
∆𝑡 = −30[ln |0.6875]|
∆𝑡 = −30(−0.3747)
∆𝑡 = 11.24 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
Note: In the above equation, it was necessary to convert the concentration of 175 ppm to a decimal
number. We did this by taking 175 and dividing by 106, with the result being 0.000175.
Simplified, it will take approximately 11.24 minutes for an area with zero concentration of acetone, and a volume
of 75,000 ft3, generating 1.4 cfm of acetone, and a Q’ of 2500 cfm to reach a concentration of 175 ppm.
Example:
Another example problem may ask you to calculate the concentration after an established timeframe. For
example, given the same data as listed in the previous problem, what would be the concentration of acetone
after 60 minutes?
MOS 5201, Safety Engineering
5
Again, we must manipulate the original equation in order to obtain the requested
(concentration).
UNITinformation
x STUDY GUIDE
We can use the original equation and manipulate it as follows:
Title
𝐺 [1 − 𝑒
𝐶2 =
(−
𝑄′ ∆𝑡
)
𝑉 ]
𝑥 106
𝑄′
Now we can insert the known data from the previous problem, as follows:
1.4 𝑐𝑓𝑚 [1 − 𝑒
𝐶2 =
(−
2500 𝑐𝑓𝑚 (60)
)
75,000 𝑓𝑡 3
]
2500 𝑐𝑓𝑚
𝐶2 =
𝑥 106
1.4 𝑐𝑓𝑚 [1 − 0.1353]
𝑥 106
2500
𝐶2 =
1.4 𝑐𝑓𝑚 (0.8647)
𝑥 106
2500 𝑐𝑓𝑚
𝐶2 = 0.000484 𝑥 106
𝐶2 = 484.2 𝑝𝑝𝑚
Given the criteria for the equation in the previous problem, you can see that after 60 minutes the concentration
of acetone will be 484.2 ppm (parts per million).
LEV – Calculating Capture Velocities
When faced with an issue of removing highly toxic materials from a single source, it is necessary to determine
the capture velocities of the exhaust system. An example of a single source, highly toxic material that needs
to be removed is a welding operation on stainless steel, which contains hexavalent chromium. The use of an
LEV with the hood close to the source of the contamination can be used to significantly remove the
contaminant. Of special note when using the following equation is that it is only accurate for a distance of 1.5
times the duct diameter.
𝑉=
𝑄
10𝑋 2 + 𝐴
Where:
V = Velocity (fpm)
Q = Flow Rate (cfm)
X = source distance from hood opening (ft) (Note: Equation is only accurate for a limited distance of
1.5 times the diameter of a round duct or the side of a rectangle or square duct)
A = Area (sq. ft.)
Example
The welding operation mentioned above requires the use of an LEV with a duct diameter of 12 inches with a
volumetric flow rate of 950 cfm, measured at a distance of 18 inches. The first thing to look at is the distance
from the hood, which is exactly 1.5 times the actual duct diameter, so the equation below can be used.
MOS 5201, Safety Engineering
6
You must first calculate the area of the duct (12-inch diameter) by using the following
equation.
UNIT x STUDY
GUIDE
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2
Title
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 3.14 𝑥 0.5 𝑓𝑡 2
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 3.14(0.25 𝑓𝑡)
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 0.79 𝑠𝑞. 𝑓𝑡.
Now, we insert the known data into the equation as shown below.
𝑉=
950 𝑐𝑓𝑚
10(1.5 𝑓𝑡)2 + 0.79 𝑠𝑞. 𝑓𝑡
𝑉=
950
23.29
𝑉 = 40.79 𝑓𝑝𝑚
Therefore, the capture velocity under this given scenario is 40.79 fpm.
A safety engineer or safety professional must also have a basic understanding of radiation. Simplified,
radiation exposures are controlled by one of three methods or a combination of both. The method of exposure
control involves time, distance, and shielding. The dose an employee received is a function of the rate
multiplied by the time of exposure, which is expressed mathematically as shown below.
Dose = Dose Rate x Time of Exposure
Example
A medical radiological technician is exposed to 28 mrem per hour for an 8-hour work period. Her dose is
calculated as shown below.
𝐷𝑜𝑠𝑒 = 28
𝑚𝑟𝑒𝑚
𝑥 8 ℎ𝑜𝑢𝑟𝑠
ℎ𝑜𝑢𝑟
𝐷𝑜𝑠𝑒 = 224 𝑚𝑟𝑒𝑚
Distance
Using the inverse square equation, we can easily determine what an employee’s exposure can be if he or she
moves outward from the source. For example, if a laboratory technician has an exposure of 35 mrem/hour at
a distance of 0.5 feet from the source and has found a way to work 2 feet from the source, we can determine
the exposure at 2 feet using the equation below.
𝐼2 = 𝐼1
(𝑑1 )2
(𝑑2 )2
Where:
I1 = intensity at distance 1
I2 = intensity at distance 2
d1 or d2 = distances at locations 1 and 2 (in feet)
MOS 5201, Safety Engineering
7
Example
UNIT x STUDY GUIDE
Title
Insert the data from above.
𝐼2 = 𝐼1
(𝑑1 )2
(𝑑2 )2
𝐼2 = 35 𝑚𝑟𝑒𝑚/ℎ𝑟
(1 𝑓𝑜𝑜𝑡)2
(2 𝑓𝑒𝑒𝑡)2
𝐼2 = 35 𝑚𝑟𝑒𝑚/ℎ𝑟
1
4
𝐼2 = 8.75 𝑚𝑟𝑒𝑚/ℎ𝑜𝑢𝑟
Shielding Calculation
The third method for controlling radiation exposure is by shielding. Various materials have different absorption
coefficients.
𝑰 = 𝑰𝒐 𝒆− µ𝒙
Where:
I = intensity after shielding
Io = original intensity
µ = linear absorption coefficient of material
x = shield thickness
Example
The same laboratory technician has determined that he cannot work 2 feet from the source, but he thinks that
shielding will work instead and make the work more productive. Given the following data, determine the
intensity of the radiation on the opposite side of the shield. The intensity at 0.5 feet from 137Cs source is 35
mrem/hr. What is the intensity at this point if a 5cm lead shield is placed between the source and the
detector? [µ for lead, (662 keV gamma ray) = 1.23 cm -1]
𝐼 = (35
𝑚𝑟𝑒𝑚 − 1.23 𝑥 5
)𝑒
ℎ𝑟
𝐼 = (35
𝑚𝑟𝑒𝑚
) (0.002)
ℎ𝑟
𝐼 = 0.075 𝑚𝑟𝑒𝑚/ℎ𝒓
MOS 5201, Safety Engineering
8
When dealing with Beta (β) particle radiation, we must factor in the radiation scatter
because it
UNIT xbuild-up
STUDYfactor
GUIDE
has charge and mass. This is done by the addition in the equation, which is shown
Title below.
𝑰 = 𝜷𝑰𝒐 𝒆−µ𝒙
Where:
I = intensity after shielding
Io = original intensity
µ = linear absorption coefficient of material
x = shield thickness
β = radiation scatter build-up factor (For radiation workers, OSHA assumes β = 1)
Note: Build-up factors can be found in American Nuclear Standard 6.4.3. Using a β of 1 would not make a
difference in this calculation.
In order to access the following resources, click the links below.
The following will help students to develop an understanding of the combined gas law through basic
experiments.
Anderson, L., Nobile, N., & Cormas, P. (2011). Teaching the combined gas law. Science Scope, 35(1), 60.
http://go.galegroup.com.libraryresources.columbiasouthern.edu/ps/retrieve.do?tabID=T002&resultList
osition=1&docId=GALE%7CA268067212&docType=Article&sort=RELEVANCE&contentSegment=&p
rodId=ITOF&contentSet=GALE%7CA268067212&searchId=R2&userGroupName=oran95108&inPS=
true
Concrete cutting in construction is a major source of exposure to respirable crystalline silica. To reduce
exposure, local exhaust ventilation (LEV) may be integrated into hand tools used in concrete cutting.
Shepherd, S., Woskie, S. R., Holcroft, C., & Ellenbecker, M. (2008). Reducing silica and dust exposures in
construction during use of powered concrete-cutting hand tools: Efficacy of local exhaust ventilation
on hammer drills. Journal of Occupational and Environmental Hygiene, 6(1), 42–51. Retrieved from
https://www-tandfonlinecom.libraryresources.columbiasouthern.edu/doi/pdf/10.1080/15459620802561471
Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit
them. If you have questions, contact your instructor for further guidance and information.
The Occupational Safety and Health Administration (OSHA) maintains a video library including several safety
topics related to specific industries. Click the following link to view several videos related to construction
safety on OSHA’s website: https://www.osha.gov/video/index.html.
MOS 5201, Safety Engineering
9
THESE (4) QUESTION
Explored Hierarchy of Controls in the areas of walking and working surfaces, tools and
machines, and materials handling. Please further explore the hazards and their controls in the
QUESTION 1
Your quality control/quality assurance manager has requested your assistance in the testing
and repair facility. One of the test procedures utilizes a radiation source that is emitting gamma
radiation at a rate of 50 mrem/hour at a distance of 1 foot. This testing is done for
approximately 2 hours per day, 2 days per week.
The Occupational Safety and Health Administration (OSHA) has a PEL of 1.25 rem per quarter
and 5 rem per year.
Determine the employee’s exposure for 1 year.
Calculate the exposure rate if the employee was moved to a distance of 3 feet from the
Calculate the exposure rate of the employee if a 5 cm lead shield was installed between the
source and the detector. The employee is working at a distance of 1 foot from the source. [µ for
lead, (662 keV gamma ray) = 1.23cm-1]
Your response must be at least 200 words in length.
QUESTION 2
There are several gas cylinders that are under pressure, which are located outside of the
maintenance department. As the safety professional, you have been asked a couple of
questions regarding this issue. Please provide the correct responses and discuss your
recommendations for any control measures.
A. The volume of the gas cylinder is 25.7 liters and exerts a pressure of 670 mmHg. If part
of the gas is used, the pressure drops to 595 mmHg. What would be the remaining
volume of gas?
B. One of the cylinder’s content is under pressure at 1900 psi (per the gauge) at 70°F. As
the day heats up because of the sun, the temperature increases to 105°F. What is the
pressure at 105°F?
Your response must be at least 200 words in length.
QUESTION 3
The maintenance manager of your facility has asked you to evaluate a particular welding booth
that utilizes a local exhaust ventilation (LEV) system to remove the contaminant at the source.
The exhaust system does not appear to be drawing enough of the welding fumes away from the
employees to reduce the iron oxide below the permissible exposure limit (PEL) of 10 mg/m3.
During your investigation, you determine that the diameter of the duct system is 8 inches and has
an air velocity of 350 fpm (feet per minute). The welding operation is performed at a distance of
1.5 feet from the exhaust opening. Determine the flow rate, and provide your recommendations
on the control measures that could improve the capture of the iron oxide contaminant.
Your response must be at least 200 words in length.
QUESTION 4
A process involves the removal of oil and other liquid contaminants from metal parts using a
heat-treat oven, which has a volume of 15,000 ft3. The oven is free of solvent vapors. The
ventilation rate of the oven is 2,100 cfm, and the safety factor (K) is 3.
The solvent used in the process evaporates at a rate of 0.6 cfm (cubic feet per minute). The
operator would like to know how long it would take the concentration to reach 425 ppm.
Your response must be at least 200 words in length.

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