9
Ventilation
In order to reduce or eliminate any hazard from the work area, there are five
methods that the safety professional must adhere to. These methods include
Elimination, Substitution, Engineering Control, Administrative Control, and
Personal Protective Equipment. As everyone knows, the best (permanent)
method is to eliminate the hazard out of the workplace. One way to reduce
or eliminate the concentration of hazardous materials from the workplace
is through ventilation. Today’s safety professional must have more than just
a general knowledge of ventilation and ventilation systems. He or she must
have an in-depth knowledge. Furthermore, the Associate Safety Professional
(ASP)/Certified Safety Professional (CSP) exam candidate will be expected
to answer numerous questions regarding ventilation on the examination. It
is for this reason that this chapter on ventilation has been included. Using
the Board of Certified Safety Professionals (BCSP) exam reference sheet1,
this chapter will walk through each equation and explain, in detail, the use
and purpose of the equation and provide examples and solutions to various
scenarios.
Purpose for Using Ventilation
The primary purposes for using ventilation are to (a) maintain an adequate oxygen supply, (b) control hazardous concentrations of chemicals,
(c) remove odors, (d) control temperature and humidity, and (e) remove
contaminants at the source, before they enter the workplace. By understanding how to use and control ventilation, the safety professional can
successfully eliminate or greatly reduce any concentration of contaminants in the air.
Types and Selection of Ventilation
Ventilation is basically divided into three categories: general, dilution, and
local ventilation.
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General Ventilation
General ventilation is primarily used for comfort, such as temperature, humidity, and odor control, such as the air conditioning and heating system in your
home.
Dilution Ventilation
Dilution ventilation is a system designed to dilute contaminants by mixing
fresh air with contaminated air. Dilution ventilation is used primarily in
health and fire protection. The components of a dilution system include the
source of air exhaust, the source of air supply, a duct system, and a method
to filter and temper incoming air.
Dilution ventilation is used to control the following:
• Contaminants of moderate toxicity
• A large number of sources
• Intermittent exposures
• Where emission sources are well distributed
Local (Exhaust) Ventilation
Local (exhaust) ventilation systems are designed to control contaminants at
the source before mixing with breathing air occurs. When selecting which
type of ventilation system to design and install (especially in the industrial
environment), use the following criteria to make the selection.
Local (exhaust) ventilation is used to control the following:
• Highly toxic substances
• Single-source emissions
• Direct worker exposures
Note: Energy is required to move air. Air is constantly in turbulence moving
at 25 ft/min.
General Concepts of Ventilation Notes
1. Air movement results from difference in pressure.
2. Difference in pressure can be attained by heating or by mechanical
means.
3. A temperature gradient contributes to the ventilation.
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Principles of Air Movement
Calculating for Volumetric Air Flow
To begin to understand ventilation, it is necessary to understand the principles of air movement. Airflow rates are calculated by multiplying the velocity by the cross-sectional area of the hood or duct in which the air flows. This
is represented mathematically as
Q VA
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 sf [square feet])
An example of this would be to calculate the volumetric flow rate of a duct,
given the following information: The duct is a round duct measuring 8 in. in
diameter and the measured flow rate is 378 fpm. The first step would be to
calculate the cross-sectional area of the duct. (Note: Remember to convert the
unit of measurement into feet.)
We determine that the radius (1/2 the diameter) is equal to 4 in. or 0.33 ft.
Using the equation for the area of a circle, the duct cross section is 0.35 sf.
Insert these data into the equation as follows:
Q 378 fpm 0.35 sf
Q = 132.3 or 132 cfm
Calculating Static Pressure, Velocity Pressure, and Total Pressure
Air moves from an area of higher pressure to that of lower pressure. Exhaust
fans work by pushing air upstream of the fan, thus lowering the pressure on
the downstream side. This pressure differential causes air to flow into the
hood or duct. This pressure created by the fan is referred to as static pressure.
Static pressure on the downstream side of the fan is positive and negative
on the upstream side. Velocity pressure is the pressure in the direction of flow
necessary to cause the air at rest to flow at a given velocity. The total pressure
on a ventilation system equals the sum of static and velocity pressure. Total
pressure is written mathematically as
TP SP VP
where
TP = total pressure (in inches, water gauge)
SP = static pressure (in inches, water gauge)
VP = velocity pressure (in inches, water gauge)
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FIGURE 9.1
Illustration of static and velocity pressures inside a duct (Source: Yates, W. David 2024).
An illustration of both static and velocity pressure can be seen in Figure 9.1.
Calculating Velocity of Air
Another relationship to understand is the relationship between velocity and
velocity pressure. Airflow velocity is important in various ways. Airflow
velocity is used to capture contaminants and overcome cross-drafts, transportation of contaminants through the duct, and balancing of “losses” (as
described previously in this chapter) in the system and in the discharge of
the contaminant from the stack. This is described as the “magnitude” of the
system, which is a function of the velocity pressure. This relationship is written mathematically as
V = 4005 VP
where
V = velocity (fpm)
VP = velocity pressure (H2O)
Example
Given a velocity pressure of 0.2″, determine the velocity of air.
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V = 4005 0.2
V 4005 0.45 V = 1802fpm
Contaminant Generation
When using dilution ventilation, it is important to understand that 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 minus the removal. The following equation is used to determine the concentration buildup that will occur over a
given period:
G QC2 Q1 t2 t1 ln V
G QC1 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 (parts per million [ppm])
V = volume of room or enclosure (ft3)
Q = flow rate (cfm)
t2 – t1 = time interval or Δ t
Example
Acetone is being generated under the following conditions: G = 1.4 cfm,
Q′ = 2500 cfm, V = 75,000 ft3, C1 = 0, and K = 2. How long will it take (in
minutes) before the concentration reaches 175 ppm?
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:
t V G QC2 ln
Q G
We can now insert the data from our example into the equation and
solve for time, as follows:
t 75, 000 ft 3 1.4 cfm 2500 cfm 0.000175 ln 2500 cfm 1.4 cfm
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0.9625 t 30 ln 1.4 t 30 ln 0.6875
t 30 0.3747 t 11.24 min
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 min 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 min?
Again, we must manipulate the original equation in order to obtain
the requested information (concentration). We can use the original equation and manipulate it as follows:
Qt G 1 e V C2 106
Q
Now, we can insert the known data from the previous problem, as
follows:
2500 cfm 60 75 , 000 ft 3 1.4 cfm 1 e 106
C2 2500 cfm
C2 1.4 cfm 1 0.1353 2500
C2 1.4 cfm 0.8647 2500 cfm
106
106
C2 0.000484 106
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C2 = 484.2 ppm
Given the criteria for the equation in the previous problem, you can see
that after 60 min, the concentration of acetone will be 484.2 ppm.
Calculating Purge Rates
Assuming the air in a room is contaminated, but the generation of new contaminants has ceased, how long will it take to reduce the air concentration to
an acceptable limit? We utilize the following equation to determine the time
required to reduce the concentration:
Q
C ln 2 t2 t1 V
C1 where
C1 = the measured concentration
C2 = the desired concentration
Q′ = (Q/K)
K = design distribution constant (a constant factor [1–10])
Q = flow rate (cfm)
t2 – t1 = time interval or Δ t
Example
Using the information provided in the acetone problem above, how
much time would be required (assuming all sources of generation have
ceased) to reduce the contamination to 50 ppm?
0.00005 ppm 2500 cfm
ln t2 t1 .
0
000175
ppm
75
, 000 ft 2
ln 0.28 0.03333 t 0.03333 t 1.2729
0.03333
0.03333
38.2 min t
Given the information above, it will take 38.2 min to reduce the concentration of acetone from 175 to 50 ppm. (Note: Remember to convert ppm to
a decimal number.)
Example
In another case, given the same information, you may be asked to determine the concentration after a given period. In order to calculate this,
you will need to manipulate the equation as follows:
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Qt C2 C1e V 106
Now, we can insert the known data to solve for the unknown, as follows:
C2 2500 cfm 60 min 75 , 000 ft 3
10 6
0.000175 ppm e 2
C2 0.000175 ppm e 106
C2 0.000175 ppm 0.1353 106
C2 = 23.7 ppm
Given the information above, the concentration of acetone after 60 min,
with no additional generation of contaminant, will be 23.7 ppm.
Steady-State Concentration
The objective of dilution ventilation is to maintain a steady-state concentration, which is less than the threshold limit value (TLV) or permissible exposure limit (PEL) for the contaminant of concern. We can calculate this using
the following equation:
Q G
C
where
Q′ = the effective rate of ventilation corrected for incomplete mixing, cfm
(Q′ = Q/K)
K = design distribution constant to allow for incomplete mixing of contaminant air (1–10)
C = concentration of gas or vapor at time t, ppm
For example, given that Q′ = 2500 cfm and G = 1.2 cfm, what is the steadystate concentration of the gas or vapor?
Inserting the known data into the equation, we have the following:
2500 cfm =
1.2 cfm
C
2500 cfm
=C
1.2 cfm
480 ppm = C
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Calculating Rate of Generation for Liquid Solvents
To calculate the rate of generation for liquid solvents, we utilize the following
equation, which also accounts for incomplete mixing of air:
Q
403 106 SG ER K
MW C
where
Q = actual ventilation rate (cfm)
SG = specific gravity of volatile liquid
ER = evaporation rate of liquid (pints/minute)
K = design distribution constant to allow for incomplete mixing of contaminant air (1–10)
MW = molecular weight of liquid
C = desired concentration of gas or vapor at time t (ppm). Note: Normally
the TLV or PEL.
Example
Methyl ethyl ketone (MEK) is evaporating from a container at a rate of 1.2
pints every 60 min. Determine the actual ventilation rate given the following information: PEL = 200 ppm, SG = 0.81, MW = 72.11, and assume
that K = 4.
Insert the known data into the equation as follows:
1.2 pints
4
60 minutes
72.11 200 ppm
403 106 0.81 Q
Q=
26 , 114 , 400
14 , 422
Q = 1810.7 cfm
Calculating Vapor or Gaseous Concentrations
To calculate the vapor or gaseous concentrations, we use the following
equation:
C
Pv 106
Pb
where
C = concentration (ppm)
Pv = pressure of chemical (mm Hg)
Pb = barometric pressure (mm Hg)
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Example
Given a chemical with a pressure of 350 mm Hg, what would be the concentration at a barometric pressure of 760 mm Hg?
Insert the data into the equation as follows:
Cppm 350 mm Hg 106
760 mm Hg
Cppm 3.5 108 mm Hg
760 mm Hg
Cppm = 460 , 526.3 ppm
While not specifically listed on the BCSP exam reference sheet, a useful
equation for calculating the concentration of vapors as a percentage is
Cpercentage Pv 100
Pb
Given a chemical with a vapor pressure of 350 mm Hg and a barometric
pressure of 760 mm Hg, what is the concentration percentage?
Cpercentage 350 mm Hg 100
760 mm Hg
Cpercentage = 46.05%
Calculating Room Air Changes
Determining the number of air changes in a room is done by utilizing the
following equation, which is NOT listed on the BCSP examination reference
sheet:
N changes =
60Q
Vr
where
Q = actual ventilation rate (cfm)
Vr = volume of room (cubic feet or ft3)
Nchanges = number of room air changes
Example
Given that Q = 2500 cfm and Vr = 75,000 ft3, calculate the number of room
air changes that is occurring.
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Nchanges 60 min 2500 cfm 75, 000 ft 3
Nchanges = 2
In this example, we calculate the air to change 2 times every 60 min or 2 air
changes per hour.
Calculating Concentration of a Contaminant with Dilution Ventilation
Now we will calculate the concentration of a contaminant, using air changes
per hour. To do this, we will utilize the following equation:
C
Nt G
1 e 60 106
Q where
C = concentration at a given time (ppm)
G = rate of generation of contaminant (cfm)
Q′ = (Q/K)
K = design distribution constant (a constant factor [1–10])
Q = flow rate (cfm)
t = time (in hours)
Note: The equation listed on the BCSP examination reference sheet does
not multiply the product of the equation by 106. Therefore, it will be necessary for you to remember this correction.
Example
As the CSP, you are asked to determine the concentration of a contaminant in a storage area having a ventilation system that provides 7 air
changes per 1 h. The volumetric flow rate is 2500 cfm, and the contaminant is being generated at a rate of 1.2 cfm. What is the concentration in
parts per million?
C
7 1 1.2 60 10 6
1
e
2500 cfm C 0.00048 1 0.89 106
C = 52.8 ppm
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Local Exhaust Ventilation
As mentioned previously in this chapter, local (exhaust) ventilation systems are designed to control contaminants at the source before mixing with
breathing air occurs. They are used primarily to control the concentrations
at the source of highly toxic contaminants. A local exhaust ventilation system consists of the following parts: (a) hood, (b) duct, (c) air-cleaning device,
(d) fan, and (e) stack. The proper design of hoods is necessary to capture
the contaminant at the source of release. The term capture velocity is defined
as the minimum velocity of hood-induced air necessary to capture the contaminant. An example of a local exhaust ventilation system is illustrated in
Photograph 9.1.​
Canopy Hood
Canopy hoods are used where hot gases and vapors are encountered and
workers do not work directly over the source of emissions. The major drawback to using canopy hoods is the potential for the worker to place himself
or herself between the contaminant and the exhaust stream.
PHOTOGRAPH 9.1
Example of local exhaust ventilation system2
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Down Draft Hood
Down draft ventilation is used where heavier than air contaminants exist,
which are not being propelled away from the source of contaminant release.
This type of ventilation system draws the air downward and away from the
worker’s breathing zone.
Enclosure Hood
An example of an enclosure hood is the laboratory hood. This hood actually
encloses the contaminant source and the air is forced in an opposite direction (upward, downward, or backward away from the worker).
Receiving Hood
Receiving hood exhaust systems are used at the point of contaminant generation. An example of this system is a welding fume exhaust system that
can be placed very close to the point of contaminant generation.
Openings
There are basically two types of hood openings. Openings can be either
flanged or plain. Flanged openings have some type of “lip” on them (see
Figure 9.2). Flanges are sometimes designed to create a certain desired
airflow.
Plain openings do not have a flange and therefore air movement is directly
into the duct. See Figure 9.3 for a diagram of a plain opening. ​
Calculating Hood Entry Losses
All hoods, regardless of the type, have significant energy loss, which must
be accounted for. The hood entry loss represents the energy necessary to
overcome the losses caused by air moving through and into the duct. Before
discussing hood entry loss calculations, it is first necessary to show how to
FIGURE 9.2
Flanged opening (Source: Yates, W. David 2024).
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FIGURE 9.3
Plain opening (Source: Yates, W. David 2024).
calculate the coefficient of entry loss, which is the square root of the ratio of
duct velocity pressure to hood static suction. This is written mathematically
as follows:
Ce =
VP
SPh
where
Ce = coefficient of entry loss
VP = velocity pressure of the duct (″wg)
SPh = static pressure of the hood (″wg)
Theoretically, if there are no losses, then SPh would equal to VP + Ce = 1.0.
However, as we have previously mentioned, all hoods have some loss.
Example
Calculate the coefficient of entry loss given an exhaust system VP of
1.25 “wg and a static pressure of the hood of 1.82 ″wg.
=
Ce
1.25
= 0.829
1.82
Understanding how to calculate for the coefficient of entry loss, we can
now discuss the equation for calculating the hood entry loss, which is
described mathematically as follows:
1 C VP
h 2
e
e
Ce2
where
he = hood entry loss (″wg)
Ce = coefficient of entry loss
VP = velocity pressure of duct (″wg)
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Example
A local exhaust system has a VP pressure of 1.37 ″wg and a Ce of 0.829.
What is the hood entry loss of this system?
1 0.829 1.37 wg
2
he 0.8292
he 0.3128 1.37
0.687
he 0.624 wg
Now that we have discussed the relationship between hood entry loss,
velocity pressure, and the coefficient of entry loss, we can calculate the
static pressure of the hood, using the following equation:
SPh VP he
where
SPh = static pressure of the hood (″wg)
he = overall hood entry loss (″wg)
VP = duct velocity pressure (″wg)
Note: When calculating for SPh, it is understood that the static pressure
of the hood is always positive. A more accurate way of writing the equation would be as follows:
SPh VP he
Example
As the safety professional, you are called upon to calculate the static
pressure of the hood with the following information: VP = 0.2 ″wg and
he = 0.1 ″wg.
Insert the data into the equation as follows:
SPh 0.2 wg 0.1wg
SPh 0.3 wg.
Calculating Airflow Velocity
Calculating airflow velocity in a local exhaust ventilation system requires
that we account for the coefficient of entry loss. We can calculate the airflow
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velocity and account for the coefficient of entry loss by using the following
equation:
V = 4005Ce SPh
where
V = velocity of air (fpm)
Ce = coefficient of entry loss
SPh = static pressure of the hood (″wg)
Example
Given that Ce = 0.829 and SP = 0.3 ″wg, what is the airflow velocity of the
system?
V 4005 0.829 0.3
V 3320 0.55 V = 1826 fpm
Calculating Capture Velocity for Plain Opening Hood
As mentioned in Calculating Hood Entry Losses, suction sources have contours and other imperfections that create hood entry losses, which further
have an impact on capture velocities. Therefore, it is necessary to discuss the
relationship between distance (from the source), the air flow, and the capture
velocity. We can calculate the capture velocity using the following equation:
V
Q
10X 2 A
where
V = velocity (fpm)
Q = flow rate (cfm)
X = source distance from hood opening (ft) (Note: The 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 (square feet [ft2])
Example
Calculate the capture velocity of a round duct measuring 10 in. in diameter with the contaminant source being 1.2 ft from the duct opening. The
flow rate is 600 cfm.
The first step would be to determine the area of the duct, using the
circle area equation.
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Acircle r 2
Acircle 3.14 0.417 2
Acircle 3.14 0.1736 Acircle = 0.55 ft 2
Next, we insert the known data into the equation as follows:
V
600 cfm
10 1.2 ft 0.55 ft 2
2
V=
600 cfm
14.95 sq.ft
V = 40.13 fpm
Therefore, the capture velocity under this given scenario is 40.13 fpm.
Ducts
Exhaust ducts are used to convey the contaminated air from the hood to
the air cleaner or stack. When air moves through exhaust ducts, a certain
amount of friction loss occurs because of the friction of the air stream with
the duct. The selection of exhaust duct size is based on minimizing friction
loss while maintaining an adequate transport velocity to keep particulate
matter from settling out. When two or more ducts branch off from a single
exhaust source, a special problem is encountered. After determining the
amount of airflow required in each branch of the system, the designer
must assure that the exhaust volume is properly proportioned between
the two branches. This is done in one of two ways: (1) balancing the two
branches by the proper sizing of ducts and fittings to assure proper distribution, or (2) blast gates, which are slide gates that can be pushed into
the duct to partially block the airflow to lower the amount of air entering
that branch.
One equation that is not listed on the BCSP examination reference sheet is
Figure 9.4.
V1 A1 = V2 A2
where
V = velocity of air (fpm)
A = cross-sectional area (ft2)
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FIGURE 9.4
Duct size variations (Source: Yates, W. David 2024).
Example
A round duct has a cross-sectional area of 0.5 ft2 on the first run with
a velocity of 300 fpm. The duct size then increases at the second run to
0.75 ft2. Determine the velocity of the second run.
V1A1 = V2 A2
300 fpm 0.5 sq.ft V2 0.75 sq.ft 300 fpm 0.5 sq. ft V2 0.75 sq. ft 0.75 sq. ft 0.75 sq. ft 200 fpm = V2
Fans
Fans generate the airflow volume (Q) of the system against airflow resistance presented by the system. In other words, they create the SP differential
needed to cause the desired airflow. There are two basic types of fans: (1)
axial and (2) centrifugal. Axial fans create airflow when the air enters and
leaves along the axis of rotation. Centrifugal fans create airflow as the air
enters along the axis of rotation and leaves perpendicularly (accelerates centrifugally) through the blades.
Calculating Static Pressure of the Fan (SPh)
To calculate the static pressure of the fan (SPh), we use the following equation:
SPfan SPout SPin VPin
Example
Calculate the static pressure of the fan, given the following information:
SPout = 1.8 ″wg, SPin = 0.3 ″wg, and VPin = 0.4 ″wg.
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SPfan 1.8 0.3 0.4
SPfan 1.1wg
Air-Cleaning Devices
Air cleaners or air pollution devices remove contaminants from the air
stream to protect the community, protect the fan, recover materials, and
enable circulation.
Ventilation Measurement Equipment
There are several different types of ventilation measuring equipment. The
most common types include pitot tubes, rotating vane anemometers, and
thermal anemometers.
Pitot Tubes
Pitot tubes are probes that are inserted into the duct system and connected
to a manometer. Pitot tubes are used to measure various pressures within
the system.
Rotating Vane Anemometers
Rotating vane anemometers are used to measure airflow through large supply and exhaust systems. It is recommended that the size of the rotating vane
anemometer should not exceed 5% of the cross-sectional area of the duct.
Thermal Anemometers
Thermal anemometers are primarily digital instruments that measure the
heat removed by an air stream as it passes over a probe, which allows for calibration of the velocity of the air stream at a given density. The probe can be
inserted directly into the air stream. This is a useful instrument, but the user
should understand that errors can be created by the movement and location
of the probe. It should be perpendicular to the air stream and maintained in
that position until such time as the measurement is stabilized. Furthermore,
the user should understand that the probe is extremely fragile; thus, precautions should be taken to protect the probe from damage.
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Key Information to Remember on Ventilation
1. Pitot tubes usage is limited to velocities at or below 600–800 fpm.
2. Blast gates are used to balance the airflow in ducts of different sizes.
3. Capture velocity is defined as the minimum velocity of hood-induced
air necessary to capture the contaminant.
4. When calculating the static pressure of the hood, remember that the
SPh is always positive.
5. Static pressure on the downstream side of the fan is positive and
negative on the upstream side.
6. Backward curved fan blades are the most efficient.
7. Centrifugal fans are the best for local exhaust ventilation systems.
8. The equation for calculating capture velocities of plain hood openings 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.
References
1. Board of Certified Safety Professionals. Comprehensive Examination Equation
Reference Sheet. Previously retrieved from: http://www​.bcsp​.org​/pdf​/ASPCSP​
/ExamRef5​.pdf. This link is no longer available but is embedded in the exam
software.
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