CHAPTER 10 SAMPLE QUIZ QUESTIONS
10.1. What material is commonly substituted for silica sand to prevent silicosis in
blasting workers?
ANS: steel shot
ref. p. 202
10.2. Ventilation is considered which of the following strategies to hazard control?
a.
b.
c.
d.
Engineering
Administrative
Work practice
Personal protective equipment
ref. p. 200
10.3. For each of the following industrial materials, suggest substitutes that are
feasible for some operations and that prevent certain hazards:
Silica (for blasting)
ANS:
Lead-based paint
Freon (as a propellant)
Volatile organic compounds
steel shot
iron oxide pigments
propane
water-based solvents
ref. p. 201-202
10.4. What is the name given to a type of ventilation system that can be likened to
"sweeping dirt under the rug?"
ANS: dilution ventilation
ref. p. 202
10.5. Which of the following types of ventilation most resembles sweeping dirt under
the rug?
a.
b.
c.
d.
e.
dilution ventilation
exhaust ventilation
push-pull ventilation
makeup air ventilation
ordinary heating and air conditioning systems
ref. p. 202-203
10.6. Which of the following solutions is preferred for the problem of toxic air
contaminants arising from the surface coatings of materials during the welding
process?
a.
b.
c.
d.
ventilation
removal of the coatings prior to welding
substitution of crimping or other joining process to replace welding
provision of respirators for the welder
ref. 200-201
10.7. One part of a manufacturing process for metal products involves tumbling the
parts in an abrasive, and the noise in the area of this operation is measured at
94 dBA. An acoustics engineering firm has designed a barrier for enclosing this
operation that engineers claim will attenuate the absolute sound pressure by
50%. If the claim is correct, calculate the new noise level in dBA generated by
this operation.
ANS: The 50% reduction is equivalent to dividing the noise in half.
equivalent to a 3dB decrease in noise level. Thus,
This is
New noise level = 94 dBA - 3 dBA = 91dBA
ref. p. 211-212
10.8. A particularly noisy process is operated by a single operator working at a
control console. The 8-hr TWA exposure level for this operator is 96 dBA. The
company has initiated an engineering project to alleviate the problem and has
two plans:
Plan A - Move the operator's control console from its current position 5 feet
from the source of the noise to a point 10 feet away.
Plan B - Enclose the noise source in an enclosure that would be effective in
reducing the absolute sound pressure level of the noise by 75%.
What would be the new decibel exposure level if
a. Plan A were carried out
ANS: Doubling the distance reduces the noise level by a factor of 4. The dB
level is thus reduced by half twice or 6 dB. New dB reading = 96 - 6 = 90 dB.
b. Plan B were carried out
ANS: Reducing noise by 75% is reduction by a factor of 4 or a 6 dB reduction.
New dB reading = 96 - 6 = 90 dB
c. Both Plan A and Plan B were carried out
ANS: Each plan would reduce noise by a factor of 4, resulting in a 16-fold
reduction or 12 dB reduction. New dB reading = 96 - 12 = 84 dBA.
ref. p. 211-218
10.9. Which of the following would be most effective in dealing with a work-related
noise exposure problem?
a. Enclose the noise source with a barrier that reduces the noise level
by 50%.
b. Position the operator at a distance twice as far from
the source of
the noise.
c. Rotate personnel so that each worker is exposed to the noise source
for only one-half shift.
d. Provide ear protection that cuts the noise level by one half.
ref. p. 211-218
10.10. Which of the following would be least effective in dealing with a work-related
noise exposure problem?
a. Enclose the noise source with a barrier that reduces the noise level
by 50%.
b. Position the operator at a distance twice as far from the source of
the noise.
c. Rotate personnel so that each worker is exposed to the noise source
for only one-half shift.
d. Provide ear protection that cuts the noise level by one half.
ref. p. 211-218; also earlier chapters that emphasized the preference of engineering
and administrative controls over personal protective equipment.
10.11. From the perspective used by the federal enforcement agency, rank the following
solutions to a worker noise exposure problem (from most effective,"1", to least
effective,"4").
Enclose the noise source with a barrier that reduces the noise level by
2
50%.
1
Position the operator at a distance twice as far from the source of the
noise.
3
Rotate personnel so that each worker is exposed to the noise source for
only one-half shift.
Provide ear protection that cuts the noise level by one half.
4
ref. p. 211-218; also earlier chapters that emphasized the preference of engineering
and administrative controls over personal protective equipment.
10.12. What is a "threshold shift?"
ANS: an indication that a worker's hearing has been damaged
ref. p. 222
10.13. A "threshold shift" is
a. a hazard of walking and working surfaces
b. an administrative control for rotating employees exposed to excessive
noise
c. a barrier to continued attenuation of a noise source
d. an indication that a worker's hearing has been damaged
ref. p. 222
10.14. A "threshold shift" is an indication that a worker's hearing has been damaged.
a. true
b. false
ref. p. 222
10.15. A "threshold shift" is a hazard of walking and working surfaces.
a. true
b. false
ref. p. 222
10.16. A "threshold shift" is an administrative control for rotating employees exposed
to excessive noise.
a. true
b. false
ref. p. 222
10.17. A "threshold shift" is a barrier to continued attenuation of a noise source.
a. true
b. false
ref. p. 222
10.18. List the elements of an effective hearing conservation program.
ANS: audiometric testing, noise monitoring, calibration of equipment, training,
warning signs for noisy areas, recordkeeping of audiometric tests and equipment
calibration
ref. p. 223
10.19. Hearing damage at what noise frequency is viewed as typical of industrial
exposures?
ANS: 4000 Hz
ref. p. 223
10.20. Which of the following noise frequencies is viewed as typical of industrial
exposures?
a.
b.
c.
d.
1000 Hz
4000 Hz
10000 Hz
20000 Hz
ref. p. 223
10.21. Gamma rays and X rays typify which of the two primary classes of radiation?
ANS: ionizing
ref. p. 223
10.22. Radio waves and microwave radiation typify which of the two primary classes of
radiation?
ANS: non-ionizing
ref. p. 223
10.23. The best exhaust ventilation systems are the "push" types.
a. true
b. false
ref. p. 203
9.24.
You are requested to design a general exhaust ventilation system to deal with a
process that leaks bromine into the plant air at the rate of one cubic inch per
hour. Calculate how much flow in cubic feet per hour is needed to keep the
bromine vapors within OSHA limits (PEL and AL). Show your calculations.
ANS: The PEL is 0.1 ppm.
for PEL:
1 ft. 3
1 in. /hr x ----12 in.
--------------------X
3
=
0.1 X = 106/1728 = 578.7 ft3/hr
X = 5787 ft3/hr
0.1
-----106
The AL is 1/2 PEL and would require a ventilation system of double the capacity
required to meet the PEL.
for AL:
x = 2 x 5787 = 11574 ft3/hr
ref. p.205-206
10.25. Which is a preferred design for exhaust ventilation systems?
(a)
(b)
ANS: b
ref. p. 202
10.26. Which is a preferred design for exhaust ventilation systems?
ANS: b
ref. p. 203
10.27. Identify the device indicated by the large arrow.
What is its purpose?
ANS: pressure switch--to switch on alarm when filter needs service
ref. p. 204
10.28. The device shown is a
a.
b.
c.
d.
tornado
cyclone
hurricane
heat pump
ref. p. 206
10.29. The purpose of the device shown is
a.
b.
c.
d.
to remove particulates
to absorb toxic gases
to act as a heat exchanger
none of the above
ref. p. 206-207
10.30. The purpose of the device shown is
a.
b.
c.
d.
to remove particulates
to absorb toxic gases
to act as a heat exchanger
none of the above
ref. p. 206
10.31. The purpose of the device shown is
a.
b.
c.
d.
to act as a heat exchanger
to remove particulates
to absorb toxic gases
none of the above
ref. p. 206
10.32. The purpose of the device shown is
a.
b.
c.
d.
to absorb toxic gases
to remove particulates
to act as a heat exchanger
none of the above
ref. p. 206
10.33. The purpose of the device shown is to
a.
b.
c.
d.
reduce
reduce
charge
remove
ionizing radiation
non-ionizing radiation
particles
toxic gases
ref. p. 207
10.34. The purpose of the device shown is to
a.
b.
c.
d.
reduce
reduce
remove
remove
ionizing radiation
non-ionizing radiation
particulates
toxic gases
ref. p. 207
10.35. The device shown is identified as
.
ANS: an electrostatic precipitator
ref. p. 207
10.36. The device shown is known as a
a.
b.
c.
d.
packed tower wet scrubber
cyclone precipitator
electrostatic precipitator
fabric collector
ref. p. 208
10.37. What is the specific purpose of the electric motor indicated by the large arrow?
ANS: drive a vibrator
ref. p. 209
10.38. Which (pressure vs. time) waveform represents a low pitched soft sound?
a
b
c
D
ANS: a
ref. p. 210
10.39. Which (pressure vs. time) waveform represents a high pitched soft sound?
a
b
c
D
ANS: b
ref. p. 210
10.40. Which (pressure vs. time) waveform represents a low pitched loud sound?
a
b
c
D
ANS: c
ref. p. 210
10.41. Which (pressure vs. time) waveform represents a high pitched loud sound?
a
b
c
D
ANS: d
ref. p. 210
10.42. Assume the worker in the diagram is equidistant from each machine and compute
noise exposure.
ANS: Combining the noise from sources A & B:
Source A (86 dBA) + Source B (86
dBA) doubles the sound exposure and increases the db level of 3 db to 89 dBA.
Then combining this resultant 89 dBA with Source C (82 dBA) we get a difference
between the sources of 89-82 = 7 db. Using Table 10.1, this results in an
incremental db of 0.8 dB. So, 89 + 0.8 = 89.8 dBA. Finally, Source D is a mere
78 dB for a difference of 89.8 - 78 = (approx) 12 dB. According to Table 10.1, a
difference of 12 dB between sources would represent a combined noise level of
89.8 + 0.2 = 90 dBA
ref. p. 212-213
10.43. The diagram illustrates what device?
ANS: sound-level meter
ref. p. 217
10.44.
The diagram illustrates
a.
b.
c.
d.
absolute noise pressure and pitch
noise enclosure effectiveness
octave band analysis
none of the above
ref. p. 218
10.45. The diagram illustrates
a.
b.
c.
d.
absolute noise pressure and pitch
noise enclosure effectiveness
octave band analysis
none of the above
ref. p. 222
10.46. You are requested to design a general exhaust ventilation system to deal with a
process that leaks hexachloroethane into the plant air at the rate of one cubic
inch per hour. How much flow in cubic feet per hour is needed to keep the
hexachloroethane vapors within OSHA limits (PEL and AL)? The PEL is 1 ppm.
ANS:
AL = 1/2 PEL = (1/2)(1 ppm) = .5 ppm
1/x = .5/106 = .5 x 10-6
x = 2.0 x 106 cu.in./hr
1 ft3 = (12 in)3 = 1728 in3
In cubic feet per hour:
x = (2.0 x 106 cu.in./hr)/(1728 in3/cubic foot)
= 1157.4 ft3/hr
ref. p. 205-206
10.47. Noise Source A has a sound power level of 99 dB, and noise Source B has a sound
power level of 82 decibels. Compute the ratio of absolute sound power between
Source A and Source B.
NOTE TO INSTRUCTOR: This examination question goes beyond the scope of the
text. It is a technical question intended for students who have an engineering
or science background. Supporting lecture notes for instructors who desire to
teach this technical extension of Chapter 10 topics are available from the
author.
ANS:
dB difference = 99dB – 82dB = 17dB = 10 log10 [KW/W]
= 10 log10 K
1.7 = log10 K
K = 50.12
10.48. An air sample bag is needed at the PEL for carbon disulfide (20 ppm). The
smallest quantity of carbon disulfide available for this experiment is 5 ml.
Calculate the minimum capacity air sample bag necessary for this experiment.
Show your calculations.
ANS:
5ml/X
=
20/1,000,000
X
=
=
5,000,000/20
=
250,000/1000
=
250,000 ml
250 liters
ref. p. 205-206 also see Chapter 9 for concepts of ppm
10.49. A certain drying process produces 5 cubic feet of methanol vapors per hour. The
PEL for methanol is 200 ppm. If general exhaust ventilation is used, calculate
how much flow in cubic feet per hour is needed to keep the methanol vapors
within OSHA limits. Show your calculations.
ANS:
PEL
---106
E
=
5(106)
-------PEL
=
=
liberated
--------exhaust
=
5(106)
------=
200
5
--E
25,000 ft3/hr
ref. p. 205-206
10.50. Explain why it seems so difficult to reduce noise decibel levels even by a
“small” amount, such as 5 or 6 decibels.
ANS.
In terms of absolute sound pressure level, a reduction in decibel level of 5 or
6 decibels is NOT really a “small” amount. Due to the logarithmic ratio scale, a
reduction of noise level by only three decibels corresponds to a reduction in
the absolute sound pressure level by a factor of two (that’s cutting the sound
pressure level IN HALF). A reduction of 6 dB corresponds to reducing the
absolute sound pressure level by 75% (or a factor of four).
ref. p. 211-212, 218
A more detailed answer follows:
ANS.
The decibel scale is deceptive because it represents a logarithmic ratio of
absolute sound pressures. The range of absolute sound pressures that the human
ear can sense is incredibly large. In terms of absolute sound pressure the
loudest sound heard is millions of times louder than the faintest audible sound.
On the decibel scale an increase of only 3 decibels represents a DOUBLING of the
absolute sound pressure. Extending this logic, increasing the decibel level by
6 decibels (3 decibels and then 3 decibels again) doubles the absolute sound
pressure TWICE, thus QUADRUPLING the absolute sound pressure. Conversely,
reducing the noise level by 6 decibels represents cutting the absolute sound
pressure in half TWICE, which is the same as reducing it by a factor of 4
(reducing it 75%). So, reducing noise levels by 5 or 6 decibels is a very
significant reduction indeed. It is no wonder that it is so difficult to reduce
decibel levels by this much.
ref. p. 211-212, 218
10.51. Explain, in terms easily understood by workers and managers, the following
physical characteristics of occupational noise:
a. the physical characteristic of noise that gives rise to the sensation of pitch.
ANS. Pitch varies as the frequency of the pressure vibrations: the higher the
frequency of the vibrations, the higher the perceived pitch.
b. the physical characteristic of noise that gives rise to the sensation of loudness.
ANS. Loudness is the sensation that comes from sensing the amplitude of the pressure
variations in the sound wave. Thus, if the oscillations of absolute pressure are
great, then the sound will be perceived as a loud sound.
c. how noise is affected by the distance to the source (in what way and at what
proportional rate).
ANS. The loudness of noise varies inversely with the square of the distance from the
source of the sound. Thus, if the source is moved away to a distance twice as
far as the original distance, the absolute pressure level received will decrease
by a factor of 4 (i.e., 2 squared).
d.
how doubling the absolute sound pressure affects decibel level (in what way and by
how much).
ANS. Doubling the absolute sound pressure results in an increase in the decibel ratio
of 3 dB. Conversely, cutting the absolute sound pressure in half reduces the
decibel ratio by 3 dB.
ref. p. 209-212 & 218-219
10.52. Describe some engineering solutions to the problem of toxic air contaminants
arising from surface coatings during the welding process. Indicate a priority to
rank the preference for the solutions to the problem.
ANS. The best solution, if feasible, would be to eliminate the welding operation by
substituting a crimping or other joining operation that does not require
welding. The second best solution would be to eliminate the source of the
contaminant by removing the surface coating prior to the welding process. If
these solutions are not feasible or are not satisfactory for the process, the
provision of a ventilation system would be a possible engineering solution to
the problem.
ref. p. 200-201
10.53. An operator is exposed to the following sources of noise:
Source A:
Source B:
Source C:
81 dBA
83 dBA
85 dBA
Perform calculations to
worker. Show your work.
determine
the
approximate
overall
noise
exposure
to
the
ANS.
Combining Sources A & B: difference = 2 dB, so add 2.1 to louder source:
83 + 2.1 = 85.1 (approximately 85 dB)
Combining with Source C: the difference is nearly zero, so add 3 dB:
85 + 3 = 88 dB
ref. pp. 212-213, especially Table 10.1, p. 212
10.54. An operator working an 8-hour shift is exposed to 85 dbA the first half of the
shift and 92 dbA the second half. Calculate the noise dose. Does the total dose
exposure exceed the OSHA PEL? Does it exceed the AL?
ANS.
From Table 10.2, p. 215, the allowable noise dose for 85 dB is 16 hours. The
allowable noise dose for 92 dB is 6.2. Computing ratios for partial doses and
adding them to calculate the total dose exposure (Eqn. 10.1, p. 214):
D = 4/16 + 4/6.2
=
.25 + .645
= .895
Since .895 < 1.0, the PEL is not exceeded.
Since .895 > .5, the AL is exceeded.
ref. pp. 214-216