Mining Ventilation (English)

1 CONTAMINANTS AND EXPOSURE STANDARDS
The basic requirement for the mine ventilation system is to provide air for people to breath and
in a state that will not cause any immediate or future ill effects.
Because of the processes of mining, if positive airflow through the workings was not provided
the air would very quickly become stale, contaminated and unfit for human consumption. The
ventilation system must therefore be sufficient to deal with the contaminants released during
mining. If they are not adequately dealt with, as they are identified, they may become at best a
discomfort to mine workers, and at worst cause serious or even fatal illness.
The prime contaminants produced during mining are
1. dust
2. heat
3. gases (including water vapour i.e. Humidity)
and the prime method for dealing with these is an effective ventilation system that
1. supplies oxygen and coolth
2. dilutes dust and gases and
3. removes the contaminants from the workplace.
It is possible that any known substance will be identified, in either the air we breathe or the food
we eat. Although the human body is equipped to reject or absorb these substances this can
only be done providing the quantities involved are not excessive.
The search to identify specific substances and their harmful concentrations is on going with
limits constantly under review. Levels of atmospheric contaminants that are “safe” in an
occupational health and safety environment are often difficult to determine. There are many
factors which must be considered including:
• Variability in response of individuals to contaminants
• Synergistic effects (i.e. combined effect of simultaneous exposure to several contaminants)
• Work rate (affects respiration rate)
• Work cycle (“compressed” work cycles give the body less recuperation time between
exposures)
• Changes in scientific understanding
As a result of the above factors, recommended maximum exposures to various contaminants
are constantly under review and the latest information should always be sought
Worksafe Australia produce a publication titled “Exposure Standards for Atmospheric
Contaminants in the Occupational Environment” which provides comprehensive information on
the present exposure standards for known hazardous substances along with information of
substances presently under review. This standard represents the concentration of substances
that should not cause undue discomfort or impair the health of persons when exposed to these
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substances. They do not however represent a “no-effect” level to all persons exposed to the
concentrations stated. They are of course based on current knowledge and research and are
subject to change. This information can be found on the Internet at http://www.nohsc.gov.au
This website tells us that when interpreting the actual “Exposure Standard” it is important to note
that they ….”have been established on the premise of an eight-hour exposure, during work of
normal intensity, under normal climatic conditions and where there is a sixteen-hour period
between shifts to permit elimination of any absorbed contaminants.”
And “ Heavy or strenuous work increases lung ventilation, thereby increasing the uptake of
airborne contaminants. Similarly, heavy physical work under adverse climatic conditions, such
as excessive humidity or heat, or work at high altitudes, may lead to an increased uptake of
contaminants. It is therefore of particular importance that any evaluation of the working
environment considers the lung ventilation rate where there is a significant airborne
concentration of contaminant.”
This website also provides descriptions of the specific exposure standards for peak, short term,
time weighted average and general excursions.
1.1 Peak Limitation
For some rapidly acting substances and irritants, the averaging of the airborne concentration
over an eight-hour period is inappropriate. These substances may induce acute effects after
relatively brief exposure to high concentrations and so the exposure standard for these
substances represents a maximum or peak concentration to which workers may be exposed.
Although it is recognised that there are analytical limitations to the measurement of some
substances, compliance with these ‘peak limitation’ exposure standards should be determined
over the shortest analytically practicable period of time, but under no circumstances should a
single determination exceed 15 minutes.
1.2 Short Term Exposure Limit (STEL)
Some substances can cause intolerable irritation or other acute effects upon brief
overexposure, although the primary toxic effects may be due to long term exposure through
accumulation of substances in the body or through gradual health impairment with repeated
exposures. Under these circumstances, exposure should be controlled to avoid both acute and
chronic health effects.
Short-term exposure limits (STELs) provide guidelines for the control of short-term exposure.
These are important supplements to the eight-hour TWA exposure standards that are more
concerned with the total intake over long periods of time. Generally, STELs are established to
minimise the risk of the occurrence in nearly all workers of:
• intolerable irritation
• chronic or irreversible tissue change, and
• narcosis to an extent that could precipitate industrial accidents.
STELs are expressed as airborne concentrations of substances, averaged over a period of 15
minutes. This short term TWA concentration should not be exceeded at any time during a
normal eight-hour working day. Workers should not be exposed at the STEL concentration
continuously for longer than 15 minutes, or for more than four such periods per working day. A
minimum of 60 minutes should be allowed between successive exposures at the STEL
concentration.
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1.3 Time-Weighted Average (TWA)
Except for short-term exposure limits, or where a peak value has been assigned, the exposure
standards for airborne contaminants are expressed as a time-weighted average (TWA)
concentration of that substance over an eight-hour working day, for a five-day working week.
During periods of continuous daily exposure to an airborne contaminant, these TWA exposures
permit excursions above the exposure standard provided they are compensated for by
equivalent excursions below the standard during the working day. However, it is not necessarily
acceptable to expose workers to concentrations significantly higher than the exposure standard
solely because the exposure is for less than an eight-hour day or because the exposure occurs
only occasionally. Permissible variations in the exposure standard for such situations are
dependent on such factors as the acute effects of short-term exposures, or on the relationship
between accumulation and elimination of the body burden of the material or its metabolites, and
should only be accepted in the light of expert advice.
1.4 Guidance on General Excursion
In practice, the actual concentration of an airborne contaminant arising from a particular
industrial process may fluctuate widely with time, with some of the major excursions giving rise
to a significant proportion of the overall exposure.
Even where the TWA exposure standard is not exceeded, there should be some control of
concentration excursions. A practical approach to control has been developed, based on
observations of the variability in concentrations observed in industrial environments. A process
is not considered to be under reasonable control if short term exposures exceed three times the
TWA exposure standard for more than a total of 30 minutes per eight-hour working day, or if a
single short term value exceeds five times the TWA exposure standard. It should be
emphasised that guidance of this type, aimed at placing some restraint on concentration
excursions, is not directly health-based and does not supersede any STEL or peak limitation
set.
Where adequate toxicological or epidemiological data allows the assignment of a STEL, the
STEL will supersede this guidance on general excursion.
1.5 Adjusting the Eight-hour Exposure Standard for Longer Periods
“Compressed” work cycles (usually in fly-in/ fly out operations) consisting of 12-hour shifts for up
to (and sometimes exceeding) 14 days duration is now commonplace. As yet, there is no
agreement on the extent to which TWA exposure standards should be reduced in response to
various work patterns. Specialist consideration and expert advice should be sought in the
specification of modified exposure standards.
NOHSC recommends the use of the Brief and Scala (1975)1 model to adjust the time-weighted
average (TWA). This method was chosen because it is a simple calculation, it is the most
conservative model developed and does not require any detailed knowledge of the substance.
16 h
Adjusted (TWA) Exposure Standard 8 (24 – h) Eight Hour Exposure Standard
×
× ×
=
Where h = hours worked per day
1 BRIEF, R., SCALA, R., “Occupational Exposure Limits for Novel Work Schedules.” American Industrial Hygiene Association
Journal. 36: pp467-469, 1975.
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Example
The eight-hour exposure standard for Carbon Monoxide is 30 ppm. The adjusted TWA
exposure for Carbon Monoxide (CO) assuming a twelve-hour working shift is calculated as
follows.
16 12
Adjusted TWA 8 (24 – 12) 30
×
× ×
= = 15 ppm
No adjustment is necessary for the excursion limits of peak concentrations or the short-term
exposure limit (STEL).
A Comparison of some Atmospheric Exposure Standards Adjusted for Shift Duration
TWA STEL
CONTAMINANT
8-hour 10-hour 12-hour Any
* Carbon Monoxide (CO) 30 21 15 400
* Carbon Dioxide (CO2) 5,000 3500 2,500 35,000
* Nitrogen Dioxide (NO2) 3 2 1.5 5
* Nitrous Oxide (NO) 25 18 12 UR
* Sulphur Dioxide (SO2) 2 1 1 5
* Hydrogen Sulphide (H2S) 10 7 5 15
# Respirable Dust 5.00 3.50 2.50 NA
# Quartz Bearing Dust 0.20 0.14 0.10 NA
# Respirable Combustible Dusts 2.00 1.40 1.00 NA
NOTES: * = ppm # = mg/m3 UR = Under Review NA = Not Applicable
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“Some mines are so dry that they are entirely devoid of water and this dryness causes the workmen
even greater harm, for the dust, which is stirred up and beaten up by digging penetrates the windpipe
and lungs and produces difficulty in breathing and the disease the Greeks call asthma. If the
dust has corrosive qualities, it eats away the lungs, and implants consumption in the body. In the
mines of Carpathian Mountains women are found to have married seven husbands, all of whom this
terrible consumption has carried off to a premature death.”
Georgius Agricola – “De Re Metallica” (1556)
2 OCCUPATIONAL HEALTH AND SAFETY
There is a surprisingly broad range of environmental hazards in underground mining, including
poisonous, asphyxiant, carcinogenic or explosive dusts and gasses and extremes of heat and
humidity. A significant number of these hazards can result in serious health problems, ranging
from long-term physical impairment (e.g. lung disease) to immediate death (e.g. carbon
monoxide poisoning, or heat stroke). It is the role of a mine’s ventilation systems to control
these hazards, but before any ventilation engineering design work is carried out, it is vital that
the hazards are well understood. This chapter describes some of the more common mine
environmental hazards.
It should be noted that the human body is very resilient and well equipped to deal with nature’s
climatic and atmospheric extremes. However, some processes involved in activities such as
mining produce contaminant levels and environmental conditions which are beyond those with
which the body’s natural defence systems can cope.
The human capacity to function at the maximum potential will reduce rapidly in atmospheres
contaminated with dust, gases, heat and
humidity and the long term effects of
exposure to contaminants may have serious
impact on general medical health.
2.1 The Respiration System
Medical science has proven the human body
to be very resilient and well equipped to deal
with nature’s climatic and atmospheric
extremes. However it has also shown that
some processes involved in today’s
manufacturing and industry may have a
detrimental effect on parts within the human
body.
Gases, vapours and dusts may enter the
human body in three ways
• Inhaled into the respiratory system
• Ingested with food and saliva into
the digestive system
• Absorbed through the skin
However it is the respiratory system which provides the major mode of entry into the body
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The respiration system is the main point of entry into the body for many of the particulate and
gaseous contaminants found in the underground environment. It is therefore important to have a
basic understanding of the principles behind the operation of the human respiratory system.
Every organ in the body requires oxygen. Oxygen is ‘captured’ in tiny air sacs (called alveoli) in
the lungs. When air is breathed in, it passes through
the walls of the alveoli and surrounding capillaries
into the blood stream. The blood transports the
oxygen to the body tissues, where it is consumed in
the process of energy production, producing carbon
dioxide as a waste product. The blood stream
carries the carbon dioxide back to the air sacs to be
breathed out. The difference between oxygen and
carbon dioxide concentrations in blood in the
capillaries and the air in the alveoli causes an
exchange of gases.
The respiratory system is basically a ventilation
system, which supplies air to the alveoli.
By the very nature of their function, the lungs are
exposed to any dusts, fumes, smoke, aerosols,
mists, gases or vapours small enough to remain
airborne.
An adult breathes between 2.0 litres and 4.0 litres of air per minute (l/m) and during times of
hard work, this can increase to about 8.0 l/min. Because more air is inhaled as work becomes
harder, it is easy to understand why workers in heavy dusty jobs such as mining and
construction are more likely to suffer from dust and other contaminant-related lung disorders.
By the very nature of their function, the lungs are exposed to any dust, fumes, smoke, aerosol,
mist, gas or vapour that is small enough to remain airborne.
The human respiratory system has a number of defence systems which are designed to deal
with the concentrations and size ranges of dust normally found in nature. These include:
• Nasal hairs. These act as pre-filters. When we inhale, the mucus coated membranes of
the nasal passages and hairs of the nose aid in capturing almost all of the coarser particles.
• Cilia. The trachea and bronchi are
lined with sticky mucus that is
wafted up these airways via the
action of cilia, which are hair-like
cells acting with a wave-like
motion. The dust is moved
upwards to the mouth, where it is
either expectorated or swallowed.
Inhalation of very coarse dust
particles causes coughing and
sneezing, which has the effect of
speeding up the removal process.
The finer dust particles will
however reach the alveoli.
• Macrophages. These are relatively large white cells (up to 10 μm diameter), which lie on
the surfaces of the alveoli. T he cells are mobile and their role is to completely engulf the
fine foreign particles which enter the alveoli. Once this has occurred, the macrophages
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usually then move up the respiratory system and are eventually expectorated or swallowed.
Certain dusts are toxic to macrophages (including free crystalline silica and asbestos) and
these substances cannot therefore be expelled from the lungs. They remain in the lung
tissue, resulting in a gradual incapacitation of the lungs.
2.2 Dust
In nature dust is mainly formed by the grinding action of wind blown particles on rocks, the
impact action of rocks on each other as they knock together during landslides, and the tumbling
action in rivers and ocean wave motion. Because this type of impact is limited, dust formed in
this way is rather coarse e.g., beach sands, muds and clays.
In mining dust is formed by powerful concentrated forces such as blasting, drilling, crushing, and
grinding and consequently forms much smaller dust particles than those formed by nature.
Dust is a result of the disintegration of matter and the size of the dust particle produced is
determined by the impact per unit area. For example striking a rock with a hammer will split the
rock into large pieces forming coarse dust particles. If we were to use the same force using a
chisel it would break only a small piece of the rock into fine dust particles because the force is
directed onto a much smaller area.
As an example of dust production, crushing 1mm3 of rock to 1μm particle sizes would yield
1,000 million dust particles. In a drill hole 3.6m deep and 32mm diameter the volume of rock
removed is 2,895,291 mm3 producing 2.89 x 1015 dust particles of 1μm diameter. If it takes 10
minutes to bore the hole and there is a ventilating airflow of 20m3/s each mm3 of air would be
contaminated with 241 particles. If the particles had a density of 3kg/m3 the eight-hour
exposure level would be 0.7 mg/m3. In reality the majority (99%) of drilling particles are much
larger than 1μm and are contained in the drilling water.
Some of the more significant sources of dust in underground mines include blasting, movement
of rock in stopes, mucking operations, mechanical rock cutting (e.g. raise drilling, road-header
etc), ore passes, rock breakers, crushers and conveyor transfer points. Dust is also liberated to
the ventilating air from by the tyres of passing traffic lifting the dust from the surface of the mine
roadway.
2.2.1 The Hazard of Dust
The long-term effect of dust inhalation is not necessarily confined to the respiratory system and
there is some research that suggests that the respiratory system merely provide the mode of
entry into the body, other modes being absorption and ingestion. The respiratory system is very
selective with respect to the size and quantity of dust retained in the lungs. It is not simply a
matter of the quantity of dust in the atmosphere that dictates the amount deposited in the lungs,
it is also the duration of exposure and the rate of deposition to the critical area.
The size of dust particles is measured and expressed in microns (μm). The smallest dust
particle that can be seen with the naked eye under good conditions (i.e. black on white with
good lighting) is around 25μm and It is generally accepted that the smallest particle visible to
the naked eye can be as large as 50μm.
Dust particles in mine airflow will settle out of laminar airflows according to Stokes Law. If the
airflow is turbulent then the motion of airborne particles is unpredictable are more likely to be
removed from the airflow by coagulation and impingement rather than settling.
2.2.1.1 Settling
The terminal velocity of an airborne dust particle is dependant upon the atmospheric drag force
holding the particle up and the gravitational force pulling it down. This is known as Stokes Law.
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Where
G = Gravitational Force
d = Geometric diameter of the sphere (m)
wS = Density of the sphere (kg/m3)
wa = Density of the air (kg/m3)
g = Acceleration due to gravity (m/s2)
F = Drag Force
v = Velocity of the particle (m/s)
η = Viscosity of the fluid (kg/m.s)
Stokes Law
Stokes Law is applied to spheres
G = F
d (w w )g 3 d v
6
1
S a
π 3 − = π η Equation 2-1 Stokes Law
Therefore ( ) S a
2
w w
18
v d g −
η
=
Brownian Motion
The Brownian motion is a random motion that occurs when dust particles collide with gas
molecules in the air with no net tendency to move downwards. This motion completely masks
observations made on the particle in question and the resistance to the particle becomes less
and they tend to “slip” past the gas molecule at a speed faster than that indicated by stokes law.
Slip Corrected Velocity
Cunningham (1910) introduced a slip correction
 

 
 λ
= +
P
C d
v v 1 2A Equation 2-2 Cunningham’s slip corrected velocity
Where
VC = Slip corrected velocity of the particle (m/s)
A = a constant which varies from 0.7 to 0.9
λ = the mean free path of the gas molecule (6.53 x 10-8m at 20 °C and 101.3kPa)
dP = diameter of the falling particle (m)
Particles having terminal settling velocities of the same order as the displacement caused by the
Brownian motion will remain suspended, even in still air.
The terminal settling velocities may be calculated from the above equations. For example a
quartz particle with a geometric diameter of 10μm will have a slip corrected settling velocity of
8.04 x 10-3 m/s and if released from a height of 4.0m would require 8.29 mins to settle to the
ground and 1μm particle would take 12.8 hours. In fact because of the turbulence associated
with mine ventilation along with the Brownian motion it would be considerably greater.
In those cases were the settling velocity is equal to or less than the Brownian displacement the
particle will never settle, even in still air.
Gravitational Force
Drag Force
πd3(wS wa )g
6
1 −
3πdηv
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Physical Properties of Airborne α-Quartz Particles at 19.0/20.0 °C and 85kPa
Equivalent
geometric
diameter of
Quartz particle
(μm)
Aerodynamic
Diameter
( = diameter of unit
density sphere)
(μm)
Stokes diameter
(from experimental
data dp/ds =
1:1.67)
(μm)
Stokes
Terminal
Velocity
(m/s)
Slip-corrected
terminal
velocity
(m/s)
Brownian
displacement
per second
(m)
250 407 417.5 4.977 4.979 4.36 x 10-7
100 162.8 167.0 0.796 0.797 6.89 x 10-7
50 81.4 83.5 0.199 0.199 9.74 x 10-7
40 65.1 66.8 0.127 0.128 1.09 x 10-7
30 48.8 50.1 7.17 x 10-2 7.19 x 10-2 1.26 x 10-6
20 32.6 33.4 3.18 x 10-2 3.20 x 10-2 1.54 x 10-6
10 16.3 16.7 7.96 x 10-3 8.04 x 10-3 2.18 x 10-6
5 8.14 8.35 1.99 x 10-3 2.03 x 10-3 3.08 x 10-6
2 3.26 3.34 3.19 x 10-4 3.33 x 10-4 4.84 x 10-6
1 1.63 1.67 7.96 x 10-5 8.69 x 10-5 6.89 x 10-6
0.5 0.814 0.835 1.99 x 10-5 2.35 x 10-5 9.74 x 10-6
0.1 0.163 0.167 7.96 x 10-7 1.52 x 10-6 2.18 x 10-5
0.05 0.081 0.084 1.99 x 10-7 5.63 x 10-7 3.08 x 10-5
0.02 0.033 0.033 3.19 x 10-6 1.77 x 10-7 4.87 x 10-5
0.01 0.016 0.017 7.96 x 10-9 8.08 x 10-8 6.89 x 10-5
2.2.1.2 Coagulation
Particles of dust in air will coagulate spontaneously and continuously, irrespective of the
substance of which they are composed. As soon as they touch they will fuse and stick together.
This process continues until the particles become large enough to settle.
The rate of coagulation increases with the turbulence of the medium in which they are
contained. As eddies and swirls are formed the velocities of the particles relative to each other
becomes greater, increasing the chance of collision and therefore increasing the rate of
coagulation.
2.2.1.3 Impingement
Impingement (impaction) of dust particles occurs when an obstruction to the flow is encountered
and the velocity of the medium is great enough. This impaction occurs when the inertia of the
particle is high enough to cause the particle to stick to the obstruction. There is also the chance
that high velocities will sweep the particles from the obstruction.
The impaction of particles on falling water droplets can be used for the removal of airborne
particulate (dust) from the airflow. The success of this method is dependant upon the number,
concentration and collision efficiency with the efficiency dependant upon size, density and
velocity of the particle and the water droplet.
The size of dust particles is measured and expressed in microns (μm). The smallest dust
particle that can be seen with the naked eye under good conditions (i.e. black on white with
good lighting) is around 25μm. It is generally accepted that the smallest particle visible to the
naked eye can be as large as 50μm.
Dust of particular concern to mining practitioners is that fraction inhaled and retained in the
respiratory system. The site of the deposition varies with the size, shape and density of the
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particles and this was first described in the proceedings from the Pneumoconiosis Conference
held in Johannesburg in 1959.2 Deposition
in at these sites was shown to depend on
three factors:
(1) the percentage removal of particles
before reaching the pulmonary lobules
as the inhaled air passed through the
nasal chamber and air passages;
(2) the fraction of tidal air volume which
reaches the pulmonary spaces;
(3) the dust collection efficiencies of the
pulmonary spaces.
In short this experimental work shows that:
(1) Particles with a diameter larger than
10μm are deposited in the nasal
passages. The proportion deposited
falling off as the size of the particle
decreases. Virtually no 1μm particles
are deposited in the nasal passages.
(2) Particles with a diameter larger than 2μm tend to be deposited in the branching ducts
leading to the lungs (bronchial tubes). For particles below 2μm there is insufficient time for
settlement to occur. Below 0.5μm, the probability for deposition increases due to the
bombardment of these very fine particles by molecules.
(3) Because of the high retention time in the lungs, the remaining particles have a high
probability of deposition.
Recommendations from the 1959 Johannesburg Pneumoconiosis Conference have formed the
basis for dust measurements monitoring in the years following. One of the most significant
recommendations adopted from this conference was:
“That measurements of dust in pneumoconiosis studies should relate to the ’respirable
fraction’ of the dust cloud, this fraction
being defined by a sampling efficiency
curve which depends on the falling
velocity of the particles and which passes
through the following points: effectively
100% efficiency at 1 micron and below, 50
% efficiency at 5 microns, and zero
efficiency for particles of 7 microns and
upwards; all sizes refer to the equivalent
diameters.
(the ‘equivalent diameter’ of a particle is
the diameter of a spherical particle of unit
density having the same falling velocity in
air as the particle in question.)”3
2 HATCH, T., “Respiratory Dust Retention and Elimination” Proceedings of the Pneumoconiosis Conference. Johannesburg 9th –
24th February 1959. p 113,132)
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A simplification of the dust deposition curve was constructed on this basis and recommended by
the International Pneumoconiosis Conference held in Johannesburg, South Africa in 1959. This
curve has since become known as The Johannesburg Curve and clearly demonstrates that
100% of particles 7μm will enter the human lung, therefore it is the dust particles less than 7μm that
are the primary concern for mines.
A number of health problems may be caused by the excessive and or prolonged inhalation of
dust. The biological response is dependent upon the physical, chemical and toxicological
properties of the dust involved.
Dusts that cause little or no chemical reaction and subsequent tissue scarring are classified as
inert, although this may be misleading, as excessive inhalation will result in accumulation in the
alveoli region causing mobilisation of the macrophages (which engulf small particles) and are
the last line of defence of the respiratory system. Continued accumulation may eventuate with
the formation of plaques on the alveoli walls causing shortness of breath and an increase in the
frequency of colds and influenza etc.
Chemically active dusts (e.g. silica) decrease the active life of the macrophages and eventually
result in permanent alteration or destruction of the alveoli. These permanent changes cause
scarring of the lung tissue and result in a reduction of the elasticity of the lung tissue and a
subsequent less efficient oxygen intake. This becomes very evident with a decreased lung
expansion, breathlessness and a lessened capacity for work.
The long term effect of dust inhalation is not necessarily confined to the respiratory system and
there is some research that suggests that the respiratory system merely provide the mode of
entry into the body, other modes being absorption and ingestion. The respiratory system is very
selective with respect to the size and quantity of dust retained in the lungs. It is not simply a
matter of the quantity of dust in the atmosphere that dictates the amount deposited in the lungs,
it is also the duration of exposure and the rate of deposition to the critical area.
2.2.2 Effects of Dust
Because the pre filters of the respiratory system are not developed to cater for the much finer
dusts produced with mining activities, some of this dust manages to enter the finer passages of
the lungs and remain there. It is this accumulation of dust in the lungs, which causes the
condition known as pneumoconiosis (from two Greek words meaning dust and lungs).
The deposition in the respiratory tract occurs according to the following mechanisms:
a) Gravitational settlement as determined by the free falling terminal velocity,
b) Impacting due to inertia,
c) Diffusion as applied to microscopic particles and,
d) Interception due to the physical size of the particles.
Of those particles, less than 5μm only 50% penetrate into the alveoli region of the lungs. About
50% of these particles (usually less than 3μm) deposit mainly due to diffusion and impaction.
The remaining 50% will remain airborne and will be exhaled.
The terminal settling velocity of airborne fibres is almost independent of length. Deposition from
gravitational settlement and impaction may be avoided in the upper respiratory tract allowing the
fibre to penetrate into the fine air passages of the lungs. In these areas the length becomes the
controlling factor for deposition. Interception becomes the mechanism. Fibres of 50μm and up
3 Recommendations Adopted by Pneumoconiosis Conference – Proceedings of the Pneumoconiosis Conference. Johannesburg 9th
– 24th February 1959. p 619
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to 200μm have been found in human lungs, the diameter of these fibres is almost always less
than 3μm. The longest fibres are usually found in the bronchi and alveoli. In short, the
probability of a fibre impacting the wall of an airway becomes greater as the fibre becomes
longer.
Fibrous materials include asbestos, calcite, mica, magnesite, apatite gypsum and talc.
A number of health problems may be caused by the excessive and or prolonged inhalation of
dust. The biological response is dependent upon the physical, chemical and toxicological
properties of the dust involved.
Dusts that cause little or no chemical reaction and subsequent tissue scarring are classified as
inert. This may be misleading, as excessive inhalation will result in accumulation in the alveoli
region thus causing mobilisation of the macrophages (which engulf small particles) and are the
last line of defence of the respiratory system. Continued accumulation may eventuate with the
formation of plaques on the alveoli walls causing shortness of breath and an increase in the
frequency of colds and influenza etc.
Chemically active dusts (e.g. silica) decrease the active life of the macrophages and eventually
result in permanent alteration or destruction of the alveoli. These permanent changes cause
scarring of the lung tissue and result in a reduction of the elasticity of the lung tissue and a
subsequent less efficient oxygen intake. This becomes very evident with a decreased lung
expansion, breathlessness and a lessened capacity for work.
2.2.3 Dust Exposure Standards
Allowable concentrations of dusts in work places have been determined from epidemiological
studies that by their very nature extend over 20 to 30 year periods, or by experimental work on
animals. This is because any physiological damage to workers is not usually detected in the
early years of exposure and respiratory problems manifest only after a number of years of
exposure. As a consequence exposure standards are constantly being reviewed and updated.
A number of health problems may be caused by the excessive and/ or prolonged inhalation of
respirable dust. The biological response is dependent upon the physical, chemical and
toxicological properties of the dust involved. Diseases caused by exposure to various types of
dust can range from lung tissue scarring to pneumoconiosis, mesothelioma and lung cancer.
Reference should be made to http://www.nohsc.gov.au/databases/ for the most up to date
information on exposure standards. Some current dust exposure standards from this internet
site which are relevant to the mining industry are:
Selected Exposure Standards (FROM HTTP://WWW.NOHSC.GOV.AU/DATABASES/)
Dust Type Exposure Standard (TWA)
Quartz 0.2 mg/m3 (respirable)
Coal Dust 3 mg/m3 (respirable)
Asbestos including: Amosite, Crocidolite, other forms or mixtures. 0.1 f/ml (respirable)
Chrysotile 1 f/ml (respirable)
Where there is no exposure standard 10 mg/m3 (inspirable*)
*refer to http://www.nohsc.gov.au for definition of “inspirable”
2.2.4 Explosive Dusts
The most common explosive dust in metalliferous mining is sulphide dust. Sulphide dust
explosions can cause considerable damage to underground facilities and will cause the release
of poisonous gasses (primarily SO2).
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Sulphide dust explosions generally occur at firing time in massive pyritic or pyrrhotitic ore bodies
and have been documented since at least the early 1920’s. The explosions are known to occur
in ore and waste rock containing 20% sulphur with reporting’s of explosions in rocks with as little
as 11%. These explosions are significantly more destructive when they occur in stope blasting
than in development headings.
Because of the large quantities, high concentrations and toxicity of the sulphur dioxide gas
produced with these explosions, re-entry into the mine is extended beyond what could normally
be expected. Because of this the recommended approach is to eliminate the explosions by
adoption of preventative measures prior to and during blasting.
The following procedures should be adopted in orebodies that are known to have the potential
to produce sulphide dust explosions:
– Adequate stemming should be used
– The use of detonating cord should be eliminated
– All personnel should be removed to a safe place before firing
Enright4 & 5. (1996) when discussing preventative measures states “Murphys Law” applies with a
vengeance in underground mining and it is inevitable that despite all reasonable precautions
sulphide dust explosions will occur.”
2.3 Heat
The adverse effects of heat range from discomfort, though to life threatening illness, such as
heat stroke. Even relatively low levels of heat can lower the workforce morale, with all the
attendant problems of high accident rates and low productivity. This simple cause and effect is
often not fully appreciated and a number of mines attempt to “soldier on” without ever facing up
to what can be a very important and significant problem which impacts on employee health,
safety and mine productivity.
The principle for the control of heat build up in a mine, is the same as the principle of a car
radiator. For example if a car is allowed to idle for extended period of time (such as when
caught in a traffic jam in the middle of summer), the car engine will quickly overheat because
there is insufficient airflow passing through the radiator. Air at normal atmospheric temperature
flowing through the radiator will absorb some heat from it. If the air is constantly flowing through
the radiator then the heat will be removed, and the radiator will be kept at a lower temperature.
Up to a point the faster the flow the faster the heat removal.
The effect is similar in humans and machinery working in a confined space such as an
underground mine. The surrounding air absorbs the heat generated by humans, machinery and
the surrounding rock to a temperature equal to the temperature of the person, rock or
machinery. Which ever is the highest?
Sources of heat in mines can be categorised as either natural (e.g. rock temperatures, ambient
air temperatures and auto compression) or artificial (e.g. diesel and electrical powered
equipment) and need to be considered at the early stages of any project.
As air passes through the mine ventilation circuit it is subject to a number of sources of heat.
These include:
4 ENRIGHT, R.J., LEONTE, M. “AMIRA – Sulphide Dust Explosions Volume 1 – Cause and Prevention in Development Headings
Project P316-P316A (1990-1994)” (11 May 1995)
5ENRIGHT, R.J., LEONTE, M. “AMIRA – Sulphide Dust Explosions Volume 2 – Detection and Preventative Measures. Project
P316A-P316B (1990 –1995)” (1996).
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Auto Compression: – As air travels down the intake airways form the surface, its elevation
decreases. There is a corresponding conversion of potential energy into enthalpy6. The
magnitude of the change in enthalpy can be estimated using the steady flow energy equation
for a flow from a higher elevation (Z1) to a lower one (Z2), assuming no heat flow and no work
done:
H2 −H1 = g(Z1 − Z2 ) Equation 2-3 Auto Compression
Where:
H = enthalpy (J/kg)
Z = elevation (m)
g = acceleration due to gravity (9.81 m/s2)
The enthalpy thus increases by 9.81 kJ/kg for every 1,000m decrease in elevation. For dry air,
the thermal capacity is 1.005 kJ/°C and the theoretical dry bulb temperature increase is 9.81
kJ/kg/1,000m ÷ 1.005 kJ/kg°C = 9.76 °C/ 1,000m. In other words, the temperature of dry air
flowing down a dry 1,000m shaft into a mine would increase by 9.76 °C (assuming there is no
heat exchange between the air and the rock surrounding the shaft). The following points should
be noted:
• Autocompression is not strictly speaking a heat source (it results from a conversion of
energy, rather than from the addition of an external heat source).
• Autocompression causes the air temperature to increase, therefore as the mining depth
increases, the ventilation air has less ability to remove heat.
• The temperature rise due to autocompression is independent of the airflow rate. In
contrast, as the airflow increases, the temperature rise due to other sources of heat
decreases.
It is also important to note that water temperature will increase with depth. If the water is
contained in pipes then this increase in temperature is in the order of 0.20C per 1000m. If
the water is free flowing then this temperature increases to 2.340C per 1000m.
Transfer from the surrounding rock:
Surface rock temperature is around the
annual average air temperature and can
provide either cooling or heating depending
on the air temperature passing over the
rock. As we get deeper it gets hotter
because the rock hasn’t transferred its heat
to the air. It does not always follow that
cold places have cold rocks. In some
places the rocks are hotter than others
simply because they are newer and have
yet to cool down and this varies
considerably. For example the rock
temperatures in Tasmania are hotter than
the rock temperatures in the Barkley
Tableland of Northern Territory.
6 Enthalpy is the energy per unit mass (in this case of the air), resulting from the random motion of the air molecules. Elthalpy
includes the thermal energy and the energy due to the pressure of the air.
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The rate of temperature increase with depth is known as the geothermal gradient. The
geothermal gradient varies depending on many factors, however typical gradients in Australia
range from 1 to 3 °C per 100m of vertical depth. The surface rock temperature (i.e. 20 – 30m
below surface, where temperatures are largely unaffected by surface climatic variations) is close
to the average annual ambient dry-bulb temperature. For dry airways, the heat flow from the
surrounding rock to the ventilation air is proportional to the difference between the virgin rock
temperature and the air temperature. The rate of heat flow from the rock to the air increases
when the airway is wet. The rock surrounding an airway has the ability to absorb and
subsequently release thermal
energy, depending on the difference
between the rock temperature and
the air temperature. This is
sometimes known as the thermal
flywheel effect. In some mines,
oxidation of exposed minerals can
also be a significant source of heat.
Ground Water Ground water
flowing from the rock into an airway
acts to transfer heat from the rock
to the air. The ground water
temperature is almost always the
same temperature as the virgin rock
temperature. The amount of heat
transfer can be limited by use of
efficient pumping and drainage
practices (e.g. water should be collected in drains and kept off declines).
Machine Heat Except in cases of hoisting, hauling or pumping (where potential energy is
raised), almost all of the output power of underground machines is used to overcome friction
(i.e. it is converted into heat). Diesel engines are thermally inefficient and generate significant
heat loads. At full power, they are about 33% efficient (i.e. 33% of the fuel energy value is
converted to flywheel power (almost all of which eventually converts to heat anyway), the other
66% is converted directly to heat). For example, a diesel truck operating on a level gradient and
producing 200 kW of engine power would emit about 600 kW of heat. In underground mines,
secondary ventilation fans are also a major source of heat. For example, a fan consuming
180kW of electrical energy does no useful work in a thermodynamic sense and hence all of the
180 kW of electrical energy is converted to heat.
Explosives Only 5% of the energy produced by blasting is used to break the rock, the
remaining 95% is released as heat. For many years it was thought that this heat was dissipated
directly to the ventilation system and removed during the re-entry period. It is now more widely
accepted that this heat is transferred to the broken rock and liberated over a much longer period
of time and is variable, depending upon the ventilation rate and the rock surface exposure.
Cement Fill Only Hydraulic Fill would add significant quantities of water and therefore heat, to
the mine environment. All of the other fill types are non-draining and, except for the flushing
water used with Paste Fill, should have minimal impact. As cemented backfill cures in the
stopes, heat is produced due to the exothermic reaction of the cement and water. Most of this
heat will be absorbed into the surrounding rock and slowly released and generally this would be
insignificant on a mine wide basis. However, it may be significant locally during the first few
days after placement. If the paste is placed at a temperature of 30ºC and rises by 10ºC then this
will raise the temperature of the surrounding rock to 40ºC. In hotter mines heat may flow from
the rock to the fill until equilibrium is reached. The radiant heat from the fill in this case would
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almost be imperceptible. In the worst case some of the ‘excess’ heat would be released to the
ventilating airflow and rejected to the exhaust ventilating system.
An appreciation of the proportion of the total
heat load contributed by the various heat
sources can be gained from the example
opposite, which is based on heat load
calculations performed for the North Broken
Hill mine. At the time the calculations were
performed, the mine was operating at a depth
of about 1,600m below the surface.
It should also be noted that surface ambient
conditions can also be an important contributor
to underground heat conditions. This is
particularly the case in northern Australia,
where high surface temperatures and humidity
can give rise to heat problems, even in
relatively shallow mines.
2.3.1 Air Temperature
Temperature should not be confused with heat. Heat is a form of energy and is calculated.
Temperature is a state and is measured. For example a bath full of water at 30° contains more
heat than a cup of water at 70°. The difference being the thermal capacity which is the ability to
raise the temperature of 1kg by 1°C.
There are two measurements of air temperature are important to ventilation practitioners, the
dry-bulb and the wet-bulb.
Dry- bulb temperature is the actual air temperature, measured with a standard thermometer and
the wet-bulb temperature is the measure of the evaporative capacity of the air. Together they
measure relative humidity. Humidity is the ratio, expressed as a percentage, of the water
vapour present in the atmosphere to the amount required to saturate the air at the same
temperature. As humidity increases, the cooling from evaporation of sweat decreases.
The ability to measure both the wet and dry bulb air temperature is essential, particularly in hot
or poorly ventilated mines, to monitor potential heat stress conditions. These two properties are
also required to accurately determine the air density.
In mining, the most widely used instrument to determine wet
and dry bulb air temperature is the whirling hygrometer (sling
psychrometer), or wet and dry bulb thermometer. It consists of
two identical thermometers (usually the mercury type)
mounted side by side in a rigid plastic frame. The frame is
attached to a handle via a spindle and is free to rotate. The
thermometers are graduated from -5°C to +50 °C. The bulb of
one thermometer is surrounded by a cotton wick that is kept
moist by a small supply of distilled water in a reservoir. This is
known as the wet-bulb.
The dry bulb thermometer provides the sensible temperature of the air. The wet bulb
thermometer provides a measure of the evaporative rate of the air. When the air is dry, the
moisture in the wet cloth will evaporate faster and the temperature will cool. When the air is
humid, very little moisture evaporates from the wet cloth, and the cooling process slows. The
smaller the difference between the two temperatures, the higher the humidity If required, the
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relative humidity can be determined by equating these two temperatures using a table or slide
rule supplied with the instrument.
Equivalent electronic instruments are available, however, they do not compete with the
simplicity and reliability of the sling psychrometer. In particular, the humidity sensors on
electronic instruments seem to cause trouble unless very good quality, very expensive units are
used.
2.3.2 Body Heat Balance
Food (fuel) is oxidised in the metabolic process and converted to energy, in the forms of:
• Metabolic heat
• Mechanical work, and
• Change in mass (body growth)
The latter is negligible (usually!) and can be ignored. Although metabolic heat production
depends primarily on muscular activity (i.e. it is related to the rate of work), it also varies with:
• Condition of the individual’s health,
• Physical fitness, and
• Emotional state
In essence, the human body is a biological engine of with low mechanical efficiency. Less than
20% (usually much less than 20%) of the available energy is converted into usable mechanical
energy. That part of the metabolic process not used to provide the mechanical work (i.e. >80%)
will always appear in the body as heat. This metabolic heat must be rejected from the body to
the environment; otherwise the body’s core temperatures will increase, possibly to life
threatening levels. In fact, if none of the metabolic heat could be rejected, the body temperature
would rise by 1°C (the maximum ISO recommended acceptable rise) in 12 minutes during
moderate exercise and 4 minutes during strenuous exercise.
As we work we release energy in the form of heat and understandably the harder we work the
higher is the metabolic heat generation process. The removal of heat generated by the human
body is reliant upon our ability to sweat and the rate at which it can be evaporated.
This heat is transferred to the external environment and, if the rate of generation is greater than
the rate of transfer the body temperature will rise. This heat storage is called the “metabolic
heat generation” or the “metabolic heat production rate”.
Heat transfer between the body and the environment occurs via:
• Respiration (breathing)
• Radiation
• Conduction
• Convection, and
• Evaporation
Respiratory Heat Exchange. In a typical hot underground environment, respiratory heat
exchange accounts for about 5% of metabolic energy production. This heat loss is generally
considered to be small enough to be ignored.
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Radiant Heat Exchange
The magnitude and direction of radiant heat transfer depends on the temperature difference
between the human body (skin) and the object. This type of heat transfer is usually only
significant in mines with hot rock temperatures, and in the vicinity of hot diesel equipment.
Conductive Heat Exchange Conductive heat transfer occurs when two bodies come into
contact. In normal mining activities, conductive heat transfer to or from the human body is
usually negligible (unless, for example personnel are wearing cooling jackets).
Convective Heat Exchange Convection occurs when a layer of cool air comes in contact with
warm skin. The air increases in temperature and its density decreases. As the air becomes
lighter, it rises, taking the heat away from the skin. As the air is replaced with cooler, more
dense air, the process of heat exchange continues.
Convective heat transfer to the surrounding air is typically 15% to 20% of total cooling. It
depends on the temperature difference between the skin and the dry bulb temperature of the
air, as well as the relative air velocity and body surface area.
Evaporative Heat Exchange Evaporation is the main cooling mechanism for the body,
contributing about two thirds of total cooling. Evaporative cooling relies on the latent heat of
vaporisation of sweat from the body. The effectiveness of this form of heat transfer is dependent
upon:
• efficiency and rate of sweating and,
• the evaporative capacity of the environment (depends mainly on air temperature, humidity,
air velocity, and type of clothing).
The ability of the body to sweat depends on physical fitness, acclimatisation (training the body
to sweat efficiently) and on the body having sufficient fluid.
There is considerable variation in the sweat rate of individuals and this is partly associated with
the degree of acclimatisation (it also depends on general health and genetic factors). Sweat
that drips from the body, serves no cooling function and it begins to drip from the skin surface
well before the body is fully wetted. This is primarily because some areas of the human body
produce more sweat than others. Sweat begins to drip when the skin surface is approximately
50% wet.
In terms of maximising the cooling effect, the most effective place for sweat to evaporate is from
the skin. Clothing (particularly clothing made of artificial fibres) can significantly reduce the
amount of heat rejected from the body via evaporative heat exchange (it also reduces
convective heat exchange).
In underground mines, the evaporative capacity of the environment can be increased primarily
by increasing the air velocity. In some instances, it becomes necessary to also lower the wet
bulb air temperature (e.g. by reducing moisture pick-up in intake airways or through the use of
refrigerative mine air cooling).
2.3.3 Thermoregulation
The core of the body must be maintained at a stable temperature of between about 35OC and
40OC. To do this, the body invokes its thermoregulation processes. These include:
• Cold – Skin blood flow is reduced to reduce heat losses to environment. Shivering
commences to raise the metabolic rate.
• Heat –Skin blood flow is increased to increase heat transfer rate. Sweating increases to
maximise evaporative heat loss.
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The effectiveness of the body’s thermoregulation system is dependent on a large number of
factors, including:
• Fitness
• Age
• Obesity results in excessive fat insulating the body, reducing heat loss
• Acclimatisation is the adaptation of the body’s thermoregulatory systems to working in hot
conditions. Most of the effects of acclimatisation are generally developed within a week of
working in hot conditions, but the process continues for at least 14 days. Conversely,
acclimatisation is lost after a period of 7 to 14 days away from hot conditions. The degree
of acclimatisation is related to the level of heat stress experienced on the job. Some
personnel are heat intolerant and can never be successfully acclimatised to hot working
conditions.
• Hydration Levels are critical. Even a small decrease in hydration levels will lead to a
substantial reduction in the ability to work effectively in heat. Significant fluid intake is
required to counter losses through sweating in hot conditions. Hydration levels can also be
compromised by diuretics (alcohol, caffeine etc) and also by illness.
• Clothing acts as insulation, reducing the body’s ability to reject heat to the environment.
The illustration below shows the Temperature Response to Heat Stress there is a tremendous
spread in the effectiveness of the body’s thermoregulation system, even across a relatively
narrow sample group (in this case, 99 acclimatised, essentially nude men working at the same
rate in the same environment).
Obtained from Chapter 20 of “Environmental Engineering in South African mines”,
published by the Mine Ventilation Society of South Africa in 1989.
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2.3.4 Heat Related Illness
It is important to be able to recognise the causes and symptoms of overexposure to heat and to
know the treatment for these illnesses. Over-exposure to heat occurs when the body’s
temperature controlling mechanism (thermoregulation) cannot cope with the thermal
environment and rate of exertion. Over-exposure to heat may quickly progress to collapse,
unconsciousness (coma) and death.
The symptoms and treatments for some of the many classifications of heat illnesses are listed
below (in order of increasing severity):
Heat Cramps result from an imbalance in the body electrolytes, caused by vigorous activity,
dehydration and high temperatures. The body loses more fluids than it is replacing. This fluid
debit causes muscles to lose their vital electrolyte balance (complex salts), thus causing
muscular contraction (cramps).
Signs and Symptoms Treatment
Pale, clammy skin.
Sweating
Cramping pains (in the limbs and/or abdomen).
Nausea.
Spasms (in the affected limb or limbs).
Rest the victim in a cool location.
Give sips of water to drink (after nausea has passed).
Don’t massage affected limbs.
Discourage any further exercise.
Heat Exhaustion occurs after prolonged moderate elevations of core temperature. Its
development is usually attributed to the inability of the circulation system to meet the demands
of thermoregulation (i.e. the diversion of significant quantities of blood flow to the skin) whilst
also maintaining sufficient blood flow to the vital organs (brain and skeletal muscle).
Signs and Symptoms Treatment
Pale, clammy skin.
Restlessness
Cramps in the limbs and/or abdomen
Nausea and/or vomiting
Headache
Weakness
Fatigue
Rest the victim in a cool location Discourage any
further exertion.
Cool down casualty by sponging. (use tepid water)
Give cool water to drink (cautiously, after nausea has
passed).
Heat Stroke is a very serious condition known as a Core Temperature Emergency. It occurs as
a result of thermoregulatory failure. If appropriate treatment is not instigated promptly, heat
stroke carries a mortality rate of up to 80%. High body temperatures associated with heat stroke
can result in irreversible damage to organs (especially the brain, kidneys and liver) and the
nervous system.
The main causes of heat stroke in the mining industry7 are overly strenuous work in hot
environments and dehydration (often associated with excessive alcohol consumption).
7 Mining Industry heat stroke in countries such as South Africa. Note that it is unlikely that there has ever been an underground
mining fatality directly attributed to heat stroke in Australia.
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Signs and Symptoms Treatment
Body core temperature above 40.5°C
Often a cessation of sweating.
Aggressive or irrational behaviour
Staggering, Dizziness or Faintness
Vomiting
Collapse and seizures
Coma
Cool down the casualty immediately by gently
splashing with cool (but not icy) water. If possible,
increase evaporative cooling by fanning patient.
Continue treatment until medical help arrives.
Give frequent small drinks of water if patient is
conscious.
Prepare to resuscitate if required.
2.3.5 Heat Stress Indices
Measurement of hot conditions can either be by measuring the climate, that is the cooling power
of the environment or by measuring the heat strain, the effect on persons working there.
The four main parameters for directly evaluating heat strain are body core temperature, heart
rate, skin temperature and, weight loss through sweating. Although all of these can be
measured it is not practical to do so in the workplace. Even if heat strain could be measured
accurately, it does not indicate a reason for the problem and hence any possible solution.
The “degree” of heat stress is a function of the parameters outlined in the table below.
Parameters Contributing to Heat Stress
Metabolic rate
Ambient and radiant temperature
Water vapour pressure
Air velocity
Barometric pressure
Amount and type of clothing
Skin surface area and “view factor” (e.g. whether sitting, standing etc)
Variability in human thermoregulatory response.
Development of a measure to determine safe levels for work in hot places has been debated
over many years. During the 1900’s over 90 heat stress indices have been developed and used
around the world with varying success and acceptance by the international community.
Generally these indices combine one or two parameters into a single number and therefore only
partially represent the complexity of the human thermoregulatory system and the climatic
conditions encountered.
Broadly heat stress indices can be classified into three types: –
• Single measurements
• Empirical methods, and
• Rational indices
2.3.5.1 Single Measurements
Although there has been some attempt to use a single measurement to determine heat stress
conditions, there is no single parameter that provides a reliable indicator of physiological
reaction.
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2.3.5.1.1 Psychometric Wet-bulb Temperature
For many years, the wet-bulb temperature has generally been considered as the easiest and
simplest measure of heat stress and has been used widely in many Australian mines as the sole
indicator for climatic acceptability.
Australian mining legislation uses a wet-bulb temperature of 25°C, 27°C or 28°C as a trigger
point for modified work conditions such as reduced work hours or the introduction of regular rest
periods. 32°C is usually set as the upper limit for work being allowed to continue.
The psychometric wet-bulb temperature is a measure of the evaporative cooling power of the
environment and is therefore of limited value especially in high air velocities and high radiant
temperatures.
It worthy to note, that in wet-bulb temperatures above 37°C, the environment is unable to
support human life for any extended period of time.
2.3.5.1.2 Dry-bulb Temperature
Dry-bulb temperature above 45°C can give a burning sensation to exposed skin and is generally
accepted as the upper limit for work being allowed to continue.
2.3.5.2 Empirical Methods
The most commonly used empirical methods have been, effective temperature (ET), Kata
thermometer, predicted four-hourly sweat rate, and wet-bulb globe temperatures index (WBGT).
2.3.5.2.1 Effective Temperature (ET)
ET was developed in 1923 as a measure of comfort by the American Society of Heating and
Ventilating Engineers and primarily for use in offices. The ET of an environment is the
temperature of a saturated environment without movement of air that would produce the same
instantaneous thermal sensation as the environment being considered. As this index is based
on subjective thermal sensation it has shortcomings in either low (3.5 m/s)
air velocities.
Although the value can be calculated, it is normal to refer to empirically constructed nomograms
such as :-
1. Basic scale developed for essentially nude men and,
2. Normal scale for lightly dressed men.
It is not possible to reduce these nomograms to a simple mathematical expression over the
entire range.
The Basic ET may be calculated to within 0.2OC from the following equation given:-
• The difference between the wet-bulb and dry-bulb temperature is less than 5OC
• The wet bulb temperature is within the range from 25OC to 35OC, and
• The velocity of air is within the range from 0.5 m/s to 3.5 m/s
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ET = 20.86 + 0.0354TWB – 0.133V + 0.07V2 + (4.12 – X1 + X2 ) / 0.4129
Where
X1 = [ 8.33 { 17X3 – ( X3 – 1.35 ) ( TWB 20 ) } ] / [ ( X3 – 1.35 ) ( TA – TWB ) + 141.6
X2 = 4.25 [ ( TA – TWB ) X3 +8.33 ( TWB – 20 ) ] / [ ( X3 – 1.35 ) ( TA – TWB ) + 141.6
X3 = 5.27 + 1.3V – 1.15e-2V
TA = Ambient dry-bulb temperature (°C)
TWB = Ambient wet-bulb temperature
(°C)
V = Velocity of air (m/s)
e = vapour pressure of the air (kPa)
Subsequent to the development of these
nomograms, it has become customary to
use the temperature of a blackened 150
mm hollow copper sphere in place of the
dry bulb temperature. The assumption is
to take into account the effects of thermal
radiation. Measurements using the
blackened sphere are termed corrected
effective temperature.
ET has been extensively used in the
European and British mining industries. In
Queensland the Coal Mining Act –
General Rules for Underground Coal
Mines Part 2.7 (1) (a) provides that no
person shall be employed where the
effective temperature in the workplace is
or exceeds 29.4°C and Part 2.7 (3)
specifies a shortened shift strategy when
the ET exceeds 27.2°C.
The use of ET as a tool for prevention of
heat related illness has a number of
problems
1. ET exaggerates the effects of
thermal sensations at high relative
humidity
2. Only accounts to climatic conditions
and does not consider work rates
3. ET scales are most accurate in warm
climates and low heats stress
conditions.
4. ET is least accurate in velocities less
than 0.5 m/s and, greater than 3.5
m/s.
2.3.5.2.2 Kata Thermometer
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2.3.6 Air Cooling Power
A complete and supportable heat stress index must account for all of the contributing
parameters. Currently, the most complete and supportable index of heat stress is the concept
of air-cooling power, which was developed by the South African Chamber of Mines. This index
was however specific to South African mining practices (in particular, it assumed personnel
were essentially nude and fully wetted by perspiration).
There have been a number of updates and modifications to the original scales and it is
important to state which air cooling power scale is being used. McPherson (1992)8 describes
how to calculate air-cooling power, referred to as the “M” (McPherson’s) Scale Air Cooling
Power.
The concept of air-cooling power relies on the quantification of the ambient environment’s ability
to remove metabolic heat from the human body. The scale used to determine the rate of
generation of heat by the human
body and also air-cooling power is
W/m2 of body (skin) surface area9.
By definition, provided the air
cooling power is equal to, or
greater than the metabolic rate,
then there will be less than a one
in one million chance of the
healthy, acclimatised and “self
pacing” individual developing
dangerous body core
temperatures (>40°C), potentially
leading to heat stroke. This is held
to be an “acceptable” risk.
A widely used thermal acceptance
criterion is a minimum air cooling
power of 115 W/m2 (i.e. if the air
cooling power is less than 115
W/m2, then an individual could not
even sustain a “moderate” work
rate without incurring an
unacceptable (greater than one in
one million) chance of a healthy
acclimatised individual suffering
from heat stroke.
Whilst air-cooling power is the
most complete heat stress index, it
requires fairly complex calculations
(ideally requiring a computer) to
solve. The calculations require
inputs including; amount and type
8 “The Generalisation of Air Cooling Power” M.J. McPherson. Fifth International Mine Ventilation Congress The Mine Ventilation
Society of South Africa, Johannesburg, 1992.
9 For reference, the average skin surface area of a 1.7m tall, 60.5 kg South African miner has been determined to be 1.8 m2. The
modified DuBois formula, relating skin surface area (As, m2) to body mass (m, kg) and height (h, m) is As = 0.217m0.425h0.725 (From
“The Mine Ventilation Practicioner’s Data Book” Second Edition, Andrew Patterson et. al. 1999 The Mine Ventilation Society of
South Africa, Johannesburg.
18 20 22 24 26 28 30 32 34 36
100
200
300
400
5 m/s
4 m/s
3 m/s
2 m/s
1.5 m/s
1 m/s
0.5 m/s
0 m/s
5 m/s
3 m/s
2 m/s
1 m/s
0.5 m/s
0 m/s
5 m/s
3 m/s 1 m/s
0 m/s
Air Cooling power (M scale) or Metabolic Heat W/m
Wet Bulb Temperature t C
Radiant Temperature = Dry Bulb Temperature
Dry Bulb Temperature = Wet Bulb Temperature + 5°C
(Note that the graph may be used without undue error
for differences between wet bulb and dry bulb temperatures of between 2 and 8 °C)
Figure From “The Generalisation of Air Cooling Power” by M.J. McPherson
Fifth International Mine Ventilation Congress The Mine Ventilation Society of South Africa, Johannesburg, 1992.
Heavy Clothing (Long Sleeved Overalls and Long Sleeved Shirt)
115 W/m Line 2
(Light Work, 5m/s Air Velocity & Light Clothing)
Light Clothing (Thin Trousers, Short Sleeved Shirt)
Unclothed
w
2
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of clothing worn, wet bulb temperature, dry bulb temperature, radiant temperature of
surroundings, air velocity, barometric pressure and the work rate of the individual. The large
number of required inputs and requirement for a computer to calculate air-cooling power has
limited its use to date as a practical tool for determining on the spot whether the thermal
conditions in an underground environment are “acceptable”. Recently this problem has to a
large extent been overcome with the development of a robust, portable heat stress meter
designed for use underground. The meter uses algorithms based on air-cooling power and is
based on an instrument that was originally developed for use by the US military in “Operation
Desert Storm” in Kuwait.
For mines which do not have access to the heat stress meter, it should be noted that normal
ranges of some of the factors mentioned above in the underground mining context have a
relatively weak effect on air cooling power. A generalised Air Cooling Power Chart that assists
in the rapid manual assessment of the acceptability of the thermal environment can be
produced on this basis. (See McPherson’s “M” Scale Air Cooling Power Chart opposite) The
following assumptions were made in order to produce such a chart:
Typical Metabolic Work-rate Classifications for Healthy Adults
Light Work Moderate Work Hard Work Very Hard Work
240 W/m2
Sleeping 40
Seated 60
Standing 70
Walking 5 km/h
Trades people
Jumbo drilling
Diesel Operator
Walking 6.5 km/h
Building brick walls
Scaling
Hand-held drilling
Shovelling
Timbering
Barometric pressure – Assume P = 100 kPa. Air cooling power is largely unaffected by normal
range of pressures found within underground mines.
Radiant temperature of surroundings is equal to the dry-bulb temperature (This is usually the
case, except near hot surfaces such as diesel radiators etc).
Dry-bulb temperature is equal to the wet-bulb temperature + 5°C (Note that the graph may be
used without undue error for differences between wet bulb and dry bulb temperatures of
between 2 and 8°C.) The majority of underground temperatures except near diesel radiators
etc will fall within this range.
Clothing – Assumptions were also made about the thermal resistance and area factor of
different clothing specifications and also regarding the body posture factor.
The above assumptions allow an air-cooling power graph to be produced, which only considers
wet-bulb temperature, air velocity (air speed over the individual), and clothing type.
􀂃 The protective clothing worn underground in Australian operations would fall somewhere
between the “Light” and “Heavy” categories.
􀂃 An example has been outlined with the heavy dashed line in the “M” scale chart. It assumes
“light” clothing, “light” work rate (115 W/m2) and air velocity of 0.5 m/s. This line projects
down to a wet-bulb temperature of 29.5°C. In other words, at the conditions and work rate
outlined, a wet bulb temperature of less than 29.5°C will ensure that there is less than a one
in one million chance of heat stroke occurring in a healthy, acclimatised individual.
In any mining operation, there will be variation in many of the variables listed earlier . It can
however be concluded that based on all the assumptions implied in McPhersons ‘M’ Scale, an
acceptable air cooling power can not be provided for even a “light” work rate at wet bulb
temperatures above 32 °C.
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It is also important to note that for a range of circumstances (e.g. heavier work rates), conditions
which could lead to the development of heat stroke could occur at wet bulb temperatures below
32°C. For the purposes of providing a simple, practical measurement to determine the
thermal “acceptability” of an underground environment, a “stop work” cut-off of 32°C wet
bulb is supported. This temperature cut-off is combined with a requirement that the air velocity
must be greater than 0.5 m/s at wet bulb temperatures over 25°C (e.g. refer to W.A.
Regulations).
2.3.7 Management of Hot Working Conditions
Human heat stress is a health hazard that can be managed by:
Instigating “working in heat” protocols
Controlling the underground ambient heat conditions
Hot working condition protocols can include the following aspects:
Worker training and education – e.g. how to recognise and treat the symptoms of heat illness,
the importance of drinking sufficient water, the need to “self-pace” according to conditions and
the mechanism of acclimatisation. “Refresher” training courses should be re-run on a regular
basis (ideally just before the onset of summer)
Hydration testing – e.g. Urine tests on personnel who work in “hot” areas of the mine. The
tests quickly show those who are not coping with hot working conditions (e.g. those not drinking
sufficient fluid, those who are ill or those who are heat intolerant). The protocols should spell
out the policy regarding those who fail the hydration test (second chance etc).
Staff selection. Some mines preclude those with certain medical conditions, those who are
overweight, personnel over a specified age and females with “child bearing capacity” from
working in thermal conditions which are beyond a specified level.
“Stop work”. It is very important that a cut-off criterion with respect to acceptable/
unacceptable underground ambient conditions is established. The criterion must be fully
supported by management and strict guidelines established to ensure that the criterion is
respected under all but exceptional circumstances (and these should be defined in the
protocol).
Establishing working in heat protocols is of fundamental importance, however it only deals with
part of the problem. In a number of deeper mines (or shallower mines hosted in hot rock), it is
very difficult to maintain ambient conditions that allow even a light rate of work without incurring
an unacceptable risk of the development of heat stroke. In these mines, it is necessary to boost
the miners’ ability to reject metabolic heat by employing one or more of the following methods:
Increase the air velocity to improve the air-cooling power. This is effective at lower wet bulb
temperatures. As the wet-bulb temperature increases, a “diminishing return” effect is apparent.
In fact, again referring to McPhersons ‘M’ Scale, it can be seen that increasing the air velocity
has very limited effect on the air cooling power, once the wet bulb temperature exceeds about
32°C. It should also be noted that there is a considerable cost penalty associated with
increasing the airflow rate:
3
Old Flow Rate
Power Cost Increase New Flow Rate  


 

α
Microclimate Cooling. This involves cooling the localised area around the miner. One example
is the use of air-conditioned cabins on mobile equipment. Another example is the use of cooling
jackets (ice vests). These use flexible frozen gel-packs (similar to those used for sports injuries),
which special pockets in an insulated vest. They are primarily intended for use in emergency
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situations (and for sports competitors), but may potentially have
some applications for more regular underground use. The gel
packs are cooled in a chest freezer (in the crib room) and have a
useful life of about 2 hours. The figure opposite is an illustration of a
Micro-climate cooling vest. (Reference http://www.steelevest.com/).
This is seldom a satisfactory “total” solution, since there is a very
real chance of something going wrong (e.g. forgetting to change the
cooling jacket gel-packs and subsequently getting “stuck” in a hot
mining area).
Refrigerative Mine Air-cooling In spite of the sometimes
considerable capital and operating costs, refrigerative cooling of
mine air is often the only supportable option for control of heat
stress in very hot mines. A simplified line diagram showing a
“typical” mine refrigeration circuit is shown below.
The refrigeration machine generally
consists of a screw compressor coupled
to plate type heat exchangers (condenser
and evaporator). The refrigerant is usually
ammonia for surface plants and R134 for
underground plants. Rated capacity of
machines used in Australian Mines ranges
from about 500 kW to 10 MW of
refrigerative cooling capacity. As a rule of
thumb, the compressor motor power draw
is 1/4 to 1/5th the nominal refrigerative
cooling capacity. The amount of (potable)
water circulated in the plants can be
considerable. For example, a 1MW plant
circulates about 25 l/s in each of the (hot
and cold) water circuits. A proportion of
this flow evaporates and some is also
dumped. Make-up water requirements for
a 1 MW plant are typically 1.5 l/s in the hot
water circuit and 1 l/s on the cold-water
circuit.
Several plant configurations are possible:
Surface Bulk Air Cooling. All of the equipment is located on the surface. The air is cooled at
the intake raise collar. This system is simple and easy to maintain. It is not well suited to mines
with complex ventilation systems (some of the cooled air may find its way to locations where
chilled is not required). Some of the “coolth” is also inevitably lost in the mine intake airways.
Surface Plant with Underground Coolers. With this system, the plant and heat rejection
towers are mounted on the surface. Chilled water is reticulated underground to cooling towers
that are close to the locations where cooling is required. Advantages of this system are that
cooling can be more precisely and more directly delivered to the required locations. The
disadvantages are high pumping costs (high volume, high head pumping although some energy
losses can be recovered with the use of a Pelton wheel turbine coupled to a generator) and
significant maintenance costs (complex system with cold and hot water storage dams insulated
chilled water pipes, batch cooling of water, large underground pump stations and Pelton wheels,
underground cooling towers which can be fouled with dust etc).
Evaporat or
Cooled Air To Mine
Hot Air Rejected t o Atmosphere
Chilled Wat er Circuit
Hot Wat er Circuit
Cooling Tower
Cooling Tower
Condenser
Refrigerat ion
Machine
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Underground Spot Cooling. All of the equipment is located underground. The refrigeration
plant and heat rejection tower are located in a return airway. The cooling tower supplies chilled
air to a specific location (often to the inlet of a secondary ventilation fan). The capacity of
refrigeration machine is restricted to a size which can be physically transported underground on
a skid (e.g. maximum of about 1MW cooling capacity). As a result, the machine generally only
has sufficient capacity to cool one or two headings. There are many disadvantages including
high maintenance costs due to arduous operating environment and difficult maintenance access
as well as logistics problems (e.g. supply of 1.5 to 2.5 l/s of potable make-up water)
Some examples of Australian mine air cooling plants are shown in the following illustrations:
600kW Refrigeration Set for Underground Spot Cooler (WMC Olympic Dam)
Cooling and Heat Rejection Towers Not Shown
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1.1 MW NH3 Surface Plant at Telfer (1 of 2)
(Plant is being gassed up with NH3) Heat rejection tower
Telfer Bulk Air Cooler Installation Near Collar of VR2 Intake Raise
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Chilled water is reticulated U/G via insulated pipes to underground cooling towers
10 MW K61 NH3 Refrigeration Plant – Mt Isa
Cold water storage dam
Heat rejection towers
Close up of one of the Mt Isa K61 Compressors
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2.4 Mine Gases
There are a surprisingly wide variety of gasses that can be found in underground mines,
abandoned workings and caves. Some of these gasses may be poisonous, irritant, asphyxiant,
radioactive or explosive and are hence of particular interest from a ventilation design viewpoint.
Mine ventilation systems are required to dilute and remove atmospheric contaminants caused
by mining operations.
2.4.1 Constituent Gases of the Atmosphere
The atmosphere (fresh air) surrounding the earth surface contains a number of different gases
combined as a mixture.
Composition of Air
GAS VOLUME (%)
Oxygen (O2) 20.93
Nitrogen (N2) 78.11
Carbon Dioxide (CO2) 0.03
Minor Inert Gases 0.93
Water Vapour Variable
Because this air is a mechanical mixture, it is possible to separate and identify each of the
gases in the mixture. Any other substance or variation of these gases contained in the
atmosphere are contaminants and are subject to Exposure Standards under legislation.
Any additional gasses or variation in the proportions of the gases normally found in the
atmosphere are regarded as contaminants. Some characteristics of the more commonly found
mine gases are discussed below.
2.4.2 Carbon Dioxide (CO2)
Carbon dioxide is colourless, has a pungent or acrid smell and a “soda water” taste. It has a
specific gravity relative to air of 1.53 (significantly heavier than air) and will not support
combustion. It doesn’t liquefy but will form dry ice at –78°C.
CO2 has a TWA exposure limit of 5,000 ppm and a STEL of 30,000 ppm.
It can be found in coal mines and in mines hosted in carbonaceous rocks, such as limestone. It
is also produced by diesel engines. The gas is about 20 times more soluble than Oxygen and
diffuses rapidly into the bloodstream. The most noticeable effect of the gas is to cause the
respiration rate to increase, which serves to alert the miner to the presence of the gas.
Physiological Effect of Carbon Dioxide
Percentage in Air Effects
0.03 None (Normal atmospheric concentration)
0.5 Respiration increases by 5%
2.0 Respiration increases by 50%
3.0 Respiration increases by 200%.
5 – 10 Violent panting, leading to fatigue. Headache
10 – 15 Intolerable panting, severe headache, rapid exhaustion and collapse
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2.4.3 Oxygen (O2)
Oxygen is a colourless, odourless, tasteless gas, with a specific gravity relative to air of 1.1.
Oxygen is the only gas whose concentration should be maintained above a recommended
value. Oxygen depletion is caused by oxidation of minerals (e.g. Sulphides and Coal).
Depletion of Oxygen also results from combustion (e.g. diesel engines, blasting etc). A
deficiency of Oxygen implies greater than normal atmospheric concentrations of other gasses
(even inert gasses). The physiological effect of various levels of Oxygen concentrations is
shown below.
Concentration
(by Vol.)in Air
%
Typical Physiological Effects (Vary with individuals and period of exposure)
20.93 Normal content of atmospheric air
17.0 to 20.0 Lowest allowable concentration (variable in mine legislation)
17.0 Noticablely faster and deeper breathing rate (equivalent to 1,500m
ASL elevation). Candle will not burn below 16%.
15.0 Dizziness, buzzing in ears, rapid heart beat
A safety lamp flame
will go out at some
point in this range.
13.0 Work is difficult. Breathing becomes rapid and lips become blue.
Nausea and headache develop slowly and may become very
severe. May lose consciousness if exposure prolonged.
Dangerous for
exposures over half
to one hour.
10.0 Liable to faint and become unconscious.
9.0 Fainting, unconsciousness.
7.0 Life endangered
6.0 Convulsive movement, probable death.
Exertion leads to
unconsciousness
70 Slow weak pulse, respiratory failure and death.
Carbon monoxide is produced by fires, the oxidation processes (e.g. blasting) and sometimes
issues from rock strata (especially in coal mines). It is also a component of combustion engine
exhaust emissions.
2.4.5 Oxides of Nitrogen (NOx)
Oxides of Nitrogen covers a mixture of gases usually found together. The most important of
these are Nitric Oxide (NO) and Nitrogen dioxide (NO2), both of which are classified as toxic.
The proportion of NO is usually small and NO also readily converts to NO2 in the presence of air
and water vapour. Consequently, NO2 is the oxide of most interest. This gas is brown in colour
and dissolves readily in water to form Nitrous (HNO2) and Nitric (HNO3) acids. In sufficient
concentration, these acids cause irritation and corrosion of the respiratory system and eyes.
The associated bleeding and accumulation of fluid in the lungs can culminate in death from
pulmonary oedema (flooding of the lungs). This can occur up to 24 hours after exposure, even
after an apparently early recovery.
Oxides of Nitrogen are primarily produced by internal combustion engines and are also a
significant constituent of blast fumes.
The TWA exposure limit for NO is 25 ppm and that for NO2 is 3 ppm. As yet an STEL for NO
has not been set, but for NO2 the STEL is 5 ppm.
Physiological Effects of NO2
Concentration ppm Typical Physiological Effects for NO2
3 TWA
50 Moderately irritating to eyes and nose
100
Irritant to respiratory passages and to the eyes. (Headache, tightness of the
chest or perhaps pain in chest and a cough). Dangerous if exposed for ½ to 1
hour.
200 Breathed for 20 minutes may cause collapse, this may be delayed for several
hours.
250 Severe pulmonary oedema, probably fatal.
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2.4.6 Sulphur Dioxide (SO2)
In massive sulphide orebodies there can be spontaneous oxidation and heating resulting in
sulphur dioxide being released to the ventilating air.
Sulphur dioxide is colourless and even at relatively low concentrations has a pungent,
suffocating sulphurous odour, and acidic taste, making it readily detectable. It is highly toxic. It
has specific gravity relative to air of 2.26 and is soluble in water, forming sulphurous acid. It is
incombustible and is also not flammable. It is associated with oxidation of reactive sulphide
ores. Of particular importance is the fact, that sulphur dioxide is produced by sulphide dust
explosions. It is sometimes also noticed in mines that have their intake airways close to smelter
stacks.
SO2 exposure limits are a TWA of 2 ppm and an STEL of 5 ppm.
Effects of Sulphur Dioxide
Concentration (ppm) Effect
3 Detectable by its odour.
100 Irritating to eyes and nose, uncomfortable to breathe.
500 Dangerous to life after only short exposures.
2.4.7 Hydrogen Sulphide (H2S)
Hydrogen Sulphide is colourless and is readily detected in small concentrations by its
unpleasant rotten eggs odour. Unfortunately, continued exposure to the gas (even for relatively
short periods of time) leads to paralysis of the olfactory nerves, meaning that the sense of smell
cannot thereafter be relied upon. The gas has a specific gravity relative to air of 1.19 and burns
in air (in concentrations ranging from 4.5% to 45%), with a bright blue flame producing Sulphur
Dioxide (SO2).
Acidic action or effects of heating on Sulphide ores produce H2S. It is also formed as a result of
the decomposition of organic compounds. It is sometimes noticed near stagnant pools of water
underground. Hydrogen Sulphide is sometimes associated with natural gas and oil reservoirs
and can migrate through strata in solution.
H2S has a TWA of 10 ppm and a STEL of 15 ppm. It is often associated with methane.
Hydrogen Sulphide Effects
Concentration (ppm) Effect
0.1 – 1 Detectable by smell
100 Irritation to eyes and respiratory tract.
200 Intense irritation of eyes and throat.
500 After 30 mins serious inflammation of eyes and throat,
coughing, palpitation, fainting, cold sweats.
600 Serious effects after a few minutes, bronchitis and chest pain.
700 Depression, stupor, unconsciousness, and death.
1,000 Paralysis of respiratory system and death.
2.4.8 Methane (CH4)
Methane is colourless, odourless and non-toxic. It has a specific gravity relative to air of only
0.55 and as a result, tends to layer against the backs in areas of low air velocity (laminar flow
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conditions). Methane is of course very dangerous in mines because it forms an explosive
mixture with air at concentrations of between 5 and 15 % (it produces an explosion of greatest
force at a concentration of 9%). It is found in all coalmines, and has been responsible for many,
many thousands of coal mining deaths directly as a result of methane explosions, or after the
explosion as a result of Carbon Monoxide poisoning (CO is a product of incomplete methane
oxidation).
Methane is often associated with other flammable/ explosive hydrocarbon gasses,
predominantly hydrogen. The presence of methane in coal mines results from chemical and
bacterial action on organic material. Methane is also surprisingly prevalent in metalliferous
mining. It is often noted during diamond drilling. The most dangerous situations arise in
metalliferous mining where large amounts of methane accumulate (e.g. the stope backs after
firing).
Coward’s Diagram shows the relationship between methane and oxygen concentration and
explosibility.
2.4.9 Coal Damps
Damp is an old miners term for gaseous products formed in coal mines to distinguish them from
pure air. Although still in use they are not commonly used in today’s mining.
2.4.9.1 Fire Damp
A combustible gas formed by the decomposition or distillation of coal or other carbonaceous
matter. Consisting mainly of Methane. (Distillation is the heating of a substance in an
atmosphere low in oxygen. This prevents oxidation if the heating was to take place in fresh air.)
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Usually lighter than air and can accumulate in unventilated mine workings. A change in
barometric pressure may cause this to be released to the ventilation system. Produced during
mine fires by the distillation of coal.
Sometimes simply referred to as “gas”.
2.4.9.2 Black Damp
An atmosphere depleted of oxygen. More specifically an atmosphere containing variable
mixtures of Carbon Dioxide (CO2) and nitrogen (N2), generally caused by the oxidation of
carbonaceous material or coal and A typical mixture would be 15% CO2 and 85% N2. In it
simplest form is the name given to CO2.
Basically Black Damp is an extinctive atmosphere, hence the term “black”.
Usually heavier than air and can accumulate in unventilated mine workings. A change in
barometric pressure may cause this to be released to the ventilation system.
2.4.9.3 Choke Damp
A mine atmosphere that causes ‘choking’ or suffocation due to insufficient oxygen. Could be
any combination of CO2 and CH4 or other gasses or products of fires, smoke included, that may
replace the oxygen content of the air.
In some places the name given to black damp.
2.4.9.4 Illawarra Bottom Gas
Any mixture of CO2 and CH4 ranging from almost 100% CO2 to almost 100% CH4 and a little N2
that is capable of forming a flammable or explosive mixture when mixed with air.
Because of its density it tends to layer at the bottom of the drive, hence its name “bottom”.
2.4.9.5 After Damp
That mixture of gasses remaining after, a fire or explosion. (Some times referred to as “after
gasses”.
2.4.9.6 White Damp
The term applied to carbon monoxide or more specifically atmospheres containing lethal
quantities of CO.
2.4.9.7 Stink Damp
Atmospheres containing hydrogen sulphide with the odour the predominate factor. It is worthy
to note that when concentrations exceed 50 ppm the sense of smell may be affected and the
odour becomes undetectable.
2.4.9.8 Fire Stink
This is the smell indicating a spontaneous combustion. The odour is often associated with
benzene.
2.4.9.9 Water Gas
A combustible mixture of gases with a typical composition being 45% each of CO and hydrogen
with smaller amounts of CO2, CH4, N2 and oxygen or after damp. Formed when water is hosed
onto incandescent masses of coal when extinguishing the fire. The water gas produced could
produce a secondary explosion.
Considered poisonous because of its high concentration of CO.
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2.4.9.10 Producer Gas
A combustible mixture of gasses formed commercially by the action of air passing through a
layer of incandescent fuel (coal, coke or charcoal). A typical mixture would be 10% CO2, 15%
CO, 74% N2 and up to 1% of other gasses including CH4.
An identical mixture may be formed in mine fires.
2.4.10 Ammonia (NH3)
Ammonia is colourless, and has a very distinctive, pungent odour (the smell is familiar to those
who have used certain disinfectant and window cleaning products). It has a specific gravity
relative to air of only 0.65 and as a result, tends to layer against the backs in areas of low air
velocity (laminar flow conditions). Ammonia is irritating or corrosive to exposed tissue, especially
the eyes and the upper respiratory system. Inhalation of ammonia vapours my result in
pulmonary oedema (flooding of the lungs) and chemical pneumonitis. Depending on the
concentration, symptoms such as burning sensations, coughing, wheezing, shortness of breath,
headache, nausea and eventual collapse may be experienced.
NH3 has a TWA of 25 ppm and a STEL of 35 ppm
Ammonia readily passes into and out of solution in water. There are two main sources of
ammonia in mines:
• As a refrigerant gas in mine cooling plants
• Chemical reaction involving ANFO explosive, cement and water.
Because of the toxic and irritant effects of ammonia, it is not used as a refrigerant in cooling
plants where there is a possibility of the gas leaking into the mine atmosphere (although it is
otherwise a particularly suitable gas for the purpose). The main source of ammonia in mines is
from a chemical reaction involving ammonium nitrate, cement and water. The ammonium nitrate
is sourced from spilt, or un-detonated ANFO. The cement source is generally shot-crete
rebound (an increasing problem with more shot-crete usage). The reaction is as follows:
Calcium Oxide (a component of the cement) reacts with water to produce an alkali – Calcium
Hydroxide: CaO + H2O → Ca(OH)2
Next, the ammonium nitrate from the ANFO reacts with Calcium Hydroxide to produce ammonia
gas (2NH3), Ca(NO3)2 and water:
2NH4NO3 Ca(OH)2 2NH3 Ca(NO3 )2 2H2O + → ↑ + +
There have been an increasing number of ammonia “fumings”. Worst affected personnel seem
to be those on charge-up (working at height, working close to face in poor ventilation, fresh
ANFO “blow-back”, working shortly after shot-crete applied). The only solutions are to remove or
isolate one or more of the components of the chemical reaction (i.e. water, ANFO or cement).
2.4.11 Radon (Rn) and Radon Daughters
Radon is a colourless and odourless gas. It is one of the isotopes10 produced by the radioactive
decay of uranium to lead.
10 Naturally occurring elements comprise a mixture of isotopes. An isotope may have the same atomic number (number of protons in
the nucleus of the atom), but different masses. For example, U238 has 92 protons and 146 neutrons, whereas the isotope U235 has
the same number of protons (92), but only 143 neutrons.
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Decay Series – Uranium to Lead (NOTE THAT THE ISOTOPES IN BOLD ARE RADON DAUGHTERS)
Nuclide Radiation Half-life
Uranium 238 α 4.5 billion yrs
Thorium 234
β
γ
24 days
Protactinium 234 β 1.2 minutes
Uranium 234 α 250,000 yrs
Thorium 230 α 80,000 yrs
Radium 226
α
γ
1,600 yrs
Radon 222 (gas) α 3.8 days
Polonium 218 (RaA) α 3 minutes
Lead 214 (RaB)
β
γ
27 minutes
Bismuth 214 (RaC)
β
γ
20 minutes
Polonium 214 (RaC’) α 160 μ seconds
Lead 210
β
γ
22 yrs
Bismuth 210 β 5 days
Polonium 210 α 140 days
Lead 206 infinite, stable
The half-life is the time taken for one half of the atoms in a radioactive substance to decay.
During the decay process, various forms of radiation are released including Alpha (α) and Beta
(β) particles as well as gamma (γ) radiation. Alpha particles are low energy, positively charged
particles that can travel a few centimetres in air. The respiratory system can be damaged by
inhaled alpha particles, but alpha particles do not penetrate the skin surface. Beta particles are
electrons. They can penetrate the skin and cause damage to the body’s cells and organs.
Gamma radiation is electromagnetic radiation that can penetrate deeply into the body.
The half-life of radon is relatively long (at 3.8 days). As a result, it doesn’t tend to expose the
lungs to a significant amount of radiation energy. It is a different story with the relatively shortlived
radon daughters (referred to radon progeny in these more politically correct times)!
Inhalation of radon daughters in sufficient concentration over a long enough period of time will
increase the likelihood of exposed personnel developing lung cancer. Radon daughters tend to
attach themselves to surfaces such as dust particles and aerosols. As a result, the control of
ambient dust (especially respirable dust) and diesel soot levels in uranium mines is of particular
importance. It is also important in order to reduce the exposure to alpha emitting dust particles.
When all else fails, mandatory respiratory protection is sometimes used as a method to limit
radon daughter exposure.
In underground uranium mines, radon daughters and radioactive dust particles are just some of
the radiation hazards. The uranium ore emanates gamma radiation, which also contributes to
the total radiation exposure of the workforce. In contrast to dust and radon daughters, Gamma
radiation exposure can’t be controlled by ventilation. Instead, it is managed by physically
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shielding the workforce from the ore (particularly high grade ore) and by mine design and shift
rotation to limit the time spent in areas where the gamma radiation levels are high.
At one, low grade underground uranium mine, the approximate contributions to the total
radiation dose are:
• Radon daughters 40%
• Gamma Radiation 40%
• Airborne alpha emitting dust 20%
Assuming the ventilation system is well designed and managed, the relative contribution to total
dose from gamma radiation will increase for mines with higher uranium grades.
It is important to note that the concept of residence time is of critical importance when designing
ventilation systems for uranium mines. Once the radon emanates from the rock, it becomes a
“time bomb”, and begins immediately to disintegrate into radon daughters. The aim of the
ventilation system design is to remove the radon gas from the underground environment as
quickly as possible after it is produced. Depending on the ore grade, ventilation design criteria
can specify underground air residence times (after exposure to rock containing uranium) of less
than 10 to 15 minutes.
Areas of low velocity, with high ore rock surface exposure (e.g. stopes and sealed off workings)
are radon daughter “breeding grounds”. Personnel should not be exposed to the resulting high
radon daughter concentrations in the air leaking or exhausted from these locations. This
contaminated air should flow to a surface exhaust airway via as direct a route as possible. The
ventilation system design should ensure that personnel work in “fresh” air that has travelled from
the surface to the working place via a direct route in low-grade (waste) rock. In order to limit
exposure of personnel to radon daughters, considerable care with mine design and scheduling
is required in uranium mines and accordingly, ventilation considerations are often the most
important mine design criterion.
The source of radon in mines is primarily via emanation from rock containing uranium and
radium. The gas is found in underground mines (not just uranium mines) and sometimes also in
caves (many of which have poor ventilation and therefore long residence times).
Determination of a “safe” or “acceptable” exposure standard for radon and radon daughters is
complex. Exposure to these gasses constitutes just one of several contributors to the total
radiation exposure. The total effective radiation dose is measured in units of Sieverts (Sv). The
Sievert is a unit of radiation derived health risk. The prevailing annual limit for radiation workers
is 20 mSv/ year, averaged over 5 yrs. Note that practice of the “ALARA” (As Low As is
Reasonably Achievable) principle is always the recommended approach with respect to
radiation doses and a number of operations have adopted internal standards which aim to
achieve dosages that are less than a third or a quarter of the prevailing limits.
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SUMMARY OF GASSES
Name of Gas Symbol Properties Smell Flammable Limits TWA STEL Physiological Effects
(ppm) (%) (ppm) (%)
Oxygen O2 Colourless
Tasteless Nil In contact with oil
or grease – 17.0 (19.0
in mines) – 13.0 Essential to maintain life
Carbon
Monoxide CO Colourless
Tasteless Nil 12.5% – 74% 50 0.005 400 0.04
Displaces oxygen in blood
• 200 ppm Drowsiness; Headache after 2 hours work
• 400 ppm Headache after 45 mins work
• 1200 ppm Palpitations after 10 mins work
• 2000 ppm Unconsciousness after 10 mins work
• >3000 ppm Death possible
Carbon
Dioxide CO2 Colourless
Soda taste
Slight
pungent Non-flammable 5000 0.5 30000 3.0 Increased respiration
Depression of breathing
Hydrogen H2 Colourless
Tasteless Nil 4.0% – 74% – – – – Non-poisonous
Asphyxiant
Hydrogen
Sulphide H2S Colourless
Sweet taste
Rotten
Eggs 4.5% – 45% 10 0.001 15 0.0015 • 100 ppm Irritation to eyes and throat : Headache
• 1000 ppm Immediate unconsciousness
Nitrogen N2
Colourless
Tasteless Nil Non-flammable – – – – Non-poisonous
Asphyxiant
Oxides of
Nitrogen NOx Firing
fumes 25 0.0 – –
• 100 ppm irritant to respiratory passage and eyes
• 200 ppm After 20 mins may cause collapse usually
delayed
• 400 ppm after 15 mins exposure may be fatal
Nitrogen
Dioxide NO2 Brown colour
Acrid taste
Firing
fumes Acrid Non-flammable 3 0.0003 5 0.0005
• 50 ppm Irritating to eyes and nose
• 250 ppm severe pulmonary oedema collapse is usually
delayed
Methane CH4 Colourless
Tasteless Nil 5.0% – 14.0%
Stop diesel engines when = or > 1.0%
Turn off electrical power when = or >
1.25%
Remove people when= or > 2.5%
Non-poisonous
Asphyxiant
Sulphur
Dioxide SO2 Colourless
Acid taste
Burning
Sulphur
Pungent
Suffocating
Non-flammable 2 0.0002 5 0.0005
Oedema
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2.5 Diesel Engines
Dr. Rudolf Diesel first patented the device and principles for a compression engine in Germany
in 1892. The diesel engine as it is now known, is used almost exclusively as the engine of
choice in Australian Mines. The exception being some electric LHD’s and trucks. Other fuels
such as hydrogen have been investigated since the 1960’s and there is one hydrogen engine
operating in Canada today but generally speaking there use remains in the laboratories until all
the safety implications have been overcome.
Diesel engines are considered to be reliable, robust and relatively easy to maintain, and
particularly efficient at partial loads. These engines do not rely on spark plugs for fuel ignition
but rely on the in-cylinder temperature, generated on the compression stroke, for ignition of the
injected fuel. For this to occur, air is injected into the cylinder and compressed to a high
pressure with a corresponding temperature rise. As the cylinder nears the top of it stroke the
temperature of the gases rise to temperatures in excess of 540°C, well above the ignition point
for diesel fuel. A fine mist of fuel is injected into the gap where it combusts. The resulting
physical and chemical processes lead to auto-ignition just prior to the cylinder reaching top dead
centre, the combustion energy released forces the piston down to bottom dead centre in the
power stroke.
Engines used in mining can be broadly classified as, either direct injection (DI), indirect injection
(IDI or otherwise referred to as pre combustion PC) they can be naturally aspirated or turbo
charged for increased power and performance. PC engines are considered to be more
acceptable for use in underground workings because they generally have lower gas emissions.
Electronic fuel management systems are progressively being introduced more extensively into
underground hard rock mines, but it is worth a note that electronic fuel management systems
have yet to be introduced into underground coal mines and this will only occur once they are
able to provide “intrinsically safe” mechanisms. Although this fuel management system has
helped reduce the more visible DPM the side effect has been the increase (albeit minor) in
levels of oxides of nitrogen (NOX).
2.5.1 Diesel Exhaust Emissions (DEE)
The production and concentrations of gases in DEE is dependent upon
• Engine type and manufacturer
• Engine speed
• Engine adjustment and maintenance
• Working load of the engine
• Type of fuel
Because of these variables it is extremely difficult to provide absolute values of the quantity and
concentrations of DEE gases.
Diesel fuels consist primarily of Carbon (84.5%) and Hydrogen (15%) with a small amount of
Sulphur (0.5%). Typically diesel fuels contain two hydrogen atoms for each carbon atom and
can therefore be represented as C12H24. Complete combustion of 1kg of diesel fuel would result
in Carbon Dioxide (CO2) water vapour (H2O) and Sulphur Dioxide (SO2) in the following
proportions
= 3.10kg
12
44
CO2 = 0.845 ×
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= 1.35kg
2
18
H2O = 0.150 ×
0.01kg
32
64
SO2 = 0.005 × =
A total of 4.46kg of gas.
For complete combustion the amount of Oxygen required is 4.46 – 1.00 = 3.46kg and as
standard air contains 23.15% Oxygen, by mass, the air required for complete combustion is
= 14.95kg
0.2315
3.46 . In other words an air to fuel ratio of 1:14.95, or alternatively each 0.0669kg of
fuel requires 1.0kg of air for complete combustion. Fuel to air ratios in normal conditions are
0.01kg at idle and 0.05kg at full throttle up to a maximum of 0.06kg. At full throttle full load this
could be as high as 0.08kg but at this ratio there is a large amount of unburnt fuel emitted.
When averaged over a full shift the fuel to air ratio would typically be between 22:1 (0.045kg)
and 25:1 (0.04kg fuel)
In reality diesel engines never operate at 100% efficiency, seldom at full load and consequently
complete combustion is never achieved. The products of this incomplete combustion are
Carbon Monoxide (CO), Hydrocarbons (including aldehydes) – (HC), Carbon (soot defined as
diesel particulate matter) – (DPM), Oxides of Nitrogen (NOX) including Nitric Oxide (NO) and
Nitrogen dioxide (NO2). Other toxic substances like polyaromatic hydrocarbons (PAH) are also
found in both the HC and DPM component of the DEE. The concentrations of gasses in DEE
are directly related to the quantity of fuel used.
Fuel consumption is related to work load on the engine (i.e. the higher the work load the greater
the fuel consumption). Maximum fuel consumption is achieved when the engine is operating at
“torque stall”, therefore maximum concentrations and emission levels of DEE gases. In the
case of mining equipment this would be when a LHD unit is bogging and when a truck is hauling
fully loaded up an incline. Ideally exhaust gas emission rates should be obtained for all normal
operating situations. For example a LHD cycle would include loading, hauling full, dumping full,
hauling empty and Idle.
Although the liquid fuel droplets in the injected mixture begin to decompose on ignition the fuel
rich zones associated with them result in incomplete combustion and the formation of CO rather
than CO2 and as the load increases the amount of excess Oxygen decreases and tends to
increase the formation of CO. Because there is a large amount of Nitrogen in the injected air
used for combustion this along with the pressure and temperature of combustion some of the
nitrogen is oxidised to NO and this then oxidises at a much slower rate to NO2 to the extent that
only 10% is oxidised by the time it is exhausted from the engine.
2.5.2 Diesel Particulates
Diesel particulate matter (DPM) is the soot
particles emitted with diesel engine exhaust.
The size of the particles is almost totally within
the respirable range. The particles contain
hundreds of adsorbed compounds, some of
which are known to be carcinogenic. DPM is
thought to be a “potential carcinogen” but no
unequivocal evidence on this matter is currently
available and the subject is as a result, still
somewhat controversial.
Metals
Carbon
Sulphate
+ Water
Polyaromatic
Hydrocarbons
(PAH)
Hydrocarbons
(HC)
COMPOSITION OF DPM
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DPM is a complex substance and has been to object of constant investigation for a number of
years. More particularly since the formation of a task force initiated by Coal Services Pty Ltd
(formally the Joint Coal Board) in 1997 to determine the management and control of DPM in
Coal mines. As yet the hard rock industry has not embraced these studies but will be keeping a
watchful eye on proceedings since the introduction of control legislation in the USA in 2001.
The American Conference on Governmental Industrial Hygienists (ACGIH) has proposed a
general exposure standard for Diesel Particulates (DP) of 0.15 mg/m3, but has failed so far to
achieve worldwide recognition. Whilst there are currently no diesel particulate exposure
standards for mines in Australia, if we were to follow the USA lead then an exposure standard of
0.4mg/m3 (eight-hour exposure) will be introduced some time in the future. The USA control
legislation is pointed at manufacturers and the reduction of DPM in the exhaust emissions. and
a standard along these lines was introduced for the USA mining industry in 2001.
It is understood that establishing a diesel particulate exposure standard for underground mines
in Australia is not on the short-term agenda. Adoption of the ALARA (As Low As Reasonably
Achievable) principle to workforce diesel particulate exposures would however be a pragmatic
and recommended response to the issue.
In recent years design strategies have concentrated on the achieving more complete
combustion and minimising the formation of in-cylinder particulate matter. The formation of NOX
is the sole function of the available oxygen and the temperature, and the higher in-cylinder
temperatures have subsequently increased the amounts of nitrogen that are oxidised to Nitric
Oxide (NO), some of which is re-oxidised to Nitrogen Dioxide (NO2).
Hence a dilemma, decreased DPM but at the expense of increased NOX emissions.
The 1990’s has seen some major advances in engine technology to reduce DEE.
• Fuel injector design have allowed manufacturers to control the rate of fuel injection resulting
in lower emissions of DPM and NOX
• Fuel injector pressure has been increased resulting in better atomisation of the fuel in the
combustion chamber resulting in decreased DPM
• Turbo-charging has resulted in better combustion and decreased DPM whilst cooling the
compressed air supplied to the intake manifold reduces the NOX that would result from the
increased combustion temperatures.
• Improved intake manifold and port configurations have achieved better in-cylinder air
distribution eliminating fuel rich spots and inturn decreasing DPM and hydrocarbons
emissions.
• Combustion chambers have been redesigned to achieve better mixing resulting in improved
combustion and decreased DPM and hydrocarbons
• Oil control has been improved significantly as prior to 1990 as much as 30% of DPM was
attributed to lubricating oil. This improvement has reduced DPM by as much as 10%.
A number of “exhaust conditioners” have been developed over the years. These conditioners
are extremely expensive and require a high level of maintenance. Some, such as the CO
catalytic converters have been highly successful whilst others are still in development stage.
The obvious problem is to get a single conditioner for all contaminants as experience has shown
that one problem is solved and another created.
Available methods for reducing underground DPM include:
• Use more efficient diesel engines (eg engines with electronic engine fuel-air mixture control
systems)
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• Pay close attention to engine maintenance (tuning, air cleaner maintenance etc)
• Use low Sulphur fuel (<0.05% Sulphur). This type of fuel is now becoming more readily
available in Australia.
• Use exhaust particulate filters especially on production equipment where high exhaust
temperatures can be maintained for long enough to burn trapped particulates off catalysed
ceramic filters. Up to 85% of particulates can be captured using particulate filters.
A recommended reference is – “Diesel Emissions in Underground Mines – Management and
Control” NSW Minerals Council, October 1999. It states, “More research into the effects of
ventilation and diesel particulate exposure is required before comprehensive guidance can be
provided on this issue” one can’t help but agree.
2.5.3 Dilution of Diesel Exhaust Emissions (DEE)
The concentrations of contaminants in the ambient air are directly related to the total quantity of
gases and particulates emitted from the diesel engine and the mine dilution ventilating air. If the
engine operating mode, emission characteristics, and the engine exhaust air volume are known,
the quantity of emissions from the engine can be calculated.
Studies have shown that diesel engines in operating mines have a relatively low effective load
factor, some times referred to as the duty cycle, and could be as little as 20% to 40% of full load
over the entire shift.
One of the questions asked of ventilation engineers is the quantity of ventilating airflow
necessary to dilute these contaminants below statutory exposure limits.
Most Australian state legislation provides some guideline for this dilution rate and in Western
Australian legislation this is very specific. It is good management to check this legislation for the
state in which the mine is located.
One equation used to calculate this dilution factor is written as
g
C C
g Q
A N
Q
Q −

= Equation 2-4 Dilution Equation
Where
Q = Quantity of airflow required for dilution (m3/s)
Qg = Exhaust gas Emission Rate (EGER) (m3/s per kW)
AC = Maximum allowable gas concentration in the workplace. (m3/s)
NC = the normal concentration of gas present in the diluting air (m3/s)
Notes: m3/s = ppm / 1,000,000
EGER = Measured concentration x volume of exhaust gasses.
Based on the NOHSC 8 hour exposure standard of 30ppm for CO, a text book exhaust
emission rate of 0.0015 m3/s per kW of installed power and an undiluted concentration of CO of
1500 ppm in the exhaust (i.e. the maximum allowable limit) the dilution airflow required for DEE
is 0.075 m3/s per kW.
– 0.0015 0.0015
0.00003 – 0.0
0.0015 0.0015
Q ×
×
=
= 0.07499m3/s
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The US Bureau of Mines (USBM) recommends the use of the following simplified formula to
calculate the dilution factor for DEE.
Y
Q = VC Equation 2-5 DEE Dilution (USBM)
Where
Q = Quantity of airflow required for dilution (m3/s)
V = Volume of gas produced (m3/s)
C = Concentration of gas in the exhaust (ppm)
Y = Maximum allowable concentration in the workplace (ppm)
i.e.
30
0.0015 1500
Q
×
=
= 0.07500 m3/s per kW
Similarly for NOX. TLV = 25ppm eight-hour exposure
25
0.0015 1000
Q
×
=
= 0.0600 m3/s per kW
From this calculation the minimum dilution airflow should be 0.075m3/s per kW of installed
diesel power.
The use of these equations is fine but only IF the emission rate is known. We can easily
determine, by measurement the concentration of the contaminants whereas the rate of emission
can be calculated from manufacturers data.
Emission rate is a function of the number and dimensions of the cylinders and the speed of the
motor. Knowing these the rate at which the gasses are forced from the engine enables the
correct calculation of the rate of emission, obviously, the greater the speed of the engine the
greater the rate of emission.
For example consider the engine data provide by a engine manufacturer shown below.
Engine: Inline 6 cylinder 4 stroke
Power Rating: 475 HP (354kW) @ 2100 RPM
Max Torque: 1550 ft.lb (2101 Nm) @ 1200 RPM
Rated Speed Max Torque Speed
Exhaust Temperature: 432 degC 518 degC
Stroke: 139mm
Bore: 130mm
From this data we can estimate the exhaust flow (emission) rate for the engine
Calculation of exhaust swept volume
106
n
S
2
2
b
D × × × = π 




Equation 2-6 Engine Displacement Volume
Where: D = Engine Displacement Volume (litres)
b = Bore diameter (mm)
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π = 3.1428
S = Stroke length (mm)
n = Number of Cylinders
Therefore 106
6
139
2
2
130
D × × × = π 




= 11.1(litres)
2000
SEV D s
×
= Equation 2-7 Swept Exhaust Volume (Four stroke Engine)
Where: SEV = Swept Exhaust Volume (m3/min)
D = Engine displacement (litres)
s = Engine speed (rpm)
2000 = Four Stroke engine factor
1000
SEV D s
×
= Equation 2-8 Swept Exhaust Volume (Two stroke Engine)
Where: 1000 = Two Stroke engine factor
The engine data provided is a four-stroke therefore
at the rated speed
2000
Swept Exhaust Volume 11.1 2100 ×
=
= 11.7 m3/min
and at max torque speed
2000
Swept Exhaust Volume 11.1 1200 ×
=
= 6.7 m3/min
Correcting for temperature and assuming pressure (P) is constant
T2
P2V2
T1
P1V1
= (Universal Gas Law)
Where: T = temperature in °K
V = Swept Volume (m3)
at the rated speed
432 273
V2
25 273
11.7
+
=
+
= 27.7 m3
and similarly for maximum torque = 17.8 m3
A further correction is then made for turbo charging and after cooling. Typically this 1.5 but will
vary from engine to engine.
Correcting for Turbo charging the emission rates now become 41.5 m3/min at the rated speed
(i.e. 27.7 x 1.5 = 41.5) and, 26.7 m3/min at maximum torque (i.e. 17.8 x 1.5 = 26.7).
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“I can remember when the air was clean and sex was dirty.”
George Burns
3 PROPERTIES OF AIR
Air is a mixture of a number gasses. Air is a perfectly mixed (homogeneous) mixture of gasses,
in spite of the fact that the various component gasses have different densities.
Composition of Air
Oxygen 20.93%
Nitrogen 78.11%
Carbon Dioxide 0.03%
Inert Gases 0.93%
Water Vapour Variable
Note that layering of gasses (e.g. methane, which is lighter than air and Carbon Dioxide, which
is heavier than air) sometimes occurs in mine airways, where the airflow is very low. If these
gasses seep slowly into an area of very low airflow (laminar flow conditions – <<1,000m below surface) mines. In
these deeper mines, it is therefore often more convenient to deal with air flow rates in terms of
mass flow (i.e. kg/s). For the majority of mines however, the change of volume due to
compression of the air is small enough to be ignored.
3.6 Moisture in Mine Air
Up to now, we have assumed “dry” air for all of our calculations. This is a simplistic assumption
since in mine airways, the air invariably also contains some water vapour.
The amount of water vapour contained in a volume of air is dependent on the temperature of the
air. At higher temperatures, the amount of water vapour that the air can contain increases.
When air contains the maximum possible amount of vapour at a given temperature, it is said to
be saturated. Saturated air has a relative humidity of 100%.
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If air containing a certain amount of water vapour is cooled, a temperature will be reached
where the air is unable to contain all of the water vapour. The vapour will condense out into
water droplets. This temperature is known as the dew point temperature.
Condensation of water vapour can cause “fogging” and can sometimes be a problem in mine
airways where warm, humid air is cooled below its dew point (e.g. by a mine air cooler).
Condensation of water sometimes also occurs in upcast ventilation shafts – in this case the
cooling of the mine air is caused by the reduction in air pressure as the air travels upwards (the
opposite of autocompression). The condensed water droplets can cause problems with surface
fan operation.
3.7 Density of Humid Air
The “Standard” density of dry air at 20°C and 101,325 Pa barometric pressure is 1.2 kg/m3
Unless otherwise stated, the performance figures supplied with mine fans assume that the fans
will operate at standard air density.
Unfortunately, many mine fans operate in locations where the fan inlet air density is significantly
different to standard air density. It is therefore necessary to understand how to calculate the
actual air density against which the fans operate. In many situations, Equation 3.4 is sufficiently
accurate, however where improved accuracy is important, it is necessary to employ a
calculation method that takes into account the effect of the moisture contained within the mine
air.
The density of air is the mass per unit volume and varies according to the water vapour
contained in the air and three input values are required to determine the density of air which
contains water vapour. They are:
• Barometric Pressure
• Dry Bulb Temperature
• Wet Bulb Temperature.
The air density can be calculated using the relationship described as
0.287045(273.15 td )
P 0.378e
+

ρ = , (kg of air/m3) Equation 3-5 Density of Humid Air
Where
ρ = Air Density (kg/m3) (Some texts describe this as vair rather than air, this is
to distinguish between dry air and air containing water vapour.)
P = Atmospheric Pressure (kPa)
tw = Wet-bulb temperature(°C)
td = Dry-bulb temperature (°C)
e = Partial Pressure of water vapour (kPa)
( ) ( )
d w
sw d w d w
371.4 0.04t 0.4t
e 371.4 0.24t 0.6t 0.24 t t P
e
+ −
+ − − −
= Equation 3-6 Partial Pressure of Water Vapour
esw = Saturated vapour pressure at the wet-bulb temperature (kPa)
= +
w
w
sw 237.3 t
17.27t
e exp. Equation 3-7 Saturated Vapour Pressure at the Wet-bulb Temperature
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EXAMPLE
What is the density of the air in a drive where the wet and dry bulb temperatures are
(respectively) 25 and 30°C and the barometric pressure is 97.5 kPa.
We will solve this problem using both the approximate method (Equation 3.4) and the more
accurate method (Equation 3.5).
Approximate Method:
Rearranging Equation 3.4…
RT
ρ = P and substituting…
kg/m3
0.2871 (273 30)
97.5
× +
ρ =
ρ = 1.121 kg/m3
In fact this is the density of dry air.
More Accurate Method:
Calculate the Saturated vapour pressure at the wet-bulb temperature
+
×
= 237.3 25
17.27 25
esw exp.
= 3.166 kPa
Calculate the Partial Pressure of water Vapour
( ( ) ( )) ( )
371.4 (0.04 30) (0.4 25)
e 3.166 371.4 0.24 30 0.6 25 0.24 30 25 97.5
+ × − ×
+ × − × − −
=
= 2.852 kPa
Use Equation 3.5 to calculate the air density (ρ):
( )
0.287045(273.15 30)
97.5 0.378 2.852
+
− ×
ρ =
ρ = 1.108 kg/m3
3.8 Air Pressure
Every person who works in an underground mine knows that the fan produces the airflow and
generally accept this fact without needing (or wanting) to know any more. The operation, use
and placement of fans will be discussed later in these notes. At this point we need to
understand that fans create a pressure differential that causes the air to flow through the
workings.
To assist with understanding consider the following, when we pump up a tyre the pressure in
the tyre becomes higher than the pressure of the atmosphere. Similarly if we were to suck the
air from a vessel (create a vacuum) the pressure of the air out side the vessel would be greater
than the pressure of the air inside the vessel. When the tyre valve is opened the air flows from
the tyre and if the vessel is opened the air will flow into the vessel. In both cases the air will
flow from the high-pressure area to the low-pressure area until the pressure inside the
vessel is the same as the pressure outside the vessel. When this happens the flow will stop.
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This is also seen on a weather map where the wind blows from the HIGH pressure to the LOW
pressure. The greater the pressure difference the faster the wind blows.
In ventilation (mines, buildings or otherwise) for air to flow from one point to another there must
be a difference in pressure between the two points. This difference in pressure is known as the
ventilating pressure and the following rules apply.
(1) Air will always flow from a high pressure point to a low pressure point and as long
as this pressure is maintained, air will continue to flow,
(2) The larger the pressure difference between these two points the greater the
quantity of air will flow. This assumes the resistance between the two points
remains unchanged.
(3) Resistance to pressure reduces the ventilating pressure, ie the pressure is used
up overcoming resistance to the airflow.
(4) If the pressure difference between two points remains the same and the
resistance to airflow between these points is increased the air quantity will
decrease
Pressure is the force applied per unit area and in ventilating terms is usually expressed in
Pascals.
One Pascal (Pa) = One Newton per square metre (N/m2)
There are may other expressions for pressure including inches of water gauge, millimetres of
mercury all of which are required to be converted to Pascals for use in mine ventilation
calculations. Of importance in to mine ventilation practitioners is the different types of pressure.
3.8.1 Atmospheric Pressure
The earth is contained in an atmosphere and
by the force of gravity, it exerts a pressure on
the surface of the earth caused by the weight
of the air above.
Atmospheric pressure at sea level is
equivalent to a mass of 10,000 kg on one
square meter (m2) of surface and is
expressed in the following ways:
• 1 Atmosphere,
• 1.013 Bar,
• 101.325 kPa,
• 1013 mb,
The pressure shown on weather chats is
usually expressed in millibars (mb) and often
referred to as the Barometric Pressure.
3.8.2 Barometric Pressure
Barometric pressure is simply a description of the instrument used to measure the pressure of
the atmosphere in which it is situated. In mine workings the pressure measured using a
barometer includes pressure changes caused by fans and therefore is not always be a true
indication of the atmospheric pressure outside the mine.
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3.8.3 Static Pressure (SP)
Static pressure is the potential or stored
energy of the air often called the “bursting”
pressure and this term does in fact make it
easier to visualise, because static pressure
is the pressure exerted by the air on the
walls of the containing vessel. For example
when the tyre valve is opened the potential
(static) pressure is then converted to kinetic
energy as it flows from the tyre. In the case of a
ventilation duct it is the pressure holding the duct
open. It is called static pressure because it is the
same as the pressure of the air, which would
exist if the air were not moving.
In ventilation ducts the static pressure is
measured with a side tube, normal to the
direction of airflow.
3.8.4 Velocity (Dynamic) Pressure (VP)
Moving air possess “kinetic energy” which is the energy associated with motion. In mine
ventilation this is termed “Velocity Pressure”. The faster the air moves the greater velocity
pressure will be and vice versa. Velocity pressure cannot be measured directly, however it can
be measured indirectly using pitot tube.
Velocity pressure is calculated from
2
VP v
ρ 2
= Equation 3-8 Velocity Pressure
Where
VP = Velocity Pressure (Pa)
ρ = Density of air (kg/m3)
v = Velocity (m/s)
3.8.5 Total Pressure (TP)
Total Pressure is the algebraic sum of the static
pressure (potential energy) and the velocity
pressure (kinetic energy) and is measured with a
facing tube, parallel to the direction of flow.
Direction of Airflow
To gauge
To gauge
Direction of Airflow
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TP= SP + VP Equation 3-9 Ventilating Pressure
Where
TP = Total Pressure (Pa)
SP = Static Pressure (Pa)
VP = Velocity Pressure (Pa)
3.8.6 Measuring Pressure in a Duct
The static, velocity and total pressure may be obtained for airflow in a duct using a tube and a
manometer.
3.8.7 Natural Ventilating Pressure
Airflow through a mine will occur
whenever there is a pressure
difference between the intake and
exhaust. Natural Ventilation Pressure
(NVP) is formed when there is a
difference in air temperature and
density between a vertical column of
air inside a mine and the
corresponding column of air outside
the mine. When these columns are
connected airflow will occur.
This is often referred to as the
chimney effect where warm air rises
up the opening displacing the colder
air above and drawing more air into
the bottom. This method for
ventilating underground was
Direction of Airflow
Total Pressure
Static Pressure
Velocity
Pressure
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C = 1.3 kg/m3 B = 1.2 kg/m3
A = 1.09 kg/m3 D = 1.15 kg/m3
600m
600m
practiced by the Greeks in 600BC when fires were lit at the bottom of shafts to create a draught
of air.
In mines this occurs when there is a difference in temperature between the workings and the
surface. Variations of temperatures will occur form day to night, summer to winter and may in
fact flow in opposite directions. For example the heat of summer may cause the air to flow into
the workings, whereas in the colder winter months the flow will be from the mine workings and
will have an impact on the operating point of the primary ventilating fans. Although this impact
on mine fan performance is limited, the potential NVP needs to be quantified.
Air will flow from the column with the lowest average temperature to the column with the highest
average temperature.
The variance of NVP is dependent upon a number of variables including
• elevation
• depth from surface
• geothermal gradient
• local climatic conditions
In its simplest form NVP can be estimated by assuming a closed circuit (intake ⇒ workings ⇒
exhaust).
NVP = ρgh Equation 3-10 Natural Ventilating Pressure
Where
NVP = Pascals
g = Acceleration due to Gravity (9.81m/s2)
h = Difference in height between start and finish of each
branch (m)
ρ = Mean True Density in each branch (kg/m3)
Example
From the data shown in the
schematic, determine the
mine Natural Ventilating
Pressure in (Pa)
Calculate the mean density in each branch
=
+
=
2
AB 1.09 1.12 1.105 (kg/m3)
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=
+
=
2
BC 1.12 1.13 1.125 (kg/m3)
=
+
=
2
CD 1.13 1.15 1.14 (kg/m3)
Calculate NVP in each branch
NVPAB = 9.81× 600 ×1.105 = 6504(Pa)
NVPCD = 9.81× 600 ×1.140 = 6710(Pa)
Total Mine NVP
= NVPAB − NVPBC − NVPCD
= 6504 – 6710
= -206(Pa)
For mines with surface intake and exhaust point at the same level, this equation can be
simplified to:
( ) NVP = gh ρMI − ρME
Where
ρMI = Mean Density of the Intake (kg/m3)
ρME = Mean Density of the Exhaust (kg/m3)
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3.9 Assignments
3.9.1
[Q]
Calculate the velocity pressure for air quantities of 10 m3/s, 20 m3/s & 30 m3/s in
ducts with diameters of 1,000 mm 1,200 mm & 1,400 mm. Assume the density of
air is 1.2 kg/m3
3.9.1
[A] Equations used: 1)…. (Pa)
2
ρv2
VP = 2)….. (m/s)
A
v = Q 3) (m )
4
πD
A 2
2
=
Density (kg/m3) 1.2
Diameter (m) 1.0 1.2 1.4
Area (m2) 0.785 1.131 1.539
Q = 10 (m3/s) v (m/s) 12.7 8.8 6.5
VP (Pa) 97 47 25
Q = 20 (m3/s) v (m/s) 25.5 17.7 13.0
VP (Pa) 389 188 101
Q = 30 (m3/s) v (m/s) 38.2 26.5 19.5
VP (Pa) 875 422 228
3.9.2
[Q]
Using a pitot static tube the following pressure readings were recorded in a flexible
duct 1,200 mm diameter.
Measuring Point 1 2 3 4 5 6 7 8
Facing tube 1825 1845 1870 1920 1880 1850 1845 1835
Side tube 1490 1545 1555 1560 1560 1555 1560 1550
If the density of the air was 1.26kg/m3 calculate
1. the average static pressure
2. the average velocity pressure
the quantity of air flowing in the duct
3.9.2
[A] Equations used: 1)…. (m )
4
πD
A 2
2
= 2)….VP=TP-SP 3)…. ρ
2VP
v = 4)….Q=VA
Area of the duct 1.132 (m2)
Measuring Point 1 2 3 4 5 6 7 8
Facing tube 1825 1845 1870 1920 1880 1850 1845 1835
Side tube 1490 1545 1555 1560 1560 1555 1560 1550
VP (Pa) 335 300 315 360 320 295 285 285
v (m/s) 23.1 21.8 22.4 23.9 22.5 21.6 21.3 21.3
Average Static Pressure 312 (Pa)
Average Velocity 22.2 (m/s)
Quantity of air 25.1 (m3/s)
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“Ventilation is a science, albeit not always a precise science, because in practice there are usually a
number of factors involved which cannot be accurately evaluated. A successful mine ventilation
engineer uses about fifty percent common sense in solving the problems he is faced with. In
addition he uses about forty percent basic knowledge and perhaps ten percent specialised
knowledge”
W. L. Le Roux
Mine Ventilation for Beginners 3rd Edition
4 FUNDAMENTALS OF AIRFLOW
The quantity of air that will flow through a “system” (duct, or mine workings) is dependent upon:
• the difference in pressure at the start of the system and the end of the system, and
• the size of the opening.
Again consider the tyre. With the valve opened the air rushes out until the pressure stops
(equalises). The more the valve is opened the faster the air escapes even though the pressure
inside the tyre was the same to start with. However there are other factors that cause more or
less air to flow. These being:
• the roughness of the wall and,
• the severity and number of times the air has to change direction.
Once something begins to move and the energy remains constant it continues to move at the
same velocity. However if the surface roughness changes it becomes difficult to continue at the
same velocity and more energy is required to maintain the velocity. The losses in energy due to
the roughness of the wall are termed “frictional pressure losses”.
When something is moving in a straight line at a constant velocity, energy is required to change
the velocity or direction. These changes occur whenever the airway in which it is flowing
changes direction, shape or size. The losses in energy due to a change in direction of the
airflow are termed “shock pressure losses”.
In a mine there is a constant conversion of energy from potential (static pressure) to kinetic
(velocity pressure) and the energy source must be maintained to ensure that the air continues
to flow. If the energy source is removed the flow will stop.
Air as we know contains water (humidity) in various
quantities. Although not strictly a fluid, there are
some similarities. Lets now consider air as a fluid
described as a substance, which deforms
continuously when subjected to shear stress.
To explain, assume the fluid consists of layers
parallel to each other and a force is applied, in a
direction parallel to its plane. This force divided by
the area of the layer is called shear stress and as
long as it is applied, the fluid will flow relative to the
other layers.
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4.1.1 Laminar Flow
At low fluid velocities the streamlines of flow are almost parallel to each other. The shear
resistance between these streamlines is caused by friction between the layers moving at
different velocities. If there is no resistance to movement, the fluid is called frictionless or ideal.
If the shear movement is resisted then the fluid is called real. In reality ideal fluids do not exist,
however in some cases the resistance is small enough to be insignificant.
4.1.2 Turbulent Flow
As the velocity is increased, the streamlines
become randomly arranged and the flow
becomes turbulent with the additional eddy
currents adding to the shear resistance.
4.2 Airflow Equation
Investigations undertaken in the 1800’s recognised that if there is no pressure difference
between the start and end of an airway, there will be no flow of air; i.e. if the pressure was equal
to zero, then the quantity of air flowing was also equal to zero. They also recognised that as the
pressure was increased then the airflow quantity also increased i.e. the pressure is proportional
to the quantity of airflow. This relationship was termed the resistance of the system and was
expressed as
Q
R = P
Where
R = Resistance (Ns2/m8)
P = Pressure (Pa)
Q = Quantity of airflow (m3)
This held true whilst the airflow was laminar however, it did not hold true for turbulent airflow
and it was noted that in fully turbulent flows that to double the quantity of air flowing then a
pressure four times the original was required. Similarly, to increase the quantity three-fold, the
pressure required was nine times the original pressure. In other words, the pressure required
increases as the square of the quantity11 (i.e. Pressure is proportional to the quantity squared).
This relationship has since been called the Ventilation Equation.
P = RQ2 Equation 11 Ventilation Equation
There can be speculation that, when air flows at low velocity through a worked-out area, or
where low velocity leakage occurs, the flow pattern is streamlined (laminar) and therefore the
11 Numerous investigations showed that the index actually lies in the range 1.8 to 2.2 for underground mines and approaching 1 in
large voids where the airflow becomes laminar. In the vast majority of cases the use of 2 is accepted as the standard.
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formula relating P, R and Q should revert to P = RQ. Although this is true, in reality the
pressure differentials are small enough to be inconsequential and it is generally accepted in
mine ventilation that all flows are turbulent and this relationship is seldom, if ever, used.
Example
Calculate the pressure loss when 4 m3/s of air flows through a duct having a resistance of 9.3
Ns2/m8
From the ventilation equation
P = R Q2
= 9.3 x 42
= 149 Pa
Example
Calculate the pressure loss when 8 m3/s of air flows through the same duct.
From the ventilation equation
P = R Q2
= 9.3 x 82
= 595 Pa
You should note that in the second example the quantity flowing is double that flowing in the
first example and that the pressure required for this increased flow is four times that required in
the first case.
4.3 Resistance (The friction factor).
In around 1850 Atkinson hypothesised that the value of R was dependant upon on certain
characteristics of the airway or duct. For example, if one airway has a small cross sectional
area and another a large cross sectional area, and all other factors (P and Q) remain constant,
air will flow more easily through the second airway. In other words, the larger the cross
sectional area (A) of the airway, the lower the resistance (R) of the airway.
Most of the early work on fluid flow was undertaken on the assumption of circular pipes and
showed that the drag force (resistance = R) of a pipe depends on the flow velocity (v), the
density of the fluid (ρ) and the cross sectional area (A) of the pipe. We can therefore say that R
= f(v, ρ, A) where f is some function of the variables. By substituting C1 for f and completing a
dimensional analysis the equation for this relationship can be written as R = C1v2ρA and would
assume C1 as a constant. Measurements on different pipes and various velocities would show
that C1 is NOT constant and there was a fourth variable, i.e. the viscosity (μ) of the fluid and the
relationship for R is rewritten as R = f(v, ρ, A, μ). Since there are four unknowns it is not
possible to obtain a unique solution. By analysis12 and substitution the equation is written as
 

 

ρ
μ
=
ρ Dv
f
D v
R
2 2 1 or  


  
μ
ρ
=
ρ
Dv
D v
R
2 2 . The non-dimensional group  


 

ρD2v2
R is referred to as
the force or Euler coefficient, denoted by Cf and the group  


 

μ
ρDv is referred to as Reynolds
Number (Re).
12 The dimensional analysis and solution is described in “Environmental Engineering in South African Mines” Chapter 1. (pp1-27)
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However in addition to Reynolds Number the surface roughness also influences the resistance
force. The surface roughness (e) also has a relationship to the area or as above to the
diameter (D), and this is described as the ‘relative roughness’ and defined as
D
ε = e .
Further evaluation and relating to Bernoulli’s Equation13
 

 

+ ρ + ρ = + ρ + ρ 2
2
2
1 2
2
1
1 g H
2
g H P V
2
P v
it is shown that 2
2
1 v
k P
ρ
= and k is described as the friction component and expressed as
D
v L
P
2
2

λ = where P is the static pressure loss due to wall friction in a parallel duct diameter D
and length L.
It is also possible to equate this to rectangular ducts if the diameter (D) is replaced with the
equivalent hydraulic diameter Dh
C
D 4A h = Equation 12 Hydraulic Diameter
Where Dh = Hydraulic diameter (m)
A = Cross sectional Area of Airway (m2)
C = Circumference of Airway (m)
The equation for friction can now be rewritten as
 


 


 

 
ρ 
λ =
h
2
D
L
2
v
P
A more complete analysis will show that the friction coefficient is a function of Reynolds Number
(Re), relative roughness (ε) and duct shape (s) where
Reynolds Number =Re =  


 

μ
ρDv
Surface roughness =
D
ε = e and
s = the shape factor of the duct cross section.
Experiments showed that the shape factor (s) had very little influence and the non-dimensional
parameters had a major influence. After further studies and experimental work on smooth and
rough walled ducts.
The equation for the actual relationship shown in this test work can be represented by
 

 

λ
= − ε +
λ Re
1 1.74 2log 2 18.7
Equation 13 Friction Factor
Where
13 Bernoulli’s Equation : Law of Conservation of Momentum is not discussed in these notes but can be found in many texts dealing
with fluid flow.
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Dh
ε = e = Relative roughness
e = Average height of surface irregularities or surface roughness
(assume 0.35 m for rock wall, 0.02 m for raisebored)
Dh = Hydraulic Diameter (m)
And
μ
ρ
Re = Dv
ρ = Density (kg/m3)
μ = Viscosity (kg/ms)
Dh = Hydraulic Diameter (m)
V = Velocity (m/s)
In mine ventilation the term λ is usually replaced by ‘k’ known as the coefficient of friction or
Atkinson’s k Factor.
4.3.1 Atkinson’s Equation
Atkinson also noted that if air has to rub against a larger area of surface in one airway than in
another, the resistance will be higher in the airway with the largest “rubbing surface”. This
rubbing surface is found by multiplying the perimeter (C) by the length (L).
A
P kCLv
2
= Equation 14 Atkinson’s Ventilation Equation (Velocity)
Where
P = Pressure (Pa)
k = Atkinson’s Friction Factor with consideration to surface roughness.
C = Circumference of the duct (m)
L = Length of duct m)
A = Cross sectional Area of the duct (m2)
v = Velocity of the fluid (m/s)
If we substitute the physical values in the Atkinson equation
m2
Left hand side Pressure is Pa = N and the 2
2
2
2
2
s
km
m
s
k m m m
A
Right hand side kCLv =
× × ×
= .
Combining both sides 2
2
2 s
km
m
N =
therefore 4
2
m
k = Ns
Substituting the dimensions of force s2
N kg m
×
= then 3
2
2 m
kg
m4
s
s
k kg m =  


 

× 



 ×
=
This is important because it shows that Atkinson’s friction factor is not dimensionless and will
vary according to the dimensions m3
kg which is the same dimensions as density. Atkinson had
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ignored density effects deriving this equation and therefore is valid only for air with standard
density.
The use Atkinson’s equation for any density it need to have a correction applied and it becomes
S
2
A
P kCLv
ρ
ρ
= ×
Where ρ = Density if air flowing in the duct (kg/m3)
ρS = Density of standard air (1.2 kg/m3)
The term
S ρ
ρ
is in Atkinson’s formula to allow for the fact that the pressure requirements
depend on the density of the air. Obviously, more pressure will be required to pass heavier
(denser) air through a system than the lighter air. In fact, the pressure (P) required is directly
proportional to the air density (ρ).
More often than not it is convenient to express Atkinson’s equation in terms of quantity of air
rather than velocity and since V = QA this equation is rewritten as;
S
3
2
ρ
ρ
A
P = kCLQ × Equation 15 Atkinson’s Ventilation Equation (Quantity)
Where
P = Pressure (Pa)
K = Atkinson’s Friction Factor (Ns2/m4)
C = Circumference of Airway (m)
L = Length of Airway m)
A = Cross sectional Area of Airway (m2)
Q = Quantity of airflow (m3/s)
ρ = Density if air flowing in the duct (kg/m3)
ρS = Density of standard air (1.2 kg/m3)
The values of ‘k’ are determined from measurement and calculation. The values below are
sighted in texts, and generally hold up to scrutiny.
Some mines may find differing results outside of these values. In practice all mine should
measure and determine their specific factor(s) because there will always be more than one
factor that could be used.
Some Typical Values for Atkinson’s Friction Factor (k) at Standard air density
Airway Type. k factor (Ns2/m4)
Smooth pipe 0.0028
Normal rigid ducting 0.0030 to 0.0035
Flexible ducting 0.0030 to 0.0065
Concrete surfaced 0.0035 to 0.0040
Rock surfaced 0.0070 to 0.0200
Raise bored 0.0035 to 0.0050
Timbered Airway 0.0400 to 0.0600
Timbered rectangular Shaft 0.0400 to 0.1000
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Example
Calculate the pressure required to overcome friction with an airflow of 60 m3/s in a 5.O m x 4.0
m haulage drive that is 450 m long and has a friction factor (k) of 0.016 Ns2/m4. Air density in
the haulage is 1.1 kg/m3.
Given that
k = 0.016 Ns2/m4
L = 450 m
ρ = 1.1 kg/m3
Q = 60 m3/s
C = 2(5 + 4) = 18 m
A = (5.0 x 4.0) = 20.0 m2
From Atkinson’s equation
S
3
2
ρ
ρ
A
P = kCLQ ×
( )
1.2
1.1
20.0
0.16 18.0 450 60
3
2
×
× × ×
=
= 53.5 Pa
Therefore, the pressure required to overcome the friction in this drive is 54 Pa
4.4 Shock Losses
Shock losses are the change in total pressure across all airway elements such as the entrance
to a system, a bend, junction, obstruction, change in section and exit from the system. In short
shock losses are the result of flow separation that occurs whenever the airflow changes
direction.
The pressure losses resulting from a change in direction can be determined from the Shock
Loss Equation
PSHOCK = X × VP
Where
PSHOCK = Pressure loss due to shock (Pa)
X = Dimensionless shock loss factor similar to Atkinson’s
friction factor and is constant only for a given set of
conditions depending upon shape, dimensions and
characteristics. (See below)
VP = Velocity Pressure of the air (Pa)
And
2
VP v
2
= ρ ×
Where
ρ = Density of the air (kg/m3)
v = Velocity of the air (m/s)
The shock loss factor X is a function of the:
1. Configuration and flow through the element.
2. Angle of the change in direction,
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3. Degree the abruptness of the change,
4. Radius of curvature,
5. Ratio of the radius to width of the airway
6. Aspect ratio between the height and width,
7. Velocity in the airway
8. Airway roughness
9. Shape of the airway
10. Airways immediately before and immediately after the change in direction,
11. The number and type of complex elements i.e. two bends, bend followed by an
expansion
The large number of variables that contribute to shock losses causes the calculation of these
losses is extremely complicated and seldom necessary in mine ventilation because of their
small magnitude and the inaccuracy of the X value.
WARNING- If airflow and pressure surveys have been used to determine the resistance of any
part of the mine workings these measurements will include the losses due to shock and any use
of additional factors will cause undetectable errors in the results.
Shock losses are generally “over used” in computer ventilation modelling and the calculation
methods shown below are approximations of the actual number taken from published texts14
and suitable for use in most computer ventilation models. Wherever possible ventilation
practitioners should determine these values by direct measurement.
14“Fan Engineering – Eight Edition” Published by Buffalo Forge Company. Buffalo, New York USA (1983) & DALY, B.B. “Woods
Practical Guide to Fan Engineering” Published by Woods of Colchester Limited UK (1978)
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4.4.1 Entrance
Shock losses occur at the entrance to any system due to the acceleration of air from zero
velocity in the surrounding atmosphere to the velocity in the airway.
4.4.2 Outlet Losses
Air exiting a system decelerates from the airway velocity to
zero velocity in the atmosphere.
The static pressure in the issuing air stream will be equal
to the surrounding atmospheric pressure. If this is taken
as a zero gauge pressure then the total pressure at the
exit will be equal to the velocity pressure.
To reduce this loss, the area of the outlet must be
increased (ie the velocity is decreased) this is normally
achieved with the use of a diffuser or evasè.
The performance of a diffuser will vary with the inlet
velocity pressure and the ratio of the outlet to inlet area,
the centreline length and the angle of divergence. In an
ideal diffuser the regain pressure would be equal to the
change in velocity pressure and the loss in total pressure
(shock loss) would be zero. In reality this will never occur.
A well-designed diffuser would have an outlet to inlet area
ratio of 4:1 with an included angle of divergence between
8° and 11°. In practice the length required to regain
pressure is excessive and generally the divergence angle is around 25°.
As a rule of thumb if the length of the divergence is four times the diameter of the smaller
airway, then the shock losses become negligible because the pressure regain of shock loss
pressures are overcome by the frictional pressure losses.
Entrance
Notched
X = 0.05
Note that the Notched entry
approximates the Bell Mouth and
apparently the vortex formed in the
notch promotes a smooth flow into the
airway.
Bell Mouth
X = 0.05
Plain
X = 0.9
Conical
X = 0.2
Flanged
X = 0.5
Plain
Flanged
Diffused
Outlets
In all cases X = 1.0 and the velocity
pressure is determined by the velocity
in the plane of the exit area
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4.4.3 Elbows
Elbows, bends and mitres are used to
guide the change in direction of flow and
restrict flow separation. The relative
losses will depend upon the abruptness
of the change in direction, Reynolds
number (not discussed in these notes)
and the wall roughness of the airway.
Other considerations include
geometrical shape i.e. circular or round,
the angle of the bend and the radius
ratio and the aspect ratio.
Those of most importance in mines are
mitre bends with curved bends having a
lesser impact due primarily to their
lengths such that shock losses are
decreased to a point of insignificance
compared to the frictional pressure
losses.
Consider the case of a typical mine
curve of radius 20 metres with a height
of 4.5 metres and a width of 4.5 metres with a friction factor “k” of 0.015 Ns2m-4. The angle of
the curve is 90° and the airflow is 200 m3/s
The calculation for “X” in a smooth bend is described by McElroy (1935)
2
m2 a0.5 90
60 . 0 X 



 θ
×
×
=
Where
m is the radius ratio = Centre line radius
Width of the Airway
= 4.5
2
= 2.25
a is the aspect ratio = Height of the Airway
Width of the Airway
= 4.5
4.5
= 1.0
Therefore
2
2 0.5 90
90
2.25 1.0
6 . 0 X 



× 
×
=
= 1.2
The velocity is 200/(4.5 x 4.5) = 9.87 (m/s)
Therefore
PSHOCK = 1.2 x (0.5 x 1.2 x 9.872)
= 70Pa
Smooth Bend
X = 0.25 x θ2
m2 a0.5 902
X = 1.15
Rectangular Mitre Bend
X = 1.15
Circular Mitre Bend
Mitre Bend other than 900
Xθ = X90 x θ
90
θ
0.315
1.5
0.350
1.0
0.425
0.5
X 0.315 0.325 0.380
R/D 2.0 2.5 3.0
R
D
Three Piece Bend 900
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4.4.4 Expansions and Contractions
4.4.5 Junctions and Splitting
These losses depend upon
1. the geometrical aspect of each airway,
2. the branching angle,
3. the branch area ratio,
4. the ratio of the downstream to upstream velocity of air and therefore the quantity of
airflow.
Gradual Contraction
A2 A1
Note: If the length of contraction is 4 x the
smaller diameter then the shock losses are
negligible
X = 0.333 [ 1 – (A2 / A1 )]
A2
Abrupt Contraction
X = 0.5 [ 1 – (A2 / A1 )]
A1
Abrupt Expansion
A1
X = ( 1 – [A1 / A2 ])2
A2
Gradual Expansion
A2 A1
Note: Expansions are complicated by the
regain in static pressure over the length of the
change.
To keep expansion losses to a minimum the
change should take place over the longest
available distance and the downstream duct
must be at least 4 times the larger diameter to
ensure full recovery of the pressure.
Deflected branch
X = (0.5(Q / Qb)2) + ((Q / Qb)-1)2 + Xb
Splitting
θ
Straight Branch
X = 0.5((Q / Qb)-1)2
Deflected branch
X = (-0.5 ((Q / Qb)-1)2.5) + Xb
Junction
θ
Straight Branch
X = 3.3((tan θ/2)–0.67).(((Q/Qb).((1/Cc)-1))2
θ
θ = 300 & v1 = v 2
v1
v2
Ideal Junction
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In the deflected branch of a split the pressure loss can be reduced considerably by using a
converging taper between the main and the deflected branch.
The recommended procedure for development of a junction is to join the upstream main branch
with the down stream main by means of a tapered convergence of at least two upstream
diameters long made to join the branch to the taper at an angle of 30° with the upstream main
branch. The idea is to maintain a constant velocity in the main branch. Obviously this is fine for
a duct system in a building but extremely difficult to achieve economically in an underground
mine.
One important aspect of junctions is that the pressure loss may in fact be negative depending of
the ratio of the branch flow to the total flow. This is due to the transfer of momentum from one
branch to the other.
In the equations for splits and junctions the following applies:
Q = Total Airflow
Qb = Airflow in the branch being evaluated
Xb = X for the deflected branch
Cc = Coefficient of Contraction
Ao
Ac
=
And Ac = Area of vena contractor
Ao = Area of orifice
4.5 Other Methods of Expressing the Shock Factor X
Some attempt to increase the friction “k” as a method to allow for shock losses. This involves
the addition of an increment to the friction factor “k” for each and every airway. The method is
tedious and prone to error due to the many variables and is not really recommended for use.
This method is inadvertently applied when the results of pressure quantity surveys are used to
“back calculate” the friction factor “k” because the pressure losses due to shock are included in
the measurements.
Use of the equivalent-length method is probably the most useful method for mine ventilation
because it can be calculated by equating the formulas for friction loss and shock loss.
To differentiate between the actual length (L) the subscript e is added to indicate equivalent
length
6.67k
Le X Dh
×
= where Dh is the hydraulic diameter of the airway.
C
Dh = 4A .
WARNING – Mine ventilation computer programmes may include preset values for shock
losses. These values are expressed as an equivalent length (Le) of the airway. It is then used
to calculate a resistance for the Le and is then added to the resistance of the airway. Some of
these values have been taken from a text that used drive sizes of only 2.0m x 2.0m and
therefore are not applicable to all mines.
“Shock losses do not lend themselves to precise calculation because of the great range of variability in
occurrence and because of a lack of understanding of their very nature.” 15
15 HARTMAN, H.L. “Mine Ventilation and Air-conditioning”
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4.6 Series Circuits
When one airway is followed by another airway and both have their own physical
characteristics, the airways are said to be in series.
This becomes obvious when we refer to the sketch below where the total quantity of air flowing
through the system is equal to the quantity of air in ‘A’, which is equal to the quantity of air
flowing in ‘B’ and likewise in airway ‘C’. i.e. QT = QA = QB = QC
The total pressure loss in the system is the sum of the pressure losses in airway ‘A’, airway ‘B’
and airway ‘C’. i.e. PT = PA + PB + PC
It then follows that the total resistance of the system is the sum of the resistance of airway ‘A’
and airway ‘B’ and airway ‘C’. i.e. RT = RA + RB + RC
This relationship is usually expressed as
RT = R1 +R2 + ….Rn
Where R = Resistance (Ns2/m8)
Example
Referring to the sketch above Airway A has a resistance of 0.10 (Ns2/m8), Airway B a resistance
of 0.07 (Ns2/m8) and Airway C a resistance of 0.17 (Ns2/m8). Calculate the pressure (P)
required to cause a flow of 100 m3/s
RT = R1 + R2
= 0.10 + 0.07 + 0.17
= 0.34 (Ns2/m8)
and
PT = RT x QT
2
= 0.34 x (100)2
= 3400 (Pa)
Many Australian mines utilise
“series ventilation” circuits in
their design. This is a term used
to describe a combination of
airways with different
resistances where the air flows
through each in turn. This is a
common practice in steeply
dipping narrow vein mines,
where the air flows from the
surface to the lowest working
Airway A Airway B Airway C
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level before being exhausted from the mine. This “reuse” of air is often confused when people
talk of recirculation of air when in fact they mean re-used (series) air. Recirculate implies that
the airs goes around in circles and never leaves the circuit.
4.7 Parallel Airway Circuits
Airways are in parallel when they have a common inlet and common outlet.
In the sketch to the right the points X and Y are common to both airways. Therefore the
pressure loss across airway A is equal to the pressure loss across airway B and equal to the
pressure loss across the system X-Y
i.e. PT = PA = PB
X Airway B Y
Airway A
Common points
The total quantity flowing in the system is equal to the quantity of air flowing in airway A and
airway B. i.e. QT = QA + QB and the overall resistance (RT) is given by the formula
T 1 2 Rn
1
….
R
1
R
1
R
1 = + +
Where R = Resistance (Ns2/m8)
Parallel systems exist in mines when air is exhausted on a number of levels.
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EXAMPLE
Three identical airways are developed in parallel. Each airway has a resistance of 0.15
(Ns2/m8). Calculate the pressure required to cause a flow of 50 m3/s through each airway.
Calculate the total airflow
QT = QA + QB + QC
QT = 50 + 50 + 50
= 150 m3/s.
Calculate the overall resistance (RT)
T A B RC
1
R
1
R
1
R
1 = + +
0.15
1
0.15
1 1
R
1
T 0.15
= + +
= 0.0167 (Ns2/m4)
Calculate the pressure
PT = RT x QT
2
= 0.0167 x 1502
= 375 Pa
4.8 Complex Circuit (Networks)
With reference to the sketch for parallel airways it is noted that whilst Airway A and Airway B
are in parallel they are also in series with the airway preceding point X and the airway following
point Y.
Lets now consider the same sketch a new airway that connects Airway A with Airway B.
Airway B
Airway A
New airway
After examining this sketch you will come to the conclusion It is not possible to reduce this
circuit to a series or parallel circuit and an analytical or reiterative technique is required.
Complex circuits or networks rely on the use of computers to undertake the numerous
calculations required.
In order to understand the complexity of network problems the following definitions apply
1. A network is any complex ventilation circuit.
2. A Junction is where two or more airways join
3. A branch is any airway between two junctions
4. A mesh is any closed path that traverses the network. Every airway must be included
once in any group of meshes.
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5. Kirchoff’s electrical laws can be restated and will apply.
Kirchoff’s 1st Law
The algebraic sum of all air quantities at a junction is zero
Kirchoff’s 2nd Law
The algebraic sum of all pressure losses in a closed mesh is zero. However if a fan is included
in the mesh then the algebraic sum of all pressure losses is equal to the pressure provided by
the fan.
Comprehensive discussion of Network analysis can be found in Chapter 8 of “Environmental
Engineering in South African Mines”.
4.9 System Resistance Curve
We have seen and discussed the
relationship of pressure (P), quantity (Q)
and resistance (R) in the previous sections.
For example the ventilation equation
(P = RQ2 ) shows the relationship between
pressure (P) and quantity (Q) is
proportional i.e. R
Q
P
2 = where R was the
resistance for a particular system with a
specific pressure and quantity.
The “system”) resistance in any duct or
mine, is determined by summation of all the
appropriate resistance values. Once this
calculation is complete we can determine
the pressure necessary to cause various quantities of air to flow. . The results of these
calculations are then shown graphically to produce the “System Resistance Characteristic”. A
system with a high resistance will be represented by a curve that is steeper than a system with
a lower resistance.
High Resistance
Low Resistance
Pressure
Quantity
0
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4.10 Assignments
4.10.1
[Q]
Calculate the pressure loss in 250 m of a rigid plastic duct with diameter of 1,200
mm & 1,400 mm for air quantities of 15 m3/s & 25 m3/s. State the k factors used and
assume the density of air is 1.2 kg/m3.
4.10.1
[A] Equations used: 1)
s
3
2
ρ
ρ
×
A
kCLQ
P = 2) (m )
4
πD
A 2
2
= 3) C = πD (m)
Determine area (A) and circumference (C) for ducts
Diameter (mm) C (m) A (m3)
1,200 3.7699 1.1310
1,400 4.3982 1.5394
Calculate pressure assuming k = 0.0035
s
3
2
ρ
ρ
×
A
kCLQ
P =
1,200 Diameter
1.2
1.2
×
1.1310
0.0035×3.7699× 250×15
P = 3
2
= 513 Pa
1,200 Diameter 1.2
1.2
×
1.1310
0.0035×3.7699× 250× 25
P = 3
2
= 1425 Pa
1,400 Diameter
1.2
1.2
×
1.5394
0.0035× 4.3982× 250×15
P = 3
2
= 237 Pa
1,400 Diameter 1.2
1.2
×
1.5394
0.0035 × 4.3982× 250× 25
P = 3
2
= 659 Pa
4.10.2
[Q]
Repeat 4.10.1 but assume an air density of 1.15 kg/m3 and 1.25 kg/m3
4.10.2
[A] 1,200 Diameter
1.2
1.15
×
1.1310
0.0035×3.7699× 250×15
P = 3
2
= 492 Pa
1,200 Diameter 1.2
1.15
×
1.1310
0.0035×3.7699× 250× 25
P = 3
2
= 1365 Pa
1,400 Diameter
1.2
1.15
×
1.5394
0.0035× 4.3982× 250×15
P = 3
2
= 227 Pa
1,400 Diameter 1.2
1.15
×
1.5394
0.0035× 4.3982× 250× 25
P = 3
2
= 632 Pa
1,200 Diameter
1.2
1.25
×
1.1310
0.0035×3.7699× 250×15
P = 3
2
= 534 Pa
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1,200 Diameter 1.2
1.25
×
1.1310
0.0035×3.7699× 250× 25
P = 3
2
= 1484 Pa
1,400 Diameter
1.2
1.25
×
1.5394
0.0035× 4.3982× 250×15
P = 3
2
= 247 Pa
1,400 Diameter 1.2
1.25
×
1.5394
0.0035× 4.3982× 250× 25
P = 3
2
= 686 Pa
4.10.3
[Q] Repeat 4.10.1 using flexible ducting.
4.10.3
[A]
Calculate pressure assuming k = 0.0065
s
3
2
ρ
ρ
×
A
kCLQ
P =
1,200 Diameter
1.2
1.2
×
1.1310
0.0065×3.7699× 250×15
P = 3
2
= 953 Pa
1,200 Diameter 1.2
1.2
×
1.1310
0.0065×3.7699× 250× 25
P = 3
2
= 2647 Pa
1,400 Diameter
1.2
1.2
×
1.5394
0.0065× 4.3982× 250×15
P = 3
2
= 441 Pa
1,400 Diameter 1.2
1.2
×
1.5394
0.0065 × 4.3982× 250× 25
P = 3
2
= 1225 Pa
4.10.4
[Q]
Calculate the pressure loss for air quantities of 13.5 m3/s, 55 m3/s & 165 m3/s in a
decline 350 m long 5.5m high and 5.0m wide. State the k factor used and assume
the density of air is 1.2 kg/m3.
4.10.4
[A]
Determine area (A) and circumference (C) for drive
Height (m) Width (m) C (m) A (m3)
5.5 5.0 21.0 27.5
Calculate pressure assuming k = 0.012
s
3
2
ρ
ρ
×
A
kCLQ
P =
1.2
1.2
×
27.5
0.012× 21.0×350×13.5
P = 3
2
= 0.7 Pa
1.2
1.2
×
27.5
0.012× 21.0×350×55
P = 3
2
= 12.8 Pa
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1.2
1.2
×
27.5
0.012× 21.0×350×165
P = 3
2
= 115 Pa
4.10.5
[Q]
Calculate the velocity (m/s) and pressure loss (Pa) for air quantities of 60 m3/s, 125
m3/s, 190 m3/s & 250 m3/s in a vertical raisebored shaft 350 m long and 4.0m
diameter. State your assumptions.
4.10.5
[A]
Assumptions: K Factor = 0.0045 Ns2/m4 Density of air 1.2 kg/m3
Determine area (A) and circumference (C) for raisebored shaft
Diameter (m) C (m) A (m3)
4.0 12.5664 12.5664
Calculate velocity A
Q
v =
12.5664
60
v = = 4.8 m/s
12.5664
125
v = = 9.9 m/s
12.5664
190
v = = 15.1 m/s
12.5664
250
v = = 19.9 m/s
Calculate pressure:
s
3
2
ρ
ρ
×
A
kCLQ
P =
1.2
1.2
×
12.5664
0.0045×12.5664×350× 60
P = 3
2
= 36 Pa
1.2
1.2
×
12.5664
0.0045×12.5664×350×125
P = 3
2
= 156 Pa
1.2
1.2
×
12.5664
0.0045×12.5664×350×190
P = 3
2
= 360 Pa
1.2
1.2
×
12.5664
0.0045×12.5664×350× 250
P = 3
2
= 623 Pa
4.10.6
[Q]
Determine the pressure losses contributed by the shock in a 90° bend of a blasted
rock airway 5.0m high and 5.0m wide for air quantities of 20 m3/s, 75m3/s 125 m3/s.
Assume standard air density.
4.10.5
[A]
Shock factor X for 90° rectangular bend = 1.15
Determine area of drive = h x w = 5 x 5 = 25.0 m2
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Calculate velocity in drive: A
Q
v =
25.0
20
v = = 0.8 m/s
25.0
75
v = = 3.0 m/s
25.0
125
v = = 5.0 m/s
Calculate velocity pressure (Pa)
2
ρv2
VP =
2
1.2×0.82
VP = = 0.4 Pa
2
1.2×3.02
VP = = 5.4 Pa
2
1.2×5.02
VP = = 15.0 Pa
Calculate shock loss PSHOCK = XVP
PSHOCK = 1.15×0.4 = 0.46 Pa
PSHOCK = 1.15×5.4 = 6.2 Pa
PSHOCK = 1.15×15 = 17.25 Pa
4.10.7
[Q]
Determine the pressure losses contributed by the shock in a 90° smooth bend with a
centre line radius of 15 m in a 1.2m diameter flexible duct with an air quantity of
m3/s. Assume standard air density.
4.10.7
[A] Shock factor X for 90° circular bend = 2
2
2 0.5 90
θ
×
m a
0.25
X =
1.2
20
m = = 16.6667
1.2
1.2
a = = 1
2
2
2 0.5 90
90
×
16.6667 1
0.25
X = = 0.0009
Determine area of duct = πr 2 = π×0.62 = 1.1310 m2
Calculate velocity in drive: A
Q
v =
1.1310
30
v = = 26.5 m/s
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Calculate velocity pressure (Pa)
2
ρv2
VP =
2
1.2× 26.52
VP = = 422 Pa
Calculate shock loss PSHOCK = XVP
PSHOCK = 0.0009 × 422 = 0.38 Pa
4.10.8
[Q]
Determine the equivalent airway length of shock losses in a 90° bend in a blasted
rock airway 5.0m high and 5.0m wide for velocities of 1.0 m/s, 2.5 m/s, 5.0 m/s,
7.5m/s & 10.0 m/s. Assume standard air density
4.10.8
[A] 6.67k
😄
Le = h 6.67×0.012
1.15×5
Le = 71.8m and will be the same irrespective of flow.
4.10.9
[Q]
Determine the pressure loss when an airway reduces abruptly in size from 6.0m
high and 6.0m wide to an airway 3.0m high and 3.0m wide for air quantities of 10
m3/s, 50 m3/s & 100 m3/s. Assume standard air density.
4.10.9
[A]
Abrupt Contraction X= 0.5 (1- (A2/A1))
A1 = 6 x 6 = 36 m2 A2 = 3 x 3 = 9 m2
X = 0.5 (1 – (9/36) = 0.5 (1 – 0.25) = 0.375
Pressure loss occurs in the entry airway
Velocity in A1 = 36
10 = 0.2778 m/s and 2
1.2×0.27782
VP = = 0.0463 Pa
Therefore PSHOCK = 0.375×0.0463 Pa
Velocity in A1 = 36
50 = 1.3889 m/s and 2
1.2×1.38892
VP = = 1.1574 Pa
Therefore PSHOCK = 0.375×1.1574 = 0.4340 Pa
Velocity in A1 = 36
100 = 2.7778 m/s and 2
1.2× 2.77782
VP = = 4.6296 Pa
Therefore PSHOCK = 0.375 × 4.6296 = 1.7361 Pa
4.10.10
[Q]
Determine the pressure loss when an airway increases abruptly in size from 6.0m
high and 6.0m wide to an airway 3.0m high and 3.0m wide for air quantities of 10
m3/s, 50 m3/s & 100 m3/s. Assume standard air density.
4.10.10
[A]
Abrupt Expansion X= (1- (A1/A2))2
A1 = 3 x 3 = 9 m2 A2 = 6 x 6 = 36 m2
X = (1 – (9/36))2 = (1 – 0.25)2 = 0.5625
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Pressure loss occurs in the entry airway
Velocity in A1 = 9
10 = 1.1111 m/s and 2
1.2×1.11112
VP = = 0.7407 Pa
Therefore PSHOCK = 0.5625 ×0.7407 = 0.4167 Pa
Velocity in A1 = 9
50 = 5.5556 m/s and 2
1.2×5.55562
VP = = 18.5185 Pa
Therefore PSHOCK = 0.5625 ×18.5185 = 10.4167 Pa
Velocity in A1 = 9
100 = 11.1111 m/s and 2
1.2×11.11112
VP = = 74.0739 Pa
Therefore PSHOCK = 0.5625 ×74.0739 = 41.6666 Pa
4.10.11
[Q]
Calculate the frictional pressure loss for the following systems. State your
assumptions for Atkinsons factors. Assume standard air density.
i)
Airflow 30 m3/s
Airway A 50m long, 5.5m high x 5.0m wide
Airway B 100m long, 4.5m high x 4.0m wide
Airway C 25m long, 4.0m high x 4.0m wide
ii)
Airflow 100 m3/s
Airway A Decline 750m long, 5.5m high x 5.0m wide
Airway B Level Development 200m long, 4.5m high x 4.0m
wide
Airway C Raisebore shaft 125m long, 4.0m diameter
4.10.11
[A]
1) Assume friction factor 0.012 Ns2/m4 for rock walled air way
Determine the resistance for each airway. A3
kCL
R = Ns2/m8
Airway A Area = 27.5 m2 Circumference = 21 m and A 3 27.5
0.012× 21×50
R = = 0.0006
Ns2/m8
Airway B Area = 18 m2 Circumference = 17 m and B 3 18
0.012×17×100
R = = 0.0035
Ns2/m8
Airway C Area = 16 m2 Circumference = 16 m and A 3 16
0.012×16× 25
R = = 0.0012
A B
C
A
B
C
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Ns2/m8
RT = RA+RB+RC = 0.0006 + 0.0035 + 0.0012 = 0.0053 Ns2/m8
Calculate frictional pressure loss
P = RQ2 = 0.0053 x 302 = 5 Pa
2) Assume friction factor 0.012 Ns2/m4 for rock walled air way and 0.0045 Ns2/m4
for raise bored airway
Determine the resistance for each airway. A3
kCL
R = Ns2/m8
Airway A Area = 27.5 m2 Circumference = 21 m and
A 3 27.5
0.012× 21×750
R = = 0.0091 Ns2/m8
Airway B Area = 18 m2 Circumference = 17 m and
B 3 18
0.012×17× 200
R = = 0.0070 Ns2/m8
Airway C Area = 12.56 m2 Circumference = 12.56 m and
A 3 12.56
0.0045 ×12.56×125
R = = 0.0036 Ns2/m8
RT = RA+RB+RC = 0.0091 + 0.0070 + 0.0036 = 0.0197 Ns2/m8
Calculate frictional pressure loss
P = RQ2 = 0.0197 x 1002 = 197 Pa
4.10.12
[Q]
Two sections of a mine are depicted in the following figure
1. Determine the airflow through each section of the mine
2. Calculate the pressure drop necessary to achieve 100m3/s.
4.10.12
[A]
Section 1 is in parallel with section two and the pressure is the same across both
sections of the mine.
Calculate the resistance
T SECTION1 RSECTION2
1
+
R
1
=
R
1
100 m3/s
0.3 Ns2/m8
0.1 Ns2/m8
Section 1
Section 2
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0.1
1
+
0.3
1
=
R
1
T
= 1.8257 + 3.1623 =4.9880
RT =
2
4.9880
1
= 0.0402 Ns2/m8
As P = RQ2 and since P is constant RTQT
2 = R1Q1
2
The airflow in Section 1 0.0402 x 1002 = 0.3 x Q1
2
2 2
1 ×100
0.3
0.0402
Q = =1,3387.75
Q1 = 13387.75 = 36.6 m3/s
Similarly the airflow in Section 2 is 0.0402 x 1002 = 0.1 x Q1
2
Q2 = 63.4 m3/s.
Pressure is P = RQ2
= 0.0402 x 1002
= 402 Pa
4.10.13
[Q]
The raise bored airway for a mine has become unstable and must be sprayed with
shotcrete to prevent it collapsing. The shaft is 3.2m diameter and 200 m in length.
The concrete will be applied to a thickness of 100mm. The current airflow is 90
m3/s. What will the airflow be once the shotecreting has been completed?. Assume
the same pressure drop.
4.10.13
[A] Determine the existing resistance. A3
kCL
R = Ns2/m8 Diameter = 3.2m
8.043
0.0045 ×10.05×200
R = = 0.0174 Ns2/m8
Determine the new resistance. A3
kCL
R = Ns2/m8 Diameter = 3.2 – (2 x 0.100) = 3.0m
7.073
0.0045 ×9.42× 200
R = = 0.0240 Ns2/m8
Since pressure is constant R1Q1
2 = R2Q2
2
2 2
2 ×90
0.0240
0.0174
Q =
Q2 = 5872.5
= 76.6 m3/s
4.10.14
[Q]
The figure below represents the workings of an underground mine.
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It is proposed to develop a new shaft (FE) and increase the mine airflow from 125
m3/s up to 240 m3/s.
It is also necessary to provide equal air to each section of the mine.
What diameter should the new shaft be? Assume the new shaft can be raisebored.
4.10.14
[A] Determine the resistance of the decline. A3
kCL
R = Ns2/m8
At 1:7 the decline will be 200 x 7 = 1,400 m long Area = 25 m2 an Circumference =
20 m.
Assume a friction factor of the decline = 0.012 Ns2/m4
AB 3 25
0.012× 20×1400
R = = 0.0215 Ns2/m8
Determine the resistance of A,B, Area 2,C
R=0.0215 + 0.0100 = 0.0315 Ns2/m8
For an equal split the resistance in both sections must be equal
RAB Area 2 C = RFE Area 1 C
Therefore RFE = RAB Area 2 C – RArea 1 =0.0315 – 0.0150 = 0.0165 Ns2/m8
Calculate the diameter
Area = π
4
πD
Area =
2
and Circumference = πD
Substituting in A3
kCL
R = 2 3
4
πD
kπDL
R =
Solving for D
s
Area 1= 0.015 Ns2/m8
Area 2 = 0.010 Ns2/m8
Shaft 4.0m diam.
200m long
Proposed Shaft
A
B C
F D
E
120 m3/s
120 m3/s
Decline 5m x 5m
gradient 1:7
240 m3/
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D =(4/R1/3π)3/5(kπL)1/5
D =(4 / 0.01651/3π)3/5(0.0045 π 200)1/5
D = 3.2339m diameter
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“Reason, Observation and Experience – the Holy Trinity of Science.”
Robert G. Ingersoll (1833-99) US Lawyer and Agnostic. The Gods.
5 MEASURING AIRFLOW
Airflow is measured in units of cubic metres per second (m3/s). To measure airflow, the airway
cross-sectional area and air velocity need to be known.
There are a large variety of air velocity measuring instruments, however the following are most
applicable to mining requirements:
• Vane Anemometer
• Hot Wire Anemometer
• Velometer
• Smoke Tube
5.1 Vane Anemometer
The instrument consists simply of a light impeller that rotates
at a speed dependent on flow rate. The impeller rotational
speed is converted into an air velocity reading using either
electronic or mechanical means. Most anemometers used in
mine ventilation survey work are capable of measuring
velocities in the range of 0.2 to 25 m/s, with an absolute
accuracy of ± 0.5 m/s, or better. They are ideally suited to
conducting primary ventilation surveys and for routine
determination of air velocity. Air velocities below 0.2 m/s
should be measured with another technique such as a smoke tube or hot wire anemometer.
Mechanical anemometers (particularly the Lambrecht 1406a) are used in the majority of
Australian mines. The mechanical instruments are preferred to the electronic ones for the
following reasons:
• They automatically engage their clutch to commence measurement about five seconds after
the mechanical start lever is depressed. This gives the operator sufficient time to move the
instrument away from the body to the position where the traverse will be commenced.
• When they are fitted with a timing mechanism, they can integrate velocity readings over a
fixed time period (usually one minute), making them ideally suited to traversing
measurement techniques. Traversing is the fastest way of averaging air velocity over an
airway cross-sectional area, with accuracy sufficient for a primary ventilation survey.
Electronic instruments will generally only provide spot readings. Some can integrate, but
this requires a switch to be held down. This can often be difficult to accomplish when the
instrument is mounted on a pole.
• Mechanical instruments are usually smaller and easier to carry. They seem to also be more
reliable and robust than their electronic counterparts.
Mechanical anemometers are delicate instruments and must be treated with care. They should
always be transported in the instrument case. Recalibration can be organised through the
CSIRO or the supplier and should be undertaken at least annually.
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5.2 Hot Wire Anemometer
The hot wire anemometer consists of an electrically heated wire at the end of an extendible
wand. In the most popular type of instrument, the wire is maintained at a constant temperature.
The amount of current required to do this is related to the speed of the air. The instrument
contains a second internal hot wire to provide compensation for changes in air temperature.
Hot wire anemometers are very useful for accurately measuring low air velocities (down to
about 0.05 m/s). Before making a velocity measurement, it is important that the flow distribution
and direction be checked first, using smoke tubes.
Hot wire anemometers are generally not designed for use in the traversing measurement
technique. They are suitable for taking spot measurements and most instruments have the
facility to electronically average a number of spot readings. They are generally not directionally
sensitive.
5.3 Velometer
The velometer is a mechanical instrument that consists simply of a hinged vane that swings in a
rectangular passage. A port in the side of the instrument casing is connected to the passage.
When the instrument is held in an airstream, airflow through the port deflects the vane against
spring pressure. The vane is connected directly to a velocity indication scale. Pitot type tubes
can be attached to the port with hosing to enable velocity to be checked in ducts.
This instrument provides fast, accurate measurements of velocities down to about 0.2 m/s. It is
well suited to routine work such as determining face velocities in secondary ventilation surveys.
The velometer is directionally sensitive and can only be used for spot surveys. It must be read
while being held by the observer and care must be taken to hold the instrument sufficiently far
from the observer to avoid interference with the reading. Velometers are not widely used in
Australian mines.
5.4 Smoke Tube
A smoke tube consists of a glass tube (similar to a gas detector tube) that emits a highly visible
smoke when air is aspirated through it via a rubber bulb. Smoke tubes give superior results to
other less satisfactory improvised alternatives such as spray paint, a handful of fine dry dust or,
cigarette smoke!
Measurements are taken by marking out a fixed distance in a drive with two paint lines on the
roadway. The distance should be that which the smoke travels in about 30 seconds (the smoke
cloud usually dissipates after this period of time). Smoke tubes are used for velocities of less
than 0.1 m/s to 0.5 m/s and a measurement distance of about 5m is usually therefore
appropriate.
The smoke is released in the centre of the airway about an arm’s length upstream of the first
mark. Two observers are required to determine the time interval between when the centre of
the smoke cloud passes the first and second mark. The distance that the smoke travelled
divided by the time taken, then multiplied by a correction factor (typically 0.8) calculates the
velocity of the air. The correction factor allows for the effects of smoke dispersal and the fact
that the velocity in the centre of the drive is usually higher than the average velocity.
It is always a good idea to check the flow distribution in the airway before commencing a smoke
tube measurement. In hot, inclined airways with low velocities, the magnitude and even
direction at the backs may be different to that on the floor.
Smoke tubes can also be used to trace air leakage (e.g. into draw-points) and to help visualise
the flow distribution.
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5.5 Pitot-Static Tube
The Pitot-static tube (often
referred to as the Pitot Tube)
consists of two concentric “L”
shaped tubes.
A pitot static tube is used to
determine pressure at a
point. A variety of designs
have been developed with
advantages between the
different designs generally
applicable to research work.
The pitot static tube can be
used successfully without
the need for corrections
given if:
a) The head is aligned to
the direction of airflow
within 15° (2.5% error in
the velocity).
b) The tube diameter must not exceed 1/25 the duct diameter in circular ducts and 1/20 the
length of the smallest side in a rectangular duct.
c) The tube diameter should not exceed 15 mm.
d) The velocity should be between 3m/s and 70m/s.
The pitot-static tube can be used to measure static, total and velocity pressure when connected
to a manometer.
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With an accurate manometer, good measurement technique and steady flow conditions,
velocities can be measured to within ± 1% for velocities above 5 m/s. The accuracy below 5 m/s
is limited by the inability of most manometers to resolve very low pressures (less than 10 Pa).
Because Pitot-static tubes are essentially a medium – high velocity measurement instrument,
they are not used for surveys in underground drives. They are most useful for the measurement
of the higher air velocities often found in ducts. Typically, a pitot-static tube would be used for
fan testing and flow measurement in secondary ventilation ducts. The most useful length for
many mines is 1.5m.
Of the above instruments, a vane anemometer, smoke tube set and pitot-static tube are
considered essential for any underground mine.
Summary – Air Velocity Measurement Instruments
Instrument Velocity Range
(m/s) Comments
Vane Anemometer 0.2 – 25 *
Versatile and accurate – the most useful ventilation
instrument – particularly for ventilation surveys. Mechanical
type is preferred.
Hot Wire Anemometer 0.05 – 15 Good for accurate spot measurement of low velocities
Velometer 0.2 – 15 Useful for quick, routine spot air velocity measurement.
Method not accurate enough for ventilation surveys.
Smoke Tube 0.05 – 0.5 Enables air flow distribution to be “seen”. Cumbersome
method to use
Pitot-Static Tube 5 – 50 Best instrument for measuring airflow in ducts.
* Instruments capable of reading up to 50 m/s are available, but not generally required for mine ventilation work.
5.6 Selection of Measurement Site
The ideal airflow-measuring site
will have a ‘smooth’ or
‘undisturbed’ airflow distribution
pattern. This is will be an area in
a drive that is:
• as straight as possible,
• of uniform shape and size
and,
• unobstructed.
As a rule of thumb, the
measurement site should be
located at least six drive
diameters downstream and a
minimum of three drive
diameters upstream of any
disturbing influence (bend,
obstruction, etc).
This is often not possible in an underground operation, and a compromise is necessary. In
many cases this will mean taking extra measurements at site that better fit the selection criteria.
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5.7 Measuring the Airway Cross Sectional Area
Several methods are available to determine ventilation station area. When conducting
ventilation surveys that require a high level of accuracy the mine surveyors should be utilised to
accurately determine the shape and area of all permanent ventilation stations.
For primary ventilation surveys, permanent ventilation stations must be clearly marked with a
paint line on the walls and backs. The station number and surveyed cross-sectional area
should also be marked. The most reliable method is to write this information on a steel or
plastic tag that is attached to the wall using surveyor’s putty.
For temporary ventilation stations a tape measure is required to determine a minimum of five
equally spaced heights (offsets). Using these measurements the area can be calculated using
Simpson’s Rule.
“The area enclosed by a curvilinear
figure divided into an even number of
strips of equal width is equal to onethird
of the width of the strip, multiplied
by the sum of the two extreme offsets,
twice the sum of the remaining offsets,
and four-times the sum of the even
offsets.”
(A 2O 4E)
3
Area = X + +
Where
X = distance between offsets. (m)
A = sum of the first and last off sets. (m)
O = sum of the remaining odd offsets. (m)
E = sum of the even offsets. (m)
For example
In this case [( ) ( ) ( )] O1 O7 2 O3 O5 4 O2 O4 O6
3
Area = X × + + + + + +
Note that the area is dived into an even number of sections i.e. an odd number of offsets are
measured.
It is very important to note that in trackless mining, the ventilation station cross-sectional areas
will change over time. This is mainly due to extra road base material being applied. In one
mine, it was found that over about 10 years, the cross-sectional areas of a number of ventilation
stations had decreased by over 10%. In recognition of this change ventilation stations
maintained in main access declines should be
remeasured at least every two years.
5.8 Traverse Velocity Measurement
The traverse method is the quickest way to determine air
velocity with sufficient accuracy for a ventilation survey.
An integrating anemometer is required for this technique
and the vane anemometer is generally the only
instrument with this facility.
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The principle of the traverse technique is to slowly traverse
the anemometer across the airway at a uniform speed and
in a pattern that ensures that the entire airway crosssectional
area is evenly represented. In order to ensure
acceptable accuracy, traverse speed should be limited to
less than 0.4 m/s (less than 0.2 m/s when measuring air
velocities below 0.75 m/s).
When using an anemometer with a 60 second timer two
readings will be required to complete one traverse – one
for each side of the opening, and the velocity for the area
will be the average of both of these readings). Multiple
traverses readings will be undertaken until a minimum of two traverse sets agree to within ±5%
of one another.
5.9 Spot Reading Method
For spot readings, the airway cross-section is subdivided
into a number of equal areas. A velocity reading is taken
in the centre of each of these smaller areas. The
velocity for the area will be the average of all spot
readings.
It should be noted that for acceptable accuracy, the size
of the squares should not be significantly larger than 1m
x 1m. This is a precise, but also time-consuming
method. Because of this, it is not commonly used for
routine work.
5.10 Single Spot Reading Method
A simple, quick, and rough form of spot reading is to take a single velocity measurement in the
centre of a drive, or duct. This reading will almost always over-estimate the average velocity,
because the fastest velocity is usually found in the centre of the airway. A correction factor
(often between 0.8 and 0.9) therefore needs to be applied. This method is useful for routine
work, but is not accurate enough for a primary ventilation survey. It also requires smooth,
unobstructed flow conditions in the airway.
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“I will now speak of ventilating machines. If the shaft is very deep and no tunnel reaches to it, or no
drift from another shaft connects with it, or when a tunnel is of great length and no shaft reaches it,
then the air does not replenish itself. In such a case it weighs heavily on the miners, causing them to
breath with difficulty and some times they are even suffocated and burning lamps are also
extinguished. There is therefore a necessity for machines which enable the miners to breathe easily
and carry on their work”
Agricola “De Re Metallica” (1556)
6 UNDERSTANDING FANS
In Chapter 3 we discussed the effect of natural ventilating pressure and how it can cause air to
flow through mine workings. Airflow through the large mechanised mining operations of today,
could not exist in the quantities required, unless there was sufficient pressure difference to
overcome the resistances in the circuit. This requires mechanical energy input provided by
fans.
The first evidence of mechanical methods for ventilating mine were described by Agricola in
1556 and involved methods such as human powered centrifugal fans, bellows connected to
wooden conduits and shaft collar deflectors, which divert winds into the mine workings.
Although large centrifugal fans were in use in the early to mid 1850’s, airflow through the
underground workings was generally provided
only by natural draft. It was not until the turn
of the 20th century that fans were used
extensively in mines.
Typically, these impellers (pictured opposite)
were fitted with backward curved paddle
blades, between 240 inch (6.0m) and 520
inch (13.0m) diameter and rotating at very
slow speeds between 0.67 revolutions per
second (rps) (40 rpm) and 1.0 rps (60 rpm)
and, driven by large steam powered engines.
One of the early makes of this type of fan was
the double inlet Walker Indestructible16.
In around 1900, Samuel Davidson developed
a forward curved multi bladed fan to handle hot air for the purpose of drying tealeaves from their
Colombo plantations in Ceylon, now Sri Lanka. The fan impeller was much smaller in diameter
with forward curved blades designed to run at significantly higher speeds. The fan was called a
Sirocco after the hot winds that blow across the Mediterranean from the coast of North Africa.
16 de la HARPE, J,H., JENNER, L.W. “The History of Mine Fans in South Africa” The Journal of the Mine Ventilation Society of
South Africa. (December 1986)
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It was one of these new generation fans
that are believed to be the first mechanical
means of ventilation installed in Australia.
This fan was installed around 1917 at on
No. 4 Shaft Zinc Mine Broken Hill and had
an airflow capacity of 45,000 cfm
(21.0m3/s) and remained in operation at
this location until it was replaced in 1921.
The replacement was a larger Sirocco fan
(pictured left), which was manufactured in
Belfast Ireland by Davidson& Co. Ltd.
This fan was fitted with a 119-inch (3.0m)
impeller designed to produce 195,200 cfm
(92m3/s) at 6 of inches water gauge (1495
Pa) when rotating at 244 rpm.
6.1 General Description of Fans
A fan is a rotating machine in which air is continuously drawn in at one pressure and delivered
at a higher pressure i.e. the mechanical energy delivered to the fan is transformed into potential
energy (pressure) and kinetic energy (velocity) by the fan. This pressure is necessary to
overcome the resistance of the particular duct or mine, in which the fan is operating.
Fans are named according to the type of impeller fitted into the casing. In mining applications
the main fan types are:
1. Axial-flow. – In an axial fan the air flows through the impeller parallel to
and at a constant distance from the axis. The pressure rise is provided by
the direct action of the blades.
2. Centrifugal or Radial-flow. – In centrifugal fans the air enters parallel to
the axis of the fan turns through 90º and is discharged radially through the
blades. The blade force is tangential causing the air to spin with the
blades and the main pressure rise is attributed to the centrifugal force.
3. Mixed flow. – In a mixed flow fan the air enters parallel to the axis of the
fan turns through an angle which may range from 30º to 90º The
pressure rise is partially by direct blade action and partially by centrifugal
action.
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6.2 Fan Terminology
The principal component of a fan is the impeller, consisting of the hub and blades. The impeller
is mounted on a shaft inside the housing, or casing, of the fan that is driven directly or indirectly
by a motor. In addition, the fan is generally equipped with an inlet cone and an outlet evasè to
improve the flow characteristics of the fan. There may also be guide vanes to control the
direction of airflow through the fan.
Centrifugal
Reproduced from Air Moving and Conditioning Association Inc. “AMCA Fan Application Manual”
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Axial
Reproduced from Air Moving and Conditioning Association Inc. “AMCA Fan Application Manual”
6.3 Axial-Flow Fans
Axial flow fans are divided into three categories
1. Free air – The impeller is not confined e.g. Desk fan
2. Tube axial – The impeller is confined in a casing
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3. Vane axial – Same as a tube axial except guide vanes are fitted inside the casing to reduce
swirl
The impeller may have two or more blades. Generally speaking, high-pressure fans have a
greater number of blades than low-pressure fans. The shape and pitch of the blades vary
considerably in different designs.
6.3.1 Construction of an Axial-Flow fan
The principal components of an axial fan are
the impeller, the motor and the casing.
The impeller comprises two parts, the hub and
the blades. Blades may be manufactured from
flat plate or cast from alloy in an aerodynamic
shape. It is also possible that the blades are
fixed or adjustable.
The number of blades on an impeller may be
expressed in terms of ‘solidity’ with 100%
solidity the maximum number of blades for that
particular design. Generally speaking the
higher the solidity the higher the pressure.
The hub to blade ratio is indicative of the
performance of the fan. For a given overall
diameter, long blades and a small diameter
hub indicates high airflow and low pressure,
obviously shorter blades and a larger hub
indicates higher pressure and lower airflow.
The clearance between the blade tips and
the casing must be kept small to prevent
pressure loss. This calls for accurate
machining of the casings and blades,
resulting in increased cost.
The air leaving the axial-flow fan usually has a rotary motion, or “swirl”. This results in a loss of
energy. This loss is reduced by the installation of vanes called guide vanes fitted to straighten
the flow of air.
Axial-flow fans are particularly suitable for mounting directly in ducts.
A second impeller, rotating in a direction opposite to the first, can also be fitted. This result in
the swirl induced by the first set of blades being neutralised by the swirl imparted by the second
set. Fans fitted with this type of additional impeller are known as contra-rotating fans
6.4 Centrifugal or Radial-Flow fans
When a stone attached to a string is swung in circles
and then let go it will fly outwards. In fact anything that
revolves tends to move away from the centre of
rotation with centrifugal force. The word centrifugal
meaning ”moving or directed away from the centre”.
The impeller of a centrifugal fan consists of a hollow
“wheel”, “rotor” or “runner”, with a number of blades
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around its circumference, similar to the paddle wheel of old steamships.
The blades are generally relatively short, and can be arranged in three ways:
• Radially
• Forward-curved, or forward-sloping
• Backward-curved, or backward sloping.
When the impeller rotates, the air between
the blades is thrown outwards by centrifugal
force and leaves the impeller at right angles
to the axis. The air is guided by the shape of
the fan casing to the fan outlet, and is then
replaced by air that enters along the direction
of the axis from either one side only, when
the fan is termed a “single-inlet” fan, or from
both sides, when the fan is termed a “doubleinlet”
fan.
6.4.1 Construction of a Centrifugal Fan
In practice the shapes of the blades differ considerably in different types of fans The
performance of the fan depend upon the design of the blades and the design of the volute
(scroll) which is the section of the fan casing along which the air flows after leaving the impeller.
It is the design of the volute that determines the amount of velocity pressure in the air after
leaving the blades that will be converted into static pressure.
On some centrifugal fans, a certain amount of control of performance is provided by the
installation of variable inlet guide vanes.
6.5 The Fan Characteristic (Performance) Curve
The performance of a fan is generally presented in the form of a graph on which is plotted the
airflow quantity (Q) along the horizontal axis and the fan pressure (P) along the vertical axis.
These points are obtained from tests where the actual fan performance is measured. A line
joins these points in the form of a curve. This curve is termed the fan characteristic curve. The
power and fan efficiency is shown for each airflow point and similarly joined to form a curve.
Fan performance curves are supplied by fan manufacturers and show the predicted airflow
volumes for a given pressure.
As the pressure requirements increase, the
airflow volume decreases. It should be noted
that these curves apply only for the particular
fan and at a particular density of air. Any
variations of density along with blade pitch
angle will result in a change to the
characteristic. These changes can be predicted
and incorporated into the design. However,
there are changes that cannot be predicted and
are caused by dirty or worn fan blades.
Fan performance curves are presented in many
differing forms. Some examples of fan curves
are shown in the following figures.
Quantity
Pressure
0
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In order to make the correct selection people interpreting these curves should be aware of the
following:
1. Two curves represent the total pressure performance whilst the other show static pressure
performance.
2. All curves show the fan shaft power expressed in different ways
3. Although only one of the curves shows the efficiency (for the pressure) drawn on the chart it
is easily interpreted by calculation from the information provided. (this is discussed later in
Section 6.13 and Section 6.14)
Some examples of fan curves are provided below
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600
800
1000
1200
1400
1600
1800
20.0 22.0 24.0 26.0 28.0 30.0 32.0 34.0 36.0 38.0 40.0
Air volume (m3/s)
Pressure (Pa)
40
45
50
55
60
65
70
75
80
Fan total pressure
Absorbed power
AL12-500
Diameter: 1200 mm
Speed: 1000 rpm (nominal)
Power: 55 kW
Single stage Axial
Ph: (08) 9455 4433
Fx: (08) 9455 1819
Density 1.20 kg/m3
Absorbed power (kW)
x 1 fan
Full bladed
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6.6 Interpreting Fan Performance Curves
The correct interpretation of fan curves is essential to ensure that the fan selection will meet the
requirements.
MULGA HILL FAN COMPANY
QWER 1250-2300-12 990 rpm 1.2kg/m3
0
500
1000
1500
2000
2500
3000
3500
0 20 40 60 80 100 120 140 160 180 200
Fan Static Pressure (Pa)
0
100
200
300
400
500
600
700
0 20 40 60 80 100 120 140 160 180 200
Volume (m3/s)
Fan Shaft Power (kW)
10
20
30 40
50
60
70
10 20 30
40
50
60
70
Reference P’nW 2345
Impeller blade
pitch setting
Pressure sometimes
given as Fan Total
Pressure
Manufacturers
reference
number
Fan performance sets
Manufacturers Code for
the fan type usually
provides, fan type, hub
diameter, impeller diameter,
number of blades, fan
Speed & air density
Fan operating point
AP = P x Q
Fan Shaft power at the
operating point
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The example curve shown above represents an adjustable pitch axial flow fan. Some points to
consider are:
1. A fan designed and manufactured to operate at setting 40 does not mean that it can operate
at settings above this point. In this case the fan would typically be fitted with a 300 kW
electric motor and if the fan pitch angle were increased the potential for the fan to stall would
be extremely high. It would be necessary to increase the size of the motor.
2. It may not be possible to increase the size of the motor because it must be small enough to
fit inside the housing and retain the same centreline.
6.7 Fan Stall
A fan will “stall” if the resistance of the system is (or becomes) excessive and as this
resistance curve becomes steeper the
intersection point with the fan
performance characteristic changes and
moves to a point of higher pressure (lower
quantity) on the fan characteristic. At this
point, both the fan pressure and airflow
quantity increase simultaneously (i.e. both
the fan characteristic and the system
characteristic have a positive slope) this is
opposed to the normal operation where
the pressure increases as the quantity
decreases. Eventually, the pressure and
quantity become unstable and fluctuate
wildly. This condition is known as
“hunting” or “stall”.
This zone is usually not shown on manufacturers curves, but it can be assumed to occur above
the highest pressure shown. If a fan is operating in this stall zone, the air tends to separate
from the blade surfaces, causing excessive vibration. These vibrations may damage the fan if it
is allowed to operate for any length of time. This zone may be very noticeable in some fans and
hardly noticeable in others.
Stall Zone Signs
1. The fan vibrates excessively
2. The fan makes unusual noises (whines because of increased pressure)
3. Input power amps show rapid oscillations or surging at a slower frequency (every couple of
seconds)
4. The quantity and pressure points alter considerably
Action to be taken
Continued operation of a fan in stall may eventually cause damage to the fan and the blades
may disintegrate. To remove the fan from this condition some air may be allowed to enter the
system at a position close to the fan (ie short circuit the system) this will have the effect of
reducing the overall system resistance). This should only be a temporary solution until the
cause of the stall is identified and rectified.
6.8 Fan Performance Control
Although fans are required for a particular duty, it should not be forgotten that mining by its very
nature is dynamic. Over the life of the mine, fans may be required to operate at a number of
Quantity
Pressure
0
Un-stable portion of curve Stable portion of curve
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specific duties. To cater for these changes, fans must have flexible characteristics. Some of
the causes for differing fan duties include;
• Increased resistance
• increased airflow requirements
• change in air density
• power saving
The output of a fan can be controlled by adjusting the system (mine resistance) or using outlet
dampers, inlet-box dampers, variable inlet vanes, variable pitch, variable speed or varying the
number of fans.
6.8.1 System Damper
Dampers (regulators) installed in fan ducts (mine openings) to act as throttling devices have the
effect of altering the system resistance. The fan will operate at different points on its
characteristic without change to the pressure, quantity and power curves.
This method is sometimes used to cause a fan to operate at a more efficient section of its
performance characteristic. In these cases efficiency may be improved but, at the cost of
decreased airflow and some times increased power. There is usually very little economic
justification for the change and is not recommended as a control system.
6.8.2 Outlet Damper
A fan outlet damper acts on the fan in the same way as a system damper i.e. it will change the
overall system characteristic by altering the resistance.
Consider the graph to the left in the figure below. Points A, B, C, D & E lie on the intersection of
the fan curve and a number of parabolic system characteristics with the same origin. These
curves represent the combined system and damper at various settings of the outlet damper
control. Point A would represent the damper wide open and the other points represent the
damper at other settings up to point E fully closed (FC). In this case the operating points D and
E are potentially unstable and it would not be recommended to operate the fan with the damper
set in this position.
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Reproduced from “Fan Engineering – Buffalo Forge Company Eighth Edition”
It is possible to represent the various settings of the damper as fan characteristic curves similar
to that shown to the right in the figure below. Obviously there are an infinite number of settings
between fully open and fully closed.
There is a caution when interpreting this graph. Consider the point C (½ open). In this case the
vertical distance from the base line to C is the pressure loss in the system and the vertical
distance C-C’ is the pressure loss caused by the outlet damper.
There is some suggestion that because the intersection points A, B, C, D, & E all lie on the
negative section of the performance characteristics then the operation will be stable even when
fully closed. There is some evidence that outlet dampers can be evaluated in this way and IF
the principle resistance is on the inlet side of the fan but it is not suitable if the principle
resistance is on the outlet side of the fan.
In essence the damper is considered to be part of the system rather than part of the fan.
6.8.3 Inlet-Box Damper
The performance of an inlet-box damper (IBD) is considered in two parts.
The first part is the pressure loss cause by the resistance of the damper, similar to the outlet
damper. The second part is due to changes caused by the inlet-box damper to the swirl of the
air onto the impeller. This pre-swirl reduces the amount of work for the impeller therefore
affecting the impeller output and resulting in a different performance characteristic.
The performance of an inlet-box damper can
be considered in the manner shown in the
figure opposite.
Here again there is some evidence to
suggest that IF significant swirl exists then
the intersection points A, B, C, D, & E all lie
on the negative section of the performance
characteristics and the operation will be
stable. There is also some evidence that if
the principle resistance is on the outlet side
even without pre-swirl the operation will be
stable.
However, if the principle resistance is on the
inlet side then the operating points should be
considered as A’, B’, C’, D’, & E’ and in this
case the points D’ & E’ lie on the positive part
of the fan characteristic and therefore the
operation is likely to be unstable. Like the
outlet damper the inlet-box damper is
generally considered to be part of the system
rather than part of the fan.
6.8.4 Variable Inlet Vanes
Variable inlet vanes (VIV’s) are similar in appearance to a camera iris and is slightly superior to
inlet damper control in respect of efficiency and stable operation.
VIV’s provide the same effect on fan performance as the inlet-box damper because they are
designed to create a pre-swirl to the air. Again they have two components that caused by the
resistance of the structure and that caused by the pre-swirl given to the air.
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VIV’s are always installed as part of the fan
and very close to the impeller. Because of
this operating points are always considered
to be points A, B, C, D, & E and highly
unlikely to operate in the unstable positive
characteristic points D’ & E’.
The use of VIV’s in Australian mines is
usually confined to primary centrifugal fans
but their use with axial flow fans is common
particularly in the deep mines of southern
Africa.
Maintenance of this component is usually
very poor and most become seized at a
particular setting. This component should
be greased weekly and exercised daily.
Exercising requires the vanes to be closed
then again open to the desired setting. This
is a very simple operation that can be
undertaken automatically, particularly given
today’s technology of PLC’s.
6.8.5 Variable Speed
Speed control can be achieved in a number of
ways
• Changing drive pulley ratios.
• Reduction gearbox drives.
• Variable speed controllers
• Number of poles in the motor
No of Poles
Motor Speed 120 Hz
×
= (can only be an
even number)
Variable speed characteristics can easily be
determined from the fan laws detailed in section
6.9. The performance characteristic curves
would be identical except that all devices to
alter the speed must have some slip to transmit
torque.
6.8.6 Variable Pitch
The figure opposite shows the performance characteristic for a fan and the operation points with
the blades set at different angles. The system is depicted by the curve drawn through points E,
D, C, B, A & Z. A fan would normally be selected to operate a point A as this would be the most
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Effect of NVP
Quantity
Pressure
0
+ NVP
– NVP
efficient use of power. Higher flow rates
would be achieved at point Z and lower
rates at points B, C, D, & E.
Altering the blade pitch is normally done
by stopping the fan, loosening the nut
holding the blade to the hub and then
turning the blade to the required pitch
setting. The nut is then again tightened
to secure the blade in its new position.
On certain fans, this adjustment can be
carried out by means of a single lever or
series of linkages when the fan is in
operation. These are termed variablepitch
in motion fans.
6.8.7 Fans in Series
If we assume that air is incompressible then the
potential performance of two identical fans is
obtained simply by multiplying by two the
pressure at a particular quantity.
In the figure opposite one fan will operate in the
system at B by adding an identical fan the
combined operating point will be at A.
Fan in series are generally well recognised as
most auxiliary development fans are configured
to operate in this manner and there are many
instances where primary ventilation fans have
two stages.
Fans are generally installed in series to
overcome an increased system resistance or
maintain a particular quantity.
In most cases fans installed in series are designed
for that purpose. However it is possible to install
unlike fans in series but this is the exception rather
than the rule.
NOTE that the NVP acts in the same way as a
series fan with the exception that NVP can have a
negative pressure that is subtracted form the fan
pressure characteristic.
When two unlike fans are installed in series then
the combined pressure generated at a given
airflow quantity will be the sum of the pressures
of the individual fans at that quantity. The
A’
One Fan
Two Fans
A
B
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A’ A
B
One Fan
Two Fans
exception being contra rotating fans when the pressure can be three times that of each
individual fan.
In practice, the overall flow is increased with the installation of a second fan in series however,
in most auxiliary fan duct installations there will be little to no increase in flow discharged at the
end of the duct, only an increase in leakage along the length of the duct.
6.8.8 Fans in Parallel
The potential performance of two identical
fans operating in parallel is obtained simply by
multiplying by two the quantity at a particular
pressure.
In the figure opposite one fan will operate in
the system at B by adding an identical fan the
combined operating point will be at A.
When fans are selected to operate in parallel,
it is essential to ensure they operate in the
stable area of the combined characteristic.
When combining parallel fan performance
curves the only the quantities are added and the
pressure will remain the same.
The important thing is to note is that both fans MUST
operate at the same pressure.
When unlike fans are installed in parallel they will
provide different quantities of air but will operate at the
same pressure.
The use of fans in parallel is not uncommon in mines
and in fact many mine circulating large quantities use
fans in parallel as it offers the opportunity for
maintaining some airflow should one fan fail. Some
mine choose to install two or even three fans and use
one as an insitu spare and only operated in an
emergency when an operating fan fails.
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The use of multiple fans in parallel may cause problems when starting. Some fan performance
curves have a decided dip in the performance and if the fan being started has to pass through
this dip or the positive slope side of the performance characteristic then it will enter the “Stall”
position and never recover.
Fan manufacturers seldom show the positive slope side of the curve and it can be misleading.
As a precaution the fan manufacturer must be questioned specifically as to the possibility of
having to pass through the dip on the stall side of the fan performance.
Fan 1
Fan 2
Fan 3
Fan 4 desired
operating position
T0 get to the desired position Fan 4 must pass
through the dip in the curve. Usually the fan enters
stall and never recovers, eventually destroying the
impeller
6.8.9 The ‘Eck’ Line
When a fan enters the unstable part of its curve it will start to stall. Operating two or more fans
in parallel, will induce stalling at a lower operating pressure. This is because succeeding fans
must start up against the pressure already being developed by the operating fan or fans.
The calculated unstable portion of the combined curve is known as the ‘Eck’ line17. The
determination of this line will show the lower “maximum useful pressure” to ensure operation in
the stable portion of the combined fan performance curve.
Consider the combined curves for one, two and three identical fans in parallel shown in the
figure below. The combined ‘Eck’ lines can be determined in the following manner.
For a single fan the unstable portion of the performance curve is shown as a dashed line and
the fan can be operated at any point to the right of the dashed line.
For two fans, the combined curve (labelled 2 Fans) shows two unstable portions.
The first is due to the combination of the two fan curves and is shown as a bold black line. Point
‘A’ on this curve is obtained by doubling the air quantity at point ‘a’ on the single fan curve.
The second unstable portion of the two-fan system is due to the unstable part of each fan and is
shown as a dotted line. This portion is constructed by marking a point to the left of the
combined curve at distance ‘a-c’ as measured for a single fan. Other points on this unstable
portion of the curve are constructed in the same manner to obtain the dotted line for two parallel
fans. Hence for two fans in parallel, the stable portion of the fans is to the right of the curve ‘a-b’.
Finally, the curve for three fans in parallel (labelled 3 Fans) has three unstable portions:
17 ‘Eck’ – In recognition of the work undertaken by Professor Bruno Eck on this subject.
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1. due to the combined P-Q curve for three fans;
2. due to the unstable portion of two fans; and
3. due to the unstable portion of a single fan.
The stable part of the curve lies to the right of the ‘Eck’ line.
0 10 20 30 40 50 60 70 80 90
Quantity (m3/s)
1000
2000
Pressure (Pa)
Eck Lines (dotted)
a A
b b
c a c
1 Fan 2 Fans 3 Fans
Resistance X
Resistance Y
Stable portion of the
combined curve
Diagrammatic only
In the figure above, two system resistance curves are plotted. Resistance X has a resistance
higher than Resistance Y.
If these fans were operated in parallel against Resistance X, the third fan would have to pass
through the unstable part of the combined curve on the run up to the system pressure. In
reality, this would not occur and the fan would stall.
If these fans were operated in parallel against Resistance Y, the third fan would run up to the
system pressure by passing underneath the ‘Eck’ line. In this case the probable maximum duty
would be 65 m3/s at 1,500 Pa.
6.8.10 Fan Diffuser
Fan diffusers ultimately cause a little extra
airflow through the fan. As shown in
Section 2 system shock losses can be
reduced by the installation of an effective
diffuser at the discharge.
The diffuser may be fitted with an internal
fairing. They vary in shape from
rectangular to round and the included
angle, the outlet to inlet ratio and the
cross sectional shape affect the efficiency.
The best pressure regain is achieved on a
diffuser without an inlet fairing, and with
an outlet to inlet area ratio of 4:1 and the
optimum included angle between 8o and
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11°. Because of the length required for this optimum in practice this angle can be much larger
(up to 25°)
PV REGAINED = η(PV INLET – PV OUTLET )
Where
η = Efficiency of the evasè (%)
6.9 Fan Laws
Fan performance curves are determined for a specific speed and density.
Unless otherwise specified the density for a fan curve can be assumed as standard air (1.2
kg/m3). Standard air is a definition of convenience that is widely adopted. The actual values of
temperature, pressure and moisture content are not important.
Should the speed or density of the fan inlet air be altered, the following fan laws will apply.
Change Density (ρ) Speed (rpm) Diameter
Quantity (Q) Q1 = Q2
2
1
2
1
rpm
rpm
Q
Q =
3
2
1
2
1
D
D
Q
Q
 

 

=
Pressure (P)
2
1
2
1
P
P
ρ
ρ
=
2
2
1
2
1
rpm
rpm
P
P
 

 

=
2
2
1
2
1
D
D
P
P
 

 

=
Power (kW)
2
1
2
1
kW
kW
ρ
ρ
=
3
2
1
2
1
rpm
rpm
kW
kW
 

 

=
5
 

  
=
2
1
2
1
D
D
kW
kW
Efficiency (%) η1 = η2 η1 = η2 η1 = η2
When considering the diameter the fans must have the same geometric dimensions, and if the
size difference is great then it can be expected
that the fan total pressure may be larger than
that given by these laws (scale effect).
It is also possible that a change in density due
to compression of the air will occur affecting the
results from the use of these laws. This is
normally only of concern at pressures above
2.5 kPa.
This is discussed more fully in “Woods Practical
Guide to Fan Engineering” or “Fan Engineering”
published by the Buffalo Forge Company.
6.10 Measuring Fan Performance
Performance testing is undertaken to determine the actual operating performance of fans
particularly at time of installation for comparison with the manufacturer’s predicted performance.
Fan performance testing is detailed in standards such as Australian Standard 2936 and British
Standard 848 and is usually specified in fan tender documents as the measure of compliance.
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Standards tests are performed using
standardised airways with devices fitted to
straighten flow and alter the operating
pressure. The installation of such
components for site measurements is
generally considered to be too costly and is
seldom (if ever) used.
Precision laboratory type testing cannot be
carried out on site. Measurements made to
determine site performance must be agreed
with the fan manufacturer at the time of the
purchase order being placed.
Aside from the regular three monthly intervals
to comply with some mining legislation the
measurement of fan performance is also undertaken:
1. For acceptance testing at time of purchase and commissioning.
2. Prior to making an operational change to the fan (i.e. altering the speed, changing a motor
etc.).
3. After making an operational change to
the fan to ensure the desired (predicted)
result has been achieved.
4. To determine if a measured decrease in
system airflow quantity is due to system
resistance or the operational performance
of the fan.
The main objective of measurements is to
accurately determine:
1. fan inlet airflow quantity
2. fan pressure
3. fan efficiency
Measurements of fan performance are made to represent the actual airflow through the fan and
must therefore be sited to ensure there is no change in the properties of the airflow between the
measuring site and the fan inlet.
Such changes may include density, caused by differing temperature and pressure as well as
airflow quantity caused by leakage into or out of the system.
Fans are primarily categorised as pressure, exhausting, in line, or diaphragm mounted.
6.10.1 Fan Total Pressure
The fan total pressure (FTP) is equal to the total pressure at the fan outlet (TPO) minus the total
pressure at the fan inlet (TPI).
FTP = TPO − TPI Equation 6-1 Fan Total Pressure
6.10.2 Fan Velocity Pressure
The fan velocity pressure (FVP) is equal to the velocity pressure at the fan outlet (VPO).
PRESSURE FAN
Free inlet, ducted outlet
EXHAUST FAN
Ducted inlet, free outlet
DIAPHRAGM FAN
Free inlet, free outlet
INLINE FAN
Ducted inlet, ducted outlet
Fan Categories
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FVP = VPO Equation 6-2 Fan Velocity Pressure
6.10.3 Fan Static Pressure
The fan static pressure (FSP) is equal to the FTP minus the fan velocity pressure.
FSP = FTP − FVP Equation 6-3 Fan Static Pressure
6.11 Pressure Fans
The figure below shows the side tube and facing tube measurements for a pressure (forcing)
fan (most mine secondary fans). From
these measurements of pressure we can
determine the Fan Pressures
By definition FTP = TPO – TPI
in this case TPI = 0
therefore FTP = 2000 – 0
= 2000 Pa
By definition FVP = VPO
From the equation for ventilation
pressure TP = SP + VP
then VPO = TPO – SP0
= 2000 –1500
= 500 Pa
= FVP
and FSP = FTP – FVP
= 2000 – 500
= 1500 Pa
Therefore, a static pressure reading downstream of
a pressure fan is equal to the Fan Static Pressure
(minus the frictional pressure losses between the
fan outlet and the measuring site).
6.11.1 Selection of Measuring Site
Satisfactory measurements can only be made in a
plane where the airflow is substantially free of swirl
and air turbulence and generally should be no closer
than 1.5 duct diameters upstream of the fan inlet or
2.5 diameters from the fan outlet.
It is preferred that airflow measurements are made
upstream of the fan (the fan intake). If it is
necessary to measure downstream (fan discharge)
this distance should be equal to the effective duct
length (described in the figure opposite). This is to
allow for the velocity profile to become evenly
distributed in the duct.
Outlet Area Discharge Duct
Fan Housing
Cutoff
Blast Area Area
25%
50%
75%
100% Effective Duct Length
To calculate the effective duct length assume a minimum 2.5
duct diameters for each 12.5 m/s and add 1 duct diameter for
each additional 5.0 m/s
Example: 25 m/s = 5 duct diameters.
If the duct is rectangular with side dimensions a and b, the
equivalent duct diameter is equal to
π
4ab
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6.12 Exhausting Fans
The figure below shows the side tube and facing tube measurements for a suction (exhausting)
fan (as installed for most mine primary ventilation). From these measurements we can
determine the fan pressures.
In this case, the static pressure at the fan
outlet (SP0) would equal zero.
Since TP = SP + VP
And SP0 = 0
Then TPO = VPO
By definition FSP = FTP – FVP
And by substitution
FSP = (TPO – TPI ) – VPO
Since TPO = VPO
Then by substitution
FSP = VPO – (TPI) – VPO
=- TPI
= – 2580 Pa
Therefore, a facing tube measuring the inlet total pressure of an exhausting fan is equal to the
FSP.
Now as SPI = TPI – VPI
And TPO = VPO = VPI (no change in diameter)
then by substitution SPI = TPI – TPO
Rearranging SPI = TPO – TPI
and as TPO – TPI = FTP
then FTP = -SPI
= -(-2000)
= 2000 Pa
Therefore, a side tube measuring the static pressure of an exhausting fan is equal to the FTP
(fan total pressure).
Examples
(Qi) The following measurements were recorded at a surface fan installation exhausting air
from a mine. Determine the FSP, FVP and FTP.
Total Pressure 1200
Static Pressure 1800
(Ai) Both measurements would be negative ie. on the suction side of a fan.
FSP = -1200 Pa
FTP = -1800 Pa
FVP = VPO = VP
= (-1200)-(-1800)
EXHAUST FAN
Ducted inlet, free outlet
TP= -2580 Pa
SP= -2600 Pa
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= -1200 + 1800
= 600 Pa
Note Velocity Pressure is always positive.
(Qii) The following measurements were made downstream of a development fan in a 1200
mm diameter duct at a site 4 duct diameters from the fan discharge. No duct was installed
upstream of the fan.
Average velocity pressure 480 Pa
Duct Static Pressure 2500 Pa
Determine the FSP.
(Aii) In a pressure system the FSP is equal to the static pressure of the duct.
FSP = SP0 = 2500 Pa
less the frictional pressure losses in the duct.
Duct area = π
2
2
1.2 




= 1.13m2
Duct Perimeter = π1.2
= 3.77m
Assume K = 0.003
Then R = ( )
(1.13)3
0.003 × 3.77× 4×1.2
= 0.037Ns2m-8
Velocity of airflow
ρ
= 2VP
1.2
2 × 480
=
= 28.3m/s
Airflow = V x A
= 28.3 x 1.13
= 32m3/s
and P = RQ2
duct losses = 0.037 x (32)2
= 37.8
= Say 40Pa
Fan SP = 2500 – 40
= 2460 Pa
6.13 Air Power
Air power is the energy required to move a specific quantity (Q) of air over a specific resistance
and can be calculated from
AP = P ×Q Equation 6-4 Air Power
Where AP = Air Power (watts)
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P = Pressure (Pa) and also = N/m2
Q = Quantity of air (m3/s)
Substituting
P x Q = N/m2 x m3/s
= Nm/s
and 1 Nm = 1 Joule and 1 Joule/second = 1 watt
therefore AP = watts
Example
A fan requiring 1500Pa to move 400m3/s of air has an air power of
AP = 1500 x 400
= 600,000watts
6.14 Fan Efficiency
Fan efficiency is important as it determines the cost of power necessary to operate the fan. For
example a primary ventilation fan providing 250m3/s at a pressure of 2000Pa will absorb 667kW
of electrical power when operating at an efficiency of 75% and only 588kW when operating at
an efficiency of 85%. Power costs at remote mine sites can be as high as $0.17/kW hour,
therefore over one year the savings are considerable. In this case $116,800 each year of
operation.
The efficiency (η) of the fan is calculated from the equation
W
Q×P
η = . In this equation “W” is
the impeller power measured with a swinging frame dynamometer. Because suitable
dynamometers are not readily available this is usually measured as the motor input power
(MIP). The multiplication of air quantity (Q) and pressure (P) results in the power exerted to the
air and is simply termed “Airpower” (AP).
The equation for fan efficiency (η) is therefore usually written as
MIP
η = AP
The efficiency of any fan can be expressed as either Fan Total Efficiency (FTE) where the FTP
is used to determine the Airpower, Fan Static Efficiency (FSE) where the FSP is used to
determine the Airpower, and the “Fan Overall Efficiency” (FOE). In the case of FOE this
includes losses in the electric driving motor, any speed change device between the motor and
the impeller and any other source of power.
Caution must be used when interpreting fan curve efficiency. Some curves show static
efficiency. Others show total efficiency.
Of particular interest to ventilation practitioners is the overall fan efficiency as this determines
the final electrical input power to the motor.
100
Motor Input Power (MIP)
FOE = Air Power ×
This is often mistaken as the Static or Total efficiencies that do not include the power losses for
gearboxes, motors etc.
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Example
Determine the FSE and FTE for a mine exhausting fan delivering 300m3/s of air. The fan inlet
total pressure has been determined as 1500Pa and the Inlet Static pressure as 2000Pa. The
motor input power is 700kW.
For an exhausting fan
FSP = Total Pressure measured at the Fan Inlet
= 1500Pa
Overall FSP efficiency
MIP
= APFSP
700
1.500 × 300
=
700
= 450
= 64%
and FTP = Static pressured at the fan outlet relative to atmosphere
= -(-2000)
= 2000Pa
Therefore the Total Efficiency
MIP
= APFTP
700
2.000 × 300
=
700
= 600
= 86%
6.15 Measurement of Airflow
The most common methods used to determine airflow in ducts is with an anemometer or a pitotstatic
tube. Anemometers are used most often in walk-in airways with velocities between
0.2m/s and 14m/s.
Fan ducts usually require the use of a pitot static tube, as velocities can be greater than 25m/s.
The velocity at a point is calculated from the equation for velocity pressure
ρ
v = 2VP
The velocity in a duct will vary from point to point and the more points measured (on a designed
basis) the more accurate is the result.
The duct is divided into equal areas and a measurement made at the centre point of each area.
The Velocity at each point is calculated and the arithmetical average of all sections is
determined and used as the duct velocity. The airflow quantity is determined by Q = v * A
The figure below shows measuring sites in circular and rectangular ducts derived from the
British Fan Test Code.
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The Australian Standard AS 2936 does not identify sites for rectangular ducts and for circular
ducts shows 8 or 10 measuring sites on 3 diameters, as opposed to the 4 diameters in the
British Standard. The 8-point measuring system is for general use with the 10 measuring point
system used for swirling flow in circular ducts
Measurement Distances for Pitot Static Tube in Circular Ducts
8 Point System 10 Point System
0.021d 0.019d
0.117d 0.077d
0.184d 0.153d
0.345d 0.217d
0.655d 0.361d
0.816d 0.639d
0.883d 0.783d
0.979d 0.847d
0.923d
0.981d
Note d = diameter
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6.16 Fan Laws
Fan performance curves are determined for a specific speed and density.
If density for a fan curve is not shown it can be assumed as standard air (1.2 kg/m3). Standard
air is a definition of convenience that is widely adopted. The actual values of temperature,
pressure and moisture content are not important.
Should the diameter, speed or density of the fan inlet air be altered, the following fan laws will
apply.
6.16.1 Density (ρ) Change
Quantity (Q) Q1 = Q2
Pressure (P)
2
1
2
1
P
P
ρ
ρ
=
Power (kW)
2
1
2
1
kW
kW
ρ
ρ
=
Efficiency (η) η1 = η2
6.16.2 Speed Change
Quantity (Q)
2
1
2
1
rpm
rpm
Q
Q =
Pressure (P)
2
2
1
2
1
rpm
rpm
P
P
 

 

=
Power (kW)
3
2
1
2
1
rpm
rpm
kW
kW
 

   =
Fan total efficiency (η) η η 1 = 2
6.16.3 Diameter Change
When the diameter (D) is increased and the fan is geometrically similar the following laws apply:
Quantity (Q)
3
2
1
2
1
D
D
Q
Q
 

 

=
Pressure (P)
2
2
1
2
1
D
D
P
P
 

 

=
Power (kW)
5
2
1
2
1
D
D
kW
kW
 

 

=
If the size difference is great then it can be expected that the fan total pressure may be larger
than that given by these laws (scale effect).
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It is also possible that a change in density due to compression of the air will occur affecting the
results from the use of these laws. This is normally only of concern at pressures above 2.5 kPa.
This is discussed more fully in “Woods Practical Guide to Fan Engineering” or “Fan
Engineering” published by the Buffalo Forge Company.
These laws should be used with caution as they are not applicable to all fans.
Example
A 1000m diameter fan tested at 1380 rpm and an inlet airflow density of 1.16 kg/m3 gave the
following results,
Quantity (Q) = 20 m3/s
Pressure (P) = 1520 Pa
Power (kW) = 40 kW
What is the expected operating performance at 1470 rpm at standard air inlet density of 1.2
kg/m3.
From equations in sections 6.16.1 and 6.16.2
Note that density change has no affect on the quantity.
2
1
2
1
rpm
rpm
Q
Q =
Therefore Q2 =
1380
20 × 1470
= 21.3 m3/s
The pressure change can be calculated by combining the effects of density and speed changes.
2
2
1
2
1
rpm
rpm
P
P
 

 

= and
2
1
2
1
P
P
ρ
ρ
=
Therefore  


 

ρ
ρ
 

 

=
2
1
2
2
1
2
1
rpm
rpm
P
P
And P2 = 







× 
1.16
1.2
1380
1520 1470
2
= 1790 Pa
The change in power change can also be calculated by combining the effects of density and
speed changes.
2
1
2
1
kW
kW
ρ
ρ
= and
3
2
1
2
1
rpm
rpm
kW
kW
 

 

=
Therefore  


 

ρ
ρ
 

 

=
2
1
3
2
1
2
1
rpm
rpm
kW
kW
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And kW2 







= ×
1.16
1.2
1380
40 1470
3
= 50 kW
6.17 Fans Failure
Fan failure is more often than not caused
by:
a) Poor preventative maintenance including
regular cleaning of guide vanes and
impellers,
b) Unrecognised increasing fan pressure
caused by increasing mine resistance,
c) Rocks and other foreign materials are
often found in fan housings and simple
things like rags, plastic etc. can become
caught on an impeller,
d) An accumulation of dust and salts on
impellers may also cause the fan to run
out of balance,
All of which cause the fan to run out
of balance. If a fan is allowed to run
for extended periods while out of
balance it can potentially cause
bearing failure and fatigue cracking in
either the impeller, the shaft or the
fan casing.
6.18 Effect of Reversal of Rotation
1. An axial flow fan will deliver a reduced volume of air in the opposite direction at a
greatly reduced pressure and efficiency, particularly if aerofoil blades are used
2. A centrifugal fan will deliver a reduced volume in the same direction at a greatly
reduced pressure and efficiency
There is often suggestion that fans can be reversed in times of emergency to reverse the
direction of flow in the mine. Although this is true the fan performance is significantly reduced.
This could be 70%.
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Before attempting this in an emergency the final affect must be completely understood and
more importantly tried, tested and evaluated long before it becomes part of any emergency
procedure.
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“It remains for me to speak of the ailments and accidents of miners and the methods by which they
can guard against these, for we should always devote more care to maintaining our health, that we
may freely perform our bodily functions, than to making profits.”
Agricola – “De Re Metallica” (1556)
7 MANAGEMENT
There has been significant change in underground metalliferous mine design and operations
over the past ten to fifteen years.
Hoisting shafts are now rare in all but very deep, high tonnage operations. In most mines, all ore
and waste is brought to the surface in diesel trucks. More diesel equipment per tonne of ore is
therefore required.
The use of ancillary diesel equipment (e.g. cable bolting, mechanical scaling, charge-up
secondary breaking and shot-creting equipment) is increasing.
There is a greater prevalence of mining methods that require little permanent development apart
from the decline. This has made the establishment of appropriately placed primary ventilation
infrastructure difficult and increased the reliance on secondary ventilation techniques.
The regulatory framework has changed and is now much less prescriptive. Instead of designing
to satisfy minimum statutory requirements, there is now much more reliance on the principle of
an employer’s duty of care.
There has been a dilution of technical expertise brought about by a number of factors, including
the recent rapid growth in the industry. As a result, there is an increasing requirement for
ventilation systems to be simple, robust and reliable so that minimal intervention or
management is required.
Mine ventilation practice has had to evolve to keep pace with these changes. And many long
established rules of thumb are now no longer appropriate.
Of all the ventilation problems encountered in mines, the vast majority are caused by higher
than acceptable concentrations of smoke, dust and heat.
Smoke reported during the shift is generally from poorly tuned and/or maintained diesel
engines. Dust is almost always from handling of dry broken rock, drilling and raiseboring
operations. Increases in temperature are due to change in ambient surface conditions or
ventilation flows being reduced (changes due to rock temperatures should be well recognised
long before they cause problems). Identification of the source of the contaminant provides the
opportunity to control the problem at this point. However, solutions of this type are more often
than not met with resistance from production and maintenance personnel. Most cases are
resolved by a compromise between removing the source and increasing the airflow.
In those situations where the source cannot be successfully controlled, for whatever reason,
then rescheduling of activities may remove the symptoms and allow work to continue.
Unfortunately it is only the most severe cases of undesirable conditions that are adequately
dealt with and the others are allowed to continue because “we will be out of there soon”. As a
consequence operators are usually only looking for a “quick fix”.
Of all the problems identified the solutions could quite easily be found in the “Top Ten Solutions”
below
1. Holes in auxiliary ventilation duct
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2. Excessive tees NOT tied off
3. Fan re-circulation
4. Poor fan selection
5. Leaking ventilation walls
6. Poor primary airflow distribution (caused by incorrectly adjusted and/or lack of controls)
7. Lack of dust control measures when loading and transporting broken dirt
8. Short-circuiting of air through stope voids, ore passes and disused vertical openings.
9. Activities and diesel equipment has outgrown the original ventilation system design.
10. Poor records of airways (particularly vertical openings), dimensions, development
method, location and purpose
All are problems that could (and should) have been avoided by better planning and
management.
7.1 Primary Airflow Requirements
There are many “Rules of Thumb” (ROT) and a few empirical equations that have been
developed to estimate the airflow requirements for mines. Some are realistic and based on
recent mechanised mines but many are unrealistic and based on older non-mechanised mines.
All are based on past experience. There use is acceptable as long as they are taken exactly for
what they are that is an idea of the magnitude of the result that will be achieved by calculation.
Some time the results will be surprisingly close to a calculated result and many times it will be
way off the mark.
Determination of the primary airflow requirement is the starting point for ventilation design. The
design airflow needs to be known before the required number, and size of surface ventilation
shafts can be estimated.
Design primary airflow requirements are almost always under-estimated, probably due to the
use of ROT or some empirical equation, but also due to a failure to allow for increases in mine
size, depth and complexity beyond initial estimates. Even then the cost of providing the
necessary airflow under-goes “pruning” in an effort to keep project capital and operating costs to
the minimum. Under-estimated (or reducing) primary airflow rates will adversely affect the
project economics in the following ways:
Operating Costs: Power costs increase dramatically with airflow rate. If airflow rates need to be
increased beyond initial estimates, there will be a substantial increase in operating costs,
particularly if airways are inadequately sized:
3
Old Flow Rate
Power Cost Increase New Flow Rate  


 

α
Capital Costs: An increase in airflow rate may require the purchase of new fans and excavation
of new airways. The cost and just as importantly the disruption to mining in doing so are usually
considerably greater than would be the case if the original airways sizes were based on realistic
primary airflow estimates.
7.1.1 Determining Primary Air Quantities.
When estimating primary airflow requirements, we need ask ourselves why ventilation is being
provided in the first place. In many cases, in mechanised mines the criterion for dilution of diesel
exhaust emissions is the overriding factor. It is however also necessary to consider equally
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relevant requirements such as control of thermal conditions and adequate dilution of dust and
mine gasses.
Analysis of airflow requirements for individual work places is usually carried out when
determining primary airflow requirements. Design airflow rates are often based on exceeding
0.04 to 0.06 m3/s of airflow per rated kW of diesel engine power in all active mining areas. This
analysis can become quite complex for larger operations and requires a thorough understanding
of the individual facets of the development and production process. The design airflow rate must
include an allowance for leakage (both primary leakage and leakage in secondary ventilation
ducts) and service areas such as ore passes, grizzlies, crushers, conveyor drives, fuel bays,
pump stations, workshops etc.
The first process is to work with the designers and schedulers to allocate air quantities and with
consideration to predicted heat loads, strata gas emissions, diesel exhaust emissions,
legislation, OH&S and Industrial relations to allocate airflow requirements for;
Location Airflow required for
Scheduled production activity
Drilling, blasting, loading, filling.
Equipment in use in each area
Dimensions and layout of access and loading routes
Scheduled development activities Equipment in use in each area
Dimensions and layout of access and loading routes
Auxiliary ventilation system
Haulage routes Equipment in use in each area
Dimensions and layout of access and loading routes
Rock handling including:
Rock breaking Crushing, Loading station,
Hoisting
Equipment in use in each area
Dust dilution/collection/removal
Operator protection
Servicing areas including:
Maintenance workshops, Electrical substations,
Pumping stations, General
maintenance and construction work areas,
Refrigeration plants
Equipment in use in each area
Dust & fume dilution/collection/removal
Fire protection
Operator protection
Storage and supply areas:
Explosive, Diesel fuel, General
consumables and stores
Equipment in use in each area
Dust & fume dilution/collection/removal
Fire protection
Operator protection
Lunch rooms and waiting areas Airflow
Temperature
Obviously not all of the above locations/activities will apply to all mining operations.
Circuit design is ultimately determined by the production schedule.
The siting of dedicated ventilation airways will be determined by mining methods and production
design layouts. The determination of total mine airflow is only the first step toward a mine
ventilation system and requires the most attention because it is at this point that the final and
cost to the operation will be determined.
Ventilation planning and design is complicated and those who are responsible for ventilation
and believe otherwise have the potential to cost their organisation a great deal of money.
Example
Lets consider a new narrow vein mine with an optimum production rate of 300,000 tonnes per
year. It has been decided that predominately one pass ventilation will be required. Ore will be
loaded from the stope to an orepass from where it will be trucked in the main decline out of the
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mine. Backfill will delivered to the stope void using a 10 tonne truck. Assume a “ROT” for
narrow vein mechanised mine to be 0.5 m3/s per 1000 tonne. (0.5 x 300 = 150 m3/s)
The production and development schedule has been evaluated and the active locations during
each month have been determined. In this case an active location is taken as any development
face, raise bore location, production drilling location, production bogging location or stope filling
location. The number of active locations during any one-month is then summed. For this
example assume the results shown graphically below.
Active Locations
0
20
40
60
80
100
120
140
160
0 6 12 18 24 30 36 42 48 54 60
Month
Number
Development Raisebore Mining Backfill
After consideration to diesel exhaust emissions, dust and heat etc. the air quantities for each
activity have been determined as:
Activity Airflow (m3/s)
Decline development 80
In ore Development 12
Production 5
Raiseboring 5
Backfill 8
This airflow was then allocated to the respective activity and summed for each month.
This particular schedule will require primary airflow ranging from 80 m3/s during initial
development peaking at 227 m3/s during the start up of production then tapering off to less than
50 m3/s toward the end of the scheduled period.
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Airflow Requirements
0
50
100
150
200
250
6 12 18 24 30 36 42 48 54 60
Month
Quantity (m3/s)
Development Raiseboring Mining Backfill
Initial Development
Production
Start up Production Production Taper
191 m3/s
157 m3/s
There are four distinct periods that require some comment.
1. Initial development (month 1 to 12) to establish the access decline and the exhaust
ventilation shaft will be ventilated with secondary fans and will not form part of the primary
ventilation requirements.
2. Production Start-up (month 13 to 18). The peak airflow during this period is 227 m3/s. This
peak at the end of initial access development and the commencement of production is
common. This occurs because of the flurry of activities necessary to bring the mine into full
production in as short a time as possible.
The difficulty of this period is the cost for providing extra air when it is not required over the
longer term. Most designers opt to bypass this period and attempt to manage their way
through it. (Not a good time to be the ventilation officer). The alternative is to purchase and
install primary fans with some form of control that will enable the airflow requirements to be
“turned down” after production settles down.
3. Main production period (month 19 to 46). In a new mine there will be ‘teething problems’
and the production and development schedules will undergo constant updating before
eventually settling down. With input from the ventilation engineer the airflow peaks and
troughs will be smoothed settling somewhere around the median of 160 m3/s. (Always
round UP its hard enough to get so don’t give it away).
4. End of the life of the project (month 47 onwards). A detailed mining schedule is only valid
until the next update. Any schedule beyond three years should be viewed with some
scepticism. Because of increasing geological information, changing economics etc most
mine schedules beyond year three do not have the same confidence level contained in the
three-year plan. Because of the very nature of mining there will be an end and the tapering
off period will vary from mine to mine, and year to year.
7.2 Primary Ventilation Fans
Primary ventilation fans are found in many different sizes, types and locations. For example
they could be any combination of
• Forcing or exhausting.
• Axial or centrifugal.
• Single or multiple fans.
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• Single or multiple shafts.
• Horizontal or vertical mounted.
• Horizontal or vertical discharge.
• Surface or underground.
Each has their advantages and disadvantages.
Some are designed to meet predicted requirements and some are found lying around and
installed in the hope that they will provide the necessary airflow.
The preference of any particular combination is usually a matter of individual choice driven by
past experience, environmental issues and capital cost.
Most fans are selected to meet the immediate short-term budget and not the long-term future
ventilation requirements of the mine. Because of this they are not suitable for the life of the
mine and have to be either upgraded or replaced both very expensive options.
7.2.1 Surface Fan Installation Arrangements
Primary ventilation fans are arranged in many different ways some of which are depicted in the
figure below.
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7.2.2 Underground Primary Fan Arrangements
Primary ventilation fans may also be
installed underground. The
underground locations could be at the
bottom of the shafts or at some
intermediate point between the
surface and the bottom of the mine.
Although underground fans
installations are predominantly the
axial flow type, centrifugal are also
used.
The use of primary fans installed
underground has generally been
avoided for a couple of reasons:
1. Excavation and preparation of the site is costly.
2. If the fan fails access for repairs is restricted because the only ventilation flow is from the
natural ventilation pressure.
3. Construction is more difficult as the use of large mobile cranes is restricted.
7.2.3 Forcing or Exhausting
The vast majority of primary fans installed in
Australian mines are located on the surface to
exhaust air from the mine.
The choice of location is usually made as a
matter of convenience rather than any
specific engineering aspect. For example a
simple mine will have one primary intake
airway (be it shaft or decline) and one primary
exhaust shaft.
Exhaust shafts are by their very nature dirty
and the air contaminated with dust and fumes
making it undesirable for the movement of
men and equipment as it is much more
desirable to enter the mine in clean intake air.
If the fan is installed in the intake airway it limits
the accessibility to the shaft for the movement
of men and equipment.
7.3 Circuit Booster Fans
Circuit booster fans are used to alter the
pressure distribution and supply ventilating air
to specific areas of the mine.
They are often necessary when a mine has
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Stope Advance
Stope Void
been developed below the mining area defined in the original mine designs.
They are also misused in many mines as a method to supply air to areas of mines in an attempt
to counteract the poor management of the primary ventilation circuit.
Poor selection of fan location and capacity has potential to create more problems that they
solve. For example they may result in recirculation of large quantities of air, as they tend to
pressurise the exhaust system forcing the out to flow back into the primary intake airflow.
7.4 Primary Ventilation Circuits
Primary ventilation circuits are designed to supply fresh air to workplaces and dilute and remove
contaminants resulted from the mining processes. The time required to flush blasting
contaminants from the mine can range from a few minutes up to many hour.
The effective and timely removal of contaminants particularly after blasting is high on the priority
lists for mine managers, as time lost on production waiting for fumes to
clear is very costly.
There are two basic circuits used
1. Parallel circuits, and
2. Series circuits.
Both have their advantages and
disadvantages and many mines use a
combination of both circuit types.
7.4.1 Parallel (One Pass) Circuits
Parallel circuits adopt a general
design philosophy of ‘one pass, flow
through’ air.
This concept ideally lends itself to
large tonnage stopes or stoping
blocks and may not be suitable for
use in some narrow vein steeply
dipping orebodies. In this situation, a
circuit is established to direct air from
the primary intake circuit over the
areas of activity and exhaust directly
to the return air circuit. It is usual for
return air circuits to be areas requiring
minimal access and therefore all
contaminants are removed directly
from the mine. This is particularly
important in the event of a fire
occurring in the work place as it
minimises the areas (and personnel)
affected by the smoke, gases and
particulates.
With multiple level accesses, it is
necessary to control the air entering
the activity area (usually by regulation
of the exhaust). The down side of this
To Primary
Exhaust
To Primary
Exhaust
TYPE A: End Access
TYPE B: Central Access
Not Longer Than 250m
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type of circuit is that short-circuiting occurs each time a stope is opened or a ventilation control
is damaged.
Whilst providing improved conditions in work areas, these circuits also decrease re-entry times
but only if the access does not form part of the return air circuit.
It is usual to provide airflow to all scheduled production locations, even when not in use. This
provides flexibility to move from one location to another without the necessity to adjust
ventilation circuits. Mine operators often plan to minimise total airflow by campaigning air from
one location to another, however experience tells us that this seldom, if ever, occurs and all
locations operate with less than desirable airflow rates.
For this type of circuit to be successful, ventilation controls must be well maintained.
7.4.2 Secondary Fans and Parallel Ventilation Circuits.
Because of the geometry of narrow vein orebodies, it may prove difficult to maintain the
philosophy of flow through ventilation without the need for extensive development and therefore
the use of secondary fans during production must be considered.
This type of circuit is normally adapted to narrow vein orebodies with long strike lengths. LHD
tramming distances are usually kept below 250m. In these circuits, air intakes via the access
(decline) and is returned to the primary exhaust at each orebody access.
7.4.3 Series Ventilation Circuits
A system most often used in narrow plunging orebodies, is a simple series ventilation circuit.
These circuits rely on secondary fans to ventilate in orebody development and production with
the return air from these activities being mixed with the primary intake airflow and re-used at the
next activity.
This type of circuit has the advantage of simplicity of control and minimal development for
ventilation.
Although the utilisation rate (total airflow to the working areas / total airflow through the mine)
can be as high as 75%, it is typically 65% with 40% not uncommon.
These circuits have disadvantages including (but not limited to):
• The need for high pressure fans (hence higher power costs),
• Low fan power efficiency for the majority of the operating period (the fan only
operates at maximum efficiency in a fully developed mine).
• Decreasing airflow with depth, and
• Contaminants from activities (and fires) affect all down stream personnel.
Although not encouraged, it is recognised that this is the system used in development areas and
in reality will continue to be used in some production areas.
7.4.4 Use of Stope Voids as Airways
There is often a misconception that stope voids will upcast airflow causing much frustration
when the air downcasts through the stope.
Because stope voids form connections between levels they create airways parallel to the ramp
or intake shaft. Generally the pressure distribution is such that the air in the ramp (intake
system) is down casting and it follows that the air in the stope will also downcast air. Obviously
if the air in the ramp is up casting then it follows that the air in the stope void will also upcast.
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If stope voids are required to up cast air, then it is necessary to design and install the
appropriate ventilation controls to cause the pressure distribution required. In general, if stope
voids are required to upcast, the primary return airway must not be lower than the top of the
stope void.
The reliance on stope voids also has the high probability of production blasting covering the
stope brow thus preventing flow through ventilation.
7.4.5 Recirculation
Recirculation occurs when air is kept within a closed circuit. It should not be confused with the
situation when air is reused, as in series ventilation circuits.
Recirculation occurs when a fan is installed in an airway in which the natural flow of air along
the airway is less than the operating capacity of the fan. In some Australian legislation there is
a requirement to install secondary fans such that the air delivered to the fan is greater than 1.3
times its open circuit capacity. This legislation was introduced in times when small drives (2.0m
x 2.0m) and low capacity fans (5 to 7 m3/s) were common. As a general Rule of thumb an
airflow of at least 1.5 times the open circuit capacity of the fan is required in airways with large
cross sections and higher capacity fans. Even then minor recirculation is possible, depending
upon the siting of the fan in the airway.
Although not recommended, some minor recirculation may be acceptable provided the, work
place temperature, contaminants in the airflow and the clearing time for blasting gasses remain
within acceptable levels.
7.5 Ventilation Controls
Regulators commonly used in Australian mine are rectangular openings in walls that are open
or closed by the placing or removal of boards. Other less common regulators are the louvre
type (either vertical or horizontal), or a sliding door (again either vertical or horizontal).
It is normal for persons to open or close a regulator in order to increase or decrease airflow to a
certain area. Sometimes this is meet with much frustration because the desired airflow cannot
be achieved. This frustration is a result of the lack of understanding of the pressure distribution
across the mine and the fact that altering one regulator effects the WHOLE mine distribution
and not just the area being adjusted.
There is often an attempt to compare the opening in a wall to an orifice (vena contracta). In a
perfect orifice the opening is circular and relatively small compared to the plate (wall), which is
very thin. As the air flows through the opening it converges and at a distance equal to half the
orifice diameter the jet is at its smallest area. This area is called the vena contracta and at this
point the velocity of the air is 1.613 times greater than in the orifice itself. At the vena contracta
the velocity pressure is 1.6132 or 2.5 times higher than the orifice.
Using the equation for velocity pressure
2
VP v
2
= ρ and knowing the velocity we
can determine the velocity pressure.
Similarly if the pressure across the plate
is known, then the velocity in the area of
vena contracta can be determined and
we can back calculate the area of the
orifice.
Considering a mine regulator, the wall
thickness can be proportionally greater
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than the thin orifice plate and the ratio of the area of the opening to the area of the wall is
different and this affects the coefficient of contraction (1.613) that becomes larger as opening
increases in size forming an increase in percentage of the opening to the airway wall. For any
calculation to be reasonably accurate the area of the regulator should be no greater than onetenth
the size of the airway in which it is constructed.
To estimate the area, quantity or pressure at a regulator the following algorithm can be used
P
A 1.2Q
ρ
= Equation 5 Area of a Regulator opening
Where A = the area of the opening (m2)
Q = Quantity of airflow (m3/s)
P = Pressure across the opening (Pa)
ρ = Density of the air flowing (kg/m3)
Like the calculation of shock losses in practice there are so many unknown factors that the
calculated result is seldom correct and further adjustment is required. In practice the airflow
through regulators is set by trial and error until the desired, or at least close to the desired result
is achieved.
It is often the case that operators, to gain increased airflow, adjust regulators. Although they
achieve their desired result it is more often than not that they have caused a detrimental effect
else where in the mine.
All adjustment to regulators should only be undertaken once the airflow distribution across the
mine has been planned and scheduled as it may be necessary to adjust more than one opening
to achieve the desired result for all areas. The suggested method for any adjustment programs
it to go to the bottom of the mine (lowest pressure area) and fully open this regulator. All other
adjustments are then made to this opening. Commence adjustment (setting the desired airflow)
at the top of the mine (highest pressure) and progress to the bottom of the mine (lowest
pressure). By the time you get to the lowest regulator a measurement of the airflow should
show that the airflow is around the quantity required.
7.6 Multiple Access Orepasses
Multiple access and uncontrolled orepasses will, when empty, form an airway parallel to the
decline potentially reducing the airflow on the decline to unacceptable levels. In addition to
short-circuiting airflows, multiple access orepasses are often a significant source of dust. As
rock falls down the pass it generates dust and acts like a pump compressing air in the pass and
forcing it out on to each open sub-level below. Efforts to prevent this by creating a positive
airflow into the ore pass from all access points are generally unsuccessful. The problem is best
managed by sealing, if necessary with temporary seals, all but one of the tipping points. ALL
other openings should be sealed or otherwise connected to the RAR system via a dedicated
airway.
7.7 Secondary Ventilation Systems
There are three predominant secondary ventilation systems used.
1. Forcing,
2. Exhausting and
3. Exhaust overlap
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Forcing ventilation system is the predominant system used in hard rock mines. These systems
provide advantages including
• The fan is located in clean intake air making it accessible for maintenance for maintenance
and repair
• Allows the use of flexible ventilation ducting (less expensive than rigid ducting)
Exhaust ventilation systems require the use of rigid ducting and because of the extra cost
involved with the purchase and installation of
rigid ducting, systems of these types are
seldom used in development headings in
hard rock mining. However, it is the accepted
practice in coalmines and many tunnelling
operations that use cutting or boring
machines.
Re-entry times may be reduced by the use of
an exhausting or exhaust overlap ventilation
system where the air is removed via the
ducting allowing immediate re-entry to the
face in fresh air. Although not commonly
used this system has benefits when
developing long drives.
In its original concept the exhausting fan
was slightly larger than the forcing fan,
was installed clean air and used rigid
ducting. This allowed air to flow in the
overlap position and prevented recirculation.
Because of the cost of rigid
ducting as opposed the less expensive
flexible type this arrangement lost
favour.
In recent times it has been adapted for
use using flexible ducting and two
exhausting fans. The use of two
exhausting fans was simply to
standardise the fans being used.
Because the fans are located in the
developing heading if the was to be a
power failure to both fans then repairs are restricted until the area is ventilated by another fan.
It is also inconvenient to move the position of the overlap as it is necessary to remove and
relocate the exhausting fans.
7.7.1 Installing Secondary Ventilation Fans
Auxiliary ventilation fans are a fact of life and used in ALL mines yet they are still installed where
it is convenient rather than where it is best suited. Poor location of auxiliary fans will result in
recirculation of air and obviously recirculation of contaminants with the resultant “less than
desirable” conditions at the working face. In most cases once the fan is installed it is seldom
relocated, no matter how “bad” conditions become. There have been instances where fans
have been poorly located to the extent that the recirculation is almost 100%. This can go
undetected simply because the fan “is running” and you can feel the breeze at the face.
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The ultimate reason that fans recirculate is the lack of air flowing passed the fan inlet. This can
occur for three reasons:
1. There is insufficient airflow in the drive,
2. The fan is to close to the exhaust
3. The seal around the fan is damaged.
One of the simplest means for detecting
recirculation is to use a smoke tube
(particularly near the backs) to check the
direction of the exhausting airflow.
Sometimes it is extremely difficult to get
close enough to the backs and it will be
necessary to undertake measurements of
air flowing in the drive and in the duct down
stream of the fan. In all cases the airflow in
the drive must be greater than the airflow in
the duct.
In general auxiliary fans should be located
approximately 20m upstream of the drive they
are ventilating, and there is a “rule of thumb”
that the airflow passed the fan should be 1.3
to 1.5 times the open circuit capacity of the
fan. If one or both of these factors does not
apply the fan must be sealed into and
appropriate airway to prevent any recirculation.
Many mine designs now incorporate the
return air rise (RAR) adjacent to the intake air
decline. This allows contaminated air to be
exhausted from the mine without being
reused. (See “Secondary Fans and Parallel
Ventilation Circuits” above). In many of these cases the truck is loaded near the stockpile bay
adjacent to the RAR and care must be taken to ensure that there is a flow of air in this area. In
the same way that the exhausting fans are sized to cause a flow in the overlap, a flow must be
caused between the fan and the exhausting point. If this is not achieved the truck loading area
very quickly becomes contaminated
with high concentrations of dust and
diesel exhaust emission.
Some options are shown in the figure
opposite.
7.7.2 The “Reuse” of Air
As discussed above, it is an acceptable
practice to “reuse” ventilating air. The
success or otherwise of this practice in
force ventilated areas lies in the correct
selection and location of secondary
fans.
Many attempts are made in
development areas to install multiple
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parallel fans in an effort to get
extra fresh air to the working
faces. In the vast majority of
these installations the airflow
passed the fans is insufficient,
resulting in the “recirculation”
of the return air through the
fans. The final result is
increased heat and
contaminants in the working
areas.
There are a number of
solutions the obvious ones
being, increase the primary
intake air, only operate one fan
at a time, construct a seal
around the fans to prevent the
recirculation. Only increasing
the primary airflow would not
restrict the flexibility of multiple
workplaces or cause some
interference with activities in
the area.
Another solution is to relocate
one of the secondary fans.
7.8 Ventilation Duct.
Flexible ventilation duct is an extremely versatile and convenient method of providing secondary
ventilation. Unfortunately, its limitations are very
poorly understood. Some practical tips for
installation include:
• The largest possible ventilation duct size
should be used to lower resistance and hence
reduce leakage. In almost all cases, using
duct size which is smaller than the fan
diameter is not acceptable.
• No one seems to get particularly concerned
with holes and rips in ventilation duct. Like
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compressed air leaks, the very high cost of these problems are often not appreciated, or are
simply ignored. It is a common experience to almost have one’s head blown off by leakage
from a ventilation duct near the fan and to then be asked why there doesn’t seem to be
much flow at the face.
• The importance of ensuring adequate clearance between mobile equipment and the
duct is also not clearly understood. This wasn’t the case in the old days when steel duct
was used! The equipment and drive size must be selected so that there is no chance of the
duct being hit.
• The importance of installation standards
is also not often appreciated. The knocker
line must be hung straight from lined up
attachment points (not from the nearest
convenient split set). Sharp corners require
the use of specially manufactured elbow
pieces. Every eyelet must be used when
connecting “Lo k®” or similar joints. Care
must be taken to ensure that “Minsup®”
clips are not left in the ends of ventilation
duct when new bag is connected. The last
one or two ventilation ducts should be
sacrificial “face bags” to avoid the rest of
the duct sustaining blast damage.
• There seems to be a common belief that an unlimited number of “T” pieces can be
connected to a ventilation duct and left open. This is incorrect. There is usually only
sufficient flow to ventilate one face, possibly two and all other “T” pieces should be tied off.
• There is a perception that “more air” can be supplied to the face by using a more powerful
fan. With electrical power costs of 10¢ per kWh, and 170 kW power draw, a fan would
consume $149,000 per annum in power costs. It is not unusual to find 1400 mm diameter
fan and duct installations, which over 500 m deliver only 7 m3/s to the face! This is
insufficient flow for even one LHD. For comparison, 23 m3/s can be delivered, with very well
installed 1,200 mm low leakage ducting using a 1,200 mm diameter, 110 kW fan.
• Use of the low leakage ducting described above can also enable large amounts of capital to
be saved. With the use of low leakage, well installed flexible ventilation ducting it is possible
to advance development headings in excess of 2000 m. This may result in the savings of
many millions of dollars by not having to excavate ventilation airways at various points along
the development heading.
7.9 Duct Leakage
Generally, the prime consideration for the purchase of ventilation ducting is cost, not only with
the original purchase price but also with the cost of transport and handling. Secondary to this is
the ease and convenience of installation and thirdly there may be occasional consideration of
the material friction factor but very seldom is any consideration given to the rate of leakage from
the duct.
Prior to the 1960’s the ventilation ducting of choice was usually a rigid type manufactured from
galvanised iron or plastics of some type.
Lay-flat flexible ventilation ducting has been used extensively in Australian Mines since around
the middle of the 1960’s. In it’s original form it was manufactured from a terylene material with
wire hoops sewn into each end. These hoops were overlapped one inside the other and
prevented from coming apart by tying a length of wire around the duct in between the hoops.
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Prior to the introduction of trackless diesel equipment auxiliary fans had relatively low airflow
rates (6 to 7 m3/s) and low pressures (2000Pa). Since the late 1960’s and the introduction of
diesel powered trackless equipment, airflow requirements for auxiliary fans have increased and
as a consequence so have fan pressures. In response to a need to lower operating costs in the
1970’s different materials were introduced as was a different type of join (spigot type). This
development process has introduced three potential areas of leakage, the material, holes
created for stitching the seams and, the type of join.
The requirement for increased airflow has grown to the extent that auxiliary ventilation fans now
provide up to 50m3/s at pressures in excess of 4000Pa resulting in increased leakage
quantities. The air that leaks from a duct is a function of many variables including; type of
material, type of joint, performance of the fan, and the number and size of holes in the duct.
The combined affect is reduced airflow discharged to the working face.
Many of the early development problems have been recognised and material coatings have
been improved. Stitching has been replaced with welding and the hoop type joint has been
resurrected and improved by holding it together with a specifically designed clamp. These
improvements have all come at increased purchase price.
The high leakage rates from some spigot low cost ducting currently available in Australia limits
its use to lengths of up to 400m in some cases this can be as little as 200 m. After this point
leakage from the system can be as high as 50% of the air produced by the fan. The leakage
rate in a well installed and well maintained low cost ducting may cause only slight
inconvenience in short (less than 400m) development headings and the use of the lower cost
stitched, ducting and the spigot joint continues to serve this purpose. Because most mines
have spare development fans any
excessive leakage is “fixed” by installing
an additional fan either in series or
parallel.
Although ventilation ducting with a high
leakage factor may be considered
suitable for short development
headings, its use in long development
headings is questionable. There are
some who believe that the introduction
of extra fan power (i.e. another fan in
the duct) will deliver the required
quantity of airflow to the face.
In Section 6.3 page 93 we showed how
we could predict the airflow through a
system using a fan performance curve
and the system resistance curve. Calculating the resistance curves for various duct lengths can
also allow us to predict the airflow achievable at various lengths. The shortfall in this
methodology is that it assumes “leakless” ducting.
In the example shown opposite a two-stage contra rotating fan in a “leakless” duct will provide
43 m3/s at 200m and 32 m3/s at 800m. In reality ducts are seldom (if ever) free of leaks and this
is particularly true of flexible, lay flat types of ducting, and as a consequence the pressure
requirement is reduced and the stall point is seldom reached. With leakage in a good quality
ducting the airflow through the fan at a distance of 800m, would be slightly higher (say 38m3/s)
and the flow to the face would be significantly less and would probably be closer to 20m3/s, and
is barely adequate for the operation of one truck and one LHD.
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The air that leaks from a duct is a function of many variables including; type of material, type of
joint, performance of the fan, and the number and size of holes in the duct. The combined
affect is reduced airflow discharged to the working face.
The other important point to note is the air through the fan has not significantly changed and is
operating very low on its performance characteristic curve. With this in mind the location of the
fan becomes important if recirculation is to be avoided. As discussed earlier a general ‘rule of
thumb’ for auxiliary ventilation fans is to locate them in airflow that is equal to 1.5 times the open
circuit capacity of the fan. In the case of a 1400mm diameter fan this is 67.5 m3/s (1.5 x 45). If
this airflow is not available then it may be necessary to install a seal around the fan to prevent
any recirculation.
Vutukuri (1984)18 developed a methodology for use in a computer program necessary to solve
the complexity of two components:
• Duct friction resistance
• Duct leakage resistance
Without leakage, determining duct friction resistance is straightforward enough. It can be
determined simply using Atkinson’s equation. Similarly, consideration of duct leakage
resistance on its own is relatively simple. The leakage resistance is often expressed as a
leakage flow per unit length for a given duct diameter at a given pressure, or more intuitively, as
an equivalent mm2 of holes per m2 of duct.
The problems commence when the interaction of the above two resistances is considered. In
particular, determining fan pressure requirements for a given flow is difficult, because the flow
rate along the duct is not constant.
The only constant in ventilation duct system is the fact that it will leak but how much is anyone’s
guess.
7.10 Velocities in Primary Airways
The economic optimum diameter of any ventilation airway can be determined by the total
development and operating costs over the useful life of the airway. Other engineering factors
include;
• dust entrainment,
• cooling power of the air,
• pressure losses caused by conveyances (cages in shafts, conveyors in drives, trucks in
declines etc.),
• effect on equipment in the airway (harmonics in rope guides in hoisting shafts),
• activities being undertaken in the airway,
• type of equipment to be used for development of the airway, or
• any combination of these and other factors.
7.10.1 Velocities in Access Drives
There is a generally accepted “guideline’ for a maximum air velocity of 5.0 m/s to be used in
declines and other access drives. The purpose of this is to ensure that dust is not entrained in
18 VUTUKURI, V.S. “Design of Auxiliary Ventilation Systems for Long Drives” Fifth Australian Tunnelling Conference, Sydney (1984)
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the primary airflow and that unnecessarily high ventilation pressures (and consequently high
electric fan power costs) are not imposed on the mine ventilation system.
Velocities above 6.0 m/s can be cost effective over short distances providing that good dust
management systems in place to ensure that:
• dirt in the trucks is sufficiently wetted, and
• roadway surfaces are sufficiently stable (wetted) to prevent dust pick up from tyres being
entrained into the ventilating air.
7.11 Water in Upcast Shafts
The processes of mining rely on the use of water for dust suppression and air-cooling. As the
ventilating air flows through the mine workings it takes up this water as vapour until it becomes,
to some degree, saturated. As this saturated air rises from the bottom to the top (surface) of the
exhaust shaft the barometric pressure is constantly decreasing and as a consequence the
temperature of the air, and water vapour mix, is also decreasing. This results with water
condensing out of the air.
This condensation process is evident by a ‘foggy’ appearance of the air. This ‘fog’ is small
droplets of water that may coagulate to form larger droplets.
A large water droplet of greater than 3mm diameter has a terminal velocity of 8.0m/s. If the
velocity of the air is less than 8.0m/s then the water droplet will fall down the shaft and if the
velocity is greater than 8.0m/s the water droplet will be pulled up the shaft. Obviously if the
velocity of the air is 8.0 m/s it stands to reason that the water droplet will neither rise nor fall and
will remain suspended at this critical velocity.
As droplets accumulate they form a ‘water blanket’ that increases the resistance in the shaft
thus lowering the quantity of air flowing. At times this may be sufficient to cause the fan to go
into stall. As the stall point is reached the water blanket drops out and the fan may recover to
recommence the cycle of water blanket formation and drop out. If allowed to continue
unchecked this may eventually cause mechanical failure of the fan.
Although 8.0m/s is seen as the critical velocity this is not an exact figure and will vary according
to a number of factors. Water droplets may coagulate to larger droplets, and the air velocity in
the shaft will vary inversely according to the density. This results in a ‘critical range’ of air
velocity between 7.0m/s and 12.0m/s in which this phenomenon occurs.
When designing exhaust ventilation shafts in deep mines with potential for saturated air the
velocity range between 7.0m/s and 12.0m/s should be avoided.
The effect is not as pronounced in shallow dry mines but the ‘critical velocity range’ should be
avoided if there is potential for ground water to be present in the shaft.
7.12 Equipment Movement in Underground Airways
Trucks (or other vehicles) travelling in underground drives can cause pressure losses up to
200Pa depending upon a number of variables such as the size of the vehicle in relation to the
cross sectional area of the drive and the direction of the airflow in the ramp. Consider the
following:
A truck having dimensions of 3.1m high x 3.5m wide is travelling at 20kmph against an airflow of
100 m3/s in a drive 5.5 m high and 5.2 m wide.
The shock loss pressure drop can be calculated from
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( ) 2
A T
Shock 1 y
y V V
2
P  


 



×
ρ
=
Where: PShock = the shock loss pressure drop (Pa)
ρ = the density of the air (kg/m3)
y =
Area of the Drive (m )
Area of the truck (m )
2
2
VA = Velocity of the air in the unobstructed drive (m/s)
VT = Velocity of the truck relative to the airflow (m/s)
Assuming standard air density of 1.2 kg/m3 then the pressure drop caused by the truck is 185
Pa.
i.e. Area of the truck = 18.85 m2
Area of the drive = 28.6 m2
y = 0.659
VA = 3.5 m/s
VT = -5.56 (- since the truck is travelling against the airflow) m/s
Substituting these values into the equation:
( ) 2
Shock 1 0.659
0.659 3.5 5.6
2
2 . 1 P 





× − −
= ×
= 185 Pa
In the same situation and only 25 m3/s the pressure drop would be 94 Pa
In some cases the airflow in the decline may in fact reverse direction as the truck travels in the
drive, usually returning to “the norm” once it has passed. This only occurs in areas with
extremely low ventilating pressure and other connections that operate in parallel with the
decline. This fluctuation of airflow may also occur in workings connected via a hoisting shaft.
The consequences of any fluctuation cause by vehicles travelling around the mine are seldom
great enough to warrant rectification. The main area for concern is when trucks travel up ramps
as they do so in an envelope (recirculation) of engine exhaust contaminants. This occurs
irrespective of the direction of airflow but when the direction of flow is UP the ramp and the truck
is travelling in the same direction there is an increase in the potential for contaminants to build
up to levels that will exceed the recommended TWA limits.
For example, if we assume trucks travel up ramps at a speed of 10 kph (2.8 m/s) and the airflow
in the 5.5 m x 5.5 m ramp is 50 m3/s then the air velocity in the ramp is 1.7m/s. This gives a
relative velocity between the truck and the air of 4.5 m/s, with the air flowing down the ramp and
1.1 m/s with the air flowing up the ramp. In the second case, it should be noted that the truck is
travelling faster than the air.
To achieve a reduction in exhaust emission contamination levels the direction of airflow should
be down the ramp. This increases the relative velocity between the air and the truck, which in
turn increases the turbulence of air around the vehicle causing the envelope to break up and
significantly reduce the level of exhaust emission and dust contamination to which the operator
is potentially exposed.
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In development headings causing air to flow down the ramp will require the installation of an
exhaust overlap ventilation system. This has the disadvantage of providing contaminated
airflow to the working face whilst the trucks (or for that matter any other diesel powered
equipment) are operating in the ramp.
Because of the relative periods of time
spent at the face and the hauling in
the ramp, forcing ventilation systems
are almost always used.
To reduce potential for exhaust
emissions to build up to
concentrations in excess of the MAC’s
while travelling up ramps it is vital to
ensure that engine performance is
matched to the work rate required of
the vehicle (i.e. the power to weight
ratio). It is also essential to ensure
that the engine is tuned and
maintained to manufacturers
specifications.
7.13 Re-entry after Blasting in a Development Heading
Blasting is an intermittent activity carried out when people are removed from the underground
areas affected by the fumes (gases and dust) generated by the blast. The concentration of
contaminants in the affected parts of the mine often greatly exceeds the maximum TWA
concentrations for a brief period of time until the fumes are diluted by the ventilation currents
and removed from the mine. These peaks generally have duration of a few minutes and occur
whenever blasting takes place.
Planning and scheduling for rates of advance are usually based upon experience of crews and
the normal geotechnical and materials handling requirements of short development headings.
For this reason estimates for long heading development rates can and are usually overstated as
most neglect to consider the extended re-entry time as a direct result of the length of the
heading and the subsequent time required to clear the blasting fumes.
For example consider a long development heading 5.0m x 5.0m x 500m, and an airflow of
25m3/s delivered in a forcing system. The velocity in the drive (Q/A) is 1.0m/s. Assuming that
all contaminants are contained in the plug that is being removed and all of the 25m3/s is
sweeping the face the minimum clearing time from the face to the primary ventilation circuit,
would be only 8.3 minutes and in the scheme of things this is not considered to be of any
concern.
Because of duct leakage, duct
discharge velocity profiles and the
distance of the duct discharge from the
face it can be expected that these times
could be increased by a minimum of 15
to 25 minutes (i.e. re-entry times from
20 minutes up to 35 minutes).
Gases produced by blasting are to a
significant and variable degree either
trapped in the blasted muck pile to be
released slowly as the muck is removed
or dissolved in water naturally present in
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the mine or used for dust suppression. It is estimated that up to 60% of the gases produced by
blasting are either trapped in the muck pile or removed by dissolution.
Gases produced from blasting include carbon monoxide (CO), carbon dioxide (CO2), ammonia
gases (NH3), sulphur dioxide (SO2) and oxides of nitrogen (NOX). The key contaminants are
considered to be CO and NOx.
Factors that impact on gas production include product manufacture and quality control, the
degree of confinement of the explosives (i.e. hole burden and spacing), moisture in the drill
holes and the effectiveness of the initiating process.
A wide range of estimates exists of the amount of gas produced per kg of explosives used. In
the case of CO references have been found to gas production rates between 1.25 litres per kg
and 50 litres per kg. In the case of oxides of nitrogen the range is from 19 litres/kg to 33
litres/kg.
Skochinsky (1969) and De Sousa (1993) have developed algorithms for calculation of gas
clearance after blasting in development headings. These algorithms are useful to gain some
idea of what to expect but should not be relied on as the sole indicator of when it is safe to
return to a blasted area. The only real indicator for safe re-entry is accurate measurement of
the contaminants involved.
Skochinsky (1969)
( )3
1
2
MExplosives VDrive
Q
t = 7.8 × × Equation 6 Blast Gas Clearing (Skochinsky (1969))
t =Time for the air to clear (seconds)
Q =Quantity of air sweeping the face (m3/s)
MExplosives=Mass of explosives used in the blast (kg)
VDrive=Volume of the excavated drive prior to the blast (m3)
EXAMPLE
Consider a development heading 5.5m high and 5.5m wide 350 metres from the secondary fan.
Assuming 200 kg of explosives used in the blast and 20 m3/s is sweeping the face then
Volume of the drive = 5.5 x 5.5 x 350 = 10587.5 m3
And ( )3
1
200 10,5872
20
t = 7.8 × ×
= 1,097 seconds
= 18.2 minutes
deSousa (1993)
( )
( )
2A Et
C Q
4Et
x vt
x,t
2
π
=
− −
Equation 7 Blast Gas Clearing (deSousa (1993))
C(x,t) = Gas concentration (multiply by 106 to get ppm)
t = Time after blast (seconds)
x = Distance to face (m)
Q = Quantity of Contaminant (m3)
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A = Cross sectional area of the drive (m2)
V = velocity of the ventilating air (m/s)
E = Dispersion co-efficient (m2/s)
r
E 28.8 v Rh λ
= × × × λ
v = Velocity of the ventilating air (m/s)
Rh = Hydraulic Radius (Area/[height + width]) (m)
λ = Friction Factor of a smooth walled drive. (0.008)
λr = Friction factor of the drive (if unknown use 0.015)
This methodology will determine the concentration of the particular gas at a particular time.
EXAMPLE
Consider the development heading above and calculate the gas concentration for NO2 19
minutes after the blast.
t = 19 x 60 = 1,140 (seconds)
x = 350 (m)
Q = 0.5147 (m3) from table below
Gas
Gas
produced /
kg of
explosives
(kg/kg)
Density
of gas
(g/m3)
Volume of gas
produced / kg
of explosives
(m3)
Mass of
explosives
used in firing
(kg)
Volume of
gas
produced
from firing
(m3)
CO 0.0163 1.25 0.01304 200 2.61
CO2 0.1639 1.977 0.082903 200 16.58
NO2 0.0035 1.36 0.002574 200 0.5147
A = 5.5 x 5.5 = 30.25 (m2)
V = 20/30.25 = 0.66 (m/s)
E = Dispersion co-efficient (m2/s)
0.015
E = 28.8 × 0.66 × 2.75 × 0.008 = 38.17 (m2/s)
Rh = (30.25/[5.5 + 5.5]) = 2.75(m)
λ = 0.008
λr = 0.015
And ( )
( )
2 30.25 3.142 38.17 1,140
C 0.5147
4 38.17 1,140
350 0.66 1,140
x,t
2
× × ×
=
× ×
− − ×
= 0.000005 (m3) or 5 (ppm)
EXAMPLE 2
This methodology was used to estimate the clearing times for nitrogen dioxide (NO2) from a
development heading 5.5m high and 5.5m wide 350 metres from the secondary fan.
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The gas NO2 has been used because of its high toxicity, low exposure standard and the fact
that it will take longer to dilute to safe levels than other gasses produced by blasting. The results
are shown graphically below.
This clearly demonstrates the importance of adequate airflow sweeping the workplace.
Howes (1982)19
The following methodology has been adopted by some Australian mining companies as a
standard to establish the earliest re-entry time.
 

 

× 



= 
R
C
C G
ln G
Q
t V
Where: t = Time to achieve the require gas concentration (seconds)
V = Volume of the gas filled space (m3/s)
Q = Quantity of air sweeping the face (m3/s)
GC = Initial gas concentration (ppm)
GR = Gas concentration required (ppm)
There are a number of steps required before the use of this equation
1. Calculate the fume throwback distance. (This is the initial volume occupied by the blast
fumes
( )
F D A
L K M
A × ×
×
=
Where: L = Length of the fume throwback (m)
K = Constant (Assume 25 for development headings)
19 HOWES. M.J., “Advanced Ventilation Workshop” (1998)
0
5
10
15
20
25
30
35
40
1 11 21 31 41 51 61 71 81 91 101 111 121 131
Time (minutes)
Concentration (ppm)
Safe re-entry level
2ppm
10 m3/s
20 m3/s
5 m3/s
15 m3/s
Blasting Fume Clearing Times
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M = Mass of explosives used. (kg)
FA= Face advance (m)
D = Density of the rock (kg/m3)
A = Area of the face (m2)
2. Calculate the Volume of gas produced (m3)
(A) (B) (C)
(A) x (B)
Gas Gas produced /
kg of explosives
(kg/kg)
Density of gas
(g/m3)
Volume of gas
produced / kg of
explosives (m3)
CO 0.0163 1.25 0.01304
CO2 0.1639 1.977 0.082903
NO2 0.0035 1.36 0.002574
3. Calculate the concentration of gas in the in the drive. Assume the drive volume after the
blast
10 (ppm)
Volume of the Drive (m )
G G (m /kg) Quantity of explosives (kg) 6
3
3
PRODUCED
DRIVE ×
×
=
4. Calculate the time taken to mix in the drive. (At this step assume perfect mixing of the
gas and the ventilating air)
Airflow (m / s)
Time for mixing (s) [Length of drive (m) – Distance of throwback zone (m)] Area of drive (m ) 3
× 3
=
5. Calculate the time for dilution to the exposure standard using the Blast Clearing
Equation above  


 

× 



= 
R
C
C G
ln G
Q
t V
6. Determine the total time for clearing in seconds
Clearing Time (s) = Time for mixing + Time for dilution
EXAMPLE
Consider a development heading 5.5m high and 5.5m wide 350 metres from the secondary fan.
Assuming the face advance is 3.0m, the rock density is 3.1 (kg/tonne), 200 kg of explosives
used in the blast and 20 m3/s is sweeping the face.
1. Calculate the fume throwback distance
( )
F D A
L K M
A × ×
×
= = 97.75 (m)
Where: L = 350 (m)
K = 25
M = 200 (kg)
FA= 3 (m)
D = 3.1 (kg/m3)
A = 30.25 (m2)
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2. Calculate the Volume of gas produced (m3/kg)
(A) (B) (C)
(A) x (B)
Gas Gas produced /
kg of explosives
(kg/kg)
Density of gas
(g/m3)
Volume of gas
produced / kg of
explosives (m3)
CO 0.0163 1.25 0.01304
CO2 0.1639 1.977 0.082903
NO2 0.0035 1.36 0.002574
3. Calculate the concentration in drive
6
DRIVE 10
10542
G 0.002574 200 ×
×
=
= 49 ppm
4. Calculate the mixing time
20
Time for mixing (s) [350 – 97.5] 30.25
×
=
=382 seconds
= 6.4 minutes
5. Calculate the time taken to dilute the gas.



×  



= 
2
ln 49
20
t 10542 C
= 2,064 seconds
34.4 minutes
6. Calculate the time taken for re-entry
Time before re-entry = 6.4 + 24.5 = 30.9 minutes
SUMMARY OF EXAMPLES
Method Result (minutes)
Skochinsky (1969) 18.2
De Sousa (1993) 25.9
Howes (1988) 30.9
Re-entry times are best established by measurement of the relevant contaminants combined
with the local knowledge and experience of the people directly involved in the blasting and reentry
process.
7.14 Gases from Sulphide Orebodies
See Section 2.2.4 Explosive Dusts page 12
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7.15 Gasses from Diesel Engines
See Section 2.5 Diesel Engines page 41
7.16 Control of Mine Gases
7.16.1 Prevention
Preventing the formation of harmful mine gasses is a simple and obvious control method.
Examples of this approach include:
Diesel exhaust gas concentrations can be reduced by using high efficiency engines with
electronic mixture control or by using alternative power sources (e.g. electric LHD’s, shaft
hoisting etc).
Ammonia formation can be controlled by reducing ANFO spillage and shot-crete rebound.
7.16.2 Extraction
The aim with this option is to extract gasses via as direct a route as possible to the surface.
Examples include:
Uranium mine – ventilation system. Radon daughter laden air is extracted from the stope void
and flows directly to the exhaust ventilation system.
Coalmine – gas drainage. Methane drainage boreholes are bored into gassy strata in advance
of the face. Gas is extracted from the boreholes.
7.16.3 Isolation
This involves separating personnel from areas where high concentrations of harmful gasses are
known to be present. Examples include:
Uranium mine and coalmine return airways.
Mine locations that may be affected by blast fumes at firing time.
7.16.4 Containment
It is sometimes possible to seal off areas (dangerous gas concentrations occur) from the rest of
the mine. An example includes:
Bulk-heading off old workings. (i.e. those in which harmful gasses are known to occur). Sealing
agents can include shot-crete or urethane coatings. It is normally wise to vent the voids to
exhaust, since even with extremely efficient sealing, it is difficult to control leakage due to
changes in barometric pressure etc.
7.16.5 Dilution
Dilution of undesirable gasses with fresh air represents a simple and effective method of control.
The required airflow rate can be calculated from the following formula:
AC NC
Q
Q g

= Equation 8 Dilution Equation
Where:
Q = The required fresh air flow rate (m3/s)
Qg = The gas flow rate (m3/s)
AC = The target gas concentration
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NC = The normal concentration of the gas in fresh air
Example:
Methane at 2% concentration has been detected in air flowing out of a stope via a drill drive.
The measured flow out of the drill drive is 2 m3/s. How much fresh air needs to be supplied to
the drill drive (with a flame-proof fan and anti-static vent duct) to bring the methane
concentration down to 1%?
Methane flow rate = 2 m / s 0.04 m / s
100
2 × 3 = 3
The normal concentration (NC) of methane in air is 0.
Substituting…
4m / s
0.01 0
Q 0.04 = 3

=
i.e. 4m3/s of fresh air is required to dilute the methane to 1% concentration.
7.17 Effect of Atmospheric Changes on Mine Strata Gases
Many underground coalmines have goaf areas (old worked out areas), which receive little or no
ventilation. These areas are usually sealed off and this allows the accumulation of gases such
as methane. As long as the gases remain within the goaf area there is no safety problem.
However, when the atmospheric conditions change and the pressure falls, the pressure on the
gas decreases and therefore the volume of the gas increases (Boyle’s Law). Because the
volume of the goaf cannot change some of the gas will be forced out into the ventilation system.
Designing the ventilation system to prevent this must include;
(a) Sufficient airflow to dilute the gases to harmless mixtures
(b) Directing them away from men and machinery.
The quantity of air will depend on:
(a) The total volume of the goaf. The greater the volume of the goaf the more gas will be
forced into the ventilation system,
(b) The rate of fall of the atmospheric pressure. The actual size of the fall is less
important than the rate at which it occurs, since the pressure drop determine the rate of
expansion of the gas.
The best way to ensure the gas is directed away from the work areas is to ensure that the goaf
is connected directly to the exhaust air system.
It is important to note that even if the goaf is sealed off they will nearly always “breath”.
7.18 Spontaneous Combustion
Although coal is considered to be a commonplace commodity it is extremely complex and differs
widely from seam to seam. Certain coal seams and carbonaceous shales have a tendency to
oxidise at normal working temperatures and others are prone to spontaneous combustion. A
great deal of work has taken place over many years in an attempt to identify the exact
components of coal that cause self-heating but as yet it is still not fully understood.
What is known is seams liable to spontaneous combustion are those,
• thick or a composite,
• are of inferior quality,
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• are low rank (i.e. relatively young),
• contain pyrites,
• readily adsorb and desorb moisture when there is a large gap in humidity between the coal
and the air, and
• have a high moisture holding capacity
The spontaneous combustion process starts with oxidation at normal temperature (say 20°C)
and the rate of oxidation is dependant upon
• the temperature,
• percentage of oxygen and,
• humidity
If the oxygen used up in the oxidation process is not replenished the rate of oxidation will slow
and eventually cease when the content fall below 2%.
The heat produced by the oxidation process causes a rise in temperature, increasing the rate of
oxidation and again increasing the temperature of the coal. The presence of fine coal increases
the rate of oxidation due to the relatively large surface area exposed to the ventilating air. Once
the temperature reaches 80 – 100°C the oxidation process rapidly increases until the
temperature reaches 650°C and ignition occurs. If sufficient heat is removed in the ventilating
air then the heating will slow and temperatures will remain below ignition temperatures.
The classic spontaneous combustion occurs in broken ‘dry’ coal of mixed size with sufficient
fines to have a large surface exposure and large lumps that create the voids for the passage of
humid air. The airflow would be in sufficient quantities to replenish the oxygen but insufficient to
remove the heat.
Spontaneous combustion is most likely the result of poor mining practices such as:
• leakage of air through the goaf (this could be the result of high ventilation pressures) or,
• a poor extraction sequence that leaves large stocks of broken coal.
There are a number of signs that indicate a heating is in progress.
1. ‘Sweating’ is the condensation of moisture on the roof, ribs or metal straps and washers
used in ground support. This results from high temperatures in the goaf area driving out the
moisture contained in the coal. This hot now humid air condenses on the cooler surfaces
outside the goaf.
2. ‘Haze’ or ‘fog’ caused when the hot humid air from the goaf condenses in the cooler
ventilating airflow.
3. A ‘smell’ comes under many names such as ‘goaf stink’, ‘fire stink’ or ‘stink damp’.
4. ‘Hot air’ from the goaf is an indication that a heating is well advanced.
5. ‘Smoke’ a fire is imminent.
In a longwall retreat goaf there is a region immediately behind the face in which the quantity of
air is sufficient to remove the heat of oxidation. Deep within the goaf the quantity of air is
insufficient to allow the oxidation process to continue and in fact the oxidation process uses up
the oxygen. Between these regions there is a region that is conducive for spontaneous
combustion. As the longwall face retreats this region also retreats and remains at the same
distance behind the face.
The circuits commonly used in Australia to prevent spontaneous combustion are shown below.
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Bleeder headings must be adopted with extreme caution as they pose a risk of promoting a
heating or conversely having an unventilated goaf allowing a build up of gases that could be
forced out by a change in atmospheric pressure.
Basic Longwall Ventilation (“R” System)
Main
Returns
GOAF
GOAF
Main
Intakes
Bleed Regulator should be kept adjacent to the face line
to minimise leakage into the goaf
R
Bleeder
Heading
Maingate
Tailgate
GOAF
GOAF
Tailgate
Return
GOAF
Back Return System
R
Maingate
Intake
Tailgate
Intake
Return GOAF
Maingate
Intake
Tailgate
Intake
“Y” Return System
7.19 Control of Dust
Once dust is liberated from the rock, it is transported to various parts of the mine in airborne
form via the mine airways. When the relative velocity between a rock surface and the
ventilating air exceeds 5.0 m/s the inspirable dust fraction adhered to the rock surface may be
dislodged and will become entrained into the air stream. In underground mines this may occur
whenever rock is moved from one point to another. The dust may come from two sources the
exposed surface of the rock being transported or, from the roadway surface when dust, picked
up by the tyres of vehicles, is entrained into the airstream. These occurrences are usually
overcome by wetting the rock on either the load being transported or the roadway surface.
Obviously in roadways with air velocities less than 5.0 m/s the problems associated with dust
from tyre pick up is much reduced.
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It is normal practice to apply water from either a water cart or overhead sprays onto
underground roadways to allay dust pick up from tyres. The consequences of these systems
are either not enough water or some times excessive water, particularly from overhead sprays,
that causes erosion of the roadway that in turn requires constant maintenance to the roadway
surface. In open-pit mining the use of additives to water has been used in attempts to reduce
the roadway maintenance requirements. In recent years a number of underground mines have
trailed these additives with varying degrees of success.
Dust suppression agents come generally as water additives and are described by their
respective manufacturers as either,
• Soil wetting agents (detergents),
• Binding agents (surfactants), or
• Compaction enhancers.
Most of these agents have been manufactured with the objective to reduce roadway surfaces
breaking up therefore reducing maintenance costs. It is important to note that all products are
susceptible to water and have the same requirements for well constructed roadways, sloped to
prevent water pooling. Some binding agents cause surfaces to become very slippery and any
use of these products must be carefully considered.
Water is the main cause of roadway erosion and poorly constructed roadways will always have
high maintenance costs irrespective of the use of any wetting or binding agent for dust
suppression.
The three main methods for controlling dust in underground mines include:
• Application of water
• Dilution Ventilation
• Localised extraction ventilation
Water. Application of water is by far the most popular (and usually the most practical) dust
control method. Wetting the dust prevents it from becoming airborne. To be effective, the water
should be applied (by spray jets) to the point at which the dust is liberated (e.g. at the drill bit, on
the roadway, at the drawpoint etc). It is important to try to ensure that the spray water is as
clean as possible. Muddy re-cycled mine water may introduce more dust into the mine when
the water evaporates. Some dust types may be difficult to wet effectively with water and wetting
agents called surfactants may need to be added to the water. A rule of thumb design criterion is
to aim to supply between 1 and 5% of water to broken rock (on a weight for weight basis). It is
not usually considered safe or advisable to apply water at ore handling facilities. This is
because of the serious danger of mud rushes occurring in vertical storage facilities if an
excessive build-up of water occurs.
Dilution. In most underground mines, dust is produced at diverse and often numerous
locations. The amounts of dust produced at the individual locations are sometimes quite small.
In these circumstances, the reduction of dust concentration by dilution with an appropriate
airflow rate is a legitimate method of dealing with airborne dust. Determining the amount of
airflow required to adequately dilute dust on a mine wide basis is surprisingly difficult. The
determination is complicated by the cyclic and dynamic nature of the mining process, the
effectiveness of watering down dust and many other factors. Some design factors, which may
assist, are listed below, however it must be emphasised that no two mines are the same and
application of the factors must be tempered with practical experience:
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• In many mines, it has been found that satisfactory dust levels can be maintained if sufficient
air is circulated to satisfy diesel exhaust dilution criteria (e.g. 0.05 m3/s per kW of rated
diesel engine power).
• In all cases, the minimum air velocity to ensure adequate mixing and dilution must be
greater than the lower limit of turbulent flow (about 0.1 m/s).
• Air velocities in intake airways and in airways in which men travel should be less than about
6 m/s. At higher velocities, increasing problems with re-entrainment of settled dust will be
experienced.
7.19.1 Dust Extraction
This involves exhausting dust-laden air from a point as close as possible to where the dust is
generated, so that dust can be prevented from entering the main intake mine airways. This type
of system is best suited to fixed facilities, such as ore passes, crushers and conveyor belt
transfer points. In underground mines, it is almost always preferable from an overall cost
(capital + operating) viewpoint to exhaust the dusty air directly to an exhaust airway. There may
be some circumstances where the dust may need to be filtered so that the extracted air can be
re-used. The most suitable filtration systems for the underground environment are generally
wet type dust scrubbers, although fabric bag-house type filtration is sometimes also used.
Dust is a result of the disintegration of matter and the size of the dust particle produced is
determined by the impact per unit area. For example striking a rock with a hammer will split the
rock into large pieces forming coarse dust particles. If we were to use the same force using a
chisel it would break only a small piece of the rock into fine dust particles because the force is
directed onto a much smaller area.
Disintegration processes occur in many ways with crushing, grinding, blasting and drilling being
the obvious ones. Other processes that cause dust particles to become airborne are those
involving the transportation of previously broken material by loaders, uncovered trucks and
conveyor belts. The liberation of dust particles to the atmosphere occurs when the velocity of
the air relative to the vehicle is sufficiently high enough to cause dust pick up.
Pick Up Velocities (m/s)
Particle Size(μm)
Quartz Coal
75 to 100 6.3 5.3
Dry 35 to 75 5.3 4.2
10 to 35 3.1 3.2
75 to 100 7.4 6.3
Semi Dry 35 to 75 6.3 5.3
10 to 35 4.2 4.2
Dust is also liberated to the atmosphere whenever broken material is transferred e.g. conveyor,
loading and unloading processes.
The initial step in the design of a ventilation extraction system requires an accurate assessment
of the volume flow required to effectively remove the hazard. If this calculation is incorrect then
all following decisions regarding the ducting, fans and filters are also likely to be incorrect.
Because a suction exhaust type system requires most of the principles of ventilation to be
applied (also that the laws of airflow apply equally to both pressure and exhaust systems) we
will focus our attention on the design considerations for an exhausting system.
Ensuring adequate ‘capture’ of the contaminant is the first consideration and must take into
account the following variables of the contaminant to be handled.
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VELOCITIES FOR CAPTURE HOODS
Type Of Emission Capture Velocity Example
Lazy Emission 0.25 to 0.5 m/sec Fume Control
Low Velocity Emissions points 05 to 1.0 m/sec Manual Activities e.g. Aspiration
Active Emissions mechanical
screens 1.0 to 2.5 m/sec Low speed Emissions e.g. Transfer Points, Vibratory
Screens.
High Velocity 1.0 to 10.0 m/sec High Speed and rotary emissions e.g. Transfer Points,
Vibratory Screens e.g. Sawing, Grinding, Polishing
Wherever possible capture hoods
should be designed to enclose the entire
process source. If this is not possible
the hood should be located as close to
the source as is practical and sited to
make use of the direction of flow of the
particles caused by the process. This
site must be chosen to take account not
only of the thermal properties of the
process but also the effects of gravity so
as to avoid the larger particles falling out
before reaching the influence of the
hood. It is also important to site the
hood in a position that will not allow
personnel to come between it and the
process source.
7.19.1.1 Hood Design
Each hood is designed to control the velocity
in all directions at a given distance from a
point of exhaust. This is illustrated in the
figure opposite that shows the rapid velocity
decrease in the vicinity of an exhaust intake.
The efficiency of exhaust hoods can be
greatly improved if they are shaped to
gradually merge into the ducting and thereby
improving the aerodynamics of the airflow.
Having determined the capture
velocity and angle of the hood the
next step is the siting of the hood and
unfortunately the economic and
engineering importance of this
procedure is usually ignored or at
least misunderstood.
As shown in the figure below, if we
were to hold a piece of string of length
X at the centre of the duct and moved
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it in any direction we would circumscribe a semi-circle and finally a hemisphere of radius X. To
maintain the capture velocity at any point on the surface of this hemisphere we must ensure that
sufficient air is drawn through the duct.
The airflow required to maintain this velocity may be calculated by the equation:
Q = VA3
Where
Q = Exhaust airflow quantity (m3/sec)
V = Required capture velocity (m/sec)
A = Area of the geometric shape (m2)
Dalla Valle in the 1930’s refined this equation to cater for circular or rectangular hood faces:
Q = V (10 x X2+ A)3
Where
Q = Exhaust airflow quantity (m3/sec)
V = Required capture velocity (m/sec)
X = The distance from the hood face to the process source (m)
A = Area of the hood face (m)
However this equation also has its limitations as shown by Fletcher in the ’80’s and should not
be used for rectangular hood face areas with an aspect ratio not equal to 1 i.e. they should only
be used for hood face areas that are either square or circular.
Length
Aspect ratio = Width
Example
What quantity of airflow is required when a hood of 400 mm diameter is placed 320 mm from a
dust source with a capture velocity of 3.0 m/sec
Q = V(10 x X2 + A)
= 3([10 x 0.322] + [3.142 x 0.22])
= 3.45 (m3sec)
If this same dust hood was repositioned 200 mm from the dust source and the capture velocity
remains unchanged then the air volume becomes:
Q = V(10 x X2+ A)
= 3([10 x 0.22 + [3.142 x 0.22)
= 1.58 (m2/sec)
These results shows that by moving the hood closer the volume of air required is halved, and
the cost of the installation is reduced as a smaller fan will be required.
In general hoods should be designed to approximately cover the area (larger never smaller) of
the contaminant source and if this source process is hot, the velocity pressure must exceed the
thermally induced draught.
Although we have shown that the closer the hood is to the source, it must never hinder the
process otherwise the operator may remove it.
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Finally the hood face should be designed to have the smallest possible area that will be
acceptable to the process operator.
7.19.1.2 Duct Design
After having decided the capture velocities and calculated the airflow requirements for each
hood the next step is to determine the optimum transport velocity inside the duct. This velocity
should be such to prevent any settling out of the particles as they are transported through the
system. It should be noted at this point that the transport velocities should be slightly higher in
main ducts than in branch ducts.
Air Velocities for Dust Transport
Type of Dust (or vapour) Branch Duct Velocity (m/s)
Paint fumes 8 to 10
Limestone 13
Metal fumes (welding) 15
Sandblasting 18 to 22
Coal 20
Lead dust 28
Either making use of system damper plates or changing the cross-sectional area of the duct can
achieve control of velocity inside ducts. The use of damper plates is very expensive as far as
system pressure losses are concerned and should be avoided wherever possible, if however it
is intended to expand either the system or the process they must be considered. The other
occasion they are used is when the installation space available prevents the use of the desired
duct configuration.
Duct material, shape, size, branching and bends are all critical as they all have the potential to
provide excessive resistance (pressure loss) and add to the size of the fan. To keep these
pressure losses to a minimum ducting should be streamlined as far as possible by avoiding
sharp bends or sudden changes of cross-section or cross-sectional area. Branch pipe sizes
must then be calculated to provide the volume of air required at each point.
Duct design should provide for maintenance after the plant has been installed and should
therefore provide sufficient inspection and cleaning openings. It is also important to take
precautions against corrosion and abrasion as particles may cause rapid wear in ducts,
particularly bends and it is therefore necessary to provide protection at these points.
7.20 Economics of Airflow
The objective of ventilation economics does not differ from the principles of any investment i.e.
to receive the maximum return on the investment.
Ventilation circuits are designed to fit the operational requirements for a specific mine. Any
economic evaluation is carried out to determine the most cost effective alternative to meet these
requirements.
Alternatives evaluated may not necessarily include the optimum economic situation. Many
times, it is a compromise between the engineering and the scheduling requirements to
determine the most cost effective alternative.
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Ventilation costs should be regularly monitored as changing cost in any of the variables may
alter the cost effectiveness of circuit design.
It is extremely difficult to provide a single set of economic factors for ventilation studies as costs
such as development and power will vary from mine to mine (due to location of the mine from
services), rock type, ground conditions, accounting procedures of the mine etc.
7.20.1 Cost of Airflow
If the cost per unit of power to operate the fans is known and the airflow and pressure losses in
an airway can be determined then it is possible to calculate the cost of power to provide the
airflow.
Example
If the cost of electrical is $0.10/kW hour, what would be the annual cost of power for a flow of
150m3/s in a 3.0m diameter 200m long concrete lined airway?
From Atkinson’s ventilation equation
2
3 Q
A
KCL
P = ×
then determine the pressure losses for the airway (for a concrete line airway assume k = 0.004
Ns2/m4)
2
3 150
7.064
0.004 9.425 200
×
× ×
=
= 481 Pa
the airpower is
AP = P x Q
= 481 x 150
= 72,150 watts
= 72 kW
and the cost of power is $0.10/kW per hour then
Cost of Power = AP x unit cost of power
= 72 x $0.10
= $7.20/hour
or = $63,072/year
If the diameter of the shaft was increased to 4.0m then the cost of power would be:
2
3 Q
A
kCL
P = ×
2
3 150
12.566
0.004 12.566 200
P ×
× ×
=
=114 Pa
and Air Power (P x Q) =114 x 150
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=1710 Watts
or = 1.710 kW.
and cost of power for 1 year would be:
= 17.1 x 0.10 x 24 x 365
= $14,979
From this it becomes obvious that the cost of
power decreases as the cross sectional area of
the airway increases. Although the cost of
power initially decreases quite rapidly it then
levels out until the cost savings become
relatively insignificant with each incremental
increase in airway cross sectional area.
7.20.2 Optimum Airway
In most mines, the airways of highest
resistance are the primary intake and exhaust
airways. The pressure loses in these airways
can be 90% of the total mine losses and
therefore require most consideration.
Besides the operating costs of power there is
also the capital cost for development of the
airway. Simplistically the larger the cross
sectional area of the airway, the more rock to
be broken and removed therefore the higher
the cost. I.e. the cost for development
increase as the cross sectional area of the
airway increases.
One of the frequently used terms associated
with ventilation economics is “optimum
diameter”, “optimum velocity” or “optimum
cross-sectional area”.
Using the variable operating (power) cost and
fixed capital (development and fans) the
airway can be evaluated to determine the lowest total cost for a specific quantity of air over the
life of the airway.
This is achieved by the simple addition of the operating costs (power) and the capital costs
(development and fans) for a number of airway diameters.
In all cases it is simply that point where the total cost over the life of the airway (or system) is at
it’s lowest.
Cost of Power
Airway Diameter (m)
Cost ($)
0
0 +
+
Capital Cost
Airway Diameter (m)
Cost ($) 0
0 +
+
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Project Cost
Airway Diameter (m)
Cost ($)
0
0 +
+
Decreasing Operating Costs
Increasing Capital Costs
Operating + Capital Cost
Economic Diameter
7.20.3 Time Value of Money
When considering the operating costs particularly over a long period of time it is important to
consider the time value of money. For example if you were to invest $1,000 at 10% it would be
worth $2,593.75 in 10 years time (excluding tax) and if you wanted to have $1,000 in 10 years
time you would have to invest $385.54 at 10% today i.e. the Present Value of $1,000 in 10
years discounted at 10% is $385.54.
This concept of Present Value makes no mention of inflation. This is because inflation
influences the purchasing power of the money not the amount of money.
The Time Value of money refers to the effect that interest rates (as returns from potential
investment opportunities) can have on the value of money and not the effect of inflation.
When considering expenditure on ventilation airways, we should consider that the money need
not be spent and it could be invested. Thus for a given project, the expenditure must return the
original and operating expenditure equivalent to interest rates which could be earned by
investing the money.
The future sum of our investment can be calculated from
F = P(1+ i)n
Where F = Future Value ($)
P = Present Value invested ($)
i = Interest rate % (rate of return on capital)
n = Number of interest compounding periods usually yearly
Example
What is the future value (F) of $1,000 at the end of 6 years if interest is 10%?
F = P(1 + i)n
= 1,000(1 + 0.1)6
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= $1,771.56
Similarly, if we know the value of a future sum we can determine the Present Value by solving
for P. i.e.
(1 i)n
F
P
+
=
( )20 1 0.20
1,000
+
=
= $26.08
This shows that $26.08 invested today for 20 years at 20% interest will grow to $1,000.
Both of these functions deal with single payment present values.
If we were to invest money as a series of equal investments, each investment draws compound
interest for a different number of periods.
The future amount of money from this type of investment is calculated by:
( )
i
F A 1 i 1
+ n −
=
Where A = Amount Invested ($)
Example
Calculate the future value (F) 6 years from now with yearly investments of $1,000 at 10%
interest.
( )
0.1
1,000 1 0.1 1
+ 6 −
= ×
= $77,156.61
This shows the future value of yearly investments of equal value.
Should we wish to know the present value (P) of a series of $1,000 payments made at the end
of each year of 6 years if the compound interest is 10% per year. We can use the following
equation.
( )
( ) 1
i 1 i
A 1 i 1
P
n
+
+
=
( )
( )n
6
0.11 0.1
1 0.1 1
1,000
+
+
= ×
= $4,355.26
Thus the Present Value of a series of $1,000 payments made each year for 6 years with a
compound interest of 10% $4,355.26.
7.21 Rule-of-Thumb Principles and Design Factors
There are a large number of ventilation rule-of-thumb (ROT) principles that have evolved over
many years, many are contradictory and many are now no longer applicable. When using any
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ROT they must be placed in the correct context and this can only be done successfully if there
is a good understanding of the basis from which the specific ROT has been derived.
Although experience should not be ignored and basing ventilation design specification on past
and present practices in comparable mines can and has led to near disastrous consequences of
over and under ventilated mines.
There is some basis for the use of ROT principles particularly when attempting to “get a feel” for
what may be required, but mine ventilation systems must be designed specifically for each and
every mine, as no two mines are identical.
One site on the Internet (www.mcintoshengineering.com) provides a comprehensive list of ROT
principles. Before using any of these values the origin and circumstances of the factor should
be fully understood.
Listed below are some factors encountered by the author over many years.
Rule-of-thumb factor Comment
Fresh air quantities
(a) 0.024 – 0.094 m3/s per tonne of ore and waste
produced in the mine (it is assumed that this
refers to a daily production rate)
(b) 0.094 – 0.47 m3/s per man underground
(c) 0.063 – 0.126 m3/s per kW of operating diesel
equipment.
Consider a modern mechanised mine producing 1,000,000
tonne of ore and waste.
From (a) the quantity of air should lie between 65 and 257
m3/s
A modern mine may have 50 people underground therefore
from (b) the quantity of air required is between 4.7 and 23.5
m3/s. However in a labour intensive operation there could be
2,000 people underground and the airflow would be between
188 and 940 m3/s.
For diesel equipment 4,000 kW will require 252 and 504 m3/s
Velocities in airways
Service Shaft 5.1 – 7.6
Production Shaft 7.6 – 10.2
Main Entry 2.6 – 7.6
Conveyor tunnels, declines 2.5 – 5.1
Inlet and exhaust rises (no access) ±10.0
Inlet and exhaust rises (access required ) 2.5 –
5.1
Exhaust mains 10.2 – 15.2
Exhaust shafts (concrete lined) 15.2 – 20.3
Exhaust shafts (rock section) 10.2 – 15.2
Avoid 7.5 to 12.5 m/s in saturated airways.
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“in today’s increasingly uncertain, competitive and fast moving world, companies must rely more
and more on individuals to come up with new ideas, to develop creative responses and push for
changes before opportunities disappear or minor irritants turn into catastrophes.
Innovations, whether in products, market strategies, technological processes or work practices, are
designed not by machines but by people.”
Rosabeth Moss Kanter – “The Change Masters”
8 NETWORKS AND COMPUTER MODELLING
8.1 Ventilation Networks
In section 4 we looked at simple series and parallel circuits. Mines consist of many airways
(including stope voids open orepasses etc.) interconnected in such a way that it is impossible to
reduce them to simple series or parallel circuits.
This network of airways can, with the use of computers be analysed to predict airflow quantities
and pressures. Consider the following figure to represent three possible stages for
development of a mine. At first a simple series circuit, expanding when a new area is opened
up creating a parallel circuit and finally the connection of the two working areas resulting in a
network of airways.
Area 1
A H
B G
D E
C F
Series circuit ABCDEFGH
X Area 2 Y
Area 1
A H
B G
D E
C F
The construction of airway DY now presents a different
problem consisting of branches, junctions and meshes.
Branches (connect junctions) and are
ABC, CD, DEF, DY, CXY, YF and FGH. (Total 6)
Note: a branch is any series of airways and may have
different dimensions and must be included at least once
in the group of meshes
Junctions (two or more branches)
C, D, Y and F. (Total 4)
Meshes (and closed circuit)
CXYDC, DYFED, CXYFEDC and ABCXYGHA. (Total 4)
X Area 2 Y
Area 1
A H
B G
D E
C F
Parallel circuits CDEF and CXYF
with common junctions C and F
Compound or Network System
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The solution of the series and parallel was explained in section 4. The solution of networks is
undertaken with the following assumptions:
1. Air is incompressible and,
2. The airflow in the network obeys Atkinson’s law
Kirchhoff’s electrical laws can be restated and applied to ventilation networks i.e.
1. The total quantity of air entering a junction is equal to the total quantity leaving the junction.
i 1
Q 0
S
i
=
Σ =
2. The sum of the pressure changes in a mesh is zero.
( ) ( )
i 1
P P NVP m 0
m
i fi
=
Σ − − =
Where density is constant
m = number of branches forming a closed mesh
Pi = frictional pressure drops.
Pfi = total pressure drop across the fan
NVP = natural ventilating pressure.
In the simplest case of a mesh without a fan and no NVP this reduces to
i 1
P 0
m
i
=
Σ =
These equations are only correct when all terms are considered at standard density
8.2 Analysis of a Network20
The solution of a network requires the determination of the characteristic of each and every
airway, the number of branches, the number of junctions and the number of meshes.
From Kirchhoff’s 1st law for a network containing ‘b’ branches and ‘j’ junctions there will be ‘b’
airflows to determine and therefore ‘b’ equations must be solved.
Similarly from Kirchhoff’s 2nd law there will be ‘j’ equations to be solved. However only j – 1 of these are
independent because the jth junctions are already defined by flows at other junctions and therefore there
are b – (j – 1) equations to construct. These are obtained by choosing a minimum of, b – j +1 meshes
and writing equation. (P P ) (NVP)m 0
m
Σ i − fi − = for each of them. This gives j –1 junction equations from
Kirchhoff’s 1st law and b – j +1 mesh equations from Kirchhoff’s 2nd law.
When written in terms of P = RQ2 a quadratic equation for each mesh can be developed to find the
solution. For a network containing m meshes the powers of Q would be 2 m -1
20 For a detailed discussion on Network analysis refer to McPHERSON, M.J., “Ventilation Network Analysis” Environmental
Engineering in South African Mines. (1989) Chapter 8. pp211 – 239
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In relatively simple networks there is an application of this method but it falls short in large networks in
much the same way as simple series and parallel circuits in networks, and other techniques should be
considered.
8.2.1 Hardy Cross
The technique of fluid network analysis involves making an initial estimate of the flow
distribution and calculating an approximate correction to be applied to the flow in each branch
and then iterating the correction until and acceptable result has been achieved.
The technique that has found its place in mine ventilation calculations is the one developed by
Professor Hardy Cross in 1936.
Consider the figure below and assume a quantity Q passing through a duct.
Pressure drop P
Quantity Q
P = RQn
Q
P
Qa Q
P = RQn
P = RQan
n is the constant for the range of flow considered. To determine Q an estimate of Qa is made
i.e. Q = Qa + ΔQ where ΔQ is the error for quantity. Similarly ΔP is the corresponding error for
pressure.
Differentiating P=RQn we find
Slope of the curve nRQn 1
Q
P = −
Δ
Δ
Or at point Pa
n 1
nRQa
Q
P = −
Δ
Δ
n 1
nRQa
Q P −
Δ
Δ =
But n
a
ΔP = RQn −RQ
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Therefore n 1
a
n
n a
nRQ
RQ RQ
Q −

Δ = which is the “out of balance” equation in a single duct.
Considering a network with b branches and applying Kirchhoff’s laws and assuming the index n = 2
for a fully turbulent flow we can derive the mesh correction factor Qm to be
Σ( )
Σ
=
=


 

 
− −
Δ =
b
i 1
i ia fi
fi
b
i 1
i ia ia
m
2R Q S
R Q Q P
Q
Qia is the absolute value and S is the slope of the fan characteristic.
The procedure for the use of this method is described as follows:
(a) Estimate the quantity of air flow through each branch
(b) Decide on a pattern of closed meshes and determine the number. The minimum
number is determined by
No. of branches – No. of junctions + 1
(c) For each mesh evaluate the mesh correction factor ΔQm.
(d) Correct the flow in each branch.
(e) Repeat steps (c) and (d) until values of ΔQm are below the prescribed value.
(f) Repeat steps (b) to (e) for each of a number of changes to the network.
8.3 Computer Modelling
Modelling of ventilation systems with computers can be an excellent tool to assist with
ventilation system design. However, ventilation simulations are not a magic panacea. They do
not replace a good working knowledge of basic ventilation theory and practice and the following
problems are very common:
• The “user friendliness” of the latest software and the ability to directly import 3D mine design
data has encouraged large numbers of inexperienced, but computer literate engineers to
have a go. Impressive graphical output is produced, but the answers are often wrong.
• Inexperienced users often use the default input values (because the user is often unsure
what the proper values should be). In one popular computer package, the supplied shock
loss equivalent lengths are grossly misleading and should never be used.
• Some computer packages add shock loss factors automatically (by default) to every mine
airway. This can cause significant problems, particularly if the model contains a large
number of short airways.
• The models are often not checked against “reality”. This appears to be due to a basic
unawareness of how to take simple ventilation pressure measurements and compare these
with simulated values.
The aim of this section is to provide some practical advice so that some of these problems can
be avoided.
The user of computer simulation software should have a very clear idea of what questions need
to be answered – before any data is entered. The main reasons for undertaking a simulation are
listed below:
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• Estimation of fan duty requirements (for primary fans and circuit fans).
• Estimation of general airflow distribution, in more complex ventilation circuits.
It should be appreciated that computer simulations are not a ventilation design “black box”
There are a number of things, which they won’t do, including:
• Standard (non thermodynamic) computer simulations will not estimate airflow distribution
where natural ventilation pressures are significant (i.e. where significant temperature
differentials exist, or where the driving pressures in individual parallel circuits are low).
• Computer simulations will not provide an accurate estimation of all airflows in every part of
the ventilation circuit. In many mines (particularly larger ones), there are too many variables,
which change on a regular basis (e.g. ore passes or stopes being bogged empty), for any
real certainty in all predicted airflows.
• Computer simulations will not design your ventilation system for you! Important aspects like
the economically optimum shaft sizes etc must be determined by separate calculations.
8.4 Input Data
The old adage of “garbage in, garbage out” well and truly applies with ventilation simulations.
Some brief information is provided here on some aspects of input data.
8.4.1 Simulation Model Layout
For many years the use of a “skeleton” model has been seen as the best approach. In these
models main airways are included, but minor ones (e.g. small escape rises etc which do not
carry significant airflow) are excluded.
There are a number of advantages in this approach including:
• The model is a more manageable size, reducing the chances for data entry error
• Quicker simulation time and faster screen manipulation for 3-D programs
• Minimal loss in accuracy if removal of unimportant airways is done with care
As always there are also a number of disadvantages
• The person who developed it owns the model, and they are generally the only person who
understands it.
• Does not allow for any predictive work other than primary fan duty to be undertaken
• Does not allow the model to be used to “problem solve” the ventilation system
Today’s software packages allow the use of a large number (20,000) of branches to be used.
This along with the ability to import data from other software used to design mine openings,
means that ventilation models can be constructed to represent every opening in the mine. This
provides the advantage of other people being able to ‘view’ the network in the same manner
they would view survey or design plans.
8.4.2 Number of Airways
Modern simulation programmes have the facility to import centre line string data from 3D mine
planning packages. Before importing the data into the simulation package, it is wise to reduce
the number of points used to define the model. For example, the raw data may use 20 or more
points in series to define a bend in a decline, whereas this could perhaps be reduced to three or
four points, whilst still retaining a presentable 3-D format. The advantages are a less unwieldy
simulation model with very little reduction in accuracy.
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8.4.3 Friction Factors
Friction factors vary for different types of airways (and different airway sizes). If at all possible,
gauge and tube pressure surveys should be undertaken to determine “typical” local friction
factors.
8.4.4 Shock Losses
Shock losses seem to be a major source of misunderstanding and error in ventilation simulation
work. Two common problems are evident:
• One program applies a shock factor to every new airway by default. If the model is
constructed from a large number of very short airways in series, the resulting combination of
default shock losses can cause the simulation model to substantially overestimate the total
mine resistance. Note that the Atkinson Friction Factor for drives, etc includes an allowance
for “normal” shock losses (decline bends etc). Shock factors generally only need to be
applied to “dog legs”, inlet losses and discharge losses in main ventilation raises etc.
• One program misleads the user into believing that an “equivalent length” shock factor can
be used to accurately describe a particular shock loss (e.g. right angle bend), regardless of
the airway size or roughness.
The problems caused by the equivalent length concept are almost universal. Some notes
outlining the problem in more detail are presented below:
The equation for the equivalent length of a shock loss is given as
6.67k
Le = XDh
Where: Le = equivalent length of shock loss (m)
X = shock factor
k = Atkinson Friction Factor). Ns2m-4
Dh = Hydraulic diameter (m) =(4 * A)/ C
A = cross-sectional area (m2)
C = drive perimeter. (m)
It can be seen from the above equation that the equivalent length is directly proportional to the
hydraulic diameter and inversely proportional to the Atkinson friction factor of the equivalent
length airway (i.e. equivalent length for a given shock loss is not constant, but depends on
airway size and roughness characteristics). For illustration, using the above formulae, the
equivalent length of a discharge loss (X=1) from a 1.2m airleg rise (k = 0.015) = 12m of 1.2m x
1.2m airleg rise. In contrast, the equivalent length of the same discharge loss (X=1) from a 4m
diameter raise bored hole (k = 0.004) = 150m of 4m raise bore. That’s quite a difference in
equivalent length!
It is important to recognise and understand the problem outlined above. Until there are some
changes made in how shock losses are dealt with in some of the simulation packages, it is
strongly recommended that all shock losses be calculated separately.
“Woods Practical Guide to Fan Engineering” Daly, B. B. (1978) is a very useful source of
information on shock factors. (Note that Daly uses ‘K’ for the shock factor, rather than ‘X’)
8.4.5 Fans
Ventilation programs have the facility to enter fan curves. Most programs require fan pressure in
terms of fan static pressure. Unfortunately, some fan suppliers only provide fan total pressure
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curves. A fan static pressure curve (including pressure losses or regains from associated fan
inlet and discharge ducting) should always be requested when ordering primary mine fans.
When simulating a new mine to determine the required primary fan duties, the required fan
static pressures can be determined by “fixing” the flow in the primary exhaust airway(s). The
required fan static pressure can be read off the simulation results.
8.4.6 Fixed Airways
Fixed airways should be used very sparingly. Excessive use of fixed airways will overly
constrain the simulation model and may result in errors or program crashes. Fixed airways are
generally only used for airways where the resistance is difficult to measure (e.g. lack of
accessibility etc). Even then this should be a temporary fix and once the resistance has been
determined the fixed airflow should be replaced with the appropriate resistance. Most of the
time, the only airways that should be fixed are airways with ventilation controls (e.g. fans or
regulators). Fixed airways should never be applied to locations where there are no ventilation
controls (e.g. the flow in a decline should never normally be fixed, unless a control such as a set
of doors or booster fan is planned).
If a large number of fixed airflows are required, it generally means that airway resistances have
been incorrectly determined and therefore need to be adjusted. (i.e. the wall roughness or the
cross sectional area needs to be looked at.)
8.5 Results:
For existing mines, “Calibration checks” against existing fan pressures, regulator pressures and
major airway flows should always be made, before using simulation models to predict future
ventilation system performances. An appreciation of the limitations in accuracy with the
technique (due to uncertainties in input values) is required. Agreement to within 15% of actual
airflows is about as good as can be expected.
BASIC MINE VENTILATION References & Further Reading
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“the best way to learn the practice of mine ventilation is to slug it out in an environmentally tough
underground mine for several years. If a person is devoted full time to the study and practice of mine
ventilation and has access to another well qualified mine ventilation engineer the average engineer
will start to become reasonably proficient in the basics of mine ventilation after a period of one to
two years”
F. Bossard – Manual of Mine Ventilation Design Practice
References & Further Reading
1. AMIRA “Mine Ventilation Measurements – A Manual of Recommended Practice”
2. BARENBURG, A.W.T. “Psychrometry and Psychrometric Charts – Third edition”
Chamber of Mine of South Africa.
3. BOSSARD, F.C. “A Manual of Mine Ventilation Design Practices Second Edition (1983)
4. BOOTH-JONES, P. A., ANNEGARN, H.J., BLUHM, S.J. “Filtration Of Underground
Ventilation Air by Wet Dust Scrubbing”. (pp209-213) Proceedings of the 3rd International
Mine Ventilation Congress. Published by The Institute of Mining and Metallurgy. London
(1984)
5. BRIEF, R., SCALA, R., “Occupational Exposure Limits for Novel Work Schedules.”
American Industrial Hygiene Association Journal. 36: 467-469, 1975
6. DALY, B.B. “Woods Practical Guide to Fan Engineering” (Published by Woods of
Colchester 1978)
7. de la HARPE, J,H., JENNER, L.W. “The History of Mine Fans in South Africa” The
Journal of the Mine Ventilation Society of South Africa. (December 1986)

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