Ventilation
Topic 1
1. The Need to Ventilate Mines
Ventilation is the process of conducting an adequate flow of pure, fresh air along airways, working
places and service points underground. The main aims of ventilation can be summarized as follows:
1.
2.
3.
4.
to provide air for mine workers and equipment,
to dilute the concentration of explosive and toxic gases, fumes and radon to environmentally
safe levels and to remove them from the mine,
to dilute the concentration of airborne dust to physiologically acceptable levels and to
remove it from the mine and
to provide a thermally acceptable environment in which persons can work without undue
discomfort or any danger of heat exhaustion and to remove heat from the mine as may be
necessary.
2. Air for Breathing
The average composition of ordinary air by volume is:
Nitrogen
Oxygen
Carbon Dioxide
Argon and other rare gases
78.09%
20.95%
0.03%
0.93%
The above figures do not take water vapour into account, which is always present in amounts varying
from 1% to 6% by volume.
Since oxygen is essential for human life, it stands to reason that a change in the composition of the
air could cause discomfort and even death below certain O2 percentages.
Respiration
Breathing is the most basic of human functions and involves removing oxygen from inhaled air and
returning carbon dioxide in return. The amount of air breathed and the amount of oxygen retained
depends largely upon the level of muscular activity.
During hard work a person absorbs about three litres of oxygen per minute compared to 0.3 l per
minute required by a sleeping person. A considerable oxygen deficiency can be tolerated:
21% - normal
17% - person still works comfortably, but breathing is faster and deeper
14% - symptoms like dizziness, buzzing in the ears, rapid heart action and headaches start to
appear
6% - death
These figures assume that the oxygen deficiency is due to displacement of oxygen by a gas such as
methane or the absorption of oxygen by the oxidation process of metals and minerals. If the
deficiency is due to carbon dioxide, the effects on a person are more severe.
Carbon dioxide acts as a stimulant which regulates human breathing. When the percentage of carbon
dioxide in atmospheric air (normally 0.03%) is increased to about 3%, the breathing rate is doubled.
At 5% the rate of breathing is quadrupled and laboured but death will usually not occur until all the
oxygen has been depleted.
1
To avoid this, each person working underground should be supplied with at least 0.015m 3/s of fresh
3
air and 0.1 m /s of air should ideally be supplied.
3. Gases and Fumes
Prolonged exposure to sufficient concentrations of certain gases will disable or kill a person. These
are known as toxic gases and can be divided into four groups:
1.
2.
3.
4.
asphyxiant gases
irritant gases
narcotic gases
protoplasmic gases
1. Asphyxiant gases
If the oxygen supply to the bloodstream and the tissues of the body decreases or fails, asphyxiation
results. The may occur in two forms:


simple asphyxia occurs when sufficient air is breathed, but the air has an oxygen
deficiency;
chemical asphyxia occurs when a substance present in the air prevents oxygen from being
utilized by the body (e.g. carbon monoxide).
2. Irritant gases
These gases irritate and inflame tissue in contact with them. Inflammation of the lungs leads to
pulmonary oedema (fluids are produced in the lungs, causing suffocation). Nitrous fumes and
hydrogen sulphide act in this way.
3. Narcotic gases
Gases such as chloroform and ether which act on the nervous system, after having been absorbed in
the blood and transported to the tissue, are narcotic gases. If large quantities are absorbed, the
normal coma will terminate in death.
4. Protoplasmic poisons
After absorption by the body, these gases destroy the living cells in contact with them. Examples
include fumes of metals such as mercury and lead.
Explosive gases
Some gases encountered in a mine, although safe to breathe may be explosive. Four conditions must
exist underground to cause a gas explosion:
1.
2.
3.
4.
the gas must be produced,
the gas concentration must reach an explosive level,
a flame or spark must ignite the gas and
there must be sufficient oxygen present in the atmosphere.
If one of these factors is absent, the explosion will not occur.
Some Less Common Gases in Mines
1. Hydrocyanic Gas (HCN)
Also known as hydrogen cyanide or prussic acid, this gas affects people in a way similar to H2S, only
it is far more poisonous. Even a gas mask is not proof against it as it can be absorbed through the
skin.
2
It is a colourless gas with a smell resembling that of bitter almonds. It is slightly lighter than air. The
gas occurs in mines only when acid water comes into contact with cyanide in sand which has been
used for sand-filling.
2. Chlorine (Cl2)
This is heavy, greenish-yellow irritant gas with a pungent odour. It is extremely poisonous. A
concentration of 0.005% can prove fatal and 0.001% is given as the MAC value. It causes coughing,
smarting of the eyes, tightness of the chest and vomiting. As in the case of nitrous fumes, oedema
(waterlogging) of the chest may occur several hours after exposure.
Several chemical tests are available for chlorine, but in practice they are not necessary as the gas is
easily recognised by its smell. The gas only occurs in mines where liquid chlorine is used to disinfect
drinking water.
3. Aldehydes
These are a series of organic compounds with a general formula C nH2nO. The lowest member of the
series is CH2O (formaldehyde). Several aldehydes, with formaldehyde predominating, are present in
exhaust gases from diesel engines. They have a pungent smell and are intensely irritating to the
eyes, mucus membranes and the skin. At high concentrations they can be suffocating. Aldehydes can
be detected by the human nose at concentrations far below that necessary to cause irritation and thus
no test is required.
Headings Under Which Mine Gases are Studied
source
properties - colour, taste, smell, relative density (RD), solubility, combustibility, toxicity
effects on humans
detection
legal
safety
Source:
To associate the presence of a gas with a particular condition.
Properties:
To ascertain whether the gas is:
detectable by physical means,
lighter or heavier than air,
can be diluted or rendered harmless by means of water,
is combustible or not,
is poisonous or not.
Effects on Humans:
Enables action to be taken in the event of gassing underground.
Detection:
Enables use of correct instruments and procedures in gas detection.
Legal Aspects:
The dangerous effects of these gases are only apparent once the legal limits have been
exceeded.
3
Relative Density
The ratio of the mass of a given volume of gas to that of an equal volume of air.
The following table reveals important source properties of various gases encountered in the
mines.
4
5
A further summary of gases are provided in the table below.
Approx.
Name of
Relative
Symbol
Explosive Range
Gas
Density
in Air
Allowable
For 8 Hrs
(TLV)
Amount in Air
Danger to Life
N2
0.97
Non-explosive
-
Above 88%
CO
0.97
12 to 72%
30 ppm
0.1% (30 Min.)
-
1.00
Non-explosive
-
-
H2S
1.18
4 to 44%
10 ppm
500 ppm
NO2
1.23
Non-explosive
3 ppm
100 ppm
Methane
CH2
0.55
5 to 15%
Visible cap
25% (Def 02)
Oxygen
O2
1.10
Non-explosive
-
Below 12%
Hydrogen
H2
0.07
4 to 74%
-
25%
Ammonia
NH3
0.6
Non-explosive
100ppm
2500ppm(30min)
Nitrogen
Carbon
Monoxide
Air
Hydrogen
Sulphide
Nitrous
Fumes
6
Characteristics
Colourless, tasteless,
odourless
Colourless, tasteless,
odourless
Colourless, tasteless,
odourless
Colourless, smell of bad eggs
Reddish brown colourdelayed effect
Colourless, tasteless,
odourless
Colourless, tasteless,
odourless
Colourless, tasteless,
odourless
Colourless, Acrid Biting
taste, strong pungent odour
The following gases and their concentrations in ambient air would affect humans in various ways such
as:
Carbon monoxide (CO)
0.02% - Slight poisoning symptoms in two hours (Canary – no reaction)
0.05% - Headache and discomfort in one hour
0.1% - Giddiness in half an hour
- Helplessness in one hour
- Death in two hours
- (Canary – No distress in 12 to 15 minutes)
0.2%
- Serious effects in 30 minutes
- (Canary – Serious effects in 3 minutes)
0.5% - Man – Serious effects in 5 to 7 minutes
- Death in 15 minutes
1.0%
- Man – Immediate danger
Hydrogen Sulphide
Low concentrations
- Headaches
- Irritates eyes and nose
- Impairs sense of smell
0.05% - Kills canary in 30 seconds
0.1%
- Quickly causes unconsciousness
0.2% - Kills man in a few minutes
To summarise some of the important gases dealt with in this section of mine ventilation the following
is given:
Hydrogen
4 to 74%
- Explosive range
Colourless
Odourless
Lightest gas
Can occur with methane
Methane
5 to 15%
- Explosive range
10%
- Most violent explosion would occur
> 15%
- Burns in this range
Colourless
Odourless
Lighter than air
The following fresh air supply would be required by man in the different conditions encountered:
a)
b)
c)
d)
Required
Shallow gold mine
Deep gold mine
Factories
3
= 0.015 m /s
3
= 0.08 m /s
3
= 0.14 m /s
2
= 1.5 l/s/m or 7.5 l/s/person
The effect of carbon dioxide inhalation would have the following effects on human breathing:
a)
b)
c)
0.03% - Normal breathing
3%
- Breathing rate doubled
5%
- Breathing rate quadrupled
7
The effect of oxygen and the lack thereof in ambient air would be the following:
a)
b)
c)
d)
21%
17%
14%
6%
- Normal ambient concentration
- Breath deeper and faster
- Dizziness, buzzing in the ears and a rapid heart rate
- Death
The following air requirement could be set for human work performed:
a)
b)
Hard work
Sleeping
- 3 litres per minute
- 0.3 litres per minute
Nitrous fumes have the following combinations:
NxOy
a) Nitric Oxide
b) Nitrogen dioxide
c) Nitrogen Trioxide
d) Nitrogen tetroxide
8
Ventilation
Topic 2
Tutorial 1
Gas Calculation
The questions below are designed to enable calculation of the concentrations in the question set.
3
A diesel vehicle travels along an airway. The total gas emission is 0.2 m /s. The exhaust gases
contain 1500 ppm of carbon monoxide and 750 ppm of oxides of nitrogen. The cross-sectional area of
2
the airway is 9 m and the air velocity is 3 m/s. What is the concentration of carbon monoxide and
oxides of nitrogen in the air after complete mixing is the vehicle travels at :
a) 8 km/h in the opposite direction to the air,
b) 6 km/h in the same direction as the air and,
c) at the same speed as the air and in the same direction?
9
Ventilation
Topic 3
Ventilation Requirements
At the start of the Industrial Revolution, mines were generally shallow and ventilation of the
workings was not a serious consideration. As mines became deeper, this situation changed.
Empirical Guidelines for Ventilation in South Africa
Gold Mines
10
A wide range of conditions exists under which gold is mined. For this reason it is difficult to specify an
overall air quantity. Typical values range from 3 to 6 kg/s per 1000 t of rock mined per month. Main
surface fan pressures of up to 9 kPa are used. This means that single shaft systems may transport up
to 700 m3/s of air and have electric power consumption for the main fans of about 8 MW.
Base Metal Mines
Base metal mines vary enormously in mining method, degree of mechanisation, depth and many
other factors. Most are highly mechanised and thus if a standard of 60 -65 m3/s of air per MW of rated
diesel capacity is adequate and if it is assumed that only 50% of the air arrives at the working place,
then about 120 -130 m3/s/MW should be supplied. Depending on the mining method, between 30 000
and 60 000 t/month can be produced from 1 MW of installed diesel capacity.
The following example calculation would be the requirement for trackless equipment utilisation in an
underground mining situation:
Trackless equipment:
LHD 1 x 100 kW diesel power unit
11
Quantity requirement:
The method of calculating the amount of ventilation air required for an operational fleet in a “multi—
heading” mining development system is the one used by the MSHA (Mine Safety and Health
Administration, USA) where:
Qt
Qt
Q1
Q2
Qn
=
=
=
=
=
100% x Q1 + 75% x Q2 + 50% x Qn
Total air quantity required
Permissible volume rate for largest rated unit (diesel)
Permissible volume rate for second largest rated unit (diesel)
Combined permissible volume rate for each additional diesel unit.
Using the MSHA formula and the example fleet as given above, the required quantity for the mine is:
Qt
=
=
(100 x 0.07) + (0 x 0.07 x 0.75) + (0 x 0.07 x 0.5)
7.0 m³/s (Section requirement)
There will be 1 diesel units operating in the heading therefore the ventilation requirements are as
calculated above. If more diesel units are to be operated in the ramp system, the air requirements
need to be adjusted accordingly.
Coal Mines
Because of the possibility of methane related problems, minimum ventilation requirements are
specified by law. The following list is taken from the old Mines and Works Act which is still in
force via the Minerals Act:
10.8.1
10.8.2
10.8.3
10.8.4
10.8.5
25 l/s x maximum tons per shift
less than 200 persons per ventilation district
longwall face ventilation velocity at least 0.25 m/s
air velocity in airways at least 0.25 m/s
2
ventilation in advance headings at least 300 l/s/m
12
Ventilation Practice in Mines
To ensure adequate ventilation of a mine, provision is made for suitable pathways (called airways) for
the air to enter the workings. Once the air has passed through the workings, airways are required to
allow it to leave the mine.
Most mines have ascensional ventilation of the workings - the fresh air is conveyed directly to the
lowest working places by means of a downcast system and then ascends through the rest of the
mine. This is generally more satisfactory than descentional ventilation in which air descends through
the workings and is then taken to surface from the lowest reaches of the mine via an upcast shaft.
13
Air always flows along the path of least resistance, but this may not be where it is required for use. To
prevent the air from returning directly to surface along upper levels connecting downcast and upcast
shafts, ventilation doors or airlocks are installed. An airlock consists of two doors arranged so that
when one is opened the other can be kept closed to prevent airflow. If a small volume flow rate of air
is required continuously in a given level, then an opening of reduced size, called a regulator, is made
in the ventilation door or door frame.
Although general airflow can be made to pass directly through the main working places, there will
usually be several places where auxiliary ventilation will be required (such as development ends, hoist
chambers, shaft loading stations, etc.). The provision of such ventilation requires an additional fan
and a suitable route for the air. This route is generally a ventilation duct.
Ventilation of Main Working Places
Stopes
A normal stope connects two levels, of which the lower usually acts as the intake airway and the
upper as the return airway. The distribution of air in a stope is controlled by closing the bottom and top
of the stope as far as possible, leaving openings only near the faces and other areas where people
may be working. This will direct most of the air towards the face
14
Development Ends
Three main systems can be used to ventilate the working face in a development end.
A) Forcing systems
The fresh air from the through ventilation is forced into the end through a ventilation duct by a fan and
returns via the development end. The intake to the duct must extend well into the upstream portion of
the intake airstream (at least 5 m) in order to prevent recirculation of used air. The fan should be
situated in that part of the duct which extends into the fresh air so that it is accessible after blasting
when the roadway itself should not be entered. The intake and delivery ends of the duct should be
fitted with wire screens to prevent any extraneous material from being drawn into the fan. The duct
should be maintained in good condition in order to deliver as much air as possible to the development
end. The delivery end of the duct must be kept as close as possible to the face (< 12-m to the face).
The end of the duct may be damaged by blasting if it is too close to the face. One method of avoiding
damage is to remove the last section or two of the ventilation duct before blasting and replace them at
the time of loading, or to use a flexible, retractable duct.
B) Exhaust Systems
Air from near the working face of the end is exhausted through a ventilation duct and is replaced by
fresh air drawn along the roadway. The duct must discharge the return air well beyond the entrance to
the development end (at least 5 m), on the downstream side to prevent the air re-entering the end.
This system is not commonly used because fresh air drawn along the roadway tends to enter the
exhaust ventilation duct directly and very little of it actually reaches the face due its low velocity. This
situation is exacerbated by the fact that the intake of the return duct cannot always be kept right up to
the face as it may be damaged by the blast. Shaft sinking operations often use this type of ventilation
method to allow for re-entry period limitations.
15
C) Overlap Systems
Force overlap system
Exhaust overlap system
In exhaust overlap systems, an auxiliary duct forces fresh air from the roadway onto the face in
addition to an exhaust ventilation duct. The length of the auxiliary duct commonly varies from 6 to 60
m. The distance from the face to the intake end of the exhaust duct is generally kept within 15 m (but
usually 5 m) of the face. It is important that the exhaust duct and the auxiliary forcing duct overlap by
at least 5 m. A common length of overlap is 10 m. This ensures that the air entering the auxiliary
forcing duct is fresh air coming up the roadway and not air returning from the face. It is essential to
ensure that the volume flow rate of air being exhausted from the face is at all-time greater (2 - 4
times) than the volume flow rate of air being forced onto the face by the auxiliary duct. This prevents
short-circuiting of air. The drives of the fans must also be interlocked so that if the main exhaust fan
stops, the in-bye forcing fan also stops.
16
The following law requirement is enforced via the South African Mining law:
This permission supersedes all previous permissions granted in this regard.
RE-ENTRY INTERVALS
In terms of Regulations 2.10.10 and 10.10.2 in force in terms of Schedule 4 of the Mine Health and
Safety Act, 1996 (Act no 29 of 1996) the intervals which must expire before persons are allowed to reenter the workings of your mine in which blasting has taken place are fixed as follows:
1.
NIL RE-ENTRY INTERVAL
No interval of re-entry needs to be observed in all workings in which through ventilation has been
established and which are ventilated by air, which does not become contaminated by blasting fumes.
Should the air in any of the workings become contaminated by blasting fumes, the general reentry interval, as set out in paragraph 2 below must be observed in those workings.
2. GENERAL RE-ENTRY - REGULATION 10.10.2 (For Blasting once in 24 Hours)
A general minimum re-entry interval of three (3) hours after the blast in all ventilation districts must be
observed, with the exception of the workings mentioned in paragraph 1 above and other workings
where special re-entry intervals have been set in terms of this permission.
Blasting more than once in 24 hours for certain stoping sections will be allowed provided that the
minimum re-entry interval of three (3) hours is maintained and that an air quality as specified in 3.1 c)
of this permission is maintained at all times.
3. THIRTY MINUTE RE-ENTRY INTERVAL - REGULATION 10.10.1(a) (For blasting more than
once in 24 Hours)
The following provisions must be made applicable to all multi-blast development ends or
shafts being sunk:
3.1 Minimum Air Quantities Required (Relative to the Air Density at the working face)
a) The minimum quantity of air forced must not be less than 0,3 cubic metres per second
for every square metre of face area, for all multi-blast development ends.
b) The quantity of air exhausted from the development end must not be less than twice more
than the quantity of air supplied by the force column referred to in paragraph a) above, the
minimum force: exhaust ratio of 1:2 must be maintained at all times.
c) The air supplied must be of a quality as prescribed in Regulation 10.6.6 and also applies to
conditions on re-entry. Dust measurement results with a tyndallometer must be less than an
AQI of 1.0 taken over a 2-minute period.
3.2.1 Ventilation Arrangements
3.2.1 Horizontal Development, Inclines, Declines and Raises.
a) An exhaust-overlap system of ventilation must be used for every end being multi-blasted.
b) The intake of the exhaust column must be carried to a point not exceeding 30 metres
from the face.
c) The discharge end of the force column providing the quantity of air prescribed must not
be more than twelve metres from the face of the development end being blasted.
d) The minimum overlap distance between the exhaust column intake and the force
column intake points must be at least 9 metres and not exceeding 15 metres.
e) Fans in the exhaust column must be positioned in such a manner that the exhaust
column remains under negative pressure thus ensuring that no exhaust fumes leak back
into the intake air flowing to the face.
f) To prevent open circuit exhaust fans from recirculating a volume of at least 0.6m /s/m , of
through ventilation, at these fan sites, must be maintained at all times.
3
17
2
g) The exhaust fans in an exhaust-overlap system, which are the primary source of
ventilation, must be interlocked with all other electrical appliances and equipment in the
end being multi-blasted. This is to ensure that in the event of the exhaust fans stopping
that all other electrical appliances and equipment will also shut down.
h) No butterfly valves to be positioned in any exhaust column in development ends.
i) In a raise the exhaust column must be kept within 30 m of the face at all times and on the
face side of any ore-pass connection.
j) The force fan must be positioned only in the overlap section of the ventilation system.
k) A water blast or other similar dust allaying mechanism must be operated during the blast
and re-entry period at a discharge point not exceeding 15 metres from the face.
l) If, at any stage, blasting fumes from the end being multi-blasted contaminates any
working places in the vicinity, then multi-blasting must cease and conventional time
lasting (as per paragraph 2) must be followed until conditions have been rectified for
multi-blasting. The ends so contaminated must also of necessity be on conventional time
blasting.
m) The dust and fumes from blasting operations must be exhausted directly to surface via an
established smoke way and must not contaminate any place where persons may be
required to work or travel
3.2.2 Shafts
a) The bank area must be kept clear of blasting fumes and the shaft must
remain downcasting, in the bank area, at all times.
b) The force column delivery must at least be to the bottom deck of the
stage during blasting.
3.3 Waiting Place in terms of Regulations 2.10.6 and 8.10.33
a) Blasting must be carried out from a place of safety demarcated by the manager. This
position must be sign posted as “Waiting Place” and also act as a contraband control
point.
b) The blasting times must be recorded and the re-entry interval must be specified and
posted on the waiting place signboard and other relevant conspicuous places.
c) Blasting initiation must be conducted electronically.
18
3.4 Miscellaneous
a) Continuously operating flammable gas measuring instruments must be
used at all drilling sites (inclusive of cover/diamond/prospect drilling
sites).
3.5 Returns to Department of Minerals and Energy (Mine Health and safety Inspectorate)
a) All returns must be sent to this Regional Office.
b) These returns must be submitted on a monthly basis per multi-blasted
development end.
c) The following is a listing of the minimum information that will required in
the returns :
Mine Information
i)
ii)
iii)
iv)
Name of Mine
Shaft complex, section and development end identification.
Development end cross sectional area in m 2.
A sketch plan of the working place showing all ventilation arrangements and volume
flow rates.
Ventilation Appliances (before blasting)
i)
ii)
iii)
iv)
Force column delivery quantity; force column distance from the face.
Exhaust column intake quantity and distance from face.
Water blast or dust allaying mechanism, distance from face.
Exhaust-overlap distance.
Environmental Conditions on Re-Entry (30 minutes after blast) – at a point 3 metres from the
intake of the exhaust column (face side).
I.
II.
Concentrations of, oxides of nitrogen, carbon monoxide and amount of flammable gas in the
atmosphere.
Concentrations of dust by means of tyndallometer readings.
Your attention is also drawn to the requirements of Regulations 8.11, 10.1.1, 10.3.1, 10.6.3,
10.6.5, 10.10.3, 10.10.4 and 10.10.6.
All regulations mentioned above are in force in terms of Schedule 4 of the Mine Health and Safety
Act, 1996 (Act no 29 of 1996).
This permission may be amended or withdrawn at any time should such action be considered
necessary.
All persons concerned must be made fully conversant with the terms of this permission, copies of,
which must be readily available to them.
Forces Causing Airflow in Mines
To cause airflow in mines a difference in pressure between the intake and return shafts is essential.
This pressure difference can be caused by either natural or mechanical ventilation
Natural Ventilating Pressure
Single Shaft
In a mine with only one shaft, there will be a natural flow of air as long as that temperature of the
workings is markedly higher than that on surface. Warm air rises and a convection current is set up
19
within the shaft which is sufficient to ventilate small, shallow workings. By the late seventeenth
century it was recognised that two interconnected shafts (or a shaft and
an adit) provided a flow of air and reduced the risk of suffocation.
Twin Shafts
Where the two shafts are of unequal depth, in winter air will flow down the shallow shaft and up the
deep one because the temperature in the shallow shaft will be lower than that in the deep shaft. In
summer the temperature in the shallow shaft will usually be higher than that in the deep shaft and so
the direction of the air current will be reversed. With two shafts of equal depths, the direction of the air
current may be determined by the direction of the prevailing wind or by one shaft being wetter (and
therefore cooler) that the other.
Natural ventilation in most modern mines results because the air in the downcast shaft is cooler and
denser than the air in the upcast. The unequal densities, although influenced by other factors, are due
mainly to the transfer of heat from the underground rock to the mine atmosphere.
With the exception of very shallow mines, underground air temperatures approach rock temperature
and remain fairly constant. Consequently, the temperature, density and total weight of the air column
in the upcast shaft will not fluctuate to any great extent. The density and total weight of the air column
in the downcast shaft, however, fluctuates with surface air temperature. Since a density change in
either of the columns will cause a change in the natural ventilating pressure, this pressure may vary
from a maximum in one direction, to zero, to a maximum in the opposite direction depending on
whether the surface air temperature is lower than, equal to or greater than the underground air
temperature. This results in an unpredictable volume flow rate which may vary from a maximum in
one direction to some value in the opposite direction. In some instances, natural ventilation may
maintain underground air quality standards, but generally is alone is not a dependable means for
continuous and effective ventilation. Thus additional ventilation has to be provided by mechanical
means.
There is still today a mine in Zimbabwe (Blanket mine) utilising natural ventilation to ventilate the
whole mine. This is only possible because of its workings depth below surface (800mbc) and the
intake shafts positions.
A) Natural ventilation in a twin shaft system where the shafts are of unequal depth.
20
B) Natural ventilation in a twin shaft system where the shafts are of unequal depth
Mechanical Ventilation
A fan takes in air at one pressure and raises this air to a higher pressure, i.e. it increases the pressure
of the air on its delivery side to a value above the pressure of the air on its intake side. Due to this
induced pressure difference, the air is able to overcome the resistance of the circuit through which it is
flowing.
Usually the main fan (or fans) is situated at the top of the upcast shaft or return opening. Its main
advantages are:
(1)
(2)
(3)
The working environment’s pressure is slightly less than the outside atmosphere pressure,
because the fan creates a negative pressure within the mine. When the fan stops,
underground pressure builds up to atmospheric pressure and this increase slows down the
liberation of strata gases and gas emissions from the goaf, and prolongs the time required for
the gas to fill active workings.
Through the use of discharge evasée, velocity pressure can be recovered enabling power
savings
No heat is added to the air supply underground
The disadvantages are:
(1)
(2)
(3)
(4)
It can only be applied if all the return air flows in the upcast shaft.
Foul air goes through the fan, making repair or maintenance difficult. Corrosive fumes and
particles cause corrosion on the fan blades and reduce effective air passage area and can
throw the fan out of balance.
Natural ventilation pressure is reduced
It is difficult to hoist men and materials in the upcast shaft due to safety requirements
In shallow mines having fractured rock, in areas of contiguous mining where there may be ground
fractures into abandoned mines, and in mines with old workings outcropping on the surface, the most
advantageous position for the fan is at the top of the downcast shaft or intake opening. Some of the
disadvantages are:
21
(1)
(2)
(3)
access to the shaft is difficult, requiring special airlocks, etc.:
it is difficult to transport men, material or ore in the shaft
heat is added to the downcast air.
From the thermodynamical point of view, the best location is at the bottom of the upcast shaft.
Because the fan adds heat to the air in the upcast shaft, natural ventilation is increased. The
other advantages are:
(1)
(2)
no heat is added to the air breathed by miners
the downcast shaft is left clear for hoisting
The disadvantages are:
(1)
(2)
(3)
the fan handles foul air
it is not accessible during fires
any leakages across doors above the fan results in recirculation of foul air.
The following paper is provided for further reading and to indicate the relevance of ventilation
systems.
22
23
24
25
26
27
Ventilation
Topic 4
Fluid Mechanics
Fluid mechanics consists of two fields of study:


fluid statics and
fluid dynamics.
Fluid statics deals with stationary fluid bodies such as the hydrostatic pressures associated with water
acting against a dam wall. The fundamental equation associated with this field describes the pressure
exerted by a column of incompressible fluid of height ‘H’ and density‘w’:
p=wgH
[1]
where:
g = gravitational acceleration
2
9.79 m/s in Johannesburg
Fluid dynamics deals with fluids subjected to a shear force and which deform continuously,
moving relative to the rest of the fluid body.
In a flowing liquid each particle changes its position with a certain velocity. The magnitudes and
directions of the particle velocities may vary with position as well as time. This may be illustrated using
streamlines. These are imaginary lines in a fluid, the tangents of which give the direction of flow at
that point.
Generally speaking, there are three types of flow as can be visualised from the above drawing,
namely plug, laminar and turbulent flow.
Plug Flow
Plug flow is a type of laminar flow and only occurs with yield stress materials moving at low velocities.
The central part of the fluid moves as a plug with the same particle velocities. A shearing effect is only
present at the pipe wall. The velocity profile therefore has a flat centre portion as shown in the
diagram below.
Laminar Flow
In laminar flow the flow pattern is smooth with fluid layers travelling straight or in gently curving lines
parallel to the conduit axis. The velocity of each layer increases towards the middle of the stream until
a maximum velocity is attained at the centre. In this type of flow there is only one component of fluid
velocity: a longitudinal component. Here the fluid shear stress resistance is caused by the sliding
action only and is independent of the roughness of the pipe.
28
Turbulent Flow
Turbulent flow only occurs at high velocities or at low fluid viscosities and is characterised by chaotic
motion of fluid particles. This random movement results in two components of velocity: a longitudinal
and a transverse component. The longitudinal velocity attempts to make the fluid flow parallel to the
conduit axis, while the transverse component attempts to move the fluid in a direction normal to the
pipe axis.
Types of Fluid
Newtonian Fluids
In this type of fluid, viscosity is constant and is only influenced by changes in temperature and
pressure. Examples include gas, oil and water.
Non-Newtonian Fluids
The viscosity of this type of fluid is proportional to the magnitude of shear stress. Examples include
drilling mud and cement slurries.
FLUID MECHANICS
Fluid mechanics consists of two fields of study:
Fluid statics, and fluid dynamics.
Fluid statics deals with stationary fluid bodies, i.e. the hydrostatics of a dam acting against the dam
wall. The basic equation of fluid statics is that the pressure exerted by a column of height 'H' of an
incompressible fluid of density 'w' can be expressed as:
P=wgH
(1)
Where:
w - fluid density
H - column height
2
g - gravitational acceleration = 9.79 m/s at altitude of Johannesburg
Fluid dynamics deals with fluids subjected to a shear force, and which deform continuously, moving
relative to the rest of the fluid body. In a flowing liquid each particle changes its position with a certain
velocity. The magnitudes and directions of the velocities of all the particles may vary with position as
well as time. This may be illustrated by using streamlines. Streamlines are imaginary lines in a fluid,
the tangents of which give the direction of flow at that point.
In laminar flow the fluid particles move along smooth paths such that the movement of all particles on
the same path is identical. This type of flow is associated with very low velocities and highly viscous
fluids.
29
Turbulent flow is characterised by fluid particles moving in irregular paths and colliding with one
another in a haphazard manner. Most practical flow problems of interest to the mining engineer are
related to turbulent flow.
Whether a particular set of conditions will lead to laminar or turbulent flow is determined by its
REYNOLD'S NUMBER
Reynolds number:
R wvD

Where:
3
w - density (kg/m )
v - velocity (m/s)
D - pipe diameter (m)
- dynamic viscosity (Ns/m )
2
R < 2 000: laminar flow; pressure directly proportional to velocity
R > 4 000: turbulent flow; pressure proportional to the square of velocity
2 000 < R > 4 000: laminar or turbulent flow depending on the roughness of the pipe and
upstream turbulence of the fluid.
30
THE LAWS OF FLUID DYNAMICS
Airflow is analysed as it passes through a fixed region in space defined by a control surface.
The properties of the air are examined as it enters and leaves the volume bounded by the
control surface.
Laminar & Turbulent flow - the REYNOLDS NUMBER of a fluid determines whether laminar
or turbulent flow will predominate.
THE LAW OF CONSERVATION OF MASS
This law states that the mass flow entering the control surface must equal the mass flow rate
leaving the control surface, i.e.:
M1=M2
Or
W1A1V1=W2A2V2
This is the CONTINUITY EQUATION;
Where:
M = mass flow rate
w = density
A = flow area measured normal to the direction of flow
V = flow velocity
If the air is of constant density, then:
A1V1=A2V2
Or
Q1=Q2
where Q is the volume flow rate
Equation (2) is known as the CONTINUITY EQUATION.
31
THE LAW OF CONSERVATION OF MOMENTUM
Newton's second law of motion states that the force 'f'’ to accelerate a body must be equal to the
product of the mass 'm' of the body and its acceleration 'a', i.e. F = ma (N)
If this law is applied to an ideal (frictionless), incompressible (constant-density) fluid, EULER'S LAW of
conservation of momentum is obtained.
PROOF- EULER'S LAW
Let a fluid move steadily through an imaginary fixed volume element xyz (Figure below)
Consider the flow across plane yz (in the x - direction)
If pressures P1 and P2 act on the opposite surfaces, the corresponding forces are:
P1y z
And
P2y z
These are surface forces. In addition to these, the fluid may be subjected to body or mass forces.
If: B denotes the mass force per unit mass in the x direction and w denotes the constant density of the
fluid, the corresponding mass force is:
Bwxyz
32
The resultant force on the fluid is thus:
F = P1yz – P2yz + Bwxyz
The fluid flows in with a velocity V and out with velocity V . If x is very small, the acceleration will be
constant and is represented by:
1
2
a = (V - V ) / t
1
2
Where:
t = the time for the fluid to flow through distance x
Since the mass of the fluid element is m = wxyz, Newton's Law (F=ma) yields:
P yz – P yz + Bwxyz = wxyz(V - V ) / t
1
2
1
2
which simplifies to:
P - P + Bwx = wx(V - V ) / t
1
2
1
2
The mean velocity over the distance x is:
(V + V ) / 2 = x / t
1
2
Substituting for x / t in the main equation yields:
P1 P2 +Bwx =w (V22- V21)
2
This is EULER'S EQUATION for the x - direction, also known as the MOMENTUM EQUATION.
33
Ventilation
Topic 5
Fluid Mechanics - Bernoulli’s Equation
Bernoulli’s Equation and Compressible Fluids
Assume that a fluid may be considered incompressible if its density does not change by more than
5%. If it is assumed further that the relationship between static pressure and density is linear, then the
acceptable change in static pressure is also 5%. To find the limit of acceptability of the Bernoulli
equation in terms of air velocity, that part of the equation stating that:
2
P = ½ wV
may be used. The absolute pressure of atmospheric air is about 105 Nm-2, 5% of which is
3
-2.
5 x 10 Nm
2
With w = 1 kg.m-3 for air, the dynamic pressure 1/2 wv equals 5% of the absolute pressure when:
2
3
V = 5 x 10 x 2
or 100 m/s.
Bernoulli’s equation therefore gives rise to appreciable errors for air velocities greater than 100 m/s
when air is at about atmospheric pressure. As these velocities are not exceeded in mine airways, the
equation can be used when dealing with air, even though it is a compressible fluid.
To find the limit in terms of height, that part of the equation stating:
P=wgH
may be used. The value of H that results in a 5% change in absolute pressure is obtained by
substituting:
3
P = 5 x 10 Nm
w = 1 kg.m-3
-2
g = 9.8 m.s
-2
This gives:
H = 510 m.
It is now apparent that for air at about atmospheric pressure with height changes of less than about
500 m and velocity changes which do not exceed 100 m/s, the application of Bernoulli’s equation
should not produce unduly great errors. In the majority of mine ventilation applications, very little error
will be introduced. For extended surveys of shafts, the height limit will be exceeded and other
methods will be employed.
Bernoulli’s Equation and Non-Frictionless Fluids
The Bernoulli and Euler equations are only valid for ideal or frictionless fluids. Real fluids are
subjected to shear resistance against relative movement of their layers.
34
Consider a fluid volume xyz subjected to a shear force ‘s’ acting on two opposite faces a distance ‘y’
apart (see diagram 1 at the end of these notes). The quotient of the force ‘s’ and the area of the plane
parallel to its direction of action (area xz) is the shear stress:
Shear stresses give rise to a continuous movement of the one face relative to the other. If the velocity
of the one layer relative to the other is dV, then the rate of shear strain is dV/y. Newton postulated
that:
The factor of proportionality, , is called the coefficient of dynamic viscosity and is expressed in
-2
-1 -1
N.s.m or kg.m .s . Viscosity depends on temperature. This is to be expected since the shear stress
(of which viscosity is a measure) results from molecular motion within the fluid and temperature is a
measure of the degree of molecular motion. The variation of viscosity with temperature is shown in
figure 2 at the end of these notes. Note that the viscosity of gases is proportional to temperature while
the viscosity of liquids is inversely proportional.
In many real fluids the effects of shear stresses are small and Bernoulli’s concept of the ideal fluid is
reasonably accurate for most practical cases concerning the relationship between pressure and
velocity changes in a fluid. The Navier-Stokes equation is similar to Euler’s equation but it allows for
viscous forces. This equation is rather complex and has been applied successfully to real fluid flow in
only a few simple cases.
5. PROOF OF BERNOULLI'S EQUATION
If the x - direction is taken as vertically upwards (Figure above), where
P1 & V1 are at height H1 and
P2 & V2 are at height H2
the distance x = H2- H1
35
The force B per unit mass is now due to gravity and corresponds to the weight per unit mass of the
fluid column between H1 and H2. The weight per unit mass is g, therefore: B = -g as the weight is
directed downwards, opposite to the direction of P1
Substituting x = H2- H1and B = -g into Euler's equation yields:
2
2
P1- P2- gw(H2- H1) = w(V1 - V2 ) / 2
Or
2
2
P1+ w(V1 / 2) + gwH1= P2+ w(V2 / 2) + gwH2
This is BERNOULLI'S EQUATION for the ideal, incompressible fluid in terms of pressure. Air is both
compressible and non-frictionless (non-ideal).
5.1
BERNOULLI'S EQUATION and compressible fluids
A fluid may be considered incompressible if it's density does not change by more than 5%. Assuming
a linear relationship between static pressure and density, the acceptable change in static pressure is
also 5%.
Bernoulli's equation may be used if air is near atmospheric pressure and: the height does not change
by more than 510 m and velocity changes do not exceed 100 m/s
5.2 BERNOULLI'S EQUATION and non-frictionless fluids
36
Bernoulli's equation and Euler's equation are only valid for ideal or frictionless fluids. In reality,
however, fluids are subjected to shear resistance against relative motions of their layers.
Consider a fluid volume xyz subjected to a shear force 'S' acting on two faces a distance 'y' apart
(Figure above)
The shear stress is thus:
If the velocity of one face relative to the other is dV, the rate of strain is dV/y.
Thus:
Where:
= coefficient of dynamic viscosity (Ns/m or kg/ms)
2
In many real fluids, the effects of shear forces are small and Bernoulli's equation may be applied.
37
6. THE BERNOULLI EQUATION & MINE VENTILATION
6.1.1 For an ideal fluid:
2
P1+ ½ w1v1 + w1g1H1= constant
[1]
where:
P1
2
½ w1v1
w1g1H1
-
represents static pressure (SP)
represents velocity pressure (VP)
represents potential energy
6.1.2 For a non-ideal fluid:
P1 w1V12 +w1g1H1  P2 w2 V22 w2 g2 H2 PL
2
2
[2]
Where:
PL is the sum of all friction losses
if H 0
P1 w 1V1 2 P2 w2 V22 PL
2
[3]
2
6.2 Absolute Pressure & Gauge Pressure
38
The absolute pressure inside the column is 110 - 0.4 = 109.6 kPa
The absolute pressure inside the column at:
x1 is 109.6 kPa
x2 is 100.6 kPa
P = 9 kPa
6.3 Pressure Loss (PL)
From equation [3]
P = P + 1/2w v 1- (P2+ 1/2w v )
[4]

[5]
2
L
1
1
2
2
2
PL = { (SP) + (VP)1} - { (SP)2+ (VP)2}
1
6.3.1 Constant diameter column
(VP)1= (VP)2
(constant diameter)
PL= (SP)1- (SP)2
[6]
PL = (+700) - (+200)
= 500 Pa
39
Note:
(TP)1 = (VP)1+ (SP)
= 200 + 700
= 900 Pa
(TP)2 = (VP)2+ (SP)2
= 200 + 200
= 400 Pa
PL = (TP)1- (TP)2
= 900 - 400
= 500 Pa
[7]
6.3.2 Variable diameter column
PL = (TP)1- (TP)2
= 600 - 500
= 100 Pa
Note: (SP)1= 500
(SP)2= 200
PL(SP)1- (SP)2
6.4 Fan Total Pressure (FTP)
The effect of adding a fan is as follows:
2
2
P1+ ½ w1v1 + FTP = P2+ ½ w2v2 + PL
[8]
40
FTP = (P2+ ½ w2v2 ) - (P1+ ½ w1v1 ) + PL
2
2
[9]
FTP = (SP2+ VP2) - (SP + VP ) + PL
[10]
FTP = (TP)2 - (TP)1+ PL
[11]
1
1
6.5 Boundary Conditions at Extremities of Columns
6.5.1 Column intakes
From equation [7]:
PL
= (TP)1- (TP)2
but PL = 0
and (TP)1
=0
thus 0
= 0 - (TP)2

(TP)2
=0
but
(VP)
= 200 Pa
and
(TP)2
(SP)2
2
= (VP)2+ (SP)2
= (TP)2- (VP)2
= 0 - 200
= -200 Pa
Therefore, at the intake:
(TP) = 0
(SP) = - (VP)
assuming no shock losses
As an example of entry losses affecting pressures within the duct, consider the previous
example but this time make allowance for an entry loss of 75% of the velocity pressure
Inlet shock losses = 0.75 x 200 = 150 Pa
PL
(TP)2
= (TP)1 - (TP)2
= (TP)1 - PL
41
but (TP)1

(TP)2
=0
= 0 - 150
= -150 Pa
(VP)2
(TP)2
(SP)2
= 200 Pa
= (SP)2+ (VP)2
= (TP)2- (VP)2
= (-150) - (200)
= -350 Pa
6.5.2 Column deliveries
(SP) = 0
(VP) = w V
2
2
(TP) = (SP) + (VP)
 (TP) = (VP)

thus at the delivery:
(TP) = (VP)
(SP) = 0
6.5.3 The global pressure change
Consider air being drawn into a column at point 1 by a fan and delivered from point 2:
From equation [10]:
FTP = (SP2+ VP2) - (SP1+ VP1) + PL
= (TP)2- (TP)1+ PL
but point 1 in an intake and point 2 is a delivery
(TP)1 = 0
(TP)2= (VP)2
42
So
FTP
= (TP)2- (TP)1+ PL
= (VP)2- 0 + PL
= (VP)2+ PL
6.6 Fan Pressures
6.6.1
Fan total pressure
a)
FTP
= (TP)2 - (TP)1
= 600 - (-240)
= 840 Pa
b)
FTP
= (SP)2- (SP)1
= 410 - (-430)
= 840 Pa
6.6.2 Fan static pressure (FSP)
FSP
= (SP)2- (TP)1
= 410 - (-240)
= 590 Pa
The effect of a fan at the inlet of a column
43
J = facing gauge
FTP
= (TP)2 - (TP)1
= 800 – 0
= 800 Pa
The measurement may be accomplished with a facing gauge at point 2.
or:
FTP
= (SP)2- (SP)1
= 600 - (-200)
= 800 Pa
Using a side gauge
The effect of a fan at the outlet of a column
FTP
= (SP)2- (SP)1
= 0 - (-800)
= 800 Pa
using a side gauge
or:
FTP
= (TP)2- (TP)1
= 200 - (-600)
= 800 Pa
using both side and
facing gauges
6.7 Evasées
This is a device which is fitted to the discharge of a system and is used to convert the wasted velocity
pressure into useful static pressure. This is done by enlarging the area at the discharge of the system,
thus decreasing the air velocity and hence the velocity pressure. This reduction in velocity pressure
represents a gain in static pressure.
44
At the discharge this 100 Pa is 'wasted'. An evasée could be fitted to reduce the velocity
pressure from 100 Pa to 50 Pa.
Now (VP) at
A
= 100 Pa
B
= 50 Pa
loss in (VP)
= 50 Pa
but (TP)
= (SP) + (VP)
From the previous equation it is clear that whatever is lost in (VP) should be gained in (SP) so that the
(TP) will remain unaltered.
loss in (VP) = 50 Pa
so gain in (SP) = 50 Pa
A certain amount of this 'saving' is used to overcome friction and shock losses within the evasée.
Assuming that these losses amount to 10 Pa, then:
theoretical gain in (SP) = 50 Pa
actual gain in (SP)
= 50 - 10 Pa
= 40 Pa
45
The efficiency of the evasée can be calculated by:
Evasée efficiency
=
actual (SP) gain
x 100
theoretical (SP) gain
= 40 x 100
50
= 80%
A static pressure measurement at the discharge of any column will read zero as all of the static
pressure is used up in overcoming the friction of the system.
But between A and B there is a gain is (SP). The (SP) is thus greater at B than at A. Since the (SP) at
B is 0, the (SP) at A must be negative. This is always the case when an evasée is fitted to the
discharge of a system.
Note:
I.
VP is constant as long as the column diameter is constant and decreases as soon as the
diameter increases in the evasée.
II.
A negative static pressure is obtained at the evasée inlet (A) of 40 Pa. There is then a gain in
SP of 40 Pa and the SP at the end of the column equals zero.
III.
The friction losses between A and B are 10 Pa. In order to measure this loss due to friction in
the evasée it would be necessary to measure TP at A and the TP at B and obtain the
difference. TP is measured using a facing gauge.
IV.
The benefit of using evasées is that the fan pressure decreases by an amount equal to the
regain in SP at the evasée. Consequently, a larger quantity of air will flow through the system.
46
6.7.1 Shock / entry losses for ventilation columns
Often quoted as a percentage of velocity pressure:
Straight inlet
Flanged inlet
90% - circular pipes
50% - circular pipes
70% - rectangular pipes
Inlet cone: 5 - 25% - depending on angle and length
Flared cone: 3 -15% - depending on radius
47
Ventilation
Topic 7
7. Airflow in Airways
Air density:
Wits:
Klerksdorp & OFS:
Sea level:
3
0.96 kg/m
3
1.00 kg/m
3
1.20 kg/m
Elementary Laws of Airflow
1. For air to flow from one point to another, there must be a difference in pressure between
these points. This is known as ventilating pressure.
2. Air flows from high pressure to low pressure.
3. The larger the ventilating pressure, the greater the quantity of air that will flow.
4. Resistance to airflow reduces the pressure (this is used to overcome the resistance).
5. If the pressure difference between two points remains the same, and the resistance
between the points is increased, the quantity of air flowing between the points will decrease
Factors creating resistance to airflow
1.
2.
3.
4.
5.
6.
Skin factors (rougher increases R)
Area normal to direction of flow (smaller area increases R)
Distance (greater distance increases resistance)
Shock losses (sudden changes in area increase resistance)
Obstructions to airflow (increase resistance)
Changes in direction (increase resistance)
7.1 Pressure Loss in Laminar Flow
This pressure loss is due to the viscous resistance of air (shear stress generated when layers flow
over one another)
· Circular ducts: Hagen-Poiseuilles Equation
P = 32Lv
2
D
48
Where:
P: pressure loss (Pa)
: dynamic viscosity of air (Pa.s)
L: airway length (m)
v: mean air velocity (m/s)
D: airway diameter (m)
7.2 Pressure Loss in Turbulent Flow
Due to shear stresses which are caused by the frictional resistance of the airway wall.
· For circular airways: D'Arcy's Equation
P = 4 fwLv
2D
2
Where:
P: pressure loss (Pa)
f : friction factor [ ]
w: air density
·
Non-circular - use equivalent diameter
(De) = 4A
C
Where:
A: cross-sectional area of airway (m 2)
C: perimeter of the airway (m)
7.3 Reynold's Number
· For circular ducts
Re = wvD = vD

= w
49
Where:
Re: Reynold's number [ ] (dimensionless)
2
: kinematic viscosity of air (m /s)
NB:
1)
Re < 2 000
2 000 < Re < 4 000
Re > 4 000
laminar
transitional
turbulent
2) of air increases with temperature
· For non-circular ducts
D = 2a b
A+b
Where:
a: breadth
b: width
7.4 D'Arcy's Friction Factor & Reynold's Number
7.4.1 Laminar
Compare Hagen-Poiseuille & D'Arcy Equation
32 Lv = 4 fwLv
2
D
2D
2
f
= 16 
WvD
f = 16/Re
this only applies to laminar flow
7.4.2 Turbulent
The friction factor (f) is a function of Re and relative roughness e.
Relative roughness = e
D
where e: average height of surface irregularities
50
Required data can not be obtained from the graph below
The Stanton and Nikuradse diagram (after Schlichting)
7.5 Atkinson's Equation
P = KCLV2
A
since v = Q/A
P = KCLQ2
A3
X W
1.2
Where:
2
4
K: Atkinson's friction factor (Ns /m )
C: perimeter of airway (m)
L: Length of airway (m)
A: Area of airway (m 2)
To account for density
P = KCLV2 X W
A



(1.2 = density of air at msl [mean sea level])
1.2
7.5.1 D'Arcy's friction factor and Atkinson's friction factor for a non-circular duct
4fwLV 2C K CLV2
2(4A)
A
X W
1.2
k = 0.6f
51
7.5.2 The resistance of an airway
P = RQ
2
R = KCL


Q = AV
P =
KCLQ2 W
3
A
laminar:
turbulent:
models:
1.2
P = RQ
2
P = RQ
n
P = RQ
7.5.3 Typical values of the friction factor (K)
52
7.6 Losses of Pressure Due to Other Causes
Two cases:
I.
II.
change of direction (bend)
change in cross-sectional area
P = (1/2 wv )
2
53
Where : loss factor
Another method entails calculation of the equivalent length, for example:
45(short radius bend) in a 0.5 m duct. Treat as a straight duct, but increase the length to account
for the increased resistance due to the bend.
L =n
for a 45bend, n = 18
D
the increase in length is L = 18 x diameter
= 18 x 0.5
= 9 m increase in length
Obstructions retard the flow of air in the following manner:
1. By causing a reduction in the cross-sectional area, and consequently an increase in the
air velocity
2. By disrupting the flow pattern of air, causing eddy currents which result in a loss of
Pressure
Measures to reduce the loss factor, , include:
1. Rounding off sharp corners and inserting guide vanes
2. Limiting angles by which the conduit increases or decreases it's cross-sectional area to 7
3. Rounding corners and inserting guide vanes in junctions
4. Chamfering the horizontal take-off from the shaft
5. Providing cages with fairings and close fitting doors.
7.7 Air Power
WD = PAL
Power =PAL
t
but volume / t = Q = AL
t
Power = PQ
Or

Power KCLQ2
w Q

=KCLQ3 w
A3
1.2

NB
Power is proportional to Q
3
54
Bends
55
The table below indicates the different values of for different bends and conditions
56
Ventilation
Tutorial 3
Airflow
PROBLEMS – AIRFLOW
QUESTION 1
(a) Describe the mechanism of pressure loss in:
(i) Laminar flow, and
(ii) Turbulent flow.
(b) Describe two ways in which obstructions retard the flow of air in airways.
(c) What are the units of Reynold’s number, resistance, and the friction factor (K).
QUESTION 2
Calculate the air density in a shaft given the following information:
Reynold’s number
37000
Air velocity
7 m/s
Shaft diameter
6.9 m
Dynamic viscosity of air
0.0016 Pa.s
QUESTION 3
Calculate the pressure required to overcome friction when 50 cubic metres per second of air
at a density of 1.2 kg per cubic metre flows through a 300 m length of 3m * 3m roadway.
QUESTION 4
A 200 m long raise has been developed using a 2 m diameter raise borer. If the Reynolds Number is
2 000 000 and the surface roughness is one centimetre, what is the expected pressure loss in the
raise for an airflow of 45 cubic metres per second?
QUESTION 5
What is the resistance of the roadway in Question 3?
QUESTION 6
Calculate the pressure loss when 4 cubic metres per second of air flows through a duct having a
resistance of 9.3 (usual units for resistance).
QUESTION 7
Calculate the air power expended when 4 cubic metres per second of air flows through a duct having
a resistance of 40.
QUESTION 8
Plot the system resistance curves of the ducts dealt with in Question 6 and 7.
57
Ventilation
Topic 8
Ventilation Networks
A mine normally consists of a large number of interconnected airways of various shapes, sizes and
lengths. In order to facilitate calculation, complex networks can be subdivided into branches,
junctions and meshes.
Branch :
Junction:
Mesh:
the airway between two junctions
a point where three of more airways meet
a closed path traversed through the network
The following picture indicates the branches, junctions and meshes in an underground
mine layout.
8.1 Solving Ventilation Networks
The following methods are available:
58
I.
II.
III.
IV.
equivalent resistances
direct analysis
ventilation analogues
iterative techniques
8.2 Equivalent Resistances
Airways are in series or parallel
8.2.1 Roadways in series
QT = QA = QB
The total pressure in the system is the sum of the pressure loss in airways A and B:
PT= PA + PB
2
since P = RQ , this can be written as follows:
2
2
2
RTQT = RAQA + RBQB
but
QT = QA = QB
RT = RA + RB
8.2.2 Graphical representation of roadways in series
Plot the resistance curve of the airways:
AB
BC
CD
2
3
RAB = 0.066 Ns /m
2
3
RBC = 0.045 Ns /m
2
3
RCD = 0.012 Ns /m
59
The curve for each airway may be plotted by following the series laws and referring to quantity
and pressure:
i.e.
PT = P1 + P2
QT = Q1 = Q2
To plot the system resistance curve of AD:
Use a table:
You know R,
choose Q and calculate P
Because the quantity of air in each airway is equal, any quantity line can be selected and the pressure
required by airway 1 can be added to the pressure required by airway 2 (both on the same quantity
line) to obtain the total pressure required by both airways.
A point can then be plotted and other points can be obtained in the same way until sufficient points
are obtained to plot a system curve representing both airways together. This is shown below.
60
The dotted curve is the combined curve for the two airways in series.
8.2.3 Roadways in parallel
Parallel roadways are side by side and have a common inlet and return
PT = PA = PB
QT = QA + QB
since P = RQ
2
2
2

2
2
RTQT = RAQA = RBQB
And
QB = ( RAQA / RB) and QT =
( RAQA2 / RT)
Substitution yields:
(RAQA / RT) = QA + (RAQA / RB)
(RA / RT) = 1 + (RA / RB)
2
2
dividing by RA :
1 / (RT) = 1 / (RA) + 1 / (RB)
61
8.2.4 Graphical representation of roadways in parallel
Airways in parallel can be solved by plotting the system resistance curves for airway. This is
done by following the laws for quantity and pressure:
PT = PS = P2
QT = Q1 + Q 2
Because the pressures are all equal, any pressure line can be selected and the quantity in airway
one added to the quantity in airway two (both on the same pressure line) to obtain the total
quantity in both airways. A point can now be plotted and the other points can be obtained
in the same way until sufficient points have been obtained to plot a system curve.
8.3 Networks: Roadways in Series and Parallel
In order to solve complex networks, the system may be simplified by observing an order of operation:
I.
parallel airways first
I.
begin with the smallest parallel circuit
II.
solve the resultant series circuit
62
9. REGULATORS & BOOSTER FANS
9.1 Regulators
Regulators are openings of specific size which are used to control airflow. This is achieved by limiting
the area through which air flows, thereby increasing the overall resistance of the systems and
reducing the volume of air in circulation. Regulators have the effect of “using up” pressure that would
otherwise have been used in overcoming friction.
Case 1
PL = 500 Pa
Over the length of airway, a pressure drop of 500 Pa is experienced due to friction.
Case 2
PL = 500 Pa
In the same airway, the installation of a regulator results in only 200 Pa being available to
overcome friction. 300 Pa is used up by the regulator. The total pressure drop remains 500
Pa. The quantity of air flowing through the system in case 2 may be determined as follows:
PQ
And
2

P1 = P2
2
2
Q1 Q2
2
2
Q2 = Q1 x P2
P1
Q2 = Q1 x ( P2 / P1 )
63
Q2 = 30 x (200 / 500)
3
Q2 = 18.97 m /s
The pressure drop is the same, but the overall resistance is increased because the quantity of air
flowing through the system has decreased:
P = RQ2

Resistance: case 1
R=P
2
Q
R = 500
2
(30)
= 0.55 Ns/m
Resistance: case 2
8
R = 500
2
(18.97)
= 1.39 Ns/m
8
Size of regulator required:
Ar = 1.1 Q (w / P)
Where:
2
Ar - area of regulator (m )
Q - quantity of air flowing through the regulator (m 3/s) (i.e. Q2)
P - pressure used up by regulator
3
w - air density (kg/m )
64
A regulator works as a sharp-edged orifice. The actual flow through a regulator is less due to losses
and the contraction of the airstream after exit from the orifice. The coefficient of discharge, Cd allows
for these effects:
Q = Cd Ar (2 H g)
3
Where Q: volume flow rate (m /s)
2
Ar: area of the regulator (m )
H: pressure producing the flow (metres of air column)
if Cd = 0.64, then
Q = 0.64 x Ar (2 H g)
from the above:
Ar =
.
Q
.
0.64 (2 P g / w g)
= 1.105 Q (w / P)
65
9.2 Booster Fans
Underground booster fan installed through a wall in a haulage
As regulators are used to decrease the quantity of air flowing in an airway, booster fans are
designed to assist the main surface fan with the in-mine pressure drop that occurs.
The sketch shows air distribution in a section of a mine. The pressure loss between A and C is 800
3
Pa. The mine has to increase the quantity of air flowing in ABC to 35 m /s. This is to be achieved by
installing a booster fan.
9.2.1 Determination of the duty of a booster fan
Pressure drop AC = Pressure drop ABC as these are parallel airways.
This pressure is supplied by fans in another part of the system.
66
At present, 800 Pa of pressure is required to cause 20 m 3/s of air to flow. An unknown pressure
3
“P2” is required to cause 35 m /s to flow.
P2
2
= P 1 x Q2
2
Q1
2
= 800 x (35)
2
(20)
= 2 450 Pa
There is already a pressure difference of 800 Pa across ABC due to the action of other fans.
booster fan pressure = 2 450 - 800
= 1 650 Pa
3
The quantity of air handled by the booster fan must be a total of 35 m /s at a minimum pressure of
1650 Pa.
3
The duty of the booster fan is thus 35 m /s at 1 650 Pa
Note: How can AC have a pressure drop of 800 Pa when the pressure drop in ABC is 2 450 Pa and
the two airways run in parallel?
The answer is simple:
The booster fan adds pressure to the air and there is thus a “fan pressure gain” in ABC. The pressure
losses due to friction in ABC are 2 450 Pa and the pressure gain due to the booster fan is 1 650 Pa.
The difference is 800 Pa.
67
Ventilation
Tutorial 5
Regulators & Booster Fans
QUESTION 1
The pressure loss in the above airway is 720 Pa and the air density is 1,058 kilograms per cubic
metre. What size regulator would be required to reduce the air quantity to 30 cubic metres per
second?
QUESTION 2
The pressure drop over the parallel airways shown in the sketch below is 900 Pa. The air density is
1,15 kilograms per cubic metre. Determine the size of the regulator required to reduce the quantity of
air flowing in airway B to 60 cubic metres per second.
QUESTION 3
The sketch below shows how air is distributed in a section of a mine. The pressure loss A-C is
900 Pa. It is desired to increase the quantity flowing in ABC to 30 cubic metres per second by
installing a booster fan. Determine the duty of the required booster fan.
68
Ventilation
Topic 10
Fan Types
10. FAN TYPES
10.1 General Description
Main ventilation fans are usually placed at the top of the upcast shaft. This has the following
advantages:
1. The working environment is maintained at a slightly lower pressure than the atmosphere. If
the fan fails, natural ventilation will occur until the pressure underground has built up to
normal atmospheric pressure.
2. Discharge evasées may be fitted, converting velocity pressure into static pressure and
thereby enabling power savings.
3. No heat is added to the underground air.
The disadvantages include:
1. The fan must handle foul air, making repair and maintenance difficult. Corrosion may become
a serious problem.
2. Natural ventilation is reduced
3. It becomes difficult to use the upcast shaft for hoisting of materials.
An alternative to this layout requires the placement of the fan at the top of the downcast shaft.
69
This presents the following disadvantages:
1. Access to the shaft is very difficult
2. Hoisting of men, materials and ore is difficult.
3. Heat is added to the downcast air.
Placing the fan at the bottom of the upcast is generally the optimal solution.
1. Although the fan still produces heat, no heat is added to the air breathed by miners
2. The up-cast shaft is clear for hoisting.
Disadvantages:
1. the fan handles foul air
2. the fan is not accessible during fires
3. re-circulation of foul air is likely unless a complex system for preventing this is installed.
10.1.1 Types of Fans
Natural ventilating pressure is not sufficient to force ventilation to flow through modern large scale
mines. In order to overcome this, various types of fans have been devised.
The compression ratio of a fan is the discharge absolute pressure divided by the inlet absolute
pressure. If this ratio is less than 1.1, the device may be referred to as a fan. If the ratio is greater than
this value, the device is referred to as a compressor. This means that the pressure increase due to a
fan is limited to 10 kPa, if the fan inlet is at atmospheric pressure. Fan outlet pressure is usually
between 200 Pa and 7.5 kPa.
A fan is a rotating machine drawing air in one side and delivering it at a higher pressure at the other.
Mechanical energy is thus converted into potential energy.
70
i) Axial flow fans the direction of airflow is parallel to the axis of the shaft
ii) Centrifugal fan or Radial flow fan air enters the fan parallel to the axis and is discharged
radially
71
10.1.2 Mine Fans
i) Main fans
These are responsible for ventilating the bulk of the mine. They may be axial or radial but are usually
3
radial. They are electrically driven and commonly move quantities of air ranging from 12 m /s to
3
320 m /s.
72
The surface main fan installation above includes two centrifugal fan installations. One fan is generally
switched off and acts as a stand-by system in case of maintenance or emergency. It can be clearly
seen that the two fans are connected in parallel in this specific system.
Another view of the same fan installation is the following:
The following surface fan installation is given to illustrate an axial fan.
73
The surface fan installation above is that of an axial main fan system. Note the grass surrounding the
installation. This could become dangerous in case of a fire in that a fire could ultimately damage the
motor and electrical equipment.
ii) Auxiliary fans
Such fans are used to ventilate areas where air would not naturally enter. Examples of these areas
include development ends and some stopes. Auxiliary fans commonly have diameters of 400 mm,
570 mm or 760 mm.
10.1.3 Fan Parts
Principle component: impeller- hub
- blades
- mounted on shaft driven directly/ indirectly by motorconcealed
by housing
inlet cone
outlet evasée
guide vanes to control the direction of air through the
cone.
74
10.2 Axial Flow Fans
The impeller consists of two or more blades. Higher pressure machines generally have a higher
number of blades. Shape and pitch of the blades vary considerably. The general configuration
consists of short stubby blades on a large hub. The size of the hub is important as it prevents re-entry
of air at the centre of the fan, thereby preventing pressure loss.
Some fans have adjustable pitch. This means that the pitch may be mechanically altered once the fan
has been stopped. Variable pitch refers to adjustment via a lever. This may occur during operation in
some models. For this type of variability, the clearance between the blades and the housing needs to
be very small. Accurate machining is required, increasing the cost.
Fans may have two or more sets of blades mounted behind one another on the same shaft
(multistage fans). These can produce higher pressure than is possible with a single stage fan. Air
leaving an axial flow fan has a rotary motion, which represents a loss of energy. This can be corrected
with guide vanes (stationary blades) or by using a multi-stage fan with a contra-rotating second
impeller. In some cases, guide vanes may also be adjustable.
10.3 Centrifugal Fans (Rotary Fans)
75
Here the impeller consists of a hollow "wheel" or runner with a large number of blades orientated
around the circumference.
The blades are generally short and have a high width to height ratio. These fans may be
arranged in three ways:
I.
radially
II.
forward curved
III.
backward curved
76
When the impeller rotates, air between the blades is thrown outwards, leaving the impeller at right
angles to the axis before being forced to the outlet by the fan casing. Air moves into the fan to replace
the evacuated air. These fans may have single or double inlets.
10.4 Fans Used in Auxiliary Ventilation
10.5 Effects of Reversal of Direction
I.
Axial flow fan: will deliver reduced quantity in the opposite direction at greatly reduced
pressure and efficiency ().
II.
Centrifugal: will deliver reduced volume in the same direction at reduced pressure and .
77
Ventilation
Topic 11
Fan Characteristics
Measures of Fan Pressure
Fan pressure is expressed in terms of pressure rise through the fan as follows:
1. Fan total pressure is the difference between the total pressures at the fan discharge and
intake.
2. Fan pressure is the fan total pressure minus the fan velocity pressure.
3. Fan velocity is the velocity pressure corresponding to the average velocity at the fan
discharge. The average velocity is calculated by dividing the volume flow rate of air by the
area of the fan discharge orifice.
Fan static pressure is often regarded as a measure of useful fan pressure.
The total pressure developed by a fan depends on the volume flow rate through it. When intake and
discharge are fully open, the fan total pressure is equal to the fan velocity pressure and the fan static
pressure is zero. As the volume flow rate decreases, the fan velocity pressure decreases while the
fan total pressure and consequently the fan static pressure increase. Theoretically, the fan total
pressure should reach a maximum value when the intake and/or discharge are closed (i.e. when there
is no airflow through the fan). Due to friction and shock losses, the fan total pressure reaches a
maximum at a point in the range of the volume flow rate. The position of this point being dependant
on the design of the fan in question.
A graph of fan pressure against the volume flow rate of air is known as the fan characteristic and is
determined by the laboratory work described below. The shape of the curve varies with the type of
fan. A typical characteristic curve is shown overleaf.
Fan Tests
Reasons for testing fans
1. To check whether the fan performance complies with the manufacture’s specifications. This
must be done as soon as possible after installation.
2. To check the fan performance after an overhaul.
3. To obtain performance figures which may not be readily available. Examples include
performance at different pitch settings and ranges not covered by existing curves.
4. To determine whether the fan is still performing satisfactorily. Prolonged use may have
resulted in inefficiency due to wear or accumulations of dirt on various parts.
Measurements Required to Assess Fan Performance
A collection of curves for a particular fan applies only for:
a) a given air density,
b) a specified fan speed and
c) a particular pitch setting (if the fan has adjustable pitch blades or guide vanes.
These measurements should all be specifies on the graph.
78
Air density is measured at the fan intake and is obtained from whirling hygrometer and barometer
readings, using psychrometric charts.
The fan speed is measured by means of an accurate rev-counter or stroboscope.
The curves are produced using the following data:
1. quantity handled by the fan,
2. pressure produced by the fan and
3. input power.
Quantities are measured with a vertical manometer.
The input power is measured with a watt-meter in preference to the ammeter and voltmeter generally
found on the control panel of the fan.
General method
To obtain different points on the characteristic curve of the fan, the resistance against which the fan is
working is altered in stages. The quantity, pressure and input power are recorded at each stage. Such
tests are conducted in laboratories.
Manometers may be employed to measure fan pressure. They are used differently, depending on the
location of the fan in the duct.
11.1.1 Forcing fan at duct inlet
11.1.2 Fan in column
i) FTP
79
ii) FTP
NB duct of uniform diameter
iii) FSP
11.1.3 Exhaust fan at pipe outlet
11.2 Calculation of Power and Efficiency
The electric power consumed by a three-phase AC motor is described by the following equation:
W = EI pf 3
80
Where W : power (kW)
E : voltage (kV)
I : current (A)
Pf : power factor (0.9)
Power delivered at the shaft = motor output power
Power delivered to fan = fan input power
= motor output x drive efficiency
Power added to air by the fan = air power
W a = PQ
1000
Where:
Wa: air power (kW)
3
Q: volume flow (m /s)
P: pressure (Pa)
Air power can also be calculated by multiplying fan input by fan efficiency.
Using the sketch, it can be seen that:
Motor efficiency = motor output power = b x 100%
power input to motor a
Drive efficiency = fan output power
= c x 100%
motor output power
b
81
Fan efficiency
= air power
fan input power
=
Overall efficiency = air power
Power input to motor
d x 100%
c
= d x 100%
a
For example:
3
A fan ventilating a mine handles 85 m /s of air at a pressure of 750 Pa. The fan is driven by a 3-phase
AC motor which consumes 65 A from a 1 000 V supply; the pf is 0.9. The motor efficiency is 95% and
the drive efficiency is 92%.
Calculate:
a) the input power
b) motor output power
c) fan input power
d) fan efficiency
e) overall fan efficiency
Answer:
a) a) input power: W
= EI pf 3
= 1x 65 x 0.9 x 1.73
= 101 Kw
b) motor output power:
= 101 x 0.95
= 96 kW
c) c) fan input power:
= 96 x 0.92
= 88 kW
d) d) fan efficiency:
= air power
fan input power
= PQ / 1000
88
= 63.75
88
= 72%
82
e)
Overall efficiency:
= air power
input power
= 63.75 / 101
= 63%
Fans with belt drives
The speed of a belt driven fan can be altered by changing the diameter of either the motor pulley
or the fan pulley.
Fan speed varies:
inversely with the diameter of the fan pulley
directly with the diameter of the motor pulley
Fan speed = motor speed x motor pulley diameter
fan pulley diameter
Example:
A fan runs at a speed of 12.3 rps. The fan pulley diameter is 195 mm and the motor pulley diameter is
150 mm. Determine the pulley sizes required to increase the fan speed to 18 rps.
From the sketch it can be seen that the fan speed may be increased by either:
decreasing the fan pulley diameter
increasing the motor pulley diameter
83
I.
new  
= old x old fan speed
new fan speed
= 195 x 12.5
18.0
= 133.25 mm
II.
new  = old x new fan speed
old fan speed
= 150 x 18
12.3
= 219.5 mm
11.4 Characteristic Curves
Quantity on the x axis is plotted against:



pressure
power
efficiency
In order to calculate the efficiency of the fan, air power must first be determined:
W a = PQ
1000
Efficiency is determined by the following expression:
= W a
x 100
input power
Example table
84
2
The resistance of the system is calculated to be 33.95 Ns .m
-3
Determine:
1. the fan operating point
2. fan efficiency
3. fan power
The fan operating point is given by the intersection of the system resistance curve and the fan
characteristic curve.
3
1. FOP = 4.7 m /s
2. = 62.5%
3. Power = 5.665 kW
11.5 Natural Ventilating Pressure
i) NVP assisted
(SOP = System Operating Point)
2
2
P = RQ P = RQ - NVP
ii) NVP opposed
2
2
P = RQ P = RQ + NVP
85
11.6 The Stall Zone
This is the section of the characteristic curve in which both fan pressure and volume increase
simultaneously, as opposed to normal conditions where volume flow decreases as pressure
increases.
If the fan is operating in the stall zone, air tends to separate from the blade surfaces, leading to
excessive vibration and possibly damage to the fan if continued for long periods. A fan should never
be operated in the stall zone.
The stall zone is very marked in some fans, but hardly noticeable in others.
Characteristics of a fan in the stall zone:




vibrates excessively
whines
input amps show rapid oscillation
quantity and pressure difference
Situation remedied by:

recirculating some of the air through the system (e.g. allowing some leakage to occur across
an airlock)
altering the characteristics of the mine so that the fan operates on a different part of its curve.

11.7 Typical Power Characteristics
I.
Overloading power characteristic
If the power curve increases continuously as volume increases (i.e. if the volume is too high due to air
short-circuiting) power requirements may be too high and the fan will burn out.
86
ii)
Non-overloading power characteristics
If the maximum power requirement occurs neither at maximum volume nor pressure, but in between,
the power requirements will rise to a point as volume flow increases, but start decreasing once this
point has been reached
11.8 Fan Evasées
11.8.1 Fitting a fan evasée to the discharge of a fan:
improved fan performance
no change in system resistance
The magnitude of the improvement in performance is dependant on the characteristics of the evasée.
Example:
2
2
An 85% efficient evasée with an inlet area of 0.5 m and an outlet area of 1.2 m is to be fitted to the
discharge of an axial flow fan. Plot the new characteristic curve. The prevailing air density is
3.
1.2 kg/m
3
Choose any quantity, say 4 m /s
2
2
VP at inlet: v w = (4/0.5) x 1.2 = 38 Pa
2
2
87
VP at outlet:
2
2
0.5v w = (4/1.2) x 1.2 = 7 Pa
2
38 - 7 = 31 Pa loss in VP = 31 Pa theoretical gain in SP
Actual gain
= 85% of 31 Pa
= 26 Pa
3
The actual gain in SP must now be added to the fan curve at Q = 4 m /s to obtain a new curve.
3
Another point must now be plotted. Try Q = 6 m /s
Actual regain in SP
= (VPin - VPout) x 
100
2
2
= 1/2 w (vin - vout ) x 0.85
2
2
= 1/2 (1.2)[(6/0.5) - (6/1.2) ] x 0.85
= 61 Pa etc.
11.8.2 Fitting an Evasée to a system discharge
reduced system resistance curve
no change in fan curve
Once again, the magnitude of the resistance reduction depends on the characteristics of the evasée.
88
Example:
2
2
An 80% efficient evasée with an inlet area of 0.5 m and an outlet area of 1.2 m is fitted to the
discharge of a vent column. The density of the air is 1.0 kg/m3. Plot the new system resistance curve.
3
Assume 8 m /s flows through the system.
2
VPin = ½ wv = 128 Pa
2
VPout = ½ wv = 22 Pa
Theoretical SP regain = 106 Pa
actual SP regain
= 0.8 x 106
= 85 Pa
The actual SP regain is now subtracted from the system resistance curve in order to obtain a point on
the system resistance curve.
3
At 8m /s,
old P
= 600 Pa
2
-3
R old = 9.38 Ns .m
new P = 600 – 85
= 515 Pa
2. -3
R new = 8.05 Ns m
89
2
Use Rnew in the formula P = RQ to plot new curve
The system and evasée are considered as a single unit and are represented by the new system
resistance curve.
11.9 Fan Laws
Fan characteristics curves are valid for a specific speed and specific air density. Fan laws may be
used to predict the outcome of changes.
1. Varying speed, but constant air density
Q n
2
P n
3
W n
= 


2. Varying density, constant speed
Q=Q
P w
W w
= 


Where: Q: quantity
n: speed
P: pressure
W: power
: efficiency
w: density
Both of these entities may change simultaneously.
90
Situation 1:
n1 = 10 rps
3
w1 = 1.2 kg/m
3
At design point: Q1 = 115 m /s
P1 = 1.37 kPa
W 1 = 225 kW
1 = 70%
Situation 2:
n1 changes to 13.3 rps
w1 changes to 1.04 kg/m
3
Q2
varying speed
=
Q1
x
varying density
n2
x
n1
P2
=
P1
x
( n2)
( n1)
W2
=
W1
x
( n2)
( n1)
2
=
1
x
1
2
3
1
x
w2
w1
x
w2
w1
x
Therefore:
3
Q2 = 152.95 m /s
P2 = 2.04 kPa
W 2 = 458.76 kW
2 = 70%
91
1
92
93
94
12. FANS IN SERIES AND PARALLEL
12.1 Fans in series
The figure below is given to illustrate multiple development end auxiliary ventilation. Each column
could have more than one fan installed, either one after the other or spaced at regular intervals along
the column.
Fans that installed one behind the other so that each fan handles the same quantity of air are said to
be in series. They may be directly behind one another or installed some distance apart. Provided they
handle the same quantity of air, they are in series.
Both fans will handle the same quantity of air, but each fan will add its own pressure, according to its
capabilities. The total amount of pressure added by the fans is the sum of the individual pressures
added by each fan.
95
The above picture illustrates a development end being ventilated by either a force or an exhaust
ventilation column and fan or both systems together.
96
Fan laws for fans in series are thus as follows:
QT = Q1 = Q2
PT = P1 + P2
If the performance curves of both fans and the resistance of the system in which they operate is
known, a combined performance curve of both fans can be obtained.
12.2 Fans in parallel
Fans installed side-by-side in such a way that they draw air from the same source and deliver it to the
same place are said to be in parallel.
In such a configuration, each fan must produce the same amount of pressure as the other as they
have common intakes and discharges (as with parallel airways). Thus the total amount of pressure
added by both fans equals the amount of pressure added by each fan.
97
The ventilation column above clearly illustrates the use of bends in a force or exhaust ventilation
system
98
The above picture illustrates the use of a force and exhaust ventilation system.
Each fan will handle a different quantity of air, depending on its characteristics. The quantity of air
handled by fans in parallel is the sum of the individual quantities handled by both fans.
The fans laws for fans in parallel may be written as follows:
QT = Q1 + Q2
PT = P1 = P2
The following pages provide the student with an overview of fans, their characteristics, basic
calculations and graphical illustrations.
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
VENTILATION ENGINEERING
FANS & AIRFLOW
Test
Date:
Time:
Venue:
Total:
2001
11h30 - 13h30
CM4
70 marks
Instructions
ANSWER ALL QUESTIONS
This test accounts for 25% of the final course mark
1. List the four classes (with any sub-classes) of toxic gases, providing an example
of each and indicating the effect of the gas on the body.
[10]
2. In twin shaft systems, natural ventilating pressure can result in significant airflows
through the mine. Briefly discuss the factors which determine the direction in
which NVP acts (i.e. from downcast to upcast or upcast to downcast).
[4]
3. How much fresh air is required to ensure that a source of 0.0005 m3/s of methane
to ensure that permissible levels are not exceeded?
[3]
4. A fluid with a density of 1.1 kg/m3 flows at a velocity of 18 m/s along a conduit of
0.25 m in radius. The coefficient of dynamic viscosity is 0.0016 Pa.s. What flow
conditions predominate?
[3]
5. A 200m long, straight airway of 1.5 m diameter interconnects with a 45(short
radius) bend. From the perspective of airway resistance, what is the equivalent
length of the airway and the bend?
[2]
6. A fan runs at speed of 10 rps with a motor pulley diameter of 100 mm and a fan
pulley diameter of 500 mm. The speed must be increased to 15 rps. Calculate the
pulley sizes required if:
a) only the fan pulley is changed
b) only the motor pulley is changed
c) both pulleys are changed
[9]
122
7. A mine ventilation fan circulates a volume of 105 m3/s of air under a pressure of
725 Pa. The three phase motor is connected to a 380 V power supply and
consumes 65 A. The efficiencies of the drive and the motor are 92% and 95%
respectively. The power factor is 0.9.
a) What is the input power?
b) What is the motor output power?
c) What is the fan efficiency?
d) What is the efficiency of the whole unit?
[6]
8. A haulage is 300 m long and is 3m high by 3m wide. The air density underground
is approximately 1.15 kg/m3. If K = 0.0031 Ns2/m4, what is the air power required
to convey a quantity of 50 m3/s along the airway?
[3]
9. What are the units of Reynold’s Number?
[1]
10. Given the following data, complete the table and plot the characteristic curve of
the fan in question:
a) If the system resistance is 33 Ns2.m-8, determine the fan operating point.
[10]
b) Due to a fall go ground in an R.A.W., the system resistance increases to
350 Ns2.m-8. Plot the new curve. Will the fan be capable of operation at
this new point? On inspection of the fan, what symptoms would you
expect to observe if the fan was operating on this part of its characteristic
curve? How could this situation be remedied?
[11
123
11. Explain the difference between sensible and latent heat, giving an example for the
sources of each.
[3]
12. Determine the total amount of heat added to air passing over a heater bank at a
barometric pressure of 85 kPa. The flow rate is 10 m3/s. Inlet air has a
temperature of 13C and a relative humidity of 90%. Outlet air has a temperature
of 30C. The humidity ratio at both points is 0.0097 kg/kg.
[5]
END
Given formulae
124
MEMO
Mark breakdown
Gases & Fumes
Gas calc
Ventilation practice
Fluid mechanics
- Bernoulli
Airflow in airways
Vent networks
Fan types
Regulators and booster
Fan characteristics
Q1
What size regulator would be required to reduce the air quantity in an airway from 80
m3/s to 20 m3/s if the air density is 1.1 kg/m3 and the pressure loss is 500 Pa?
[5]
Ans:
P1 = P2
Q12
Q22
500 = P
6400 400
P = 31.25 Pa
P used up by regulator must be 500 - 31.25 = 468.75
A = 1.2 Q  (w/P)
= 1.2 x 20 x  (1.1/468.75)
= 1.162 m2 (1.065 also acceptable)
Gases & Fumes
1. List the four classes (with any sub-classes) of toxic gases, providing an example of
each and indicating the effect of the gas on the body.
[10]
Ans
1. Asphyxiant gases
If the oxygen supply to the bloodstream and the tissues of the body decreases or
fails, asphyxiation results. The may occur in two forms:
125
simple asphyxia occurs when sufficient air is breathed, but the air has an oxygen
deficiency;
chemical asphyxia occurs when a substance present in the air prevents oxygen
from being utilized by the body (e.g. carbon monoxide).
2. Irritant gases
These gases irritate and inflame tissue in contact with them. Inflammation of the
lungs leads to pulmonary oedema (fluids are produced in the lungs, causing
suffocation). Nitrous fumes and hydrogen sulphide act in this way.
3. Narcotic gases
Gases such as chloroform and ether which act on the nervous system, after having
been absorbed in the blood and transported to the tissue, are narcotic gases. If large
quantities are absorbed, the normal coma will terminate in death.
4. Protoplasmic poisons
After absorption by the body, these gases destroy the living cells in contact with
them. Examples include fumes of metals such as mercury and lead.
Q2.
In twin shaft systems, natural ventilating pressure can result in significant airflows
through the mine. Briefly discuss the factors which determine the direction in which
NVP acts (i.e. from downcast to upcast or upcast to downcast).
[4]
unequal depth
- air down shallow shaft and up deep shaft as T in shallow lower than that in deep. In
Summer, T in shallow may be higher and direction will reverse.
equal depth
- Direction determined by prevailing wind or by one shaft being wetter thus cooler
modern mines
air in downcast = cooler & denser than in upcast. Fluctuations in downcast column
may lead to complete reversal of NVP.
126
Q3
How much fresh air is required to ensure that a source of 0.0005 m3/s of methane to
ensure that permissible levels are not exceeded?
[3]
answer: permissible level = 1% (10 000 ppm)
Q = Q1 x 10^6
MAC - N
Q = quantity of fresh air required (m3/s)
Q1 = quantity of gas (m3/s)
MAC = max allowable concentration (ppm
N = conc of gas in normal air
= 0.0005 x 10^6
10 000 - 0
= 0.05 m3/s
Q4
3
A fluid with a density of 1.1 kg/m flows at a velocity of 18 m/s along a conduit of 0.25
m in radius. The coefficient of dynamic viscosity is 0.0016 Pa.s. What flow
conditions predominate?
[2]
answer:
R = wvD
μ
= 1.1 x 18 x 0.5
0.0016
= 6187.5
= Turbulent flow
Q5
A 200m long, straight airway of 1.5 m diameter interconnects with a 45(short radius)
bend. From the perspective of airway resistance, what is the equivalent length of the
airway and the bend?
[3]
127
Answer:
L=n
D
for a 45bend, n = 18
i.e. L = 18 x 1.5
= 27 m
Total length = 227 m
Q6
A fan runs at speed of 10 rps with a motor pulley diameter of 100 mm and a fan
pulley diameter of 500 mm. The speed must be increased to 15 rps. Calculate the
pulley sizes required if:
d) only the fan pulley is changed
e) only the motor pulley is changed
f) both pulleys are changed
[9]
answer:
fan speed = motor speed x motor pulley diameter
fan pulley diameter
original fan speed = 10 = motor speed x 100 / 500
motor speed = 50 rps
new fan speed = 15 = 10 x 100 / new fan pulley
new fan pulley = 66.6 mm
or
new fan speed = 15 = 10 x new motor pulley / 500
new motor pulley = 750 mm
128
Q7
A mine ventilation fan circulates a volume of 105 m3/s of air under a pressure of 725
Pa. The three phase motor is connected to a 380 V power supply and consumes 65
A. The efficiencies of the drive and the motor are 92% and 95% respectively. The
power factor is 0.9.
e) What is the input power?
f) What is the motor output power?
g) What is the fan efficiency?
h) What is the efficiency of the whole unit?
[7]
Answer 7a:
W = EI pf 3
= .38 x 65 x 0.9 x 3
= 38.5 Kw
Answer 7b:
Input power x efficiency of motor
Answer 7c:
Air power
[2]
= 38.5 x .95
= 36.6 kW
[1]
= p x Q / 1000 = 105 x 725 / 1000
= 76.125 kW
Fan input power = motor output x drive efficiency = 36.6 x .92 = 33.7 kW
Fan efficiency = air power / fan input power x 100%
= 76.125 / 33.7 x 100
= 233.67 %
[3]
Answer 7d:
Overall fan efficiency = air power / motor input power x 100%
= 76.125 / 38.5 x 100
= 197.73%
[1]
129
Q8
A haulage is 300 m long and is 3 m high by 3 m wide. The air density underground is
approximately 1.15 kg/m3. If K = 0.0031 Ns2/m4, what is the air power required to
convey a quantity of 50 m3/s along the airway?
[3]
answer:
P = KCLQ2 x w
A3
1.2
= 0.0031 x 12 x 300 x (50)2 x 1.15
3
9
1.2
= 36.68 Pa
Q9
What are the units of Reynold’s Number?
[1]
Answer: dimensionless
Q10
Given the following data, complete the table and plot the characteristic curve of the
fan in question:
a. If the system resistance is 33 Ns2.m-8, determine the fan operating point.
[10]
b. Due to a fall go ground in an R.A.W., the system resistance increases to 350
2
-8
Ns .m . Plot the new curve. Will the fan be capable of operation at this
new point? On inspection of the fan, what symptoms would you expect to
observe if the fan was operating on this part of its characteristic curve?
How could this situation be remedied?
[11]
130
Answer:
[2]
If the system resistance is 33 Ns2.m-3, determine the fan operating point.
Q = 5.15 m3/s
P = 875 Pa
Eff = 78%
[8]
Due to a fall go ground in an R.A.W., the system resistance increases to 350
Ns2.m-8. Plot the new curve. Will the fan be capable of operation at this new
point? On inspection of the fan, what symptoms would you expect to observe
if the fan was operating on this part of its characteristic curve? How could
this situation be remedied?
no - stall zone.
[6]
Symptoms:
Vibrates
Whines
Input amps show rapid oscillation
[3]
Remedy:
Allow some leakage across an airlock (re-circulation)
Altering characteristic of mine so that fan operates on different part of curve.
[2]
131
Q11
i) Explain the difference between sensible and latent heat, giving an example for
the sources of each.
[3]
ii) Determine the total amount of heat added to air passing over a heater bank at a
barometric pressure of 85 kPa. The flow rate is 10 m3/s. Inlet air has a
temperature of 13C and a relative humidity of 90%. Outlet air has a temperature
of 30C. The humidity ratio at both points is 0.0097 kg/kg.
[5]
Ans (i)
Sensible Heat
Any heat which is added or taken away from the air and resulting in a change in the
temperature of the air is called sensible heat. This is the heat added by lights. solar
radiation, conduction, machinery, heaters, etc.
Latent Heat
When heat added to or removed from a substance results in a phase change with a
temperature change is called latent heat. A latent heat process always involves a
change in the amount of water vapour present in air. Latent heat is added to air from
sources such as the human body, outside air being introduced into an air-conditioned
space and any other appliance which involves water (kettles, evaporators etc.)
(ii)
3
Flow rate = 10m /s
From chart:
3
Specific volume of air at intake: 0.982 m /kg, mass flow rate: 10/.982 = 10.1823 kg/s
[2]
Sigma heat at pt 1. = 37.5 kJ/kg at pt 2. = 55.0 kJ/kg S = 17.5 kJ/kg
[2]
Total heat added = 17.5 x 10.1823 = 178.207kW
And
Q=MaCaTa
= 10 x 1.005 x (30 – 13)
= 170 W
132
[1]
Heat & Refrigeration
Topic 1
Deep Level Mining
1.Introduction to Psychrometry
Psychrometry is concerned with the thermodynamic properties of mixtures of air and water vapour.
This is of particular importance in mine ventilation as air undergoes large changes in moisture content
as it flows through a mine. Information on the thermodynamic properties of air is vital in understanding
phenomena as diverse as the flow of air in an upcast shaft and the interaction between heated air and
chilled water sprays.
Air
Air is mixture of many gases. The principle components are Nitrogen, Oxygen, Argon and Carbon
dioxide. Omitted from this list is water vapour. The term “air” is frequently used to describe a mixture
of the above gases and water vapour, but some confusion exists as it is also employed to describe
just the above gases. In order to avoid this confusion, the terms “air” and “dry air” may be adopted,
where the former includes water vapour.
Gas relationships
Calculations of the properties of gases are effected using the mol. This is the quantity of a gas which
is numerically equal to the molecular mass of the gas. Thus a mol of oxygen (which has a molecular
mass of 32) has a mass of 32 g, and a mol of carbon dioxide, with a molecular mass of 44.01 has a
mass of 44.01 g. A mol of any gas contains a fixed number of molecules, given by L, the Avogadro
constant (6.02283 x 1026 molecules per mol).
Gas Laws
Boyle's Law
If a constant mass of gas is kept at a constant temperature, then its pressure is inversely
proportional to its volume:
P1V1 = P2V2 = P3V3 =…. etc.
133
A certain mass of air occupies a volume of 16 m 3 when the pressure is 100 kPa. If the temperature of
the air remains constant, what volume will the same mass of air occupy at a pressure of 80 kPa?
Charles' Law
If a given mass of gas is kept at a constant pressure, the volume of the gas is directly proportional to
its absolute temperature:
V1 = V2 = V3
T1 T 2
T3
3
8 m /s of air at 26C dry bulb enters an underground hoist chamber. The pressure of the air remains
constant and there is no change in the amount of water vapour in the air. If the air leaves the chamber
at 33C dry bulb, what is the volume of the air leaving the hoist chamber?
The Universal Gas Law and the Ideal Gas Relationship
Combining Boyle's and Charles' Laws yields the Universal Gas Law:
P1V1  P2V2 
T1
T2
P3V3
T3
As the pressure of real gases tends to zero, their behaviour approximates that of ideal
gases1. The behaviour of ideal gases is described by the ideal gas relationship:
PV = mRT
Where:
P = absolute pressure (kPa)
3
V = volume (m )
m = mass of gas (kg)
R = gas constant (kJ/kg K)
T = absolute temperature (K)
134
Avogadro’s hypothesis states that at a given pressure and temperature, the number of molecules per
unit volume is constant. Applying this fact to the above equation yields the Universal Ideal Gas
Equation:
PV  m R* T
M



Where:
M = molecular mass
R* = universal gas constant (kJ/kg mol K). This has a numerical value of 8.31436 kJ/kg mol K
The ratio m/M (= n) is the number of mols of the gas and the universal ideal gas equation is
PV = nR*T
Gases of interest in mine ventilation:
Table 1 Gasses and their mol mass
Gas Mixtures
The universal ideal gas equation applies both to pure gases and to mixtures of gases. In order to
apply the equation to mixtures of gases, Dalton’s law of partial pressures is required. This states that
the pressure exerted by a mixture of gases present in a given volume, and at a given temperature is
equal to the sum of the pressures exerted by the individual components of the mixture if each alone
occupied the volume at the same temperature.
For k gases, the universal ideal gas equation can be written as follows:
(P1 + P2 +...+ Pk)V = (n1 + n2 +...+ nk)R*T
135
The composition of a gas mixture can be expressed in terms of mole fractions (x i) for each
component (i)
xi = ni =
ni
.
n
(n1 + n2 +...+ nk)
2. Psychrometric Properties
The gas constant for a gas or gas mixture is R or R*/M (where R* is the universal gas constant and M
is the molecular mass). For standard dry air, the gas constant has the value:
8.31436
28.9664
= 0.287035 kJ/kg K
Dry air may have a composition which differs appreciably from the standard. The air may also have a
water vapour component. In order to predict changes in air, it is necessary to factor these differences
into calculations. This is done by means of psychrometric equations.
1.
Moisture Content (r)
This is the mass of water vapour associated with a unit mass of dry air.
r  mw M wPw M w P w
m a M aP a
.
M a P  P w 
P is the absolute pressure, which is equal to the sum of the dry air pressure and the water vapour
pressure.
2. Specific Volume ()
This is the volume occupied by the air expressed per unit mass of dry air.
v = = R* T =
ma
Ma Pa
R* T .
Ma (P - Pw)
136
3. Density (w)
The mass of air (dry air plus water vapour) per unit volume.
In contrast to the first two properties, density is the mass of dry air and water vapour per unit
volume.
4. Enthalpy (H)
This property is measured in J/kg. It was once known as total heat or total energy.
H = Ha + H w
The enthalpy of dry air is given by:
Ha = Cpa tdb
= 1.005 tdb
Where tdb is the dry bulb temperature (C) and Cpa is the thermal capacity at constant pressure for dry
air (1.005 kJ/kgC). The enthalpy of the water vapour component (Hw) is the enthalpy of the water
vapour associated with a unit mass of dry air. For the purposes of most mine ventilation calculations,
Hw may be approximated by the following empirical relationship:
H’w = 1.80 tdb + 2 501 (J/kg)
This is the linear equation form. Two other equations in the quadratic and cubic related forms could
be used as provided in the textbook (p. 442 EESAM handbook).
Use of the linear equation expression results in an enthalpy expressed per unit mass of dry air.
137
5. Sigma Heat (S)
This property is the enthalpy of the air less the enthalpy of the water vapour associated with
the unit mass of dry air. The latter is calculated as if the water vapour was present as liquid
at the wet bulb temperature. Sigma heat is expressed per unit mass of dry air:
S = H - rH’wl
[kJ/kg]
Where:
H’wl = 4.18 twb
and
r = moisture content
6. Relative Humidity ()
Relative humidity is the ratio of the actual vapour pressure to the saturated vapour pressure at dry
bulb temperature:
= Pw x 100%
P”ws
Where:
P”ws = 0.6105 Exp. (17.27 tdb / 237.3 + tdb)
[kPa]
Pw is usually given
7. Dew point temperature (tdp)
This is the temperature to which air must be cooled for condensation to begin. At this point the air
becomes saturated. As there is no change in moisture content during the cooling process, the vapour
content remains constant. The dew point is thus the temperature at which the saturated vapour
pressure is equal to the actual vapour pressure.
Dew point can be obtained from tables relating saturated vapour pressure to temperature.
tdp 237.3 x
℃
17.27 x
138
Where:
3. Psychrometric charts
In order to determine the properties of an air-water mixture, the following information is required:
barometric pressure (kPa)
wet bulb temperature (C)
dry bulb temperature (C)
The procedure for analysis is as follows:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Get wet bulb temperature
Find appropriate chart (pressure related)
Determine state of air-water mixture
Plot starting point using wet & dry bulb temperatures
Read off apparent specific humidity
Read off apparent specific volume
Calculate: apparent density = 1 / apparent specific volume
Read off vapour pressure
Calculate relative humidity
Calculate enthalpy
Calculate sigma heat content
Calculate dew point
The mass of dry air is given by:
M=Qxw
Or
M = Q / 

As an example, figure 1 below is provided whereby all the information could be determined from such
a chart. Please note that this information only relates to the 100 kPa barometric pressure level. All
psychrometric charts are available in Barrenburg’s book for psychrometry.
139
Special softwares for PC’s are also available from suppliers as well as within software packages such
as Environ 2.5 and the latest windows based VUMA package. The intension here is not to use such
packages but rather to familiarise the student with the parameters and related formula and actually
calculate the psychrometric values of importance.
140
Figure 1: Psychrometric chart at a pressure of 100 kPa
Please note: Pages 459 to 463 (E.E.S.A.M.) are to be studied as examples when using
psychrometric calculated data. Air mass flow and heat flow calculations are important
in this context.
The following important steps are to be followed in order for the student to be able to read the
psychrometric charts.
141
142
143
144
145
146
147
148
149
The following is the first page of Barrenburg psychrometric charts as an example. The
student should familiarise him/ herself with the use of these charts.
150
Example of the sigma-heat chart found in Barrenburg’s book of charts
151
Another example of the enthalpy chart also found in Barrenburg’s book of charts
152
4. Heat Sources
1. Natural Sources of Heat
1.1 Heat transfer from rock strata
The temperature of virgin rock increases with depth below the earth’s surface as a result of the flow of
heat from the interior of the planet (Where does this interior heat originate?). The temperature of the
rock at depth is thus reliant on the thermal conductivity of the rock and on the long-term average
surface temperatures. Whenever the rock surface in an excavation is at a higher temperature than the
ventilating air, heat will flow from the rock into the air (Heat flows from hot to cold as airflow would
occur from a high pressure to a low pressure). This may occur by convection (fissure water) or
conduction.
1.2. Oxidation of rock strata
The oxidation of certain minerals, such as pyrite, is an exothermic reaction and results in the release
of substantial amounts of heat. Mt. Isa mine in Australia has experienced difficulties as a result of this,
but fortunately, the problem does not occur in South African gold mines.
2. Autocompression
This phenomenon is not strictly speaking a source of heat. It applies to compressible fluids and
results in an increase in temperature when potential energy is converted into enthalpy (enthalpy is
total heat). Autocompression is important in the design of mine ventilation systems as such systems
rely heavily on circulating fluids (i.e. chilled water) to remove excess heat within the mine. The
enthalpy increase associated with autocompression serves to reduce the amount of heat that the fluid
is able to remove. The increase in enthalpy due to autocompression is typically 9.79 kJ/kg for every
1 000 m.
Therefore, the temperature increase effect as air or any fluid changes elevation is independent of the
frictional impact. The steady flow equation gives the following:
H2 – H1 = (Z1 – Z2)g +q12
[J/kg]
Where:
H = Enthalpy (J/kg)
153
Z = height above datum point (m)
2
g = Gravitational acceleration (m/s ) and
q = surrounding heat (J/kg)
The subscripts 1 and 2 refer to the inlet and outlet positions respectively.
The increase in temperature is sometimes referred to as the adiabatic laps rate. When ignoring
external heat influences (q = 0), and calculating the temperature difference between two vertically
different positions, the equation through manipulation of equalities become:
 T ( Z1 Z2) g
Cp
℃

3. Artificial Sources of Heat
These are sources of heat arising from normal day to day activities taking place in the mine.
3.1 Electrical machinery
Hoist room with electric motors, etc.
In many mines, electricity is the principle source of power. All equipment either converts this energy
into heat energy or useful work. In the thermodynamic sense, “useful work” involves increasing the
potential energy of an object or fluid by lifting it to a higher elevation against the force of gravity.
154
Keeping this definition in mind, it becomes apparent that fans and lamps convert all the electrical
energy supplied to them into heat. Winches, hoists, locomotives and other motors convert a greater or
lesser amount of their electrical energy into heat, depending on the application and efficient design.
Increases in enthalpy and heat dissipated would differ in each specific design because of differences
in motor power utilised and the application required (Refer to pages 598 to 605 in EESAM for
example calculations).
Loco on maintenance
3.2 Diesel equipment
Underground diesel operated machinery
155
As with electrically powered machinery, heat is dissipated when the diesel engine is not employed in
“useful work”(as defined above). LHDs and similar vehicles seldom traverse large gradients and thus
the greater portion of energy consumed by the engine is release as heat. To provide an example,
internal combustion diesel engines compares to about one third of the efficiency of that of electrical
equipment. Therefore, diesels will produce about three times more heat than a mechanically similar
electrical powered machine. The range of heat production could be between 30 to 500 kW depending
upon the equipment rated power and work load performed. A calculation could easily be done once
the diesel fuel consumption and fuel calorific value is known.
Underground diesel LHD operating
3.3 Compressed Air
This represents a safe and convenient source of power for rock drills, pumps and fans. The principle
difference between this and other sources of energy is that the exhaust air enters the mine’s
ventilation with a low enthalpy (the air has provided mechanical power). This acts as a source of
coolth and may cancel the heat generated by the device. Rock drills may release a negative amount
of heat (-3 kW) as a result of this phenomenon.
156
3.4 People
People working underground
People function by internally combusting various foodstuffs. This results in the emission of
heat. Typical values are:
at rest 90 - 115 W
light work rate 200 W
moderate work 275 W
hard work rate 470 W
For a shift of 1000 workers, this may result in a substantial source of heat.
157
3.5 Explosives
Explosives being put into the drilled hole
Large amounts of heat are liberated when explosives are used. There is no evidence to suggest that
the heat liberated during blasting is completely removed during the re-entry period. This heat most
probably enters the broken rock and is released over a long period of time:



ammon gelignite 4650 kJ/kg
ammon dynamite 4030 kJ/kg
anfex 3820 kJ/kg
4.
Heat & Psychrometry
Sensible Heat
Any heat which is added or taken away from the air, resulting in a change in the temperature of the air
is called sensible heat. This is the heat added by lights. solar radiation, conduction, machinery,
heaters etc. Sensible heat is removed by passing the air over a colder surface. In a pure sensible
heating process, no water is evaporated or condensed and thus the humidity ratio of the air does not
change. Sensible heating takes place along a horizontal line on the psychrometric chart.
158
Air at the intake of an electric heater bank of a plant at sea is at a temperature of 15C with a relative
3
humidity of 90%. 8 m /s of air pass over the heater bank. After the bank, the air is at a temperature of
30C. The humidity ratio at both points is 0.0097 kg/kg. The amount of heat added to the air can be
determined by calculation or from psychrometric charts. The specific volume of air at the intake (from
3
the psychrometric chart) is 0.829m /kg and the mass flow rate is 8 / 0.829 or 9.65 kg/s. The total heat
can be calculated using the formula:
q = MCT
= 9.65 x 1.012 x (30 - 15)
= 146.5 kW
(Use a psychrometric chart to verify this. Note the barometric pressure at sea level 101.3kPa).
Latent Heat
When heat is added to or removed from a substance and results in a phase change without a
temperature change, it is called latent heat. A latent heat process always involves a change in the
amount of water vapour present in air. Adding latent heat to air implies an increase in the humidity
ratio (evaporation) whilst removing latent heat reduces the humidity ratio (condensation). Latent heat
is added to air from sources such as the human body, outside air being introduced into an airconditioned space and any other appliance which involves water (kettles, evaporators etc.)
Consider a humidifier with electric elements for water heating. This is a process where only, the
absolute humidity (or moisture content) of the air is increased. Air passing through such a system has
an initial temperature of 30C and a final temperature of 30C. The initial relative humidity is 37% and
the final relative humidity is 50%. There is no sensible heat increase as the dry bulb temperature
remains constant. The initial absolute humidity is 0.0097 kg/kg and the final absolute humidity is
0.0134 kg/kg. 0.0037 kg of water has thus been added to each kilogram of air passing over the
humidifier. If the amount of air is 9.65kg, the total amount of water added to the air is 9.65 x 0.0037 =
0.03571 kg. To evaporate one kg of water requires 2 450 kJ, thus 0.03571 x 2 450 = 87.5 kW is
required (Verify this using a psychrometric chart).
Heat - Geothermal Gradients
The quartzites of the Wits system are good conductors (Thermal conductive values of between 5 to 7
W/mK) and therefore rock temperatures do not increase rapidly with depth in the central Wits area
(Geothermal gradient).
In the Free State (and some other areas), the quartzites are overlain by Karoo shales and
Ventersdorp Lavas which are not good conductors of heat (Generally < 4 W/mK). The VRT in these
areas increases more rapidly with depth.
The earth constantly drifts on crustal plates. These plates are relatively thin, i.e. 90km in
thickness, with relation to the earth’s radius or diameter. Mining activity (depth) only takes
place in a very small upper-layer portion in the drifting plates. The geothermal flow of heat
emanating from the earth’s core and passing through that skin layer has an average value of
0.05 to 0.06W/m2. The heat flux can naturally differ and therefore be much higher depending
on the geological region, thereby referring to thermal conductivity of the local strata.
The mean heat flow of continents and oceans are 0.065 and 0.101 W/m2 respectively, which
when weighted yield a global mean of 0.087 W/m2. More than half of the Earth’s heat loss
comes from oceanic lithosphere (Pollack et al, 1993). The heat flow in geological terms are
generally expressed in units of milli-watt per square metre (mW/m2) The data collected by
Pollack et al constituted some 10,337continental measurements and 9,684 oceanic
measurements.
Figure 3 depicts the measurement stations around the world from which the data was
complied.
159
Figure 3 Geographic distribution of heat flow measurement sites
From measurement done by Jones (Jones, 1988, p. 3252) the Witwatersrand Basin heat
flow has a mean value of 0.051 W/m2 (standard deviation [SD] of 6mW/m2).The heat flow
in the Witwatersrand Basin is higher than the worldwide mean for Archaean cratons which is
0.041 W/m2 (sd of 11 W/m2). The reason (Jones, 1988) for the excess heat flow in the
Witwatersrand Basin is because of radioactive heat produced in the basin’s strata which
measures about 0.0075 W/m2. Measurements from borehole data were also listed (p. 3248)
as compiled from various sites around South African mining operations. The heat flow was
determined to be between 0.041 W/m2 (Avondrust AR1 borehole) and 0.072 W/m2
(Hartebeesfontein-HB5d-borehole). Further discussions on thermal gradients are provided in
sub-section
2.5.6.
Table 3 indicates some international heat flow values as given by Jones (1988, p.3255)
Table 3. International heat flow trends (Jones, 1988)
160
When Fourier’s law is then applied to a heat flux value of say 0.06 W/m 2 for each square
metre of land surface to indicate the variation in temperature, , with respect to depth, D,
then
𝑑𝜃
𝑑𝐷
=
0.06
𝑘
℃/m
Eq.2.2.2
The increase of virgin rock temperature (VRT) with depth is known as the geothermal gradient. The
inverted use of equation 2.2.2 is also often used thereby providing integer values in units of m/°C. The
inverted use of the equation is then referred to as the geothermal step
𝑑𝐷
𝑑𝜃
. However, in normal
mining terms, geothermal gradient is the general expression used (C/m) (McPherson, 1992).
Watermeyer and Hoffenberg (1932, pp. 595 – 597) explained that in the coal mines of Great Britain
the geothermal gradient averages a rise of 0.556C for every 21 m in depth (0.0265C/m), whereas
on the Witwatersrand in South Africa, the gradient is more forgiving, i.e. 0.556C for every 67 m in
depth (0.0083C/m). This rather flat geothermal gradient of the Witwatersrand area is because of the
high thermal conductivity of the strata.
Table 4 indicates some relevant thermal gradient values for different geological areas in
South Africa
Table 4 Mean thermal gradients
VRT calculation
As a rough estimate, the following linear related formula could be applied to determine the VRT at
depth. Care should be taken not to generalise these two equations as the geology at different
locations within the Witwatersrand do change rapidly and would influence the VRT accordingly. The
best method of VRT determination is to measure it in-situ, preferably within a freshly opened up area.
Witwatersrand area:
18 + (9.3 x d)
Free State area:
20 + (14.5 x d)
Where:
d = depth [km]
161
The following graph shows the VRT with relation to depth for various countries/areas over the world.
The graph above was constructed with data available. Some of the lines in the graph for the different
regions were extrapolated further than actual data available, i.e. Hungary, Bushveld complex, etc.
These data points indicating VRT values should ultimately be measured to verify or correct the
extrapolated values obtained. The VRT values for the Witwatersrand and Free State for instance are
only valid up to a depth of around 3500 mbc.
5. Heat - Mixing Airstreams
Use:
Sigma heat content
Moisture content
(S)
(r)
If two airstreams A and B mix to form airstream C, the sigma heat content of the mixture (Sc) can be
calculated from:
Sc = (MA x SA) + (MB x SB)
(MA + MB)
162
Note:
1. (MA x SA) = total kW heat from A
2. (MB x SB) = total kW heat from B
3. (1.) + (2.) = total kW heat in C
4. (MA + MB) = total mass flow in C
5. Sc = sigma heat content of C
Similarly the moisture content of the mixture (rc) can be calculated from:
rc = (MA x rA) + (MB x rB)
(MA + MB)
Where:
1.
2.
4.
5.
6.
(MA x rA) = the total moisture from A
(MB x rB) = the total moisture from B
(1.) + (2.) = the total moisture in C
(MA + MB) = the total mass flow in C
rc moisture content of the mixture C
Once sc and rc have been determined, the temperature of the mixture can be read off at the
INTERSECTION of these two lines on the relevant psychrometric chart.
6. Heat - Mixing Airstreams:
The straight line method
Procedure:
1. Plot the temperature of A
2. Plot the temperature of B
3. The temperature of C lies on the straight line between these two points, closer to the
point with the higher mass of air. The exact position of the “mixture point” is
determined by the inverse ratio of the mass of either original point to the total mass.
(This ratio is multiplied by the distance between the original two points)
163
Total mass flow
= 62.6 + 29.4
= 92.0 kg/s
B’s proportion
= 29.4 / 92.0
= 0.32
A’s proportion
= 62.6 / 92.0
= 0.68
Therefore C lies (1 - 0.68) x length AB from A
or C lies (1 - 0.32) x length AB from B
Note: C lies closer to A because A has the larger mass
Example
3
3
42.8 m /s of air at a temperature of 25.0/27.2 C is mixed with 23.3 m /s of air at temperature
of 31.7/32.5 C. The barometric pressure is 115 kPa.
Determine the temperature of the mixture. The answer is provided in the note below.
7. Heat Impacts & Acclimatisation
164
Effect of Heat on Workers Production
Experiments conducted by COMRO (now Miningtek) show that a worker’s production is affected by
heat and air velocity. At wetbulb temperatures of less than 27.2C, a person’s work output is not
affected, even in environments with low air velocities. The combined effect of wetbulb temperature
and air velocity is shown below:
Table 5 Heat & Productivity relationship
Effect of Heat on Health
Physiological Effect
As the temperature of work rate increases, the body tries to compensate by sweating more. The
evaporation of sweat has a cooling effect. Under extremely hot conditions, little cooling results.
Working at a high rate leads to:
cramps
heat exhaustion
heat stroke
Psychological Effect
More mistakes are made when people work under hot conditions. Errors lead to accidents and low
morale.
165
Why people are acclimatised
Acclimatised men work harder in hot conditions with a greatly reduced chance of heat stroke.
The need for acclimatisation of underground workers in deep, hot mines has been recognised
for some time
7.2 Practical aspects of human heat stress (Chapter 21, Environmental Engineering in SA
Mines - EESAM)
7.3 Heat Balance of a Working Man (Chapter 20, Environmental Engineering in SA Mines)
166
8. Refrigeration
The Refrigeration Cycle
Refrigeration is the process of cooling. Heat is removed from a substance at one temperature level
and then discharged at a higher temperature level. The removal of this heat produces a condition
called “cold”. This term, together with “hot”, is relative and makes no reference to the absolute amount
of heat in a substance.
A refrigeration machine uses mechanical work to absorb heat at one temperature and to reject heat to
a sink at a higher temperature. Work is thus expended to produce coolth, a phenomenon in
accordance with the laws of thermodynamics which state that work and heat are mutually convertible.
Large cooling plants are used to cool water either on surface or underground. This water is then
circulated to working places to be used as service water, or it is used to bulk-cool downcast air. It may
also be circulated through the sections of the mine and used to cool the air in water-to-air heat
exchangers.
There are several types of refrigeration system available. This course will describe the commonly
used vapour compression system. In machines of this type, a substance known as the refrigerant
circulates continuously. It enters the evaporator as a liquid and is evaporated, the heat required for
this being obtained from the substance being cooled. The resultant vapour then enters the
compressor, where it is compressed to a higher pressure and then discharged to the condenser. Here
the refrigerant is cooled and then condensed to a liquid, the latent heat released being transferred to
the circulating condenser water. The liquid refrigerant then leaves the condenser and passes through
an expansion valve before once again entering the evaporator.
The Pressure-Enthalpy (P-h) Diagram
This is a useful means for examining and analysing the operation of a vapour compression system.
The diagram shows the various properties of the refrigerant
167
The main features are:



the sub-cooled or liquid region: lines of constant temperature in this region will be vertical,
extending upwards from the saturated liquid line.
the wet region: this represents the condition of the refrigerant while undergoing a change of
phase from liquid to vapour during the addition of latent heat at any given pressure. When a
liquid changes into a vapour or vice versa at a constant pressure, the temperature remains
constant while latent heat is added or removed. Temperature lines in this region are therefore
horizontal.
the super-heated region
Single stage compression
On leaving the condenser, the refrigerant is at point A. The liquid then passes through the expansion
valve where the pressure is reduced from the condenser pressure to the evaporator pressure (A to B).
No work is done during this process and thus the enthalpy remains constant. The refrigerant leaving
the expansion valve consists of a mixture of saturated liquid at condition F and saturated vapour at
condition C. This mixture condition is given by point B. Heat is now added to the liquid refrigerant by
the evaporator, causing it to evaporate so that the refrigerant leaves the evaporator as a vapour, point
C on the saturated vapour line. An ideal compression is represented by a line of constant entropy (line
CE).
168
Due to inefficiencies in the compressor, further heat is added and the compression path follows line
CD. Point D represents the condition of hot gas as it enters the condenser from the compressor. In
order to cool the gas, cooling must first be provided to reduce its temperature to the saturation point at
L. Further cooling causes the gas to condense until the final liquid state is reached at point A.
Cycle Variations
In order to improve efficiencies and performance, several modifications to the vapour compression
cycle can be made. These include using economisers with multi-stage compressors and sub-cooling
the refrigerant when it leaves the condenser.
1. Sub-cooling
Sub-cooling of the liquid refrigerant from the condenser may be obtained from an external cooling
medium such as air, water or some other fluid at a temperature below that of the liquid refrigerant.
When sub-cooling occurs, the amount of flash gas produced after the expansion valve is reduced.
Thus less refrigerant has to be circulated for a fixed cooling duty. The nett benefit of sub-cooling is to
increase the refrigeration effect without affecting the work of compression. This benefit is usually
small.
2. Economisers
The properties of some refrigerants are such that in large cooling plants, several stages of
compression are necessary to cover the pressure change from the evaporator to the condenser.
When two or more stages of compression are used, the gas can be cooled after each stage by the
addition of flash gas from the corresponding expansion valve. The same number of expansion valves
and stages of compression are required.
Flash gas from the first expansion valve after the condenser is separated from the remaining liquid
refrigerant in the economiser and is piped back to enter the compressor between the first and second
stages. This arrangement enables the hot gas from the first stage of compression to be cooled by the
flash gas. The liquid refrigerant in the economiser is circulated to the evaporator in the usual manner.
The result of this process is an increased refrigeration effect.
169
A liquid boils at constant temperature, provided the pressure remains the same. The heat added to
the liquid is utilised in increasing the internal energy of the molecules until they no longer can remain
in their liquid state, thus the molecules burst free to form a vapour or a gas.
If the pressure is raised, additional heat is required to vaporize the fluid. Therefore, the boiling
temperature will increase.
170
- Do example 1 on page 616 (EESAM)
- Performance of a refrigeration cycle
COP definition
The coefficient of Performance of a cooling plant is defined as the ratio of the cooling duty to
the input power, i.e. the ratio of what is got out to what is put in.
Steady flow energy equation.
Compressor: W 12 = H2 – H1 (J/kg)
Condenser: q23 = H3 – H2 (J/kg)
Exp. Valve: O = H4 – H3 (J/kg)
Evaporator: q41 = H1 – H4 (J/kg)
And: W 12 + q23 + q41 = 0
W 12 = mechanical energy added to compressor.
q23 = Heat exchange in condenser
q41 = Heat added to refrigerant
H = Enthalpy.
Thus: W 12 + q41 = -q23
Properties of refrigerants
Page 622 (EESAM)
Ideal refrigerant:
a)
b)
c)
d)
e)
f)
g)
h)
Cheap
Not affect the compressor lubricant
Non-corrosive
Non-toxic
Non-flammable
Non-explosive
Leakages easily detectable
Must have a high latent heat of evaporation so that flow rates can be kept as low as
possible
171
The vapour compression cycle (McPhersons, 1993)
The following is a brief description of the components of a refrigeration plant:
1)
Compressor
The compressor is the device where mechanical work is input to the system
Types of compressors:
a) Reciprocating
b) Screw and
c) Centrifugal types.
2)
Evaporator
The evaporator is a heat exchanger. It is typically of the shell-and-tube type. In large units, the
refrigerant liquid is on the outside of the tubes while the medium to be cooled, i.e. water,
passes through the tubes. Within the evaporator, the pressure remains relatively low,
therefore the boiling point of the refrigerant is low, i.e. R12 boils at 4C at a pressure of 351
kPa. The heat required to maintain the boiling point is extracted from the liquid passing on the
other side of the tube walls. Therefore, the liquid is cooled. The refrigerant is now vaporised
and passes on to the compressor.
3)
Condenser
The refrigerant vapour comes from the compressor at a relatively high pressure and
temperature. The condenser could be of a similar heat exchanger type used for the
evaporator, i.e. shell-and-tube type, Heat is now removed from the refrigerant by water or air
and the refrigerant cools and condenses back to a liquid state. The pressure is high therefore
the condensing temperature is relatively high, i.e. R12 at a pressure of 1217 kPa condenses
at 50C.
172
4)
Expansion valve
The fourth component, the expansion valve, duty is to reduce the pressure of the refrigerant
back to the evaporator condition. At the exit from the expansion valve, the refrigerant is at a
low pressure and corresponding low boiling point and the cycle repeats.
Measure of plant performance
COP = useful
cooling effect (evaporator heat transfer)
work input from compressor
= q41
W 12
Thus:
q41
q23q41
COP =
.
The area under the ideal Ts diagram represents heat.
Thus
q41 = area 41 ab
-q23 = area 23 ba
-q23 - q41 = area 1234
Now: W 12 = area 1234
And
Cannot COP =
area 41ab
area 1234
= T1 Sa -
Sb
 Sa - Sb
=
T1
or
 T 2-T1
T4
 T3 T4
Example:
Evaporator Temp = 4C
Condenser Temp = 50C
What is the maximum possible co-efficient of performance of this unit?
COP = 273,15
+4
50 - 4
= 6,025
173
Actual COP
COP actual is lower than ideal COP.
REASON:
a) In actual compressors and expansion valves there is an increase in entropy.
b) As refrigerant passes through a condenser and evaporator there is a slight change in
pressure and temperature.
c) Point 1 on the diagram lies between the two phase region (liquid and vapour). The
pressure of liquid droplets would cause some erosion of the compressor impeller. A
few degrees of super hear are imparted to the vapour before it leaves the evaporator.
COP = q41
W12
= H1
- H4
H2 - H1
Cycle efficiency =
actual COP
Carnot COP
Example
Water flow rate: 50 l /s
Water inlet temp: 20C
Water outlet temp: 10C
Evaporator
Refrigerant Pressure: 363 kPa
refr. inlet Temp: 7 C
refr. inlet Temp: 65 C
Compressor
(A) Refer to Pressure-Enthalpy diagram (R12)
Analyse the performance characteristic of the unit.
t evap = t4 = 5 C
can be measured
t cond = t 3 = 45 C
B) Temperature of refrigerant entering compressor = 7C.
Thus 7-5 = 2C of superheat
Point 1 = 7C
Point 2 = 65C
Etc.
Thus: Compressor. Inlet
Compressor. Outlet and inlet
Condenser. Outlet
Evaporator. Inlet
H1 = 356 kJ/kg
H2 = 386 kJ/kg
H3 = 244 kJ/kg
H4 = 244 kJ/kg
174
(C) Carnot COP
=
T4
t3 – t4
=
273,15 + 5
45 - 5
= 6,954
= H1 -
H4
H2 H1
= 356 – 244
386 – 356
Actual COP
= 3,733
3,733
x 100
6,954
= 53,7%
Cycle eff:
=
(D) Useful cooling effect (or evap. Duty, qevap)
qevap = Mw,evap Cw Tw,evap (kW)
(kW)
= 50 x 4,187 x (20-10)
= 2093,5 kW
or
(E)
qevap = MR (H1 – H4)
MR =
2093,5
356 - 244
= 18,69 kg/s
(G) Condenser duty:
-qcond = qevap + W 12
= 2093,5 + 560,7
= 2654,2 Kw
-qcond = Mw,cond Cw Tw,cond
175


Tw,cond =
2654,2
140x4,187
= 4,53 C
The overall co-efficient of performance is:
Overall COP =
heat transfer in evap
Total energy consumption of refr unit
Some pictures indicating refrigeration plants on surface and underground
Underground plant (3,5 MW capacity)(E.E.S.A.M)
Surface refrigeration plant installation (38 MW capacity)
176
Surface plant machine room
Diagrammatic illustration of the full refrigeration cycle
177
9.
Cooling Plant (Further calculations)
The following measurements were obtained from a plant with a single-stage compressor.
The refrigerant used is Refrigerant 12.




Evaporating pressure:
Condensing pressure:
Compressor inlet temperature:
Compressor outlet temperature:
363 kPa (abs)
225 kPa (abs)
7C
65C

Flow rate:
140 l/s
The following measurements were obtained from a plant with a single-stage compressor.
The refrigerant used is Refrigerant 11.

plant room barometric pressure:
100 kPa

evaporating pressure:
=54 kPa vacuum
= 46 kPa (abs)

condensing pressure:
=125 kPa gauge
= 225 kPa (abs)


compressor inlet temperature:
compressor outlet temperature:
=3.4C
=70C
The evaporator duty is calculated from measurements made in the chilled water circuit and is
1 600 Kw
Calculation of performance and duty:
All enthalpy, temperature and specific volume can be read directly from the chart.
Refrigeration effect:
= HC – H B
= 224.5 - 76
= 148.5 kJ/kg
178
Heat exchange at condenser:
= HD - H A
= 260.5 - 76
= 184.5 kJ/kg
Work of compression:
= HD - H C
= 260.5 - 224.5
= 36.0 kJ/kg
Ideal work of compression:
= HE - H C
= 253.0 - 224.5
= 28.5 kJ/kg
Compressor efficiency:
= HE - HC x 100
HD - HC
= 28.5 x 100
36.0
= 79.2%
Percentage flash gas:
= HB - HF x 100
HC - HF
= 39 x 100
187.5
= 21%
Coefficients of performance (COP)
Carnot C.O.P.
= T1
.
T2 - T1
= 273.15 + 3.20
48.0 - 3.20
= 6.2
The Carnot COP is the ratio of cooling duty to the input power (the ratio of what has gone out to what
is put in) as mentioned before. The Carnot cycle is one in which heat is absorbed and rejected at
constant temperature, and where both compression and expansion processes occur adiabatically.
This cycle is the most efficient possible, hence the maximum possible COP is the Carnot COP.
179
Actual C.O.P.
= refrigeration effect
work of compression
= 148.5
36.0
= 4.1
Actual COP is always less than Carnot COP
Cycle efficiency
= Actual C.O.P. x 100
Carnot C.O.P.
= 4.1 x 100
6.2
= 66%
Power/cooling ratio
= work of compression
refrigeration effect
= 36.0
148.5
= 0.24
Refrigerant mass flow through evaporator
= Mr
= evaporator duty
refrigeration effect
= 1 600
148.5
= 10.8 kg/s
Inlet volume flow rate
= Mr x vc
= 10.8 x 0.36
3
= 3.89 m /s
Power consumed by compressor
= Mr (HD - HC)
= 10.8 x 36.0
= 389 kW
180
10. Cooling Plant
DUTY
From the first law of thermodynamics, the energy change of the water must equal the energy change
of the air, due allowance being made for evaporation. The energy balance is approximated by:
q = Mw.Cpw.tw
= Ma.Sa
In this equation only partial allowance is made for evaporation and a small additive term is omitted
from the right hand side. The magnitude of this correction is 3 to 4% and its omission is generally
justified.
Assume that a heat ‘imbalance’ of greater than 10% is unacceptable, requiring instruments to be
checked and readings to be re-taken.
WATER EFFICIENCY
From the second law of thermodynamics, the water cannot leave the tower at a temperature lower
than that of the entering air wet-bulb temperature. The wet-bulb temperature of the air leaving the
tower cannot exceed the temperature at which the water enters the tower.
two
>
twbi
twbo
<
twi
W
= (twi - two)/(twi - twbi)
i.e.: (actual) / theoretical max.
THERMAL CAPACITY FACTOR (R)
This is the ratio of the thermal capacity of the water stream to that of the air stream.
R = (Mw.Cpw)/(Ma.C’a)
181
For a particular installation, the water efficiency depends on the flow rates and temperature conditions
that are imposed on the installation. R is the parameter used to measure these conditions.
C’a is the EQUIVALENT SPECIFIC HEAT OF HUMID AIR for a process in which the moisture content
of the air varies. It is evaluated from the wet-bulb temperature and the sigma heats associated with
the inlet air and inlet water conditions.
C’a = (Swi - Sai)/( twi - tai)
(Specific heat of water = specific heat of air at that temperature)
FACTOR OF MERIT
This is a means of quantifying the basic performance characteristics of a heat exchanger.
The factor of merit is the value of the water efficiency when R = 1
The factor of merit is determined using figure 23.13 (EESAM p 635) if R (thermal capacity ratio) and
the water efficiency are known.
The factor of merit is dependent upon the design of the installation. It is fairly independent of water
and air temperatures, the barometric pressure and (within reason) of water and airflow rates.
Once the F-value of a cooling tower is known it is possible to reliably predict the performance of the
tower under a wide range of operating conditions.
11. Cooling plant tutorial
1. Typical cooling tower problem (p 636, Q 1 & 2, EESAM)
Barometric pressure:
112.5 kPa
AIR CIRCUIT
Quantity:
Temperature inlet:
outlet:
3
213 m /s
30/34 C
40.9/40.9 C
182
WATER CIRCUIT
Flow rate:
Temperature entering:
leaving:
1.
536 l/s
43.4
36.1
Calculate the following performance characteristics of the tower:
a)
b)
c)
d)
duty
water efficiency
thermal capacity ratio
factor of merit
2.
Predict the entering and leaving temperatures for the above tower when, as a result of the
installation of two new refrigeration machines, the amount of heat to be rejected increases to
23 259 kW. Assume that the factor of merit (F), the air and water quantities and the inlet air
temperature remain as given and calculated above.
11.
Cooling towers, coils and calculations
Q = Mw x Cpw x Tw (Water)
Or
Q = Ma x Sa (Air)
If condensation occurs on the outside of the coils, m aS would give slightly incorrect answers
as it does not take into account the heat removed from the system by the condensate (latent
heat).
A “rule of thumb” to be remembered is:
If both calculations were done and the results differ by more than say 7%, use the
“water side” calculation. What could be a prominent practical reason for deviation of
the two calculated results?
183
Single heat exchanger (coil type)
Another way of expressing heat transfer is the overall heat transfer coefficient U.
q = UA (tma - tmw)
(W)
UA product is a measure of the effectiveness of an indirect heat exchanger.
UA of a clean coil is between 10 – 25 KW/C.
What does a decline in UA mean?
184
Direct heat exchanger: Cooling tower
In a cooling tower, how does the heat transfer take place?
Heat is transferred from the water to the air (cooling tower) by two methods
(a) Convection (sensible heat), and
(b) Evaporation (latent heat)
Air to water (warm air @ cold/chilled water).
Several common factors that influence the amount and efficiency of direct contact heat exchangers.
(a) H2O flowrate.
(b) H2O supply temp.
(c) H2O mass flowrate.
(d) Air condition at inlet
185
(e) Duration of contact (air to water) factor of the design.
Between air and water
Size and concentration of the droplets.
Water and air efficiency: (EESAM, p. 634)
For a perfect heat transfer in a cooling tower:
Maximum thermal capacity of water = m w Cw (tw,in – tg,in) (J)
Max thermal capacity of air stream = Ma (Sw.in – Sin) (J)
Tower capacity factor (R)
Factor of merit.
Factor of merit is a measure of the basic performance of a unit (same as UA).
(See EESAM, figure 23.13 p. 635).
186
Do EESAM, Example 5 page 636:
INDIRECT HEAT EXCHANGE
187
EESAM, Chapter 19 (p. 465)
(1)
Heat Transfer
(a) Conduction
q = k A (t1 – t2)/b
Heat flux rate = thermal cond x area x temp diff thickness
Why does an insulation product / material have to attain a certain thermal conductive value?
( not > 0.05 WmK)
The ability of a material to conduct heat.
(2) Radiation
q = hR A (t1 – t2) Fev
Radioactive heat transfer
(3) Convection
(a) Boundary layer (Dominating effect of hear transfer)
The faster the air velocity, the thinner the layer, the more heat is transferred.
qc = hc A(t1 – t2)
Graphs predicting hc (p 477)
Overall heat transfer coefficient (U)
r
=
A2
A3
A4
Kins
Kpipe
h2
L
h4
=
=
=
=
=
=
=
=
hR
=
radius
2r2L
2r3L
2r4L
Thermal conductivity of insulation material
Thermal conductivity of pipe material
connective heat transfer coefficient between fluid and inside pipe surface
Length of pipe
connective heat transfer coefficient between ambient air and the outside surface of
the insulation.
radiative heat transfer coefficient 0.95 (Constant)
188
A2 A3
2
A34 = A 3 A 4
2
A23 =
Determine heat transfer coefficients from charts (pages 477 to 479)
(a) hr (page 473)
(b) h2
(c) h4
D4 = Inside pipe diameter + 2 (Pipe wall thickness) + Insulation material thickness
Now the heat gain by fluid per length is:
q = UA.(tdb - tw).L
W/length

q (heat flow) is now determined as UA, tdb and tw are known parameters.
And
q = Mf x Cf x t
M = Qf x f
kg/s

t =
q
=
Mf Cf
q
QF.p. f. C f
.
or
189
Insulated pipe section with indicating parameters
Heat transfer at wet surfaces
Do example 10 page 486
(1)
Condensation takes place (i.e. latent heat transfer)
So q = qc + qr + qL
Latent heat of condensation: = 2501 - 2,3830 T
190
When do you think will condensation take place?
Up to page 489:
Pipe Sizing
For a 2 m/s flow-rate
For new pipes f = 0.0039, however because of the ageing process this factor could increase
between 20 to 50%. UPVC (plastic type) f = 0.0032.
For bends and pipe fittings the frictional pressure loss could be increased by 10% to allow
for these additional pressure loss that would occur.
Energy and temperature changes within water systems
and
Therefore
H = Cw t
H = g.Z
(p 399.) (p 403 etc).
t = gZ
Cw
or roughly pressure drop: P = w.g.Z = Frictional pressure drop
191
Auto-compression
Page 403 (Know this relationship)
Similar to page 407 for water systems
12. MINE COOLING SYSTEMS (Review)
Three sets of heat transfers involved
(a) Transfer of heat from work areas to evaporators of Refr, units.
(b) Transfer of heat from evaporator to condensers in Refr, unit.
(c) Transfer of heat from cond, to free atmosphere (Surface).
Heat Exchanger
A device that promotes the transfer of thermal energy from one solid or fluid system
to another
- Indirect heat exchangers
- Direct heat exchangers
Indirect heat exchanged across a solid medium that separates two fluids.
Direct Direct contact between the two fluids.
PERFORMANCE CALCULATIONS FOR INDIRECT
HEAT EXCHANGERS (Review)
Equilibrium heat gained by one fluid = heat lost by other fluid.
Thus: rate of heat transfer
Another way of expressing heat transfer is
- Overall heat transfer co-efficient, U(W/m2 C)
And difference between mean temp, of air and water.
Temp of air and water avarg. in a logarithmic manner rather than a linear fashion, the more accurate
formula is:
UA is a measure of the effectiveness of an heat exchanger (similar to the pipe heat exchanged
formula indicated previously).
192
2
h = heat transfer co-efficient (W/m C)
x = thickness of tube walls
Kt = thermal cond of tube material (W/mC)
i = insider tube surface
o = outside tube surface
m = mean of inner to outer surface
hfi & hFo = heat transfer co-efficient associated with fouling
New tubes hFi & fFo = infinite
13. DIRECT HEAT EXCHANGERS
Cooling Tower
Range = tw,in – tw,out (C)
Approach = tw,out – ta,in (2C is acceptable)
Balance equation
Ma (Sout - Sin) = Mw Cw (tw,in – tw,out) (W)
The Perfect Tower
- Water would leave at inlet air wet bulb temp
- Air would leave at the incoming water temp
tw.out = ta,in
and
193
ta,out = tw,in
Water efficiency
w =
actual heat loss from water
theoretical max heat loss from water
Similarly the air efficiency:
DO Examples:
Heat exchange across walls of pipes and ducts
Pipe Sizing, Insulation etc.
1) Direct & Indirect heat exchange systems
Definition of each
2)
a) Why do we cool deep underground mines?
To reduce the heat load in the mine
b) Where do these heat load come from?
194
a) Rock (Primary source)
b) Machinery (fans, diesel, sub-station, hoists, etc.)
195
Diesel machinery could induce heat to the surroundings by as much as 2 to 2.5 times its
rated output power
c) Fissure Water (Major contributor)
d) Explosions
196
e) Men
c) How do we now reduce the heat load?
(Direct & Indirect types).
197
3) Direct heat exchangers (Cooling Towers & spray chambers type)
Indirect heat exchangers (coils etc.)
Performance calculations
(a) Equilibrium: heat gain by one fluid = heat loss by the other fluid.
Thus:
So, either the water or the air circuit could easily be measured and used for the calculation.
The water measurements are normally more accurate and the tendency therefore
Typical Spray Chamber problem
Barometric pressure:
115 kPa
AIR CIRCUIT
Quantity:
temperature
3
inlet:
outlet:
24.7 m /s
33.1/34.7 C
22.1/22.1 C
WATER CIRCUIT
Flow rate
Temperature
entering:
leaving:
35.3 l/s
18.8 C
28.6 C
1. Calculate the following performance characteristics of the spray chamber:
a) Duty
b) Water efficiency
c) Thermal capacity factor
d) Factor of merit
198
Project: Refrigeration
1. Handing in date : 31 October 2001
2. Project to be done in readable own handwriting and not typed.
3. All calculations to be shown clearly. (hand-written)
4. Assumptions, if any, to be clearly indicated as such.
Project Objective
a) To understand a Refrigeration plant operation
b) To understand the reticulation system Components etc.
c) To demonstrate the students ability in performing some basic calculations
involved in a refrigeration and reticulation system design.
Project
The following plant and chilled water distribution system is given.
Plant information
Compressor: Refrigerant inlet temperature: 6 C
Refrigerant outlet temperature: 58 C
Condenser: Water flow rate : 110/s
Refrigerant Pressure: 1150Kpa
Evaporation: Water inlet Temperature: 23.8 C
Water outlet Temperature : 6,5 C
Refrigerant Pressure : 380 kPa
Water Flow-rate: 55 /s
Use the chart supplied and analyse the unit Performance (show all
calculations)
The following information is provided for the mine distribution system
1) Pump efficiency =78%, Energy recovery unit efficiency =75%
2) Surface BAC (Direct contact system) This is the only direct contact energy
exchange system in the system.
(a) Duty =18MW
3) Pre- Cooling Tower (PCT) outlet water Temperature =21 ℃ with a factor
of merit of 0,45
4) Surface BAC (Bulk air cooler) distance to return water dam 1 = 320 m
5) Surface BAC water flow = 395 l/s with a factor of merit of 0,5
6) Coils efficiency = 65%
7) PRV operating head = 210 m
8) Plant leaving water flow and temperature = 710 l/s at 4.5oC.
9) Shaft bottom BAC uses 90 l/s.
199
Determine:
1.
2.
3.
4.
5.
6.
The temperature in each of the dams (1 – 5)
All pipe sizes (The mean water velocity in each pipe is 2 m/s)
The PCT Duty (MW)
The overall plant capacity (MW)
The return water flow rate to the PCT with the temperature
The water temperature to the plant.
Using the first horizontal intake level to the workings. what influence
could full pipe insulation have on the coil arriving water temperature ?
Indicate on your calculation sheet and on the flow sheet provided the
temperatures as the water flows down and up the reticulation system.
All reasonable assumption to be stated clearly. Where data is not given
and is required in the calculation procedure, assumption must be made.
Total project marks = 25 points.
200
201
202
203
204
205
206
207
208
209
210
211
212