1. science
2. ecology
3. pollution

MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
Sample Questions & Worked Out Examples
For
Min 03018
MINE ENVIRONMENTAL ENGINEERING
B Tech. (First Year)
Mining Engineering
MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION
QUESTIONS AND ANSWERS
MIN - 03018
MINE ENVIRONMENTAL ENGINEERING
(PART I)
B.Tech. First Year
Mining Engineering
1
CHAPTER 1
GASES
1.* In order to control the hazards from the gases, there should be rules, regulations and
laws limiting the concentration of these gases in the mine atmosphere. Based on the
international practices, briefly discuss legislative issues for mine gases.
2.** Various kinds of gases may naturally occur in underground mines. It is advantageous
to be well-informed about the mine gases. Write short accounts on the qualities and
effects of the following gases:
(a) Oxygen
(b) Nitrogen
(c) Carbon dioxide
(d) Carbon monoxide
(e) Hydrogen sulfide
3.** Mercury has been used to some extent for extracting gold in Myanmar artisanal
mining. However, improper use may lead to workers' health and environmental
problems. Write a short account on mercury poisoning symptoms
4.** Mercury and cyanide were used to some extent for extracting gold in Myanmar
artisanal mining. However, improper use of them may lead to workers' health and
environmental problems. Write a short account on cyanide and its physiological
effects on human.
5.*** Engineers in mines have to encounter some gases naturally, and some by performing
mining operations with certain equipment. In order to control the hazards from the
gases, they had better know the sources of gases. Now write down some controlling
methods for these gas sources in a moderate length.
2
CHAPTER 2
DUST
1.* Based on the international practices, briefly discuss legislative issues for mine dust.
2.* Describe the general nature of hazards caused by dust particles and then focus on
biological effects on human beings.
3.** There are several methods to handle the dust problems. Of them, give a short
discussion on:
(a) Source control
(b) Dilution
4.** Of several methods to handle the dust problems, make short discussions on:
(a) Isolation
(b) Dust collection
5.*** Discuss the sources of dust, which play a vital role in controlling dust hazards.
3
CHAPTER 3
NOISE
1.* Write short accounts on Sound intensity and power.
2.* Write short accounts on "Decibel".
3.* Write short accounts on typical noise levels.
4.** Write short accounts on hearing damage and acceptable levels
5.** Define noise. At a moderate length, discuss different velocities of sound in various
materials.
4
CHAPTER 4
MINING AND IMPACTS ON THE ENVIRONMENT
1.* Identify the wastes from mining operations.
2.* Identify the wastes from milling operations.
3.* Identify the wastes from metallurgical operations.
4.** Discuss the areas environmentally impacted by mining activities as in:
(a) Energy consumption
(b) Air
(c) Health and safety
5.** Discuss the areas environmentally impacted by mining activities as in:
(a) Water
(b) Land
(c) Air
6* Write some account on desertification of land that could be caused by mining activities.
7* Write some account on chemical pollution of underground water that could be caused
by mining activities.
8* Write a short account on chemical pollution of the atmosphere that could be caused by
mining activities.
9** Write a brief description on physical pollution of surface public water bodies that
could be caused by mining activities.
10*** Write a moderate account on chemical pollution of public water bodies that could
be caused by mining activities.
5
CHAPTER 5
TECHNOLOGIES TO IMPROVE ENVIRONMENTAL PERFORMANCE
1. * Describe some environmental impacts possibly caused by pre-combustion cleaning of
coal.
2. * Describe some environmental impacts possibly caused by transportation of coal.
3. * Make a short discussion on post combustion technologies for coal.
4** New developments are continually made in the mineral industry to abate the
environmental impacts. Write brief accounts on the following:
(a) The Penetrating Cone Fracture (PCF) Technique
(b) Laser water jet (LWJ), and
(c) The Cardox tube.
5. ** New developments are continually made in the mineral industry to abate the
environmental impacts. Write brief accounts on the following:
(a) Biological oxidation
(b) Water gun, and
(c) Plasma blasting
6. ** Write a short explanation on semi-dry tailings disposal, which is important in
protecting environment in mining industries.
7. ** Make a short discussion on combustion enhancing technologies for coal.
8. ** Write a short explanation on co-disposal of coarse and fine residues, which is
important in protecting environment in mining industries.
9. ** Make a brief discussion on waste disposal in underground old workings.
10. *** Make a brief discussion on storing of wastes in underground caverns.
11. *** Generally, there are two common ways to solve acid mine drainage (AMD)
problems, active and passive treatment systems. Explain the latter.
12. *** Mineral base is essential for industrialization programs. However, mining
activities are generating considerable disruption to the environment. Identify the
ecosystem and socio-cultural environment that may be affected by pollutions and
wastes from mining operations.
6
CHAPTER 6
MINE AND QUARRY REHABILITATION
1* Write a moderate description of landscaping or the reshaping and grading of a site,
which is an essential aspect of rehabilitation.
2** Control of wind and water erosion is important during mining and rehabilitation.
Write a short account on controlling wind erosion on disturbed soils.
3** Describe some of the land use possibilities or reclamation opportunities that a mined
land can be transformed.
4** Make a moderate discussion on the selection of species, which is important in
reclaiming a mined land.
5*** Explain at suitable length the practice and advantages of soil amendment without
fertilizers.
6***Explain at suitable length the practice and advantages of soil amendment by using
fertilizers.
7*** Discuss, at moderate length, different methods of planting on a mined land.
7
Example Questions and Answers
1. ** Various kinds of gases may naturally occur in underground mines. It is
advantageous to be well-informed about the mine gases. Write short accounts on the
qualities and effects of the following gases:
(a) Oxygen
(b) Nitrogen
(c) Carbon dioxide
(d) Carbon monoxide
(e) Hydrogen sulfide
Answer:
Oxygen
Oxygen is colorless, odourless and tasteless and slightly soluble in water. The main
hazard of oxygen is not poisoning but by the lack of it, i.e. oxygen deficiency.
Oxygen %
Effects
21
Normal concentration
17
Breath faster and deeper
14
Dizzy sensation, buzzing in the ear
10
Increased heartbeat
7
Confusion, loss of consciousness and
death
Nitrogen
Nitrogen is colorless, odourless and tasteless and practically insoluble in water. It is
chemically very inactive and its main hazard is that it dilutes the oxygen in the
atmosphere, i.e. causes oxygen deficiency.
8
Carbon dioxide
Carbon dioxide (CO2) is colorless, odourless and has a faint acid taste. It is denser than air
and does not burn or support combustion. Its main hazard is that it dilutes oxygen in the
atmosphere, i.e. causes oxygen deficiency. It also acts as a stimulus to the breathing rate of
a person, as indicated in the following table.
Carbon dioxide
Increase in respiration rate
%
0.5
Slight
2.0
50%
3.0
100%
5.0
300%
10.0
Intolerable for extended periods
First aid measures are to remove the patient to fresh air, give oxygen or artificial
respiration.
Carbon monoxide
Carbon monoxide (CO) is colorless, odourless and tasteless. It does not support
combustion but will burn and explode at concentration between 12.5 and 74%.
Preferentially fastens to haemoglobin in the blood in place of oxygen. Brain and heart
muscles, the greatest oxygen consumers are most affected. Above 4000 ppm, coma can
occur without warning. 500-1000 ppm can cause headache, first breathing, nausea,
weakness, dizziness, mental confusion and hallucination.
Hydrogen sulfide
It can be produced if acid contacts metal sulfides. And if traces of arsenic and phosphorus
are present, arsenic and phosphene can also be generated, both of which are toxic. It is
more toxic than hydrogen cyanide. It irritates eyes and respiratory tract at low
concentrations. Higher concentrations cause respiratory paralysis with asphyxia,
convulsions and coma.
9
2. ** Mercury has been used to some extent for extracting gold in Myanmar artisanal
mining. However, improper use may lead to workers' health and environmental
problems. Write a short account on mercury poisoning symptoms.
Answer:
Mercury
Description: Silvery liquid; metallic element (Hg)
Constants: At.wt. = 200.61; mp = 38.89° C; bp = 356.9° C; d = 13.546.
TLV : 0.1 milligram per cubic meter of air (ACGIH)
Hazard: Dangerous; when heated it emits highly toxic fumes
Countermeasures: Ventilation control, storage and handling, and shipping
regulations.
Mercury has disastrous results when inhaled as gas. In the lungs, it slowly becomes
soluble as methyl mercury. It is then absorbed into the blood stream, penetrates the blood/
brain barrier, and causes acute mental and muscle decline.
After absorption, it circulates in the blood and stored in the liver, kidneys, spleen and
bone. It is eliminated in the urine, feces, sweat, saliva and milk. In industrial poisoning,
the chief effect is on the central nervous system, mouth and gum. Colitis has been reported
frequently and nephritis is rarely reported.
Other symptoms of mercury poisoning are stomatitis, tremors, and psychic disturbances.
The first complaints are of excessive salivation and pain on chewing. In severe cases, there
may be loosening of teeth, and a dark line on the gum margins. In slow poisoning, the
salivation may be absent, and the only complaint is dryness of the throat and mouth. The
tremor is of intention type. It can be seen when the patient spread the outstretched fingers,
protrudes tongue, or attempts to perform specified movements. Muscles of the face, hands
and arms are chiefly affected. In more severe cases, there may be convulsive movements.
Writing is illegible. The psychic disturbances include loss of memory, insomnia, lack of
confidence, irritability, vague fear and depression.
The dermatitis produced by fulminate of mercury takes the form of small, discrete ulcers
on the exposed parts, and usually accompanied by inflammation of mucous membranes of
the nose and throat.
3. *** Discuss the sources of dust, which play a vital role in controlling dust hazards.
10
Answer:
Sources
The main source of dust in mines stems from the rock itself. The mining operations
consist of breaking the rock in situ to sizes convenient for transportation. It is inevitable
that large quantities of dust will be produced. One cubic millimeter of rock crushed to
two-micron particles would yield some 200 million particles. Drilling a hole 30 mm in
diameter and 1.8 m deep has a volume of over 1 million cubic millimeters. Fortunately
most of the dust particles produced when drilling are much larger than two micron and
very few of them become airborne. Generally this is true of most sources of dust if suitable
control methods are applied.
Blasting
Blasting produces very large quantities of dust. Some of the fine dust is carried
away by the air stream but a large amount of dust is trapped within the broken rock. Large
particles tend to settle out as do fine particles which agglomerates into lager masses. The
amount of dust formed by a blast depends to a certain extent on the pattern of the round,
the method of blasting, the type of tamping material and the type of explosive used.
Mechanical Loading
The blasted rock contains dust particles both within the rock pile and also on the rock
surfaces. Mechanical loading of the rock will release all the dust contained within the rock
pile and some of the dust on the rock surfaces. The greater the agitation of the broken
rock, the greater will be the amount of dust released. It has been demonstrated that the rate
of dust production is directly related to the hoe in scraping. One of the problems with
mobile loading equipment is ensuring that the exhaust is not directed onto rock surfaces or
the footwall and so stir up the settled dust. The advent of high capacity Load Haul Dump
equipment has created additional problems because their rate of removing broken rock
exceeds most dust allaying procedure.
Drilling
When introducing the sources of dust in mining, mention was made of the possible dust
production when drilling. Supplying water at the correct pressure and flow rate ensures
that the rock surface is wet at all times, and in fact the rock is broken under a film of
water. Leakage of compressed air into the water can however break down this film and
11
allow dusty air to escape from the hole. Also the compressed air may assist in evaporating
water which in its own right contains large amount of dust. Water is normally supplied to
the drill steel via a water tube in the drill itself. This water is delivered to the bottom of the
hole through the drill steel. It is generally not very effective when collaring the hole.
Transportation
Whenever rock changes direction it is normally subject to an acceleration followed
by a rapid deceleration. Dust particles adhering to the surface of the rock may then be
dislodged and become airborne. Typical situations are conveyor transfer points, loading
trucks, discharging trucks, etc.
Miscellaneous Sources
Diesel engines produce particulates in addition to gases in their exhausts. The
concentrations depend on the mode of operation of the engine and can vary between 25
and 250 mg/ m3 when idling and at full load respectively. Typical average emissions are of
100 mg/ m3. The particulate formed is very fine, probably less than 0.5 microns in size,
which agglomerates into particles with a mean size of between a quarter and half a micron.
Other miscellaneous sources of dust are blowing out with compressed air or the misuse of
compressed air in general. Welding can produce copious amounts of fine dust. Most other
sources are the results of making airborne previously settled dusts. Intake shafts and fans
close to cement storage silos and dry tailings dams can result in high intake dust
concentrations.
4. ** There are several methods to handle the dust problems. Of them, give a short
discussion on:
(a) Source control
(b) Dilution
Answer:
Control
The most important method of suppression and control of dust is to avoid creating
them. This simple approach can usually result in maximum benefits at minimum cost and
can be termed source control. If a dust source cannot be eliminated, then the second
method of control is to isolate it i.e. remove the personnel or alternatively remove the dust
12
as it is formed. A third method is to dilute it as it is formed, with fresh air. If the dust
source is very large, it is often desirable and necessary to filter it in order to prevent its
dispersion to the general atmosphere.
(a) Source Control
When dealing with dust formed as a result of rock breaking, source control is best
achieved by the application of water. The basic principle is to prevent the dust becoming
airborne. It is important to recognize that the water forces (quantity and pressure) must be
in equilibrium with the forces creating the dust at the point of impact. If they are not, the
application of water will not be fully effective or there will be a wastage of water.
Maintaining a moisture content of 1 % particularly when transporting broken rock through
airways, consequently a 5 % moisture content should be the target value.
The excessive use of water can result in wastage, flooding of drainage and of ore
passes, overloading pump stations and dirty water settlers. To prevent wastage when
wetting down, a suitable spray nozzle should be used. This should ensure an adequate
spread of water over a large area and prevent the settled dust being stirred up. The water
should be as clean as possible to prevent the formation of dust from the evaporation of
dirty water. Clean water contains up to 10 million particles of dust per milliliter and dirty
water can contain ten or twenty times as much. The use of this water can add to the
general dustiness of the air.
(b) Dilution
In many situations, the dust produced at the source is not of a high concentration
and there may be numerous sources. In these situations, the supply of fresh air to dilute the
dust is advocated. the quantity of fresh air required can be obtained from the dilution
formula given Chapter 2:
Qg
Q = ---------------- - Qg
AC - B
Where Q = the required dilution rate m3 / s
Qg = gas emission rate m3 / s
AC = allowable gas concentration
B = concentration present in normal air
13
In this case, the rate of formation of the dust and its concentration must be known.
There are four limiting factors to the use of dilution as a method of dust control.
1.
The quantity of dust generated must not be excessive or the quantity of fresh
air for dilution will be impractically large.
2.
Workers must be far enough away from the dust source to allow the dilution
to take place before they are exposed to the mixture.
3.
The health hazard of the dust should be low otherwise again the quantity of
fresh air for dilution becomes impractically large.
4.
The formation of dust should be uniform and not sporadic.
In many underground situations, dilution is the only practical method of
controlling the levels of dust encountered.
5. * Write short accounts on Sound intensity and Decibel.
Answer:
Sound Intensity
Sound was defined, and illustrated to be, a succession of traveling pressure waves
moving away from a source. Because they are traveling, the pressure is being transmitted
from the source. This is the same as saying that force is being transmitted or moving.
Intensity I is the amount of energy passing through unit area in unit time and is usually
expressed as Watts/ m2. Intensity is directly proportional to the square of the pressure.
Since sound measuring instruments measure pressure it is possible to calculate the total
acoustic power produced by a source. This is important because the sound or acoustic
pressures measured depend on the source and its surroundings. Moving the same source
into different surroundings and measuring the pressure at the same distance need not result
in the same values.
The intensity at a point can be obtained by:
p2
I = --------wc
where I = intensity (W/ m2)
p = pressure (Pa)
w = density (kg/ m3)
14
c = velocity of sound (m/ s)
Decibels
The ear is remarkably sensitive and responds to intensity i.e. pressure squared
rather than pressure. The lowest intensity the ear can normally detect is approximately 1012
watts/ m2. The threshold of pain is approximately 10 watts/ m2. This very large spread is
compressed into a logarithmic scale and related to a datum value to avoid negative values
and to be more meaningful. If the datum is taken as the threshold of hearing i.e. 2 x 10-5 Pa
the unit of sound becomes
(pressure)2
log
-------------(2 x 10-5)2
This is known as a Bel and a further modification made to avoid excessive use of
decimals is to multiply it by 10. This then results in the decimal or dB. With respective to
acoustic pressure the units becomes:
(pressure)2
Sound Pressure Level SPL = 10 log -------------- dB
(2 x 10-5)2
With respect to acoustic power the reference level is 10-12 watts and the units become:
Power
Sound Power Level SPL = 10 log -------------- dB
10 -12
6. ** Write short accounts on hearing damage and acceptable levels.
Answer:
Hearing Damage and Acceptable Levels
Hearing damage results from damage to the hear cells caused by excessive vibrations. This
is normally progressive and can take place over extended periods of time i.e. 10 to 15
years.
15
Acceptable levels depend on the criteria being used for assessment. From the point of view
of the underground ventilation engineer the usual criteria is the avoidance of hearing
damage. The following table gives two different permissible noise exposures, both of
which are recognized by various international bodies.
Table. Permissible Noise Exposures
Duration in hours per day
Noise level in dBa
8
90
85
4
95
88
2
100
91
1
105
94
½
110
97
¼
115
100
A second criterion that may occur underground is speech and signal interference. This
could occur at telephones, shaft signals at stations or vehicle warning sirens at ventilation
doors.
7. ** Discuss the areas environmentally impacted by mining activities as in:
(a) Water
(b) Land
(c) Air
Answer:
Air
Surface mines may produce dust from blasting operations and haul roads.
Many coalmines release methane, which is a greenhouse gas. However, methane is
generally captured, where it is economically feasible to do so. Since some coal plants
use CFCs, HCFCs and HFCs, there is the potential for the release of these ozonedepleting substances, but such releases are tiny. Tailing dams, if not vegetated or
capped, may also be a source of dust. Radiation is emitted from tailings dams where
16
radioactive elements are found in the ore. Smelter operations with insufficient
safeguards in place have the potential to pollute the air with heavy metals, sulfur
dioxide, and other pollutants.
Water
The mining industry uses large quantities of water, though some mines are able to reuse
much of their water intake. Mining brings sulfur-containing minerals into the presence of
air, where they are oxidized and react with water to form sulfuric acid. This, together with
various trace elements, impacts ground water, both from surface and underground mines.
Tailings dams and waste rock heaps are also sources of acidic drainage water, affecting
surface and underground water. The chemical deposits left behind by explosives are
usually toxic, and they contaminate and increase the salinity of mine water. In-situ
mining, in which a solvent is allowed to percolate through unmined rock, leaching
minerals directly, has the potential to contaminate ground water. Artisanal mining may
impact water where mercury is used to process gold.
Land
Mining moves large quantities of rock. In surface mining, land impacts are very large.
These impacts may be temporary where the mining company returns the rock and
overburden to the pit from which they were extracted. Many copper mines, for
example, extract ore that contains less than 1 % copper. For many nonferrous metals,
all of the mined ore thus becomes waste. Artisanal mining, alluvial mining for gold
and diamond often has an impact far greater than the size of the operation. Many areas
are marked by thousands of small holes, which have been indiscriminately dug in
search of precious minerals. Trenches that scar the landscape are problems in some
places. These activities may lead to erosion and the localized destruction of
riverbanks.
8. * Identify the wastes from mining and metallurgical operations.
Answer:
Milling Industry
Solid Wastes: Solid wastes constitute the major disposal problem of the milling industry.
The wastes consist of the host rock for the minerals that have been removed by various
concentration processes. The wastes are known as tailings and they are deposited in tailing
ponds. The solids are allowed to settle in the settling basin, and the effluent is decanted
17
from the cleanest portion. The solid residues pile themselves, if not properly handled,
constitute sources of dust and sediment to be distributed to streams by runoff.
Liquid Wastes: The major liquid waste disposal problem confronting the milling industry
consists of water decanted from the tailing ponds. The decanted water contains some
suspended solids and sometimes low concentrations of cyanide and other dissolved ions.
One challenging area of research for the milling industry and associated government and
academic institutions consists of the development reagents that are either nontoxic or at
least have minimum toxicity, particularly with respect to aquatic species.
Gaseous Wastes: With the exception of a few wet scrubbers and dust derived from the
tailings piles, the milling industry has negligible pollution problems of this type. Gaseous
wastes are essentially nonexistent in the milling industry.
Metallurgical Industry
Solid Wastes: The term metallurgical industry is broad and covers many industries. It is
difficult to define precisely. Moreover, many operations are connected with mines or mills
or both. The solid wastes from metallurgical processes range from innocuous slag from
blast furnaces to unstable solids produced by the washing of gaseous wastes and metallic
products. The most common are those that are the result of air pollution control devices
such as scrubbers and cyclones used for dust control. These solid wastes are usually in the
form of sludge from wet collecting devices, and they are frequently discarded in landfills.
Liquid Wastes: Liquid wastes from metallurgical industry range from wash water
of the raw material to effluent from wet scrubbers used for dust control.
Gaseous Wastes: Gaseous waste control is a continuing problem to the metallurgical
industry because all metallurgical processes produce gaseous emissions of one kind or
another. They may contain only particulate matter, they may contain only gases, or they
may contain both. The current practice is wet scrubbing to remove particulate matter,
followed by appropriate additional treatment to remove whatever gas happens to be
present. In the case of sulfur dioxide gases from copper and lead smelting operations, the
gas emitted is sulfur dioxide. The standard technique for handling sulfur dioxide is
conversion to sulfuric acid, which is now used in fertilizer production.
9. ** New developments are continually made in the mineral industry to abate the
environmental impacts. Write brief accounts on the following:
(a)
The Penetrating Cone Fracture (PCF) Technique
18
(b)
Laser water jet (LWJ), and
(c)
The Cardox tube.
Answer
(a) The Penetrating Cone Fracture (PCF) Technique
This technology uses a non-explosive chemical propellant to generate very high
transient gas pressures at the bottom of a short drill hole. The injector is a gun-like device
in which a propellant charge is burned in a combustion chamber external to the working
face. A propellant cartridge is inserted into a breech and fired using conventional
technology. The burnt propellant gases are directed to the bottom of a shallow drill hole by
a barrel to create a rapid but non-explosive pressure pulse at the hole bottom. It generates
very low velocity fly rock and size distribution of rock may be controlled. While much
smaller quantities of rock are broken per blast than with explosives, PCF may be used
continuously since the area does not have to be cleared in order to allow fumes to be
purged.
(b) Laser Water Jet
This is the controlled application of pulsed laser energy for hard rock excavation. It
does not involve rock drilling, and its main advantage over conventional techniques is the
ability to continuously excavate small particles from the rock surface. A cutter head fitted
to a jumbo would be supplied with laser energy via optic fibre cable from a remote laser
unit, and with a high-pressure water supply. The ore chips produced would be removed by
vacuum and pumped via slurry line to the process plant. The high-pressure water jet
provides a consistently clear path to the rock face through the dust, fog and other debris
normally found at mining face. The excavation mechanism involves vaporizing micronthick layers of materials from the surface in nanosecond (10-9 sec) time intervals. Provided
the expanding gases are held shortly against the rock surface with water, pressure
approaching 1,000,000 psi is generated. These pressures impart instantaneous shock wave
to the rock in much the same way as the traditional explosives.
(c) The Cardox Tube
The tube consists of an alloy steel tube containing a charge of liquid CO2, with a
discharge head one end, and a firing head at the other end. At the firing end is a filling
valve through which the CO2 is loaded and low-tension electrical connections to detonate a
chemical energizer, which is also loaded into the tube. At the discharge end is carefully
19
designed steel rupture disc to contain the charge and external retaining pawls to hold the
tube in the rock. A hole is drilled in the material to be broken, and the loaded tube inserted.
The increased heaving mass of gas is discharged into the surrounding material which
breaks down along the least line of resistance with negligible amount of vibration and dust.
10. Describe some of the land use possibilities or reclamation opportunities that a mined
land can be transformed.
Answer
Reclamation Opportunities
It is a common perception that mining destroys land values. However, while
mining alters the shape of the land, it often provides a reconfigured piece of land that is of
more interest and value than it was in original state. Some of the land use possibilities are
as follow:
1.
Crop-land
2.
Pasture land or source of hay
3.
Grazing land
4.
Forestry use
5.
Residential use
6.
Industrial / commercial use
7.
Recreation
8.
Fish and wildlife habitats
9.
Developed water resources, and
10.
Undeveloped land or no current use.
The planners should encompass the whole range of possibilities. Obviously, there
are several considerations. The overriding consideration is that the option be practical. The
second is that a demand should exist for the product of restoration. The third is that the
product should yield an economic or social return. Options for restoration of different
types of wasteland are shown in the following Table:
Options for restoration of different types of wasteland
x
Dry pit
x
x
x
x
x
x
x
Mine waste
x
Refuse tip
x
Housing
Industry
x
?
xx
Sport
Industry
Forestry
Wet pit
x
?
?
xx
x
x
xx
xx
Disposal
x
Waste
x
conservati
x
Nature
x
Sport
xx
Amenity
Neglected
Housing
New use
Agricultur
Old use
Water
20
xx
x
x
x
x
xx
xx
x
x
xx
xx
x
xx
x
x
x
x
x
xx: likely new use; x : possible new use; ? : Possible new use depending on circumstance.
************************* END ***********************
MINISTRY OF SCIENCE AND TECHNOLOGY
GOVERNMENT TECHNOLOGICAL COLLEGE
DEPARTMENT OF MINING ENGINEERING
QUESTIONS AND ANSWERS
ON
Min. 03018 Mine Environmental Engineering
Part II
B Tech. (Mining), First Year
1
(16 marks for each question)
CHAPTER 1- Fundamentals of Mine Ventilation
1- *
A quantity of 30 m3/s of saturated air at 25°C w.b,d.b. and 120 KPa air pressure is passed
through a cooling plant and cooled to 15° C w.b.,d.b. How much heat and water is removed?
2-*
A quantity of 400 m3/s of air enters the collar of a large shaft at 10°C w.b,d.b., 100 KPa
pressure. At the bottom of the vertical shaft, 2000 m. deep, and temperature are 16°, 18° C at
115 KPa pressure. After passing through intake haulage ways and stopes, the air reaches a
point where it is 27°, 29°C at 115 KPa. When it reaches the top of the upcast shaft, it is 22°C
w.b.,d.b. at 100 KPa. How much heat and water is removed in each section?
3-**
(a) What is mine ventilation?
(b) Explain the purposes of mine ventilation.
(c) Write the short notes of Psychrometry.
4-**
(a) What are the purposes of mine ventilation?
(b) A quantity of 10 m3/s of air at 20 °C w.b. 27 °C d.b. and 120 KPa. air pressure enters a
hoist chamber in which a 95 percent mechanically efficient 3000 KW motor is running. What
will be the w.b. and d.b. of air leaving chamber? (*** Need Psychrometry Chart)
5-**
(a) Define the term of Psychrometry and discuss of its importance in underground mining.
(b) Differentiate between Enthalpy and Sigma heat with numeric examples.
6-**
Write a summary of humidity and dew point of air
7--** (a) Find the Enthalpy & Sigma heat of the following conditions.
Air mass, M
= 1.0 Kg.
Water mass, m = 0.007 Kg.
Temperatures
= 13.5 ºC wb. & 20.0 ºCdb.
(b) How much heat is required to raise the temperature of 0.1 m3 of bath water from 15 to
45ºC? (It should be noted that the density of water is 1000 kg/m3.)
8-*** If 300,000 tons of rocks are broken per 30-day month in a mine at a depth where the average
virgin rock temperature is 40°C and by the time this rock reaches surface in a skip it has been
cooled to 20°C, what is the average rate at which heat is released by the broken rock?
If this mine is ventilated by means of 700 m3/s of air at a density of 1.2 kg/m3, by how much
could this heat increase the temperature of the air?
2
9-*** (a) At what rate is heat removed from 3 m3/s of dry air at sea level density (1.2 kg/m3), which
enters a cooler at 35ºC and leaves it at 20ºC?
(b) If 300 000 tons of rock are broken per 30-day month in a mine at a depth where the
average virgin rock temperature is 40ºC and by the time this rock reaches surface in a skip it
has been cooled to 20ºC, what is the average rate at which heat is released by the broken
rock? If this mine is ventilated by means of 700 m3/s of air at a density of 1.2 kg/m3, by how
much could this heat increase the temperature of the air?
(Thermal capacity of rock = 0.837 KJ/Kg ºC)
10-*** A quantity of 400 m3/s of air enters the collar of a large shaft at 10°C w.b,d.b., 100 KPa
pressure. At the bottom of the vertical shaft, 2000 m. deep, and temperatures are 16°, 18°C at
120 KPa pressure. After passing through intake haulage ways and stopes, the air reaches a
point where it is 27°, 29°C at 120 KPa. When it reaches the top of the upcast shaft, it is 22°C
w.b.,d.b. at 100 KPa. How much heat and water is removed in each section?
Temp.
( °C)
10°, 10°
16°, 18°
27°, 29°
22°, 22°
Pressure
(KPa)
100
120
120
100
Moisture content
(g/Kg)
7.729
8.740
18.186
16.883
App. Sp. Vol.
(m3/kg)
0.816
0.706
0.743
0.870
Sigma heat
(KJ/Kg-°C)
29.202
39.654
73.544
63.466
3
CHAPTER 2- FUNDIMENTALS OF AIRFLOW
1- *
Classify whether a fluid is laminar or turbulent when its flows at 10 m/s in 10 cm. diameter
circular pipe. Assume the dynamic viscosity of the fluid is 0.2750 Ns/m2 and density is 1100
Kg/m3.
2-*
A straight tunnel is 1000 m. long and its dimension is 4m. × 3 m. Determine the resistance of
the tunnel and frictional pressure drop along it when an air flow of 20 m3/s is passing and k =
0.01 Kg/m3.
3- ** (a) Derive the pressure form and head form of the Bernoulli's equations from the first
principle.
(b) Find the total pressure of the following condition.
Density of mine air
= 1.25 Kg/m3
Density of liquid
= 1000 Kg/m3
Difference in liquid level = 9 mm.
Velocity of air current
= 20 m/s
4-**
(a) Classify whether a fluid is laminar or turbulent when its flows at 10 m/s in 10 cm.
diameter circular pipe. Assume the dynamic viscosity of the fluid is 0.2750 Ns/m2 and
density is 1100 Kg/m3.
(b) A straight tunnel is 1000 m. long and its dimension is 4m. × 3 m. Determine the resistance
of the tunnel and frictional pressure drop along it when an air flow of 20 m3/s is passing and
k = 0.01 Kg/m3.
5-** An airway of 800 m. with no bends 3m.× 2m.dimension. Determine the resistance of airway
and frictional pressure drop along it when an airflow of 30 m3/s is passing and K = 0.0158
Kg / m3.
6-**
(a)Derive the pressure form and head form of the Bernoulli's equations from the first
principle
.
(b) Discuss the application of these equations in mining industry.
7- ** (a) To reduce the air flow of air through a 3 m. × 3.5 m. airway to 20 m3/s. It is necessary to
install a regular with pressure drop of 500 N/m2. Calculate the regulator area for a box
regulator with C = 0.62.
(b) A 0.61 m. diameter duct 300 m. long has an intake volume rate of 1.70 m3/s and a
delivery volume rate of 1.42 m3/s. Calculate the resistance coefficient of leakage paths per
100 m. of duct. The friction factor for the duct is 0.0038 Ns2/m8.
4
8-**
(a) Explain the various forms of pressure and technique of pressure measurement in
underground mine ventilation system.
(b) Find the total pressure of the following condition.
Density of mine air
= 1.25 Kg/m3
Density of liquid
= 1000 Kg/m3
Difference in liquid level = 9 mm.
Velocity of air current
= 20 m/s
9-***A rectangular airway 1 km in length and 3m wide by 2m high is unsupported, unlined and has
irregular conditions.
Determine the resistance of the airway and the frictional pressure drop along it when 25 m3/s
of air is passing.
There is an isolated right angle bend in an identical airway (x = 1.4) and an obstruction
caused by mine cars (x = 2.5).
Calculate the equivalent length of each of these obstructions and the frictional pressure drop
along the airway when 30 m3/s of air is passing (assume standard air density of 1.2 kg/m3)
10- *** Find Air Flow Rate Q, in m3/s through a rock tunnel by using the following data
Friction factor, k = 0.01 Ns2/m4
Length of tunnel, L = 1 km
Power cost, e = 100 $ per watt-year
Driving cost, M = 10 $/m3
Efficiency of electricity, η = 80%
Capital repayment, z = 3%
Number of mine workers in the tunnel = 40
Friction factor of a mine worker, x = 0.14 Ns2/m4
Optimum diameter, d = 5.128 m.
d = [ 2064. k ( l + leq.) .Q3.e / M.η.l.z ]1/7
11- ***(a) Derive the pressure form and head form of the Bernoulli's equations from the first
principle.
(b) Discuss the application of these equations in mining industry.
.
(c) Find the total pressure of the following condition.
Density of mine air
= 1.25 Kg/m3
Density of liquid
= 1000 Kg/m3
Difference in liquid level = 9 mm.
Velocity of air current
= 20 m/s
5
CHAPTER 3- VENTILATION OPERATION COST
1-* (a) To reduce the air flow of air through a 3 m. × 3.5 m. airway to 20 m3/s. It is necessary to
install a regular with pressure drop of 500 N/m2. Calculate the regulator area for a box
regulator with C = 0.62.
2-* A 0.61 m. diameter duct 300 m. long has an intake volume rate of 1.70 m3/s and a delivery
volume rate of 1.42 m3/s. Calculate the resistance coefficient of leakage paths per 100 m. of
duct. The friction factor for the duct is 0.0038 Ns2/m8.
3-** If it is required to exhaust 7 m3/s from the end of a 400 m. long, 0.76 m. diameter duct,
determine what volume rate a fan at the other end will have to handle and what pressure it
will have to produce. The friction factor for the duct is 0.0038 Ns2/m8. The resistance
coefficient of leakage paths per 100 m of duct is 10,000 Ns2/m8.
4-*** What is the maximum length of 0.61 m diameter duct which will deliver 1.42 m3/s at its
inbye end if a fan with the characteristic given in Table A is connected at the other end? The
friction factor for the duct is 0.0038 Ns2/m8. The resistance coefficient of leakage paths per
100 m of duct is 10,000 Ns2/m8.
Table A Fan characteristics
Q, (m3/s)
p (Pa)
4
5
6
1100
800
500
6
CHAPTER 4- NETWORK ANALYSIS
1-** Four airways are arranged in parallel with a total quantity of 47.20 m3/s flowing through
them. Resistances of the airways are given in the table. Find the head loss for the parallel
airways and the quantity of air flowing through each.
AirwayResistance R (N.s2/m8)
1
2
3
4
Quantity (m3/s)
2.627
0.150
0.349
0.397
4.50
18.78
12.34
11.58
1
47.20 m3/s
2
3
4
2-** For the simple ventilation circuit shown in figure, the following values of resistance for the
individual airways have been determined with R in units of N.s2/m8 throughout:
Fan
10
1
5
4
2
6
3
7
9
8
R1 = 0.0559
R2 = 0.1342
R3 = 0.1118
R4 = 0.0838
R5 = 0.1399
R6 = 0.1453
R7 = 0.1062
R8 = 0.1677
R9 = 0.1509
R10 = 0.0447
Determine the equivalent resistance for the entire system and the mine static head, given that
the fan is exhausting air at the rate of 47.19 m3/s.
7
3-** Two airways are connected in parallel across a downcast and upcast shaft. The airways have
a resistance of 0.6 Ns2/m8 and 0.7 Ns2/m8.
(a) Calculate the equivalent resistance of these airways in parallel.
(b) If the downcast shaft has a resistance of 0.2 Ns2/m8 and the upcast shaft of 0.3 Ns2/m8,
Calculate the fan pressure required circulating 30 m3/s of air around the network.
(c) Calculate the airflow in each of the two parallel branches connected across the shaft.
Total Fan Pressure
R = 0.3
R = 0.2
A
R = 0.7
R = 0.6
B
(Unit of R = Ns2/m8)
4-*** The network ABCD has 5 branches, 4 junctions and no fans or natural ventilating
pressures, Use only two meshes, (ABD & ABCD) to perform two iterations using the Hardycross technique and hence calculate the resulting airflow in each branch.
3
B
50 m3/s
A
20 m3/s
0.9
30 m3/s
0.3
C
2
20 m3/s
0.6
1
30 m3/s
0.6
10 m3/s
1.2
50m3/s
D
8
5-***
Down Cast Shaft
Up Cast shaft
Exhaust Fan
A
B
E
G
C
D
F
The pressure drop and airflow of the different points are given below.
Point A Point B Point C Point D Point E Point F Point G -
P = 20
P = 10
P = 15
P = 10
P = 20
P = 30
P = 15
Pa,
Pa,
Pa,
Pa,
Pa,
Pa,
Pa,
Q=
Q=
Q=
Q=
Q=
Q=
Q=
30 m3/s
5 m3/s
10 m3/s
15 m3/s
10 m3/s
20 m3/s
30 m3/s
Find the total pressure developed by the exhausting fan at the top of the up-cast shaft.
9
CHAPTER 5- NATURAL VENTILATION
1-* Explain the use of natural ventilation pressure in various types of mine workings such as
simple underground mine, hill site opening, and tunnelling.
2-** Calculate the natural ventilation pressure by using the following data.
Depth
Rock
°
m
Temp. C
Pressure
D/C
Mb
Shaft
Density U/C
Density
Shaft
°
Temp.°C
Temp. C
Top shaft,0
18
810
10/10
0.99
22/22
0.95
Mid shaft,1000
27
920
16/16
1.10
27/27
1.05
Bottom,2000
Mean
36
27
1030
22/22
16/16
1.21
1.10
32/32
27/27
1.15
1.05
3-***Given the mine ventilation circuit, calculate the magnitude and direction of the natural
ventilation pressure. Omitted data must be assumed.
P1
H1
T1
Temperature of air, Tair
P5
Hair
P4
Temperature of U/G T2
P3
H2
P2
A
B
Diagram of an underground mine.
Where,
H1 = H2 + Hair
H1
= 914.4 m
H2
= 609.6 m
Hair
= 304.8 m
T1
= 32.2 °C
T2
= 21.1 °C
Tair
= 4.5 °C
Ra
= The gas constant of air = 287.1 J/Kg °K
G
= The gravitational constant = 9.81 m/s2
P1 - P4 = Absolute barometric pressure 100 KPa
10
CHAPTER 6- MINE FANS
1-*
What are fan characteristic curves? Illustrate your answer with diagrams.
2-* (i) State the "fan laws"
(ii) The operating conditions of a fan are as follows.
Fan speed
True density of air
Quantity flowing
Operating pressure
Power requirement
= 600 rpm
= 1.2 kg/m3
= 115 m3/s
= 1370 Pa
= 225 KW
If the fan speed is increased to 798 RPM and true density of air falls to 1.04 kg/m3. Calculate
the new operating conditions of the fan.
3-**
(a) Draw the diagram of the construction of a centrifugal fan and explain the various types of
centrifugal fans.
(b) Draw the diagram of the construction of an axial flow fan and explain its operating
mechanisms.
4-** (a) What are the fan characteristic curves and mine system curve? Illustrate your answer with
diagrams.
(b) Draw the fan characteristics curves and explain the important reasons of fans in series and
fans in parallel.
(c) Calculate the theoretical head for a back-ward curved centrifugal fan under the following
conditions:
Q = 11.8 m3/s, n = 300 rpm, D = 1.83 m, b = 0.61 m, θ = 50°, ρ = 1.2 kg/m3.
11
Min. 03018 Mine Environmental Engineering
Part “II”
Sample Questions and Answers
Question 1-* An airway of 800 m. with no bends 3m.× 2m.dimension. Determine the resistance of
airway and frictional pressure drop along it when an airflow of 30 m3/s is passing and K = 0.0158
Kg / m3.
Answer 1R
= KCL/A3
= 0.0158 ×10 × 800 / 6 × 6 × 6
= 0.585 Ns2/m8
P = RQ2
= 0.585 ×30 × 30
= 526.5 N/m2
Question 2-** (a) What is mine ventilation?
(b) Explain the purposes of mine ventilation.
(c) Write the short notes of Psychrometry..
Answer 2
(a) Mine ventilation is the process of conducting an adequate flow of pure fresh air along air- ways,
working places and service-points underground.
(b) Purposes of Mine Ventilation
1- To supply fresh air in necessary quantities for the respiration of underground mine
workers.
2- To dilute the concentration of the explosive, inflammable or toxic gases, fumes and radon
to environmentally safe levels in the underground works.
3- To dilute the concentration of the airborne dust to physiologically acceptable levels and to
remove from the mine.
4- To provide a thermally acceptable environment in which person can work without undue
discomfort or any danger of exhaustion from heat and to remove heat from the mine as may
be necessary.
12
(c) Psychrometry is the study of the air, oxygen, nitrogen and water vapour mixture under conditions
of varying moisture content, temperature and pressure. Water vapour is usually contained in normal
air about 2 to 3% by weight, but it can largely change the underground mine environment.
Pressure
Temperature
1 atm.
100 ºC
1 atm.
0 ºC
Vapour pressure
1 atm.
0.005 atm.
Wet bulb and dry bulb temperatures can give the correct water vapour content.
Question 3-** (a) Find the Enthalpy & Sigma heat of the following conditions.
Air mass, M
= 1.0 Kg.
Water mass, m = 0.007 Kg.
Temperatures
= 13.5 ºC wb. & 20.0 ºCdb.
(b) How much heat is required to raise the temperature of 0.1 m3 of bath water from
15 to 45ºC? (It should be noted that the density of water is 1000 kg/m3.)
Answer 3 (a)
1- Dry air,
1.0 Kg. × 20 × 1.005 = 20.10 KJ.
2- Liquid water,
0.007 × 13.5 × 4.187 = 0.40 KJ.
3- Evaporation,
0.007 × 2460 = 17.22 KJ.
4- Super heat of water, 0.007 × 6.5 × 1.884 = 0.09 KJ.
Enthalpy
= 20.1 + 0.40 + 17.22 + 0.09 = 37.81 KJ.
Sigma heat = 20.1 + 0
+ 17.22 + 0.09 = 37.41 KJ.
Answer 3 (b)
Heat required
= mass × temperature rise × thermal capacity
= (0.1× 1000) × (45-15) × 4.187
= 100 × 30 × 4.187
= 12561 kJ.
Question 4-**Calculate the natural ventilation pressure by using the following data.
Depth
Rock
Pressure
D/C
Density U/C
Density
13
Temp.°C
m
Mb
Shaft
Shaft
°
Temp.°C
Temp. C
Top shaft,0
18
810
10/10
0.99
22/22
0.95
Mid shaft,1000
27
920
16/16
1.10
27/27
1.05
Bottom,2000
36
1030
22/22
1.21
32/32
1.15
Mean
27
16/16
1.10
27/27
1.05
Answer 4•
0m
•
10/10
22/22
1000 m . • 16/16
27/27
22/22
32/32
2000m
.
•
Calculate NVP
Pressure at base of downcast shaft
= P 2000 + (2000 × 1.10 × 9.81)
= P 2000 + 21582
Pressure at base of upcast shaft
= P 2000 + (2000 × 1.05 × 9.81)
= P 2000 + 20601
∆p = 980 N/m2 from downcast to upcast
By rule of thumb method
(7.5 N/m2 / 5.6 °C / 30 m )
= 7.5 × 11/5.6 × 2000 / 30
= 982 N/m2
•
•
14
Question 5-***
(a) At what rate is heat removed from 3 m3/s of dry air at sea level density (1.2 kg/m3), which enters
a cooler at 35ºC and leaves it at 20ºC?
(b) If 300 000 tons of rock are broken per 30-day month in a mine at a depth where the average
virgin rock temperature is 40ºC and by the time this rock reaches surface in a skip it has been cooled
to 20ºC, what is the average rate at which heat is released by the broken rock?
If this mine is ventilated by means of 700 m3/s of air at a density of 1.2 kg/m3, by how much could
this heat increase the temperature of the air?
(Thermal capacity of rock = 0.837 KJ/Kg ºC)
Answer 5(a) :
Heat removed
= (3 × 1.2) × (35-20) ×1.005
= 3.6× 15 × 1.005
= 54.3 kilo joules per second (kJ/s)
= 54.3 kW
Answer 5(b) :
Heat removed by rock
= 300 000 × 1 000 × (40-20) × 0.837 kJ/month
= 300 000 × 1 000 × 20 × 0.837 / 30 × 24 × 60 × 60 kJ/s
=1938 kJ/s
=1938 kW
Heating 700 m3/s of dry air by 1ºC requires
700 × 1.2 × 1 ×1.005
= 844 kJ/s
= 844 kW
The air temperature would thus increase by 1938 / 844 × 1 = 2.3ºC.
***** END *****