mine ventilation
TRANSCRIPT
MINE VENTILATION
Purpose of ventilation is
to supply fresh air to workers
to dilute or disperse harmful gases to acceptable level
to cool the deeper mines
Mine ventilation is mainly related to the quantity control of air, its movement and its
distribution.
Conditioning functions and processes commonly used in mines consist of the following:
1) Quality Control
a) Gas Control
b) Dust Control
2) Quantity Control
a)Ventilation
b)Auxilary or face ventilation
c)Local exhaust
3) Temperature-Humidity Control
a) Cooling
b)Heating
c)Dehumidifaction
Processes may be applied individually or jointly. We concentrate here an mine
ventilation as the most important and universally used process. The engineering principles of
mine air quality control are
1) Prevention
2) Removal
3) Suppression
4) Cantoinmat?
5) Dilution
PROPERTIES AND CHARACTERISTICS OF MINE GASES
Hydrogen:
It rarely occurs in mine gases. Its characteristics are:
Chemical symbol…………………………..H2
Specific gravity……………………………0.070
Sources…………………………………….Charging of batteries, fires and explotions.
Action of strong acid water on iron.
Effect on life……………………………….has no effect when breathed
Explosive range…………………………….from 4.1% to 74%
Colorless, odorless, tasteless
Methane:
Methane is the most explosive gas in mines. The properties of methane are:
Chemical symbol…………………………..CH4
Specific gravity…………………………….0.555
Source……………………………………...from coal and adjoining strata.
Effect on life……………………………….nonpoisonous
Explosive range…………………………….5% to 15%
Colorless, odorless, tasteless
Methane is lighter than air and will accumulate in the face of advance workings, in
pockets in the roof and other places where there is not enough air in circulation.
A proper mixture of methane and air is explosive and can be ignited when there is 5%
of methane present in the air. The maximum explosiveness of a methane and air mixture is
reached when the methane present is about 10% of volume. If there is more than 15% of
methane present, the mixture is not explosive due to insufficiency of oxygen.
MAC for CH4: 1% stop machines if CH4: 1.5% leave the mine if it is <2%
Nitrogen:
This is the most abundant gas in air. Its properties are:
Chemical symbol……………………………N2
Specific gravity……………………………...0.967
Source……………………………………….Normal constituent of air
Effect on life………………………………...will not support life
Explosive range……………………………..non explosive
Carbon Monoxide:
It is the most common poisonous gas encountered in mines and is known as wnite
dump.
Chemical symbol…………………………CO
Specific gravity…………………………...0.967
Source…………………………………….Mine fires, explosions, blasting.
Effect of life………………………………very poisonous
Explosive range…………………………..12.5% to 74%
CO is the product of incomplete combustion. Considerable carbon monoxide is found in the
air near mine fires and explosions. It is also produced during blasting operations using
explosives.
Carbon Dioxide:
Chemical symbol………………………….CO2
Specific gravity……………………………1.529
Source……………………………………..normal constituent of air
Effect on life………………………………slightly poisonous in large quantity
Explosive range…………………………...non explosive
Colorless, odorless, slight acid taste in quantities.
Its presence in mine air is increased by breathing of men, decaying of timbers. It is a product
of complete combustion.
Determination of Dilution Requirements
The quantity of fresh air needed to dilute harmful gas is determined by the following
relationship:
Where Qg: harmful gas inflow rate (m3/sec)
MAC: maximum allowable concentration of gas
B: concentration of gas in normal in take air (%)
Example:
Calculate the air quality necessary to dilute methane gas in a coal mine to its MAC when the
inflow rate is (0.118 m3/sec) and its concentration in the in take air 0.1%.
According to mine safety regulation, MAC for methane in a coal mine is 1%. Therefore
QUANTITY CONTROL
Quantity control in mine ventilation is concerned with supplying air of the desired
quantity and in the desired amount to all working places throughout the mine. Air is necessary
not only for breathing but to disperse chemical and physical contaminants (gases, dusts, heat
and humidity).
Breathing requirements are easily met [0.01 m3/sec per person usually is enough].
However, higher minimum quantity of air per person should be 1.4 m3/sec at face, 4.3 m3/sec
in the lost crosscut of a coal mine or a minimum velocity at the coal face should be 0.3 m/sec.
The basis for specifying the amount of air flow in working places is the velocity or
quantity needed to disperse or dilute contaminants. If the quantity needed for dilution is not
enough for effective dispersion, then the velocity or quantity is increased to a certain level.
There is a proportional relationship between the quantity and the velocity as follows:
Q=V*A
Where Q: quantity (m3/sec)
V: velocity (m/sec)
A: cross sectional area of the opening (m2).
Example:
Calculate the quantity of air needed to dilute 0.038 m3/sec of carbon dioxide in the
stops of a metal mine and to maintain a critical face velocity of 0.76 m/sec. The cross-
sectional area of each stop is 4.5*60 m and the CO2 content of the in take air is 0.03%. The
MAC for CO2 is 0.5%.
for needed dilution.
The quantity for effective dispersion
Q=V*A=0.76*4.5*6.0=21.24 m3/sec
Specify the critical quantity as the larger value
Q=21.24 m3/sec
Once quantity requirements in all the working places have been specified, it is then necessary
to create a pressure difference in the mine to provide desirable air flow. For this purpose,
natural ventilation or mechanical ventilation is used. Natural ventilation is based on the
pressure difference resulting from thermal energy. Mechanical ventilation uses rotational
energy of a fan as energy resource. In modern mines only fans are used to provide a pressure
difference for ventilation. Natural ventilation is too variable, unreliable and insufficient to be
used.
Fans can be divided into two groups: centrifugal fans and axial fans. They are installed
in either the blower, booster or exhaust position. (Underground booster fans are strictly
prohibited in coal mines).
PRESSURE LOSSES
Pressure which forces air to move, continuously drops down while air travels through
airways. Total pressure loss in a mine consists of sum of static pressure loss, Hs and velocity
pressure loss Hv
Mine HT=mine Hs+Mine Hv
Static pressure loss is due to frictional loss (Hf) and shock losses (Hx). Frictional loss is
mainly due to the friction between the surface of airway and air.
Frictional loss depends on length of airway, perimeter, cross-sectional area and friction
factor. Frictional loss can be calculated by the following equation
where Hf: pressure loss in Pa (mmwater)
L: length of the airway, m.
C: perimeter of the airway, m.
A: cross-sectional area of the airway, m2
K: friction factor, Ns2/m4
This equation is also known as Atkinson equation. Since V=Q/A, the Atkinson equation can
be written as follows
where is the resistance of airway and
The Atkinson equation is reduced to the following simple form
H=R.Q2
Shock losses:
Shock losses mainly occur due to changes in direction and changes in cross-sectional area.
Shock losses can be calculated by directly
Hx=x.Hv
where x: shock loss factor
Hv: velocity pressure
Or by equivalent length method (le):
The latter one is the most frequently used one and in this method, shock loss is expressed in
terms of equivalent length of a straight airway. For this purpose the following table can be
used:
Source of shock loss equivalent length, m
Entrance 1
Discharge 22
Construction gradual 0.4
Construction abrupt 3
Band, acute, sharp 50
Velocity pressure is given by
where γ : density of air
V: velocity of air
Velocity pressure is not considered in mine ventilation calculations only static pressure is
used. Therefore, total pressure loss, HT is equal
Example:
Given a rectangular mine opening of cross-sectional dimensions 1.8*6, length 2250 m, and
friction factor 0.00835, calculate the pressure difference for an airflow of 66.1m3/sec.
C=2[1.8+6]=15.6m
A=1.8*6=10.8m2
MINE VENTILATION CIRCUITS
Mine ventilation circuits arranged in series or parallel or as combination series-parallel
circuits. The combination circuits are called networks.
In series circuits, Q is constant
H and R are cumulative.
Total resistance
Total pressure loss (head)
Quantity
In parallel circuits, Q is cumulative
H is constant
R is inversely proportional to
A B
C D
E F
(the equivalent circuit resistance)
Example:
Four airways are arranged as a series circuit. Their resistances are 2.627, 0.151, 0.349
and 0.397. If the quantity is 47.2 m3/sec, find the pressure difference across the circuit.
= (2.627+0.151+0.349+0.397)*47.22=7851 Pa.
= 7.85 kPa.
Example:
If the airways of the previous example are rearranged as s parallel circuit, calculate the
pressure difference across the circuit and the quantity of air flow in airway 1.
Example:
Given the following data and figure, calculate the total pressure loss and total
resistance if the quantity through the fan is 18.88 m3/s.
Airway Width, m Height, m Length, m Friction Factor
AB (Shaft) 3.66 diameter 182.88 0.0046
AB C
1
2
D
BC 6.10 1.83 609.60 0.0093
CD 6.10 1.83 914.40 0.0093
DE (Shaft) 3.66 diameter 182.88 0.0046
Shock loss condition Equivalent length, m
Entrance (A) 0.914
Gradual Expansion (B) 0.305
Obtuse bend, sharp (C) 4.572
Sharp bend, right (D) 21.346
Discharge (Fan) 0
E
D C
B
+--------------
Σ HT=64.05Pa
mine resistance.
NATURAL SPLITTING
In natural splitting, the air is allowed to flow freely in the branches of the circuit. The
air flow always tends to take the easiest way, e.g. the airway with the smallest resistance. In
natural splitting, the direction of flow and magnitude of H and Q for each airway are not
known. They must be calculated algebric or graphical solutions.
Example:
For the following network, find the quantity of air flowing in each airway. In addition,
calculate mine HT if the mine Q is 15 m3/s.
Mine HS=RT*Q2=0.219*(15)2 = 49.3 Pa.
MINE CHARACTERISTIC CURVE
R=0.052
R=0.871
R=0.0267
R=0.0178 R=0.494 R=0.0168
A
B
C D
E
F
The H*Q relationship for an airway is termed its characteristic; for the entire mine, it
is called the mine characteristic. A graphical plot of pressure difference vs. quantity is
referred as the mine characteristic curve.
Example:
Calculate and plot the characteristic curve of the mine with resistance R=0.0239.
Using the H=R*Q2 equation, the following table can be constructed:
H=0.0239Q2
Q1 (m 3 /sec) H (Pa)
0 0
20 9.56
40 38.24
60 86.04
80 152.96
100 239
For performance may also be expressed graphically. A fan characteristic curve plots its H-Q
relationship.
Example:
H
Q
200
150
100
50
Mine characteristic curve
20 40 60 80 100
Points on the characteristic curve of a centrifugal fan are as follows.
Q1 (m 3 /sec) H (Pa)
0 149
20 164
40 157
60 137
80 100
100 52
Characteristic curves are used mainly to determine the system operating point when a
fan is connected to a mine. The operating point is the combination at which fan and the mine
are in equilibrium. When the fan and mine characteristic curves are plotted on the same graph,
the operating points lies at their connection. Operating point is a useful characteristic to
employ for a selection of a fun.
Example:
Determine the operating point graphically using the data of two previous examples.
H
Fan characteristic curve
Q
20 40 60 80 100
200
100
In selecting a fan for a given mine, it may be the case that no fan curve intersects the mine
curve at the required pressure difference and quantity. The fan’s performance may be
modified by changing its steep.
CONTROLLED SPLITTING
Sometimes, it may be necessary to deliver air in desired amount and in desired
quantity to the working places connected in parallel. This requires to control the direction of
flow and quantity. There are two ways of controlling them:
i) positive regulation
ii) negative regulation
In the first method, the resistance of an airway which requires greater airflow is
reduced. The second method uses regulators (scale doors). In mine ventilation practice, the
second method is the worst used technique. In this method, the airway with the highest
pressure loss is known as the free split and no resistance is added to this branch. The flow is
controlled by regulators which are placed in other branches with lower pressure drop. The
regulators create shock losses and due to shock loss, the ventilation cost increases.
Example:
Given the three airways with resistance R1=5.585, R2=2.793 and R3=8.378 arranged in
parallel, with the total quantity Q=47.195m3/s, find system H and R and the quantities Q1,Q2,
and Q3.
200
150
100
50
H
Q
Operating point [70,100]
20 40 60 80 100
Example:
Suppose for airways in the previous example that the required air distribution is
Q1=11.80m3/s, Q2=16.52m3/s and Q3=18.88 m3/s calculate the pressure loss and determine
amount of regulation in each airway.
R1,Q1
R2,Q2
R3,Q3
Q Q
Pressure losses in the airways
Pressure loss in the airway 3 is the highest. Therefore, this airway is taken as free split.
Regulators are needed for airway 1 and airway 2 to achieve desired airflows with Q1=11.80
and Q2=16.52
Shock losses in airway 1 and airway 2
Airway 1: 2986.08-778.87=2207.21 Pa
Airway 2: 2986.08-761.45=2224.63Pa
Airway 3: no shock loss
CALCULATION OF THE SIZE OF REGULATORS
A number of formulas are available to calculate the size of regulators. The equivalent
orifice formula is the simple but accurate one and is given as follows
where A: the equivalent orifice, m2
Q: quantity, m3/s
Hx: shock loss created by regulator, Pa
Q1=11.80
Q2=16.82
Q3=18.8
Example:
NATURAL VENTILATION
Natural ventilation is the term used to describe airflow resulting from a pressure
difference caused by natural means. This pressure difference is due to difference between air
densities in intake and return shafts. Source of unequal densities is difference in elevation and
temperature in the intake and the return shafts.
Only temperature difference is not enough to produce N.V.
No natural ventilation, because air densities in two shafts are equal.
Temperature and elevation difference can produce N.V.
tm tm
tS
tS
tS:surface temperature
tm:temperature in mine
tS>tm : in summer
tS<tm : in winter
In summer, temperature of air column in the shaft AB is lower then the temperature of air
column in the shaft EC because:
and
The colder column of air will push up the warmer column of air and the air will flow from AB
to CE
, therefore
Cold air
tm
tm
tsA
BC
D
Warm air
In summer tm<tS
In winter, the situation is a bit different. This time, temperature of air column in the
shaft AB is greater than that of air column in the shaft EC.
In winter, the surface temperature (ts) is lower than the temperature inside the mine (tm)
ts<tm
Therefore the air column of EC is heavier than the air column of AB and the airflow will be
from EC to BA
flow from AC to BA.
NATURAL VENTİLATION PRESSURE (NVP)
Several approaches have been developed to calculate nvp. The first formula is given as
follows:
1)
Where Hn: natural ventilation pressure, Pa.
B. the average absolute pressure in Pa.
L: the vertical height of the air columns in m
T1 and T2: the average absolute temperatures of columns in 0C
Warmair
tm
tm
ts
A
BC
D
ts
In winter tm>tS
2)
3)
where g: gravitational acceleration (=9.8)
average density of the air column in the in take shaft (kg/m3).
average density of the air column in the return shaft.
L : length of the air column, m.
NPV is independent of the quantity flowing. It depends only on the temperature difference
and elevation difference. NPV can be seen as a fan with characteristic parallel to the quantity
axis.
Therefore, the amount of air flowing through a mine due to NPV is a function of the mine
characteristic only.
Example: The results of a temperature survey are shown in the following figure find NPV, Q
and direction of flow when RT=2.91
H
Q
Mine characteristic curve
NPV
Q1
Point Barometric Pressure(kPa) t( 0 C) γ(kg/m 3 )
E 84.67 37.8 0.925
A 87.84 35 0.960
B 94.53 29.4 1.069
C 94.53 23.9 1.093
D 87.67 29.4 0.954
Average Barometric Pressure:
Average Temperatures :
304.8m
609.6m
E
35
29.4
37.8 29.4
23.9
914.4m
D 1520m
B 8000m C
Average Densities :
Average density of air in two columns :
Take the highest value as NPV
NPV=349 Pa
Direction of air flow is from C to B
MINE TRANSPORTATION AND HAULAGE
The principal aim of an underground transportation system is to move ore from the
face to the outside the mine. In addition to the movement of ore, a transportation system must
also move supply materials inside; it must move men in and out of the mine; it must move
rock and other debris out of the mine for disposal. These operations can be summarized as
follows:
Ore and rock
Inside the mine supply materials Outside the mine
Men
Mine haulage is divided into three categories based on the various areas of the mine: Face
haulage, intermediate or secondary haulage and main haulage:
Mine Haulage
H
Q
Mine characteristic
NPV
Operating point [349,11]
5 10 15 20
525
350
175
Face haulageUsed at face from the mining machine to the intermediate transport system
Intermediate or secondary HaulageUsed between face haulage and main haulage
Main HaulageFrom secondary haulage to the outside the mine
We will basically focus on the main haulage system which mainly uses rail haulage.
A transportation system can also be divided into two categories:
Continuous transportation and discrete system:
Rail haulage and rope haulage are examples of discrete system. The continuous system uses
chutes and conveyors.
The lecture notes will mainly concentrate on the rail haulage but give a brief summary on
Chutes and conveyor haulage.
A Chute: the loading arrangement that uses gravity flow in moving broken rock from a higher
level to a lower level.
Ordinary Chute
The inclination angle at which coal starts to slip on various types of ground is given as
follows:
Ground on which coal slips Angle
Footwall 350-380
Wool 300-350
Sheet-iron 170-250
The chutes are means of transportation used in face haulage. They are used in transportation
of ramble material and coal in nearly dip seams.
Spiral Chutes
They are chutes that allow a vertical transportation of material and coal. They control
the slapping velocity of material and prevent them from being breaking up small pieces.
Friction coefficient between material and slipping surface when material is inactive
(motionless) is greater than that of when it is in motion.
For this reason, material gets faster and faster after motion. On the other hand, the
speed of material reduces due to centrifugal forces and reaches a balance. The speed of
material depends on inclination of the spiral. For example, when the inclination angle is 220
The speed is around 1.5 m/sn.
Operational Principles
The following figure shows a simple spiral.
Material at location A follows various routes of slipping; inner route, middle route and
outer route (The following figure)
d
h
D
h
α1α2 α3
d A1 A2 A3
Dm D
A
The speed of the material is the highest at plane AA1 and lowest AA3.
Suppose that material at location A slips at the middle of the spiral chut on the plane AA2. In
addition, let m be mass of material and μ, friction coeeficient. Forces at an inclined plane are
as follows:
Friction force against motion,
Force for motion,
Force in motion,
Acceleration rate :
Using the above relationship, we obtain
and speed of the material, V
where l is length over which material travels.
and finally
m.g
m.g.sinα2m.g.cosα2
R
motion
α2
given α2=250, m=1 kg and μ=0.4, h=3000mm
Haulage capacity of the spiral chuts, Q
where F; cross-sectional area of haulage
ρ; specific density of material, ton/m3
V; speed, m/s
Usage of the spiral chuts
The height of the spiral chut is 250 m ore more. Material size to be hauled should be 8-
10 cm. The spirals are made of sheet iron which is covered by material resistant to abrasion.
Haulage Chimneys
(Ore Posses, Raises and Winzes)
Haulage Chimneys are vertical or nearly vertical mine openings. They are drived into
rock or ore, especially in metal mining. The material (ore or rock) are haulade from an upper
level to a lower level by means of gravitational flow.
Cross-sectional area of haulage chimneys depends on two parameters: haulage capacity in
volume and size of material to be hauled. The capacity of haulage chimneys are given as
follows:
where Q : haulage capacity (ton/mm)
A : cross-sectional area of opening, m2
V : average velocity of material, m/s
Ρ : specific gravity of material, t/m3
A gate is used to control the flow. The gate can move upwards or downward.
Blocking can occur because of greater size of material. Blocking is an undesirable
situation because it prevents broken material to move down. One way of handling with
blocking is to divide the haulage way into two sections: manway and oreposs.Man way is
used for transporting man and oreposs is used for transporting broken are and other material.
The plates:
Another way of controlling flowing speed of material is to use plates. They are located at
interval of 1.5-2.5 m. In fact there is a relationship between dropping velocity and distance as
follows: V=1.35*a
gate
Main haulage drive
a
Mine car
Forces Needed to Move the Rolling Stock
To move a trip, a mine locomotive must overcome the following resistances for both
itself and the mine cars: (1) rolling resistance, (2) acceleration resistance, (3) grade resistance
and (4) curve resistance. The effort that mine locomotive makes to move the trip is called
tractive-effort and it is a function of the weight of the locomotive and its ability to adhere to
the track.
The following figure shows the relationship between mine locomotive and its trailing loads.
Forces Needed to Move a Mine Car
To move a mine car, the following resistances must be overcome:
1) Rolling resistance: This is equal to weight, in tons, of the mine car (including payload, if
any) multiplied by a frictional coefficient, in kg per ton.
where FR: rolling resistance (kg)
T: frictional coefficient (kg/ton)
Wc: weight of the mine car (including payload) (ton).
Forces in effect on the wheel of the mine car are shown as follows:
Locomotive
Tracticeeffort
Train resistance(Trailing Loads)
= { Grade resistanceCurve resistanceRolling resistanceAcceleration resistance
It is possible to express T in terms of the above parameters as follows
Where μ is frictional coefficient between the wheel and wheelbase (axle)
The parameter h characterizes quality of touch between wheels and rail. When the rails are
dirty and dusty, the value of h is high, otherwise low. Ideally h takes values between 2 and 15
r/R is most often equal 0.1
μ takes value between 0.2 and 0.01 depending on the quality of the mine cars.
From the values of the parameters, it can easily be seen that T takes values between 2
and 15.
2) Acceleration resistance, FA
According to Newton law.
where F : force to accelerate a body (kg)
m : mass of the body (ton)
a : acceleration rate (cm/sn2)
RFR
Wc
h
3) Grade resistance
1000.sinα is grade I expressed in mm/m. for example, on a 1%grade, a ton must be
raised 1 m for every 100m that the mine car advances, making the grade resistance 10 kg per
ton for each 1% of grade. For example, the pull (force) required to overcome a grade 3% for a
10-ton car is 300 kg. (10 kg per ton for each 1% of grade x 3% grade x 10 tons)
where Fg : grade resistance (kg)
Wc : weight of the mine car (ton)
i : grade (mm/m)
it grade is in favor of loads, then the sign of I is negative, that is,
Fg=Wc*(-i)
4) Curve resistance
A railway may not follow a straight line. For example, it may be necessary to curve the
railway as follows:
Fg
Wc
α
If grade is against loads
Fg=Wc.sinα
If the unit of Fg is kg and the
unit of Wc is toni then
Fg=Wc.1000.sinα
Wc Fg
In such cases, an additional resistance arises. Let us study the forces under curved railway:
The centrifugal force is the most effective one for curved railway. This force is expressed by
the following formulation.
where F: centrifugal force (kg)
m: mass of the mine car (ton)
V: speed of the car (m/s)
R: radius of curvature.
G F
h
Gage, e
BA
When the mass is expressed as weight and unit of spreed is km/h
Centrifugal force leads to deformation of the railway and overturn of mine car. When
there is no reason for over turn of mine car. Let us study this condition and take tangent of
these angles:
and when AB=e/2 and AG=h
This condition determines the speed of mine car without overturning.
For example, when R=15m, e=0.6m and h=1m.
If the speed of the mine car exceed this limit, overturn occurs thread of overturn can be
eliminated by means of super elevation shown in the following figure:
D: super elevation
E: gage of the track
using sinα
By super elevation, centrifugal force is reduces or even eliminated
Wc
d
e
using
From the above equation:
Super elevation obtained in this way is called theoretical super elevation and yields higher
values. For example, when e=0.6m, R=15m and V=25km/h
dt=20cm.
This value is too great for elevation. Instead of using the above equation, the following one is
used:
From this, we obtain
Curve resistance can be calculated as follows:
C is a function of the radius of curvature and gage of the track, wheel base and diameterand is
calculated using the following equation.
where L=the wheel base
R=radius of curvature
E=gage of the track.
L
e
When we sum all these resistances up, we obtain train resistance for a single mine car.
where FT is the train resistance. For the single mine car. If the number of mine car is n, then
the train resistance or the trailing load is
Forces Needed to Move Locomotive
Similar to movement of a mine car, the forces needed to move the locomotive itself is given
as follows:
where
FL: force needed to move the locomotive (kg)
WL: weight of the locomotive (ton)
TL: frictional coefficient (kg/ton)
a: acceleration rate (cm/s2)
i: grade (mm/m)
c: curvature constant
Tractive Effort (TE)
Tractive effort is the force exerted by the locomotive to overcome all the resistances
[resistance of the locomotive + trailing loads]. It is calculated as follows:
Where TE:tractive effort
WL: weight of the locomotive (ton)
Ώ: adhesive value.
Adhesive values for locomotives
Unsanded rails Sanded rails
Clear dry rails, starting
And accelerating………. 0.30 0.40
Clear dry rails, continuous
Running…………… 0.25 0.35
Clear dry rails, locomotive
Breaking………… 0.20 0.30
Wet rails…………… 0.15 0.25
To move a trip, the following condition must be satisfied
Breaking Conditions
Problem:
Determine the size of locomotive required to pull a 300-ton train load from a level at a
maximum acceleration of 20 cm/s2 if the friction coefficient with 15 kg/ton is employed on all
equipment. Use an adhesive factor of 0.3
i=0, n*Wc=300 ton
C=0
Problem
Using the same size of locomotive and trip calculated in the previous problem, determine the
maximum grade against which this locomotive can pull the train without accelerating.
i=18 mm/m
i=1.14%