Locomotive Haulage in Mines
The locomotives are very flexible in nature & they can be used for men transport also. The rail transport finds its main application as gathering and haulage in the underground mines and tunnels. The rope haulage system works on track, and so is the locomotive haulage.
The locomotive haulage is best suited as a long distance haulage with a gradient in the range of 1 in 200 to 300.
However, gradient up to 1 in 30 for a short distance can also be negotiated by this system. The system is flexible, comparing to rope and belt conveyor systems.
Tractive effort
It is the total force delivered by the motive power of the locomotive, through the gearing, at wheel treads. When this force is greater than the product of locomotive weight and the coefficient of adhesion between the wheels and rails, the wheels will slip, i.e. it will roll. This can be numerically expressed as:
Total or maximum tractive effort
TE = Wʟ C
Whereas: C is the coefficient of adhesion whose value depends upon the condition of track, and whether it is sanded or not. Wʟ is the weight of locomotive.
Drawbar pull
This is the force exerted on the coupled load by a locomotive through its drawbar, or coupling, and is the sum of the tractive resistance of the coupled load. The drawbar pull that a locomotive is capable of developing is determined by subtracting the tractive effort, from the sum of the tractive resistance of the locomotive. This resistance is offered by the several sources: rolling resistance, which the entire train offers, is equal to weight of the train in tons. (i.e. weight of locomotive + weight of mine cars with pay load) multiplied by a frictional coefficient m which could be 10-15 kg/ton.; Curve resistance which can be ignored gradient resistance and the force required to provide acceleration to the motion (as given in the formulae specified below).
Drawbar pull
= R₀Wᴛ
Wᴛ is weight of trailing load, i.e. weight of train in tons. T.
R₀ is running resistance/t in kg.
Example
An 8 t locomotive has a coefficient of adhesion 0.18 and a coefficient of rolling resistance 100 N/tonne mass. The maximum tractive force available to haul the trailing load in kN along a straight track is ____
Solution
W = mg = (8 x 1000) x 9.8 N = 78400 N
Total roller resistance = 8 x 0.18 = 800 N
Draw bar pull (DBP) = Tractive effort - pull required to move the loco
= μW - pull = 0.18 x 78400 – 800
= 13312 N = 13.3 kN
Example
A 10,000 kg locomotive runs at a speed of 3 m/s along a level track. The coefficient of adhesion at the track is 0.23, and the rolling resistance of the locomotive is 90 N per 1000 kg. The maximum power available for hauling the train in kW is ____
Solution
W = mg = 10,000 x 9.8 = 98000 N
V = 3 m/s
μ = 0.23
Rolling resistance = 90 N/1000 kg
Hence, Rolling resistance for 10,000 kg locomotive = 90 x 10 = 900 N
Tractive effort (TE) = μW
= 0.23 x 98000 = 22540 N
Drawbar pull (DBP) = TE - pull (or rolling resistance)
= 22540 – 900 = 21640 N
Now, Power = DBP x V
= 21640 x 3 = 64920 W ≈ 65 kW
In case of rope haulage, the power to move the load is available from external fixed motor to the haulage. While in case of locomotive haulage, the driving unit i.e. locomotive is coupled to a train due to which more safety can be attained.
The locomotives are very flexible in nature & they can be used for men transport also. The rail transport finds its main application as gathering and haulage in the underground mines and tunnels. The rope haulage system works on track, and so is the locomotive haulage.
The locomotive haulage is best suited as a long distance haulage with a gradient in the range of 1 in 200 to 300.
However, gradient up to 1 in 30 for a short distance can also be negotiated by this system. The system is flexible, comparing to rope and belt conveyor systems.
Tractive effort
It is the total force delivered by the motive power of the locomotive, through the gearing, at wheel treads. When this force is greater than the product of locomotive weight and the coefficient of adhesion between the wheels and rails, the wheels will slip, i.e. it will roll. This can be numerically expressed as:
Total or maximum tractive effort
TE = Wʟ C
Whereas: C is the coefficient of adhesion whose value depends upon the condition of track, and whether it is sanded or not. Wʟ is the weight of locomotive.
Drawbar pull
This is the force exerted on the coupled load by a locomotive through its drawbar, or coupling, and is the sum of the tractive resistance of the coupled load. The drawbar pull that a locomotive is capable of developing is determined by subtracting the tractive effort, from the sum of the tractive resistance of the locomotive. This resistance is offered by the several sources: rolling resistance, which the entire train offers, is equal to weight of the train in tons. (i.e. weight of locomotive + weight of mine cars with pay load) multiplied by a frictional coefficient m which could be 10-15 kg/ton.; Curve resistance which can be ignored gradient resistance and the force required to provide acceleration to the motion (as given in the formulae specified below).
Drawbar pull
= R₀Wᴛ
Wᴛ is weight of trailing load, i.e. weight of train in tons. T.
R₀ is running resistance/t in kg.
Example
An 8 t locomotive has a coefficient of adhesion 0.18 and a coefficient of rolling resistance 100 N/tonne mass. The maximum tractive force available to haul the trailing load in kN along a straight track is ____
Solution
W = mg = (8 x 1000) x 9.8 N = 78400 N
Total roller resistance = 8 x 0.18 = 800 N
Draw bar pull (DBP) = Tractive effort - pull required to move the loco
= μW - pull = 0.18 x 78400 – 800
= 13312 N = 13.3 kN
Example
A 10,000 kg locomotive runs at a speed of 3 m/s along a level track. The coefficient of adhesion at the track is 0.23, and the rolling resistance of the locomotive is 90 N per 1000 kg. The maximum power available for hauling the train in kW is ____
Solution
W = mg = 10,000 x 9.8 = 98000 N
V = 3 m/s
μ = 0.23
Rolling resistance = 90 N/1000 kg
Hence, Rolling resistance for 10,000 kg locomotive = 90 x 10 = 900 N
Tractive effort (TE) = μW
= 0.23 x 98000 = 22540 N
Drawbar pull (DBP) = TE - pull (or rolling resistance)
= 22540 – 900 = 21640 N
Now, Power = DBP x V
= 21640 x 3 = 64920 W ≈ 65 kW
Example
A locomotive is negotiating a level curved path with radius of curvature 30 m. The gauge of the rails is 0.75 m. If the maximum elevation of the track is 1/8 of the track gauge, the maximum turning speed in m/s of the locomotive is ____.
Solution
Super elevation h = AV2/gR
Given h = (1/8)A
g = 9.8 m/s2
R = 30 m
Putting values
(1/8)A = AV2/(9.8 x 30)
∴ V = 6.06 m/s
Example
A 12 tonne diesel locomotive of 60 kW is plying in an underground haulage roadway. The coefficient of adhesion is 0.25 and the maximum gear efficiency is 80 %. The speed in m/s at which it will haul a train at its full power is _____.
Solution
P = 60 kW = 60,000 W
η = 0.8
W = mg = 12000 x 9.8 N = 117,600 N
μ = 0.25
Tractive effort
TE = μW = 0.25 x 117,600 = 29400 N
Draw bar pull
DBP = TE - rolling resistance
As rolling resistance is not given, consider it as zero.
∴ DBP = TE = 29400 N
Using P = DBP x V
(60,000/0.8) = 29400V
∴ V = 2.55 m/s
Example
A 24 t locomotive pulls 300 t of trailing load from a level siding. If antifriction bearings having specific rolling resistance of 100 N/t is fitted on all equipment and the coefficient of adhesion is 0.3, the maximum possible acceleration of the train in m/s2 will be ____.
Solution
W = mg = (24 x 1000) x 9.8 N
TE = μW = (0.3 x 24 x 1000 x 9.8) N
DBP = TE - rolling resistance
= (0.3 x 24000 x 9.8) - 100 x 24
= 68160 N
Mass of trailing load = 300 x 1000 kg
Now Force (i.e. DBP) = ma
68160 = (300 x 1000) x a
∴ a = 0.22 m/s2
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