Mining Machinery - Questions from DGMS, GATE and PSU Exams

135 ton of coal is to be hauled by direct rope haulage in a shift of 6 hours hauling time where the length of haul is 1100 m. The average speed of rope is 6 km/hr. Set changing time is 1 min & tub capacity is 1.5 ton. Calculate the number of tubs per set?

(a) 6*

(b) 9

(c) 12

(d) 15

Solution: Explanation: In one cycle tub will cover = 1100+1100 - 2200 m = 2.2 km

Time taken in one cycle - 2.2/6 = 0.37 hr = 22 min

Total time taken in one cycle including set changing time = 22 + 1 + 1 = 24 min

In 6 hours i.e. 6 x 60 min number of cycles = 360/24 = 15

Let there be N number of tubes

So15 x N x 1.5=135

∴ N = 6 tubs

For a direct rope haulage installation in a mine drift if the total rope tension is 115 kN, the moment of inertia of the 3 m diameter drum is 9000 kg/m², and the speed of. the rope is 12 m/s, the total torque of the drum in kNm is

(a) 172.5*

(b) 175.2

(c) 176.1

(d) 179.1

Solution: Torque = F x R

= 115 x (3/2) = 172.5 kNm

The centre lines of winding drum and winding pulley are parallel and 0.6 m apart. When one cage is at pit bottom, its winding rope is positioned at 1.8 m from the drum centre. The plan distance between the winding drum and the winding pulley axes is 60 m. The fleet angle of installation is ____

(a) 1.15°*

(b) 1.35°

(c) 1.42°

(d) 1.82°

Solution: See figure.

Distance of wire rope from drum centre = 1.8 - 0.6 = 1.2 m

Distance A = 60 m

Fleet angle θ = tan^-1(1.2/60) = 1.15⁰

A cord hung over a flat pulley supports a weight W₁ at one end and W₂ at the other end If the coefficient of friction is 0.3, find the greatest ratio of W₁ to W₂ possible without the cord slipping. Angle of contact is 180°.

(a) 2.56*

(b) 2.85

(c) 2.36

(d) 3.25

Solution: T₁/T₂ = e^μ(^θ/57.3)

= 2.7^{0.3(180/57.3)}

= 2.56

An 8 t locomotive has a coefficient of adhesion 0.18 and a coefficient of rolling resistance 100 N/te mass. The maximum tractive force available to haul the trailing load in kN along a straight track is

(a) 0.64

(b) 13.33*

(c) 14.13

(d) 14.93

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

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

(a) 40.7

(b) 65.0*

(c) 67.1

(d) 67.7

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

An AFC of length 114 m runs at a speed of 0.535 m/s. The mass of chains and flights is 54 kg/m, and the coefficient of friction between the chain and trough is 0.33. What is the power needed to drive the empty conveyor?

(a) 6.94 kW

(b) 10.68 kW*

(c) 21.36 kW

(d) 25.74 kW

Solution: W = mg = 54 x 9.8 = 529.2 N

F = μW = 0.33 x 529.2 = 174.64 N

Power = FV = 174.64 x 0.535 = 93.43 Watt

Total power to drive the conveyor

= Power per m x total length

= 93.43 x 114

= 1065 Watt = 10.65 kW

A conveyor transported the coal at the rate of 220 te/hr is 600 m long moving up a gradient of 1 in 60. If the belt width (W) is 0.75 m, and cross-sectional area of the material is 0.1 W², determine the belt speed. The bulk density of coal 0.8 te/m³.

(a) 0.04 m/s

(b) 0.96 m/s

(d) 1.80 m/s

Solution: T = 220 t/hr = 0.061 t/s

W = 0.75 m

a = 0.1W² = 0.1 x 0.75² = 0.056 m²

b = 0.8 t/m³

T = abV

Hence, 0.061 = 0.056 x 0.8 x V

From this, V = 1.36 m/s

A cable belt conveyor is required to carry coal of bulk density 0.8 t/m³ at a rate 600 t/hr up an inclined length of 1200 m. The total vertical lift is 300 m. The power required to raise the material against gravity in kW will be

(a) 490.5*

(b) 392.4

(c) 120.0

(d) 1765.8

Solution: Carrying capacity T = 600 t/hr

= (600/3600) t/s

= (1/6) t/s = 1000/6 kg/s

So if t = 1 s, m = 1000/6 kg

Force F = mg = (1000/6) x 9.8 N

Velocity = vertical distance/time = 300/1 = 300 m/s

Power = Fv = (1000/6) x 9.8 x 300

= 490,000 W

= 490 kW

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

(a) 5.1

(b) 5.8

(d) 6.5

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

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

(a) 2.548*

(b) 2.448

(c) 2.038

(d) 1.630

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

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

(a) 0.12

(b) 1.30

(c) 0.22

(d) 1.20

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

For a 150 m long chain conveyor, the component of power required to move coal against gravity along an inclination 15° is 5 kW. Hourly capacity of the chain conveyor will be

(a) 47.2 te*

(b) 37.8 te

(c) 13.1 te

(d) 3.4 te

Solution: P = 5000 Watt

L = 150 m

Angle = 15⁰

Force, F = mg sinθ

P = Fv = F x (distance/time)

For 1 sec

P = F x distance

= mg sinθ x distance

∴ 5000 = m x 9.8 x sin15⁰ x 150

Giving m = 13.14 kg in one second.

Hourly capacity/hr = Mass handled/hr

= 13.14 x 3600 = 47310 kg

= 47.3 te

In a conveyor belt drive the tension on the tight side is double that on the slack side. If the value of µ is 0.3. find the angle of lap required if the belt is not to slip on its driving drum.

(a) 132° *

(b) 135°

(c) 136°

(d) 150°

Solution: T₁/T₂ = e^μ⁽^θ^/57.3)

where

T₁ = tension on the tight side

T₂ = tension on the slack side

μ = coefficient of friction

lap θ = wrapping angle or the angle of lap in degrees

Here T₁/T₂ = 2

2 = e^0.3^(θ/57.3⁾

By trial and error,

θ = 132⁰

Or from the given options, 132⁰ satisfies the equation.

The approximate head generated by a single-stage centrifugal pump of 250 mm diameter impeller running at 1440 rev/min with manometric efficiency of 0.7 is

(a) 25 m*

(b) 36 m

(c) 72 m

(d) 144 m (GATE, 1998)

Solution:

v = πDN/60

= 3.14 x 0.25 x 1440/60

= 18.84 m/s

H = v²/2g = 18.84²/2 x 9.8

= 18.11 m

Now Manometric head (the head to which pump has to work)

= H/η = 18.11/0.7 = 25.87 m

The resistance to motion of a dumper on a horizontal track is (35 + 0.1 V²) N per tonne, where v is the velocity in m/s. Its tractive effort is 325 N/te. The acceleration in m/s² at a velocity of 10 m/s is

(a) 0.14

(b) 7.14

(c) 0.28

(d) 28.00

Solution: As mass is not given, we will consider it as 1 tonne i.e. 1000 kg.

R = 35 + 0.1V²

= 35 + 0.1 x 10² = 45 N

Tractive effort (TE) = 325 N

Pull = TE - Resistance

= 325 - 45 = 280 N

F = ma

280 = 1000 x a

∴ a = 0.28 m/s²

The production rate of a bucket wheel excavator is 400 m³/hr. in a material of digging resistance 25 N/ mm. Assuming digging resistance inversely varies with square root of production capacity of the excavator, the production rate in m³/hr. in another material of digging resistance of 45 N/mm will be

(a) 1296.0

(b) 298.1*

(c) 222.2

(d) 123.5

Solution: R ∝ 1/Q

i.e. R₁/R₂ = Q₂²/Q₁²

25/45 = Q₂²/400²

Giving Q₂ = 298.1 m³/hr

Dozer blade has the dimension of 4500 mm x 2000 mm operates on level ground. The output capacity of the dozer is

(a) 11 m³ approx.)*

(b) 15 m³ (approx.)

(c) 20 m³ (approx.)

(d) 8 m³ (approx.) (DGMS FMC 2015)

Solution: Consider following figure.

The output of the dozer blade = (1/2) x h² x L

= 0.5 x 2 x 2 x 4.5 = 9.0 m³

Consider a 20% bulging of the material in front of the blade.

The output of the dozer blade = 9 x 1.2 = 10.8 m³

A centrifugal pump rotating at a speed of 1450 rpm discharges 360 m³/hour, the head of 85 m. The driving power of the pump is 115 KW. If the pump rotates at 960 rpm, its power will be

(a) 25 KW (approx)

(b) 34 KW (approx)*

(c) 30 KW (approx)

(d) 45 KW (approx) (DGMS SMC 2015)

Solution: P₂/P₁ = N₂³/N₁³

P₂/115 = (960/1450)³

Giving P₂ = 33.37 kW

Two belt conveyors load a ground bunker, each at a rate of 400 tph which is initially filled with 10000 t of coal. Coal is discharged from the bottom of the ground bunker onto a belt conveyor at a rate of 1200 tph. The time elapsed in hours before the bottom conveyor starts to operate below its rated capacity is

(a) 6.5

(b) 8.5

(c) 12.5

(d) 25.0* (GATE 2009)

Solution: No. of belt conveyors = 2

Loading rate = 400 tph

Initially quantity in bunker = 10000 t coal

Rate of Coal discharged from ground bunker onto belt conveyor = 1200 tph.

Capacity of 1st and 2nd belt conveyor (q) = 400 tph

Capacity of 3rd belt conveyor (Q) = 1200 tph

Initial mass of coal in the bunker (M) = 10000 tonne

Doing mass balance

M + q x T + q x T = Q x T

10000 + 2 x 400 x T = 1200 x T

∴ T = 25 hr.

A 1100 V, 3f power supply system of a mine draws a load of 185 kW. The ammeter reading shows 115 A. The power factor of the system is

(a) 0.84*

(b) 0.73

(c) 0.64

(d) 0.45 (GATE 2009)

Solution: P = √3 VIcosφ

∴ 185 x 1000 = √3 x 1100 x 115 cosφ

∴ cosφ = 0.86

A cage of floor area 5 m² suspended in a shaft has a drag coefficient 2.5. If the velocity of air in the shaft is 6 m/s, the drag force (N) experienced by the cage is

(a) 120

(b) 170

(c) 200

(d) 270* (GATE 2008)

Solution: F_D = C_dρV²A/2

= 2.5 x 1.2 x 6² x 5/2

= 270 N

A direct rope haulage pulls 8 tubs loaded with coal through an incline of length 500 m having an inclination of 1 in 6. Consider the following additional data.

Capacity of tub = 1.0 tonne

Tare weight of tub = 500 kg

Hauling speed = 9 km per hour

Coefficient of friction between wheel and rail = 1/60

Coefficient of friction between rope and drum = 1/10

Mass of rope per meter = 1.5 kg

The minimum power required to haul the tubs in kW is

(a) 345.50

(b) 348.60

(c) 350.10

(d) 365.50 (GATE 2014)

Solution: Mass of rope = 1.5 x 500 = 750 m

Mass of one tub = 1000 + 500 = 1500 kg.

Mass of 8 tubs = 8 x 1500 = 12000 kg.

Gravity component of tub = 12000 x 9.8 x (1/6) = 19600 N

Gravity component of rope = 750 x 9.8 x (1/6) = 1225 N

Force required to overcome friction of tub = (1/60) x 19600 = 326.7 N

Force required to overcome friction of rope = (1/10) x 1225 = 122.5 N

Total force = 21274.2 N = 21.3 kW

Velocity = 9 km/hr = 2.5 m/s

Power = FV = 21.3 x 2.5 = 53.25 kW

Hence, we see that given options are not correct.

In a direct rope haulage operating along a 1500 m long incline, 180 tonne of coal is hauled in 7 hours. The average rope speed is 7.5 km/h. The set changing time for the tubs is 2 minutes each at the top and bottom of the incline. If the tub capacity is 1.0 tonne, the number of tubs in the set is ––––. (GATE 2018)

Solution:

Direct rope haulage operating along a 1500 m long incline

180 tonne of coal is hauled in 7 hours

The average rope speed is 7.5 km/h = 2.08 m/s

The set changing time for the tubs is 2 minutes each at the top and bottom of the incline.

The tub capacity is 1.0 tonne

The number of tubs in the set = ?

The tub covers 2.08 m in 1 second

Time taken by tubs to cover 1500 m distance is = 1500 / 2.08 = 721.15 s

= 12 minutes

So the total time taken = 12 + 2 x 2 = 16 min = 0.27 hrs

180 tonne of coal is hauled in 7 hours

Coal transported in 0.27 hr = ( 0.27 x180) / 7 = 6.94 tonne = 7

An SDL of 1.0 tonne capacity operates with a cycle time of 6 minutes. The dimension of the face is 4 m x 3 m. Five blasts are conducted per shift with an average pull of 1.2 m. If the density of blasted coal is 0.8 tonne/m³, the time required by the SDL in the shift to lift all the prepared coal in hours is ––––. (GATE 2018)

Solution: SDL capacity = 1.0 tonne

Cycle time = 6 minutes

Dimension of the face is 4 m x 3 m

Five blasts are conducted per shift with an average pull of 1.2 m

Density of blasted coal = 0.8 tonne/m³

Time required by the SDL in the shift to lift all the prepared coal in hours = ?

Volume of coal blasted per shift = 5 x 1.2 x 4 x 3 = 72 m³

= 72 x 0.8 = 57.6 tonne

SDL takes 6 minutes to transport = 1 tonne

Time taken by SDL to transport 57.6 tonne is

= 57.6 x 6 = 345.6 min = 5.76 hrs

A flat belt conveyor is carrying coal of bulk density 1 tonne/m³ at a rate of 400 tonne/h. The belt speed is 3 m/s. Coal is spread over the belt covering 80% of the belt width in a shape of a triangle. If the pile height is 1/4 of the belt width, the width of the belt in mm is

(a) 1109

(b) 909

(c) 709

(d) 609* (GATE 2011)

Solution: T = abV

T = 400/3600 t/s

a = area of triangle of base 0.8W and height (W/4)

= (1/2)(0.8W)(W/4)

b = 1 t/m³

V = 3 m/s

∴ 400/3600 = (1/2)(0.8W)(W/4) x 1 x 3

Giving W = 0.609 m = 609 mm

Data for a centrifugal pump discharging water from a sump to the surface are given.

Head, m : 180

Discharge rate, m³/hr : 320

Operating hours per day for 270 days in a year : 14

Operating hours per day for remaining 95 days : 20

Overall efficiency of the pumping system : 0.70

Specific weight of mine water, kN/m³ : 10.20

The annual electrical power consumption in GWh, due to pumping operation, is ____

______. (round off up to 2 decimals).(GATE 2024)

Solution: Operating hours per year

= 270 x 14 + 95 x 20 = 5680 hrs

Head, h = 180 m

Q = 320 m³/hr = 320/3600 m³/s

η = 0.7

w = 10.20 kN/m³ = 10200 N/m³

Power = ?

Power = wgQH/η

= 10200 x 9.8 x (320/3600) x 180/0.7

= 2284800 W = 2284.8 kW

Total power consumption in one year

= 2284.8 x 5620

= 1279488 kWh

= 1279488/(1000 x 1000)

= 1.28 GWh

A 35.0 kW motor transmits power to a pulley of 600 mm diameter, which rotates at 400 rpm to drive a flat belt. The tension in the tight side is 2.5 times of the slack side. Neglect all transmission losses. If the maximum allowable tension is 8.0 N per mm of belt width, then the minimum width of the belt in mm, is –––– . (round off up to 2 decimals). (GATE 2024)

Solution: V = πDN/60

= 3.14 x 0.6 x 400/60

= 12.56 m/s

T₁/T₂ = 2.5

T₁ = 8W Newton (W is width of belt in mm)

∴ T₂ = T₁/2.5 = 8W/2.5 = 3.2W

Power = FV

∴ 35,000 = (8W - 3.2W) x 12.56

∴ W = 580 mm

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