In a Lang’s lay wire rope:

the wires in the strands are laid in the same direction as the strands

the wires in the strands are laid in the opposite direction to the strands

the wires in the strands are laid in the same direction as the strands

the outer wires are shaped to interlock

Which of the following statements about the laws of comminution are correct? (Select all that apply.)

Rittinger’s law, energy proportional to new surface created, fits fine grinding best

Kick’s law fits coarse crushing best

Bond’s law uses the 80%-passing sizes of the feed and the product

Rittinger’s law, energy proportional to new surface created, fits fine grinding best

Bond’s law states that energy is proportional to the logarithm of the reduction ratio

Kick’s law fits coarse crushing best

Bond’s law uses the 80%-passing sizes of the feed and the product

Which of the following statements about rock–tool interaction and mechanical cutting are correct? (Select all that apply.)

Drag picks are suited to coal and soft to medium-strength rock

Specific energy falls as the depth of cut increases

Drag picks are suited to coal and soft to medium-strength rock

Specific energy falls as the depth of cut increases

A disc cutter breaks rock mainly by dragging a sharp edge through it in shear

A blunt pick lowers the specific energy of cutting

Air compressors for mines are built multistage with intercooling mainly because it:

brings the compression work closer to the isothermal minimum

raises the delivery temperature to dry the air

brings the compression work closer to the isothermal minimum

eliminates the need for an air receiver

increases the volumetric flow of free air per stage

Mining Machinery: Mechanical, Hydraulic and Pneumatic Power, Wire Ropes and Hoisting, Haulages and Conveyors, Face Machines, Pumps, Comminution and Rock Cutting

The second chapter for Section 4 of the GATE Mining Engineering (MN) paper, for its Mining Machinery sub-heading. The syllabus names the generation and transmission of mechanical, hydraulic and pneumatic power; materials handling — wire ropes, haulages, conveyors, face and development machinery, hoisting systems and pumps; comminution methods and machinery; and rock–tool interaction with mechanical cutting. Every one of these ends in the same few equations — power is force times speed, a rope on a drum obeys the capstan law, a pump lifts ρgQH, a crusher spends energy against a size reduction — and the chapter is built round them, with the machines described in enough detail to know what each equation is being applied to.

1. Mechanical, hydraulic and pneumatic power

Mine machines are driven by electric motors, diesel engines, hydraulics and compressed air. Mechanical transmission uses shafts, couplings, gearboxes, chains and belts; power is P = Tω = Fv, and a gear train trades speed for torque at nearly constant power, less its losses. Hydraulic power transmits force by an incompressible fluid: a cylinder of bore D at pressure p exerts F = p π D²/4, and the power delivered is P = p Q. So 20 MPa on a 200 mm bore gives 628 kN — the reason powered roof supports and hydraulic excavators are hydraulic. Pumps (gear, vane, piston) generate the flow, valves control it, and actuators (cylinders, motors) use it.

Pneumatic power — compressed air — is flameproof and simple, and drives rock drills, loaders and small hoists, but it is the least efficient of the three, since most of the compression work leaves as heat. The minimum (isothermal) work to compress air from p₁ to p₂ is W = p₁V₁ ln(p₂/p₁): 1 m³ of free air at 100 kPa to 700 kPa needs 100 × ln 7 = 194.6 kJ. Real compression is nearer adiabatic and needs more, so compressors are built multistage with intercooling, which brings the work towards the isothermal value; air receivers smooth the demand and drop out moisture, and pipe friction and leakage are the main distribution losses.

2. Wire ropes and hoisting systems

A wire rope is wires twisted into strands and strands laid round a core (fibre or steel). Its designation, such as 6 × 19 or 6 × 37, gives strands × wires per strand; more, finer wires make a more flexible rope, fewer, coarser ones a more abrasion-resistant one. In regular (ordinary) lay the wires in each strand are twisted opposite to the strands, making a stable rope that resists kinking; in Lang’s lay they are twisted the same way, giving a longer bearing length of each wire and better wear resistance but a tendency to untwist, so both ends must be fixed. Locked-coil ropes, with interlocking shaped outer wires and a smooth surface, are used for winding and guide ropes. A rope’s factor of safety is its breaking load over the maximum static load it carries, and regulations prescribe minimum values that fall as the depth of wind increases.

A drum winder coils the rope on a cylindrical (or bi-cylindro-conical) drum; a friction (Koepe) winder passes the rope over a friction-lined wheel with a conveyance on each end and a tail rope below to balance the rope weight. The Koepe drive holds only while T₁/T₂ ≤ e^μθ; with θ = π for a single wrap and a lining μ of about 0.25, e^0.25π = 2.19, so the heavy-side to light-side ratio, including inertia forces while accelerating, must stay below that. Skips carry rock, cages carry people and materials. The steady hoisting power is P = m g v/η: lifting a net 10 t at 12 m/s at 90% efficiency needs 10 000 × 9.81 × 12/0.9 = 1308 kW. Output is payload × trips per hour; an 80 s cycle with 8 t skips gives 45 × 8 = 360 t/h. Safety devices include overwind and overspeed protection, safety catches on cages, and brakes applied on power failure.

⚠️ θ in radians, and the heavier side on top

e^μθ needs θ in radians — 180° is π, not 180. And T₁ is always the tight (heavier) side: the check is heavy-side tension over light-side tension, which is greatest at the start of the wind when the loaded skip is accelerating upward.

3. Haulages, conveyors, face and development machinery, and pumps

Rope haulages move tubs or cars by a rope from a stationary engine: direct (a single rope on an incline, gravity returning the empties), main-and-tail (two ropes for undulating roads) and endless (a continuous rope with tubs clipped on). Locomotive haulage needs a tractive effort equal to the sum of rolling, gradient and acceleration resistances: TE = W × (rolling resistance per tonne) + W g sin θ + W a, limited by adhesion, μ × weight on driving wheels. A 50 t train with a rolling resistance of 10 kg per tonne (98.1 N/t) up a 1 in 200 gradient needs 4905 + 2452.5 = 7357.5 N at constant speed.

A belt conveyor carries a load cross-section A at belt speed v, so its capacity is Q = 3600 A v ρ t/h (A in m², v in m/s, ρ in t/m³): 0.1 m² at 3 m/s of 0.9 t/m³ coal is 972 t/h. The drive pulley transmits an effective tension Te = T₁ − T₂, limited by the capstan law T₁/T₂ ≤ e^μθ; so the slack-side tension must be at least T₂ = Te/(e^μθ − 1), and the drive power is Te × v. Wrap is raised with snub pulleys or tandem drives, μ with lagging. Chain conveyors — the AFC and the stage loader — suit the face, where they are pushed over with the supports. Face and development machines: the shearer and plough on longwall faces, the continuous miner in bord and pillar, the road header and the tunnel boring machine for drivages, drill jumbos, LHDs and SDLs, shuttle cars and roof bolters.

Mine pumps dewater the workings. The hydraulic power is ρ g Q H and the input P = ρ g Q H/η: 0.1 m³/s against a 200 m head at 75% efficiency needs 1000 × 9.81 × 0.1 × 200/0.75 = 261.6 kW. Centrifugal pumps, often multistage for high heads, dominate; reciprocating and positive-displacement pumps handle high heads at low flow and dirty water. For a centrifugal pump of fixed size the affinity laws give Q ∝ N, H ∝ N², P ∝ N³, and the pump must have enough net positive suction head (NPSH) to avoid cavitation, which limits suction lift.

4. Comminution, rock–tool interaction and mechanical cutting

Comminution reduces the size of run-of-mine rock in stages. Crushers — jaw and gyratory for primary crushing, cone and roll crushers for secondary and tertiary, and impact crushers and hammer mills for softer, less abrasive rock such as coal and limestone — work mostly by compression. Grinding mills — rod, ball, autogenous and semi-autogenous — tumble the charge and break it by impact and attrition. The reduction ratio is feed size over product size. The energy needed is given by three laws: Rittinger’s, energy ∝ new surface created (∝ 1/P − 1/F), fitting fine grinding; Kick’s, energy ∝ log of the reduction ratio, fitting coarse crushing; and Bond’s, in between and the one used in design: W = 10 Wi (1/√P80 − 1/√F80) kWh/t, with P80 and F80 the 80%-passing sizes in μm and Wi the work index, the energy to reduce a tonne from a very large size to 80% passing 100 μm.

Rock–tool interaction decides how a cutting machine performs. A drag pick (chisel or conical point-attack pick) is dragged through the rock and breaks it in tension and shear ahead of the tip; it cuts coal and soft to medium rock efficiently but wears fast in hard, abrasive rock. A disc cutter on a tunnel boring machine is rolled under a very high normal force; it indents and crushes a zone under the edge, and the cracks from neighbouring grooves meet to release chips between them. The specific energy — energy per unit volume cut, SE = cutting power/production rate — is the efficiency measure: 100 kW producing 50 m³/h is 2 kWh/m³. SE falls as the depth of cut increases (bigger chips for the same crushing), and rises sharply with a blunt tool; cuttability tests measure it to select the machine and the tool spacing.

🧠 Bond in one line

With Wi = 12 kWh/t, F80 = 10 000 μm and P80 = 100 μm: W = 10 × 12 × (1/10 − 1/100) = 120 × 0.09 = 10.8 kWh/t. The factor 10 is √100 — it is why Wi is defined at 100 μm — and sizes must be in micrometres.

Key takeaways

Hydraulic force = p × area and power = pQ; isothermal air compression work = p₁V₁ ln(p₂/p₁), approached by multistage compression with intercooling.
Lang’s lay wears better, regular lay resists kinking; a Koepe winder holds while T₁/T₂ ≤ e^μθ, θ in radians.
Hoist power = mgv/η; output = payload × trips per hour; tractive effort = rolling + gradient + acceleration resistance.
Belt capacity = 3600 A v ρ t/h; slack-side tension T₂ = Te/(e^μθ − 1); pump input = ρgQH/η, with Q ∝ N, H ∝ N², P ∝ N³.
Bond: W = 10 Wi(1/√P80 − 1/√F80) with sizes in μm; specific energy of cutting falls with depth of cut and rises with a blunt tool.

Practice questions (15)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A winder raises an unbalanced net load of 10 t at a steady rope speed of 12 m/s. The overall efficiency of the drive is 90%. Taking g = 9.81 m/s², the power required, in kW, is ____.
Numerical answer — type the value.
Show answer
Answer: 1308
P = m g v/η = 10 000 × 9.81 × 12/0.9 = 1 177 200/0.9 = 1 308 000 W = 1308 kW. Multiplying by the efficiency instead of dividing gives 1059.5 kW; leaving out g gives kg·m/s, not watts.
For a friction (Koepe) winder with the rope passing over half the wheel and a lining coefficient of friction of 0.25, the maximum ratio of heavy-side to light-side rope tension before slip, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.19
T₁/T₂ = e^μθ with θ = π: e^0.25π = e^0.7854 = 2.19. The ratio must hold during acceleration, when the heavy side is heaviest; 1 + μθ = 1.79 is only the first two terms of the exponential.
A skip hoist carries 8 t per trip and completes one trip every 80 s. Its hoisting capacity, in t/h, is ____.
Numerical answer — type the value.
Show answer
Answer: 360
Trips per hour = 3600/80 = 45, so capacity = 45 × 8 = 360 t/h. In a balanced two-skip system one skip rises as the other falls, and the 80 s cycle already counts each trip once; doubling it to 720 counts the descending skip as a second payload.
A locomotive hauls a 50 t train at constant speed up a gradient of 1 in 200. The rolling resistance is 10 kg per tonne of train (take it as 98.1 N/t) and g = 9.81 m/s². The tractive effort required, in N, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 7357.5
Rolling: 50 × 98.1 = 4905 N. Gradient: W g sin θ ≈ 50 000 × 9.81/200 = 2452.5 N. At constant speed there is no acceleration term, so TE = 7357.5 N. Down the same gradient the gradient force would help, leaving 2452.5 N.
A belt conveyor carries coal of bulk density 0.9 t/m³ with a load cross-section of 0.1 m² at a belt speed of 3 m/s. Its capacity, in t/h, is ____.
Numerical answer — type the value.
Show answer
Answer: 972
Q = 3600 A v ρ = 3600 × 0.1 × 3 × 0.9 = 972 t/h. Without the 3600 the figure is 0.27 t/s; using the solid density of coal (about 1.4 t/m³) instead of the loose bulk density overstates the capacity.
A conveyor drive pulley has an angle of wrap of 210° and a coefficient of friction of 0.35 with the belt. The effective tension to be transmitted is 30 kN. The minimum tight-side tension, in kN, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 41.51
θ = 210° = 3.6652 rad, e^μθ = e^1.2828 = 3.6068. T₂ = Te/(e^μθ − 1) = 30/2.6068 = 11.51 kN, and T₁ = Te + T₂ = 41.51 kN. Answering 11.51 gives the slack side; using θ in degrees makes the exponential meaningless.
A mine pump delivers 0.1 m³/s of water against a total head of 200 m with an overall efficiency of 75%. Taking ρ = 1000 kg/m³ and g = 9.81 m/s², the input power, in kW, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 261.6
Water power = ρ g Q H = 1000 × 9.81 × 0.1 × 200 = 196.2 kW; input = 196.2/0.75 = 261.6 kW. Quoting 196.2 kW gives the hydraulic output, and multiplying by 0.75 (147.2 kW) reverses the efficiency.
One cubic metre of free air at 100 kPa is compressed isothermally to 700 kPa. The work done, in kJ, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 194.59
W = p₁V₁ ln(p₂/p₁) = 100 kPa × 1 m³ × ln 7 = 100 × 1.94591 = 194.59 kJ. Using the pressure difference (600 kJ) treats compression as happening at constant pressure; log₁₀ 7 in place of ln 7 gives 84.5 kJ.
A hydraulic prop cylinder has a bore of 200 mm and works at 20 MPa. The force it exerts, in kN, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 628.32
F = p × πD²/4 = 20 × 10⁶ × π × 0.2²/4 = 20 × 10⁶ × 0.031416 = 628 318 N = 628.32 kN. Using the bore as a radius gives four times as much (2513 kN), and leaving it in mm mixes units by a factor of a million.
Ore with a Bond work index of 12 kWh/t is ground from an 80%-passing feed size of 10 000 μm to an 80%-passing product size of 100 μm. The specific energy required, in kWh/t, to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 10.8
W = 10 Wi (1/√P80 − 1/√F80) = 10 × 12 × (1/10 − 1/100) = 120 × 0.09 = 10.8 kWh/t. Dropping the feed term gives 12, the work index itself; using sizes in mm makes the result a thousand-fold wrong.
In a Lang’s lay wire rope:
1. the wires in the strands are laid in the opposite direction to the strands
2. the wires in the strands are laid in the same direction as the strands
3. there is no central core
4. the outer wires are shaped to interlock
Show answer
Answer: B — the wires in the strands are laid in the same direction as the strands
Lang’s lay twists wires and strands the same way, so each outer wire runs diagonally along the rope with a long bearing surface — better wear, but a tendency to untwist. Opposite twist is regular (ordinary) lay, and interlocking shaped wires describe a locked-coil rope.
The speed of a centrifugal mine pump is increased by 10%. By the affinity laws its head increases by about:
1. 10%
2. 21%
3. 33%
4. 5%
Show answer
Answer: B — 21%
H ∝ N², so H₂/H₁ = 1.1² = 1.21, a 21% rise. 10% is the change in flow (Q ∝ N), and 33% is roughly the change in power (P ∝ N³, 1.1³ = 1.331) — which is why running a pump faster is an expensive way to gain head.
Which of the following statements about the laws of comminution are correct? (Select all that apply.)
1. Rittinger’s law, energy proportional to new surface created, fits fine grinding best
2. Bond’s law states that energy is proportional to the logarithm of the reduction ratio
3. Kick’s law fits coarse crushing best
4. Bond’s law uses the 80%-passing sizes of the feed and the product
Show answer
Answer: A — Rittinger’s law, energy proportional to new surface created, fits fine grinding best; C — Kick’s law fits coarse crushing best; D — Bond’s law uses the 80%-passing sizes of the feed and the product
Rittinger (surface area) suits fine sizes and Kick (log of reduction ratio) suits coarse ones, with Bond between them, written in the 80%-passing sizes. The logarithmic law in the second option is Kick’s, not Bond’s; Bond’s depends on 1/√P80 − 1/√F80.
Which of the following statements about rock–tool interaction and mechanical cutting are correct? (Select all that apply.)
1. Drag picks are suited to coal and soft to medium-strength rock
2. Specific energy falls as the depth of cut increases
3. A disc cutter breaks rock mainly by dragging a sharp edge through it in shear
4. A blunt pick lowers the specific energy of cutting
Show answer
Answer: A — Drag picks are suited to coal and soft to medium-strength rock; B — Specific energy falls as the depth of cut increases
Drag picks cut coal and softer rock efficiently but wear quickly in hard abrasive rock, and deeper cuts release bigger chips for the same crushed zone, so energy per unit volume falls. A disc cutter rolls under a high normal force, indenting and chipping rather than dragging; a blunt tool crushes more rock into fines and raises the specific energy.
Air compressors for mines are built multistage with intercooling mainly because it:
1. raises the delivery temperature to dry the air
2. brings the compression work closer to the isothermal minimum
3. eliminates the need for an air receiver
4. increases the volumetric flow of free air per stage
Show answer
Answer: B — brings the compression work closer to the isothermal minimum
Cooling the air between stages removes the heat of compression, so each later stage compresses cooler, denser air and the total work moves from the adiabatic towards the isothermal value. Delivery temperature falls rather than rises, and a receiver is still needed to smooth demand and drop out moisture.