Mine hoists (extractors)

A passenger lift moves a one-tonne car at 1 to 4 m/s. A production mine hoist (in French, an extracteur) may raise thirty tonnes of ore from more than a kilometre down at 15 to 18 m/s. The stored energy, the mass of the rope itself and the consequences of a failure make it the most demanding vertical drive in industry. This lesson is course section 3.4.

Drum or Koepe

Two mechanical arrangements dominate.

• Drum winder. The rope winds onto a rotating drum. A single drum carries one conveyance, and the rope is stored on the drum. A double drum winds one rope on while the other pays out, which balances the two sides naturally.
• Koepe (friction) winder. The rope is not stored. It passes over one driving sheave and is moved by friction against the sheave lining, with a conveyance on one side and a counterweight or a second conveyance on the other. A balance rope hanging beneath both conveyances compensates the weight of the head rope, so the imbalance stays almost constant during the whole wind.

The conveyance is a cage (people, materials, mine cars) or a skip, a self-dumping bucket for ore or coal. Skips give the highest production because loading and dumping are automatic at fixed positions.

The balanced hoist

As for the lift, the static load is the out-of-balance mass. With payload $m_p$, conveyance mass $m_c$ and counterweight $m_{cw}$, the force on the rope and the torque at a sheave or drum of radius $R$, through gear ratio $i$, are

$$F_{ub} = \big[(m_c + m_p) - m_{cw}\big]\, g, \qquad T_\text{sheave} = \frac{F_{ub}\, R}{i} \cdot \frac{1}{\eta^{\pm 1}} \tag{3.8}$$

with $+1$ when motoring and $-1$ when generating, exactly as in lesson 1. A well-designed hoist balances the conveyance plus half the payload, $m_{cw} = m_c + \tfrac12 m_p$: the peak motoring and generating torques are then equal, and the drive rating is the smallest.

In a deep shaft the rope is a major mass: a 40 mm rope 1200 m long weighs several tonnes, and its weight moves from the conveyance side to the sheave side as the skip rises. The Koepe balance rope, or a tapered drum rope, deals with it. The dynamic torque follows the same law as always, $T_\text{dyn} = J_\text{eq}\, d\omega/dt$, and with such inertia and forces, acceleration and jerk must be limited.

The winding cycle

Every trip follows a programmed speed-time profile, the winding diagram: a slow pull-away, controlled acceleration to full speed, a full-speed run, deceleration, a short creep at low speed as the skip approaches the dump or the landing, and a precise stop. The creep makes the final position independent of the length of the wind, rope stretch and load. For a trapezoidal profile with acceleration $a$, full speed $v_m$ and hoisting distance $H$:

$$t_\text{cycle} = \frac{H}{v_m} + \frac{v_m}{a} + t_\text{creep} + t_\text{rest} \tag{3.9}$$

Production is inversely proportional to $t_\text{cycle}$, so each term is an optimisation target, limited by the rope's safety factor, the motor rating and the jerk. Modern drives replace the trapezoid with a jerk-limited S-curve (lesson 3).

Hoisting distance H

1,000 m

Full winding speed v_m

15.0 m/s

Acceleration a

0.80 m/s²

Loading and dumping t_rest

15 s

Creep time t_creep

5 s

Single drumKoepe with balance rope

• Hoist speed (m/s)
• Out-of-balance rope force (kN)

Cycle time

105.4 s

H/v_m + v_m/a

66.7 + 18.8 s

Trips per hour

34.2

Production (20 t skips)

683 t/h

t_cycle = H / v_m + v_m / a + t_creep + t_rest

Predict first

In the widget's deep-shaft hoist, you double the acceleration from 0.8 to 1.6 m/s². How much shorter is the cycle?

About halfAbout 9 s, less than 10 %It does not change

Drives: from Ward-Leonard to regeneration

Mine hoists have pushed variable-speed drives forward for a century: they need full torque at zero speed, smooth control from creep to full speed, and a way to absorb energy whenever a loaded skip descends or a heavy counterweight overhauls the motor.

1. Ward-Leonard DC systems: the DC hoist motor is fed by a motor-generator set, whose generator field controls the armature voltage smoothly in all four quadrants. Excellent control and natural regeneration, at the cost of a large, inefficient rotating plant.
2. Thyristor converters (from the 1970s): two anti-parallel thyristor bridges feed the DC motor straight from the grid. The dual converter reverses torque instantly and returns descent energy to the supply through its inverting bridge.
3. AC vector drives (today's standard): a cage or wound-field synchronous motor on a PWM inverter under field-oriented control, with an active front end for full regeneration. The largest hoists still use cycloconverter-fed synchronous motors, gearless and direct-drive, for very high torque at low speed.

In every generation the requirement is the same: true four-quadrant operation with the energy recovered. A large hoist regenerates megawatts on each descending wind; burning that in resistors would be both wasteful and a thermal problem.

Hoist safety

The same fail-safe philosophy as the lift (lesson 5), on a larger scale:

• overspeed protection: the winding speed is checked against the programmed profile, with position-dependent limits near the shaft ends;
• overwind protection: independent detection of travel beyond the normal landing, backed by mechanical catch gear and arrestor blocks;
• slack-rope and rope-tension monitoring: a jammed conveyance, or a Koepe rope losing its grip on the sheave;
• fail-safe brakes: spring-applied, power-released disc or drum brakes, sized to stop the fully loaded descending conveyance.

Together with the shaft gate contacts, these form the hoist safety circuit, supervised by a safety-rated PLC or a dedicated hoist monitor.

FoundationStart here if this is new to you

A mine hoist is a lift for rock, a kilometre deep and ten times faster. Everything in this chapter still applies (counterweight, quadrants, S-curve, fail-safe brake), but the numbers are huge: one skip of ore weighs as much as twenty cars, and the rope alone can weigh several tonnes.

ExplorerGo deeper: derivations and open questions

Rope weight in a drum winder. In the widget, choose the single drum. The out-of-balance force falls during the wind as the hanging rope shortens. For a 6 kg/m rope in a 1000 m shaft, by how much does the motor torque change from the bottom to the top? How does a double drum change this?

Optimum speed. For a fixed motor power, a higher $v_m$ means a lower torque available for acceleration. Write the cycle time as a function of $v_m$ with $a$ limited by the available torque, and find the speed that maximises production.