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100+ Free GCC Mines & Works Plant Engineering Exam Practice Questions
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Sample GCC Mines & Works Plant Engineering Exam Practice Questions
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1A deep-level gold mine shaft uses a single-drum cylindrical winder to lift rock from a depth of 1,800 m. The steel wire rope has a mass of 8.5 kg/m and the suspended skip empty mass is 6,000 kg. If the maximum payload of rock per trip is 10,000 kg, calculate the static safety factor of the rope at the headgear sheave when the loaded skip is at the bottom of the shaft, assuming a nominal rope breaking force of 2,400 kN. (Take g = 9.81 m/s²).
A.7.82
B.8.50
C.6.56
D.9.21
Explanation: 1. Suspended rope mass = 1,800 m × 8.5 kg/m = 15,300 kg.
2. Total suspended mass at sheave = 15,300 kg (rope) + 6,000 kg (skip) + 10,000 kg (payload) = 31,300 kg.
3. Total static load force = 31,300 kg × 9.81 m/s² = 307,053 N = 307.05 kN.
4. Static Safety Factor = Breaking Force / Static Load = 2,400 kN / 307.05 kN = 7.816 ≈ 7.82.
2According to South African Mine Health and Safety Act regulations for rock winding in a vertical shaft, what is the minimum permitted static safety factor (S) for a new winding rope at the headgear sheave for a suspended rope length of L = 2,000 m, using the standard formula S = 8.5 - 0.001 × L (subject to absolute minimum limits)?
A.6.50
B.8.50
C.5.00
D.4.50
Explanation: 1. Formula for rock winding static safety factor: S = 8.5 - 0.001 × L.
2. Substitute L = 2,000 m: S = 8.5 - (0.001 × 2,000) = 8.5 - 2.0 = 6.50.
3. Since 6.50 exceeds the statutory minimum limit of 4.5, the minimum required static safety factor is 6.50.
3A Koepe (friction) winder has a drum angle of wrap θ = 190° (3.316 rad) and the friction coefficient between the steel rope and the polyurethane lining is μ = 0.25. Calculate the maximum static tension ratio (T1/T2) before rope slip occurs across the friction wheel.
A.2.29
B.1.85
C.3.12
D.4.75
Explanation: 1. The limiting ratio of rope tensions for friction winders is given by Eytelwein's formula: T1/T2 = e^(μ × θ).
2. Convert angle of wrap to radians: θ = 190° × (π / 180°) = 3.3161 rad.
3. Exponent product = μ × θ = 0.25 × 3.3161 = 0.8290.
4. Limiting tension ratio T1/T2 = e^(0.8290) = 2.291 ≈ 2.29.
4A double-drum winder has a drum width of 2.4 m and is positioned at a horizontal distance of 45 m from the headgear sheave wheel centerline. Calculate the maximum fleet angle when the rope is at the extreme outer flange of the drum.
A.1.53°
B.3.05°
C.0.76°
D.2.45°
Explanation: 1. Half width of drum (distance from drum center to flange) = 2.4 m / 2 = 1.2 m.
2. Fleet angle α is calculated as tan(α) = (Half drum width) / (Distance to sheave) = 1.2 m / 45 m = 0.02667.
3. α = arctan(0.02667) = 1.528° ≈ 1.53°.
Note: This is right at the standard engineering maximum limit of 1.5° to prevent rope scrubbing.
5A cylindrical drum winder has a total equivalent moment of inertia of 120,000 kg·m² at the drum shaft. The drum diameter is 4.5 m. If the winder accelerates the conveyances at a rate of 1.2 m/s², calculate the torque required at the drum shaft strictly to overcome rotational acceleration.
A.64.0 kN·m
B.128.0 kN·m
C.32.0 kN·m
D.270.0 kN·m
Explanation: 1. Drum radius R = 4.5 m / 2 = 2.25 m.
2. Angular acceleration α = linear acceleration / radius = 1.2 m/s² / 2.25 m = 0.5333 rad/s².
3. Acceleration torque T_acc = I × α = 120,000 kg·m² × 0.5333 rad/s² = 64,000 N·m = 64.0 kN·m.
6Under South African DMRE regulations, what is the minimum required braking torque capacity for mechanical emergency brakes on a rock-winding drum engine relative to the maximum static out-of-balance torque?
A.2.0 times maximum out-of-balance static torque
B.1.5 times maximum out-of-balance static torque
C.1.0 times maximum out-of-balance static torque
D.3.0 times maximum out-of-balance static torque
Explanation: DMRE regulations specify that the mechanical braking system of a winding engine must be capable of holding the conveyance under maximum out-of-balance conditions with a safety factor, producing a braking torque of at least 2.0 times the maximum static out-of-balance torque.
7What primary safety function does an Ormerod or King type detaching hook serve in a vertical mine shaft headgear?
A.It detaches the winding rope from the conveyance during an overwind and suspends the conveyance on catch bells/plates.
B.It automatically releases the skip payload into the headgear bin when reaching the surface bank.
C.It absorbs shock loads caused by emergency braking at the shaft bottom.
D.It balances rope tension in multi-rope Koepe friction winding installations.
Explanation: A detaching hook (such as Ormerod or King design) is designed to open when pulled through a copper/steel catch plate in the headgear during an overwind. It releases the winding rope to prevent it breaking and simultaneously locks into the catch plate, holding the conveyance safely suspended.
8A skip carrying 12,000 kg of ore has a skip mass of 7,000 kg and suspended rope mass of 10,000 kg. If the winder accelerates upward at 1.8 m/s², calculate the dynamic tension load on the rope at the sheave and the corresponding dynamic safety factor if rope breaking force is 3,200 kN. (g = 9.81 m/s²).
A.Dynamic load = 336.7 kN, Dynamic Safety Factor = 9.50
B.Dynamic load = 284.5 kN, Dynamic Safety Factor = 11.25
C.Dynamic load = 388.2 kN, Dynamic Safety Factor = 8.24
D.Dynamic load = 420.0 kN, Dynamic Safety Factor = 7.62
Explanation: 1. Total suspended mass M = 12,000 kg + 7,000 kg + 10,000 kg = 29,000 kg.
2. Total upward acceleration = g + a = 9.81 m/s² + 1.8 m/s² = 11.61 m/s².
3. Dynamic tension load F_dyn = 29,000 kg × 11.61 m/s² = 336,690 N = 336.69 kN ≈ 336.7 kN.
4. Dynamic Safety Factor = Breaking Force / Dynamic Load = 3,200 kN / 336.69 kN = 9.504 ≈ 9.50.
9What is the key mechanical advantage of a bi-cylindro-conical (BCC) winder drum compared to a plain cylindrical drum in deep-shaft winding?
A.It reduces peak motor torque during initial acceleration by starting winding on the small diameter while unwinding off the large diameter.
B.It eliminates the need for brake paths on the winder drum sides.
C.It allows rope fleet angles exceeding 5 degrees without rope wear.
D.It increases maximum winding velocity by a factor of three during mid-wind.
Explanation: Bi-cylindro-conical (BCC) drums feature small end drums connected via scroll cones to a large center drum. Starting the wind of the loaded conveyance on the small diameter minimizes the torque required during acceleration, while the descending empty conveyance unwinds from the large diameter, balancing peak power requirements.
10In deep vertical shafts, fixed steel guide systems (top-hat or rectangular sections) are often preferred over rope guides under which of the following operating conditions?
A.High winding speeds (> 15 m/s) with small shaft clearances between conveyances.
B.Ultra-deep shafts (> 2,500 m) where rope guide weight becomes negligible.
C.Shafts with high seismic ground movement where rigid steel guides never deform.
D.Installations where initial capital cost must be minimized.
Explanation: Fixed steel guide systems provide rigid positioning and tight tolerance tracking, preventing conveyance sway and collision when operating at high winding speeds (> 15 m/s) in shafts with small clearances.
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