Question 6 of 6: Shaft Hoist Design — Rope, Drum, Cycle Time and Motor Power

Question 6: Shaft Hoist Design — Rope, Drum, Cycle Time and Motor Power (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The rope diameter (6.01), rope weight (6.02), drum diameter (6.03), winds/hour and cycle time (6.04), the velocity–time diagram (6.05), steady-state hoisting velocity (6.06), maximum drum rpm (6.07), average linear and angular acceleration (6.08–6.09), motor speed and gearbox ratio (6.10–6.11), maximum static rope load (6.12), and the steady-state, acceleration and maximum horsepower (6.13–6.15).

Approach. Select the smallest available rope diameter that clears the assumed 7.5:1 static safety factor against its own weight plus the loaded skip (6.01–6.02); size the drum from the given D/d ratio (6.03); derive winds/hour directly from the required production rate and cycle time from the reciprocal (6.04); close the trapezoidal velocity–time profile against the shaft depth to solve for the steady-state hoisting velocity (6.05–6.06), from which every kinematic quantity (6.07–6.09) and, via the motor's synchronous speed, the gearbox ratio (6.10–6.11) follow; then combine the maximum static rope load (6.12) with the steady velocity and the average acceleration to get the three horsepower figures (6.13–6.15).

1. Part 6.01–6.02 — select the rope diameter and find its weight. For each candidate diameter, the breaking load and unit weight follow from the given formulas, and the maximum static load is the loaded skip (22 t) plus the full 450 m of suspended rope:

   Rope selection (breaking load / static load, tonnes)

   d (mm)	BL = 0.07625d² (t)	w = 0.00577d² (kg/m)	Rope wt over 450 m (t)	Static load, skip+rope (t)	Factor of safety
   47.6	172.8	13.07	5.88	27.88	6.20
   50.8	196.8	14.89	6.70	28.70	6.86
   54.0	222.3	16.83	7.57	29.57	7.52
   63.5	307.5	23.27	10.47	32.47	9.47

   The two smaller ropes fall short of the assumed 7.5:1 minimum; the 54.0 mm (2.125 in) rope is the smallest that clears it, at SF = 7.52. $$\boxed{d = 54.0\ \text{mm}, \qquad w = 0.00577(54.0)^2 = 16.83\ \text{kg/m}}$$
2. Part 6.03 — drum diameter. $$D_{drum} = 108\,d = 108 \times 0.0540 = \boxed{5.832\ \text{m}}$$
3. Part 6.04 — winds per hour and cycle time. Each wind delivers the 10 t payload, so at 500 t/hr: $$\text{winds/hr} = \dfrac{500}{10} = 50, \qquad t_{cycle} = \dfrac{3600\ \text{s}}{50} = \boxed{72\ \text{s}}$$
4. Part 6.05 — velocity–time diagram. The cycle is a trapezoid: linear acceleration for 12 s, constant velocity V for the remaining travel time, linear deceleration for 12 s, then the 10 s decking dwell at zero velocity before the next wind begins.
   Velocity–time profile for one 72 s hoisting cycle: 0–12 s accelerate, 12–50 s constant V = 8.5 m/s, 50–62 s decelerate, 62–72 s decking dwell.
5. Part 6.06 — steady-state hoisting velocity. The shaft depth equals the area under the trapezoid: accel and decel each cover $\tfrac12 V t_{acc}$, and the constant-speed portion covers $V(t_{cycle} - t_{acc} - t_{dec} - t_{deck})$: $$D_{shaft} = \tfrac12 V t_{acc} + \tfrac12 V t_{dec} + V(t_{cycle}-t_{acc}-t_{dec}-t_{deck}) = V(t_{cycle} - t_{acc} - t_{dec} - t_{deck} + t_{acc})$$ Substituting $t_{cycle}=72$, $t_{acc}=t_{dec}=12$, $t_{deck}=10$ simplifies the bracket to $(72 - 12 - 10) = 50$ s of effective travel time: $$V = \dfrac{425}{50} = \boxed{8.5\ \text{m/s}}$$ Check: accel + decel distance $=2\times\tfrac12(8.5)(12)=102$ m; constant-speed distance $=8.5\times(72-12-12-10)=8.5\times38=323$ m; total $=102+323=425$ m, matching the shaft depth exactly.
6. Part 6.07 — maximum drum rpm. At V = 8.5 m/s the drum surface speed equals V, so $$n_{drum} = \dfrac{V}{\pi D_{drum}} \times 60 = \dfrac{8.5}{\pi(5.832)}\times 60 = \boxed{27.84\ \text{rpm}}$$
7. Part 6.08 — average linear acceleration. $$a = \dfrac{V}{t_{acc}} = \dfrac{8.5}{12} = \boxed{0.708\ \text{m/s}^2}$$
8. Part 6.09 — average angular acceleration of the drum. $$\alpha = \dfrac{a}{D_{drum}/2} = \dfrac{0.708}{2.916} = \boxed{0.243\ \text{rad/s}^2}\ \ (\approx 2.32\ \text{rpm/s})$$
9. Part 6.10 — motor speed at steady state. With 8 pole pairs (16 poles) on a 60 Hz Canadian supply, the motor's synchronous speed — and, to a first approximation, its steady running speed — is $$n_{motor} = \dfrac{120f}{\text{poles}} = \dfrac{120(60)}{16} = \boxed{450\ \text{rpm}}$$
10. Part 6.11 — gearbox ratio at steady state. $$\text{ratio} = \dfrac{n_{motor}}{n_{drum}} = \dfrac{450}{27.84} = \boxed{16.2:1}$$
11. Part 6.12 — maximum static load on the rope. The maximum static tension occurs with the loaded skip at the shaft bottom and the full 450 m of rope suspended, exactly the load used to select the rope in Step 1: $$F_{static} = (12+10+7.57)\ \text{t} = 29.57\ \text{t} = 29{,}570\ \text{kg} \times 9.81 = \boxed{290.1\ \text{kN}}$$
12. Part 6.13 — horsepower at steady state, HP(M)₃. At constant velocity the motor need only overcome gravity on the maximum static load found in Step 11 (the worst case, skip at the bottom of its travel): $$P_3 = F_{static}\,V = (290{,}100)(8.5) = 2{,}465.8\ \text{kW}$$ $$\boxed{HP(M)_3 = \dfrac{2{,}465.8}{0.7355} = 3{,}352.6\ \text{HP(M)}}$$
13. Part 6.14 — horsepower to accelerate the maximum static load, HP(M)₁. The inertial force needed to accelerate the full static mass at the average linear acceleration, evaluated at the instant it reaches full speed V (the point of maximum accelerating power): $$F_{inertia} = m\,a = (29{,}570\ \text{kg})(0.708\ \text{m/s}^2) = 20{,}946\ \text{N}$$ $$P_1 = F_{inertia}\,V = (20{,}946)(8.5) = 178.0\ \text{kW}$$ $$\boxed{HP(M)_1 = \dfrac{178.0}{0.7355} = 242.1\ \text{HP(M)}}$$
14. Part 6.15 — estimated maximum horsepower, HP(M)_max. The motor sees its greatest demand at the instant it is still accelerating the maximum static load and has just reached full speed — it must supply the steady-state (gravity) power of Step 12 and the inertial power of Step 13 simultaneously: $$P_{max} = P_1+P_3 = 178.0+2465.8 = 2{,}643.9\ \text{kW}$$ $$\boxed{HP(M)_{max} = HP(M)_1+HP(M)_3 = 242.1+3{,}352.6 = 3{,}594.6\ \text{HP(M)}}$$