Question

Ore is hoisted from 620 m depth using a single skip of 7 tonne pay load. The skip winding system has constant acceleration/deceleration of 1 m/s\(^2\) and a constant speed of 10 m/s. The skip loading time and unloading time are 120 s and 60 s, respectively. Considering the overall utilization of the skip as 70%, the maximum daily capacity of the winding system, in tonne, is \(\underline{\hspace{1cm}}\). (round off to the nearest integer)

Hint:

For calculating maximum daily capacity, account for cycle times and skip utilization.

Answer 1

Correct Answer: 1250

The maximum hoisting speed \( V = 10 \, \text{m/s} \), and the hoisting depth \( H = 620 \, \text{m} \). The time to reach constant speed is given by:
\[ t_{\text{accel}} = \frac{V}{a} = \frac{10}{1} = 10 \, \text{s} \] The time to decelerate is the same: \( t_{\text{decel}} = 10 \, \text{s} \). So, the total time for the hoist is:
\[ t_{\text{hoist}} = \frac{H}{V} + t_{\text{accel}} + t_{\text{decel}} = \frac{620}{10} + 10 + 10 = 62 + 20 = 82 \, \text{s} \] The total time per cycle (loading + unloading):
\[ t_{\text{cycle}} = t_{\text{hoist}} + t_{\text{loading}} + t_{\text{unloading}} = 82 + 120 + 60 = 262 \, \text{s} \] The number of cycles per day (assuming 24 hours of operation):
\[ N_{\text{cycles}} = \frac{24 \times 3600}{t_{\text{cycle}}} = \frac{86400}{262} \approx 330.5 \] The total daily hoisted tonnage is:
\[ \text{Daily Capacity} = N_{\text{cycles}} \times \text{Pay Load} \times \text{Utilization} \] \[ = 330.5 \times 7 \times 0.70 = 1617.45 \, \text{tonnes} \] Rounded to the nearest integer:
\[ \boxed{1617} \, \text{tonnes} \]