Question

The information for a dragline operation is given:
Bucket capacity: 20 \( \text{m}^3 \)
Bucket fill factor: 0.9
Digging and filling time: 20 s
Swinging (to and fro) time: 32 s
Dumping time: 10 s
Dragline utilization: 90%
The output of the dragline, in \( \text{m}^3\,\text{h}^{-1} \), is . (rounded off to one decimal place)

Hint:

Add up the digging, swinging and dumping times for one full cycle, find cycles per hour, then multiply by the effective bucket load and the utilization factor.

Answer 1

Correct Answer: 940.6

Step 1: Find the effective load per bucket.
The bucket rarely fills all the way to its rated capacity, so the fill factor scales it down to the load actually dug each cycle: \[ V_{eff} = 20 \times 0.9 = 18\ \text{m}^3 \]

Step 2: Find the total cycle time.
One full dragline cycle is digging and filling, then swinging out to the dump point and back, then dumping. The swing time given, 32 s, is already the full to and fro swing, so it is used as it is, not doubled: \[ t_{cycle} = 20 + 32 + 10 = 62\ \text{s} \]

Step 3: Find the theoretical number of cycles per hour.
There are 3600 s in an hour, so the number of cycles possible in one hour is \[ n = \dfrac{3600}{62} \approx 58.06\ \text{cycles/h} \]

Step 4: Find the theoretical and actual output.
The theoretical hourly output is cycles per hour times the effective bucket volume: \[ Q_{theo} = 58.06 \times 18 \approx 1045.16\ \text{m}^3/\text{h} \] The dragline is not digging every minute of the hour, only 90% of the time (utilization), so the actual achievable output is: \[ Q = 1045.16 \times 0.90 \approx 940.6\ \text{m}^3/\text{h} \]

Final Answer:
The dragline delivers about 940.6 cubic metres of material per hour. \[ \boxed{940.6\ \text{m}^3/\text{h}} \]