Question

In a bord and pillar panel six shuttle cars, each of 10 tonne capacity, are deployed to transport coal produced by two continuous miners to a belt conveyor. Each shuttle car on an average carries 80% of its rated capacity and makes 7 round trips in an hour. The belt conveyor has a capacity such that the effective material cross section area is of 0.09 m$^{2$ and runs at a speed 1.1 m/s. The broken coal has a bulk density of 1.2 tonne/m$^{3}$. The ratio between the production and the capacity of the belt conveyor, in percent is}

• A. 65.46
• B. 71.42
• C. 78.56
• D. 82.46

Hint:

Belt capacity in tonne/hour = cross-sectional area × speed × bulk density × 3600.

Answer 1

The Correct Option is C

Step 1: Production from shuttle cars.
Each shuttle car capacity = 10 tonne.
Effective load due to 80% filling = $10 \times 0.8 = 8$ tonne/trip.
Trips per hour = 7.
Six shuttle cars total hourly production:
$P = 6 \times 8 \times 7 = 336$ tonne/hour.

Step 2: Belt conveyor capacity.
Cross-sectional area = 0.09 m$^{2}$.
Speed = 1.1 m/s.
Volumetric flow = $0.09 \times 1.1 = 0.099$ m$^{3}$/s.
In tonne/s: $0.099 \times 1.2 = 0.1188$ tonne/s.
Convert to tonne/hour:
$0.1188 \times 3600 = 427.68$ tonne/hour.

Step 3: Ratio of production to belt capacity.
Ratio = $\dfrac{336}{427.68} \times 100 = 78.56%$.
Thus, the required ratio is 78.56%.