Question

A rake of wagons, with inter-wagon gap of 100 cm, is moving at a speed of 0.4 km per hour under a silo loading system. The dimensions of the wagon are 8 m (L) \(\times\) 3 m (W) \(\times\) 3 m (H). The silo stops discharging the material between the wagons. Considering the fill factor of the wagon as 0.95 and the bulk density of coal as 1.2 tonne per cubic meter, the loading rate of the silo, in tonne per hour, is . (rounded off to nearest integer)

Hint:

Work out the coal mass one wagon carries, then the time one wagon length plus the inter-wagon gap takes to pass the silo at the given speed.

Answer 1

Correct Answer: 3648

Step 1: Find the coal loaded into one wagon.
The wagon box has volume \( 8 \times 3 \times 3 = 72\ \text{m}^3 \), but it is not filled to the brim, only to 0.95 of that volume, so the actual coal volume per wagon is \( 72 \times 0.95 = 68.4\ \text{m}^3 \). Multiplying by the bulk density gives the mass loaded per wagon: \[ m = 68.4 \times 1.2 = 82.08\ \text{tonne} \]

Step 2: Find how long it takes one wagon to pass the loading point.
Because the silo stops discharging over the gap between wagons, the time available to load each wagon is the time it takes for one wagon length plus one gap to pass under the silo. This distance is \( 8 + 1 = 9\ \text{m} \) (the 100 cm gap is 1 m). The rake moves at \( 0.4\ \text{km/h} = 400\ \text{m/h} \), so the time per wagon is: \[ t = \dfrac{9}{400} = 0.0225\ \text{h} \]

Step 3: Find the loading rate.
The loading rate is the mass loaded per wagon divided by the time that wagon takes to pass: \[ \text{Rate} = \dfrac{82.08}{0.0225} = 3648\ \text{tonne/hour} \]

Final Answer:
The silo loads coal at a rate of about 3648 tonnes per hour. \[ \boxed{3648\ \text{tonne/hour}} \]