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: Understanding the Question:
A silo drops coal into wagons as they roll past underneath. It cannot load anything while the gap between two wagons is passing, so we need the rate at which mass is delivered averaged over a full wagon plus gap cycle.

Step 2: Key Formula or Approach:
Find how many wagon-plus-gap lengths pass the silo in one hour, then multiply by the mass carried in each wagon. This gives $\text{Rate} = (\text{wagons passing per hour}) \times (\text{mass per wagon})$.

Step 3: Detailed Explanation:
One wagon plus its trailing gap covers $8\ m + 1\ m = 9\ m$ of track (the gap of 100 cm equals 1 m).
The rake speed is $0.4\ km/h = 400\ m/h$.
So the number of such 9 m units passing the silo in one hour is $400 / 9 = 44.44$ per hour.
Each wagon carries a coal volume of $8 \times 3 \times 3 \times 0.95 = 68.4\ m^3$, filled to only 0.95 of the box, and at a bulk density of 1.2 tonne/m3 this is a mass of $68.4 \times 1.2 = 82.08$ tonne.
Multiplying the wagons per hour by the mass each one carries, $\text{Rate} = 44.44 \times 82.08 = 3648$ tonne/hour.

Step 4: Final Answer:
The silo delivers coal into the rake at about 3648 tonnes per hour.