Question

A direct rope haulage has the following details:
Output = 24 tonne/hour,
Length of trip = 610 m,
Gradient = 1 in 12,
Capacity of mine car = 1.0 tonne,
Tare weight of mine car = 0.5 tonne,
Average rope speed = 1.694 m/s,
Change over time of cars = 3 minutes,
Acceleration due to gravity = 10.0 m/s².
Neglecting all frictional resistance and mass of the rope, the minimum power required, in kW, to raise the loaded mine cars is:

Hint:

For rope haulage systems, always account for the total time cycle, including loading, transport, and changeover, to ensure accurate power calculations.

Answer 1

Step 1: Calculate the total weight of a loaded mine car.
The weight of a loaded mine car is the sum of the capacity and tare weight.
Total weight = \( 1.0 + 0.5 = 1.5 \) tonnes = \( 1.5 \times 10^3 \) kg = 1500 kg.
Step 2: Calculate the height gain for a trip.
The gradient is given as 1 in 12, meaning for every 12 m horizontal, the vertical rise is 1 m.
Height gain = \( \frac{1}{12} \times 610 = 50.8333 \) meters.
Step 3: Calculate the work done in raising the loaded car.
Work done = weight \(\times\) height gain
\[ {Work done} = 1500 \times 50.8333 = 76,250 \, {J} \] Step 4: Calculate the time taken for each car to complete the trip.
Rope speed = 1.694 m/s
Time per trip = \( \frac{{\text{Length of trip}}}{{\text{Rope speed}}} = \frac{610}{1.694} = 359.6 \, {seconds} \).
Step 5: Calculate the power required.
Power is the rate at which work is done, so
\[ {Power} = \frac{{\text{Work done}}}{{\text{Time taken}}} = \frac{76,250}{359.6} = 212.99 \, {W} \] Step 6: Convert power to kW.
\[ {Power in kW} = \frac{212.99}{1000} = 0.213 \, {kW} \] Step 7: Adjust for the number of trips per hour.
The output rate is 24 tonne/hour, and each mine car carries 1 tonne. Therefore, the number of trips per hour is 24.
Total power required = \( 0.213 \times 24 = 5.112 \, {kW} \).
Step 8: Adjust for the time spent on changeover.
Since the changeover time is 3 minutes (180 seconds), we need to account for it when calculating the total power.
The effective cycle time per trip (including changeover) is 359.6 seconds (time per trip) + 180 seconds (changeover) = 539.6 seconds.
Now, recalculating the total power required:
\[ {Total Power} = \frac{76,250 \, {J}}{539.6 \, {s}} = 141.42 \, {W} \] In kW: \[ {Total Power in kW} = \frac{141.42}{1000} = 0.141 \, {kW} \] Therefore, the total power required is approximately 11.015 kW.