Question

The following are the various parameters of the Shovel
Capacity of the bucket – 20 m\(^3\)
Fill factor (f) = 0.8
Swell factor (s) = 0.5
Cycle time (t) = 60 sec
The Volume of the material carried by the Shovel per hour is

• A. 300 m\(^3\)
• B. 388 m\(^3\)
• C. 400 m\(^3\)
• D. 480 m\(^3\)

Hint:

Be very careful with equipment productivity problems, as definitions of factors can vary.
Here, the calculation path to the answer shows that the required output is Bank Cubic Metres (BCM) and the "swell factor" is used as \( V_{bank} / V_{loose} \).
The standard formula for shovel output in LCM is: \( Q = (V \times f \times 3600) / t \). This would give 960 LCM/hr.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the hourly production of a shovel in terms of volume, given its operational parameters. The final answer will depend on whether the output is measured in loose volume or bank (in-situ) volume.

Step 2: Key Formula or Approach:
1. Calculate the actual loose volume per bucket cycle.
2. Calculate the number of cycles per hour.
3. Calculate the hourly production. The use of the swell factor suggests the question may be asking for the bank volume.
- Loose Volume per cycle = Bucket Capacity \( \times \) Fill Factor - Bank Volume per cycle = Loose Volume per cycle \( \times \) Swell Factor (Here, swell factor is defined as Bank Volume / Loose Volume, or \(1 / (1 + \text{swell \%})\)). - Cycles per hour = 3600 / Cycle time (in sec) - Production per hour = Volume per cycle \( \times \) Cycles per hour

Step 3: Detailed Explanation:

1. Calculate Loose Volume per Cycle:
The fill factor (0.8) indicates that the bucket is not completely full.
\[ \text{Loose Volume per Cycle} = 20 \, \text{m}^3 \times 0.8 = 16 \, \text{m}^3 \text{ (LCM)} \]

2. Apply Swell Factor:
The swell factor (0.5) relates the bank (in-situ) volume to the loose volume. The calculation that leads to the correct answer implies a definition where Bank Volume = Loose Volume \( \times \) Swell Factor. This is an unconventional definition but required to solve the problem as posed.
\[ \text{Bank Volume per Cycle} = 16 \, \text{m}^3 \times 0.5 = 8 \, \text{m}^3 \text{ (BCM)} \]

3. Calculate Cycles per Hour:
\[ \text{Cycles per hour} = \frac{3600 \, \text{sec/hr}}{60 \, \text{sec/cycle}} = 60 \, \text{cycles/hr} \]

4. Calculate Hourly Production:
The question asks for the volume of material carried per hour. Based on the calculation path, this refers to the bank volume.
\[ \text{Hourly Production (Bank Volume)} = 8 \, \text{m}^3/\text{cycle} \times 60 \, \text{cycles/hr} = 480 \, \text{m}^3/\text{hr} \]

Step 4: Final Answer:
The volume of the material (in bank measure) carried by the shovel per hour is 480 m\(^3\).
This corresponds to option (D).