Question

A coal heading is of 4 m wide and 2.5 m height has an advance of 1 m per cycle. The amount of explosive used is 5 kg per blast. Taking specific gravity of coal as 1.2 t/m\(^3\). The powder factor is

• A. 1.55 te/kg
• B. 2.40 te/kg
• C. 2.99 te/kg
• D. 3.32 te/kg

Hint:

Pay close attention to the units.
The desired unit for the powder factor is tonnes/kg.
Ensure your mass of coal is in tonnes and your mass of explosive is in kg before dividing.
Also, be aware that "specific gravity" given in units of t/m\(^3\) is effectively the density.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the powder factor for a blasting operation in a coal heading, given the dimensions of the heading, the advance per blast, and the amount of explosive used.

Step 2: Key Formula or Approach:
Powder Factor (PF) is a measure of the efficiency of explosive use.
It is defined as the mass of rock or coal broken per unit mass of explosive used.
\[ \text{Powder Factor} = \frac{\text{Mass of coal broken (in tonnes)}}{\text{Mass of explosive used (in kg)}} \] First, we need to calculate the volume of coal broken, and then its mass.

Step 3: Detailed Explanation:

1. Calculate the volume of coal broken per blast:
Volume = Width \( \times \) Height \( \times \) Advance
\[ \text{Volume} = 4 \, \text{m} \times 2.5 \, \text{m} \times 1 \, \text{m} = 10 \, \text{m}^3 \]

2. Calculate the mass of coal broken:
The specific gravity of coal is given as 1.2 t/m\(^3\).
This is the density of the coal.
Mass = Volume \( \times \) Density
\[ \text{Mass} = 10 \, \text{m}^3 \times 1.2 \, \text{t/m}^3 = 12 \, \text{tonnes (te)} \]

3. Calculate the Powder Factor:
Mass of explosive used = 5 kg
\[ \text{Powder Factor} = \frac{12 \, \text{te}}{5 \, \text{kg}} = 2.4 \, \text{te/kg} \]

Step 4: Final Answer:
The powder factor is 2.40 te/kg.
This corresponds to option (B).