Question

In a bench blasting, as shown, 49 kg of explosive is put in each hole. The detonation time of holes, in millisecond, and the associated scale distance (SD) vs. peak particle velocity (PPV) plot for the given blast pattern is shown.

The PPV at a distance of 100 m, in \( \text{mm}\,\text{s}^{-1} \), is . (rounded off to two decimal places)

Hint:

Find the largest charge weight that detonates in the same delay from the firing time diagram, use it to scale the distance, then plug into the given PPV versus scaled distance equation.

Answer 1

Correct Answer: 7.79

Step 1: Read the delay pattern from the figure.
The layout shows two rows of holes. The row nearer the free face fires at 0, 25, 50 and 75 ms after initiation, and the back row fires at 50, 75 and 100 ms. Every hole carries the same charge of 49 kg.

Step 2: Find the maximum charge firing at any single instant.
Ground vibration is governed by the largest charge that detonates within the same delay interval, since that is what produces the highest single vibration pulse. Looking at the delay numbers, two holes share the 50 ms mark (one from each row) and two holes share the 75 ms mark, so at those instants 49 + 49 = 98 kg of explosive fires together. This 98 kg is the maximum charge per delay, \( W \).

Step 3: Calculate the scaled distance.
The scale distance combines the monitoring distance and the charge per delay as \[ SD = \dfrac{D}{\sqrt{W}} = \dfrac{100}{\sqrt{98}} \approx \dfrac{100}{9.90} \approx 10.10\ \text{m}\,\text{kg}^{-0.5} \]

Step 4: Apply the given attenuation law.
The plot gives the site's vibration prediction equation, \( PPV = 250\,(SD)^{-1.5} \). Substituting the scaled distance: \[ PPV = 250 \times (10.10)^{-1.5} \approx \dfrac{250}{32.11} \approx 7.79\ \text{mm/s} \]

Final Answer:
The peak particle velocity expected 100 m from the blast is about 7.79 mm/s. \[ \boxed{7.79\ \text{mm/s}} \]