Assume the relation PPV = K [ (D / Q) ^ β ]. The PPV in mm/s, at a distance of 200 m from the blast face, is __________ (round off up to 2 decimals).

Steps to Solve

1. Identify the Given Values

We have the following values from the problem:

• Distance from the blast face for Location A, $D_A = 100 , m$ with PPV $= 112.5 , mm/s$.
• Distance from the blast face for Location B, $D_B = 300 , m$ with PPV $= 20.3 , mm/s$.
• Maximum charge per delay, $Q = 1200 , kg$.
• We need to calculate PPV at a distance $D = 200 , m$.

2. Establish the Relation

The Peak Particle Velocity (PPV) is given by the equation:

$$ PPV = K \left( \frac{D}{\sqrt{Q}} \right)^\beta $$

We can set up two equations using the known PPV values for Locations A and B to find the constants $K$ and $\beta$.

3. Setting Up the Equations for Location A

For Location A, substituting the known values into the formula, we get:

$$ 112.5 = K \left( \frac{100}{\sqrt{1200}} \right)^\beta $$

4. Setting Up the Equations for Location B

For Location B, substituting the known values gives:

$$ 20.3 = K \left( \frac{300}{\sqrt{1200}} \right)^\beta $$

5. Dividing the Equations to Eliminate K

We can divide the equation for Location A by the equation for Location B to eliminate $K$:

$$ \frac{112.5}{20.3} = \left( \frac{100}{\sqrt{1200}} \right)^\beta / \left( \frac{300}{\sqrt{1200}} \right)^\beta $$

Simplifying this leads to:

$$ \frac{112.5}{20.3} = \left( \frac{100}{300} \right)^\beta $$

6. Solving for β

By simplifying further:

$$ \frac{112.5}{20.3} = \left( \frac{1}{3} \right)^\beta $$

Now take the logarithm of both sides to solve for $\beta$:

$$ \log \left( \frac{112.5}{20.3} \right) = - \beta \log(3) $$

Thus,

$$ \beta = - \frac{\log \left( \frac{112.5}{20.3} \right)}{\log(3)} $$

7. Calculating K

Now we will substitute $\beta$ back into the equation for Location A to find $K$.

8. Calculating PPV for D = 200 m

Finally, substitute the values of $K$ and $\beta$ into the PPV equation for $D = 200 , m$ to find the final result:

$$ PPV_{200} = K \left( \frac{200}{\sqrt{1200}} \right)^\beta $$

9. Final Calculation

Perform the calculations to find the exact value and round off to 2 decimals.