Illustrative Example 5.2: Calculation of b 52

 Illustrative Example 5.2:

The breakage function for a crusher machine is given by:

$$B(x:y) = 0.3 + 0.7 \left(\frac{y}{x}\right)^{0.45} \exp\left(-3.2 \frac{y}{x}\right)$$

Given that size class 2 has mesh size boundaries of 13.4 cm and 9.5 cm, and size class 5 has boundaries of 7.50 cm and 5.60 cm, calculate $b_{52}$.

Illustrative Example 5.2: Calculation of $b_{52}$

Problem Description

The breakage function for a crusher is defined as:

$$B(x:y) = 0.3 + 0.7 \left(\frac{y}{x}\right)^{0.45} \exp\left(-3.2 \frac{y}{x}\right)$$

Here:

• Size class 2 has mesh size boundaries of 13.4 cm (upper limit) and 9.5 cm (lower limit).
• Size class 5 has mesh size boundaries of 7.50 cm (upper limit) and 5.60 cm (lower limit).

The goal is to calculate $b_{52}$ using the relationship:

$$b_{ij} = B(D_{i-1}:d_{pj}) - B(D_i:d_{pj})$$

Step-by-Step Solution

1. Determine the Representative Particle Size for Size Class 2 ($d_{p2}$):
   The representative size is the geometric mean of the boundaries:

   $$d_{p2} = \sqrt{D_{i-1} \cdot D_i} = \sqrt{13.4 \times 9.5} = 11.28 \, \text{cm}$$

2. Breakage Function for Size Class 5 ($b_{52}$):
   The formula for $b_{52}$ is given by:

   $$b_{52} = B(D_4:d_{p2}) - B(D_5:d_{p2})$$

   Substitute the breakage function for $B(x:y)$:

   $$B(x:y) = 0.3 + 0.7 \left(\frac{y}{x}\right)^{0.45} \exp\left(-3.2 \frac{y}{x}\right)$$

   For $B(D_4:d_{p2})$:

   • $D_4 = 7.5 \, \text{cm}$, $d_{p2} = 11.28 \, \text{cm}$:
   $$B(D_4:d_{p2}) = 0.3 + 0.7 \left(\frac{7.5}{11.28}\right)^{0.45} \exp\left(-3.2 \frac{7.5}{11.28}\right)$$

   Calculate step by step:

   $$\frac{7.5}{11.28} = 0.6651$$ $$(0.6651)^{0.45} = 0.8271, \quad \exp(-3.2 \times 0.6651) = \exp(-2.1283) = 0.1196$$ $$B(D_4:d_{p2}) = 0.3 + 0.7 (0.8271)(0.1196) = 0.3 + 0.0693 = 0.4393$$

   For $B(D_5:d_{p2})$:

   • $D_5 = 5.6 \, \text{cm}$, $d_{p2} = 11.28 \, \text{cm}$:
   $$B(D_5:d_{p2}) = 0.3 + 0.7 \left(\frac{5.6}{11.28}\right)^{0.45} \exp\left(-3.2 \frac{5.6}{11.28}\right)$$

   Calculate step by step:

   $$\frac{5.6}{11.28} = 0.4964$$ $$(0.4964)^{0.45} = 0.6748, \quad \exp(-3.2 \times 0.4964) = \exp(-1.5885) = 0.2041$$ $$B(D_5:d_{p2}) = 0.3 + 0.7 (0.6748)(0.2041) = 0.3 + 0.0962 = 0.3964$$

3. Calculate $b_{52}$:
   Subtract $B(D_5:d_{p2})$ from $B(D_4:d_{p2})$:

   $$b_{52} = 0.4393 - 0.2934 = 0.1459$$

Final Answer:

$$b_{52}=0.1459$$