The breakage function Numerical Example

 

Problem Statement:

The breakage function in a crusher machine is given by:

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

If size class 2 has mesh sizes boundaries 13.4 cm and 9.5 cm and size class 5 has boundaries 7.50 cm and 5.60 cm, calculate b52

Solution:

Calculate $b_{52}$, the breakage distribution from size class 5 to size class 2, where:

• Size class 2 has mesh size boundaries $D_2 = 13.4 \, \text{cm}$ and $D_3 = 9.5 \, \text{cm}$,
• Size class 5 has mesh size boundaries $D_5 = 7.5 \, \text{cm}$ and $D_6 = 5.6 \, \text{cm}$.

The formula for calculating $b_{ij}$ is:

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

Where:

• $d_{p2} = \frac{13.4 + 9.5}{2} = 11.28 \, \text{cm}$ (the average size in size class 2).

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Solution:

Substitute the given breakage function into the formula for $b_{52}$:

$$b_{52} = 0.3 \left(\frac{7.5}{11.28}\right)^{0.45} + 0.7 \left(\frac{7.5}{11.28}\right)^{3.2} - 0.3 \left(\frac{5.6}{11.28}\right)^{0.45} - 0.7 \left(\frac{5.6}{11.28}\right)^{3.2}$$

Let us calculate this step by step.

The calculated value of $b_{52}$ is approximately 0.1403.

This represents the fraction of particles in size class 5 that break into size class 2.