Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 2 / i

Solution

Write the distributed fragment power-law size distribution as

n (D_{f}) = K_{f} D_{f}^{- α_{f}}

. A spherical fragment has mass

π ρ D_{f}^{3} /6

. Since this population contains one half of the target mass,

\frac{1}{2} \frac{π ρ D _{t}^{3}}{6} = \frac{π ρ K _{f}}{6} \int_{D_{m i n, f}}^{D_{m a x, f}} D_{f}^{3 - α_{f}} d D_{f},

and hence

K_{f} = \frac{D _{t}^{3}}{2} \frac{4 - α _{f}}{D _{m a x, f}^{4 - α_{f}} - D _{m i n, f}^{4 - α_{f}}} .

The distributed fragments have total geometric cross-section

σ_{dist} = \frac{π K _{f}}{4} \int_{D_{m i n, f}}^{D_{m a x, f}} D_{f}^{2 - α_{f}} d D_{f}

= \frac{π D _{t}^{3}}{8} \frac{4 - α _{f}}{α _{f} - 3} \frac{D _{m i n, f}^{3 - α_{f}} - D _{m a x, f}^{3 - α_{f}}}{D _{m a x, f}^{4 - α_{f}} - D _{m i n, f}^{4 - α_{f}}} .

The single fragment containing the other half of the mass has diameter

2^{- 1/3} D_{t}

and cross-section

π D_{t}^{2} / (4 2^{2/3})

. Therefore the exact result within the model is

σ_{tot, 1} = \frac{π D _{t}^{2}}{4 2 ^{2/3}} + σ_{dist} .

For

D_{m a x, f} ≫ D_{m i n, f}

and

3 < α_{f} < 4

, area is dominated by the smallest distributed fragments while mass is dominated by the largest, so normally the first term is negligible and

σ_{tot, 1} ≃ \frac{π D _{t}^{3}}{8} \frac{4 - α _{f}}{α _{f} - 3} D_{m i n, f}^{3 - α_{f}} D_{m a x, f}^{α_{f} - 4} .

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