Write the distributed fragment <power-law size distribution> as $n(D_f)=K_fD_f^{-\alpha_f}$. A spherical fragment has mass $\pi\rho D_f^3/6$. Since this population contains one half of the target mass,
$$
\frac12\frac{\pi\rho D_t^3}{6}
=\frac{\pi\rho K_f}{6}
\int_{D_{\min,f}}^{D_{\max,f}}D_f^{3-\alpha_f}\,dD_f,
$$
and hence
$$
K_f=\frac{D_t^3}{2}
\frac{4-\alpha_f}
{D_{\max,f}^{4-\alpha_f}-D_{\min,f}^{4-\alpha_f}}.
$$
The distributed fragments have total <geometric cross-section>
$$
\sigma_{\rm dist}
=\frac{\pi K_f}{4}
\int_{D_{\min,f}}^{D_{\max,f}}D_f^{2-\alpha_f}\,dD_f
$$
$$
=\frac{\pi D_t^3}{8}
\frac{4-\alpha_f}{\alpha_f-3}
\frac{D_{\min,f}^{3-\alpha_f}-D_{\max,f}^{3-\alpha_f}}
{D_{\max,f}^{4-\alpha_f}-D_{\min,f}^{4-\alpha_f}}.
$$
The single fragment containing the other half of the mass has diameter $2^{-1/3}D_t$ and cross-section $\pi D_t^2/(4\,2^{2/3})$. Therefore the exact result within the model is
$$
\boxed{\sigma_{{\rm tot},1}=
\frac{\pi D_t^2}{4\,2^{2/3}}+\sigma_{\rm dist}}.
$$
For $D_{\max,f}\gg D_{\min,f}$ and $3\lt\alpha_f\lt4$, area is dominated by the smallest distributed fragments while mass is dominated by the largest, so normally the first term is negligible and
$$
\boxed{\sigma_{{\rm tot},1}\simeq
\frac{\pi D_t^3}{8}
\frac{4-\alpha_f}{\alpha_f-3}
D_{\min,f}^{3-\alpha_f}D_{\max,f}^{\alpha_f-4}}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-316/2/i/solution