Applications of the Linear Mass-Sectional Breakage Population Balance to Various Milling Process Configurations

ABSTRACT

After an initial grounding discussion, in which the linear mass-sectional population balance is described, and models for reducing the number of its parameters are discussed, this modeling approach is applied to a wide range of processes including batch mills, single-pass continuous mills, mills in series, mills with recycle, milling circuits including classification, and recycled-batch milling. The linearity of the model allows its straightforward inclusion in calculations for all of these processes. The sensitivity of the model to its inputs (order of rate kernel in size, fragment distribution) is explored for batch operations. Then, the effects of key continuous process design variables are explored including the (a) number of mill passes in pendulum milling (mills in series), (b) classifier properties in circuits with classification, and (c) ratio of feed tank-to-mill residence time in recycled-batch operations.

APPENDIX

The following fills in the gaps in developing the solution for the time-dependent mill discharge size distribution in getting from Eq. 46 to Eq. 47. Eq. 46 can be simplified to:

$$ {\overline{\mathbf{m}}}_{\mathbf{r}}(s)=\frac{{\mathbf{m}}_{\mathbf{o}}}{s}+\frac{\left(s+\frac{1}{\tau_T}\right)}{s\left(s+\frac{1}{\tau_M}+\frac{1}{\tau_T}\right)}{\mathbf{EDE}}^{-\mathbf{1}}{\overline{\mathbf{m}}}_{\mathbf{r}}(s) $$

(52)

Multiplying by the inverse of the eigenvalue matrix (i.e., by E⁻¹) and collecting terms leads to:

$$ \left(\mathbf{I}-\frac{s+\frac{1}{\tau_T}}{s\left(s+\frac{1}{\tau_M}+\frac{1}{\tau_T}\right)}\mathbf{D}\right){\mathbf{E}}^{-\mathbf{1}}{\overline{\mathbf{m}}}_{\mathbf{r}}(s)=\frac{{\mathbf{E}}^{-\mathbf{1}}{\mathbf{m}}_o}{s} $$

(53)

We define an inverse diagonal matrix via:

$$ {\overline{\mathbf{L}}}^{-\mathbf{1}}(s)=\left(\mathbf{I}-\frac{s+\frac{1}{\tau_T}}{s\left(s+\frac{1}{\tau_M}+\frac{1}{\tau_T}\right)}\mathbf{D}\right)\kern1.25em ;\kern1em {\overline{\mathbf{L}}}^{-\mathbf{1}}(s)={l}_{jk}^{-1}={\delta}_{jk}\times \left(1+\frac{s+\frac{1}{\tau_T}{S}_{br,k}}{s\left(s+\frac{1}{\tau_M}+\frac{1}{\tau_T}\right)}\right) $$

(54)

Therefore:

$$ {\displaystyle \begin{array}{c}{\overline{\mathbf{L}}}^{-\mathbf{1}}(s){\mathbf{E}}^{-1}{\overline{\mathbf{m}}}_{\mathbf{r}}(s)=\frac{{\mathbf{E}}^{-\mathbf{1}}{\mathbf{m}}_o}{s}\kern0.75em \Rightarrow \kern0.75em {\mathbf{E}}^{-\mathbf{1}}{\overline{\mathbf{m}}}_{\mathbf{r}}(s)=\frac{\overline{\mathbf{L}}(s)}{s}{\mathbf{E}}^{-\mathbf{1}}{\mathbf{m}}_o\Rightarrow \kern0.75em {\overline{\mathbf{m}}}_{\mathbf{r}}(s)=\mathbf{E}\left(\frac{\overline{\mathbf{L}}(s)}{s}\right){\mathbf{E}}^{-1}{\mathbf{m}}_o\\ {}\overline{\mathbf{L}}(s)={l}_{jk}={\delta}_{jk}/\left(1+\frac{\left(s+\frac{1}{\tau_T}\right){S}_{br,k}}{s\left(s+\frac{1}{\tau_M}+\frac{1}{\tau_T}\right)}\right)\end{array}} $$

(55)

This is then utilized in Eq. 47.

In a similar vein, the expression below fills in the gaps between substituting the time-dependent mill discharge size distribution into the mill feed equation, transforming to the Laplace domain, and rearranging to formally invert to the time domain with the final line being the result given in Eq. 49.

$$ {\displaystyle \begin{array}{c}{\mathbf{m}}_{\mathbf{f}}(t)={\mathbf{m}}_{\mathbf{o}}{e}^{-t/{\tau}_T}+\frac{1}{\tau_T}{\int}_0^t{e}^{-\left(t-\theta \right)/{\tau}_T}{\mathbf{m}}_{\mathbf{r}}\left(\theta \right) d\theta \kern0.75em \Rightarrow \kern0.75em {\overline{\mathbf{m}}}_{\mathbf{f}}(s)=\frac{1}{s+\frac{1}{\tau_T}}\left({\mathbf{m}}_{\mathbf{o}}+\frac{{\overline{\mathbf{m}}}_{\mathbf{r}}\left(\mathrm{s}\right)}{\tau_T}\right)\\ {}{\overline{\mathbf{m}}}_{\mathbf{f}}(s)=\frac{1}{s+\frac{1}{\tau_T}}\left(\mathbf{I}+\frac{1}{\tau_T}\mathbf{E}\left(\frac{\overline{\mathbf{L}}(s)}{s}\right){\mathbf{E}}^{-\mathbf{1}}\right){\mathbf{m}}_{\mathbf{o}}\\ {}{\mathbf{m}}_{\mathbf{f}}(t)=\left({e}^{-t/{\tau}_T}+\mathbf{EK}(t){\mathbf{E}}^{-\mathbf{1}}\right){\mathbf{m}}_{\mathbf{o}}\kern1em ;\kern1em \mathbf{K}(t)={\mathrm{L}}^{-1}\left\{\frac{1}{1+{\tau}_Ts}\frac{\overline{\mathbf{L}}(s)}{s}\right\}\end{array}} $$

(56)