Introduction & Context

The Cost per Ton of Size Reduction metric is a fundamental economic indicator in process engineering that normalizes all annualized operating and capital expenses associated with comminution equipment to the mass of material treated. It enables direct comparison among competing size-reduction technologies (e.g., disc mills, ball mills, roller presses) irrespective of their absolute throughput or installed power. The metric is routinely used in:

Process selection during pre-feasibility studies
Operating-budget forecasting for existing plants
Energy-efficiency benchmarking against industry best-practice curves
Life-cycle cost (LCC) optimization when integrating renewable-energy credits or carbon-pricing schemes

Methodology & Formulas

All monetary values are expressed in base-currency units; energy in kWh; throughput in t a⁻¹; time in calendar years. The derivation proceeds in four stages:

Annual Energy Cost
\[ C_{\text{energy}} = E_{\text{spec}} \cdot P_{\text{energy}} \cdot \dot{m} \] where
$E_{\text{spec}}$ = specific energy consumption (kWh t⁻¹)
$P_{\text{energy}}$ = unit energy price (currency kWh⁻¹)
$\dot{m}$ = nominal throughput (t a⁻¹)
Annual Labor Cost
\[ C_{\text{labor}} = R_{\text{labor}} \cdot H_{\text{labor}} \] where
$R_{\text{labor}}$ = hourly labor rate (currency h⁻¹)
$H_{\text{labor}}$ = scheduled labor hours per year (h a⁻¹)
Annual Depreciation
\[ C_{\text{dep}} = \frac{I_{\text{cap}}}{L} \] where
$I_{\text{cap}}$ = initial capital expenditure (currency)
$L$ = depreciable life (a)
Total Annualized Cost
\[ C_{\text{annual}} = C_{\text{energy}} + C_{\text{maintenance}} + C_{\text{wear}} + C_{\text{labor}} + C_{\text{dep}} \]
Unit Cost of Size Reduction
\[ \text{Cost per ton} = \frac{C_{\text{annual}}}{\dot{m}} \]

Validity Regimes
Parameter	Constraint	Action if Violated
Throughput $\dot{m}$	$\dot{m} > 0$	Clamp to 1 × 10⁻⁹ t a⁻¹ to avoid division by zero
Lifespan $L$	$L > 0$	Clamp to 1 a

The final comparison between two technologies A and B is expressed as the percentage difference:

\[ \Delta = \frac{\text{Cost}_{\text{B}} - \text{Cost}_{\text{A}}}{\text{Cost}_{\text{B}}} \times 100\% \]

A positive $\Delta$ indicates that technology A is cheaper on a per-ton basis.

Sum all direct energy, media, and liner costs for the period.
Divide by the tons of fresh feed that actually reached the target size (not total throughput).
Include ancillary power (classifier, pumps, conveyors) if they scale with tonnage.
Express in $/t of new minus-mesh material to keep comparisons fair across shifts or campaigns.

Low mill utilization: every idle hour spreads fixed energy and media charges over fewer tons.
Coarse recycle loads also raise specific energy; aim for steady-state circulating load below design limit.

Normalize to the same P80 and same local power price before comparing.
Use the Bond/CEEC database for hard-rock SABC circuits; use the VRM database for cement applications.
Adjust for ore work index: multiply reference cost by (Wi_plant ÷ Wi_ref)² to remove ore-hardness bias.

Not always; high-chrome costs more upfront but wears slower.
Run a 30-day trial: log consumption rate, power draw, and final tonnage.
Calculate cost per ton for both media types; include downstream benefits such as less scrap iron reporting to flotation.

Worked Example – Cost per Ton of Size Reduction for a Disc Mill vs. a Ball Mill

A mineral processing plant is comparing two grinding options for a 365,000-ton-per-year feed: a disc mill and a ball mill. The goal is to determine the annualized cost per ton of material processed for each mill and the percentage cost advantage of the disc mill.

Known Input Parameters

Energy price = 0.1 $/kWh
Equipment lifespan = 10 years
Disc mill – Energy consumption = 100 kWh/ton
Disc mill – Maintenance cost = 5,000 $/year
Disc mill – Wear parts cost = 3,000 $/year
Disc mill – Labor rate = 20 $/hour
Disc mill – Labor hours = 2,000 hours/year
Disc mill – Initial capital cost = 100,000 $
Ball mill – Energy consumption = 150 kWh/ton
Ball mill – Maintenance cost = 8,000 $/year
Ball mill – Wear parts cost = 6,000 $/year
Ball mill – Labor rate = 25 $/hour
Ball mill – Labor hours = 2,500 hours/year
Ball mill – Initial capital cost = 150,000 $
Throughput (both mills) = 365,000 tons/year

Step-by-Step Calculation

Calculate the annual energy cost for each mill: \[ \text{Energy Cost} = \text{Energy Consumption} \times \text{Energy Price} \times \text{Throughput} \]
- Disc mill: $100 \times 0.1 \times 365{,}000 = 3{,}650{,}000$ $
- Ball mill: $150 \times 0.1 \times 365{,}000 = 5{,}475{,}000$ $
Calculate the annual labor cost: \[ \text{Labor Cost} = \text{Labor Rate} \times \text{Labor Hours} \]
- Disc mill: $20 \times 2{,}000 = 40{,}000$ $
- Ball mill: $25 \times 2{,}500 = 62{,}500$ $
Determine straight-line depreciation (capital cost spread over lifespan): \[ \text{Depreciation} = \frac{\text{Initial Cost}}{\text{Lifespan}} \]
- Disc mill: $100{,}000 / 10 = 10{,}000$ $ per year
- Ball mill: $150{,}000 / 10 = 15{,}000$ $ per year
Sum all annual cost components (energy, labor, depreciation, maintenance, wear parts) to obtain the total annual cost: \[ \text{Annual Cost} = \text{Energy Cost} + \text{Labor Cost} + \text{Depreciation} + \text{Maintenance} + \text{Wear Parts} \]
- Disc mill: $3{,}650{,}000 + 40{,}000 + 10{,}000 + 5{,}000 + 3{,}000 = 3{,}708{,}000$ $
- Ball mill: $5{,}475{,}000 + 62{,}500 + 15{,}000 + 8{,}000 + 6{,}000 = 5{,}566{,}500$ $
Calculate the cost per ton of material processed: \[ \text{Cost per Ton} = \frac{\text{Annual Cost}}{\text{Throughput}} \]
- Disc mill: $3{,}708{,}000 / 365{,}000 = 10.159$ $/ton
- Ball mill: $5{,}566{,}500 / 365{,}000 = 15.251$ $/ton
Determine the percentage cost difference between the two mills: \[ \%\text{Diff} = \frac{\text{Cost per Ton}_{\text{Ball}} - \text{Cost per Ton}_{\text{Disc}}}{\text{Cost per Ton}_{\text{Disc}}} \times 100 \] \[ \%\text{Diff} = \frac{15.251 - 10.159}{10.159} \times 100 = 33.387\% \]

Final Answer

Disc mill cost per ton = 10.159 $/ton
Ball mill cost per ton = 15.251 $/ton
Disc mill is 33.387 % cheaper on a per-ton basis than the ball mill.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle

Parameter	Constraint	Action if Violated
Throughput \(\dot{m}\)	\(\dot{m} > 0\)	Clamp to 1 × 10⁻⁹ t a⁻¹ to avoid division by zero
Lifespan \(L\)	\(L > 0\)	Clamp to 1 a