Leach Tank Sizing Calculator
Sizes leach tanks for a stated residence time from the slurry flow, the required residence time τ, the number of tanks in series and a live-volume fraction — flow entered directly as Q or derived from a solids basis using the house slurry-density relation — giving total live volume, per-tank volume and optional flat-bottom-cylinder geometry.
Plain tank-volume arithmetic for a stated residence time, in two flow-input modes. Enter the slurry volumetric flow Q directly, or work from a solids basis — dry solids t/h, % solids by mass, solids SG and liquor SG — and the slurry density and flow are derived using the same relation the slurry-density calculators publish, with every step shown. Then the total live volume is Q·τ and the per-tank volume divides that across the tanks in series and the live-volume fraction. Optional flat-bottom-cylinder geometry returns the per-tank diameter and height from an H/D ratio. SI primary throughout. The required residence time comes from metallurgical testwork for the specific ore and conditions; this calculator sizes vessels for a stated residence time — it does not predict leach performance.
Calculator
From metallurgical testwork for the specific ore and conditions.
Working volume ÷ geometric volume. Default 0.85 — a labelled assumption to verify.
Flat-bottom cylinder basis. Default 1.0 — a labelled assumption. No freeboard beyond f_live.
The required residence time comes from metallurgical testwork for the specific ore and conditions; this calculator sizes vessels for a stated residence time — it does not predict leach performance.
Audit trail
- Slurry flow Q = 187.04 m³/h (entered directly)
- V_live_total = Q · τ = 187.04 × 24 = 4488.96 m³
- V_tank = V_live_total / (N · f_live) = 4488.96 / (6 × 0.85) = 880.188 m³
- D = (4·V_tank / (π·(H/D)))^(1/3) = 10.3871 m; H = (H/D)·D = 10.3871 m (flat-bottom cylinder; no freeboard/agitation allowance beyond f_live)
Worked example (solids basis): 100 t/h dry solids, 40 % w/w, solids SG 2.70, liquor SG 1.000 → slurry SG 1.337, slurry mass 250 t/h, Q ≈ 187.0 m³/h; τ = 24 h, N = 6, f_live = 0.85 → V_tank ≈ 880 m³ (D = H ≈ 10.4 m at H/D = 1.0). The same case loads by default above.
Slurry density is derived via the house relation shared with Slurry density · Percent solids (mass ↔ volume) · Slurry mass balance. Related: Kiln residence time · Leach reagent consumption
Formulas
Diagram
Worked example
Solids basis: 100 t/h dry solids at 40 % w/w, solids SG 2.70, liquor SG 1.000. Required residence time 24 h across 6 tanks in series, live-volume fraction 0.85, H/D = 1.0.
- 01Solids mass fraction Xs = 40 / 100 = 0.40; ρ_solids = 2700 kg/m³, ρ_liquor = 1000 kg/m³
- 02Slurry density = 1 / (0.40/2700 + 0.60/1000) = 1 / 0.00074815 = 1336.6 kg/m³ (slurry SG 1.337)
- 03Slurry mass flow = dry solids / Xs = 100 / 0.40 = 250 t/h
- 04Q = 250 t/h × 1000 / 1336.6 = 187.0 m³/h
- 05Total live volume V_live = Q · τ = 187.0 × 24 = 4488.9 m³
- 06Per-tank volume V_tank = V_live / (N · f_live) = 4488.9 / (6 × 0.85) = 880.2 m³
- 07Geometry at H/D = 1.0: D = (4 × 880.2 / π)^(1/3) = 10.39 m; H = 1.0 × D = 10.39 m
Each of the 6 tanks holds about 880 m³ of live volume — a flat-bottom cylinder roughly 10.4 m diameter by 10.4 m high at H/D = 1.0; total live volume 4489 m³.
FAQ
Does this calculator predict leach recovery or kinetics?
What is the live-volume fraction and why 0.85?
How is the solids-basis flow derived?
Why do the two flow modes give the same size?
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