💧 Calculators

Pulp Dilution Water

Computes how much water must be added to take a pulp from one mass percent solids to a lower one, Water = M × (C₁/C₂ − 1), where M is the incoming pulp mass (or mass flow). It follows from the mass balance: the solids mass does not change on dilution, so the final pulp mass is M·C₁/C₂ and the difference is water. This is the most routine operation in a mineral processing plant — grinding, desliming, flotation and thickening each demand their own percent-solids range, and getting the dilution water wrong throws off residence time, viscosity and reagent consumption. Both percentages are by mass (weight of solids per weight of pulp), and the result comes out in the same unit entered for M. Enter the pulp mass or flow, the current percent solids and the target percent solids.

Pulp mass or mass flow M (t/h) Current percent solids C₁ (%) Target percent solids C₂ (%)

Result

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Dilution water: how much to add to drop percent solids

In a mineral processing plant every unit wants its own percent-solids range: grinding runs a dense pulp, rougher flotation wants something far more fluid, the thickener takes what's left and sends clean water back. Whenever feed grade shifts or the circuit gets reconfigured, someone has to tell the operator how far to open the dilution valve. Guessing costs money in both directions: too little water leaves the pulp viscous, hurts reagent dispersion and chokes the pump box; too much shortens residence time, thins the collector and sends solids over the cyclone overflow.

The formula is Water = M × (C₁/C₂ − 1). M is the incoming pulp mass or mass flow, C₁ the current percent solids and C₂ the target, both by mass. It drops straight out of the balance: solids mass stays put when water goes in, so M·C₁ = M_final·C₂, the outgoing pulp weighs M·C₁/C₂, and the gap is water. With the screen defaults — 100 t/h at 65% heading to 35% — solids total 65 t/h, final pulp rises to 185.714 t/h and 85.714 t/h of water goes in. Useful ranges when picking a target: ball mill discharge 70 to 78%, cyclone feed 50 to 65%, rougher flotation 28 to 40%, thickener underflow above 55%.

Both percentages have to be mass based, dry solids weight over total pulp weight. If the reading came from an instrument calibrated in volume percent, convert it first using solids density — in iron ore the gap runs past thirty points. The calculation assumes clean dilution water; where a plant recirculates process water carrying fines, the solids it brings belong in the balance and the answer here comes out short. Diluting is all it does: asking for a C₂ above C₁ means thickening, and the screen refuses rather than handing back a negative. Output units follow whatever unit M went in as.

Frequently asked questions

Where do the 85.714 t/h from the default values come from?

Pulp arrives at 100 t/h carrying 65% solids, which is 65 t/h of solid and 35 t/h of water. Dilution neither creates nor destroys solid, so those same 65 t/h now have to make up 35% of the outgoing pulp: 65 ÷ 0.35 = 185.714 t/h. The missing water is 185.714 − 100 = 85.714 t/h, exactly what the screen prints to three decimals. Notice that total water in the stream jumps from 35 to 120.714 t/h, nearly quadrupling, and that only the C₁/C₂ ratio drives the answer: any pair holding 65/35 = 1.857 calls for the same water per tonne of pulp fed.

Can I use this to thicken the pulp instead?

Not on this screen. It rejects any C₂ above C₁ and shows the check-your-values warning, since the formula would hand back a negative number that an operator might read as water to add. The mirror calculation does exist and stays simple: to climb from C₁ to C₂, water to remove equals M × (1 − C₁/C₂). Taking those same 100 t/h from 35% up to 65% means pulling out 46.154 t/h of water, the return leg of the default example. That water leaves through a thickener, cyclone or filter, though, never through a valve, so sizing it becomes a different job.

Is percent solids measured by mass or by volume?

By mass, always: dry solids weight divided by total pulp weight. Confusion creeps in because nucleonic gauges usually ship calibrated in pulp density, and hydrocyclone spreadsheets sometimes work on a volume basis. To convert, use mass% = (vol% × ρs) ÷ (vol% × ρs + (100 − vol%) × ρwater). In an iron ore with ρs = 4.8 t/m³, 30% by volume becomes 67.3% by mass — typing 30 into the field would call for a water flow wildly out of scale, and the operator would only find out at the pump box.

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Solve the dilution equation C₁V₁ = C₂V₂. Provide any three values to compute the fourth.

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Computes the silt density index of ASTM D4189, the test that measures the fouling potential of the feed water of a reverse osmosis membrane: the sample is filtered through a 0.45 µm membrane at 207 kPa, the time to collect 500 mL is measured at the start and again at the end of the test, and the index is the percentage of flow loss divided by the duration. The expression is SDI = (1 − initial time ÷ final time) × 100 ÷ duration, and the number tells how much the filter plugged per minute of test: membrane makers usually require under 5 for a spiral wound element and under 3 for extended warranty, and water above that forces stronger coagulation or filtration upstream. The test is only valid if the flow loss stays below 75%; if the filter plugs before that, run it again with a shorter duration. The duration was adopted as an input instead of fixing 15 minutes, the standard value, precisely because poor water demands repeating at 10 or 5 minutes — results from different durations are not comparable, which is why the value is recorded as SDI₁₅, SDI₁₀ or SDI₅. Enter the initial time, the final time and the test duration.

The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.