Reflux / gold

There is gold in this water and you cannot see it

Gold ore is mostly nothing. A few grams in a tonne of rock, which is too little to pick out, so a mine does not pick it out. It crushes the rock to flour, mixes it with water and a few hundred milligrams a litre of cyanide, bubbles in oxygen, and the gold dissolves. It is then in the liquid at about two milligrams a litre, clear as tap water, and a thousand cubic metre tank of it holds two kilos of gold you cannot see. The dangerous part of this plant is the water, not the gold.

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01 A few grams in a tonne 02 The tank, and what is actually in the water 03 The carbon, and the number people get wrong 04 The loop that keeps the plant from gassing itself 05 What a fixed recovery block never sees
01

A few grams in a tonne

Most Australian gold ore is under five grams of gold per tonne of rock, and the median for the USGS deposit models is 2.9 g/t. Five grams is the mass of a small coin spread through a tonne of stone, and it is not in one place: it is in flecks along veins and, in a refractory ore, locked inside sulphide grains you cannot even see it in.

So the mine does not separate the gold from the rock. It dissolves the gold and leaves the rock behind, which is a completely different problem and a much easier one.

4 Au + 8 CN- + O2 + 2 H2O → 4 Au(CN)2- + 4 OH- Elsner's equation. Two cyanides per gold atom, and the oxygen is the electron acceptor, which is why a leach with no oxygen in it does nothing at all. The product is the dicyanoaurate ion, and that ion is what is in the water.
02

The tank, and what is actually in the water

Crush the rock to flour, make it a slurry, dose a few hundred milligrams a litre of sodium cyanide, and sparge air or oxygen in at the bottom. The agitator is not decoration: if the solids settle, the surface area stops being available and the leach stops.

milled ore NaCN 300 to 500 mg/L NaCN oxygen in 2 mg/L gold take the rock out clear as tap water 12 m 1,000 m3 of liquor = 2 kg of gold
A leach tank in section: milled ore and water in, a few hundred milligrams per litre of cyanide, oxygen through the sparger ring, and two pitched blade impellers keeping the solids up. The gold leaves the rock as the dicyanoaurate ion and rides in the liquid at about 2 mg/L. Take a sample and filter the rock out of it and what is left is clear: 2 mg/L is one part in five hundred thousand.

The arithmetic that makes the hook work is one line and the units cancel exactly. Grams per tonne of ore, times tonnes of ore per cubic metre of pulp, is grams per cubic metre, which is milligrams per litre.

What is dissolved in your tank

Drag the head grade. Everything else is a plant number you can change.

Dissolved gold2.00mg/L
In the tank2.00kg of gold
At today's price, roughly$0USD

Clear as tap water Two milligrams a litre is about one part in five hundred thousand. There is nothing to see, and that is the point.

Recovery into solution is taken at 95%. The price used is a round 110 USD per gram, so the dollar figure is an order of magnitude, not a quote.

03

The carbon, and the number people get wrong

Dissolved gold at two milligrams a litre is not something you can pour into a furnace. It has to be concentrated first, and the way a modern plant does that is activated carbon: charcoal granules, mostly pore, moving through the slurry counter-current to it.

pulp carbon moves the other way 6 tanks x 4 h 6 tanks x 1 h leach load the carbon
Six leach tanks of four hours, then six adsorption tanks of one hour. The pulp runs left to right and the carbon is pumped right to left against it, so the most loaded carbon meets the richest solution.

Six leach tanks of four hours, then six adsorption tanks of an hour each. The pulp flows one way and the carbon is pumped the other, so the most heavily loaded carbon meets the richest solution and the barren carbon meets the solution on its way out. The carbon leaves at around three thousand grams a tonne, from ore that was three.

inside one pore it sticks in the pores 3 g/t in the rock on the carbon 012345678901234567890 012345678901234567890 012345678901234567890 012345678901234567890 , g/t x1,000 richer
Activated carbon is mostly pore. The dissolved gold crosses to the pore mouths and stays, taking the carbon from nothing to thousands of grams per tonne while the solution it came from is still clear.

Here is the number people get wrong. The textbook equilibrium loading at that solution concentration is around twelve thousand grams a tonne. A plant runs near three. If you size a circuit off the isotherm you will buy a quarter of the carbon you need and wonder why your tails are high.

The reason is not that the isotherm is wrong. It is that the carbon is never at equilibrium. It is in contact for a finite time and loading is a rate, so what you get is the equilibrium value times how far along the approach you got.

Why the plant loads a quarter of what the book says

Drag the total carbon contact time. The dashed line is equilibrium; the solid line is what the carbon actually leaves with.

Equilibrium12,000g/t
What you get3,000g/t
Fraction of equilibrium25%of the book

A quarter of the book Six hours of contact gets you a quarter of the way to equilibrium. To reach ninety per cent you would need about forty eight hours of it.

Equilibrium is Freundlich, q* = K C0.6, with K set so that q*(2 mg/L) is 12,000 g/t. The approach is first order, q = q*(1 − e−kt), with k = 0.0479 per hour, which is the rate constant that puts six hours at a quarter.

From there it is unit operations in a row. Hot caustic strips the gold back off the carbon in an elution column, electricity plates it onto steel cathodes, and the cathode sludge is smelted into a bar. The carbon goes back round.

04

The loop that keeps the plant from gassing itself

Cyanide in water is a weak acid system. Let the pH fall and the cyanide ion takes a proton and becomes hydrogen cyanide, which is a gas at room temperature and comes out of the liquid. Half of it has already done that at pH 9.3.

7 8 9 10 11 12 pH the plant runs here gas comes off here 11.2 8.4 10.4 pH falling 7 8 9 10 11 12 more acid more alkaline stays in the liquid leaves as gas half and half at 9.3 hydrogen cyanide gas probe control lime in never sleeps
The real acid dissociation of hydrogen cyanide, and the same two zones on a scale you can read a direction off. Half the cyanide is already a gas at pH 9.3, which is why the plant holds the slurry above 10 on a lime loop that has no manual mode: letting the pH fall is letting it make a gas.

What is in your headspace

Drag the pH. This is the acid dissociation, not a rule of thumb.

As cyanide ion94%stays in the liquid
As hydrogen cyanide6%wants to leave it

Normal operation Above pH 10 the great majority stays as the ion, which is why a plant holds it there and never lets it drift.

f(CN-) = 1 / (1 + 10(pKa − pH)), pKa = 9.31 at 25 C.

That is why lime addition is on a control loop that never sleeps, and why the loop is one of the few things in the plant with no manual mode. Below about 9.3 the plant is making a gas it cannot afford to make.

And what leaves the plant does not leave as cyanide. The usual answer is the sulphur dioxide and air process: SO2 and air over a copper catalyst oxidise cyanide to cyanate, which is far less toxic, down to under a milligram a litre. The word is cyanate. Nothing is destroyed; it is converted, and what is left is measured.

The reason any of this is written down in the detail it is: in January 2000 a tailings pond in Romania overtopped and sent roughly a hundred thousand cubic metres of water carrying up to a hundred and twenty tonnes of cyanide into the Tisza river system. The EU task force put the fish kill at 1,240 tonnes. The gold was never the problem.

beach tailings pond embankment January 2000 up to 120 t of cyanide over 1,000 t of fish
A tailings impoundment: a beach of settled tailings, an embankment of coarse rock with a crest and a downstream face, and the pond behind it. In January 2000 one of these overtopped its own crest.
05

What a fixed recovery block never sees

In a steady state flowsheet a leach is often one block with a recovery on it. Give it ore and it gives you back a percentage, and it will do that at any residence time, any carbon inventory and any pH, because it has none of them.

one envelope rock water cyanide carbon gold 0 3,000 6,000 9,000 12,000 g/t on the carbon mg/L gold in solution textbook 12,000 plant about 3,000 the gap
The mass balance closes on rock, water, cyanide and carbon. The loading curve is where the model and the plant disagree.

The model that is worth having is a mass balance on rock, water, cyanide and carbon that carries the two rates the plant is actually limited by: how fast gold goes into solution, and how fast it comes back out onto the carbon. Then the carbon inventory, the number of stages and the elution frequency are outputs rather than assumptions, and the pH loop has somewhere to live.

That is the layer being built at Reflux: the model that knows which curve it is on, checks the end of it, and tells you what it verified.

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Sources: ore grades from Geoscience Australia's Australian mineral facts for gold and from USGS Open File Report 2014-1074, whose 2.9 g/t is a deposit model median rather than a head grade; the 300 to 500 mg/L cyanide dose and the HCN speciation from the International Cyanide Management Code and its June 2021 mining guidance; the six by four hours and six by one hour circuit, the 3 g/t to about 3,000 g/t loading and the equilibrium near 12,000 from Modeling of Gold Cyanidation and from Kemix, Activated Carbon in Gold Recovery; the sulphur dioxide and air process and its under 1 mg/L residual from the INCO process description; and the January 2000 Baia Mare figures, up to 120 tonnes of cyanide and 1,240 tonnes of fish, from the EU Baia Mare task force report. The loading rate constant used in the calculator above is fitted to put six hours of contact at a quarter of equilibrium, which is the plant number; it is a teaching value, not a measurement of any one carbon.

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