So according to my understanding, to get one tonne of neodymium, we need 49/(18.5%*4.1%*50%) = 12920 tonnes. Is that correct? That would be a lot of mining for a tonne of neodymium.
Ultimately, I am interested in the amount of earth that has to be moved/mined to get one tonne of copper/other metals (which of course depend of course on the mine and the ore grade).
Thanks a lot :-). I am trying to make sense of the numbers to get a better understanding of the footprint of certain technologies (e.g. given that there are 2 tonnes of neodymium in a modern wind generator, we need to remove 26,000 tonnes of earth to get the neodymium for this (if my numbers are correct).
Best regards, ProGeologist.
[1] Rare Earth mining: https://ui.adsabs.harvard.edu/abs/2013JOM....65j1327T/abstract - tables on page 1329 and 1331
the stripping ratio (which is usually larger than 3:1 for copper, i.e. 3 tonnes of waste for one tonne of copper ore) and the ore grade (which is <5% for copper, i.e. out of a tonne of copper, you can extract only 50kg copper). Is my interpretation correct?
You are partially correct. In this scenario the amount of copper that would be delivered to the processing plant (also known as a mill or concentrator) in mined ore is 50 kg, i.e the amount of metal in the mill feed is 50 kg.
To get usable copper metal from the ore, after mining, it needs to treated in a processing plant to concentrate the copper bearing minerals and to remove the gangue minerals. The copper concentrate is then sent to a smelter to produce copper metal. The waste from the concentrator is sent to the tailings dam/pond/storage-facility.
If you account for the recoveries of these two processes, typically 80% for concentrating and 90% for smelting, the amount of copper metal produced from 50 kg of copper in mill feed would be 36 kg (50 x 0.8 x 0.9).
How do they stack up in the mining process? Using a 1% ore grade and 3:1 stripping ratio, does that mean that to get 1 tonne of copper, we need to move 400 tonnes of earth, as we need 100 tonnes copper ore that comes along with 300 tonnes of waste?
Yes. Such a scenario would give you 1 tonne of copper in mill feed.
Unfortunately, the paper you refer to was most likely written by primary metallurgists, or possibly chemical engineers, and not mining engineers. The emphasis of the authors is on the amount of metal recovered from mining and the amount of energy required to produce usable metal from various mining projects, not the overall quantities of material mined.
Regarding your question about the Bayan Obo mining district in China your calculations are not correct. Also, from the paper you refer to you cannot obtain the total amount of earth mined, ore and waste, because there is no reference to stripping ratio.
As you state, from Table I of your reference, the proportion of neodymium in REM (REM being rare earth metal)is 18.5%. Table II states Tonnes mined/Tonnes REM is 59. This usually refers to the tonnes of ore mined and supplied to the mill, not the total tonnes of earth mined (ore and waste). Additionally, as stated in Table II and reiterated in Table IX, the recovery from the primary processing plant is 50%.
Now, 1 tonne of REM supplied to the mill contains 0.185 t (185 kg) of neodymium. With 59 tonnes of ore mined per tonne of REM, the amount of ore mined and supplied to the mill to get 1 tonne of neodymium is 59/0.185 = 318.9 tonnes.
The mill only recovers 50% of the metal supplied to it. Therefore, for the mill to recover 1 tonne of neodymium 637.8 (318.9/0.5) tonnes of ore must be supplied to the mill.