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多少地球必须搬到1吨铜的吗

我试图理解开采矿产和金属的过程。

我发现两个措施相关的浪费:剥采比(通常大于3:1为铜,即3吨废物一吨铜矿石)和矿石品位(< 5%,铜,即一吨铜,可以提取只有50公斤铜)。我的理解正确吗?
采矿过程中如何累计?使用1%的矿石品位和3:1剥采比,这是否意味着1吨铜,我们需要400吨的地球,我们需要100铜矿石300吨废物随之而来?
做同样的逻辑也适用于稀土开采吗?我发现以下数字中国鄂博矿区(见[1]):
- 钕含量:18.5%
- REO(稀土氧化物)等级:4.1%
- 回收率:50%
- 每吨/吨REO开采:49

所以根据我的理解,1吨的钕,我们需要49 /(18.5% * 4.1% * 50%)= 12920吨。那是正确的吗?这将是一个很多矿业一吨钕。

最终,我感兴趣的地球已经移动/开采1吨的铜和其他金属(当然当然取决于矿山和矿石品位)。

谢谢:-)。我试图理解得到一个更好的理解的数字足迹的某些技术(例如,考虑到有2吨的钕在现代风力发电机,我们需要删除26000吨地球的钕(如果我的数字是正确的)。

最好的问候,ProGeologist。

[1]稀土开采:https://ui.adsabs.harvard.edu/abs/2013JOM....65j1327T/abstract——表在1329和1331页

回答

你的问题关于剥采比只适用于露天矿山。他们生产更比地下矿山废石。我通常比矿矸石。地下矿山更手术,根据情况的一些他们在开采生产用作回填废石(斯特普)[2]。>剥采比(通常大于3:1为铜,即3吨废物一吨铜矿石)和矿石品位(< 5%的铜,也就是一吨铜,可以提取只有50公斤铜)。我的理解正确吗?你是部分正确。在这种情况下的铜量将会交付给加工厂(有时也称为一个轧机或集中器)在开采矿石是50公斤,我。e金属轧机饲料的数量是50公斤。得到有用的铜金属从矿石,采矿后,它需要在一个加工工厂,精矿含铜矿石和删除(煤矸石)[1]矿物质。然后发送到冶炼厂铜精矿生产铜金属。废物的集中器发送到尾矿坝/池塘/储存设施。 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. [1]: https://en.wikipedia.org/wiki/Gangue [2]: https://www.britannica.com/technology/stope

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\ begingroup美元完美,这是有意义的。非常感谢。只有一个小问题:0.0925是从哪里来的?这是钕的数量我们得到638吨或稀土开采(当然不包括剥离率)? \ endgroup美元
- - - - - -ProGeologist
2021年4月4日上19:15

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