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Studies on the regularities of the steelmaking zinc-bearing dusts leaching in ammonium chloride solutions

机译:氯化铵溶液浸出炼钢铜粉尘的规律研究

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摘要

Dust from steelmaking production, in particular electric arc furnace dust (EAFD), is a rich secondary source of zinc. Different approaches to the EAFDs processing with the production of zinc oxide and metal phases as a final product have been studied. Ammonium chloride leaching is selective for zinc, does not require special acid-resistant equipment, and is relatively cheap. The advantages of this method are the possibility of obtaining metallic zinc as a commercial product and the absence of the need for additional solution purification from iron and chlorine ions. Chemical and mineral characterization of Waelz-oxide after the first stage of calcination are presented, the main identified phase is zinc oxide (ZnO), a small amount of Zn associated with iron (Zn_2Fe_2O_4) in the ferrites form and lead in the oxide phase (PbO) are also determined. The thermodynamic analysis of the Zn(II) - NH_4CI - NH_3 - H_2O system in the HYDRA software medium was carried out, the main zinc compounds were determined under leaching conditions. A matrix of changing conditions is compiled with using the full three-factor experiment method. Based on the obtained results, in the Statistica-10 program, three-dimensional dependences of zinc extraction on varied parameters were constructed and equations describing these dependencies were obtained. The optimal leaching parameters were determined: the concentration of ammonium chloride 4 mol/dm~3, the concentration of ammonia 4 mol/dm~3, the ratio L:S = 15.
机译:炼钢生产的灰尘,特别是电弧炉灰尘(EAFD),是一种丰富的锌源。研究了通过生产氧化锌和金属相作为最终产品的ESFDS处理的不同方法。氯化铵浸出是锌的选择性,不需要特殊的耐酸设备,并且相对便宜。该方法的优点是将金属锌作为商业产品获得金属锌,并且不需要来自铁和氯离子的另外的溶液纯化。展示第一阶段后的氧化物氧化物的化学和矿物表征,主要鉴定的相是氧化锌(ZnO),少量Zn与铁氧体中的铁(Zn_2Fe_2O_4)相关,并在氧化物相中引出( PBO)也被确定。 Zn(II) - NH_4CI - NH_3 - H_2O系统的热力学分析进行了Hydra软件介质,在浸出条件下测定了主要的锌化合物。使用全三因素实验方法编制改变条件的矩阵。基于所得结果,在统计数据中,构建了三维锌提取对变化参数的依赖性,并获得了描述这些依赖性的等式。确定了最佳浸出参数:氯化铵4mol / dm〜3的浓度,氨浓度为4mol / dm〜3,比例L:S = 15。

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