Reflector melting principle

When using a method to treat the copper, it is usually carried out in a fixed reverberatory furnace. Therefore, in fact, the reverberatory furnace performs both smelting and refining.

The principle of copper reverberatory furnace refining is essentially the same as that of ore copper. However, due to the high impurity content (sometimes up to 4%) of secondary copper, it has unique characteristics in operation. The copper is processed in a reverberatory furnace. At the time, the entire refining process includes operations such as melting, oxidation, reduction, slag removal, and casting. The core of the entire operation is oxidation and reduction. The following mainly describes the oxidation and reduction.

The basic principle of copper-dioxide oxidation refining is that most of the impurities present in copper have greater affinity for oxygen than copper-to-oxygen affinities, and the oxides of most impurities have low solubility in the copper liquid, so when smelting into the melt, In the air, the impurities are preferentially oxidized and removed, but the copper in the melt accounts for the vast majority, and the amount of impurities is very small. Therefore, when oxidizing, the copper is first oxidized.

4Cu+O2=2Cu2O

The resulting Cu2O dissolves immediately in the copper solution and reacts with impurities in the copper solution to oxidize the impurities.

[Cu2O]+[Me]=2[Cu]+(MeO)

Where: [] indicates the concentration of the substance in the copper liquid;

() Indicates the substance concentration in the slag phase;

Me is an impurity metal.

The equilibrium constant for this reaction is:

The main body in the copper liquid is metallic copper, and the concentration is large because the impurity amount is relatively small, so although the impurities are oxidized by Cu2O, [Cu] can be considered to be substantially constant (that is, constant). At the same time, since the solubility of the impurity oxide (MeO) in the copper liquid is very small and can quickly reach saturation, in most cases, when the temperature is constant, [MeO] is considered to be a constant value, so the equilibrium constant of the reaction The following formula can be used:

K'=[Cu2O][Me]

This shows that at a certain temperature (ie, K is a constant) the content of impurities in the copper liquid is inversely proportional to the content of Cu2O, the larger the [Cu2O], the smaller the [Me], ie, the more un-oxidized impurities remaining in the copper liquid. Less, the more complete the refining operation. Practice has shown that in order to remove the impurities in the copper liquid more quickly and thoroughly, the process of strengthening oxidation should be stressed to make the concentration of Cu2O in the copper liquid saturated.

The solubility of Cu2O in copper liquid increases with increasing temperature:

Temperature °C 1250

Solubility % 5 8.3 12.4 13.1

When the dissolution of Cu2O exceeds the solubility at this temperature, the melt will be divided into two layers. The lower layer is a copper solution saturated with Cu2O, and the upper layer is a copper-saturated Cu2O phase. This relationship can be derived from the Cu ¢ O system phase diagram. see it clearly. The solubility increase in the copper liquid is very small, and the melt appears stratified, so that part of Cu2O enters the slag layer, and excessive oxidation causes the reduction process to increase, and at the same time, more reducing agent is consumed, so that excessive oxidation of the copper liquid is avoided. Require that the oxidation period be maintained at 1150 to 1170 °C.

The main impurities in the oxidation refining process are briefly described as follows:

iron. The affinity of iron for oxygen is much greater than the affinity of copper for oxygen, so iron is easily oxidized and slag formation is removed. The iron oxidation reaction proceeds as follows:

Cu2O+Fe=2Cu+FeO

According to thermodynamic estimation, iron can be reduced to one hundred thousandth in the refining process.

nickel. Nickel is a difficult-to-remove impurity. Nickel and copper can generate a series of solid solutions. Although nickel is oxidized during both the melting period and the oxidation period, it is both slow and incomplete, and the NiO generated during the oxidation period is distributed in the copper solution and slag. between. NiO dissolved in the slag can form NiO · Fe2O3 which is insoluble in the copper liquid and dissolved in the slag phase. This part of nickel can be removed. Thermodynamic calculation shows that when the copper liquid contains 16% of nickel, nickel can be removed to 0.25%. .

When the copper solution contains both nickel and arsenic and antimony, the removal of nickel is more difficult. Because the NiO dissolved in the copper liquid can form nickel mica (6Cu2O · 8NiO · 2As2O3 or 6Cu2O · 8NiO · 2Sb2O3) dissolved in copper liquid with Cu, As or Sb. In order to remove nickel, only alkaline flux is used to decompose the nickel mica.

Zinc. Zinc and copper are completely miscible in the liquid state. The boiling point of zinc is 906 °C. During the refining, most of the zinc is volatilized in the metal phase during the melting phase, and then is oxidized by the oxygen in the furnace gas to ZnO and discharged with the furnace gas. Collected in the dust system, the remaining zinc is oxidized to ZnO in the early oxidation stage, and forms zinc silicate (2ZnO · SiO2) and zinc ferrite (ZnO · Fe2O3) into the slag. When refining high-zinc mixed copper materials (yellow copper, etc.), it accelerates the volatilization of zinc, increases the furnace temperature (usually maintained at 1300 to 1350 °C) during the melting period and the oxidation period, and covers the melt surface. Layer charcoal or sulfur-free coke particles, so that zinc oxide is reduced to metal zinc and volatilization, so as to avoid the formation of zinc oxide crust to prevent the process of zinc vapor.

lead. Solid lead is insoluble in copper, and it is rarely dissolved in the liquid state. However, in the oxidizing period, when lead is oxidized to lead oxide, because the density (9.2) is higher than the copper density (8.9), it sinks to the bottom of the furnace. In the case of an acid furnace bottom, PbO will react with SiO2 in the furnace material to produce lead silicate (XPbO.YSiO) with a low density. It floats to the surface of the bath and is removed. If the bottom of the furnace is a basic refractories, the removal of lead is difficult. In this case, the silica flux must be blown into the melt to increase the air volume and maintain a high furnace temperature (about 1250 °C), so that PbO and SiO2 can be used. , Produce lead silicate. Quartz slag removal method takes a long time, copper loss into the slag, in order to improve the lead removal effect, to overcome the shortcomings of the law, you can change the phosphorus copper, so that lead in the form of phosphate removal. Boron oxide can also be used as a flux to strip lead in the form of lead borate.

tin. When dealing with bronze material, the material contains high tin, and the tin and copper are mutually soluble in the liquid state. Tin oxide in the reverberatory furnace produces stannous oxide (SnO) and tin oxide (SnO2). SnO is weakly alkaline and can slag with SiO2. , but also part of the volatile. SnO2 is weakly acidic and soluble in copper solution. At this time, basic solvent (soda or limestone) must be added to make slag, and sodium stannate (Na2O · SnO2) or calcium stannate (CaO) that does not melt in the copper solution is formed. · SnO2). It has been proved that by adding a mixed flux consisting of 30% calcium oxide and 70% sodium carbonate, the tin content in copper can be reduced from 0.029% to 0.002%. The use of a mixture of 50% Fe2O3 and SiO2 flux can also quickly reduce the tin content to 0.005% and remove some of the lead.

arsenic. From As-Cu phase diagram, it can be seen that arsenic and copper are mutually soluble in the liquid state. During oxidation, arsenic can be oxidized to volatile As2O3, which is discharged along with the furnace gas, but a small amount of arsenic is also oxidized to As2O5, and copper arsenate is formed ( Cu2O·XAs2O5) is soluble in the copper solution. When nickel is present in the copper solution, arsenic can also form nickel mica with copper and nickel. This all adds difficulty to arsenic removal.

antimony. Tantalum and copper are infinitely soluble in the liquid state, and Cu and Sb2 can be generated from copper and niobium. Like arsenic, antimony also produces volatile Sb2O3 during oxidation, and Cu2O · Sb2O3 and Cu2O · Sb2O5 dissolved in copper. Therefore, when dealing with mixed copper containing high As and Sb, the oxidation and reduction processes must be repeated several times to reduce the nonvolatile As2O5 and Sb2O5 to volatile As2O3 and Sb2O3, non-volatile As and Sb, and alkaline fluxes. deal with.

Gold and silver. Gold and silver are completely concentrated in the anode copper and enter the anode mud during the electrolytic refining, and the anode mud is further treated to be recovered.

When all impurities are removed, the oxidation period ends and the process shifts to the reduction period. The first effect of the reduction is to reduce the peroxidized copper oxide to metallic copper, and to remove the gas dissolved in the copper solution. At the end of the oxidation, about 8% of Cu2O is still present in the copper solution. Too much oxygen will make copper brittle and its ductility and electrical conductivity will decrease, so it must be reduced. In the reduction period, when using heavy oil, insert wood, etc., the main chemical reactions that occur are as follows:

6Cu2O+2C2Hm=12Cu+2Co+mH2+2CO2

When reduced with NH3, the following reactions occur:

Cu2O+2NH3 6Cu+N2+3H2O

If natural gas is used as a reducing agent, so-called “reforming” must be performed on the natural gas. Otherwise, the methane CH4 component of natural gas decomposes and produces a large amount of H2 at 1000 °C. Although it can strengthen reduction, it also increases the adsorption of hydrogen by copper.