WO2019153730A1 - Method for preparing titanium alloy - Google Patents

Abstract

Disclosed is a method for preparing a titanium alloy, the method being carried out by means of the following steps: (1) preparing a solvent, a reactant and a reducing agent, which act as raw materials, wherein the solvent is cryolite or titanium-containing cryolite, the reactant is an oxide, and the reducing agent is aluminum powder; (2) mixing all the raw materials and grinding and pressing same into a block mass, placing same in a reduction furnace for an aluminothermic reduction reaction, and after the reaction is finished, cooling same to room temperature, so that an aluminum-based alloy melt and a cryolite melt are respectively solidified in a lower layer and an upper layer; and (3) preparing the aluminum-based alloy into a powder, then mixing same with a sodium fluorotitanate powder and grinding and pressing same into a block mass, placing same in a reduction furnace for an aluminothermic reduction reaction to produce a titanium alloy and a titanium-containing cryolite, then carrying out distillation under vacuum conditions to obtain a porous titanium alloy.

Description

Method for preparing titanium alloy Technical field

The invention belongs to the technical field of metallurgy, and in particular relates to a method for preparing a titanium alloy.

Background technique

Titanium alloy is widely used in various fields due to its mildness, light weight, good corrosion resistance and high heat resistance, especially in the aerospace field and ship transportation. At present, titanium alloy is in titanium sponge or titanium powder. Adding one or several other alloying elements, obtained by vacuum melting, especially when titanium alloy is prepared from sponge titanium, in order to obtain a uniform titanium alloy, two or more times are required. The remelting process has a long process flow and high cost.

Summary of the invention

An object of the present invention is to provide a method for preparing a titanium alloy, which comprises an oxide of an alloying element other than titanium and aluminum and cryolite, or an oxide of an alloying element other than titanium and aluminum, and a cryolite containing titanium. a raw material, an aluminum-based alloy obtained by removing aluminum alloy elements other than titanium, or an aluminum-based alloy containing titanium and alloying elements, and then reducing sodium fluorotitanate with aluminum in the aluminum-based alloy to obtain titanium Alloys simplify the process while reducing production costs.

The method of the invention proceeds as follows:

1. Preparing a solvent, a reactant and a reducing agent as raw materials, wherein the solvent is cryolite or titanium-containing cryolite, the reactant is an oxide, the reducing agent is aluminum powder having a particle size of ≤100 mesh; and the oxide is M _x O _y , wherein the element M is an alloying element other than titanium and aluminum in the prepared titanium alloy; the ratio of NaF/AlF ₃ in the cryolite is 1.5 to 3.0;

2. All the raw materials are added to the mixing mill for mixing and then pressed into agglomerates; the agglomerates are placed in a reduction furnace, and an aluminothermic reduction reaction is carried out under an argon atmosphere at 1000 to 1500 ° C to make an oxide M. _x O _y is reduced to a metal element by aluminum and dissolved in an aluminum melt to form an aluminum-based alloy melt containing element M, or an aluminum-based alloy melt containing titanium and element M, and the alumina formed by the reduction reaction is dissolved. In the cryolite melt, a cryolite melt in which alumina is dissolved is formed; after the aluminothermic reduction reaction is completed, the reduced product is cooled to a normal temperature, and the aluminum-based alloy melt and the cryolite melt are solidified in the lower layer and the upper layer, respectively, and the lower layer is Separating the aluminum-based alloy from the upper layer of alumina-containing cryolite;

3. The aluminum-based alloy is made into a powder with a particle size of ≤100, and then mixed with sodium fluorotitanate powder to be pressed into agglomerates, and the agglomerates are placed in a reduction furnace in an argon atmosphere and 900 to 1200. The aluminothermic reduction reaction is carried out under the condition of °C, and the aluminum in the agglomerate reduces the titanium in the sodium fluorotitanate to form a titanium alloy and a titanium-containing cryolite; after the aluminum thermal reduction reaction is completed, it is carried out at 1000-1200 ° C under vacuum. Distillation, the titanium-containing cryolite was separated by distillation to obtain a sponge-like titanium alloy.

The titanium-containing cryolite distilled in the step 3 of the above method can be used as the raw material of the step 1.

In the above method, the reaction formula of the aluminothermic reduction reaction in the step 2 is:

M _x O _y + 2y / 3Al = xM + y / 3Al ₂ O ₃ (1).

In the above method, the titanium-containing cryolite comprises a titanium-containing component and a cryolite component, wherein the cryolite component has a NaF/AlF ₃ molar ratio of 1.5 to 3.0, and the titanium-containing component is TiF ₃ , Na ₃ TiF ₆ , Na. ₂ TiF ₆ and / or titanium metal.

In the above method, the main reaction formula of the aluminothermic reduction reaction of step 3 is:

3Na ₂ TiF ₆ +4Al=3Ti+Na ₃ AlF ₆ +3NaAlF ₄ (2)

Or 3Na ₂ TiF ₆ +4Al=3Ti+Na ₃ AlF ₆ +6NaF+4AlF ₃ (3);

The reaction formula of the side reaction is:

Na ₂ TiF ₆ = 2NaF + TiF ₄ (4),

3TiF ₄ +Al=3TiF ₃ +AlF ₃ (5)

And/or TiF ₃ +3NaF=Na ₃ TiF ₆ (6).

In the above method, in the process in which the ferric TiF _{3 of the} trivalent titanium produced by the aluminum thermal reduction in step 3 is distilled together with the cryolite, part of the TiF ₃ may undergo disproportionation reaction to form TiF ₄ and Ti, and the reaction formula is:

4TiF ₃ =3TiF ₄ +Ti (7)

And TiF ₄ +2NaF=Na ₂ TiF ₆ (8);

Therefore, in the above method, the titanium-containing cryolite distilled in the step 3 contains a titanium-containing component and a cryolite component, and the titanium-containing component is mainly present in the form of TiF ₃ , Na ₃ TiF ₆ , Na ₂ TiF ₆ and metal Ti, wherein The Ti and Na ₂ TiF ₆ are mainly formed by the disproportionation reaction of low-valent fluoride of titanium.

In the above method, the amount of aluminum in the step 1 is calculated according to the mass percentage of the m alloy and the alloying element to be obtained, and the amount of aluminum in the step 1 includes the following parts of aluminum. The sum of demand:

1) reduction of titanium oxide in a given mass of alloying elements other than titanium and aluminum by chemical equation (1) the desired amount of aluminum Al ₁

2) If the titanium alloy elements aluminum, the amount of this portion of the aluminum with Al ₀ is calculated as:

Al ₀ = a ₀ m (9);

Where a ₀ is the mass percentage of the alloying element Al in the m-mass titanium alloy, and the unit is %.

3) for the chemical reaction equation (2) acquired by aluminothermic reduction of Aluminum alloy mass m is calculated as Al ₂ required:

Al ₂ = (108 × m _Ti ) / (144 × η) (10);

Where η is the actual yield of titanium in the process of preparing titanium by reducing Na ₂ TiF ₆ according to reaction formula (2); m _Ti is the mass of titanium in the titanium alloy to be prepared with mass m

m _Ti in the formula (10) is calculated as follows:

m _Ti =m–m(a ₀ +a ₁ +a ₂ +a ₃ +...) (11);

Where a ₀ , a ₁ , a ₂ , a ₃ , ... are the mass percentages of the aluminum element in the titanium alloy and other alloying elements other than titanium and aluminum in the titanium alloy, respectively, in units of %.

When the titanium-containing cryolite is used as the solvent in the step 1, the aluminum-based alloy obtained in the step 2 is an aluminum-based alloy containing titanium and other alloying elements, and the content of the metal Ti in the aluminum-based alloy is m _Ti ⁰ . Then m _Ti in the formula (11) is calculated as follows:

m _Ti = m - m (a ₀ + a ₁ + a ₂ + a ₃ + ...) - m _Ti ⁰ (12);

4) When the titanium-containing cryolite is used as the solvent in the step 1, the mass percentage of the titanium element contained in the titanium-containing cryolite is r, and the amount of the titanium-containing cryolite is W, which is contained in the titanium-containing cryolite. The formula for calculating the titanium content is:

m _Ti ⁰ =rW (13);

According to the mechanism of chemical reaction formula (4)-(8) aluminum thermal reduction side reaction and the mechanism of disproportionation reaction of side reaction products, it can be considered that titanium in titanium-containing cryolite is mainly composed of trivalent fluoride of titanium and trivalent of titanium. disproportionation reaction product consisting of fluoride; it is considered that trivalent titanium rW by calculation, calculating the thermal reduction of aluminum content of titanium in the titanium aluminum cryolite Al ₃ may be used in the chemical reaction formula (14) (15) is calculated and :

TiF ₃ +Al=Ti+AlF ₃ (14);

Then the calculation formula of Al ₃ is:

Al ₃ =9m ⁰ _Ti /16=9rM/16 (15);

The calculation formula for the amount of all aluminum is:

Then Al = Al ₀ + Al ₁ + Al ₂ + Al ₃ (16).

In the above method, the compounding amount Q of Na ₂ TiF ₆ in the step 3 is calculated according to the chemical reaction formula (2):

Q = 13 m _Ti / 3η (17).

In the above method, in the step 1, when the cryolite is used as the flux, the molecular ratio of the cryolite is 1.5 to 3.0, and the amount of the cryolite is adjusted to ensure that the alumina formed by the reaction of the reaction formula (1) is dissolved in the cryolite. In the case of a melt, the concentration is less than the saturation concentration of alumina at the reduction temperature.

In the above method, due to the presence of raw material impurities and the aluminothermic reduction of sodium fluorotitanate in step 3, the influence of certain factors may not be fully quantitatively predicted, and the possible portion of Na ₂ TiF ₆ in the high temperature evaporation process. The amount of loss is unpredictable, which may cause errors in the amount of aluminum compounded according to the above theory and the amount of sodium fluorotitanate. This may cause errors in the mass percentage of the product components and the expected results; The ingredients can be appropriately modified to achieve the quality and composition of the product to the standard.

The sponge-like titanium alloy obtained by the method of the present invention can be ball-milled to form a powdered titanium alloy.

Detailed ways

The sodium fluorotitanate used in the examples of the present invention is a reagent grade sodium fluorotitanate.

The aluminum powder used in the embodiment of the present invention is a commercially available industrial grade aluminum air blown aluminum powder.

The vanadium oxide, zirconium oxide and molybdenum oxide used in the examples of the present invention are commercially available pure grade products.

Titanium in the titanium-containing cryolite used in the embodiment of the present invention exists in the form of TiF ₃ , Na ₃ TiF ₆ , Na ₂ TiF ₆ and metal Ti.

The cryolite used in the embodiment of the present invention is a cryolite for commercial aluminum electrolysis industry.

The mixing mill apparatus used in the embodiment of the present invention is a ball mill equipped with corundum alumina balls.

The pressing device used in the embodiment of the present invention is a 5-ton molding machine device for laboratory use.

The sponge-like titanium alloy obtained in the examples of the present invention is Ti-6Al-4V alloy and Ti-6.5Al-2Zr-1Mo-1V alloy, and the mass fraction of the alloying elements is given according to industry standards.

The vacuum condition in the embodiment of the present invention is that the degree of vacuum is less than 0.3 Pa.

The size of the agglomerate used for the thermal reduction of aluminum in each step in the examples of the present invention is Φ50 mm×50 mm.

Example 1

Using V ₂ O ₅ as the reactant, the cryolite with a molar ratio of NaF/AlF3 of 2.2 was used as the flux, and the aluminum-based vanadium alloy containing metal vanadium was prepared by the aluminothermic reduction method, and then the aluminum-based alloy was used as the reducing agent for the aluminum heat. Reduction of sodium fluorotitanate to prepare 100kg Ti-6Al-4V alloy; Ti-6Al-4V alloy is a titanium alloy widely used in the aerospace industry; Al in titanium alloy is between 5.5 and 6.8%, average 6%, V is between 3.4 and 4.5%, and the average value is 4%; the yield of titanium in the process of preparing aluminum alloy by heat-reducing Na ₂ TiF ₆ is 90%.

1) First, according to the alloy composition, determine 100kg Ti-6Al-4V alloy,

Al is 100 × 6% = 6kg (Al ₀ )

V is 100×4%=4kg

Ti is 100-6-4=90kg

2) calculated according to the formula V ₂ O ₅ in an amount 4kgV ₂ O ₅ V corresponding to 7.14kg;

3) Calculate the amount of aluminum Al ₁ required to reduce 7.14 kg of V ₂ O ₅ with aluminum, and calculate according to the following chemical reaction formula:

3V ₂ O ₅ +10Al=6V+5Al ₂ O ₃

Calculated that aluminum required to reduce 7.14kg V ₂ O ₅ aluminum Al ₁ is 3.53kg, the amount of Al ₂ O ₃ produced is 6.67kg;

4) Aluminothermic reduction of sodium fluorotitanate to produce 90 kg of titanium required aluminum amount of Al ₂ , calculated according to the actual yield of titanium of 90%; calculated by formulas (10) and (11): Al ₂ = 75 kg;

5) Calculated from the above: the amount of aluminum Al is compounded by using V ₂ O ₅ as a raw material and cryolite as a flux to obtain an aluminum-based alloy containing vanadium-containing alloying elements:

Al = Al ₀ + Al ₁ + Al ₂ = 6 + 3.53 + 75 = 84.53 kg;

6) The amount of cryolite solvent:

From the literature on aluminum electrolysis, the mass concentration of saturated concentration of Al ₂ O ₃ in cryolite with a molar ratio of NaF/AlF3 of 2.2 is 10% at a temperature of 1100 ° C, and is generated by thermal reduction of V ₂ O ₅ by aluminum. The amount of Al ₂ O ₃ is calculated to be 6.67 kg, and thus the amount y of the cryolite flux having a NaF/AlF3 molecular ratio of 2.2 should be determined as:

y>6.67kg/0.1=66.7kg,

Taking the actual amount of cryolite which has a molar ratio of NaF/AlF3 of 2.2 to 100 kg;

7) taking the cryolite composition as 100kg; arranging 84.53kg of aluminum powder with a particle size of less than 100 mesh and 100kg of cryolite and 7.14kg of V ₂ O ₅ powder with a molar ratio of 2.2 in a blender and pressing it into a pellet In the furnace, the aluminothermic reduction is carried out at 1100 ° C under an argon atmosphere; after the completion of the reduction reaction, the heating and cooling are stopped, and after cooling, the aluminum-based alloy containing vanadium is separated from the solid cryolite dissolved with Al ₂ O ₃ to obtain a vanadium-containing aluminum-based alloy;

8) forming an aluminum-based alloy containing vanadium into a powder having a particle size of less than 100 mesh;

9) Calculate the ingredient Q of Na2TiF6 required to obtain Na ₂ TiF ₆ by using this aluminum-based alloy powder to obtain Ti-6Al-4V alloy; according to formula (17), it can be obtained:

Q=433kg;

10) After mixing the aluminum-based alloy powder containing vanadium with 433 kg of sodium fluorotitanate, agglomerates are prepared, and then the agglomerates are placed in an aluminothermic reduction furnace, and the inert gas is protected and 1000~ The aluminothermic reduction reaction is carried out at a temperature of 1100 ° C, and after the completion of the reaction, vacuum distillation is carried out at a temperature of 1100 ° C to distill the cryolite formed by the reduction reaction; the distillate produced cryolite has a low by-produced The product of titanium fluoride and low-temperature titanium fluoride disproportionation reaction; the product left after vacuum distillation is spongy Ti-6Al-4V alloy, the alloy composition is Al (6±0.5)%, V(4± 0.5)%, the remainder is Ti, and the mass is 100±2kg.

Example 2

400kg of titanium-containing cryolite is present, and the mass percentage of titanium in the titanium-containing cryolite is 2.5%. The cryolite and V ₂ O _{5 are used} as raw materials for aluminum thermal reduction to prepare an aluminum-based alloy containing titanium and vanadium, and then The aluminum-based alloy is used as a reducing agent for aluminothermic reduction of sodium fluorotitanate to obtain 100 kg of Ti-6Al-4V alloy. The alloy composition is Al: 5.5-6.8% by mass, 6% average, and V: 3.4-4.5%. The average is 4%, the remainder is Ti, and the actual yield of titanium in the process of aluminum thermal reduction of sodium fluorotitanate is 90%.

1) First determine the 100kg Ti-6Al-4V alloy according to the alloy composition:

Al is 100 × 6% = 6kg

V is 100×4%=4kg

Ti is 100-6-4=90kg

2) Calculate 4kgV, the corresponding amount of V ₂ O ₅ is 7.14kg;

3) Calculate the amount of aluminum Al ₁ and the amount of Al ₂ O ₃ required to form a metal V by aluminothermic reduction of 7.14 kg of V ₂ O ₅ according to the following aluminothermic reduction equation:

3V ₂ O ₅ +10Al=6V+5Al ₂ O ₃

Drawn aluminum reduction 7.14kg V ₂ O ₅ required amount of aluminum Al ₁ is 3.53kg, the resulting Al ₂ O ₃ is 6.67kg;

4) Calculate the amount of aluminum required to reduce the titanium in the titanium-containing cryolite according to formula (15) Al ₃ = 5.625kg;

5) Calculate m _Ti ⁰ = rW = 400 × 2.5% = 10 kg according to formula (13);

6) Calculate m _Ti =100-6-4-10=80kg according to formula (12)

7) Calculate the required amount of aluminum Al ₂ = 66.7 kg according to formula (10);

8) From the above calculation, the aluminum content of aluminum of V ₂ O ₅ and the aluminum-based alloy containing V and Ti prepared by using the above titanium-containing cryolite as the raw material and aluminum as the reducing agent is:

Al = Al ₀ + Al ₁ + Al ₂ + Al ₃ = 6 + 3.53 + 66.7 + 5.6 = 81.38 kg;

9) 81.38kg-100 mesh aluminum powder, 7.14kg V ₂ O ₅ and 400kg titanium-containing cryolite are mixed and ground, pressed into agglomerate, placed in aluminothermic reduction furnace, protected at 1100 ° C and argon gas Aluminothermic reduction is carried out under the atmosphere; under this temperature condition, the metal titanium reduced by the reduction of V ₂ O ₅ by the aluminum and the titanium of the titanium containing the titanium cryolite are reduced into the aluminum liquid; The Al ₂ O ₃ formed by the reduction of V ₂ O ₅ is dissolved in the molten cryolite; after cooling at the furnace temperature, the cooled reaction product is taken out from the furnace, and the solid cryolite containing alumina is mixed with V and Separation of Ti-based aluminum-based alloy ingots;

10) forming an aluminum-based alloy ingot containing V and Ti into a particle size of -100 mesh powder to obtain an aluminum-based alloy powder;

11) Calculate the amount of sodium fluorotitanate required to prepare 100 kg of Ti-6Al-4V alloy by reducing the sodium fluorotitanate with this aluminum-based alloy powder, and calculate according to formula (17):

12) Mixing 385kg of sodium fluorotitanate with aluminum-based alloy powder containing titanium and vanadium, pressing into agglomerates, placing them in an aluminothermic reduction furnace, and completing aluminum under argon protection and temperature of 1100 °C. After the thermal reduction reaction, vacuum distillation is carried out at a temperature of 1,100 ° C to distill off the cryolite formed by the reaction and the low-valent fluoride of the titanium produced by the side reaction, and the distilled cryolite is titanium-containing cryolite; The remaining reaction product of cryolite and titanium low-valent fluoride is the sponge-like Ti-6Al-4V alloy to be prepared, and the mass is 100 kg.

Example 3

ZrO ₂ , MoO ₃ and V ₂ O _{5 were used} as raw materials, and aluminum-based Zr, Mo, V alloys were prepared by using cryolite with a molar ratio of NaF/AlF3 of 2.2 as the flux, and then the aluminum-based alloy was used as a reducing agent for aluminum heat. Reducing sodium fluorotitanate to obtain 100kg of titanium alloy with material number TA15 and foreign similar grade Ti-6.5Al-2Zr-1Mo-1V. The mass percentage of Al in this alloy is between 6.0 and 7.0%, with an average of 6.5. %, Zr has a mass percentage of 1.7 to 2.3%, an average of 2.0%, a mass percentage of Mo of 0.7 to 1.2%, an average of 1%, a mass percentage of V of 0.7 to 1.3%, and an average of 1%; The yield of titanium in the process of reducing titanium fluorotitanate to obtain titanium is 85%.

Proceed as follows:

1) Determine the mass of each component in 100kg Ti-6.5Al-2Zr-1Mo-1V alloy according to the alloy composition:

Al: 100 × 6.5% = 6.5 kg,

Zr: 100 × 2% = 2kg,

Mo: 100 × 1% = 1 kg,

V: 100 × 1% = 1 kg,

Ti: 100-6.5-2-1-1 = 89.5 kg;

2) The mass of the oxide corresponding to the mass of Zr, Mo, V alloy is:

The mass of ZrO2 corresponding to 2kgZr is 2.7kg.

The mass of MoO3 corresponding to 1kgMo is 1.5kg.

The corresponding V ₂ O ₅ mass of 1kgV is 1.78kg;

3) Calculate the amount of aluminum Al ₁ required to reduce ZrO2 with aluminum, reduce MoO3 and V ₂ O ₅ to metal Zr, Mo, V, and the amount of Al ₂ O ₃ formed according to chemical reaction equation (1);

The amount of aluminum required to reduce 1.78 kg of V ₂ O ₅ is 0.89 kg, and the mass of Al ₂ O ₃ produced is 1.67 kg.

The amount of aluminum required to reduce 1.5 kg of MoO3 is 0.56 kg, and the mass of Al ₂ O ₃ produced is 1.06 kg.

The amount of aluminum required to reduce 2.7 kg of ZrO2 is 0.79 kg, and the mass of Al ₂ O ₃ produced is 1.5 kg.

From the above calculation, Al ₁ = 0.89 + 0.56 + 0.79 = 2.24 kg;

Al ₂ O ₃ mass generated by the thermal reduction of the above three oxides = 1.67 + 1.06 + 1.5 = 4.23 kg;

4) Calculate the amount of aluminum Al ₂ required for the thermal reduction of Na ₂ TiF ₆ to form 89.5 kg of titanium according to formula (10), according to which Al ₂ = 79 kg;

5) Calculate the amount of aluminum required to obtain ZrO ₂ , MoO ₃ and V ₂ O ₅ as raw materials with cryolite as flux and aluminum to reduce the aluminum alloy containing metal alloys of Zr, Mo and V. Al is:

Al = 6.5 + 2.24 + 79 = 87.74 kg;

6) The amount of cryolite flux

From step 3), the total amount of Al ₂ O ₃ produced by the aluminothermic reduction of ZrO ₂ , MoO ₃ and V ₂ O ₅ is:

1.67+1.06+1.5=4.23kg;

From the aluminum electrolysis work, the saturation concentration of Al ₂ O ₃ was 10% in the cryolite with a molar ratio of NaF/AlF ₃ of 2.2 at a temperature of 1100 ° C, and the solution was dissolved to 4.23 kg of Al ₂ O ₃ . The amount of cryolite required y:

y>4.23/0.1=42.3kg;

In order to enable the alumina to be fully dissolved in the cryolite, the amount of cryolite is 100 kg;

7) 100 kg of cryolite with a molar ratio of NaF/AlF3 of 2.2 and 87.74 kg of -100 mesh aluminum powder, 1.78 kg of V ₂ O ₅ , 1.5 kg of MoO ₃ and 2.7 kg of ZrO ₂ are mixed and pressed to form a dough, which is placed in a reduction furnace. In the thermal reduction of aluminum at 1000-1300 ° C and argon protection; aluminum thermal reduction produces an aluminum-based alloy containing alloying elements Zr, Mo and V, sinking into the lower part of the cryolite melt in which alumina is dissolved After the reduction reaction is completed, the furnace body is cooled, and after the reduction furnace is cooled to room temperature, the aluminum-based alloy is separated from the cryolite to obtain an aluminum-based alloy containing Zr, Mo, and V;

8) an aluminum-based alloy containing Zr, Mo and V prepared by aluminothermic reduction to prepare a powder having a particle size of less than 100 mesh;

9) Calculate the Na ₂ TiF ₆ compound required for the reduction of Na ₂ TiF ₆ to obtain a Ti-6.5Al-2Zr-1Mo-1V titanium alloy by using the above aluminum-based alloy powder; the yield of titanium in the reduction process is 85%. , the formula Q of Na ₂ TiF ₆ can be calculated according to formula (17):

Q=456.3kg

10) The aluminum-based alloy powder obtained in the step 8) is mixed with 456.3 kg of Na ₂ TiF ₆ , and then pressed into agglomerates, and the agglomerates are placed in an aluminothermic reduction furnace at a temperature of 1100 ° C and argon gas. Under the condition of protection, the aluminothermic reduction is carried out, and after the aluminothermic reduction reaction is completed, vacuum distillation is carried out at a temperature of 1100 ° C; the cryolite formed by the reaction and the low-valent fluoride formed by the side reaction are distilled off to obtain Ti-6.5Al-2Zr-1Mo. -1 V of titanium alloy 100 ± 2 kg, the product after vacuum distillation is titanium-containing cryolite.

Example 4

400 kg of titanium-containing cryolite, the mass percentage of titanium in the titanium-containing cryolite is 4%, and Zr, Mo and V are prepared by using ZrO ₂ , MoO ₃ and V ₂ O ₅ and the titanium-containing cryolite as raw materials. Aluminum-based alloy, and then using the aluminum-based alloy and Na ₂ TiF ₆ as raw materials to prepare 100 kg Ti-6.5Al-2Zr-1Mo-1V titanium alloy; setting aluminum during the process of preparing titanium by thermal reduction of sodium fluorotitanate The yield was 85%.

1) Determine the amount of alloying elements in 100kg Ti-6.5Al-2Zr-1Mo-1V:

Al: 100 × 6.5% = 6.5 kg,

Zr: 100 × 2% = 2kg,

Mo: 100 × 1% = 1 kg,

V: 100 × 1% = 1 kg,

Ti: 100-6.5-2-1-1 = 89.5 kg;

2) Calculate the oxide quality corresponding to the mass of Zr, Mo and V alloy elements:

The mass of ZrO ₂ corresponding to 2kgZr is 2.7kg.

The mass of MoO ₃ corresponding to 1kgMo is 1.5kg.

The corresponding V ₂ O ₅ mass of 1kgV is 1.78kg;

3) according to formula (1) is calculated aluminum reduction ZrO _2, MoO ₃ and V ₂ O ₅ is reduced to metals Zr, Mo and V, the amount of aluminum needed Al ₁ = 2.24kg;

4) According to formula (15), the amount of aluminum required to reduce the amount of aluminum in the titanium-containing cryolite is Al ₃ = 9 kg;

5) Calculate m _Ti ⁰ = rW = 400 × 4% = 16kg according to formula (13);

6) Calculate m _Ti = 89.5-16 = 73.5 kg according to formula (12);

7) Calculate Al ₂ =65.04kg according to formula (10);

8) From the above calculation, 2.7 kg of ZrO ₂ , 1.5 kg of MoO ₃ , 1.78 kg of V ₂ O ₅ and 400 kg of titanium-containing cryolite containing 4% of titanium are used as raw materials, and aluminum is thermally reduced to obtain aluminum containing Ti, Zr, Mo and V. The amount of aluminum required for the base alloy is:

Al = Al ₀ + Al ₁ + Al ₂ + Al ₃ = 6.5 + 2.24 + 65.04 + 9 = 82.78 kg;

9) 82.78kg-100 mesh aluminum powder and 400kg titanium-containing cryolite and 2.7kg ZrO ₂ , 1.5kg MoO ₃ and 1.78kg V ₂ O _{5 are} mixed and pressed into agglomerates, which are placed in an aluminothermic reduction furnace at 1000~ An aluminum-based alloy containing Ti, Zr, Mo and V is obtained by heat reduction at 1300 ° C under argon gas protection. After cooling at the furnace temperature, the cooled reaction product is taken out from the furnace to dissolve the solid of alumina. Cryolite is separated from an aluminum-based alloy ingot containing Ti, Zr, Mo, and V;

10) preparing the obtained aluminum-based alloy into a powder of -100 mesh to obtain an aluminum-based alloy powder;

11) Reducing Na ₂ TiF ₆ with the above aluminum-based alloy powder to prepare Na ₂ TiF ₆ furnish Q required for Ti-6.5Al-2Zr-1Mo-1V titanium alloy, and calculating according to formula (17):

Q=375kg;

12) 375g Na ₂ TiF ₆ and aluminum-based alloy powder were mixed and ground to form agglomerate, then the agglomerate was placed in an aluminothermic reduction furnace, and the aluminothermic reduction was completed under the condition of 1100 ° C and argon gas protection. After vacuum distillation at a temperature of 1100 ° C, the distilled cryolite contains titanium and titanium fluoride, called titanium-containing cryolite, and the remaining product of distillation is the sponge-like titanium alloy Ti-6.5Al to be prepared. -2Zr-1Mo-1V100kg.

Claims (9)

1. A method for preparing a titanium alloy, which is characterized by the following steps:

   (1) preparing a solvent, a reactant and a reducing agent as raw materials, wherein the solvent is cryolite or titanium-containing cryolite, the reactant is an oxide of an alloying element of a titanium alloy, and the reducing agent is aluminum powder having a particle size of ≤100 mesh; The oxide of the alloying element of the alloy is M _x O _y , wherein the element M is an alloying element other than titanium and aluminum in the prepared titanium alloy; the molar ratio of NaF/AlF ₃ in the cryolite is 1.5 ～ 3.0;

   (2) Adding all the raw materials to the mixing mill for mixing and pressing to form agglomerates; placing the agglomerates in a reduction furnace, performing an aluminothermic reduction reaction under an argon atmosphere and at 1000 to 1500 ° C to make oxides M _x O _y is reduced by aluminum to a metal element dissolved in an aluminum melt to form an aluminum-based alloy melt containing element M, or an aluminum-based alloy melt containing titanium and element M is formed, and the alumina formed by the reduction reaction is dissolved to In the cryolite melt, a cryolite melt in which alumina is dissolved is formed; after the aluminothermic reduction reaction is completed, the reduced product is cooled to a normal temperature, and the aluminum-based alloy melt and the cryolite melt are solidified in the lower layer and the upper layer, respectively, and the lower layer is Separating the aluminum-based alloy from the upper layer of alumina-containing cryolite;

   (3) The aluminum-based alloy is made into a powder with a particle size of ≤100, and then mixed with the sodium fluorotitanate powder to be pressed into a briquettes, and the agglomerate is placed in a reduction furnace in an argon atmosphere and 900~ The aluminothermic reduction reaction is carried out at 1200 ° C. The aluminum in the agglomerate reduces the titanium in the sodium fluorotitanate to form a titanium alloy and titanium-containing cryolite; after the thermal reduction reaction of aluminum is completed at 1000-1200 ° C under vacuum conditions Distillation was carried out, and the titanium-containing cryolite was distilled and separated to obtain a sponge-like titanium alloy.
2. The method for preparing a titanium alloy according to claim 1, wherein the reaction formula of the aluminothermic reduction reaction in the step (2) is:

   M _x O _y + 2y / 3Al = xM + y / 3Al ₂ O ₃ (1).
3. The method for preparing a titanium alloy according to claim 1, wherein the titanium-containing cryolite according to the step (1) is composed of a titanium-containing component and a cryolite component, wherein a molar ratio of NaF/AlF _{3 in the} cryolite is It is 1.5 to 3.0, and the titanium-containing component is TiF ₃ , Na ₃ TiF ₆ , Na ₂ TiF ₆ and/or metal titanium.
4. The method for preparing a titanium alloy according to claim 1, wherein in the above method, the main reaction formula of the secondary aluminothermic reduction reaction of the step (3) is:

   3Na ₂ TiF ₆ +4Al=3Ti+Na ₃ AlF ₆ +3NaAlF ₄ (2)

   Or 3Na ₂ TiF ₆ +4Al=3Ti+Na ₃ AlF ₆ +3NaF+3AlF ₃ (3);

   The reaction formula of the side reaction is:

   Na ₂ TiF ₆ = 2NaF + TiF ₄ (4),

   3TiF ₄ +Al=3TiF ₃ +AlF ₃ (5)

   And/or TiF ₃ +3NaF=Na ₃ TiF ₆ (6).
5. The method for preparing a titanium alloy according to claim 1, wherein the Na ₃ AlF ₆ and NaAlF ₄ formed in the aluminothermic reduction reaction in the step (3) are respectively referred to as NaF/AlF ₃ molar ratios. Cryolites of 1.0 and 3.0; or cryolites having different molar ratios of NaF/AlF ₃ .
6. The method for preparing a titanium alloy according to claim 1, characterized in that part of the TiF in the process of distilling out the ferric TiF _{3 of the} trivalent titanium formed by the aluminothermic reaction of the step (3) together with the cryolite ₃ Disproportionation reaction produces TiF ₄ and Ti, and its reaction formula is:

   4TiF ₃ =3TiF ₄ +Ti (7)

   And TiF ₄ + 2NaF = Na ₂ TiF ₆ (8).
7. The method for preparing a titanium alloy according to claim 1, wherein the amount of aluminum in the step (1) is determined according to the mass percentage of the m alloy and the alloying element to be obtained. Calculation; the amount of aluminum Al in the step (1) includes the sum of the aluminum requirements of the following parts:

   1) reduction of titanium oxide in a given mass of alloying elements other than titanium and aluminum by chemical equation (1) the desired amount of aluminum Al ₁

   2) If the titanium alloy elements aluminum, the amount of this portion of the aluminum with Al ₀ is calculated as:

   Al ₀ = a ₀ m (9);

   Where a ₀ is the mass percentage of the alloying element Al in the m-mass titanium alloy, and the unit is %.

   3) for the chemical reaction equation (2) acquired by aluminothermic reduction of Aluminum alloy mass m is calculated as Al ₂ required:

   Al ₂ = (108 × m _Ti ) / (144 × η) (10);

   Where η is the actual yield of titanium in the process of preparing titanium by reducing Na ₂ TiF ₆ according to reaction formula (2); m _Ti is the mass of titanium in the titanium alloy to be prepared with mass m.

   m _Ti in the formula (10) is calculated as follows:

   m _Ti =m–m(a ₀ +a ₁ +a ₂ +a ₃ +...) (11);

   Where a ₀ , a ₁ , a ₂ , a ₃ , ... are respectively the mass percentage of the aluminum element in the titanium alloy and other alloying elements other than titanium and aluminum in the titanium alloy, and the unit is %. ;

   When the titanium-containing cryolite is used as the solvent in the step (1), the aluminum-based alloy obtained in the step (2) is an aluminum-based alloy containing titanium and other alloying elements, and the content of the metal Ti in the aluminum-based alloy is m _Ti ^0, the formula (11) m _Ti is calculated in the following manner:

   m _Ti = m - m (a ₀ + a ₁ + a ₂ + a ₃ + ...) - m _Ti ⁰ (12);

   4) When titanium-containing cryolite is used as the solvent in the step (1), the mass percentage of the titanium element contained in the titanium-containing cryolite is r, and the titanium-containing cryolite is W, and the titanium-containing cryolite is contained. The formula for calculating the titanium content is:

   m _Ti ⁰ =rW (13);

   According to the mechanism of chemical reaction formula (4)-(8) aluminum thermal reduction side reaction and the mechanism of disproportionation reaction of side reaction products, it can be considered that titanium in titanium-containing cryolite is mainly composed of trivalent fluoride of titanium and trivalent fluoride. disproportionation reaction product composition; rW can therefore be considered to trivalent titanium calculation, calculating the thermal reduction of aluminum containing titanium, aluminum Al cryolite amount of titanium may be used in the chemical reaction formula ₃ (14) and (15) calculates the formula for:

   TiF ₃ +Al=Ti+AlF ₃ (14);

   Then the calculation formula of Al ₃ is:

   Al ₃ =9m ⁰ _Ti /16=9rW/16 (15);

   Then the calculation formula of the total aluminum dosage in step (1) is:

   Then Al = Al ₀ + Al ₁ + Al ₂ + Al ₃ (16).
8. The method for preparing a titanium alloy according to claim 1, wherein the compounding amount Q of Na ₂ TiF ₆ in the step (3) is calculated according to the chemical reaction equation (2):

   Q = 13 m _Ti / 3η (17).
9. The method for preparing a titanium alloy according to claim 1, wherein in the step (1), when the cryolite is used as a flux, the molecular ratio of the cryolite is 1.5 to 3.0, and the cryolite is blended. The amount should be such that when the alumina formed by the reaction of the reaction formula (1) is dissolved in the cryolite melt, the concentration thereof is smaller than the saturation concentration of the alumina at the reduction temperature.