材料科学基础-第7章2合金相图_ppt课件

11o 二元共晶相图o Pb-Sb、 Pb-Sn、 Ag-Cu、 Al-Si、 Zn-Sn等合金相图都属于共晶相图，在 Mg-Al及 Fe-Fe3C相图中也包含有共晶部分。o 共晶反应；共晶组织（共晶体）Section 7.4 二元共晶相图及其合金的结晶图 7-13 Pb-Sn二元相图22o 点、线、区分析o 、 固溶体；最大溶解度；室温下溶解度；熔点，共晶点o 三相共存水平线（区）o 边际（端部）固溶体合金；共晶合金；亚共晶合金；过共晶合金o 相律应用7.4.1 二元共晶相图分析图 7-13 Pb-Sn二元相图33o Solvus - A solubility curve that separates a single-solid phase region from a two-solid phase region in the phase diagram.o Isopleth - A line on a phase diagram that shows constant chemical composition.o Hypoeutectic alloy - An alloy composition between that of the left-hand-side end of the tie line defining the eutectic reaction and the eutectic composition.o Hypereutectic alloys - An alloy composition between that of the right-hand-side end of the tie line defining the eutectic reaction and the eutectic composition.The Eutectic Phase Diagram44o 边际（端部）固溶体，二次结晶 o 共晶合金o 亚共晶合金o 过共晶合金7.4.2典型合金的平衡结晶过程及室温平衡组织55Figure the lead-tin equilibrium phase diagram.6788Determine (a) the solubility of tin in solid lead at 100oC, (b) the maximum solubility of lead in solid tin, (c) the amount of that forms if a Pb-10% Sn alloy is cooled to 0oC, (d) the masses of tin contained in the and phases, and (e) mass of lead contained in the and phases. Assume that the total mass of the Pb-10% Sn alloy is 100 grams.Example Phases in the LeadTin (Pb-Sn) Phase Diagram99Figure Solidification, precipitation, and microstructure of a Pb-10% Sn alloy. Some dispersion strengthening occurs as the solid precipitates.1010Example SOLUTION(a) The 100oC temperature intersects the solvus curve at 5% Sn. The solubility of tin (Sn) in lead (Pb) at 100oC therefore is 5%.(b) The maximum solubility of lead (Pb) in tin (Sn), which is found from the tin-rich side of the phase diagram, occurs at the eutectic temperature of 183oC and is 97.5% Sn.(c) At 0oC, the 10% Sn alloy is in a + region of the phase diagram. By drawing a tie line at 0oC and applying the lever rule, we find that:1111Example SOLUTION (Continued)(d) The mass of Sn in the phase = 2% Sn 91.8 g of phase = 0.02 91.8 g = 1.836 g. Since tin (Sn) appears in both the and phases, the mass of Sn in the phase will be = (10 1.836) g = 8.164 g.(e) Mass of Pb in the phase = 98% Sn 91.8 g of phase = 0.98 91.8 g = 89.964 gMass of Pb in the phase = 90 - 89.964 = 0.036 g. 1212Figure Summary of calculations1313Figure Solidification and microstructure of the eutectic alloy Pb-61.9% Sn.1414Figure The cooling curve for a eutectic alloy is a simple thermal arrest, since eutectics freeze or melt at a single temperature.1515Figure (a) Atom redistribution during lamellar growth of a lead-tin eutectic. Tin atoms from the liquid preferentially diffuse to the plates, and lead atoms diffuse to the plates. (b) Photomicrograph of the lead-tin eutectic microconstituent (x400).1616Example Amount of Phases in the Eutectic Alloy(a) Determine the amount and composition of each phase in a lead-tin alloy of eutectic composition. (b) Calculate the mass of phases present. (c) Calculate the amount of lead and tin in each phase, assuming you have 200 g of the alloy.Example SOLUTION(a) The eutectic alloy contains 61.9% Sn.1717Example SOLUTION (Continued)(b) At a temperature just below the eutectic:The mass of the phase in 200 g of the alloy = mass of the alloy fraction of the a phase = 200 g 0.4535 = 90.7 gThe amount of the phase in 200 g of the alloy = (mass of the alloy mass of the a phase) = 200.0 g 90.7 g = 109.3 g1818Example SOLUTION (Continued)- Mass of Pb in the phase = mass of the a phase in 200 g (concentration of Pb in ) = (90.7 g) (1 0.190) = 73.467 g- Mass of Sn in the phase = mass of the a phase - mass of Pb in the a phase = (90.7 73.467 g) = 17.233 g- Mass of Pb in phase = mass of the b phase in 200 g (wt. fraction Pb in ) = (109.3 g) (1 0.975) = 2.73 g- Mass of Sn in the phase = total mass of Sn mass of Sn in the phase = 123.8 g 17.233 g = 106.57 g1919Figure Summary of calculations 2020Figure The solidification and microstructure of a hypoeutectic alloy (Pb-30% Sn).2121Figure (a) A hypoeutectic lead-tin alloy. (b) A hypereutectic lead-tin alloy. The dark constituent is the lead-rich solid , the light constituent is the tin-rich solid , and the fine plate structure is the eutectic (x400).2222Example Determination of Phases and Amounts in a Pb-30% Sn Hypoeutectic AlloyFor a Pb-30% Sn alloy, determine the phases present, their amounts, and their compositions at 300oC, 200oC, 184oC, 182oC, and 0oC.2323Example SOLUTION2424Example Microconstituent Amount and Composition for a Hypoeutectic AlloyDetermine the amounts and compositions of each microconstituent in a Pb-30% Sn alloy immediately after the eutectic reaction has been completed.Example SOLUTIONAt a temperature just above the eutecticsay, 184 oCthe amounts and compositions of the two phases are:2525Figure The cooling curve for a hypoeutectic Pb-30% Sn alloy.2626o Eutectic Colony Sizeo Interlamellar Spacingo Amount of Eutectico Microstructure of the EutecticStrength of Eutectic Alloys2727Figure (a) Colonies in the lead-tin eutectic (x300). (b) The interlamellar spacing in a eutectic microstructure.2828Figure The effect of growth rate on the interlamellar spacing in the lead-tin eutectic.2929Example Design of a Directional Solidification ProcessDesign a process to produce a single grain of Pb-Sn eutectic microconstituent in which the interlamellar spacing is 0.00034 cm.Example SOLUTIONWe could use a directional s