| Research shows that adding Ti, Zr and B in the aluminum alloy, and rational allocation of their proportional share can improve the heat resistance of aluminum alloy, while maintaining good electrical conductivity and strength. But the theory research is less.The product and the effect on the microstructure of aluminium alloy of adding Ti, Zr and B in pure aluminium are researched by scanning electron microscopy and Transmission electron microscope. Experiments show that there are hexagon TiB2, ZrB2and cubic Al3Ti, Al3Zr in the presence of3different samples of the atomic number ratio (Ti+Zr):B>1:2; while only hexagon TiB2, ZrB2is found in3different samples of atomic number ratio (Ti+Zr):B=1:2. In addition the metallographic results show that grain size of3samples of atomic number ratio (Ti+Zr):B>1:2is generally less than that of3samples of the atomic number ratio (Ti+Zr):B=1:2. When the atomic number ratio (Ti+Zr):B=1:2, refining effect is not obvious, electron scattering is less, the conductive property is less influenced.By the bond length difference method (referred to as BLD) of EET, the bond energy and covalent electron number of TiB2, ZrB2, Al3Ti and Al3Zr were calculated. The results show that the energy of the strongest bond of TiB2, the covalent electron number of the strongest bond is far greater than the energy of the strongest bond of Al3Ti, the covalent electron number of the strongest bond, so TiB2has priority, when B is consumed, Ti and Al bind to Al3Ti The energy of the strongest bond of ZrB2, the covalent electron number of the strongest bond is far greater than the energy of the strongest bond of Al3Zr, the covalent electron number of the strongest bond, so ZrB2has priority, when B is consumed, Zr and Al are combined into Al3Zr. Hexagon lattice structure of TiB2, ZrB2prior to the cubic lattice structure of Al3Ti and Al3Zr. This explains the results of scanning electron microscope:when the atomic number ratio (Ti+Zr):B=1:2, only hexagon lattice existence (TiB2, ZrB2) is found; when the atomic number ratio (Ti+Zr):B>1:2, there are six square lattice (TiB2, ZrB2) and cubic lattice (Al3Ti, Al3Zr).Surface electron density differences of α-Al with TiB2, ZrB2, Al3Ti an Al3Zr were calculated. Surface electron density difference (1.11%) of Al3Ti (111) and α-Al (111) is far less than that(40.88%) of TiB2(0001) and α-Al (110); Surface electron density difference (20.05%) of Al3Zr (111) and α-Al (111) is far less than that (87.09%) of ZrB2(0001) and α-Al (110). The lower the surface electron density is, the better the coherent relationship between the two substances, the more heterogeneous nucleation in α-Al, the better the effect of refinement. So the grain size of the atomic number ratio (Ti+Zr):B>1:2is generally less than that of the atomic number ratio (Ti+Zr):B=1:2. This is consistent with the result of metallographic experiment. And the smaller the grain, the electron scattering from grain boundary is more serious, and the conductivity of the alloy is reduced. Therefore the electrical conductivity of aluminum alloy is better when the atomic number ratio is (Ti+Zr):B=1:2and only TiB2and ZrB2generate.The Peierls-Nabarro forces of α-Al, TiB2, ZrB2, Al3Ti and Al3Zr were calculated. The results are as follows:0.05002,14.03272,7.20934,0.43760,0.61224. The Peierls-Nabarro forces of TiB2, ZrB2, Al3Ti and Al3Zr are so large that dislocations can not cut them and can only take the bypass mechanism. When a large of fine TiB2, ZrB2, Al3Ti and Al3Zr disperse in the aluminum matrix, dislocation will be pinned well and recrystallization temperature will be improved and aluminum alloy will have good heat resistance.After analyzing the influence of conductivity and heat-resistance of aluminum alloy of adding different content Ti, Zr and B, we should make the atom ratio of Ti, Zr and B (Ti+Zr):B=1:2. Then the aluminum alloy can keep good heat-resistance and does not reduce the conductivity. |
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