Question

求碳质量分数0.77%＜w(C)＜2.11%和碳质量分数2.11%＜w(C)＜4.3%的铁碳合金在室温下的各组织的含量。

Answer 1

【答案】：含碳质量分数0.77%＜X₃＜2.11%铁碳合金室温下的组织为P+Fe₃C(珠光体+渗碳体)。
w(P)=(6.69-X₃)/(6.69-0.77)×100%
w(Fe₃C)=(X₃-0.77)/(6.69-0.77)×100%碳质量分数2.11%＞X₄＞4.3%铁碳合金在室温下组织为(P+Fe₃C)+L_d(P+C_m+Ld)
第一步：w(A)=(4.33-X₄)/(4.33-2.11)×100%
w(L_d)=(X₄-2.11)/(4.33-2.11)×100%
第二步：w(P)=[(6.69-2.11)/(6.69-0.77)]×[(4.33-X₄)/(4.33-2.11)]×100%
w(Fe₃C)=(2.11-0.77)/(6.69-0.77)×100%