已知18MnMoNb中,ω(C)=0.17%~0.23%,ω(Mn)=1.35%~1.65%,ω(Si)=0.17%~0.27%,ω(Mo)=0.45%~0.65%,ψ(Nb)=0.025%~0.050%,求碳当量?

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问题 已知18MnMoNb中,ω(C)=0.17%~0.23%,ω(Mn)=1.35%~1.65%,ω(Si)=0.17%~0.27%,ω(Mo)=0.45%~0.65%,ψ(Nb)=0.025%~0.050%,求碳当量?

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答案ω(CE)=ω[C+Mn/6+(Cr+Mo+V)/5+(Ni+Cu)/15] =[(0.17—0.23)+(1.35~1.65)/6+(0.45~0.65)/5+0/15]% =(0.485~0.635)%。 答:碳当量为(0.485~0.635)%。

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