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Solidification and cast structures

Suppose that the secondary dendrite arm spacing λ₂ in an aluminum alloy casting is related to the cooling rate Ṫ by λ₂ = A·Ṫ^(−1/3). In a region cooled at 1.0 K/s, λ₂ = 50 µm. Which is closest to λ₂ in a region cooled at 125 K/s?

  1. AAbout 4.5 µm
  2. BAbout 10 µm
  3. CAbout 0.40 µm
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Answer: B. About 10 µm

λ₂ is inversely proportional to the cube root of the cooling rate, so when the cooling rate increases 125-fold, λ₂ becomes 1/125^(1/3) = 1/5, giving 50/5 = 10 µm. About 0.40 µm assumes that λ₂ is inversely proportional to the cooling rate and divides by 125, and about 4.5 µm uses an exponent of 1/2 and divides by √125 ≈ 11.2. The exponent is about 1/3 because coarsening during solidification, in which thin arms dissolve and are absorbed into thicker ones, proceeds in proportion to the cube root of the local solidification time. λ₂ serves as a guide to the cooling rate in each part of a casting, and the smaller λ₂, the finer the interdendritic segregation, microshrinkage and intermetallic compounds, which improves the tensile strength, elongation and fatigue strength.

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  1. Segregation in castings and ingots includes microsegregation, on the scale of the spaces between dendrite arms, and macrosegregation, on the scale of the whole ingot. Which statement about the difference between them is correct?
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