High-temperature materials and creep

When an alloy was oxidized in air at high temperature, it followed the parabolic law x² = k_p t, in which diffusion through the dense oxide scale controls the rate, and the scale was 2.0 µm thick after 100 h. Which is closest to the scale thickness after 900 h of oxidation at the same temperature?

  1. AAbout 18 µm
  2. BAbout 6.0 µm
  3. CAbout 4.2 µm
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Answer: B. About 6.0 µm

Under the parabolic law the thickness is proportional to the square root of time, so x = 2.0 × √(900/100) = 2.0 × 3 = 6.0 µm. As the scale thickens, the distance that ions and electrons must travel through it grows and growth slows, giving this form (Wagner's theory). About 18 µm is calculated with the linear law, in which thickness is proportional to time; the linear law is seen when the scale is porous or cracked and not protective, or when a reaction at an interface controls the rate. About 4.2 µm is calculated with the cubic law, in which thickness is proportional to the cube root of time. The smaller the parabolic rate constant k_p of a scale, the more protective it is, and this serves as a guide in choosing alloys.

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  1. The Pilling–Bedworth ratio of a metal's oxide scale (PBR, the ratio of the volume of oxide to the volume of metal used to form it) is calculated as PBR = (M_ox ρ_M)/(n M_M ρ_ox) (M is the molar mass, ρ is the density, and n is the number of metal atoms in one formula unit of the oxide). For magnesium (24.31 g/mol, 1.74 g/cm³) and MgO (40.30 g/mol, 3.58 g/cm³, n = 1), which combination of PBR and scale behavior is correct?
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