What will be the depth of penetration of a nonferrous alloy heated from 30–600°C at 10,000 Hz. The resistivity of the alloy is same as that of steel in the previous problem.

Question

Accepted Answer

For the concerned alloy [latex]μ_γ[/latex] = 1

[latex]\delta=5030 \sqrt{\frac{ ho}{1 imes 10^4}}=50.30 imes \sqrt{ ho}[/latex]

For 30°C

[latex]\delta_{30}=50.30 imes \sqrt{16 imes 10^{-6}} \ =0.2 \ cm[/latex]

Similarly

[latex]\begin{aligned}& \delta_{200}=0.276\ \mathrm{~cm} \& \delta_{400}=0.353\ \mathrm{~cm} \& \delta_{600}=0.440\ \mathrm{~cm}\end{aligned}[/latex]

It can be seen that the depth of penetration increases very slowly and can be treated as constant at some mean value.
In the last example, the depth of penetration, affected by changes in resistivity and permeability shows an abrupt change at the Curie point (760°C).

Question 10.2: What will be the depth of penetration of a nonferrous alloy ......

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