Determine the depth of penetration for 0.4% carbon steel heated from 30–900°C at 1000 Hz. The resistivity and permeability at various temperatures is given below.

Temperature °C | Resistivity \rho \times 10^{-6} ohm.cm | Permeability \mu \gamma |

30 | 16 | 40 |

200 | 29.6 | 36 |

400 | 49.3 | 30 |

600 | 76.6 | 15 |

760 | 104.2 | 1 |

900 | 114.6 | 1 |

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\delta=5030 \sqrt{\frac{\rho}{\mu_\gamma f}} \ \mathrm{~cm} \\ For \ 30°C \\ f = 1000 \ Hz, \ \rho = 16 \times 10^{-6} \ ohm.cm, \ \mu_\gamma = 40

Substituting

\begin{aligned}\delta_{30} & =5030 \sqrt{\frac{16 \times 10^{-6}}{40 \times 10^3}} \\& =0.101 \ \mathrm{~cm}\end{aligned}Similarly

\begin{aligned}& \delta_{200}=0.135\ \mathrm{~cm} \\& \delta_{400}=0.174 \ \mathrm{~cm} \\& \delta_{600}=0.219 \ \mathrm{~cm} \\& \delta_{760}=1.62 \ \mathrm{~cm} \\& \delta_{900}=1.70 \ \mathrm{~cm}\end{aligned}Question: 10.5

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