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anonymous

  • one year ago

Aluminum cylinder: length 1 m, diameter 1 cm Current: 1000 A Conductivity: σ=35 Sm/〖mm〗^2 Linear temperature coefficient: α=3.8∙〖10〗^(-3) 1/K Quadratic temperature coefficient: β=1.3∙〖10〗^(-6) 1/K^2 Heat transfer coefficient to the ambient air: λ=1 K/(Wm^2 ) 1)• Temperature of the aluminum rod with linear temperature dependency? 2).• Temperature of the aluminum rod with quadratic temperature dependency?

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  1. Michele_Laino
    • one year ago
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    is the current inside the aluminum rod a cause or is it an effect?

  2. Michele_Laino
    • one year ago
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    if it is a cause, then inside the aluminum rod we have this electric field \({\mathbf{E}}\): \[{\mathbf{j}} = \sigma {\mathbf{E}}\] where \({\mathbf{j}}\) is the current density vector, and \(\sigma\) is the conductivity of aluminum. Furthermore, the power over volume, dissipated in such rod of aluminum is given by the subsequent formula: \[W = \frac{{{{\mathbf{j}}^2}}}{\sigma }\] provided that we are in static condition of electricity

  3. anonymous
    • one year ago
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    Hint: The electrical losses of the aluminum rod P_A transferred via heat to the ambiance can be computed with the temperature at the surface ϑ_S, the ambient temperature ϑ_A, and the surface A with the following equation: P_A=(ϑ_S-ϑ_A)/λA

  4. Michele_Laino
    • one year ago
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    so, thermal energy is: \[dQ = Wdt = \frac{{{{\mathbf{j}}^2}}}{\sigma }dt\] where the magnitude of \({\mathbf{j}}\), is: \[\frac{I}{S} = \frac{I}{{\pi {r^2}}}\] and: \(2r=1\;cm\)

  5. Michele_Laino
    • one year ago
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    I meant \(dQ\) is thermal energy over volume, of course

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