Question

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?

Answer 1

is the current inside the aluminum rod a cause or is it an effect?

Answer 2

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

Answer 3

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

Answer 4

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: \(2r=1\;cm\)

Answer 5

I meant \(dQ\) is thermal energy over volume, of course