Answer:
The value is [tex]T_m = 435.2 \ K[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 200 \ A[/tex]
The radius is [tex]R = 0.001268 \ m[/tex]
The length of the wire is [tex]L = 0.91 \ m[/tex]\
The resistance is [tex]R = 0.126 \ \Omega[/tex]
The outer surface temperature is [tex]T _o = 422.1 \ K[/tex]
The average thermal conductivity is [tex]\sigma = 22.5 W/mK[/tex]
Generally the heat generated in the stainless steel wire is mathematically represented as
[tex]Q = \frac{Power}{ \pi r^2L}[/tex]
[tex]Q = \frac{I^2 R}{ \pi r^2L}[/tex]
=> [tex]Q = \frac{200^2 * 0.126}{3.142 * (0.001268)^2 * 0.91}[/tex]
=> [tex]Q = 1.096*10^{9}\ W/m^3[/tex]
Generally the middle temperature is mathematically represented as
[tex]T_m = T_o + \frac{Q * r^2 }{ 6 * \sigma }[/tex]
[tex]T_m = 422.1 + \frac{1.096*10^{-9} * 0.001268^2}{6 * 22.5}[/tex]
[tex]T_m = 435.2 \ K[/tex]
