Question

An electric current of 200 A is passed through a stainless steel wire having a radius R of 0.001268 m. The wire is L = 0.91 m long and has a resistance R of 0.126 Ohm. The outer surface temperature Tw is held at 422.1 K. The average thermal conductivity is K = 22.5 W/mK. Calculate the center temperature. Power = I^2R (Watts) Where I = current in nmps and R = Resistance in ohms I^2R (Watts) = q pi(Radius)^2L

Answer 1

Answer:

The value is   $T_m = 435.2 \ K$

Explanation:

From the question we are told that  

   The  current is  $I = 200 \ A$

   The radius is  $R = 0.001268 \ m$

   The  length of the wire is  $L = 0.91 \ m$\

    The  resistance is  $R = 0.126 \ \Omega$

    The  outer surface temperature is  $T _o = 422.1 \ K$

    The average thermal conductivity is  $\sigma = 22.5 W/mK$

   

Generally the heat generated in the stainless steel wire is mathematically represented as  

    $Q = \frac{Power}{ \pi r^2L}$

     $Q = \frac{I^2 R}{ \pi r^2L}$

=>   $Q = \frac{200^2 * 0.126}{3.142 * (0.001268)^2 * 0.91}$

=>   $Q = 1.096*10^{9}\ W/m^3$

Generally the middle temperature is mathematically represented as

      $T_m = T_o + \frac{Q * r^2 }{ 6 * \sigma }$

       $T_m = 422.1 + \frac{1.096*10^{-9} * 0.001268^2}{6 * 22.5}$

       $T_m = 435.2 \ K$