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

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

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Sapo 2 months 2021-07-26T18:12:28+00:00 1 Answers 13 views 0

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    2021-07-26T18:13:45+00:00

    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

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