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JCECE Medical JCECE Medical Solved Paper-2008

  • question_answer
    The temperature of the two outer surfaces of  a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness \[x\]and \[4x,\] respectively are \[{{T}_{2}}\]and \[{{T}_{1}}({{T}_{2}}>{{T}_{1}}).\] The rate of heat transfer through the slab, in a steady state is\[\left( \frac{A({{T}_{2-}}{{T}_{1}})K}{x} \right)f,\] with \[f\] equals to

    A)  1               

    B)  1/2

    C)  2/3             

    D)  1/3

    Correct Answer: D

    Solution :

     Let the temperature of common interface be \[\text{T}{{\,}^{\text{o}}}\text{C}\text{.}\]Rate of heat flow \[H=\frac{Q}{t}=\frac{K\,A\,\Delta T}{l}\] \[\therefore \] \[{{H}_{1}}={{\left( \frac{Q}{t} \right)}_{1}}=\frac{2K\,A(T-{{T}_{1}})}{4x}\] and \[{{H}_{2}}={{\left( \frac{Q}{t} \right)}_{2}}=\frac{KA({{T}_{2}}-T)}{x}\] In steady state, the rate of heat flow should be same in whole system ie, \[{{H}_{1}}={{H}_{2}}\] \[\Rightarrow \] \[\frac{2\,KA(T-{{T}_{1}})}{4x}=\frac{KA({{T}_{2}}-T)}{x}\] \[\Rightarrow \] \[\frac{T-{{T}_{1}}}{2}={{T}_{2}}-T\] \[\Rightarrow \] \[T-{{T}_{1}}=2{{T}_{2}}-2T\] \[\Rightarrow \] \[T=\frac{2{{T}_{2}}+{{T}_{1}}}{3}\] ?(i) Hence, heat flow from composite slab is \[H=\frac{KA({{T}_{2}}-T)}{x}\] \[\frac{KA}{x}\left( {{T}_{2}}-\frac{2{{T}_{2}}+{{T}_{1}}}{3} \right)=\frac{KA}{3x}({{T}_{2}}-{{T}_{1}})\] ?(ii) [From Eq. (i)] Accordingly, \[H=\left[ \frac{A({{T}_{2}}-{{T}_{1}})K}{x} \right]f\]?(iii) By comparing Eqs. (ii) and (iii), we get


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