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JCECE Engineering JCECE Engineering 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 \[2K\] \[\text{and}\] 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\[{{T}^{o}}C\]. 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,                 \[{{H}_{1}}={{H}_{2}}\] \[\Rightarrow \]               \[\frac{2KA(T-{{T}_{1}})}{4x}=\frac{KA({{T}_{2}}-T)}{x}\] \[\Rightarrow \]               \[\frac{T-{{T}_{1}}}{2}={{T}_{2}}-T\] \[\Rightarrow \]               \[T-{{T}_{1}}=2{{T}_{2}}-2T\] \[\Rightarrow \]               \[T=\frac{2{{T}_{2}}+{{T}_{1}}}{3}\]                                         ? (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)] By comparing Eqs. (ii) and (iii), we get \[\Rightarrow \]                               \[f=\frac{1}{3}\]


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