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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • 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)  ½

    C)  2/3                                       

    D)  1/3

    Correct Answer: D

    Solution :

                    Let the temperature of common interface be\[T{}^\circ C\]. Rate of heat flow \[H=\frac{Q}{t}=\frac{KA\Delta T}{l}\] \[\therefore \]  \[{{H}_{1}}={{\left( \frac{Q}{t} \right)}_{1}}=\frac{2KA(T-{{T}_{1}})}{4x}\] 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)] Accordingly, \[H=\left[ \frac{A({{T}_{2}}-{{T}_{1}})K}{x} \right]f\]                     ...(iii) By comparing Eqs. (ii) and (iii), we get . \[\Rightarrow \]               \[f=\frac{1}{3}\]


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