InterviewSolution
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InterviewSolution
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One end of a rod of length L and crosssectional area A is kept in a furnace at temperature `T_(1)`. The other end of the rod is kept at a temperature `T_(2)`. The thermal conductivity of the matrieal of the rod is `K` and emissivity of the rod is e. It is gives that `T_(!)=T_(s)+DeltaT`, where `DeltaTltlt T_(s),T_(s)` is the temperature of the surroundings. If `DeltaTprop(T_(1)-T_(2))` find the proportional constant, consider that heat is lost only by rediation at the end where the temperature of the rod is `T_(1)`. |
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Answer» Rate of heat conduction through rod=rate of the heat lost from right end of the rod. `(KA(T_(1)-T_(2)))/(L)`=`eAsigma(T_(2)^(4)-T_(s)^(4))` (i) Given that `T_(1)`=`T_(s)+DeltaT` `T_(2)^(4)`=`(T_(s)+DeltaT)^(4)`=`T_(5)^(4)(1+(DeltaT)/(T_(s)))` Using binomial expansion, we have `T_(2)^(4)`=`T_(s)^(4)(1+4(DeltaT)/(T_(s)))` (as `DeltaTltlt T_(s))` `T_(2)^(4)`=`-T_(s)^(4)`=`4(DeltaT)(T_(s)^(3))` Subsituting in Eq. (i), we have `(K(T_(1)-T_(S)-DeltaT))/(L)`=`4esigma T_(s)^(3)DeltaT` `(K(T_(1)-T_(s)))/(L)`=`(4esigmaT_(s)^(3)+(K)/(L))DeltaT` `DeltaT`=`(K(T_(1)-T_(s)))/((4esigma LT_(s)^(3)+K))` Comparing with the given relation, proportionality constant =`(K)/(4esigmaLT_(s)^(3)+K)` |
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