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One end of rod of length `L` and cross-sectional area `A` is kept in a furance of temperature `T_(1)`. The other end of the rod is kept at at temperature `T_(2)`. The thermal conductivity of the material of the rod is `K` and emissivity of the rod is `e`. It is given that `T_(2)=T_(S)+DeltaT` where `DeltaT lt lt T_(S)`, `T_(S)` being the temperature of the surroundings. If `DeltaT prop (T_(1)-T_(S))`, find the proportionality constant. Consider that heat is lost only by radiation at the end where the temperature of the rod is `T_(2)`. |
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Answer» From the figure it is clear that emission takes place from the surface at temperture `T_(2)` (circular cross section). Heat conduction and radiation through laeral surface is zero. Heat conducted through rod is `Q=(KA(T_(1)-T_(2))Deltat)/(l)` Energy emitted by the surface of the rod in the same time `Deltat`, is `E=epsilonsigmaA(T_(2)^(4)-T_(S)^(4))Deltat` Since rod is at thermal equilibrium `:. E=Q` hence, `(KA(T_(1)-T_(2)))/(l)=epsilon sigma A(T_(2)^(4)-T_(S)^(4))Deltat` `implies T_(1)-T_(2)=(epsilon sigma(T_(2)^(4)-T_(S)^(4)))/(K)` Here `T_(1)-T_(2)=DeltaT` and `T_(s) gt gt Delta T` `T_(1)-(DeltaT+T_(S))=(epsilonsigmal)/(K)[(DeltaT+T_(s))^(2)-T_(S)^(4)]` `T_(1)-(DeltaT+T_(S))=(epsilonsigmal)/(K)xxT_(S)^(4)[(1+(DeltaT)/(T_(S)))^(4)-1]` `T_(1)-(DeltaT+T_(S))=(epsilonsigmal)/(K)xxT_(S)^(4)[(1+(4DeltaT)/(T_(S)))^(4)-1]` or `T_(1)-(T_(S)+DeltaT)=(4epsilonsigmal)/(K)xxT_(S)^(3)DeltaT` or `T-T_(s)=(4epsilonsigmalT_(s)^(3))/(K)DeltaT+DeltaT` or `T_(1)-T_(s)=((4epsilonsigmalT_(s)^(3))/(K)+1)DeltaT` `:. The proportionality constant`= `(1+(4epsilonsigmalT_(s)^(3))/(K))` |
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