Saved Bookmarks
| 1. |
Oneend of a rodof lengthL andcross - sectioal area A is keptin a furanceof temperature T_(1) . Theotherend of the rod is keptat a tempearute T_(2). Thethermal conducitivity of the material of the rod isK and emissivityof rodis K and emissivityof rodis e. It isgiven that T_(2) = T_(s)+ DeltaT, where Delta T lt lt T_(s), T_(s)beingthe temperature of surroudings. If Delta T prop (T_(1) - T_(s)) , find the proportionalityconstant. Considerthat heat is lostonly by radiation at theendwherethe temperature of the rod is T_(2).. |
Answer» SOLUTION :Rateof Heatgained by rodthroughconduction , `Q = (KA(T_(1) - T_(2)))/(L)`  Rateof Heatlostby rodthrough radiation, `Q_(2) = e A SIGMA (T_(2)^(4) - T_(s)^(4))` In steady state `Q_(1)= Q_(2)` `therefore (KA(T_(1)-T_(2)))/(L) = e A sigma(T_(2)^(4) - T_(s)^(4))..........(1)` Given `T_(2) = T_(s) + DeltaT "" therefore T_(2)^(4) = (T_(s) + DeltaT)^(4) = T_(S)^(4) + (1 + (DeltaT)/(T_(s)))^(4)` As `Delta T lt lt T_(s)` , so usingbionmial theorem . ` T_(2)^(4) = T_(s)^(4) ( 1+(4Delta T)/(T_(s))) rArr T_(2)^(4) - T_(s)^(4) = 4T_(s)^(3) DeltaT ` substitutingthis valuein (1) , we get`(KA(T_(1) - T_(2)))/(L) = e A sigma. (4T_(s)^(3) DeltaT) ` Also`T_(2) = T_(s) + Delta T` `therefore(KA(T_(1) - T_(s) - DeltaT))/(L) = e A sigma (4T_(s)^(3) DeltaT) rArr (K(T_(1) - T_(s)))/(L)- (K)/(L) DeltaT = 4esigma T_(s)^(3) Delta T` (or) `(K(T_(1) - T_(s)))/(L)= (4esigma T_(s)^(3) + (K)/(L)) DeltaT , Delta T = (K(T_(1) - T_(S)))/((4 e sigmaL T_(s)^(3) + K))` Butgiven `Delta T = C(T_(1) - T_(s))` `therefore ` Constant of proportioanlityC is `(K)/(K + 4 e sigma L T_(s)^(3))` |
|