A cylinder rod of length 1, thermal conductivity K and area of cross section A has one end in the furnace at temperature `T_1` and the other end in surrounding at temperature `T_2`. Surface of the rod exposed to the surrounding has emissivity e. Also `T_2=T_s+DeltaT and T_sgt gt DeltaT`. If `T_1-T_s prop DeltaT`,find the proportionality constant.

Answer

A, C, D

Explanation

From the figure is clear that emission takes place form the
surface at temperature `T_2`(circular cross section). Heat
conduction and radiation through lateral surface is zero.
Heat conducted through rod is
`Q=(KA(T_1-T_2)Deltat)/l`
Energy emitted by surface of rod in same time `Deltat,` is
`E=esigmaA(T_2^4-T_s^4)Deltat`
Since rod is at thermal equilibrium therefore `E=Q`
hence, `(KA(T_1-T_2)Deltat)/l =esigmaA(T_2^4-T_s^4)Deltat`
`rArrT_1-T_2=(esigmaA(T_2^4-T_s^4)l)/K`
Here `T_2-T_s=DeltaT and T_sgt gt DeltaT`
`T_1-(DeltaT+T_s)=(esigmal)/K[(DeltaT+T_s)^4-T_s^4]`
`T_1-(DeltaT+T_s)=(esigmal)/KxxT_s^4[(1+(DeltaT)/(T_s))^4-1]`
`(DeltaT+T_s)=(esigmal)/KxxT_s^4[1+(4DeltaT)/(T_s)-1]`
or `T_1-(T_s+DeltaT)=(4esigmal)/KT_s^3DeltaT`
or `T_1-T_s=((4esigmalT_s^3)/K+1)DeltaT`
`:. The proportionality constant=((1+4esigmalT_s^3)/K)`