If the thermal conductivity of the material of a conductor is 375 `W m^(-1)K^(-1)`, then calculate the thermal resistance of a 20 mm thick and 1.5 m length of the rod. If an identically shaped rod but of thermal conductivity 600`Wm^(-1)K^(-1)` is connected in (1) series and (2) parallel, then calculate the effective conductivities of the combinations.

Explanation

For a series combination `A_1=A_2=A`
`d'=d_1+d_2=2d`
`K_1=375Wm^(-1)K^(-1)` & `K_2=600Wm^(-1)K^(-1)`
`A=(3.142xx(20xx10^(-3))^2)/4`
`A=3.142xx10^(-4)m^2`
`K_s=(2K_1K_2)/(K_2+K_2)=(2xx375xx600)/((375+600))=(450000)/(975)=461.5Wm^(-1)K^(-1)`
For a parallel combination
`A'=2A,d'=d_1=d_2`
hence `K_p=(K_1+K_2)/(2)` (refer to list of formulae)=`(375+600)/(2)=487.5Wm^(-1)K^(-1)`
and effective thermal resistance in series l'=2e
`R_s=(l)/(K_sA)=(2xx1.5xx4)/(46.1xx3.142xx(2.0xx10^(-3))^2)=2.069xx10^(-5)xx10^6`
`=2.069xx10^(-5)xx10^6=20.69KW^(-1)` (kelvin wat`t^(-1))`