Three cylindrical rods A, B and C of equal lengths and equal diameters are joined in series as shown if Fig. Their thermal conductivities are 2K, K and 0.5K, respectively. In steady state, if the free ends of rods A and C are at `100^@C` and `0^@C`, respectively, calculate the temperature at the two junction points. Assume negligible loss by radiaiotn through the curved surface. What will be the equivalent thermal conductivity?

Explanation

As the rods are in series, `R_(eq)=R_A+R_B+R_C` with `R=((L)/(KA))`
`R_(eq)=(L)/(2KA)+(L)/(KA)+(L)/(0.5KA)=(7L)/(2KA)` ..(i)
and hence, `H=(dQ)/(dt)=(Deltatheta)/(R )=((100-0))/((7L)/(2KA))=(200KA)/(7L)`
Now in series, rate of flow of heat remains same i.e.,
`H=H_A=H_B=H_C`.
So for rod A, `[(dQ)/(dt)]_A=[(dQ)/(dt)]`
i.e., `((100-theta_(AB))2KA)/(L)=(200KA)/(L)`
`theta_(BC)=((400)/(7))=57.1^@C`
Furthermore, if `K_(eq)` is equivalent thermal conductivity,
`R_(eq)=(L+L+L)/(K_(eq)A)=(7L)/(2KA)`
`K_(eq)=(6//7)K`