Question

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?

Answer 1

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`