Question

Three cylindrical rods $A,B$ and $C$ of equal lengths and equal diameters are joined in series as shown in the diagram. Their thermal conductivities are $2K,K$ and $0.5K$ respectively. In steady-state, if the free ends of rods $A$ and $C$ are at $100{}^{\circ }C$ and $0{}^{\circ }C$ respectively. Assume negligible loss through the curved surface. What will be the equivalent thermal conductivity?

• A. $K_{\text{eq }}=\left(\frac{6}{7}\right)K$
• B. $K_{\text{eq }}=\left(\frac{4}{5}\right)K$
• C. $K_{\mathrm{eq}}=\left(\frac{8}{7}\right)K$
• D. $K_{\mathrm{eq}}=\left(\frac{7}{6}\right)K$

Answer 1

Answer: (A)

As the rods are in series, $R_{eq}=R_A+R_B+R_C$ with $R=(L/KA)$
i.e., $R_{\mathrm{eq}}=\frac{L}{2\mathrm{KA}}+\frac{L}{\mathrm{KA}}+\frac{L}{0.5\mathrm{KA}}=\frac{7L}{2\mathrm{KA}}$
Furthermore if $K_{\text{eq }}$ is equivalent thermal conductivity,
$R_{\mathrm{eq}}=\frac{L+L+L}{K_{\mathrm{eq}}A}=\frac{7L}{2\mathrm{KA}}$ [from Equation (i)]
$[$ from Equation (i)
i.e., $\Rightarrow \frac{3L}{K_{\text{cq }}}=\frac{7L}{2KA}$
$\Rightarrow K_{eq}=\frac{6}{7}K$