Question

Radius of a metallic rod varies with x (distance from end A) as $r=(x+1)^{\frac{3}{2}}$m. End A and B of rod are maintain at $100^\circ C$ and $0^\circ C$ respectively. Thermal conductivity and length of rod are $\frac{3}{\pi}$ W/m-K and 2m respectively. Assume steady state flow of heat in rod and rod is covered with adiabatic coating:

• A. Rate of heat flow through the rod is 600 W.
• B. Value of x at which temperature is $50^\circ C$, is 70 cm
• C. Temperature is approximately $15.6^\circ C$ at x = 1m
• D. Ratio of thermal resistance of rod between (x = 0 & x = 1) and (x = 1 & x = 2) is $\frac{27}{5}$

Answer 1

Answer: (C), (D)

C, D

Answer 2

The rate of heat flow $H$ is given by $H = -KA(x) \frac{dT}{dx}$. The area $A(x) = \pi r^2 = \pi (x+1)^3$. Integrating Fourier's law from $x=0$ to $x=2$ and $T=100^\circ C$ to $T=0^\circ C$ yields $H=675$ W. The temperature distribution is $T(x) = 100 - 112.5 \left(1 - \frac{1}{(x+1)^2}\right)$. At $x=1$m, $T(1) = 100 - 112.5(1 - 1/4) = 100 - 84.375 = 15.625^\circ C$. The thermal resistance of a segment is $R_{th} = \int \frac{dx}{KA(x)}$. $R_{th}(0,1) = \frac{3}{8K\pi}$ and $R_{th}(1,2) = \frac{5}{72K\pi}$. The ratio is $\frac{27}{5}$.