Question

One end of a long rod is inserted into furnace while the other end projects ambient air. Under steady state the temperature of the rod is measured at two points 75 mm apart and found to be ${125}^{\circ }C$ and ${88.6}^{\circ }C$ respectively while the ambient temperature is ${20}^{\circ }C$. if the rod is 25 mm in diameter and heat transfer coefficient for air is $23.36W/m^{2}K,$ the thermal conductivity of rod material in W/mK is

• A. 115.2
• B. 101. 1
• C. 113.2
• D. 150

Answer 1

The correct option is A 115.2


$\begin{array}{c}{T}_{\infty }={20}^{\circ }C,d=25mm=0.025m\\ h=23.36W/m^2-K\\ \text{We know the formula:-}\\ \frac{T-T_{\infty }}{T_0-T_{\infty }}=e^{-mx}\\ atx=0.075m,T={125}^{\circ }C\\ \frac{125-20}{T_0-T_{\infty }}=e^{-0.075m}\cdots \cdots \left(1\right)\\ atx=0.075+0.075=0.15m,\\ T={86.25}^{\circ }C\\ \frac{88.5-20}{T_0-T_{\infty }}=e^{-0.15m}\cdots \cdots \left(2\right)\\ \text{Equation (2)}÷\text{Equation (1)}\\ \frac{\frac{88.5-20}{T_0-T_{\infty }}}{\frac{135-20}{T_0-T_{\infty }}}=\frac{e^{-0.15m}}{e^{-0.075m}}\\ 0.633=e^{-0.15m+0.075m}\\ 0.633=e^{-0.075m}\\ \text{Taking logarithm on both sides}\\ 0.075m=6.4567\\ m=5.695\\ \sqrt{\frac{ph}{kA}}=5.495\Rightarrow \frac{ph}{kA}=32.43\\ \frac{\pi .d\times h}{k\times \frac{\pi }{4}{d}^2}\\ \frac{4h}{kd}=32.43\\ \frac{4\times 23.36}{k\times 0.025}=32.43\\ k=\mathrm{11.5.23}Wm/k-K\end{array}$