Answer to Question #217157 in Mechanical Engineering for Shivu

Question

A 2 cm diameter glass rod(k=0.8W/mk) is 6 cm long.I t has base temperature at 120°C and is exposed to the air at 20°C. The temperature at the tip of the rod is measured as 35°C. What is the convection heat transfer coefficient.H ow much heat is lost by the rod?

Answer 1

Part 1

Perimeter of the rod $P= \pi d= \pi*0.02 = 0.0628 m$

Cross sectional area $=\frac{\pi d^2}{4}=\frac{\pi 0.02^2}{4}=3.14 *10^{-4}m^2$

$m= \sqrt{\frac{hp}{kA}} = \sqrt{\frac{h*0.0628}{0.8*3.14*10^{-4}}} \\ \implies m= 15.8\sqrt{h}\\$

$125 h=-15.8 \sqrt{h}C_1e^{-15.8\sqrt{h}*0.06}+15.8 \sqrt{h}C_2e^{15.8\sqrt{h}*0.06}\\ 15= C_1e^{-0.06\sqrt{h}}+C_2e^{-0.948\sqrt{h}}\\ \implies C_1 = 82.3\\ C_2=0.8994\\ \therefore h=7.615 W/m^2C$

Part 2

$q= \sqrt{hpkA}*tanhmL\\ q=\sqrt{h*0.0628*0.8*8.14*10^{-4}}*83.144 tanh(15.8*\sqrt{h}tan0.06\\ q=0.010960*82.314\\ q=0.9021 W$