Question #217157

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?


1
Expert's answer
2021-07-16T03:55:37-0400

Part 1

Perimeter of the rod P=πd=π0.02=0.0628mP= \pi d= \pi*0.02 = 0.0628 m

Cross sectional area =πd24=π0.0224=3.14104m2=\frac{\pi d^2}{4}=\frac{\pi 0.02^2}{4}=3.14 *10^{-4}m^2

m=hpkA=h0.06280.83.14104    m=15.8hm= \sqrt{\frac{hp}{kA}} = \sqrt{\frac{h*0.0628}{0.8*3.14*10^{-4}}} \\ \implies m= 15.8\sqrt{h}\\

125h=15.8hC1e15.8h0.06+15.8hC2e15.8h0.0615=C1e0.06h+C2e0.948h    C1=82.3C2=0.8994h=7.615W/m2C125 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=hpkAtanhmLq=h0.06280.88.1410483.144tanh(15.8htan0.06q=0.01096082.314q=0.9021Wq= \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


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