Part 1
Perimeter of the rod P = π d = π ∗ 0.02 = 0.0628 m P= \pi d= \pi*0.02 = 0.0628 m P = π d = π ∗ 0.02 = 0.0628 m
Cross sectional area = π d 2 4 = π 0.0 2 2 4 = 3.14 ∗ 1 0 − 4 m 2 =\frac{\pi d^2}{4}=\frac{\pi 0.02^2}{4}=3.14 *10^{-4}m^2 = 4 π d 2 = 4 π 0.0 2 2 = 3.14 ∗ 1 0 − 4 m 2
m = h p k A = h ∗ 0.0628 0.8 ∗ 3.14 ∗ 1 0 − 4 ⟹ m = 15.8 h m= \sqrt{\frac{hp}{kA}} = \sqrt{\frac{h*0.0628}{0.8*3.14*10^{-4}}} \\
\implies m= 15.8\sqrt{h}\\ m = k A h p = 0.8 ∗ 3.14 ∗ 1 0 − 4 h ∗ 0.0628 ⟹ m = 15.8 h
125 h = − 15.8 h C 1 e − 15.8 h ∗ 0.06 + 15.8 h C 2 e 15.8 h ∗ 0.06 15 = C 1 e − 0.06 h + C 2 e − 0.948 h ⟹ C 1 = 82.3 C 2 = 0.8994 ∴ h = 7.615 W / m 2 C 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 125 h = − 15.8 h C 1 e − 15.8 h ∗ 0.06 + 15.8 h C 2 e 15.8 h ∗ 0.06 15 = C 1 e − 0.06 h + C 2 e − 0.948 h ⟹ C 1 = 82.3 C 2 = 0.8994 ∴ h = 7.615 W / m 2 C
Part 2
q = h p k A ∗ t a n h m L q = h ∗ 0.0628 ∗ 0.8 ∗ 8.14 ∗ 1 0 − 4 ∗ 83.144 t a n h ( 15.8 ∗ h t a n 0.06 q = 0.010960 ∗ 82.314 q = 0.9021 W 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 q = h p k A ∗ t anhm L q = h ∗ 0.0628 ∗ 0.8 ∗ 8.14 ∗ 1 0 − 4 ∗ 83.144 t anh ( 15.8 ∗ h t an 0.06 q = 0.010960 ∗ 82.314 q = 0.9021 W
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