Explanation Given information: Width of trench s = 5 μm Base temperature of nanowire T 8 = T s = 300K and, Nanotube Temperature T 1 = 324.5 K for ξ 1 = 1.5 μm Nanotube Temperature T 2 = 326.4 K for ξ 2 = 3.5 μm Temperature of surrounding T s u r r = 300 K Thermal conductivity K c n = 3100 W/m.K Diameter D = 14 nm Calculation: The thermal conductivities for both the laser irradiation locations are shown below: The circuit of irradiation on the left half of the nanotube is shown below: The equations regarding irradiation of the nanotube is as follows: q = q l , 1 + q r , 1 Here, q l , 1 = T max , 1 − T ∞ R t , c , l + ξ 1 k c n A c n q l , 1 = T max , 1 − 300 R t , c , l + 1.5 3100 × 1.54 × 10 − 16 q l , 1 = T max , 1 − 300 R t , c , l + 3.14 × 10 12 _ _ _ _ _ _ _ _ _ ( 1 ) q r , 1 = 324.5 − 300 5 / 2 3100 × 1.54 × 10 − 16 + R t , c , r q r , 1 = 24.5 5.24 × 10 12 + R t , c , r _ _ _ _ _ _ _ _ _ _ _ ( 2 ) q r , 1 = T max , 1 − T ∞ s / 2 − ξ 1 k c n A c n + s / 2 k c n A c n R t , c , r q r , 1 = T max , 1 − 300 5 / 2 − 1.5 3100 × 1.54 × 10 − 16 + 5 / 2 3100 × 1.54 × 10 − 16 + R t , c , r q r , 1 = T max , 1 − 300 2.1 × 10 12 + 5.24 × 10 12 + R t , c , r _ _ _ _ _ _ _ _ _ _ ( 3 ) And, the circuit of irradiation of the right half of the nanotube is shown below: q = q l , 2 + q r , 2 q l , 2 = T max , 2 − T ∞ s / 2 k c n A c n + ξ 2 − s / 2 k c n A c n + R t , c , l q l , 2 = T max , 2 − 300 5 / 2 3100 × 1

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ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3, Problem 3.120P

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The two contact resistances at the left and right ends of the nanotube.

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Chapter 3, Problem 3.119P

Chapter 3, Problem 3.121P

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The two contact resistances at the left and right ends of the nanotube.

Solution Summary: The author explains the thermal conductivities for both the laser irradiation locations.

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Chapter 3 Solutions