14_InstSolManual_PDF_Part10

14_InstSolManual_PDF_Part10 - kA 1 T H 2 T C 2 H 5 1 0.024...

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14.57. Set Up: For water, Solve: The heat conducted by the rod in 10.0 min is Reflect: The heat conducted by the rod is the heat that enters the ice and produces the phase change. 14.58. Set Up: For copper, For steel, is the same for both sections of the rod. Solve: (a) For the copper section, (b) For the steel section, 14.59. Set Up: Let the temperature of the fat-air boundary be T . A section of the two layers is sketched in Fig- ure 14.59. A Kelvin degree is the same size as a Celsius degree, so and are equivalent units. At steady state the heat current through each layer is the same, equal to 50 W. The area of each layer is with Figure 14.59 Solve: (a) Apply to the fat layer and solve for For the fat layer (b) Apply to the air layer and solve for For the air layer and Reflect: The thermal conductivity of air is much lass than the thermal conductivity of fat, so the temperature gradi- ent for the air must be much larger to achieve the same heat current. So, most of the temperature difference is across the air layer. L 5
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Unformatted text preview: kA 1 T H 2 T C 2 H 5 1 0.024 W 21 4 p 21 0.75 m 2 2 1 29.6C 2 2.7C 2 50 W 5 9.1 cm T C 5 2.7C. T H 5 T 5 29.6C L 5 L air . H 5 kA T H 2 T C L T 5 T H 2 HL kA 5 31C 2 1 50 W 21 4.0 3 10 2 2 m 2 1 0.20 W 21 4 p 21 0.75 m 2 2 5 31C 2 1.4C 5 29.6C T H 5 31C. T C 5 T . H 5 kA T H 2 T C L 2.7 C 31 C T L air air k 5 0.024 W / m K 4.0 cm k 5 0.20 W / m K fat r 5 0.75 m. A 5 4 p r 2 , W / m # C W / m # K L 5 kA D T 1 Q / t 2 5 1 50.2 W / m # K 21 4.00 3 10 2 4 m 2 21 65.0C 2 0C 2 5.39 J / s 5 0.242 m. Q t 5 kA D T L . Q t 5 kA D T L 5 1 385 W / m # K 21 4.00 3 10 2 4 m 2 21 100C 2 65.0C 2 1.00 m 5 5.39 J / s. Q / t k s 5 50.2 W / m # K. k c 5 385 W / m # K. k 5 1 Q / t 2 L A 1 T H 2 T C 2 5 1 4.73 W 21 0.600 m 2 1 1.25 3 10 2 4 m 2 21 100 C 2 5 227 W / m # K. Q t 5 kA 1 T H 2 T C 2 L . Q t 5 2.84 3 10 3 J 600 s 5 4.73 W. Q 5 mL f 5 1 8.50 3 10 2 3 kg 21 3.34 3 10 5 J / kg 2 5 2.84 3 10 3 J. L f 5 3.34 3 10 5 J / kg. 14-10 Chapter 14...
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