Thermodynamics HW Solutions 394

# Thermodynamics HW Solutions 394 - 1 40 1 s 3600 2 s m...

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Chapter 4 Transient Heat Conduction 4-125 A thick wall is exposed to cold outside air. The wall temperatures at distances 15, 30, and 40 cm from the outer surface at the end of 2-hour cooling period are to be determined. Assumptions 1 The temperature in the wall is affected by the thermal conditions at outer surfaces only. Therefore, the wall can be considered to be a semi-infinite medium 2 The thermal properties of the wall are constant. Properties The thermal properties of the brick are given to be k = 0.72 W/m. ° C and α = 1.6 × 10 -7 m 2 /s. Analysis For a 15 cm distance from the outer surface, from Fig. 4-23 we have 25 . 0 1 70 . 0 ) s 3600 2 )( s / m 10 (1.6 2 m 15 . 0 2 98 . 2 C W/m. 0.72 ) s 3600 2 )( s / m 10 (1.6 C) . W/m 20 ( 2 6 - 2 6 - 2 = = × × α = ξ = ° × × ° = α T T T T t x k t h i Air 2 ° C L =40 cm Wall 18 ° C C 14.0 ° = ⎯→ = T T 25 . 0 2 18 2 1 For a 30 cm distance from the outer surface, from Fig. 4-23 we have 038
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Unformatted text preview: . 1 40 . 1 ) s 3600 2 )( s / m 10 (1.6 2 m 3 . 2 98 . 2 C W/m. 0.72 ) s 3600 2 )( s / m 10 (1.6 C) . W/m 20 ( 2 6-2 6-2 = − − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = × × α = ξ = ° × × ° = α ∞ ∞ T T T T t x k t h i C 17.4 ° = ⎯→ ⎯ = − − − T T 038 . 2 18 2 1 For a 40 cm distance from the outer surface, that is for the inner surface, from Fig. 4-23 we have 1 87 . 1 ) s 3600 2 )( s / m 10 (1.6 2 m 4 . 2 98 . 2 C W/m. 0.72 ) s 3600 2 )( s / m 10 (1.6 C) . W/m 20 ( 2 6-2 6-2 = − − − ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = × × α = ξ = ° × × ° = α ∞ ∞ T T T T t x k t h i C 18.0 ° = ⎯→ ⎯ = − − − T T 2 18 2 1 Discussion This last result shows that the semi-infinite medium assumption is a valid one. 4-105...
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