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Thermodynamics HW Solutions 281

# Thermodynamics HW Solutions 281 - Chapter 3 Steady Heat...

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Chapter 3 Steady Heat Conduction ( b ) The total thermal resistance with the fiberglass insulation is C) W.m/ 6048 . 0 ( C/W 02253 . 0 ) m C)(16.8 W/m. 036 . 0 ( C/W 02253 . 0 4 2 4 conv,2 glass fiber 3 2 1 total ° + ° = ° + ° = + + + + = L L R R R R R R The thickness of insulation is determined from cm 7.7 m 0.077 = = ⎯→ ° + ° ° = ⎯→ = 4 4 total 2 1 ins C W.m/ 6048 . 0 ( C/W 02253 . 0 C ) 8 23 ( W 9 . 99 L L R T T Q & The outer surface temperature of the wall is determined from C 8.3 ° = ⎯→ ° ° = ⎯→ = 2 2 conv 2 2 ins C/W 00350 . 0 C ) 8 ( 9 . 99 T T R T T Q & Discussion The outer surface temperature is same for both cases since the rate of heat transfer does not change. 3-170 Cold conditioned air is flowing inside a duct of square cross-section. The maximum length of the duct for a specified temperature increase in the duct is to be determined. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Steady one-dimensional heat conduction relations can be used due to small thickness of the duct wall.
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