11/14/2019 Solution Manual Incropera Heat Mass Transfer
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PROBLEM 1.1 KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid extruded insulation. FIND: (a) The heat flux through a 2 m × 2 m sheet of the insulation, and (b) The heat rate through the sheet. SCHEMATIC: A = 4 m2 k = 0.029 W m ⋅K qcond T1 – T2 = 10˚C T1 T2 L = 20 mm x ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties.
ANALYSIS: From Equation 1.2 the heat flux is q′′x = -k T -T dT =k 1 2 dx L Solving, q'x = 0.029 q′′x = 14.5 W 10 K × m⋅K 0.02 m W m2 L, the effect is negligible. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.
Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. PROBLEM 1.9 KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that through a composite wall of prescribed thermal conductivity and thickness. FIND: Thickness of masonry wall. SCHEMATIC: ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) Onedimensional conduction, (3) Steady-state conditions, (4) Constant properties. ANALYSIS: For steady-state conditions, the conduction heat flux through a onedimensional wall follows from Fourier’s law, Eq.
Fundamentals Of Heat Transfer Solutions
1.2, q ′′ = k ΔT L where ΔT represents the difference in surface temperatures. Since ΔT is the same for both walls, it follows that L1 = L2 k1 q ′′ ⋅ 2. K2 q1′′ With the heat fluxes related as q1′′ = 0.8 q ′′2 L1 = 100mm 0.75 W / m ⋅ K 1 = 375mm. × 0.25 W / m ⋅ K 0.8.
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