1. The problem statement, all variables and given/known data 1) A typical wall construction for a single-family residence might consist of 0.375 inches of outer sheathing (insulating board, hardwood siding, etc.), 3.5 inches of mineral fiber insulation and then a 0.375 inches inner surface (gypsum, etc.). Typical thermal conductivities for these materials are 0.1 W/m-C for the siding, 0.046 W/m-C for the insulation and 0.17 W/m-C for the inner surface. Determine the total thermal resistance (hr-ft2-F/Btu) and total wall conductance (Btu/hr-ft2-F). Also find the percent of total resistance for the three different materials. Provide step-by-step calculation procedures. Discuss/comment to include as a minimum things like assumptions, effects of each wall-layer on heat flow, etc (2.5 points). 2) Part 1 neglected thermal resistances due to convection heat transfer on the inner and outer wall surfaces. Recalculate the total thermal resistance (along with percent) and the wall U-factor including convection heat transfer using an outer wall convection heat transfer coefficient of 6.0 Btu/hr-ft2-F (15 mph wind in the winter) and an inner wall value of 1.46 Btu/hr-ft2-F. Provide step-by-step calculation procedures. Discuss/comment to include as a minimum things like assumptions, comparisons, etc (2.5 points). 2. Relevant equations Rth (thermal resistance ,R-value) =1/U-value=L/k (h∙ft2∙F/Btu) U-value (thermal conductance) = k/L (Btu/h∙ft2∙F) Fourier’s Law: qcond= kA (T1–T2)/L where, k=thermal conductivity(Btu∙in/h∙ft²∙F or W/m∙K) A=area through which conduction occurs T1= higher temperature T2= lower temperature L= thickness of material 3. The attempt at a solution Part 1) Siding 0.375in thickness or 0.375in×2.54cm/1in×1m/100cm=0.009525 m k=0.1 W/(m×℃) Insulation 3.5in thickness or 3.5in×2.54cm/1in×1m/100cm=0.0889 m k=0.46 W/(m×℃) Inner Surface 0.375in thickness or 0.375in×2.54cm/1in×1m/100cm=0.009525 m k=0.17 W/(m×℃) Note: As the common SI unit for Thermal Resistance is (m^2×℃)/W it is necessary to convert the thickness value into meters. Thermal Resistance (R) 〖R_th〗_siding=L/k=0.009525m/(0.1 W/m℃)=0.09525 (m^2×℃)/W 〖R_th〗_insulation=L/k=0.0889m/(0.046 W/m℃)=1.9326 (m^2×℃)/W 〖R_th〗_inner=L/k=0.009525m/(0.17 W/m℃)=0.056029 (m^2×℃)/W 〖R_th〗_total=〖R_th〗_siding+ 〖R_th〗_insulation+ 〖R_th〗_inner=(0.09525 (m^2×℃)/W )+(1.9326 (m^2×℃)/W)+(0.056029 (m^2×℃)/W)=2.08 (m^2×℃)/W or… 2.08 (m^2×℃)/W ×1W/(0.2931 Btu/h)×(10.76ft^2)/(1m^2 )×(9℉)/(5℃)=137 (hr×ft^2×℉)/Btu Percentage of R value per material: Siding: ((0.09525 (m^2×℃)/W)/(2.08 (m^2×℃)/W))×100=4.6% Insulation: ((1.9326 (m^2×℃)/W)/(2.08 (m^2×℃)/W))×100=92.9% Inner Surface: ((0.056029 (m^2×℃)/W)/(2.08 (m^2×℃)/W))×100=2.7% Note: Both the siding, and the inner wall surface have provide very little R-value. The insulation allows for thinner walls, as it would take an abundant amount of either siding or inner surface to achieve an acceptable level of thermal resistance. Wall Conductance (C) C=1/R=1/137 Btu/(hr×ft^2×℉) Note: Wall conductance is simply the inverse of the R-value, often expressed as U-value. It measures the rate of heat flow though a unit area of a material. This should not be confused with thermal conductivity (k), which is an inherent property of the material. Part 2) I have been staring at this for hours trying to understand what I'm not understanding! I know it must have something to do with the heat loss on the inside being equal to the heat gain on the outside, but I just cannot figure out how to get started on this problem without knowing the temperature difference. Any help, tips, or hints would be greatly appreciated!