Q4: Conduction and \"R\" value k So 9AT A R /k Your house has 0.5\" of gypsum wa

Question

Q4: Conduction and "R" value k So 9AT A R /k Your house has 0.5" of gypsum wallboard, 4" of fiberglass insulation and 1" of wood between your toasty warm living room (20C) and the Indiana winter (-20C). Use the table below to calculate the rate of heat loss per square meter (q/A). Material Thickness In Inches R-Value : ft' hr F/Btu Concrete Gypsum Wood Tectum Fiberboard Perlite Phenolic Foam Fiber Glass Polvisocvanurate Polyisocyanurate Composite Polystyrene Bead Board 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Note that m'KW fi hr F/Btu/5.6786 0.30 0.60 0.91 2.00 2.78 2.78 8.30 3.90 5.56 4.17 3.57 6.88 3.57 1.5 Spraved Polyurethane Foam*** 1.0 1.0 Cork

Explanation / Answer

The thermal resistance in series adds up, so the effective value of R will be:

R = R(gyp) + R(fib) + (Rw)

R = 0.3 x 0.5 + 4 x 3.9 + 1 x 0.91 = 16.66

q/A = (T2 - T1)/R

q/A = (20 - (-20))/16.66 = 2.4

Hence, q/A = 2.4

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