10 m 10 m 3 m 36 Define Physics Assumptions and Relationships 3 m 10 m 10 m
13 37 Step 4: Solve the Problem 3 m 10 m 10 m 38 Step 5: Check Your Answer Is is reasonable? Units? Have the assumptions affected our result? 3 m 10 m 10 m 39 Thermal Energy and Power; Transforming Energy; Energy in Humans; Temperature; Heat; Conduction; Home Heating • Materials are well characterized by thermal conductivity constants or Thermal Resistance: R- Value : R = d/k where d is the thickness of the material and k, the thermal conductivity. • Heat conduction using R-value: • Given in imperial units. R T T A t Q H ) ( 1 2 − = Δ Δ =
14 40 1.32 Sheathing (0.5 in. thick) 0.45 Drywall (0.5 in. thick) 0.17 Free stagnant air layer 1.01 Air space (3.5 in. thick) 1.54 Insulating glass (0.25 in space) 0.89 Flat glass (0.125 in thick) 3.7 Cellulose fiber(1 in. thick) 4.35 Fiberglass board (1 in. thick) 18.8 Fiberglass batting (6 in. thick) 10.9 Fiberglass batting (3.5 in. thick) 1.93 Concrete block (filled cores) 4 Brick (4 in. thick) 0.87 Wood shingles (lapped) 0.91 Hardwood siding (1 in. thick) R value(ft2·°F·hr/BTU) Material 41 Thermal Energy and Power; Transforming Energy; Energy in Humans; Temperature; Heat; Conduction; Home Heating • Units: (ft 2 °F hr/BTU) • (BTU A British Thermal Unit (BTU) is the amount of heat energy needed to raise the temperature of one pound of water by one degree F. This is the standard measurement used to state the amount of energy that a fuel has as well as the amount of output of any heat generating device.) • I BTU=1055 Joules • 1 R = 0.093 m 2 ·0.55 ° C·3600s/1055J = 0.17 m 2 ° K s/J = 0.17 m 2 ° K/W 42 Thermal Energy and Power; Transforming Energy; Energy in Humans; Temperature; Heat; Conduction; Home Heating • Usefulness of R-Value : – While the units of the R-value are a bit awkward to convert, the concept of a thermal resistance is especially useful in composite walls. Differences in materials and thicknesses are “built in”. – For instance it can be shown that for thermal conduction through an insulated wall, the R-values simply add: R tot = R 1 + R 2 . R 1 R 2 Fiberglass inside Brick outside tot R T T A t Q H ) ( 1 2 − = Δ Δ =
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