Conductive Heat Loss from a House

AI Thread Summary
The discussion focuses on calculating the effective thermal conductivity (k_eff) of a house's walls and ceiling, given specific dimensions and materials. The interior temperature is set at 20°C, while the exterior temperature is 0°C, with wooden beams and fiberglass insulation used in the structure. The effective thermal conductivity must account for the actual dimensions of the beams and their spacing. The calculation requires determining the area associated with each material and integrating the thickness of the walls and ceiling into the k_eff formula. This analysis is crucial for understanding conductive heat loss in residential buildings.
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Homework Statement


assuming that the temperature inside is T_in = 20 C and the temperature outside is T_out = 0 C. The walls and uppermost ceiling of a typical house are supported by 2*6-inch wooden beams (k_wood = 0.12 W/m /K) with fiberglass insulation (k_ins = 0.04 W/m /K) in between. The true depth of the beams is actually 5 *5/8 inches, but we will take the thickness of the walls and ceiling to be L_wall = 18 cm to allow for the interior and exterior covering. Assume that the house is a cube of length L = 9.0 m on a side. Assume that the roof has very high conductivity, so that the air in the attic is at the same temperature as the outside air. Ignore heat loss through the ground.

Homework Equations



The first step is to calculate k_eff, the effective thermal conductivity of the wall (or ceiling), allowing for the fact that the 2*6 beams are actually only 1*5/8 wide and are spaced 16 inches center to center.
Express k_eff numerically to two significant figures in watts per kelvin per meter.
 
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The numbers that you have for k_wood and k_ins have units proportional to 1/m, but there are two spatial quantities in k. One is the thickness and one is the area. You need to find the definition for k and figure out how much area is associated with each of the materials in your problem to come up with the k_eff.
 
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