conductivity

beanryu

In this problem you will estimate the heat lost by a typical house, assuming that the temperature inside is 20 degrees and the temperature outside is 0 degress . The walls and uppermost ceiling of a typical house are supported by 2*6-inch wooden beams with fiberglass insulation 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=18cm to allow for the interior and exterior covering. Assume that the house is a cube of length L=9.0m 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.

The first step is to calculate , 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.

please help me understand the question and find the effective.

Physics Monkey

I think it's clear that once you know the thermal conductivity of the wall, the rest of the problem is easy. The key to finding the effective thermal conductivity is to decide what fraction of the wall is wood and what fraction is fiberglass. Can you do this with the information given? How is the effective conductivity then obtained?

beanryu

hi, i don't even know the construction of the house. the conductivity of the wood is 0.12W/m/K and the conductivity of the fiberglass insulation is 0.04 W/m/K

beanryu

could u help me to solve this problem now?

Physics Monkey

Yes, you do know the construction of the house. You're told in detail how the walls and ceiling are constructed. In particular, you're told how wide the wood beams are and how far apart they are spaced. You are also told that fiberglass fills the space in between the beams. So what fraction of the total cross sectional area of a wall is covered by wood and what fraction is covered by fiberglass?