How Does Wall Composition Affect Heat Loss in a House?

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Homework Help Overview

This discussion revolves around estimating heat loss in a house, focusing on the impact of wall composition on thermal conductivity. The scenario involves a house modeled as a cube, with specific dimensions and materials described, including wooden beams and fiberglass insulation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of effective thermal conductivity, questioning how to determine the fractions of wood and fiberglass in the wall structure. There is also a focus on understanding the construction details provided in the problem.

Discussion Status

The discussion is ongoing, with some participants seeking clarification on the problem setup and others attempting to guide them toward understanding the construction details necessary for calculating thermal conductivity. There is no explicit consensus yet, but there are indications of productive dialogue regarding the fractions of materials involved.

Contextual Notes

Participants note the specific thermal conductivities of wood and fiberglass, and there is an emphasis on the importance of the wall's construction details for solving the problem. Some participants express uncertainty about the information provided.

beanryu
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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.
 
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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?
 
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
 
could u help me to solve this problem now?
 
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?
 

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