Heat Loss in a House: Estimating k_eff

In summary, the problem involves estimating the heat lost by a house with given internal and external temperatures. The house has walls and ceilings supported by 2 x 6-inch wooden beams with fiberglass insulation in between. The thickness of the walls and ceiling is taken to be 18 cm, and the house is a cube with a length of 9.0m on each side. The task is to calculate the effective thermal conductivity of the walls or ceiling, taking into account the fact that the wooden beams are actually only 1 and 5/8 inches wide and spaced 16 inches center to center.
  • #1
gmarc
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0

Homework Statement



In this problem you will estimate the heat lost by a typical house, assuming that the temperature inside is T_in= 20degrees C and the temperature outside is T_out = 0degrees C. The walls and uppermost ceiling of a typical house are supported by 2 x 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 and 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 = 9.0m 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 k_eff, the effective thermal conductivity of the wall (or ceiling), allowing for the fact that the 2 x 6 beams are actually only 1and 5/8 wide and are spaced 16 inches center to center.

The Attempt at a Solution



I know that k_eff = f(k_wood-k_ins)+k_ins, so I have everything except f, which is the fraction of the total wall/ceiling area in which the heat is conducted by wood. I'm having trouble finding this...
 
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  • #2
Hi gmarc,

If I'm reading the problem correctly, I would first start by focusing on a small section of a wall, perhaps starting with the right edge of one beam and ending with the right edge of the next beam. What fraction of that length (or area depending on how you are looking at it) is wood?
 
  • #3


To find the fraction of the wall/ceiling area in which heat is conducted by wood, we need to calculate the total surface area of the walls and ceiling and then determine the area occupied by the wooden beams.

Total surface area of walls and ceiling = 6*(9.0m)^2 = 486m^2

Area occupied by wooden beams on walls = 2*0.05m*9.0m + 2*0.05m*9.0m = 0.9m^2

Area occupied by wooden beams on ceiling = 0.05m*9.0m = 0.45m^2

Total area occupied by wooden beams = 0.9m^2 + 0.45m^2 = 1.35m^2

Fraction of total wall/ceiling area occupied by wooden beams = 1.35m^2 / 486m^2 = 0.0028

Therefore, f = 0.0028

Substituting this value into the equation for k_eff, we get:

k_eff = 0.0028*(0.12 W/m/K - 0.04 W/m/K) + 0.04 W/m/K = 0.0404 W/m/K

Therefore, the effective thermal conductivity of the wall/ceiling is 0.0404 W/m/K. This value can be used to estimate the heat loss in the house using the equation Q = k_eff*A*(T_in - T_out)/L, where A is the total surface area of the walls and ceiling and L is the thickness of the walls and ceiling.
 

1. What is heat loss in a house?

Heat loss in a house refers to the amount of heat that escapes from the interior of the house to the outside environment. This can happen through various means such as conduction, convection, and radiation.

2. Why is it important to estimate k_eff for heat loss in a house?

k_eff, or the effective thermal conductivity, is a measure of how well a material can conduct heat. By estimating k_eff, we can determine the potential for heat loss in a house and make adjustments to improve energy efficiency and reduce heating costs.

3. How is k_eff calculated for a house?

To calculate k_eff for a house, the thermal conductivity of each material used in the construction of the house is multiplied by its corresponding area and thickness, and then all of these values are summed together. This gives an overall value for the house's effective thermal conductivity.

4. What factors can affect k_eff in a house?

Some factors that can affect k_eff in a house include the type and thickness of insulation, the type of construction materials used, air leakage, and the design and orientation of the house. Climate and weather conditions can also impact k_eff.

5. How can k_eff be improved to reduce heat loss in a house?

To reduce heat loss in a house, k_eff can be improved by using high-quality insulation materials, sealing any air leaks, and choosing construction materials with a lower thermal conductivity. Proper insulation and weatherization techniques can also help improve k_eff and reduce heat loss.

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