Conductive Heat Loss from a House

[kenau_reveas]

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.

 

[OlderDan]

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.

 

[m2003]

To calculate the effective thermal conductivity, we can use the formula:

k_eff = (k_wood * L_wood) / (L_wood + L_ins)

Where k_wood is the thermal conductivity of the wooden beams, L_wood is the thickness of the wooden beams, and L_ins is the thickness of the insulation.

Plugging in the values given in the problem, we get:

k_eff = (0.12 W/mK * 1.5875 inches) / (1.5875 inches + 7.5 inches) = 0.0107 W/mK

To two significant figures, this is 0.011 W/mK.

Next, we can calculate the surface area of the walls and ceiling using the formula:

A = 6 * L^2

Where L is the length of one side of the cube.

Plugging in the value given in the problem, we get:

A = 6 * (9.0 m)^2 = 486 m^2

Now, we can calculate the total conductive heat loss from the house using the formula:

Q = k_eff * A * (T_in - T_out) / L_wall

Plugging in the values given in the problem, we get:

Q = 0.011 W/mK * 486 m^2 * (20 C - 0 C) / 0.18 m = 1,210 W

Therefore, the total conductive heat loss from the house is 1,210 watts. This means that 1,210 joules of heat energy will be lost every second from the house due to conduction.

It is important to note that this calculation only takes into account the heat loss through the walls and ceiling. Other factors such as windows, doors, and ventilation will also contribute to the overall heat loss from the house. Additionally, this calculation assumes that there is no heat loss through the ground, which may not always be the case in real life scenarios.

Overall, this calculation gives us an estimate of the conductive heat loss from the house based on the given parameters. To reduce the heat loss and improve the energy efficiency of the house, it may be beneficial to increase the thickness of the insulation or use materials with higher thermal conductivity.