# Heat Loss through a Wall

## Example – Heat Loss through a Wall

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m2). The wall is 15 cm thick (L1) and it is made of bricks with the thermal conductivity of k1 = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22Â°C and -8Â°C, and the convection heat transfer coefficients on the inner and the outer sides are h1 = 10 W/m2K and h2 = 30 W/m2K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use expanded polystyrene insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.03 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newtonâ€™s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m2K

The heat flux can be then calculated simply as:

q = 3.53 [W/m2K] x 30 [K] = 105.9 W/m2

The total heat loss through this wall will be:

qloss = q . A = 105.9 [W/m2] x 30 [m2] = 3177W

1. composite wall with thermal insulation

Assuming one-dimensional heat transfer through the plane composite wall, no thermal contact resistance and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.03 + 1/30) = 0.276 W/m2K

The heat flux can be then calculated simply as:

q = 0.276 [W/m2K] x 30 [K] = 8.28 W/m2

The total heat loss through this wall will be:

qloss = q . A = 8.28 [W/m2] x 30 [m2] = 248 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

## Heat Losses – Thermal Insulation

In order to minimize heat losses in industry and also in construction of buildings, thermal insulation is widely used. The purpose of thermal insulation of is to reduce the overall heat transfer coefficient by adding of material with low thermal conductivity. Â Thermal insulation in buildings is an important factor to achieving thermal comfort for its occupants. Thermal insulation reduces unwanted heat loss and also reduce unwanted heat gain. Therefore thermal insulation can decrease the energy demands of heating and cooling systems. It must be added, there is no material which can completely prevent heat losses, heat losses can be only minimized.

Similarly as for clothing, thermal insulation is based on low thermal conductivity materials and on its geometry (e.g. double-pane windows). The insulating properties of these materials come from the insulating properties of air. Many insulating materials (e.g. glass wool) function simply by having a large number of gas-filled pockets which prevent large-scale convection. Geometry of these materials plays also crucial role. For example, Â increasing the width of the air layer, such as using two panes of glass separated by an air gap, will reduce the heat loss more than simply increasing the glass thickness, since the thermal conductivity of air is much less than that for glass.

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References:
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