# What is the main source of convection

## What is heat loss - definition

### Example - heat loss through a wall

A major source for**Heat losses** walls are made of a house. Calculate the heat flow rate through a wall with an area of 3 mx 10 m (A = 30 m^{2} ). The wall is 15 cm thick (L._{1} ) and consists of bricks with a thermal conductivity of k_{1} = 1.0 W / mK (poor heat insulator). Assume that the inside and outside temperatures are 22 ° C and -8 ° C, and the convection heat transfer coefficients on the inside and outside are h_{1} = 10 W / m^{2} K and h_{2} = 30 W / m^{2}K each. Note that these convection coefficients are particularly dependent on the ambient and indoor conditions (wind, humidity, etc.).

- Calculate the heat flux (
**Heat loss**) through this uninsulated wall. - Let's take one now
**Thermal insulation**on the outside of this wall. Use a 10 cm thick expanded one**Polystyrene insulation**(L_{2}) with a thermal conductivity of k_{2}= 0.03 W / mK and calculate the heat flow (**Heat loss**) through this composite wall.

**Solution:**

As has been written, many of the heat transfer processes involve composite systems and even a combination of conduction and convection. With these network systems, it is often useful to use a**Overall heat transfer coefficient to work,** the as**U factor is known** . The U-factor is defined by an expression corresponding to the**Corresponds to Newton's law of cooling** :

The**Total heat transfer coefficient** is related to the total thermal resistance and is dependent on the geometry of the problem.

**bare wall**

Assuming a one-dimensional heat transfer through the flat wall and without taking radiation into account, the**Total heat transfer coefficient as follows** be calculated:

The**Total heat transfer coefficient** is then:

U = 1 / (1/10 + 0.15 / 1 + 1/30) = 3.53 W / m^{2} K.

The heat flux can then simply be calculated as:

q = 3.53 [W / m^{2} K] × 30 [K] = 105.9 W / m^{2}

The total heat loss through this wall is:

q_{loss} = q. A = 105.9 [W / m^{2} ] × 30 [m^{2} ] = 3177 W.

**Composite wall with thermal insulation**

Assuming a one-dimensional heat transfer through the flat composite wall, without**Thermal contact resistance** and without taking radiation into account, the**Total heat transfer coefficient as follows** be calculated:

The**Total heat transfer coefficient** is then:

U = 1 / (1/10 + 0.15 / 1 + 0.1 / 0.03 + 1/30) = 0.276 W / m^{2} K.

The heat flux can then simply be calculated as:

q = 0.276 [W / m^{2} K] × 30 [K] = 8.28 W / m^{2}

The total heat loss through this wall is:

q_{loss} = q. A = 8.28 [W / m^{2} ] × 30 [m^{2} ] = 248 W.

As can be seen, the addition of a thermal insulator causes a significant reduction in heat losses. It has to be added, adding the next layer of thermal insulator does not result in such high savings. This can be better seen from the thermal resistance method with which the heat is transferred through**Composite walls** can be calculated. The speed of continuous heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between these two surfaces.

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