**Thermal insulation**is the process of reduction of heat transfer between objects in thermal contact or in range of radiative influence.

**Thermal insulations** consist of low thermal conductivity materials combined to achieve an even lower system thermal conductivity.

Thermal insulation can be achieved with specially engineered methods or processes, as well as with suitable object shapes and materials.

See also: Thermal conductivity

From microscopic point of view, transport of thermal energy in solids may be generally due to two effects:

**the migration of free electrons****lattice vibrational waves (phonons)**

When electrons and phonons carry thermal energy leading to conduction heat transfer in a solid, the thermal conductivity may be expressed as:

k = k_{e} + k_{ph}

**Metals** in general have **high electrical conductivity** and **high thermal conductivity**. These properties originate especially **from** the fact that their **outer electrons (free electrons) are delocalized**. Their contribution to the thermal conductivity is very high and is referred to as the **electronic thermal conductivity, k**** _{e}**. As a result, metals are very good thermal conductors instead of thermal insulators.

For **nonmetallic solids**, **k** is determined primarily by **k**** _{ph}**, which increases as the frequency of interactions between the atoms and the lattice decreases. In fact, lattice thermal conduction is the dominant thermal conduction mechanism in nonmetals, if not the only one. In solids, atoms vibrate about their equilibrium positions (crystal lattice). The vibrations of atoms are not independent of each other, but are rather strongly coupled with neighboring atoms. The regularity of the lattice arrangement has an important effect on

**k**

**, with crystalline (well-ordered) materials like**

_{ph}**quartz**having a higher thermal conductivity than amorphous materials like glass.

It must be added, **thermal insulation** is primarily based on the **very low thermal conductivity of gases**. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g., in a **foam-like structure**). Air and other gases are generally good insulators. But the main benefit is in the **absence of convection**. Therefore, many insulating materials (e.g.,polystyrene) function simply by having a large number of **gas-filled pockets** which **prevent large-scale convection**. Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

It must be noted, heat losses from hotter objects occurs by three mechanisms (either individually or in combination):

So far we have not discussed **thermal radiation** as a mode of **heat losses**. **Radiation heat transfer **is mediated by **electromagnetic radiation **and therefore it does not require any medium for heat transfer. In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum. Any material that has a temperature above absolute zero gives off some **radiant energy**. Most energy of this type is in the **infra-red region** of the electromagnetic spectrum although some of it is in the visible region. In order to decrease this type of heat transfer, materials with low emissivity should be used. The **emissivity, ε**, of the surface of a material is its effectiveness in emitting energy as **thermal radiation** and varies between 0.0 and 1.0. In general, polished metals have very low emissivity and therefore are widely used to reflect radiant energy back to its source as in case of **first aid blankets**.

## Thermal Insulator

As was written, **thermal insulation** is based on the use of substances with **very low thermal conductivity**. These materials are known as **thermal insulators**. Common thermal insulators are wool, fiberglass, rock wool, polystyrene, polyurethane, and goose feather etc. These materials are very poor conductors of heat and are therefore good thermal insulators.

It must be added, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g., in a **foam-like structure**). Air and other gases are generally good insulators. But the main benefit is in the **absence of convection**. Therefore, many insulating materials (e.g.,polystyrene) function simply by having a large number of **gas-filled pockets** which **prevent large-scale convection**. In all types of thermal insulation, evacuation of the air in the void space will further reduce the overall thermal conductivity of the insulator.

Alternation of gas pocket and solid material causes that the heat must be transferred through **many interfaces** causing rapid decrease in heat transfer coefficient.

In case of **thermal radiation insulation**, reflective insulations can be used. Reflective insulations are usually composed of multilayered, parallel foils of high reflectivity, which are spaced to reflect thermal radiation back to its source.

## 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 m^{2}). The wall is 15 cm thick (L_{1}) and it is made of bricks with the thermal conductivity of k_{1} = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h_{1} = 10 W/m^{2}K and h_{2} = 30 W/m^{2}K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

- Calculate the heat flux (
**heat loss**) through this non-insulated wall. - Now assume
**thermal insulation**on the outer side of this wall. Use expanded**polystyrene insulation**10 cm thick (L_{2}) with the thermal conductivity of k_{2}= 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 **Newton’s law of cooling**:

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

**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/m^{2}K

The heat flux can be then calculated simply as:

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

The total heat loss through this wall will be:

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

**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/m^{2}K

The heat flux can be then calculated simply as:

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

The total heat loss through this wall will be:

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