## Extruded Polystyrene – XPS

Generally, **polystyrene** is a synthetic aromatic polymer made from the monomer styrene, which is derived from benzene and ethylene, both petroleum products. **Polystyrene** can be solid or foamed. **Polystyrene** is a colorless, transparent thermoplastic, which is commonly used to make foam board or beadboard insulation and a type of loose-fill insulation consisting of small beads of polystyrene. Polystyrene foams are 95-98% air. Polystyrene foams are good thermal insulators and are therefore often used as building insulation materials, such as in insulating concrete forms and structural insulated panel building systems. **Expanded polystyrene** and **extruded polystyrene** are both made from polystyrene, but EPS is composed of small plastic beads that are fused together and XPS begins as a molten material that is pressed out of a form into sheets. XPS is most commonly used as foam board insulation.

**Extruded polystyrene **(XPS) is also a thermoplastic polymer. **Extruded polystyrene** has a closed cell structure and is often stronger, with a higher mechanical performance and is, in principle, often more expensive than EPS. Its density range is about 28–45 kg/m^{3}. XPS is produced from the same base materials as EPS and therefore also has crude oil at its basis. The production process of **e****xtruded polystyrene** is only slightly different from expanded polystyrene.

Similarly as EPS, XPS has a wide variety of applications. It can be used for the insulation of buildings, roofs and concrete floors. Extruded polystyrene material can be also used in crafts and model building, in particular architectural models.

Although both expanded and extruded polystyrene have a closed-cell structure, they are permeable by water molecules and can not be considered a vapor barrier. In expanded polystyrene there are interstitial gaps between the expanded closed-cell pellets that form an open network of channels between the bonded pellets. If the water freezes into ice, it expands and can cause polystyrene pellets to break off from the foam.

## Thermal Conductivity of Extruded Polystyrene

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer, and hence the greater the insulation’s effectiveness. **Typical thermal conductivity values** for **extruded polystyrene** are between **0.025 and 0.040W/m∙K**.

In general, 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., **extruded 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.

## Example – Extruded Polystyrene Insulation

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**extruded polystyrene insulation**10 cm thick (L_{2}) with the thermal conductivity of k_{2}= 0.028 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 + 0.1/0.028 + 1/30) = 0.259 W/m^{2}K

The heat flux can be then calculated simply as:

q = 0.259 [W/m^{2}K] x 30 [K] = 7.78 W/m^{2}

The total heat loss through this wall will be:

q_{loss} = q . A = 7.78 [W/m^{2}] x 30 [m^{2}] = 233 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.