**Polyisocyanurate foam (PIR)**, also referred to as **PIR**, **polyiso**, or **ISO,** is very similar asis very similar tot is also a closed cell thermoset polymer formed by reacting a di- or polyisocyanate with a polyol. The thermal conductivity can be lower than of polyurethane foam. **Polyisocyanurate foam** is typically used for metal-faced panels, roof boards, cavity wall boards and pipe insulation. PIR foam panels laminated with pure embossed aluminium foil are used to fabricate a pre-insulated duct used for heating, ventilation, and air conditioning systems. On the other hand, PIR cannot be used to insulate existing cavity walls since no PIR foam can be injected into existing cavity walls. It offers even better thermal stability and flammability resistance than polyurethane foam.

## Thermal Conductivity of Polyisocyanurate Foam

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of a given thickness (in meters) 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 **polyisocyanurate foams** are between **0.022 and 0.035W/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 make 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., **polyisocyanurate foam**) 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 – Polyisocyanurate Foam 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 the area (A = 30 m^{2}). The wall is 15 cm thick (L_{1}), and it is made of bricks with 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**polyisocyanurate foam****insulation**10 cm thick (L_{2}) with the thermal conductivity of k_{2}= 0.022 W/m.K and calculate the heat flux (**heat loss**) through this composite wall.

**Solution:**

As was written, many heat transfer processes involve composite systems and even involve a combination of conduction and convection. It is often convenient to work with an** overall heat transfer coefficient, **known as a **U-factor **with these composite systems. 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 + 0.1/0.022 + 1/30) = 0.207 W/m^{2}K

The heat flux can be then calculated simply as:

q = 0.207 [W/m^{2}K] x 30 [K] = 6.21 W/m^{2}

The total heat loss through this wall will be:

q_{loss} = q . A = 6.21 [W/m^{2}] x 30 [m^{2}] = 186 W

As can be seen, adding a thermal insulator causes a significant decrease in heat losses. It must be added that adding the next layer of the 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.