Nick Connor

Example – Glass Wool 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 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 22°C and -8°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 glass wool insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.023 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.

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:

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

2. 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 + 0.1/0.023 + 1/30) = 0.216 W/m²K

The heat flux can be then calculated simply as:

q = 0.216 [W/m²K] x 30 [K] = 6.48 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 6.48 [W/m²] x 30 [m²] = 194 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.

See also:

Insulation Materials

Thermal Conductivity of Glass Wool

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 glass wools are between 0.023 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. glass wool) 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.

See also:

Insulation Materials

Glass Wool

Glass wool (originally known also as fiberglass) is an insulating material made from fibres of glass arranged using a binder into a texture similar to wool. Glass wool and stone wool are produced from mineral fibres and are therefore often referred to as ‘mineral wools’. Mineral wool is a general name for fiber materials that are formed by spinning or drawing molten minerals. Glass wool is a furnace product of molten glass at a temperature of about 1450 °C. From the melted glass, fibres are spun. This process is based on spinning molten glass in high-speed spinning heads somewhat like the process used to produce cotton candy. During the spinning of the glass fibres, a binding agent is injected. Glass wool is then produced in rolls or in slabs, with different thermal and mechanical properties. It may also be produced as a material that can be sprayed or applied in place, on the surface to be insulated.

Applications of glass wool include structural insulation, pipe insulation, filtration and soundproofing. Glass wool is a versatile material that can be used for the insulation of walls, roofs and floors. It can be a loose fill material, blown into attics, or, together with an active binder sprayed on the underside of structures. During the installation of the glass wool, it should be kept dry at all times, since an increase of the moisture content causes a significant increase in thermal conductivity.

Thermal Conductivity of Glass Wool

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 glass wools are between 0.023 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. glass wool) 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 – Glass Wool 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 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 22°C and -8°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 glass wool insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.023 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.

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:

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

2. 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 + 0.1/0.023 + 1/30) = 0.216 W/m²K

The heat flux can be then calculated simply as:

q = 0.216 [W/m²K] x 30 [K] = 6.48 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 6.48 [W/m²] x 30 [m²] = 194 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.

See also:

Insulation Materials

Example – Stone Wool 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 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 22°C and -8°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 stone wool insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.022 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.

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:

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

2. 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/m²K

The heat flux can be then calculated simply as:

q = 0.207 [W/m²K] x 30 [K] = 6.21 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 6.21 [W/m²] x 30 [m²] = 186 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.

See also:

Insulation Materials

Thermal Conductivity of Stone Wool

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 mineral wools are between 0.020 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. stone wool) 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.

See also:

Insulation Materials

Stone Wool – Rock Wool

Stone wool, also known as rock wool, is based on natural minerals present in large quantities throughout the earth, e.g. volcanic rock, typically basalt or dolomite. Next to raw materials, also recycled rock wool can be added to the process as well as slag residues from the metal industry. It combines mechanical resistance with good thermal performance, fire safety and high temperature suitability. Glass‐ and stone wool are produced from mineral fibres and are therefore often referred to as ‘mineral wools’. Mineral wool is a general name for fiber materials that are formed by spinning or drawing molten minerals. Stone wool is a furnace product of molten rock at a temperature of about 1600 °C, through which a stream of air or steam is blown. More advanced production techniques are based on spinning molten rock in high-speed spinning heads somewhat like the process used to produce cotton candy.

Applications of stone wool include structural insulation pipe insulation, filtration, soundproofing, and hydroponic growth medium. Stone wool is a versatile material that can be used for the insulation of walls, roofs and floors. During the installation of the stone wool, it should be kept dry at all times, since an increase of the moisture content causes a significant increase in thermal conductivity.

Thermal Conductivity of Stone Wool

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 mineral wools are between 0.020 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. stone wool) 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 – Stone Wool 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 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 22°C and -8°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 stone wool insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.022 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.

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:

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

2. 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/m²K

The heat flux can be then calculated simply as:

q = 0.207 [W/m²K] x 30 [K] = 6.21 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 6.21 [W/m²] x 30 [m²] = 186 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.

See also:

Insulation Materials

How Emergency Thermal Blanket Works

How Emergency Thermal Blanket Works

As was written, thermal insulation is based on the use of substances with very low thermal conductivity. But sometimes the most effective way to protect persons or devices from leakage or gain of heat is to use a thermal radiation insulator. Thermal radiation does not require any medium for energy transfer. Moreover, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum.

In contrast to heat transfer by conduction or convection, which take place in the direction of decreasing temperature, thermal radiation heat transfer can occur between two bodies separated by a medium colder than both bodies. For example, solar radiation reaches the surface of the earth after passing through cold layers of atmosphere at high altitudes.

In order to insulate thermal radiation, space blankets (depending on the function, also known as a Mylar blanket, emergency thermal blanket or safety blanket) can be used. Space blankets were first developed by NASA in 1964. The highly reflective insulators are often included in emergency kits (therefore emergency thermal blankets) and are also used by long-distance runners after finishing a race to avoid a large swing in body temperature. Space blankets are designed to reflect heat back to body or deflect heat when used as a shelter from the sun, they can be used to insulate everything from the Mars rovers to marathon runners, from satellites to sun shields, and from rockets to residences. It is one of the simplest, yet most versatile spinoffs to come out of the Agency. Highly reflective foils in radiant barriers and reflective insulation systems reflect radiant heat away from persons and living spaces, making them particularly useful in very cold climates.

In case of emergency thermal blankets, the blankets are used to prevent/counter hypothermia. Their compact size before unfurling and light weight makes them ideal when space is at a premium. They may be included in first aid kits and also in camping equipment. Their design reduces the heat loss in a person’s body which would otherwise occur especially due to:

• Thermal radiation. As was written, radiation heat transfer This is achieved by minimisation of radiant heat transfer rate, q [W/m²], from a body to its surroundings is proportional to the fourth power of the absolute temperature. In this case, the emissivity of a material plays also very important role. 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. The surface of a blackbody emits thermal radiation at the rate of approximately 448 watts per square metre at room temperature (25 °C, 298.15 K). Real objects with emissivities less than 1.0 (e.g. aluminium foil) emit radiation at correspondingly lower rates (e.g. 448 x 0.07 = 31.4 W/m²). See also: Kirchhoff’s Law of thermal radiation
• Water evaporation and large scale convection. Thermal blankets are usually made from BoPET (Biaxially-oriented polyethylene terephthalate) airtight foil, causing them waterproof and windproof. This prevents large scale convection and heat losses caused by evaporation of perspiration.

The blackbody emissive power, E_b [W/m²], from a blackbody to its surroundings is proportional to the fourth power of the absolute temperature and can be expressed by the following equation:

E_b =  σT⁴

where σ is a fundamental physical constant called the Stefan–Boltzmann constant, which is equal to 5.6697×10^-8W/m²K⁴ and T is the absolute temperature of the surface in K.

See also:

Heat Losses

Space Blanket – Emergency Thermal Blanket

As was written, thermal insulation is based on the use of substances with very low thermal conductivity. But sometimes the most effective way to protect persons or devices from leakage or gain of heat is to use a thermal radiation insulator. Thermal radiation does not require any medium for energy transfer. Moreover, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum.

In contrast to heat transfer by conduction or convection, which take place in the direction of decreasing temperature, thermal radiation heat transfer can occur between two bodies separated by a medium colder than both bodies. For example, solar radiation reaches the surface of the earth after passing through cold layers of atmosphere at high altitudes.

In order to insulate thermal radiation, space blankets (depending on the function, also known as a Mylar blanket, emergency thermal blanket or safety blanket) can be used. Space blankets were first developed by NASA in 1964. The highly reflective insulators are often included in emergency kits (therefore emergency thermal blankets) and are also used by long-distance runners after finishing a race to avoid a large swing in body temperature. Space blankets are designed to reflect heat back to body or deflect heat when used as a shelter from the sun, they can be used to insulate everything from the Mars rovers to marathon runners, from satellites to sun shields, and from rockets to residences. It is one of the simplest, yet most versatile spinoffs to come out of the Agency. Highly reflective foils in radiant barriers and reflective insulation systems reflect radiant heat away from persons and living spaces, making them particularly useful in very cold climates.

In case of emergency thermal blankets, the blankets are used to prevent/counter hypothermia. Their compact size before unfurling and light weight makes them ideal when space is at a premium. They may be included in first aid kits and also in camping equipment. Their design reduces the heat loss in a person’s body which would otherwise occur especially due to:

• Thermal radiation. As was written, radiation heat transfer This is achieved by minimisation of radiant heat transfer rate, q [W/m²], from a body to its surroundings is proportional to the fourth power of the absolute temperature. In this case, the emissivity of a material plays also very important role. 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. The surface of a blackbody emits thermal radiation at the rate of approximately 448 watts per square metre at room temperature (25 °C, 298.15 K). Real objects with emissivities less than 1.0 (e.g. aluminium foil) emit radiation at correspondingly lower rates (e.g. 448 x 0.07 = 31.4 W/m²). See also: Kirchhoff’s Law of thermal radiation
• Water evaporation and large scale convection. Thermal blankets are usually made from BoPET (Biaxially-oriented polyethylene terephthalate) airtight foil, causing them waterproof and windproof. This prevents large scale convection and heat losses caused by evaporation of perspiration.

See also:

Heat Losses

Building Insulation – Home 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 materials with low thermal conductivity.  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. 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,  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.

Insulation Materials

As was written, thermal insulation is based on the use of substances with very low thermal conductivity. These materials are known as insulation materials. Common insulation materials 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 insulation 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.

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

• Heat Conduction
• Heat Convection
• Thermal Radiation

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 (high reflectivity) should 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. 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.

Types of Insulation – Categorization of Insulation Materials

For insulation materials, three general categories can be defined. These categories are based on the chemical composition of the base material from which the insulating material is produced.

In further reading, there is a brief description of these types of insulation materials.

Inorganic Insulation Materials

As can be seen from the figure, inorganic materials can be classified accordingly:

• Fibrous materials
  • Glass wool
  • Rock wool
• Cellular materials
  • Calcium silicate
  • Cellular glass

Organic Insulation Materials

The organic insulation materials treated in this section are all derived from a petrochemical or renewable feedstock (bio-based). Almost all of the petrochemical insulation materials are in the form of polymers. As can be see from the figure, all petrochemical insulation materials are cellular. A material is cellular when the structure of the material consists of pores or cells. On the other hand, many plants contain fibres for their strength, therefore almost all the bio-based insulation materials are fibrous (except expanded cork, which is cellular).

Organic insulation materials can be classified accordingly:

• Petrochemical materials (oil/coal derived)
  • Expanded polystyrene (EPS)
  • Extruded polystyrene (XPS)
  • Polyurethane (PUR)
  • Phenolic foam
  • Polyisocyanurate foam (PIR)
• Renewable materials (plant/animal derived)
  • Cellulose
  • Cork
  • Woodfibre
  • Hemp fibre
  • Flax wool
  • Sheeps wool
  • Cotton insulation

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 22°C and -8°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 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.

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:

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

2. 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:

q_loss = 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.

See also:

Heat Losses

Applications of Thermal Insulators

Clothing

The purpose of clothing is similar. The insulating properties of clothing come from the insulating properties of air. 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. Without clothes, our bodies in still air would heat the air in direct contact with the skin and would soon become reasonably comfortable because air is a very good insulator. It must be added, also in this case air will be flowing due to natural convection. In natural convection, air surrounding a body receives heat and by thermal expansion becomes less dense and rises. Thermal expansion of air plays a crucial role. In other words, heavier (more dense) components will fall, while lighter (less dense) components rise, leading to bulk air movement.

In the wind, the warm air surrounding our body would be replaced by cold air, thus increasing the temperature difference and the heat loss from the body. Clothes constitute a barrier to this moving air. The main benefit is in the absence of large-scale convection. Moreover, clothes are made from materials, which are generally good insulators. Many insulating materials (e.g. wool) 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.

Building 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 is to reduce the overall heat transfer coefficient by adding materials with low thermal conductivity.  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. 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,  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.

See also:

Heat Losses