Thermal Engineering

Thermal Conductivity of Cork Insulation

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 cork are between 0.035 and 0.043W/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. cork insulation) 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

Example – Cork 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 cork 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.038 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 + 0.1/0.038 + 1/30) = 0.343 W/m²K

The heat flux can be then calculated simply as:

q = 0.343 [W/m²K] x 30 [K] = 10.29 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 10.29 [W/m²] x 30 [m²] = 308 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

Cotton Insulation

Cotton insulation is made from post-industrial recycled cotton textiles. Cotton insulation consists of 85% recycled cotton and 15% plastic fibers. Similarly as for cellulose, in order to make the cotton insulation flame retardant, cotton must be impregnated.

Thermal Conductivity of Cotton Insulation

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 cotton insulation is around 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 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. cotton insulation) 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 – Cotton 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 cotton insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.04 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 + 0.1/0.040 + 1/30) = 0.359 W/m²K

The heat flux can be then calculated simply as:

q = 0.359 [W/m²K] x 30 [K] = 10.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 10.78 [W/m²] x 30 [m²] = 323 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 Cotton Insulation

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 cotton insulation is around 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 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. cotton insulation) 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

Example – Cotton Insulation 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 cotton insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.04 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 + 0.1/0.040 + 1/30) = 0.359 W/m²K

The heat flux can be then calculated simply as:

q = 0.359 [W/m²K] x 30 [K] = 10.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 10.78 [W/m²] x 30 [m²] = 323 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

Foam Glass – Cellular Glass

Cellular glass (foam glass) is a glass foam material, which is formed from a reaction between glass and carbon at high temperatures. Foam glass has a cellular structure and it is impermeable. It has good thermal properties and therefore it can be used a thermal insulation. Its high impermeability makes it ideal as a barrier against soil humidity. Because foam glass is entirely made of inorganic materials, it is non‐flammable. Foam glass has also a high compressive strength, which makes the material very suitable for the insulation of flat roofs which are covered with bitumen of other heavy substances. On the other hand, foam glass is not suitable for the insulation of wooden flooring, because of its high impermeability.

Thermal Conductivity of Foam Glass

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 foam glass are between 0.038 and 0.055W/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. foam glass) 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 – Foam Glass

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 foam glass insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.04 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 + 0.1/0.040 + 1/30) = 0.359 W/m²K

The heat flux can be then calculated simply as:

q = 0.359 [W/m²K] x 30 [K] = 10.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 10.78 [W/m²] x 30 [m²] = 323 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 Foam Glass

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 foam glass are between 0.038 and 0.055W/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. foam glass) 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

Example – Foam Glass

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 foam glass insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.04 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 + 0.1/0.040 + 1/30) = 0.359 W/m²K

The heat flux can be then calculated simply as:

q = 0.359 [W/m²K] x 30 [K] = 10.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 10.78 [W/m²] x 30 [m²] = 323 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

Aerogel

Aerogel is a synthetic porous ultralight solid material derived from a gel, in which the liquid component of the gel has been replaced with a gas (during a supercritical drying process). Aerogels can be made from a variety of chemical compounds, but the base material for aerogel is usually silicon. Aerogel has a very low thermal conductivity of 0.013 W/m∙K. Its density is also very low, about 150 kg/m³. These are remarkable thermal insulative properties. It must be noted, aerogels may have lower thermal conductivity than that of the gas (air has about 0.025 W/m∙K) they contain. This is caused by the Knudsen effect, a reduction of thermal conductivity in gases when the size of the cavity encompassing the gas becomes comparable to the mean free path.

Thermal Conductivity of Aerogel

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 aerogels are around 0.013W/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. aerogel) 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 – Aerogel

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 aerogel 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 + 0.1/0.013 + 1/30) = 0.125 W/m²K

The heat flux can be then calculated simply as:

q = 0.125 [W/m²K] x 30 [K] = 3.76 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 3.76 [W/m²] x 30 [m²] = 113 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

Small Modular Reactor – SMR

Small modular reactors (SMRs) are nuclear fission reactors which are smaller than conventional reactors. The term “small” in the context of SMRs refers to design power output. As per the International Atomic Energy Association (IAEA) classification, small modular reactors are defined as reactors which produce power output of less than or equal to 300 MWe. It must be noted, most of the commercial reactors operating around the world are large reactors with power output ranging between 1000 MWe and 1600 MWe. The term “modular” in the context of SMRs refers to its scalability and to the ability to fabricate major components of the nuclear steam supply system (NSSS) in a factory environment and then transported to the site. Its scalability means that some of SMRs are to be deployed as multiple-module power plants.

• Light Water Reactors
• High Temperature Gas Cooled Reactors
• Liquid Metal Cooled Reactors

According to their promoters, its scalability, modularity, robust design and enhanced safety features of the SMR offers great advantages over large commercial reactors. It must be noted, this reactor design is currently (2018) at the stage of development, however their technology is similar to proven naval reactors.

SMRs have potential in expanding countries for enhancing the energy supply security both and embarking countries, which have inadequate infrastructure or less established grid system (not suited for large commercial reactors). However, SMRs have potential to become a key part of the energy mix even in developed countries that often face problems with the construction of large commercial reactors.

See more: ADVANCES IN SMALL MODULAR REACTOR TECHNOLOGY DEVELOPMENTS, a supplement to ARIS, IAEA, 2014.

Advantages and Disadvantages of Small Modular Reactors

Small modular reactors are very specific. Their size and modularity offer many advantages. On the other hand they have some disadvantages, which must be taken into account during decision making.

Possible Advantages

Enhanced safety and security

Lower thermal power of the reactor core, compact architecture and employment of passive concepts have the potential for enhanced safety and security compared to earlier designs and large commercial reactors. The passive safety systems are very important safety feature in the SMR. Therefore there is less reliance on active safety systems and additional pumps, as well as AC power for accident mitigation. These passive safety systems are able to dissipate heat even after loss of offsite power. The safety system incorporates an on-site water inventory which operates on natural forces (e.g. natural circulation). In reactor engineering, natural circulation is very desired phenomenon, since it is capable to provide reactor core cooling without coolant pumps, so that no moving parts could break down.

Modularity

As was written, the term “modular” in the context of SMRs refers to its scalability and to the ability to fabricate major components of the nuclear steam supply system (NSSS) in a factory environment and then transported to the site. This can help limit the on-site preparation and also reduce the construction time. This is very important, since the lengthy construction times are one of key problems of the larger units. Moreover, the in-factory fabrication and completation of major parts of the nuclear steam supply system can also facilitate implementation of higher quality standards (e.g. inspections of welds).

Construction time and financing

Size, construction efficiency and passive safety systems (requiring less redundancy) can reduce a nuclear plant owner’s capital investment due to the lower plant capital cost. In-factory fabrication of major components of nuclear steam supply system can significantly reduce the on-site preparation and also reduce the construction time. This in turn can lead to easier financing compared to that for larger plants.

Possible Disadvantages

Large-scale Production

Most of the economic benefits (especially lower capital cost) stated are valid for n-th unit produced. In order to achieve these economic benefits, large-scale production of SMRs and initial orders for tens of units is required.

Licensing

One of very important barriers is licensing of new reactor designs. For example, in regulating the design, siting, construction, and operation of new commercial nuclear power facilities, the NRC currently employs a combination of regulatory requirements, licensing, and oversight. Historically, the licensing process was developed for large commercial reactors. The licensing process for new reactor designs is a lengthy and costly process.

See also:

Type of Reactors