What is Inorganic Insulation Material – Definition

As can be seen from the figure, inorganic materials can be classified accordingly. Glass wool is typical example of inorganic insulation materials. Thermal Engineering

Insulation Materials

Thermal Insulators - ParametersAs 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.

 

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.

Insulation Materials - Types

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:

Example of Insulation – Glass Wool

Glass wool - thermal insulationGlass 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.

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Heat Losses

We hope, this article, Inorganic Insulation Material, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Example – Cork Insulation Calculation – Definition

Example – cork insulation calculation. Calculate the heat flux (heat loss) through insulated wall. Use cork insulation 10 cm thick. Compare it with a bare wall. Thermal Engineering

Example – Cork Insulation

heat loss through wall - example - calculationA 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:

u-factor - overall heat transfer coefficient

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:

overall heat transfer coefficient - heat loss calculation

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:

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

  1. 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:

overall heat transfer coefficient - thermal insulation calculation

cork insulationThe 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:

qloss = 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.

thermal resistance - equation

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Example – Cork Insulation Calculation, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Cotton Insulation – Definition

otton 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 Engineering

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.

 
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.

Insulation Materials - Types

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

Other Insulation Materials

  • Cellular Glass
  • Aerogel
  • Vacuum Panels

Thermal Conductivity of Cotton Insulation

Thermal Insulators - ParametersThermal 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

heat loss through wall - example - calculationA 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:

u-factor - overall heat transfer coefficient

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:

overall heat transfer coefficient - heat loss calculation

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:

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

  1. 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:

overall heat transfer coefficient - thermal insulation calculation

cotton insulationThe 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:

qloss = 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.

thermal resistance - equation

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Cotton Insulation, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Thermal Conductivity of Cotton Insulation – Definition

Thermal conductivity of cotton insulation. Typical thermal conductivity values for cotton insulation is around 0.035W/m∙K. Thermal Engineering

Thermal Conductivity of Cotton Insulation

Thermal Insulators - ParametersThermal 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.

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Thermal Conductivity of Cotton Insulation, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Example – Cotton Insulation Calculation – Definition

Example – cotton insulation calculation. Calculate the heat flux (heat loss) through insulated wall. Use cotton insulation 10 cm thick. Compare it with a bare wall. Thermal Engineering

Example – Cotton Insulation Insulation

heat loss through wall - example - calculationA 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:

u-factor - overall heat transfer coefficient

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:

overall heat transfer coefficient - heat loss calculation

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:

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

  1. 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:

overall heat transfer coefficient - thermal insulation calculation

cotton insulationThe 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:

qloss = 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.

thermal resistance - equation

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Example – Cotton Insulation Calculation, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Foam Glass – Cellular Glass – Definition

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. Thermal Engineering

Foam Glass – Cellular Glass

foam glass - cellular glassCellular 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.

 
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.

Insulation Materials - Types

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

Other Insulation Materials

  • Cellular Glass
  • Aerogel
  • Vacuum Panels

Thermal Conductivity of Foam Glass

Thermal Insulators - ParametersThermal 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

heat loss through wall - example - calculationA 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:

u-factor - overall heat transfer coefficient

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:

overall heat transfer coefficient - heat loss calculation

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:

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

  1. 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:

overall heat transfer coefficient - thermal insulation calculation

Foam Glass - Cellular GlassThe 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:

qloss = 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.

thermal resistance - equation

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Foam Glass – Cellular Glass, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Thermal Conductivity of Foam Glass – Definition

Thermal conductivity of Foam Glass. Typical thermal conductivity values for foam glass (cellular glass) are between 0.035 and 0.040W/m∙K. Thermal Engineering

Thermal Conductivity of Foam Glass

Thermal Insulators - ParametersThermal 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.

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Thermal Conductivity of Foam Glass, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Example – Foam Glass Calculation – Definition

Example – foam glass calculation. Calculate the heat flux (heat loss) through insulated wall. Use foam glass insulation 10 cm thick. Compare it with a bare wall. Thermal Engineering

Example – Foam Glass

heat loss through wall - example - calculationA 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:

u-factor - overall heat transfer coefficient

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:

overall heat transfer coefficient - heat loss calculation

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:

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

  1. 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:

overall heat transfer coefficient - thermal insulation calculation

Foam Glass - Cellular GlassThe 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:

qloss = 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.

thermal resistance - equation

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Example – Foam Glass Calculation, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Aerogel – Definition

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. Thermal Engineering

Aerogel

Aerogel - thermal insulation
A flower is on a piece of aerogel which is suspended over a flame from a Bunsen burner. Aerogel has excellent insulating properties, and the flower is protected from the flame. Source: wikipedia.org License: Public Domain

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.

 
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.

Insulation Materials - Types

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

Other Insulation Materials

  • Cellular Glass
  • Aerogel
  • Vacuum Panels

Thermal Conductivity of Aerogel

Thermal Insulators - ParametersThermal 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

heat loss through wall - example - calculationA 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:

u-factor - overall heat transfer coefficient

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:

overall heat transfer coefficient - heat loss calculation

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:

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

  1. 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:

overall heat transfer coefficient - thermal insulation calculation

aerogel insulationThe 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:

qloss = 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.

thermal resistance - equation

 
References:
Heat Transfer:
  1. Fundamentals of Heat and Mass Transfer, 7th Edition. Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera. John Wiley & Sons, Incorporated, 2011. ISBN: 9781118137253.
  2. Heat and Mass Transfer. Yunus A. Cengel. McGraw-Hill Education, 2011. ISBN: 9780071077866.
  3. U.S. Department of Energy, Thermodynamics, Heat Transfer and Fluid Flow. DOE Fundamentals Handbook, Volume 2 of 3. May 2016.

Nuclear and Reactor Physics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
  6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See also:

Insulation Materials

We hope, this article, Aerogel, helps you. If so, give us a like in the sidebar. Main purpose of this website is to help the public to learn some interesting and important information about thermal engineering.

What is Small Modular Reactor – SMR – Definition

Small modular reactors (SMRs) are nuclear fission reactors which are smaller than conventional reactors. Small modular reactors have power output of less than 300 MWe. Thermal Engineering

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.

SMR designs include:
  • 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.

 
References:
Reactor Physics and Thermal Hydraulics:
  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN: 978-0415802871
  6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN: 978-3-319-13419-2
  7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition, John Wiley & Sons, 2006, ISBN: 978-0-470-03037-0
  8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010, ISBN 978-1-4020-8670-0.
  9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.

See also:

Type of Reactors

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