What is Polyisocyanurate Foam – Definition

Polyisocyanurate foam (PIR), also referred to as PIR, polyiso, or ISO, is a closed cell thermoset polymer formed by reacting a di- or poly-isocyanate with a polyol. Thermal Engineering

Polyisocyanurate Foam

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

 
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 Polyisocyanurate Foam

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 polyisocyanurate foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus 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. polyisocyanurate foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Polyisocyanurate Foam Insulation

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 polyisocyanurate foam insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.022 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

polyisocyanurate foam insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.022 + 1/30) = 0.207 W/m2K

The heat flux can be then calculated simply as:

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

The total heat loss through this wall will be:

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

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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, Polyisocyanurate Foam, 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 – Polyurethane Foam Insulation Calculation – Definition

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

Example – Polyurethane Foam 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 polyurethane foam insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.028 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

polyurethane foam insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.028 + 1/30) = 0.259 W/m2K

The heat flux can be then calculated simply as:

q = 0.259 [W/m2K] x 30 [K] = 7.78 W/m2

The total heat loss through this wall will be:

qloss = q . A = 7.78 [W/m2] x 30 [m2] = 233 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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 – Polyurethane Foam 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 Thermal Conductivity of Polyurethane Foam – Definition

Thermal conductivity of Polyurethane Foam. Typical thermal conductivity values for polyurethane foam are between 0.022 and 0.035W/m∙K. Thermal Engineering

Thermal Conductivity of Polyurethane Foam

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 polyurethane foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus 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. polyurethane foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

 
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 Polyurethane Foam, 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 Polyurethane Foam – Definition

Polyurethane foam insulation is available in closed-cell and open-cell formulas. Polyurethane foam can be used as cavity wall insulation or as roof insulation. Thermal Engineering

Polyurethane Foam

Polyurethane foam - thermal insulationPolyurethane foam (PUR) is a closed cell thermoset polymer. Polyurethane polymers are traditionally and most commonly formed by reacting a di- or poly-isocyanate with a polyol. Polyurethane foam insulation is available in closed-cell and open-cell formulas. Polyurethane foam can be used as cavity wall insulation or as roof insulation, floor insulation, pipe insulation, insulation of industrial installations. Insulating panels made from PUR can be applied to all elements of the building envelope. Another important aspect is that PUR can also be injected into existing cavity walls, by using the existing openings and some extra holes.

 
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 Polyurethane Foam

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 polyurethane foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus 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. polyurethane foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Polyurethane Foam 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 polyurethane foam insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.028 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

polyurethane foam insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.028 + 1/30) = 0.259 W/m2K

The heat flux can be then calculated simply as:

q = 0.259 [W/m2K] x 30 [K] = 7.78 W/m2

The total heat loss through this wall will be:

qloss = q . A = 7.78 [W/m2] x 30 [m2] = 233 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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, Polyurethane Foam, 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 – Extruded Polystyrene Insulation Calculation – Definition

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

Example – Extruded Polystyrene 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 extruded polystyrene insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.028 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

extruded polystyrene insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.028 + 1/30) = 0.259 W/m2K

The heat flux can be then calculated simply as:

q = 0.259 [W/m2K] x 30 [K] = 7.78 W/m2

The total heat loss through this wall will be:

qloss = q . A = 7.78 [W/m2] x 30 [m2] = 233 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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 – Extruded Polystyrene 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 Thermal Conductivity of Extruded Polystyrene – Definition

Thermal conductivity of extruded polystyrene – XPS. Typical thermal conductivity values for extruded polystyrene are between 0.025 and 0.040W/m∙K. Thermal Engineering

Thermal Conductivity of Extruded Polystyrene

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 extruded polystyrene are between 0.025 and 0.040W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. extruded polystyrene) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

 
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 Extruded Polystyrene, 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 Extruded Polystyrene – XPS – Definition

Extruded polystyrene (XPS) is also a thermoplastic polymer. Extruded polystyrene has a closed cell structure and is often stronger, with a higher mechanical performance and is, in principle, often more expensive than EPS. Thermal Engineering

Extruded Polystyrene – XPS

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

Extruded polystyrene - thermal insulationExtruded polystyrene (XPS) is also a thermoplastic polymer. Extruded polystyrene has a closed cell structure and is often stronger, with a higher mechanical performance and is, in principle, often more expensive than EPS. Its density range is about 28–45 kg/m3. XPS is produced from the same base materials as EPS and therefore also has crude oil at its basis. The production process of extruded polystyrene is only slightly different from expanded polystyrene.

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

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

 
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 Extruded Polystyrene

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 extruded polystyrene are between 0.025 and 0.040W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. extruded polystyrene) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Extruded Polystyrene Insulation

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 extruded polystyrene insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.028 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

extruded polystyrene insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.028 + 1/30) = 0.259 W/m2K

The heat flux can be then calculated simply as:

q = 0.259 [W/m2K] x 30 [K] = 7.78 W/m2

The total heat loss through this wall will be:

qloss = q . A = 7.78 [W/m2] x 30 [m2] = 233 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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, Extruded Polystyrene – XPS, 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 – Expanded Polystyrene Insulation Calculation – Definition

Example – Expanded Polystyrene Insulation Calculation. Calculate the heat flux (heat loss) through insulated wall. Use expanded polystyrene (EPS) insulation 10 cm thick. Compare it with a bare wall. Thermal Engineering

Example – Expanded Polystyrene 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 expanded polystyrene insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.03 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

expanded polystyrene insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.03 + 1/30) = 0.276 W/m2K

The heat flux can be then calculated simply as:

q = 0.276 [W/m2K] x 30 [K] = 8.28 W/m2

The total heat loss through this wall will be:

qloss = q . A = 8.28 [W/m2] x 30 [m2] = 248 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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 – Expanded Polystyrene 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 Thermal Conductivity of Expanded Polystyrene – Definition

Thermal conductivity of expanded polystyrene. Typical thermal conductivity values for expanded polystyrene are between 0.030 and 0.040W/m∙K. Thermal Engineering

Thermal Conductivity of Expanded Polystyrene

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 expanded polystyrene are between 0.030 and 0.040W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. expanded polystyrene) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

 
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 Expanded Polystyrene, 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 Expanded Polystyrene – EPS – Definition

Expanded polystyrene (EPS) is a rigid and tough, closed-cell foam. Building and construction applications account for around two-thirds of demand for expanded polystyrene. Thermal Engineering

Expanded Polystyrene – EPS

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

expanded polystyrene - thermal insulation

Expanded polystyrene (EPS) is a rigid and tough, closed-cell foam. Building and construction applications account for around two-thirds of demand for expanded polystyrene. It is used for the insulation of (cavity) walls, roofs and concrete floors. Due to its technical properties such as low weight, rigidity, and formability, expanded polystyrene can be used in a wide range of applications, for example trays, plates and fish boxes.

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

 
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 Expanded Polystyrene

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 expanded polystyrene are between 0.030 and 0.040W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. expanded polystyrene) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Expanded Polystyrene 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 expanded polystyrene insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.03 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

expanded polystyrene insulationThe overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.03 + 1/30) = 0.276 W/m2K

The heat flux can be then calculated simply as:

q = 0.276 [W/m2K] x 30 [K] = 8.28 W/m2

The total heat loss through this wall will be:

qloss = q . A = 8.28 [W/m2] x 30 [m2] = 248 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

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

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