Cellulose Insulation – Paper Wool

Cellulose Insulation

Cellulose insulation is made from recycled paper products, primarily newspapers and has a very high recycled material content. The obtained cellulose fibres have a wool like structure (therefore paper wool). In order to make the cellulose fibres moisture and flame retardant, boric acid or ammonium sulfate are added. Cellulose insulation is used in wall and roof cavities to insulate, draught proof and reduce free noise. Cellulose insulation is used in both new and existing homes, usually as loose-fill in open attic installations and dense packed in building cavities. Cellulose and the other loose-fill materials can be blown into attics, finished wall cavities, and hard-to-reach areas.

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 Cellulose 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 cellulose insulation 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. cellulose 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 – Cellulose 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 cellulose 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.


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

cellulose 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

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:

  7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
  9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.

Advanced Reactor Physics:


See above:

Insulation Materials