# 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

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m2). The wall is 15 cm thick (L1) and it is made of bricks with the thermal conductivity of k1 = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h1 = 10 W/m2K and h2 = 30 W/m2K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use expanded polystyrene insulation 10 cm thick (L2) with the thermal conductivity of k2 = 0.03 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

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

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m2K

The heat flux can be then calculated simply as:

q = 3.53 [W/m2K] x 30 [K] = 105.9 W/m2

The total heat loss through this wall will be:

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:

The overall heat transfer coefficient is then:

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

The heat flux can be then calculated simply as:

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

The total heat loss through this wall will be:

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.

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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