Heat Losses

Heat Losses

zeroth-law-of-thermodynamics-heatWhile thermal energy refers to the total energy of all the molecules within the object, heat is the amount of energy flowing from one body to another spontaneously due to their temperature difference. Heat is a form of energy, but it is energy in transit. Heat is not a property of a system. However, the transfer of energy as heat occurs at the molecular level as a result of a temperature difference. When a temperature difference does exist heat flows spontaneously from the warmer system to the colder system, never the reverse. This direction of thermodynamic processes is given by the second law of thermodynamics.

As a result, any object hotter than the surroundings must continuously lose a part of its thermal energy. This is a natural behaviour of all objects. When the flow of heat stops, they are said to be at the same temperature. They are then said to be in thermal equilibrium. Heat losses from hotter objects occur by three mechanisms (either individually or in combination):

  • Heat Conduction. Heat conduction, also called diffusion, occurs within a body or between two bodies in contact. It is the direct microscopic exchange of kinetic energy of particles through the boundary between two systems. When an object is at a different temperature from another body or its surroundings
  • Heat Convection. Heat convection depends on motion of mass from one region of space to another. Heat convection occurs when bulk flow of a fluid (gas or liquid) carries heat along with the flow of matter in the fluid.
  • Thermal Radiation. Radiation is heat transfer by electromagnetic radiation, such as sunshine, with no need for matter to be present in the space between bodies.

Heat Losses – Clothing

The purpose of clothing is similar. The insulating properties of clothing come from the insulating properties of air. 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. Without clothes, our bodies in still air would heat the air in direct contact with the skin and would soon become reasonably comfortable because air is a very good insulator. It must be added, also in this case air will be flowing due to natural convection. In natural convection, air surrounding a body receives heat and by thermal expansion becomes less dense and rises. Thermal expansion of air plays a crucial role. In other words, heavier (more dense) components will fall, while lighter (less dense) components rise, leading to bulk air movement.

In the wind, the warm air surrounding our body would be replaced by cold air, thus increasing the temperature difference and the heat loss from the body. Clothes constitute a barrier to this moving air. The main benefit is in the absence of large-scale convection. Moreover, clothes are made from materials, which are generally good insulators. Many insulating materials (e.g. wool) function simply by having a large number of gas-filled pockets which significantly decrease material’s thermal conductivity. Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Heat Losses – Thermal Insulation

expanded polystyrene - thermal insulation
Expanded polystyrene insulation

In order to minimize heat losses in industry and also in construction of buildings, thermal insulation is widely used. The purpose of thermal insulation of is to reduce the overall heat transfer coefficient by adding of material with low thermal conductivity.  Thermal insulation in buildings is an important factor to achieving thermal comfort for its occupants. Thermal insulation reduces unwanted heat loss and also reduce unwanted heat gain. Therefore thermal insulation can decrease the energy demands of heating and cooling systems. It must be added, there is no material which can completely prevent heat losses, heat losses can be only minimized.

Similarly as for clothing, thermal insulation is based on low thermal conductivity materials and on its geometry (e.g. double-pane windows). The insulating properties of these materials come from the insulating properties of air. Many insulating materials (e.g. glass wool) function simply by having a large number of gas-filled pockets which prevent large-scale convection. Geometry of these materials plays also crucial role. For example,  increasing the width of the air layer, such as using two panes of glass separated by an air gap, will reduce the heat loss more than simply increasing the glass thickness, since the thermal conductivity of air is much less than that for glass.

Example – Heat Loss through a Wall

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

thermal insulation - expanded polystyreneThe 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

Example – Heat Loss through a Window

Example - Heat Flux - Thermal ConductionHeat loss through windows

A major source of heat loss from a house is through the windows. Calculate the rate of heat flux through a glass window 1.5 m x 1.0 m in area and 3.0 mm thick, if the temperatures at the inner and outer surfaces are 14.0°C and 13.0°C, respectively. Calculate the heat flux through this window.

Solution:

At this point, we know the temperatures at the surfaces of material. These temperatures are given also by conditions inside the house and outside the house. In this case, heat flows by conduction through the glass from the higher inside temperature to the lower outside temperature. We use the heat conduction equation:

Heat loss through window - equation

We assume that the thermal conductivity of a common glass is k = 0.96 W/m.K.

The heat flux will then be:

q = 0.96 [W/m.K] x 1 [K] / 3.0 x 10-3 [m] = 320 W/m2

The total heat loss through this window will be:

qloss = q . A = 320 x 1.5 x 1.0 = 480W

It must be added, 15°C is not very warm for the living room of a house. But this temperature does not correspond to the interior temperature, but correspond to the surface temperature. Due to the finite convective heat transfer coefficient, there is always a considerable temperature drop between the interior temperature and the window surface temperature. Note that, both convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

See also: Convective Heat Transfer Coefficient

Similarly as for clothing, thermal insulation is based on low thermal conductivity materials and on its geometry (e.g. double-pane windows). The insulating properties of these materials come from the insulating properties of air. Many insulating materials (e.g. wool) function simply by having a large number of gas-filled pockets which prevent large-scale convection. Geometry of these materials plays also crucial role. For example, increasing the width of the air layer, such as using two panes of glass separated by an air gap, will reduce the heat loss more than simply increasing the glass thickness, since the thermal conductivity of air is much less than that for glass.

 
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).
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  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 above:

Heat Transfer