Equipment for Heat Transfer: Radiation
Figure 1:
Equipment for Heat Transfer: Radiation
black metal sheet
white metal sheet
E glass
normal glass
heat lamp (halogen light) with stand
normal lamp (regular lightbulb)
propane torch
temperature probes or
thermocouple


  Heat Transfer:
  Radiation

      Heat transfer occurs by three mechanisms: conduction, convection, and radiation. We have discussed conduction in the past two lessons. In this lesson, we will discuss radiation. We will start by looking again at the electromagnetic spectrum.



Objectives [At the end of this lesson students will be able to...]


Start-up questions

  1. How does heat (or energy) come to us from the sun?
  2. What is in the space between planet earth and the sun?
  3. Do all bodies produce heat?
  4. Do you know how night vision goggles work?

Figure 2: The Electromagnetic Spectrum

Thermal Energy Transfer: Radiation

      Radiative heat transfer is the only way to transfer heat from one place to another that does not require a medium. Let's look again at the electromagnetic radiation spectrum and the various aspects and applications of radiation. Notice the relative energies of photons in the radio/TV, microwave, IR (infrared, heat), optical (visible), UV (ultraviolet), x-ray, and gamma ray bands.


      We will demonstrate the transmission of infrared radiation (heat) through different glasses using a light meter and two sources of light (normal and heat lamp) by feel and thermocouple.

Comparison of transmission through different glasses

Light meter reading without glass:

Temperature reading without glass:

  Visible Light
(visual and light meter)
IR (heat)
(feel and thermocouple)
Normal glass    
E glass (energy efficient)    
Stain glass    
Dark double glass    


      Let's describe the basic concept of energy (or heat) transfer by radiation. We will develop the basic equation for radiative heat transfer, Stefan's law. First, some definitions:

Radiation is the energy emitted from a surface as particles or waves. The rate of radiative emission per unit time and per unit surface area , or radiative flux, is represented as Q. Note that this "Q" is actually power per unit area, or P/A; it has different units than the "Q" that we used for conduction. Again, to prevent confusion, just follow the units. In the MKS unit system, radiative flux is given in joules per second (watts) per square meter (W/m²).

Radiative flux on the [4th power] temperature of the emitting surface:

Q [temperature]4 {where T is measured in Kelvins}

Emissivity () is the ratio of a surface's ability to emit radiant energy compared with the ability of a perfect black body of the same area at the same temperature. Emissivity is a dimensionless constant having values between 0 and 1. Perfect reflectors would have = 0. Perfect absorbers would have = 1.
A material with high emissivity is efficient in both absorbing radiation energy as well as emitting it. Therefore a good absorber is also a good emitter.

Q emissivity () {0 < < 1}

      Now that we have the parameters, we need a constant to balance the equation. This is Stefan's constant (s) [also known as the Stefan-Boltzmann constant]. It has a value in the MKS unit system of:

5.67×10-8 W/(m²·K4).

Putting this all together, we get Stefan's law (or the Stefan-Boltzmann law), which states that the radiative flux (Q) is equal to the emissivity times the fourth power of temperature in Kelvins times Stefan's constant:

Qemitted = sT4

This could also be written as:

Qemitted = P/A = sT4

      There are some important points to keep in mind when working problems dealing with radiative flux. First, conservation of energy dictates that if you have any reflection (an emissivity of less than 1), then:

Qincoming = Qabsorbed + Qreflected
Also, since emissivity affects radiation absorption in the same way that it affects emission:
Qemitted = Qabsorbed = Qincoming
Note that this means if you are in space (where you only deal with radiative heat transfer) and you want to determine final temperature of an object with a given incoming radiative flux, the emissivity of the object does not matter:
Qemitted = Qabsorbed = Qincoming = sT4
so that: Qincoming = sT4
Finally, keep in mind the possibility that your surface area for absorption may differ from your surface area for emission. For example, if I have flat panel that absorbs radiation from the sun (likely to occur only for one side of the panel at a time), the panel will emit radiation from both sides (here we are assuming that the panel is thin enough so that you only need to deal with two sides).

      To illustrate surfaces with different emissivities, we will place a black sheet and a white sheet near a heat lamp and feel the difference by hand and by measurement. Notice the importance of selecting a thermally efficient external color to energy conservation. A significant amount of heat in the summer in Alabama houses is generated through radiation absorption by the roof.

Emissivity of Some Common Materials

Materials Emissivity Values
(no unit, dimensionless)

Carbon

0.85-0.95

Aluminum

0.11

Brass (oxidized)

0.61

Brass (unoxidized)

0.030

Copper (oxidized)

0.60

Copper (unoxidized)

0.020

Black gloss paint

0.90

Gold (polished)

0.020

Fire brick

0.75


Example problems

  1. I have a system in space that is generating a lot of waste heat which I need to get rid of by radiation into deep, cold space. The power I need to dump is 1.0×10³ W. My external radiator has a surface area of 1.0 m by 2.0 m, has useful emission from only one side, and has an emissivity of 0.99 (pretty good absorber). What is the equilibrium temperature of my radiator in °C?

    Solution:
    Since I am given the amount of radiation that I need to emit, I can determine Qemitted without having to worry about Qincoming. From the equation
    Qemitted = P/A = 1.0×10³ W/(1.0 m × 2.0 m) = 500 W/m²
    Qemitted = P/A = sT4 so that
    T = [Qemitted/s] = {500 W/m²/[0.99 × 5.67×10-8 W/(m²·K4)]} = 307.2 K
    We were asked for °C, so convert from K to °C: T = 307.2 - 273.15 = 34°C

  2. You work part time as a contractor building houses (to supplement your teaching income). Your client asks you what difference there is (if any) between using a light color roof and a dark color roof in the south where the summer is very hot (we all know that). The homeowner wants to use very dark shingles with an emissivity of 0.90. The roof has a 6.0"-thick fiberglass insulation and the attic temperature is always maintained at 80.0°F (26.7°C= 299.8 K). Thermal conduction through the roof to the attic and radiative loss are the main sources of heat transfer. Determine the equilibrium temperature in the day time for the shingles if the solar power is 1.0 kW/m².

    Solution:
    Energy absorption rate = 1.0 kW/m² × 0.90 = 9.0 kW/m².
    Heat loss through conduction = (0.040 W/m·K) / (6.0×0.0254 m) DT
                       = 0.26 (T- 300) W/m²
    Heat loss through radiation = sT4 = 0.90 × 5.67×10-8 W/(m²·K4) × T4
                       = 5.6×10-8 T4 W/(m²·K4)

    Balancing heat absorption and loss yields:

    900 = 0.26T - 78 + 5.6×10-8 T4
    5.6×10-8 T4 + 0.26T - 978 = 0 (NOW WHAT DO WE DO????)

    You can either go and consult a mathematician to solve the equation or make some approximations. If we are lucky, we can get pretty close to the correct answer.

    Assume that the 5.6×10-8 T4 term is small compared to the other two terms (this is true for small T). Then the solution is simply T = 3760 K. This is a pretty high temperature and violates the assumption of small T. Actually if we substitute this solution into the original full equation, we have:
    1.12×107 + 978 - 978. This certainly is not equal to zero.

    Another assumption is that 0.26T is small; then we have as an approximation of the equation: 5.6×10-8 T4 - 978 = 0 or T = 364 K (a much more reasonable assumption). To check, we can substitute this solution back into the original equation and we have:
    978 + 94.5 - 978 which indicates that our answer is a little high, but no more than about 10% off. Quite acceptable in this approximation.


Homework Only -- for Printing Solution

Homework

After teaching high school science for 20 years, you decided to change your career and became a space thermal scientist for NASA. You were charged with the design of a solar panel in space (the only heat transfer there is radiation, no heat conduction at all). Your solar panel receives 1.0 kW/m² of solar energy and its emissivity (also its absorptivity) is 0.60 (or 60.%). The solar energy is received on only one side at a time, but emission occurs on both sides. Determine the equilibrium temperature of the panel. Repeat the calculation for an emissivity of 0.30 and see what you find.