Radiation

1. Radiation

2. Goals: By the end of today’s lecture, you should be able to: • define the three fates of radiation striking matter • define black body radiation • define and use the Stefan-Boltzman Law • define view factor and describe when this is required • define refractory surfaces and interchange factors • apply view factors to problems involving radiation between two bodies • compute heat loss from a system via parallel mechanisms (conduction, convection, radiation) • describe conditions under which radiation may be a significant mechanism of heat transfer

3. Practical Radiation Problem

4. Furnace Technology

5. Fundamentals of radiation Up to this point, we have discussed heat transfer mechanisms where the energy is transferred via collisions between molecules; that is by pure conduction, or conduction combined with convection. There is another extremely important heat transfer mechanism - Radiant heat transfer. Any surface which has a temperature above absolute zero radiates energy in the form of electromagnetic waves. This does not mean, however, that the amount of thermal radiation emitted is always significant.

6. The intensity and wavelength of the radiation emanating from a surface are functions of both the temperature and the surface properties. In our discussions, we will sum up the contributions over all wavelengths and evaluate the total energy transferred as a function of temperature. When considering heat transfer by radiation, we must be concerned not only with the emission of energy from a surface, but also what happens when radiation strikes a solid surface.

7. Radiation striking a solid surface has one of three fates: 1. 2. 3. How are these properties related ? Absorption absorptivity (a) Transmission transmissivity (t) Reflection reflectivity (z) a + t + z = 1

8. Two special cases require definition: If all of the energy is either reflected or absorbed (no transmitted radiation), we define the body as If all of the energy striking a surface is absorbed, we define the body as In general, these physical properties vary with the wavelength of the incident radiation. It is the difference in the wavelength of reflected light that allows the eye to perceive colors. For heat transfer calculations, we often assume that the properties a, t, and r are independent of wavelength. When this assumption is made we say that we have gray surfaces. Opaque a + z = 1 Black body a = 1

9. Let us return to the subject of radiation emitted by a surface. Total emissive power is defined as the total amount of energy leaving the surface per unit time per unit area: W = energy/area-time [Btu/hr-ft2 or W/m2] Note: Emissive power is a function of wavelength. The important wavelengths for heat transfer are 0.5 -50 µm. For temperatures above 1500°F, the important wavelength range is between 0.5 and 5 µm . In our analysis, we will use the average values over all wavelengths.

10. Emissivity The emissivity is the ratio of the emissive power of a surface compared to the maximum emissive power. How does the emissivity relate to the absorptivity (a) at thermal equilibrium? Although this strictly applies at thermal equilibrium, we normally assume that it applies at all temperatures. e = a

11. Finally, we must ask how the emissive power of a black body is related to temperature. The answer is provided by the single most important equation in radiative heat transfer. Stefan-Boltzman Law W = sT4 where s = 0.1714 x 10-8 Btu/hr-ft2-°R4 (Stefan-Boltzman constant) = 5.676 x 10-8 W/m2-K4 For an object that is not a black body (i.e., not a perfect radiator), we can write the following expression: T is absolute temperature W = esT4

12. To calculate the heat transfer rate by radiation, we must include terms for energy output and energy received from the surroundings. Energy Energy output: input: Making the usual assumption that e = a, and multiplying by area yields: This is the expression for an object totally enclosed by surroundings at T∞.

13. Example Problem -- Radiation Abody directly exposed to the night sky will be cooled below ambient temperature because of radiation to outer space. This effect can be used to freeze water in shallow trays well insulated from the ground. Estimate the maximum air temperature for which freezing is possible, neglecting evaporation. Assume ewater’s surface = 0.95. For cooling atmospheric air at ordinary temperatures by a horizontal surface facing upward, the heat transfer coefficient is given by: h = 0.2(Tair - Twater)1/4

14. View factors

15. Previously, we found that for a body totally enclosed by its surroundings, the net rate of heat transfer by thermal radiation is given by the following expression: q = esA(Ts4 - T24) The equation for q given above is one of the most important and commonly used results, however, it does not cover all situations.

17. The calculation of view factors is a straightforward exercise in calculus as shown in the figure on the preceding page. For each point on the surface A1, we consider rays of thermal energy emanating out equally in all directions. The fraction of these rays (actually, the total solid angle) which strikes A2 gives the fraction of energy reaching that surface. Integrating over all points on surface A1 and averaging gives the view factor F12. The following relationship is true: A1F12 = A2F21

18. What is the energy transfer rate from 1 to 2 and vice versa? q1->2 = q2->1 =

19. What is the net heat transfer: q = or q = Note: These equations are for black bodies.

20. Sometimes two heat exchanging surfaces T1 and T2 are connected by a series of surfaces through which no heat is transferred. An example might be the floor and ceiling of a building connected by well insulated walls. If there is no heat flow through the walls, what do we know about the relationship between absorption and emission? Such surfaces are called refractory surfaces. When refractory surfaces are present, the view factor must be adjusted to account for this extra radiation. The modified view factor is called an interchange factor and indicated by the Symbol F. The heat transfer equation is now written as:

24. Gray surfaces: When gray surfaces are present (i.e., surfaces for which e < 1), the calculation of the appropriate form of the view factor or interchange factor is substantially more complicated. The complication arises because the energy coming off a surface involves both emitted radiation and reflected radiation. The final result is straightforward, but the details of the derivation are tedious.

27. Radiation Calculation Decision Tree Nonblack Surface Yes (gray body) e ≠ 1.0 No (black body) e =1.0 Emitting into surroundings or near –infinite space ? Refractory / Reradiating Surface Present e≠ 1.0 Yes Yes No No

29. Tair 0

32. Example Problem -- temperature measurement A thermocouple (see figure below) is measuring the temperature of hot air flowing in a pipe whose walls are maintained at Tw = 400 K,. The true temperature of the gas Tg = 465 K. Calculate the temperature Tm indicated by the thermocouple. The emissivity of the probe is assumed to be e = 0.6 and the convective heat-transfer coefficient hc = 40 W/m2 K.