1. ERT 216 HEAT & MASS TRANSFER
NURADIBAH MOHD AMER
School of Bioprocess Engineering
University Malaysia Perlis

2. HEAT TRANSFER (3) RADIATION

3. Radiation Thermal Radiation
Radiation  Energy transfer through a space by electromagnetic radiation.
There are many types of electromagnetic radiation. But we just discuss the electromagnetic radiation that is propagated as a result of a temp. difference.
Thermal Radiation
(electromagnetic radiation emitted by a body as a result of its temp.)

4. Physical Mechanism
All type of radiation were propagated at the speed of light (C), 3 x 108 m/s
This speed of light is
equal to the product of wavelength (ʎ) & frequency (v) of the radiation
Unit wavelength (ʎ)  can be in cm, angstroms (1A = 10 cm) or micrometers (1μm = 10 m)
Diagram shown a portion of the electromagnetic spectrum
Thermal radiation lies in the range from 0.1 to 100 μm, while visible-light portion of the spectrum is very narrow (0.35 to 0.75 μm)

5. Black body absorbs all radiation incident upon it.
Physical Mechanism
Black Body Radiation
Black body absorbs all radiation incident upon it.
Called black body radiation because materials appear black to the eye. They appear black because they do not reflect any radiation.
“STEFAN-BOLTZMANN LAW”
Eb : The emissive power of a blackbody (unit  watt)
σ : Stefan-Boltzmann constant
T : Absolute Temp.

6. Total Radiation over all wavelength
Total Radiation, Eb = Eb (0 - ∞) =  between zero & infinity wavelength (over all the wavelength)

7. Radiation emitted between wavelength zero to ʎ1
Radiation emitted between wavelength ʎ1 to ʎ2
Radiation emitted between wavelength ʎ2 to ∞

8. Radiation energy emitted between wavelength ʎ1 & ʎ2  Eb (ʎ1-ʎ2)
Would be calculated using:
Where 
so that,
get from Figure 8.6 or Table 8.1. Used ʎT relation
Value of  and

9. Figure 8.6: Fraction of blackbody radiation in wavelength interval

10. qin = Area x Eb (ʎ1-ʎ2) Incident radiation heat transfer (qin)
Incident radiation (qin) can be calculated between two wavelengths ʎ1 & ʎ2 :
qin = Area x Eb (ʎ1-ʎ2)

11. Radiation Properties
When radiant energy strikes a material surface, part of the radiation if reflected, part is absorbed and part is transmitted, as shown above.
Reflection radiation (qrefl.)  reflectivity, ρ = qrefl /qin
Absorbed radiation (qabs.)  absorptivity, α = qabs./qin
Transmitted radiation (qtrans.)  transmissivity, ז = qtrans./qin
Most solid bodies do not transmit thermal radiation, so that transmissivity may be take as zero (ז=0)

12. Radiation Properties qrefl. = qin x ρ qabs.= qin x α qtrans.= qin x ז
Incident radiation heat transfer (qin)
Example  on glass plate
Can break down in to the following radiation classification:
Therefore:
qrefl. = qin x ρ
qabs.= qin x α
qtrans.= qin x ז
So that, Total incident radiation 
where 
Kirchhoff’s Identity (emissivities = absorption)  discussed the total properties of the particular material over all wavelengths

13. (Transmission & absorption in glass plate)
Example 8.1
(Transmission & absorption in glass plate)
A glass plate 30 cm square is used to view radiation from a furnace. The transmissivity of the glass is 0.5 from 0.2 to 3.5 μm. The emissivity may be assumed to be 0.3 up to 3.5 μm and 0.9 above that. The transmissivity of the glass is zero except in the range from 0.2 to 3.5 μm. Assuming that the furnace is a blackbody at 2000 °C, calculate the energy absorbed in the glass and the energy transmitted.

14. Radiation Shape Factor (F12)
In order to calculate the net radiant exchange (q rad) between 2 surface  it is necessary to find shape factor (F12) from provided graphs. Then q rad can be calculate using equation:
F12: Fraction of energy leaving surface 1 that reaches surface 2
&
So that,

15. Value of F12 Parallel Rectangles
F12 obtained from the value of ratio (X/D)
Parallel Disks
F12 obtained from the value of ratio (d/X)
Perpendicular Rectangles
F12 obtained from the value of ratio (Z/X)

16. (Heat Transfer between Black Surfaces)
Example 8.2
(Heat Transfer between Black Surfaces)
Two parallel black plates 0.5 by 1.0 m are spaced 0.5 m apart. One plate is maintained at 1000 °C and the orther at 500 °C. What is the net radiant heat exchange between the two plates?

17. Solution

18. Heat Exchange between Non Black Bodies
The calculation of radiation heat transfer:
Black Surface/ Bodies is relatively easy because  all the radiant energy which strikes a surface is absorbed. The main problem to determine geometric shape factor (F12), but once this is accomplished, the calculation heat exchange is very simple using
Non Black Body more complex because  for all the energy striking a surface will not be absorbed (part will be reflected back to another heat transfer surface & part may be reflected out of the system entirely). The radiant energy can be also reflected back and forth between the heat transfer surfaces several times.

19. Heat Exchange between Non Black Bodies
So, the net radiation heat exchange between two non black bodies/surfaces:
&
where 
Emissivity, ϵ  relates the radiation on the ‘gray surface’