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

HEAT TRANSFER (3) RADIATION

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.)

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)

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.

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

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

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

Figure 8.6: Fraction of blackbody radiation in wavelength interval

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)

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)

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

(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.

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,

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)

(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?

Solution

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.

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’