RADIANT HEATING WITH INFRARED - Watlow

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It is interesting to note that Figure 10 indicates that much of the short wavelength infrared and visible radiation is transmitted through the water. This is consistent with the observation that water is largely transparent to light (you can see the bottom of a swimming pool). 10 Emissivity is often the parameter that is the most difficult to accurately determine when making radiant calculations. In some cases, emissivity can be approximated with a constant value. For example, (Figure 9) a surface with an emissivity of 0.7 will absorb 70 percent (70%) of the radiant energy that strikes it. Such a surface is called a gray body. The emissivity of many real surfaces varies with the product thickness, temperature, and the wavelength of the incident energy. The task of calculating the percentage of radiant energy absorbed by the product can be simplified if the emissivity of the product can be approximated as a constant. Absorption % 100 80 60 40 20 0 0 Spectral Absorption Curve For Water (Fig. 10) 2 4 Figure 10 illustrates that a film of water thicker than 0.05 mm will absorb most of the infrared radiation longer than two microns. Planck’s curves (Figure 4) indicates that a heater operating in the 1000˚F-1500˚F range produces almost all its energy in wavelengths longer than two microns. If calculating the radiant energy absorption of a water film thicker than 0.002 inches (0.05 mm) that is being heated by a heater operating under 1500˚F, the emissivity of the water can be considered a constant 0.93. Emissivity Values Some common examples of emissivities that can be considered constant (gray bodies) in the medium to long wavelength infrared region are shown below: 6 Wavelenght (microns) Wavelength (microns) Water - 0.05 mm thick Emissivity Values (Fig.11) Water - 0.01 mm thick Material Emissivity Material Emissivity Blackbody 1 Most Paint 0.9 Wood 0.9 Fire Brick 0.8 Rubber 0.9 Polished Silver 0.01 Aluminum Polished 0.03 Glass 0.8-0.9 Aluminum Oxidized 0.24 Steel Polished 0.07 Water 0.93 Steel Oxidized 0.80 Watlow Radiant Panel 0.85 Lamp Black 0.96 8 10 12
Heater Emissivity Just as the surface condition of the product determines how well it absorbs radiant energy, the surface conditions of the heater determines how well it emits radiant energy. The Stefan Boltzman equation computes the energy radiation from a blackbody (Emissivity = 1). All real surfaces radiate less. To calculate the watts radiated from a real surface, correction for the emissivity of that surface must be made: Watts Radiated = Stefan-Boltzman Equation x Emissivity Heater The emissivity of the heater (Eh) is a constant 0.85. If the emissivity of the product (Ep) can also be considered a constant, the wattage absorbed by the product can be determined with the following equation: Watts Absorbed = Stefan-Boltzman Equation x Eh x Ep x View Factor This represents the simplest case. If the product is opaque and somewhat reflective, then the reflected energy will bounce back to the heater, and some will be reflected back to the product for a second chance at absorption. In this case, E effective is used instead of Eh x Ep. 1 E effective = 1 + 1 - 1 Eh Ep E effective is always greater than Eh x Ep If the emissivity of the product cannot be considered a constant, an emissivity spectrum should be obtained from the material supplier. Most plastic films and sheets fall into this category. Figure 12 is a transmission spectrum for polyethylene 0.1 mm thick (0.004 in). Transmission Spectrum for Polyethylene (Fig. 12) Transmission Spectrum for Polyethylene (Fig. 12) Transmission% 100 80 60 40 20 Polyethylene 0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 Wavelength (microns) The Watlow radiant panels are manufactured with a high emissivity, hot-face coating with an emissivity of 0.85. They radiate 85 percent (85%) as much as a blackbody at the same temperature. In the infrared region, this emissivity is relatively wavelength independent, making Watlow panels effective gray body radiators. Note: This does not mean that the heater is only 85 percent (85%) efficient. It means the heater will operate somewhat hotter than a blackbody radiating the same wattage. 11
Page 1 and 2: ed W A T L O W RADIANT HEATING WITH
Page 3 and 4: The Advantages of Radiant Heat Elec
Page 5 and 6: 10 6 X 10 1.4 X 10 1.2 X 10 0.4 0.7
Page 7 and 8: Radiated Power vs. Wavelength (Fig.
Page 9 and 10: W/in 2 Radiated Net In a typical ra
Page 11: Looking at Figure 8, the new view f
Page 15 and 16: This information is useful for sele
Page 17 and 18: Step 3: Compute the Absorbed Wattag
Page 19 and 20: E = 1 = 1 = 0.70 1 + 1 - 1 1 + 1 -
Page 21 and 22: 2. Determine the wattage required t
Page 23 and 24: D. The required oven length can now
Page 25 and 26: C. Use the Stefan Boltzman equation
Page 27 and 28: Controlling Radiant Heaters The fol
Page 29 and 30: Usually this sensor temperature doe
Page 31 and 32: Tips On Oven Design Efficiency The
Page 33 and 34: Temperature Uniformity: Edge Losses
Page 35 and 36: Wiring Do not use copper wire or ri
Page 37 and 38: (Fig. 18) Time vs. Temperature Pane
Page 39: 14 15 13 12 11 10 9 8 6 4 2 7 3 5 1

RADIANT HEATING WITH INFRARED - Watlow

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