RADIANT HEATING WITH INFRARED - Watlow

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The view factor tells us the percentage of the radiant energy leaving the heater that is intercepted by the product. One could say that the view factor tells us how good a view the product has of the heater. View factor graphs exist for physical configurations other than parallel planes, such as concentric cylinders and non-parallel planes. Consult <strong>Watlow</strong> in these cases. 8 M = Heater Width Distance to Product N = Heater Length Distance to Product Note: M and N are interchangeable The View Factor The Stefan-Boltzman equation calculates the net energy radiated by the heater. It doesn’t calculate how much of the radiated energy actually hits the product. There is a geometric relationship between the size of the heater and product, and the distance between them, that determines how much of the radiated energy the product intercepts. This relationship is called the View Factor. The view factor can be determined graphically. Using Figure 8, the view factor can be determined for a flat rectangular heater radiating toward a parallel flat rectangular product. Consider a radiant heater, 20 inch X 40 inch, radiating towards a product of the same size and 10 inches away. First calculate M and N: M = Heater Width = 20 in = 2 Distance to Product 10 in View Factor F N = Heater Length = 40 in = 4 Distance to Product 10 in 1.0 0.8 0.6 0.4 0.2 0.1 0.08 0.06 0.04 0.02 N= 0.01 0.1 0.2 0.4 0.60.8 1 2 M 4 6 8 10 20 Looking at Figure 8, the view factor (F) is: F = 0.5 This indicates that 50 percent (50%) of the energy radiated by the heater will strike the product. To examine the relationship distance plays, assume the same radiant heater and product are now five inches apart, where: 10 M = 20 in 5 in = 4 N = 40 in 5 in = 8 View Factor (Fig. 8) 4 2 1 0.6 0.4 0.2 0.1
Looking at Figure 8, the new view factor (F) is: F = 0.7 Now 70 percent (70%) of the energy radiated by the panel will reach the product. By moving the heater from 10 inches to five inches from the product, the product receives 40 percent (40%) more energy The closer the heater is to the product, the more efficient the heat transfer will be. The heater to product distance will be determined by physical constraints of the equipment or by the uniformity of the radiant heat source. In applications where the heater to product distance is large, the efficiency can be improved by adding side reflectors of polished aluminum. Emissivity and Absorptivity Not all the radiant energy that reaches the product is absorbed. Some may be reflected and some transmitted right through. Only absorbed energy will serve to heat the product. Total Incident Energy = Absorbed + Reflected + Transmitted Absorbed Transmitted Reflected It is the physical nature of the product that determines how well the product absorbs the radiant energy that strikes it. This same physical property determines how well a surface emits radiant energy. At the same temperature and wavelength, absorptivity and emissivity are equal. From here on, the term “emissivity” will be used for the numerical value that describes either the ability to absorb or emit radiant energy. What is a Blackbody? A perfect absorbing surface will absorb all the radiant energy that strikes it. Such a surface has an emissivity equal to one (1) and is called a blackbody. All real surfaces have emissivities less than one (1). In general, non-metallic surfaces have good emissivities (close to 1) and shiny metallic surfaces have low emissivities (close to 0). Shiny metal surfaces are good reflectors. Thin films may have low emissivities because they transmit a large part of the incident energy. Absorption Absorption vs. vs. Wavelength Wavelenght (Fig. 9) (Fig. 9) Absorption % % 100 80 60 40 20 Transmitted Paper Paper 0 0 2 4 6 8 10 Wavelength Wavelenght (microns) Blackbody E = 1.0 E= 1.0 Gray Body E= 0.7 Grey Body E = 0.7 9
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: W/in 2 Radiated Net In a typical ra
Page 13 and 14: Heater Emissivity Just as the surfa
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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