Technical

Technical Information Radiant Infrared Heating - Theory & Principles

Infrared Theory Emissivity and an Ideal Infrared Source — Reflectivity — Materials with poor emissivity The ability of a surface to emit radiation is frequently make good reflectors. Polished gold Infrared energy is radiant energy which passes defined by the term emissivity. The same term with an emissivity of 0.018 is an excellent through space in the form of electromagnetic is used to define the ability of a surface to infrared reflector that does not oxidize easily. waves (Figure 1). Like light, it can be reflected absorb radiation. An ideal infrared source Polished aluminum with an emissivity of 0.04 and focused. Infrared energy does not depend would radiate or absorb 100% of all radiant is an excellent second choice. However, once on air for transmission and is converted to energy. This ideal is referred to as a “perfect” the surface of any metal starts to oxidize or heat upon absorption by the work piece. In black body with an emissivity of unity or 1.0. collect dirt, its emissivity increases and its fact, air and gases absorb very little infrared. The spectral distribution of an ideal infrared effectiveness as an infrared reflector decreases. As a result, infrared energy provides for emitter is below. efficient heat transfer without contact between Table 1 — Approximate Emissivities the heat source and the work piece. Spectral Distribution of a Blackbody Metals Polished Rough Oxidized Figure 1 at Various Temperatures Aluminum 0.04 0.055 0.11-0.19 !Peak Wavelength Brass 0.03 0.06-0.2 0.60 Copper 0.018-0.02 — 0.57 1400°F Gold 0.018-0.035 — — Wien Displacement Curve Steel 0.12-0.40 0.75 0.80-0.95 Stainless 0.11 0.57 0.80-0.95 ! Lead 0.057-0.075 0.28 0.63 1200°F Nickel 0.45-0.087 — 0.37-0.48 Silver 0.02-0.035 — — Tin 0.04-0.065 — — Infrared heating is frequently missapplied and Zinc 0.045-0.053 — 0.11 capacity requirements underestimated due to Radiant Energy 1000°F Galv. Iron 0.228 — 0.276 a lack of understanding of the basic principles Miscellaneous Materials of radiant heat transfer. When infrared energy 800°F Asbestos 0.93-0.96 from a source falls upon an object or work 400°F Brick 0.75-0.93 piece, not all the energy is absorbed. Some of Carbon 0.927-0.967 12 345 6789 10 Glass, Smooth 0.937 the infrared energy may be reflected or Oak, Planed 0.895 transmitted. Energy that is reflected or Wavelength (Microns) Paper 0.924-0.944 transmitted does not directly heat the work Plastics 0.86-0.95 Note — As the temperature increases, the piece and may be lost completely from the Porcelain, Glazed 0.924 peak output of the source shifts to the left of process (Figure 2). Quartz, Rough, Fused 0.932 the electromagnetic spectrum with a greater Refractory Materials 0.65-0.91 percentage of the output in the near infrared Rubber 0.86-0.95 Figure 2 range. This is referred to as the Wien Water 0.95-0.963 Displacement Curve and is an important factor Paints, Lacquers, Varnishes Black/White Lacquer 0.8-0.95 in equipment selection. Enamel (any color) 0.85-0.91 Oil Paints (any color) 0.92-.096 Emissivity — In practice, most materials and Aluminum Paint 0.27-0.67 surfaces are “gray bodies” having an emissivity or absorption factor of less than Transmission — Most materials, with the 1.0. For practical purposes, it can be assumed exception of glass and some plastics, are that a poor emitter is usually a poor absorber. opaque to infrared and the energy is either For example, polished aluminum has an absorbed or reflected. Transmission losses emissivity of 0.04 and is a very poor emitter. It can usually be ignored. A few materials, such is highly reflective and is difficult to heat with Another important factor to consider in as glass, clear plastic films and open fabrics, infrared energy. If the aluminum surface is may transmit significant portions of the evaluating infrared applications is that the painted with an enamel, emissivity increases amount of energy that is absorbed, reflected incident radiation and should be carefully to 0.85 - 0.91 and is easily heated with evaluated. or transmitted varies with the wave length of infrared energy. Table 1 lists the emissivity of the infrared energy and with different materials some common materials and surfaces. Controlling Infrared Energy Losses — Only and surfaces. These and other important the energy absorbed is usable in heating the variables have a significant impact on heat Absorption — Once the infrared energy is work product. In an unenclosed application, energy requirements and performance. converted into heat at the surface, the heat losses from reflection and re-radiation can be travels into the work by conduction. Materials excessive. Enclosing the work product in an Infrared Emitters & Source Temperatures — such as metals have high thermal conductivity The amount of radiant energy emitted from a oven or a tunnel with high reflective surfaces and will quickly distribute the heat uniformly will cause the reflected and re-radiated energy heat source is proportional to the surface throughout. Conversely, plastics, wood and temperature and the emissivity of the material. to be reflected back to the work product, other materials have low thermal conductivity eventually converting most of the original This is described by the Stefan-Boltzmann Law and may develop high surface temperatures which states that radiant output of an ideal infrared energy to useful heat on the work long before internal temperatures increase product. black body is proportional to the fourth power appreciably. This can be an advantage when of its absolute temperature. The higher the using infrared heating for drying paint, curing temperature, the greater the output and more coatings or evaporating solvents on non-metal efficient the source. substrates.

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Technical Information Radiant Infrared Heating - Source Evaluations

Evaluating Infrared Sources For process heating, it is recommended that 2. From These Points, move left to read the the infrared source have a peak output corresponding percentages (29% and Commonly available infrared sources include wavelength that best matches the selective 51%). heat lamps, quartz lamps, quartz tubes, metal absorption band of the material being heated. sheath elements, ceramic elements and When the major absorption wavelengths of the 3. The Difference between these two values ceramic, glass or metal panels. Each of these material being heated are known, the chart (22%) is the percentage of radiant energy sources has unique physical characteristics, below provides guidance in selecting the most emitted by the element within selected operating temperature ranges and peak energy efficient heat source. The relative percentage wavelengths limits. wavelengths. (See characteristics chart of radiant energy emitted by specific source 4. To Obtain the maximum percentage of the below.) and falling in a particular wavelength range energy emitted by a given element in the can be determined from the chart. Source Temperature & Wave Length desired wavelength band, multiply the Distribution — All heat sources radiate Example — Plastic materials are known to percentage in 3 above by the conversion infrared energy over a wide spectrum of have high infrared absorption rates in efficiency for the selected element wavelengths. As the temperature increases for wavelengths between 3 and 4 microns. Select (comparison chart 56% x 22% = 12.2%). any given source: a source which provides the most effective In this example, a high temperature source output to heat plastics in the 3 and 4 micron (quartz lamp 4000°F) with a peak in the 1.16 1. The total infrared energy output increases range. with more energy being radiated at all micron range, while more energy conversion wavelengths. 1. Enter Bottom of Chart at 3 and 4 microns, efficient, would not be as effective as a lower 2. A higher percentage of the infrared energy read up to corresponding points on temperature metal sheath or panel heaters is concentrated in the peak wavelengths. selected element curve (use 1400°F metal with a peak in the 2.8 to 3.6 micron range. ° 3. The energy output peak shifts toward the sheath in this example). Quartz tubes (1600 F) would provide similar shorter (near infrared) wavelengths. peak wavelengths.

The peak energy wavelength can be deter- 100 Percentage Increment of Radiant Energy 90 Falling Below any Wavelength mined using Wien’s Displacement Law. for a Black Body at Temperature T Peak Energy 5269 microns/°R 80 = A (Microns) Source Temp. (°F) + 460 70 Source 5269 microns/°R 60 = = 2.83 microns C 1400°F 1400°F + 460 50 D B ° 40 Source = 5269 microns/ R = 5.49 microns 500°F 500°F + 460 30 20 Absorption by Work Product Materials in A - 4000°F Source Temperature ° Process Applications — While most materials Below Wavelength % Radiant Energy 10 B - 1600 F Source Temperature C - 1400°F Source Temperature absorb long (far) infrared wavelengths D - 1000°F Source Temperature uniformly, many materials selectively absorb 0 0.4 0.72 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 6.4 6.8 7.2 short (near) infrared energy in bands. In process heating applications this selective Wavelength - Microns absorption could be very critical to uniform 3900 2150 1400 1000 740 550 410 310 and effective heating. Temperature (T) degrees F = Radiation Peaks

Characteristics of Commercially Used Infrared Heat Source Tungsten Filament Nickel Chrome Resistance Wire Wide Area Panels Infrared Source Glass Bulb T3 Quartz Lamp Quartz Tube Metal Sheath Ceramic Ceramic Coated Quartz Face Source Temperature (°F) 3000 - 4000°F 3000 - 4000°F Up to 1600°F Up to 1500°F Up to 1600°F 200 - 1600°F Up to 1700°F Brightness Intense white Intense White Bright Red to Dull to Bright Dark to Dull Dark to Dark to Cherry Dull Orange Red Red Cherry Red Red Typical Configuration G-30 Lamp 3/8" Dia. Tube 3/8 or 1/2" Tube 3/8 or 1/2" Tube Various Shapes Flat Panels Flat Panels Type of Source Point Line Line Line Small Area Wide Area Wide Area Peak Wavelength (microns) 1.16 1.16 2.55 2.68 3 - 4 2.25 - 7.9 2.5 - 6 Maximum Power Density 1 kW/ft2 3.9 kW/ft2 1.3 - 1.75 kW/ft2 3.66 kW/ft2 Up to 3.6 kW/ft2 3.6 kW/ft2 5.76 kW/ft2 Watts per Linear Inch N/A 100 34 - 45 45 - 55 N/A N/A N/A Conversion Efficiency 86% 86% 40 - 62% 45 - 56% 45 - 50% 45 - 55% 45 - 55% Infrared Energy Response Time Heat/Cool Seconds Seconds 1 - 2 Minutes 2 - 4 Minutes 5 - 7 Minutes 5 - 8 Minutes 6 - 10 Minutes Color Sensitivity High High Medium Medium Medium Low to Medium Low to Medium Thermal Shock Resistance Poor Excellent Excellent Excellent Good Good Good Technical Mechanical Ruggedness Poor Fair Good Excellent Good Good Fair Chromalox Model — QR QRT RAD, URAD RCH CPL, CPLI, CPH CPHI

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Technical Information Radiant Infrared Heating - Process Applications

Application Parameters Element Response Time — Some applica- Drying Surface tions, such as continuous web heating of & Temp Time Typical industrial applications of radiant paper or plastic film, require quick shutdown Heating Substrate (°F) W/In2 (sec) heating include curing or baking (powders, of heaters in case of work stoppage. In these Glass Bottles — 104 6.4 30 paints, epoxies, adhesives, etc.), drying applications, residual radiation from the Adhesives Paper — 3.2 30 (water, solvents, inks, adhesives, etc.) and infrared heaters and associated equipment Heating product heating (preheating, soldering, shrink must be considered. Residual radiation from PVC Shrinking — 300 3.2 60 fitting, forming, molding, gelling, softening, the element is a function of the operating ABS Forming — 340 9.7 — and incubating). The following are general temperature and mass. Quartz lamps and Deriving Time-Temperature Information from guidelines that can be used in evaluating and tubes have relatively low mass and the Empirical Testing — If specific information is resolving most radiant heating problems. infrared radiation from the resistance wire not readily available for a particular work Unfortunately, the process is so versatile and drops significantly within seconds after product, a simple but effective test will usually its applications so varied that it is not feasible shutdown. However, the surrounding quartz provide enough preliminary data to proceed to list solutions to every problem. envelope acts as a secondary source of with a design. Place one or more radiant radiation and continues to radiate considerable heaters in a position with the radiation To determine heat energy requirements and energy. Metal sheathed elements have more directed at a work product sample. The select the best Chromalox infrared equipment mass and slightly slower response time. Wide distance between the face of the heater and for your application, it is suggested the area panels have the most mass and the the sample should approximate the expected problem be defined using a check list similar slowest response time for both heat up and spacing in the final application. Position the to below. Several of the key factors on the list cool down. The following chart shows the sample so that it is totally within the radiated are discussed on this and following pages: average cool down rate of various sources area. Energize the heater(s) and record the after shutdown. Actual cool down of the 1. Product to be heated time necessary to reach desired temperature. source and work product will vary with 2. Physical dimensions and weight/piece Calculate the W/in2 falling on the work piece equipment design, product temperature, 3. Surface coating or solvents, if any using the exposed area of the work product ambient temperature and ventilation. 4. Infrared absorption characteristics and the maximum kW/ft2 at the face of the 5. Production rate (lbs/hr, pieces/hr, etc.) Source Temperature Vs. Time heater as listed in the product catalog page. If 6. Work handling method during heating 1600 the data is not available and a sample test can (continuous, batch or other) Source Temperature not be performed, the following table provides 1500 After Shutdown 7. Element response time (if critical) a few suggested watt densities as guidance.

2 F) 1400 8. Power level requirements in kW/ft based ° A. Quartz Lamp 3/8” Dia. on Time/Temperature relationship (if 1300 B. Quartz Tube 1/2” Dia. 1200 C. Metal Sheath 2 known) D. Wide Area Panel W/In on Work 9. Starting work temperature 1100 E. Ceramic Heater Application Heat Up Hold 10. Final work temperature 1000 Paint Baking 4-6 1 - 2 11. Ventilation (if present or required) 900 Metal Dry Off 15 8 12. Available power supply Thermoforming 10 - 15 — 800 D 13. Space limitations C Fusing or Embossing 700 (plastic films) 5-6 — Infrared Absorption Characteristics — As 600 E Silk Screen Drying 5-6 — Sheath or Surface Temperature ( Sheath or Surface Temperature previously discussed, many materials, 500 Contact your Local Chromalox Sales particularly plastics, selectively absorb A B office for further information or assis- 400 tance in determining time/temperature infrared radiation. The following chart provides 300 requirements for a particular application. data on some common plastic materials and 0 30 60 90 120 150 180 the recommended source temperatures for Elapsed Time (Sec.) thermoforming applications. Power Level or Radiation Intensity — In Time-Temperature Relationship — A critical most process applications, more than one Absorption Ideal Source step in the evaluation of a radiant heating radiant heater is needed to produce the Band(s) Temperature application is to determine the time necessary desired results. When heaters are mounted Plastic (microns) (°F) to develop work piece temperature and the together as close as possible, the net radiant LPDE 3.3 - 3.9 877 - 1170 elapsed time needed to hold temperature in output of the array is defined as the maximum HDPE 3.2 - 3.7 950 - 1170 order to obtain the desired results (curing or power level or radiation intensity. The catalog PS 3.2 - 3.7 950 - 1170 drying). The following chart shows time/ (6.4 - 7.4) 245 - 355 pages for radiant heaters indicate the temperature relationships for several typical 2 PVC 1.65 - 1.8 2440 - 2700 maximum kW/ft at the face of each heater. (2.2 - 2.5) 1625 - 1910 infrared applications and materials. Typical ranges for radiation intensity (power PMMA 1.4 - 2.2 1910 - 3265 level) are as follows: PA-66 1.9 - 2.8 1405 - 2285 Surface Time ° 2 (3.4 - 5) 585 - 1075 Curing Substrate Temp ( F) W/In (min) Cellulose 2.2 - 3.6 990 - 1910 Alkyd Paint Steel 320 3.9 3 Radiant Intensity Heater Output or Power Level (kW/Ft2) Acetate (5.2 - 6) 440 - 545 Epoxy Paint Steel 356 8.1 5 Acrylic Paint Steel 392 8.1 2 Low 1 - 2 Powder Coat Steel 400 13 6 Medium 2 - 3 High Over 3

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Technical Information Radiant Infrared Heating - Process Applications Determining kW Required — It is difficult Infrared Heating Equations — Infrared energy For drying, use the following equation. to develop simple calculations for radiant requirements can also be estimated by using WP + QS + QLH heating applications because of the many equations and nomographs developed kW = Q ⑀ variables and process unknowns. Design specifically for infrared applications. 3412 Btu/kW x Efficiency(RE) x VF x data gained from previous installations or Where: from empirical tests is frequently the most Product Heating — For product heating, the following equation can be used QWP = Btu required by work product to raise reliable way of determining installed kW the temperature from initial to vaporiza- requirements. Total energy requirements ∆ ° kW = Lbs/hr x Cp x T F tion temperature can be estimated with conventional heat 3412 Btu/kW x Efficiency(RE) x VF x ⑀ Qs = Btu required by solvent to raise the loss equations. The results of conventional temperature from initial to vaporization equations will provide a check against data Where: temperature Lbs/hr = Pounds of work product per hour obtained from nomographs or empirical QLH = Btu required for the latent heat of the ° testing. As a minimum, conventional CP = Specific heat in Btu/lb/ F vaporization of the solvent ∆T = Temperature rise in °F equations should include the following. Efficiency (RE) = Combined efficiency of the Efficiency (RE) = Combined efficiency of the source and reflector 1. Calculate the Sensible Heat required to source and reflector bring work to final temperature. Base VF = View Factor for enclosed ovens, use a VF = View Factor is the ratio of the factor of 0.9. For other applications, refer calculations on specific heat and pounds of infrared energy intercepted by the work material per hour. to the view factor table. product to the total energy radiated by ⑀ = Absorption (emissivity) factor of the work 2. Determine Latent Heat of Vaporization the source. For enclosed ovens, use a product (when applicable). Latent heat of factor of 0.9. For other applications, refer vaporization is normally small for solvents to the view factor table. Controls — Most control systems for infrared ⑀ in paints and is frequently ignored. = Absorption (emissivity) factor of the work process heating can be divided into two However, when water is being evaporated, product categories, open loop or manual systems and closed loop, fully automatic systems. the kilowatt hours required may be quite Drying & Solvent Evaporation — Removing significant. solvent or water from a product requires Open Loops or Manual Systems — The 3. Ventilation Air (when applicable). The rise raising the product temperature to the simplest and most cost effective control in air temperature for work temperatures, vaporization temperature of the solvent and system is an input controllers (percentage 350°F or less, can usually be estimated as adding sufficient heat to evaporate it. To timer) such as the Chromalox VCF Controller 50% of final work temperature rise. For calculate heat requirements for solvent operating a magnetic contactor. The timer higher work temperatures, assume air and evaporation, the following information must be cycles the radiant heaters on and off for short work temperature are the same. known. periods of time (typically 15 - 30 seconds). This control system works best with metal 4. Conveyor Belt or Chain Heat 1. Pounds of solvent to be evaporated per sheath heaters, which have sufficient thermal Requirements. Assume temperature rise hour mass to provide uniform radiation. It can be of conveyor to be the same as work 2. Pounds of work product per hour used with quartz tube or quartz lamp heaters temperature rise. 3. Initial temperature of product and solvent by using special circuitry to switch from full to 4. Specific heat of product half voltage rather than full on and full off. 5. Wall, Floor and Ceiling Losses for 5. Specific heat of solvent Enclosed Ovens. For uninsulated metal 6. Vaporization temperature of solvent Closed Loop or Automatic Systems — Since surfaces, refer to Graph G-125S. For (ie: water = 212°F) infrared energy heats the work product by insulated walls, refer to Graph G-126S. 7. Heat of vaporization of solvent direct radiation, closed loop control systems 8. Source/reflector efficiency that depend on sensing and maintaining air 6. Oven End Losses. For enclosed ovens, this 9. View factor temperature are relatively ineffective (except in will depend on shape of end area and 10. Absorption factor (emissivity) totally enclosed ovens). In critical applications whether or not air seals are used. If where temperature tolerances must be closely silhouette shrouds are used, a safety factor WARNING — Hazard of Fire. Flammable held, non-contact temperature sensors of 10% is acceptable. solvents in the atmosphere constitute a fire operating SCR control panels are recom- hazard. When flammable volatiles are released mended. Non-contact temperature sensors 7. The Sum of The Losses calculated in 1-6 in continuous process ovens, the National Fire above will be the minimum total heat can be positioned to measure only the work Prevention Association recommends not less product temperature. Properly positioned, energy requirement based on conventional than 10,000 ft3 of air be removed from the heat loss equations. non-contact temperature sensors and SCR oven per gallon of solvent evaporated. control panels can provide very accurate Reference NFPA Bulletin 86 "Ovens and radiation and product temperature control. Furnaces", available from NFPA, P.O. Box 9101, Quincy MA 02269. Technical

I-31 Technical

Technical Information Radiant Infrared Heating - Process Applications

Nomograph for Product Heating — For Estimating Total Kilowatts for Product Heating product heating, the nomograph at the right can be used. The nomograph does not take 200 250 300 350 400 100 90 80 70 60 50 40 into account heat energy requirements for air 175 ventilation. To estimate the kW for total Temperature Rise °F Overall Efficiency product heating: 150 30 1. Determine pounds of material per hour to 125 be heated (A) 100 20

2. Read across to the specific heat of the 75 C material (B) D 50 10 3. Read up to desired temperature rise in °F 25 (C) E 4. Read across to overall efficiency (D). 10 30 50 70 90 110 130 150 Kilowatts of Infrared Required Overall efficiency = Product Absorption Specific Heat 200 Factor x View Factor x Source Efficiency. of Material 300 Determine Product Absorption Factor 500 (surface emissivity) of the work product 800 (ie: ⑀ = 0.85 for enamel sheet metal). 1.0 1000 0.8 B 0.6 Determine View Factor (use 0.9 as a view A 2000 factor for well designed or enclosed 0.4 0.3 3000 ovens). Determine Source efficiency. 0.2 5000 Pounds of Material Per Hour Typical Source/Reflector efficiencies are: .0.05 8000 Quartz Lamps 0.70 to 0.80 0.1 10000 Quartz Tubes 0.60 to 0.70 Metal Sheath 0.55 to 0.65 5. Read down to Kilowatts required (E).

Nomograph for Drying — The nomograph to Estimating Infrared Kilowatts for Drying the right can be used to estimate Kilowatts 60 55 50 40% 100 90 80 70 60 50% 40 required to evaporate water from the surfaces 65 Source/Reflector of work product. Graph is based on an initial 70 Efficiency starting product temperature of 70°F. It does Work Product Absorption Factor 30 not take into account heat energy require- 80 (Emissivity x View Factor) ments for air circulation or ventilation.

1. Determine pounds of water (solvent) per B C 20 hour to be evaporated (A) 2. Read up to Source/Reflector efficiency (B). Depending on the configuration and 10 cleanliness of the reflector, typical Source/ Reflector efficiencies are: Quartz Lamps 0.70 to 0.80 A D Quartz Tubes 0.60 to 0.70 100 80 60 40 20 0 20 40 60 80 100 120 140 Metal Sheath 0.55 to 0.65 Pounds of Water Evaporated Per Hour Kilowatts of Infrared Energy Required 3. Read across to Work Product Absorption Note — To evaporate solvents other than water, calculate the energy required to heat the solvent Factor (C). This value is based on the to vaporization temperature using the weight, specific heat and temperature rise. Calculate the emissivity of the work product surface latent heat of vaporization and add to the energy required to heat the solvent to vaporization (ie: ⑀ = 0.85 for enameled sheet metal) and temperature. the view factor of the oven or space. Use 0.9 as a view factor for well designed or enclosed ovens. 4. Read down to Kilowatts required (D).

I-32 Technical

Technical Information Radiant Infrared Heating - Process Applications

Baking & Curing — The nomograph to the Estimating Watt Density for Curing or Baking right can be used to determine the watt 0.90 density required on the work product for 0 Ventilation Air 0.75 baking and curing of paints and coating. Feet Per Minute 0.65 50 (Air at 80°F) 0.50 Lacquers are cured primarily by evaporation of 150 the solvent and can be cured by infrared in 0.35 2 - 15 minutes. Enamels are cured primarily 300 by polymerization and require a longer time C (15 - 20 minutes). Varnishes, japans and house paints cure mainly by oxidation but can D usually can be accelerated by infrared heating. Absorption Factor (Emissivity) To find approximate watt density needed for baking: 1. Locate temperature product is to reach in E five minutes (A) 01 2 3 4 5 6 7 2. Read across to line representing gauge of Watt Density Required on Product (W/In2) 100 the material being heated (B) 12 Gauge Sheet Steel 3. Read up to ventilation air in feet per 200 minute over surface of the product (C). If Wood not known, estimate feet per minute based B 300

on cubic feet per minute of ventilation or 16 Gauge be Attained in F) To A 400 °

circulating air divided by the the approxi- 24 Gauge Five Minutes mate cross sectional area of the oven. In 500 applications with no forced ventilation, use 28 Gauge 20 Gauge 2 - 5 fpm. ( Temperature 600 4. Read right to the absorption factor for the work product surface or coating (ie: ⑀ = 0.85 for enameled sheet metal) (D) 5. Read down to watt density required on the product surface (E).

Determining Heater Fixture Spacing — Intensity Vs. Spacing — Point & Line Infrared Sources Having determined the total required kilowatts 20 2 A = QRT and the desired W/in on the work product, the L 18 (3/8" Dia. Quartz Tube) next step is to deterine the spacing and the B = RAD/QRT 12" number of heaters. In most conveyor type 16 3-11/16" (1/2" Dia. Quartz Tube) oven applications, a 12" spacing from the face C = S-RAD of the heater to the work product produces D = RADD/U-RAD/QR 14 (3/8" Dia. Quartz Tube) uniform distribution of the radiation. The E = U-RP 12 graph to the right shows centerline to E on Work Product on Work centerline spacing of Chromalox radiant 2 10 heaters to obtain various intensities on the D 8 work based on a spacing of 12" from the face C of the heater to the work product. Specific Average W/In Average 6 applications may require the distance to be A B increased or decreased. 4

The graph is applicable to line or point infrared 2 sources installed in reflectors. Refer to view 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 factor charts for ceramic heaters and flat panel Heater Center to Center Distance = L (In.) infrared sources. Technical

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Technical Information Radiant Infrared Heating - Process Applications View Factor for Flat Panels — While the View Factor for Two Parallel Surfaces radiation pattern from line and point infrared 1.0 10 8 sources can be controlled by reflectors, the 4 radiation pattern from flat panels is diffused 0.7 2 and the infrared energy is emitted from a large S1 L area. Consequently, the shape of the source 0.5 1 0.4 S2 and the target are a significant factor in Y X 0.6 determining the Watt density falling on the 0.3 work product. For parallel surfaces in 0.4 applications such as thermoforming or web 0.2 heating, the incident energy falling on the work product is determined by a “View Factor”. View 0.2 factor is defined as the percentage or fraction 0.10

of infrared energy leaving the surface of a flat Y/L panel (source) which is intercepted by the 0.07 0.1 surface of the work product (target). The view View Factor 0.05 factor for parallel surfaces (rectangles) can be 0.04 determined from the graph. Example — Find 0.03 the view factor for a 12 by 24" panel heater L = Distance Between Surfaces 0.02 mounted 4" from a continuous web infrared X = Length of Rectangle drying application. X/L = 24" ÷ 4" = 6, Y = Width of Rectangle Y/L = 12" ÷ 4" = 3. Read left from the intercept 0.01 of X/L = 6 and Y/L = 3 with a view factor of 0.7. 0.1 0.2 0.3 0.5 1 2 3 4 5 10 20 8 X/L

Radiant Oven Heating Example — A Heat required to heat solvent to 70°F Total Heat Absorbed — manufacturer of 66 gallon electric water ° heaters wishes to bake the paint on sheet 1.2 gph x 7.25 lb/gal x 0.34 Btu/lb x (170-70 F)= 0.1 kW 6.5 kW + 0.5 kW + 8.8 kW = 15.8 kW ° 3412 Btu/kW metal jackets (open top and bottom) at 350 F. Heat Losses — Heat losses from oven surface The jackets weigh 33 lbs, are 26" in diameter Heat required to vaporize solvent with 2 inches of insulation (Graph G-126S) = 2 by 45" high with an outside area of 25.5 ft . 12 W/ft2. Assume inside surface temperature The process requires 20 jackets be painted per 1.20 gph x 7.25 lb/gal x 156 Btu/lb = 0.4 kW of wall and ceiling = 250°F, ∆T = 180°F) hour. The jackets will be suspended from a 3412 Btu/kW Wall area 7 ft x 8 ft x 2 ft = 112 ft2 conveyor chain on 9 ft centers and will be 2 Heat absorbed by solvent = 0.1 + 0.4 = 0.5 kW Ceiling and floor area 8 ft x 4 ft x 2 ft = 64 ft rotated as they move. The chain weighs 12 Open tunnel ends = 3 ft x 6 ft x 2 ft = 36 ft2 lbs/ft. The heaters will be installed in a tunnel 3. Heat Required by Ventilation Air — (NFPA oven with 2 inches of insulation and reflective recommendation is a minimum of 10,000 Heat loss from outside surfaces of oven walls. The oven is 8 ft long, 4 ft wide and 7 ft cubic feet per gallon of solvent evaporated.) 176 ft2 x 12 W/ft2 high and has end openings 3 ft by 6 ft. 3 = 2.1 kW/hr Densiity of air = 0.080 lbs/ft 1000 W/kW Preliminary test results show the jackets must Specific heat of air = 0.240 Btu/lb/°F be baked for six minutes for a satisfactory Heat loss from open oven ends (assume the finish. The paint weighs 7.25 lbs/gal, contains Note — Ventilation air is heated by re- open ends are equal to an uninsulated metal 2 50% volatiles and covers 212 ft per gallon. radiation and convection from the work, oven surface under the same conditions as the oven ° Assume a room temperature of 70 F walls, etc. Air temperature is always less than surfaces) (See Graph G-125S.) ° Specific heat of steel = 0.12 Btu/lb/ F the work temperature. Assume a 200°F air ° 2 2 ° Boiling point of solvent = 170 F temperature. 36 ft x 0.6 W/ft x 180 F = 3.89 kW/hr Specific heat of solvent = 0.34 Btu/lb/°F 1000 W/kW Latent heat of vaporization = 156 Btu/lb Volume = 1.20 gph x 10,000 ft3 = 12,000 ft3/hr Total Heat Losses — 2.1 kW + 3.98 kW = 5.99 Heat Required for Operation — 12,000 ft3/h x 0.08 lb/ft3 x 0.24 Btu/lb/°F x (200-70°F) kW 3412 Btu/kW 1. Heat Absorbed by Jackets — Total Heat Capacity Required for Operation — (20 jackets/hr x 33 lbs = 660 lbs/hr) Heat absorbed by ventilation air = 8.78 kW 15.8 kW + 5.99 kW = 21.8 kW/hr 660 lbs/hr x 0.12 Btu/lb/°F x (350 - 70°F) = 6.5 kW 4. Conveyor Chain & Hangers — Normally As with any process heat calculation, it is not 3412 Btu/kW the conveyor chain is outside the possible to account for all the variables and radiation pattern of the heaters and is unknowns in the application. A safety factor is 2. Heat Absorbed by Solvent — Solvent heated by convection from air in the tunnel. volume Since the heat absorbed by the air has recommended. For radiant heating applications, a safety factor of 1.4 is suggested. 2 already been accounted for, the heat 25.5 ft x 20 jackets/hr x 50% = 1.20 gal/hr 2 absorbed by the conveyor may be ignored. 212 ft /gal Total Heat Required = 21.8 x 1.4 = 30.5 kWh (Conveyor speed should provide 6 minutes in the 8 foot heated area.)

I-34 Technical

Technical Information Radiant Infrared Heating - Comfort Heating

Indoor Spot Heating 11.5' Infrared spot heating of work stations and 60° Reflector personnel in large unheated structures or 22° Tilt areas has proven to be economical and 10' satisfactory. The following guidelines may be used for spot heating applications (areas with L W length or width less than 50 feet). 1. Determine the coldest anticipated inside Figure 1 — Design Area Figure 2 — Tilted Infrared ambient temperature the system must Fixtures for Spot Heating overcome. If freeze protection is provided by another heating system, this W/2 temperature will be 40°F. Fixture Mounting Height Angle of Location L 2. Determine the equivalent ambient Uniform temperature desired (normally 70°F is the Radiation nominal average). W/2 W 3. Subtract 1 from 2 to determine the W theoretical increase in ambient temperature Width of Radiation (∆T ) expected from the infrared system. If Pattern Figure 3 — Pattern Area drafts are present in the occupied area (air movement over 44 feet per minute (0.5 mph) velocity), wind shielding or 8. Divide the design area (Step 4) into the Indoor Area Heating protection from drafts should be pattern area (Step 7). considered. In many industrial environments, area heating Pattern Area Q = (areas with length or width greater than 50 ft) 4. Determine the area to be heated in ft2. This Design Area can be accomplished economically with is termed the “design or work area” (AD) (Fig. 1). If the pattern area is equal to or greater multiple infrared heaters. For quick estimates, than the design area, quotient (Q) will be determine the minimum inside temperature 5. Multiply the design area by one watt per equal to or greater than 1 and coverage is and use a factor of 0.5 watts per square foot of square foot times the theoretical adequate. If Q is less than 1, the design design area for each degree of theoretical temperature increase (∆T) desired as area exceeds the pattern area of individual temperature. If the calculated heat loss of the determined in Step 3 (minimum of 12 fixtures. Adjust the heater locations and structure, including infiltration or ventilation watts per square foot). The design factor of patterns or add additional fixtures with air, is less than the quick estimate, select the one watt per square foot density assumes a patterns overlapping as necessary, to lower value. Locate heaters uniformly fixture mounting height of 10 feet. Add 5% ensure adequate coverage. throughout the area with at least a 30% overlap for each foot greater than 10 feet in mounting height. Avoid mounting fixtures 9. Multiply quotient (Q in Step 8) by the in radiation pattern. ∆ below 8 feet. increase in theoretical temperature ( T of Step 3) by the design area (AD of Step 4) to Outdoor Spot Heating 6. Determine fixture mounting locations determine the amount of radiation to be a) In areas where the width dimension is installed. The same guidelines outlined under Indoor 25 feet or less, use at least two fixtures Spot Heating should be followed except that ∆ mounted opposite each other at Radiation (Watts) = Q x T x AD watts per square foot for each degree of the perimeter of the area and tilted at an 10. Many Types of radiant heaters are theoretical ambient temperature increase angle. This provides a greater area of available for comfort heating applications should be doubled (approximately 2 watts per exposure to the infrared energy by including ceiling, wall and portable floor square foot for each 1°F). This factor applies to personnel in the work area. Tilt the standing models. Choose specific fixtures outdoor heating applications with little or no fixtures so that the upper limit of the from the product pages. It is preferred that wind chill effect on personnel. If wind velocities fixture pattern is at approximately six feet half the wattage requirements be installed above the center of the work station area are a factor in the application, determine the on each side of the work station in the equivalent air temperature from the Wind Chill (Figure 2). design area. b) When locating fixtures, be sure to Chart in NEMA publication HE3-1971 or other allow adequate height clearance for large Controls — Manual control by percentage information source. moving equipment such as cranes and lift timers may be adequate for a small installation. Note — Increasing the infrared radiation to trucks. To provide better control of comfort levels in massive levels to offset wind chill can create c) Avoid directing infrared onto outside varying ambient temperatures, divide the total walls. discomfort and thermal stress. In outdoor heat required into two or three circuits so that exposed applications, a wind break or shielding 7. Estimate (tentatively) the radiated pattern each fixture or heating element circuit can be is usually more effective. area. Add length of fixture to the fixture switched on in sequence. Staging can be Technical pattern width (W) to establish pattern accomplished by using multistage air thermo- length (L). Pattern Area = L x W (Fig. 3). stats set at different temperatures. I-35